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The Secondary Agriculture course aims to:
1.0.9 INTRODUCTION TO AGRICULTURE
1.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) define agriculture;
b) state the main branches of agriculture;
c) describe farming systems;
d) describe farming methods;
e) explain the role of agriculture in the economy.
1.2.0 Content
1.2.1 Definition of agriculture.
1.2.2 Branches of agriculture.
1.2.3 Systems of fanning; Extensive, Intensive, Large scale, and Small sea Study these under the following headings; Meaning, Advantages and
Disadvantages.
1.2.4 Methods of farming: Mixed farming; Nomadic Pastoralism; Shifting cm Organic farming; Agro-forestry.
1.25 Roles of agriculture in the economy: Food supply; Source of employment Foreign exchange earner; Source of raw materials for industries; Provision market for industrial goods; Source of capital.
2.0.0 FACTORS INFLUENCING AGRICULTURE
2.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) explain the human factors influencing agriculture;
b) explain biotic factors influencing agriculture;
c) explain how climatic factors influence agriculture;
d) define soil;
e) describe the process of soil formation g) determine soil constituents;
h) classify soils by physical characteristics;
i) explain chemical properties of soils;
j) relate crop and livestock distribution to soils in different regions.
2.2.1 Content
2.2.2 Human factors: Levels of education and technology; Health — HIV/AIDS and health in general; Economy (include liberalization); Transport and communication; Market forces (local and international); Government policy; Cultural and religious beliefs.
2.2.3 Biotic Factors: Pests, Parasites, Decomposers, Pathogens, Predators, Pollinators
Nitrogen fixing bacteria
2.2.4 Climatic Factors
– How temperature influences crop and livestock production.
Note: – Each factor to be discussed with respect to Land potentiality, Crop production, Livestock production, Crop and livestock distribution in Kenya.
2.2.5 Edaphic factors: Definition of soil, Soil formation, Soil profile (definition, characteristics of different soil layers, difference between soil formed in situ and depositions, Soil depth and its influence on crop production).
3.0.0 SOIL AND WATER CONSERVATION
3.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) define soil erosion;
b) explain the various factors that influence erosion;
c) list the agents of erosion;
d) describe various types of erosion;
e) describe various methods of erosion control;
1) carry out soil erosion control measures;
g) describe water harvesting and conservation techniques;
h) describe micro-catchments and their uses;
i) design and construct a micro-catchment.
3.2.0 Content
3.2.1 Soil erosion – definition
3.2.2 Factors influencing erosion: land use and ground cover, topography-gradient and
length of slope (horizontal and vertical intervals).
3.2.3 Agents of erosion: Water, Wind, Human beings and Animals.
3.2.4 Types of erosion: Splash/rain drop, Sheet, Rill, Gully (gully formation, types of gullies), River bank, Solifluction, Landslides.
3.2.5 Biological/cultural control: Grass strips, Cover crops, Contour farming and strip cropping, Mulching, Afforestation/forestation.
.2.6 Physical/structural control: Stone lines, Filterstrips, Trashlines, Terraces (level, graded, broad-based, narrow-based. Bench, fanya juu, fanya chini), Bunds, Cutoff-drains/Diversion ditches, Gabions/porous dams, Ridging.
3.2.7 Water harvesting: Roof catchments, Rock catchments, Weirs and dams, Ponds, Retention ditches/Level terraces.
3.2.8 Micro-catchments: Types, Uses, Laying out and construction methods.
4.0.0 WATER SUPPLY, IRRIGATION AND DRAINAGE
4.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) state the sources of water for the farm;
b) describe collection, storage, pumping and conveyance of water;
c) describe water treatment and explain its importance;
d) define irrigation;
e) explain the importance of litigation;
1) describe methods of irrigating land;
g) list the equipment used in litigation;
h) grow a crop through irrigation;
1) cart: out maintenance oil drilling equipment and facilities;
j) define drainage;
k) explain the importance of drainage;
1) describe the methods of drainage;
m) explain how agricultural activities pollute water and how this can be prevented;
4;2;0 Content
4;2;l Water supply: Sources of water4 Collection and storage of water Pumps and pumping, Conveyance of water (Piping types of pipes Choice of pipes, Canals, Transportation in containers), Water treatment (Meaning, Methods, Importance), Uses of water on the farm
4;2;2 litigation: Definition, Importance (include irrigation as a method of land reclamation) Methods (surface4 subsurface, overhead, drip).
Note.’- the advantages and the disadvantages of each.
Maintenance practices of each irrigation system.
4;2i Project on crop production through any method of Irrigation.
414 Drainage: Definition, Importance (include as a method of land ret lamatioii)5
Methods of drainage (surface, sub-surface, pumping, planting of appropriate trees);
4;2; Water Pollution: Meanings Agricultural practices that pollute water, Methods of pollution prevention and control.
SOIL FERTlLITY I (Organic Manures)
Specific Objectives
By the end of the topic5 the learner should be able to:
a) define soil fertility;
b) explain how soil fertility can be maintained; describe how soil loses fertility;
l) define and distinguish organic matter manure and humus;
e) explain the Importance of organic matter In the soil
Soil fertility Definition4 How soil loses fertility, Maintenance of soil fertility. Organic Manure Organic matter and humus4 Importance of organic matter in the soil, types of organic manures – green manure, Famyard manure Compost manure
Note; For each type, describe its preparation, advantages and disadvantages and u4c.
5.2.3 Compost manure: Meaning, Materials used and materials to avoid, Preparation methods and procedure (Heap and Pit).
6.0.0 SOIL FERTILITY: II (INORGANIC FERTILIZERS)
6.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) list the essential elements;
b) classify’ the essential elements;
c) state the role of each macro-nutrients, micro-nutrients;
d) describe the deficiency symptoms of the macro-nutrients, micro-nutrients;
e) identify and classify fertilizers;
1) describe the properties of various fertilizers;
g) describe soil sampling and testing procedures;
h) use appropriate methods of fertilizer application;
i) calculate fertilizer application rates;
j) explain how soil acidity and alkalinity affect crop production.
6.2.0 Content
6.2.1 Essential elements
• Macro-nutrients: carbon, hydrogen and oxygen, fertilizer elements (N.PK.), liming elements (Ca, Mg), Sulphur, Role of macro-nutrients in crops,
Deficiency symptoms of macro-nutrients ts in crops
• Micro-nutrients: Role of micro-nutrients in crops, Deficiency symptoms of micro-nutrients in crops,
6.2.2 Inorganic fertilizers: Classification of fertilizers, Identification of fertilizers, Properties of fertilizers, Methods of fertilizer application, Determination of fertilizer rates..
6.2.3 Soil sampling: Meaning, Soil sampling methods and procedures, Sites to avoid, Preparation and Procedure of sending soil for testing.
6.2.4 Soil testing: Meaning, Importance, Testing for pH, How soil pH affects crop production.
Note Learners to make a table showing optimum pH range for various crops.
7.0.0 FARM TOOLS AND EQUIPMENT
7.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) identify various farm tools and equipment;
b) name parts of various farm tools and equipment;
c) describe the use of various tools and equipment
d) carry out maintenance practices on tools and equipment.,
7.2.0 Content
7.2.1 Garden tools and equipment
7.2.2 Workshop tools and equipment: Woodwork tools and equipment, Metalwork tools and equipment.
7.2.3 Livestock production tools and equipment
7.2.4 Plumbing tools and equipment
7.2.5 Masonry tools and equipment.
Note: Study the above tools under the headings: Name and uses, Parts and uses, Maintenance practices
See the appendix for the list of tools and equipment to be studied.
8.0.0 CROP PRODUCTION I (LAND PREPARATION)
8.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) explain the importance of land preparation;
b) describe the various types of cultivation;
c) relate cultivation operation to correct tools and or implements;
d) prepare a piece of land ready for crop production.
8.2.0 Content
8.2.1 Land preparation: Definition, Importance.
8.2.2 Operations in land preparation: Clearing land before cultivation (importance include clearing as a method of land reclamation; Methods and equipment.
Note: For each operation: give reasons and explain how it is carried out.
• Sub-soiling: Meaning, Importance, Equipment used.
8.2.3 Minimum tillage: Definition, Importance, Practices.
9.0.0 CROP PESTS AND DISEASES
9.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
a) define pest and disease;
b) state the main causes of crop diseases;
c) describe the harmful effects of crop pests and diseases;
d) identif’ and classify some of crop pests and diseases;
e) carry out general disease and pest control measures.
9.2.0 Content
9.2.1 Pests: Definition, Classification of pests (mode of feeding, Crops attacked, Stage of growth of crop attacked, Field and storage pests), Identification of common pests, Harmful effects of pests, est control measures.
9.2.2 Diseases: Definition, Classification of diseases according to cause, Identification of common diseases, Disease control, Harmful effects of diseases, Disease control measures.
10.0.0 CROP PRODUCTION II (PLANTING)
10.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
10.2.0 Content
10.2.1 Types of planting materials
Suckers, Tubers, Vines, Cuttings and setts.
10.2.2 Selection of planting materials: Suitability to ecological conditions (use maize hybrids and coffee varieties as examples), Purity, Germination percentage,
Certified seeds.
.2.3 Preparation of planting materials: Breaking dormancy, Disease and pest control/seed dressing, Seed inoculation, Chitting.
Note: Give appropriate crop examples for each practice.
10.2.4 Planting:
note: Give appropriate crop examples for each method
Plant population:
110,0 CROP PRODUCTION III (NURSERY PRACTICES
11,1,0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe a nursery bed;
(b distinguish between a nursery bed, a seedling bed and a seed bed;
(c) state the importance of a nursery bed;
(d) select a suitable site for a nursery
(e) prepare a nursery bed;
(0 establish a nursery bed
(g manage a nursery bed;
(h) transplant crops from a nursery;
(1) bud a seedling;
Q) graft a seedling;
(k) explain the importance of budding, grafting, layering and tissue culture;
(I) describe damage caused by animals on tree seedling and how to prevent it
11 2,0 Content
Nursery bed:
Definition, Difference between a nursery bed, seedling bed and a seed bed, importance, Site selection, nursery establishment (vegetable nursery, tree nursery, vegetative propagation nursery (tea as an example)
use of sleeves and other innovations for growing young plants making and using seedling boxes for growing young plants preparation of rooting medium
preparation of cuttings.
11,22 Routine management in raising seedlings: Seed drilling, Mulching, Watering, Shading, Pricking out, Hardening off, Weed control, Pest control, Disease control,
I I,23 Budding: Meaning, Methods and procedure, Appropriate plants, Appropriate tools and materials.
Note: Learners to practise budding of orange scions on lemon root-stocks or other appropriate plants,
II l4 Grafting: Meaning, Methods and procedure, Appropriate plants, Appropriate tools
and materials.
Note: Learners to practice grqlIing on appropriate fruit trees.
1124 Importance of budding and grafting.
11.2.6 Layering: Methods, Importance, Appropriate crops/plants tbr layering, Materiah used in layering.
11.2.7 Tissue culture fbr crop propagation
11.2.8 Transplanting of vegetable seedlings from nursery to seedbed: Timing, Procedure and precautions
11.2.9 Transplanting of tree seedlings: Timing, Digging appropriate holes, Planting including firming and watering, Protecting the seedlings after transplanting
– Shading
– Damage caused by animals on tree seedlings and how to prevent it.
12.0.0 CROP PRODUCTION IV (FIELD PRACTICES I
12.IS Specific Objectives
By the end of the topic the learner should be able tot
(a) define crop rotation;
(I,) state the importance of crop rotation;
(c) draw a crop rotation programme;
(d) distinguish terms used in crop fanning;
(e) state the importance of mulching in crop production;
(f) describe the importance of various routine field practices In crop production;
(g) catty out various field practices;
(h) state the correct stage rot harvesting various crops;
(i) describe harvesting practices for various crops
122O Content
12.2.1 Crop rotation: Definition, Importance, Factors influencing crop rotation, Rotational programmes.
12.12 Terms used in crop production; Monocropping, intercropping, Mixed cropping
12.2.3 Mulching; Meaning, Importance, Types of mulching materials (organic, inorganic), Advantages and disadvantages of mulching materials.
12.24 Routine field practices: Thinning, Rogueing; Gapping, Training Pruning( Coffee single and multiple stem, capping, de-suckering, changing cycles; banana stool management; pyrethrum – cutting back), Earthing up, Crop protection (weed control pests and disease control
Note:- Study the importance and timing of each activity and the appropriate kite crops.
12.2.5 Harvesting: Stage and timing of harvesting, Methods of harvesting, Precautions during harvesting
12.2.6 Post = harvesting practices: Threshing/shelling, Drying, Cleaning, Sorting and grading, Dusting, Packaging.
12.2.7 Storage: Importance, types of storage, Preparation of *tore.
13.0.0 CROP PRODUCTION V (VEGETABLES)
13.1.1 Specific Objectives
By the end of the topic, the learner should be able to:
(a) grow a vegetable crop from nursery establishment to harvesting;
(b) keep a crop production records;
(c) market farm produce.
13.2.0 Content
13.2.1 Vegetable crops: Tomatoes (use varieties that require pruning and staking), Carrots, Onions, Cabbages/Kales.
14.0.0 CROP PRODUCTION VI (FIELD PRACTICES II
By the end of the topic, the learner should be able to:
(a) describe management practices in crop production;
(b) carry out management practices for a given crop;
(c) explain how crop production can be an economically lucrative activity.
14.2.0 Content
14.2.1 Production of: Maize/millet/sorghum, Beans
• Discuss the above crops under the following:-
– Meaning of hybrids, composites and cultivars
– Selecting best hybrids, composites or cultivars for given climatic regions.
– Raising of a maize/sorghum/millet and bean crop from seed bed preparation to harvesting.
– Keeping records in production of maize/sorghum millet and beans.
14.2.2 Rice production: Land preparation, Water control, Use of flooding in rice-field, Fertilizer application, Weed control.
14.2.3 Harvesting of the following crops: Cotton, Pyrethrum, Sugar cane , Tea, Coffee Under the following: Stage of harvesting; ; Method and procedure of harvesting; Precautions in harvesting.
Note: Compare cost of production with value of product for maize/sorghum/ millet and beans
15.0.0 FORAGE CROPS
15.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) define and classify pastures;
(b) identify forage crops;
(e) describe the ecological requirements of forage crops;
(d) describe the establishment and management of pastures and fodder;
(e) describe forage utilization and conservation.
15.2.0 Content
15.2.1 Pastures: Definition, Classification, Establishment, Management, Utilization – grazing systems -rotational grazing, herding, zero grazing.
15.2.2 Fodder crops: ; Napier/bana grass; Guatemala grass; Sorghum; Kale; Edible cana Lucerne; Clovers; Desmodium; Mangolds; Agro-forest trees/bushes used as fodder.
Nb. Study the above crops under the following: Ecological requirements, Establishment and management, Production per unit area, Utilization.
15.2.3 Forage conservation: Hay making, Silage making, Standing hay.
16.0.0 WEEDS AND WEED CONTROL
16.1.0 Specific Objectives
By the end of the topic, the learners should be able to:
(a) define a weed;
(b) identify weeds;
(e) classify weeds;
(d) explain the characteristics which make the weeds competitive;
(e) describe ways of controlling weeds;
(f) state harmful effects of weeds;
(g) control weeds;
(h) exercise safety measures to oneself, to crops and the environment while controlling weeds.
16.2.0 Content
16.2.1 Weeds: Definition of a weed, Weed identification and classification, Competitive ability of weeds (Appropriate examples for each ability), Harmful effects of weeds (appropriate examples for each effect).
Note:- See appendix Resources B on weeds to be studied.
16.2.2 Weed control methods: Chemical weed control (classes of herbicides, methods of application and safety measures in use of chemicals), Mechanical weed control, Cultural weed control, Biological weed control, Legislative control.
17.0.0 AGRO-FORESTRY
17.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) define agro-forestry;
(b) state the importance of agro-forestry;
(c) describe various forms of agro-forestry;
(d) explain the importance of trees;
(e) select appropriate trees for different uses;
(f) describe tree nursery management and transplanting;
(g) explain routine tree management;
(h) select appropriate sites for trees in the farm and other areas;
(i) describe various methods of tree harvesting.
17.2.0 Content
17.2.1 Definition of agro-forestry: Forms of agro-forestry
17.2.2 Importance of agro-forestry
17.2.3 Importance of trees and shrubs: Important trees and shrubs for particular purposes; Trees and shrubs to avoid at certain sites and reasons.
17.2.4 Tree nursery: Types of nurseries, Seed collection and preparation, Nursery management, Transplanting.
17.2.5 Care and management of trees: Protection, Pruning and training, Grafting old trees.
17.2.6 Agro-forestry practices: Alley cropping, Woodlots in farms.
17.2.7 Sites for agro-forestry trees: Boundaries, River banks, Terraces, Slopes, Homestead.
17.2.7 Tree harvesting methods.
18.0.0 LIVESTOCK PRODUCTION I (COMMON BREEDS)
18.1.0 Specific Objectives
By the end of the topic the learner should be able to:
(a) state the importance of livestock;
(1,) name various livestock species;
(e) define the terms livestock, breed and type;
(d) describe the various breed characteristics;
(e) state the origin of various livestock breeds;
(f) classify the various breeds into types;
(g) name the external parts of the various livestock species.
18.2.1 Content
18.2.2 Importance of livestock
18.2.3 Livestock species: Cattle (exotic and indigenous), Goats, Sheep, Pigs, Poultry (chicken), Rabbits, Camels.
Discuss each under the following: Breed, origin and characteristics, Type of each breed, External parts of each livestock species, Typical conformation
18.2.4 Terms used to describe livestock in different species by age, sex and use.
19.0.0 LIVESTOCK PRODUCTION III (SELECTION AND BREEDING)
19.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe reproduction and;
(b) reproductive systems;
(c) select breeding stock;
(d) describe breeding systems;
(e) identify signs of heat in livestock;
(f) describe methods used in serving livestock;
(g) describe signs of parturition in cattle, pigs and rabbits.
19.2.0 Content
19.2.1 Reproduction and reproductive systems: Cattle and Poultry.
19.2.2 Selection: Meaning; Factors to consider in selecting a breeding stock- Cattle, Sheep, Goats, Pigs, Camels; Methods of selection – mass selection , contemporary comparison, progeny testing.
19.2.3 Breeding: Meaning; Terms used in breeding – dominant and recessive genes, Heterosis (hybrid vigour), Epistasis; Breeding systems – Cross-breeding, Upgrading, Inbreeding, Line breeding, Out-crossing
Note: Discuss under the headings: Definition, Advantages and Disadvantages
19.2.4 Signs of heat in Cattle, Pigs and Rabbits.
Note: Study the oestrus cycle of each of the above.
19.2.5 Methods of service in livestock: Natural mating, Artificial insemination, Embryo transplant.
Note: Discuss advantages and disadvantages of each
19..2.6 Signs of Parturition in Cattle, Pigs and Rabbits.
Note: Learners to handle livestock in appropriate caring manner.
LIVESTOCK HEALTH I (INTRODUCTION TO LIVESTOCK HEALTH)
20.1.0 Specific Objectives
By the end of the topic the learner should be able to:
(a) define health and disease;
(b) describe signs of sickness in animals;
(c) state the predisposing factors of livestock diseases;
(d) categorize animal diseases;
(e) carry out disease control practices;
(fl state the importance of maintaining livestock healthy;
(g) describe appropriate methods of handling livestock.
20.2.0 Content
20.2.1 Health and disease: Definitions; Importance of keeping livestock healthy; Predisposing factors of livestock diseases; Signs of ill-health in livestock.
20.2.2 Classification of livestock diseases by cause.
20.2.3 General methods of disease control
20.2.4 Appropriate methods of handling livestock.
21.0.0 LIVESTOCK HEALTH II (PARASITES)
21.1.0 Specific Objectives
By the end of the topic the learner should be able to:
(a) describe host parasite relationship
(b) identify different parasites;
(c) describe the life-cycle of parasites;
(d) state signs & symptoms of attacks;
(e) explain methods of parasite control in livestock.
21.2.0 Content
21.2.1 Host: Parasite relationship; Effects of parasites on hosts.
21.2.2 External parasites: Ticks, Tsetse flies, Mites, Lice, Fleas, Keds
21.2.3 Internal parasites: Roundworms (Ascaris spp); Tapeworms (Taenia spp); Flukes (Fasciola spp).
Note: The parasites should be studied under the following: -Identification, Livestock species attacked, Part(s) of livestock attacked or inhabited and mode of feeding.
21.2.4 Signs and symptoms of attack.
21.2.5 Describe the life cycles of the following:Roundworm (Ascaris spp); Tapeworm (Taenia spp); Liver fluke (Fasciola spp); Ticks (appropriate example one host, two host , three host)
Note: Indicate whether soft or hard tick
21.2.6 Methods of parasite control giving appropriate example of a parasite for each method.
2.0.0 LIVESTOCK HEALTH III (DISEASES)
22.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe causes and vectors of main livestock diseases;
(b) state the incubation period;
(c) describe the signs of each disease;
(d) state the predisposing factors where applicable;
(e) carry out simple control measures of livestock diseases;
(fl state the measures taken to avoid environmental pollution.
22.2.0 Content
22.2.1 Protozoan diseases: East coast fever; Anaplasmosis; Coccidiosis; Trypanosomiasis (Nagana).
22.2.2 Bacterial diseases: Fowl typhoid; Foot rot; Contagious abortion (Brucellosis); Scours; Blackquarter; Mastitis; Anthrax; Pneumonia
22.2.3 Viral diseases: ; rinderpest; Foot and mouth ; Newcastle; Fowl pox; Gumboro; African Swine fever.
22.2.4 Nutritional diseases: Milk fever; Bloat.
All the above diseases should be studied under the following:
23.0.0 LIVESTOCK PRODUCTION II (NUTRITION)
23.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) identify and classify livestock feeds;
(b) describe digestion;
(c) define terms used to express feed values;
(d) compute a livestock ration;
(e) prepare balanced ration for various livestock;
(0 describe the appropriate livestock handling techniques while feeding.
23.2.0 Content
23.2.1 Livestock nutrition: Feeds and Feeding (identification, classification of feeds, terms used in expressing feed values, computation of livestock rations, preparation of livestock rations); Digestive systems (ruminant eg cattle, and non- ruminant eg pig and poultry); Digestion in cattle, pig and poultry.
23.2.2 Appropriate livestock handling techniques while feeding.
24.0.0 LIVESTOCK PRODUCTION IV (LIVESTOCK REARING PRACTICES)
24.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe livestock rearing practices;
(b) carry out livestock rearing practices;
(c) describe appropriate handling techniques of livestock during routine management.
24.2.0 Content
24.2.1 Routine livestock rearing practices: Feeding practices (flushing, steaming up, creep feeding); Parasites and Disease control practices (vaccination, dehorning, hoof trimming, docking, dipping/spraying, dusting); Breeding practices (crutching, tupping and serving, raddling, ringing); Identification; Debeaking; Tooth clipping; Culling: Describe general methods and carry out practicals on cattle, poultry; Castration (open, closed, caponization);
Management during parturition:- pigs, cattle, sheep, goats and rabbits.
24.2.2 Bee Keeping (Apiculture): Importance; Colony; Siting of the apiary and hive; Stocking a bee hive; Management — feeding and predator and pest control; Honey harvesting and processing.
24.2.3 Fish Farming (aquaculture): Importance; Types of fish kept in farm ponds; Management; Harvesting; Processing and preservation.
24.2.4 Appropriate handling of livestock during routine management.
25.0.0 LIVESTOCK PRODUCTION VI (CATTLE)
25.1.0 Specific objectives
By the end of the topic, the learner should be able to:
(a) raise young stock;
(b) describe milk by its components;
(c) describe milk secretion and let-down;
(d) milk using correct procedure and technique;
(e) describe marketing of beef cattle and milk;
25.2.0 Content
25.2.1 Raising young stock: ; Feeding; Weaning ; Housing; Routine practices.
25.2.2. Milk and milking: Milk composition,; Milk secretion and let down; Clean milk production (equipment and materials (include milking machine), cleanliness of the milk man /milk woman, milking procedure (by hand and by machine), Milking techniques); Dry cow therapy.
25.2.3 Marketing of milk
25.2.4 Marketing beef cattle.
26.0.0 LIVESTOCK PRODUCTION V (POULTRY)
26.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) identify parts of an egg;
(b) select eggs for incubation;
(e) describe conditions necessary for artificial incubation;
(d) identify, suitable sources of chicks;
(e) describe broodiness and natural brooding; (fl describe brooder and brooder management;
(g) describe rearing systems;
(h) describe the feeding for each age and category of poultry;
(i) identify stress and vices;
0) state the causes of stress and vices;
(k) state the effects of vices and stress in poultry;
(1) state control measures of vices and stress;
(m) describe marketing of eggs and poultry meat;
(n) select, sort and grade eggs for marketing;
(o) explain how poultry production can be an economically lucrative activity.
26.2.0 Content
262.1 Parts of an egg
26.2.2 Incubation: Meaning; Selection of eggs for incubation; Natural incubation (Signs of broodiness in poultry, Preparation and management of natural incubation); Artificial incubation (management of the incubator).
26.2.3 Sources of chicks
26.2.4 Brooding: ; Meaning; Natural brooding; Artificial brooding (brooder and brooder management, conditions equipment, management of layers and broilers.
26.2.5 Rearing systems: Extensive (free range); Semi-intensive (fold system); Intensive (deep litter and battery cage system.)
Note: Include advantages and disadvantages of each system.
26.2.6 Chicken feeding: Broilers and Layers.
26.2.7 Stress and vices in chicken: Identification; Causes; Control.
26.2.8 Marketing: Eggs —(-include grading of eggs for marketing) and meat.
27.0.0 FARM STRUCTURES
27.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe parts of a building;
(b) identify materials for construction;
(c) describe various farm structures and their uses;
(d) describe siting of various structures;
(e) construct and maintain farm structure.
27.2.0 Content
27.2.1 Farm building and structures: Siting; Parts of a building (foundation, wall, roof
27.2.2 Livestock buildings and structures: Crushes; Dips; Spray race; Dairy shed/parlour; Calf pens; Poultry houses and structures (deep litter, Coops, folds/arks, Runs, battery cages); Rabbit hutches/Rubbitry; Piggery/pigs sty; Fish ponds; Silos (for silage); Zero grazing unit; Bee hives.
27.2.3 Farm stores: Feed; Farm produce; Chemical; Machinery; Tools
27.2.4 Green house: Meaning; Construction materials used; Uses.
27.2.5 Fences in the farm: Types of fences and materials used; Uses — advantages and disadvantages; Gates and passes in fences; Fence reinforcement.
Note: Construct any of the following structures: a crush, a beehive, a hutch
28.0.0 FARM POWER AND MACHINERY
28.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) describe various sources of power in the farm;
(b) describe various systems of a tractor;
(c) describe the various tractor drawn implements, their uses & maintenance;
(d) describe the various animal drawn implements, their uses and maintenance;
(e) describe tractor service and maintenance practices.
282.0 Content
28.2.1 Sources of power in the farm: Human ; Animal; Wind; Water; Biomass (wood/charcoal, biogas); Fossil fuel (coal, petroleum, natural gas); Electrical (hydro, geothermal, nuclear, storage battery); Solar.
28.2.2 Tractor Engine: four stroke cycle engine (diesel and petrol); Two stroke cycle engine
28.2.3 Systems of the tractor: Fuel system; Electrical; Ignition; Cooling; Lubrication; Transmission (clutch, gears, differential, final drive).
28.2.4 Tractor service and maintenance
28.2.5 Tractor drawn implements, their uses and maintenance: Attachment methods (one point hitch – draw bar, three point hitch — hydraulic and power take off- P. T. 0); Implements (trailer, disc plough, mould board plough, harrows – disc [plain, notched], spike toothed, spring tined, sub-soilers, ridgers); Rotary tillers; Mowers (Gyro, reciprocating, planters and seeders); Cultivators/weeders; Sprayers; Harvesting machines (grain, root crops, forage); Shellers.
28.2.6 Animal drawn implements, uses and maintenance: ploughs; carts; ridgers.
29.0.0 AGRICULTURAL ECONOMICS I (BASIC CONCEPTS AND FARM RECORDS)
29.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) define economics and agricultural economics;
(b) explain basic concepts of economics;
(c) describe the importance of agricultural economics;
(d) explain the importance of farm records;
(e) describe the different types of farm records;
(0 keep farm records.
29.2.0 Content
29.2.1 Definition: Economics and Agricultural Economics.
29.2.2 Basic concepts of economics: Scarcity; Preferences and choice; Opportunity cost.
292.3 Uses of farm records
29.2.4 Types of farm records: Breeding; Feeding; Production; Health; Field operations; Inventory; Labour; Marketing.
30.0.0 AGRICULTURAL ECONOMICS II (LAND TENURE AND LAND REFORM)
30.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) define the term tenure;
(b) describe tenure systems;
(c) describe land reforms
30.2.0 Content
30.2.1 Land tenure: Definition; Tenure systems – (I) individual (types, advantages and disadvantages) and (ii) Collective (description, advantages and disadvantages).
30.2.2 Land reforms: definition; types of reform and reasons for each (fragmentation, consolidation, adjudication, registration (emphasize the importance of a title deed); Settlement and resettlement.
31.0.0 AGRICULTURAL ECONOMICS III (PRODUCTION ECONOMICS)
31.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) explain various parameters of national development;
(b) relate national development to agricultural production;
(c) state the factors of production and explain how each affects production;
(d) describe how the law of diminishing returns relates to agricultural production;
(e) describe agricultural planning and budgeting in a farming business;
(0 state sources of agricultural support services;
(g) describe risks and uncertainties in farming;
(h) explain ways of adjusting to risks and uncertainties.
31.2.0 Content
31 .2.1 National income: Household-firm relationship; Gross Domestic Product (GDP); Gross National Product (GNP); Per Capita Income; Contribution of agriculture to national development.
31.2.2 Factors of production: Land (definition and methods of acquisition); Labour (definition, types, measures of labour, ways of increasing labour efficiency; Capital (definition, types and sources); Management (definition, role of a farm manager)
31.2.3 Production function: Increasing returns; Constant returns; Decreasing returns
3 1.2.4 Economic laws and principle: The law of diminishing returns; The law of substitution; The law of equimarginal returns; Principle of profit maximization.
31.2.5 Farm planning: Meaning; Factors to consider; Steps
31.2.6 Farm budgeting: Definition; Importance; Types (partial and complete)
3 1.2.7 Agricultural services available to the farmer
3 1.2.8 Risks and uncertainties in farming: Meaning; Common risks and uncertainties; Ways of adjusting.
32.0.0 AGRICULTURAL ECONOMICS IV (FARM ACCOUNTS)
32.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) state the importance of farm accounts;
(b) distinguish and describe the various financial documents and their uses;
(c) identify various books of accounts and their uses;
(d) prepare and analyse financial statements.
32.2.0 Content
32.2.1 Financial documents and books of accounts: Financial documents (Invoices,
Statements, Receipts, Delivery notes, Purchase orders); Books of Accounts
(Ledger, Journal, Inventory, Cash book); Financial statements; Cash analysis;
Balance sheet; Profit and loss account.
33.0.0 AGRICULTURAL ECONOMICS V (AGRICULTURAL MARKETING AND ORGANIZATIONS)
33.1.0 Specific Objectives
By the end of the topic, the learner should be able to:
(a) define market and marketing;
(b) describe the various types of markets;
(c) describe how the law of supply and demand affects the prices of agricultural products;
(d) state various marketing functions, agents and institutions;
(e) identi problems in marketing of agricultural products;
(f) list various agricultural organizations;
(g) describe the role of each of the agricultural organizations.
33.2.0 Content
33.2.1 Market and marketing
33.2.2 Types of markets
33.2.3 Demand, supply and price theory
33.2.4 Marketing functions
33.2.5 Problems of marketing agricultural products and possible solutions
33.2.6 Marketing boards, agents and institutions
33.2.7 Co-operatives: Formation; Functions
33.2.8 Associations and unions: Agricultural society of Kenya (ASK); Young Farmers Clubs (YFC); Kenya National Farmers Union (KNFU); Agricultural based Women groups.
APPENDIX RESOURCES
A TOOLS AND EQUIPMENT TO BE STUDIED
1. GARDEN TOOLS AND EQUIPMENT TO BE STUDIED
Panga Knap-sack sprayer
Axe Sprinkler
Mattock/pick axe Hose pipe
Jembe/hoe Garden shear
Fork jembe Pruning saw
Spade Pruning knife
Wheelbarrow Meter rule
Watering can Secateurs
Rake Garden fork
Tape measure Pruning-hook
Soil auger Levelling boards.
Elastrator Stir-up pump
Burdizzo Milk churn
Syringes and needles Strainer/sieve
Thermometer Rope
Halter Milking stool
Hoof trimmer Weighing balance
Strip cup Hot iron
Trochar and canula Tooth clipper
Hard broom Drenching gun
Wool shears Dosing gun
Ear notcher Bolus gun
Bull ring and leading stick Dehorning wire
Bucket Chaff cutter
Cross cut saw Jack plane
Tenon/back saw Scrapper
Coping saw Try square
Compass saw/key hole saw Wood clamp
Rip saw Sash clamp
Bow saw G-clamp
Hack saw Mallet
Wood chisel Soldering gun
Cold chisel Tin-snip
Mes and rasps Claw hammer
Divider Sledge hammer
Center punch Wire strainer
Spoke share Pliers
Screw drivers Brace and bits
Spanners Hand drills and bits
Pipe wrench Riveting machine
Pipe cutter Crow bar
Levelling rod Masons’ square
Spirit level Plumb bob
Mason’s trowel Metal float
Wood float Shovel
Meter rule
COMMON NAME BOTANICAL NAME
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 3 | 1 | Introduction to computers | By the end of the lesson, the learner should be able to:
i) define the terms: -computer -data -information -information communication system ii) explain the parts of a computer |
– Discussion
method – Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa 1/Ed (2004) P 1-4 Nairobi Longhorn Publishers
Aid Chart on main parts of a computer system
|
|
| 2 & 3 | Classification of computers | By the end of the lesson, the learner should be able to:
i) describe the following computers – supercomputer – mainframe -minicomputer -microcomputer ii) classify computer according to purpose |
– Brief lecture method
– Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa 1/Ed (2004) P 4-8 Nairobi Longhorn Publishers
Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah 4/Ed (2005) P 9-14 Nairobi, SunLitho |
||
| 4 | 1 | Classification according to functionality | By the end of the lesson, the learner should be able to:
i) explain of digital computers ii) explain of analog computers iii) explain on hybrid computers
Assignment Attempt review questions 1.1 No. 1,2…..8 |
– Discussion
method – Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa 1/Ed (2004) P 8-9 Nairobi Longhorn Publishers |
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 4 | 2 & 3 | Development of computers | By the end of the lesson, the learner should be able to:
i) name the advantages of using computers ii) outline where computers are used iii) explain the five generation of computers namely: – first generation – second generation – third generation – forth generation -fifth generation |
– Brief lecture method
– Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 11-15 Longhorn Publishers Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 6-8
|
|
| 4 | The computer laboratory | By the end of the lesson, the learner should be able to:
i) outline the factors to be considered when preparing computer laboratory ii) explain on safety precaution and practices |
– Discussion
method – Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 17-19
|
||
| 5 | 1 | Practical hands-on skill | By the end of the lesson, the learner should be able to:
i) define the terms:
ii) explain on post process iii) explain the procedure of shutting down the computer |
– Question and
answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 21-22
|
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
FORM : ONE
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 5 | 2 & 3 | Keyboard layout | By the end of the lesson, the learner should be able to:
i) identify the keys: -alphanumeric keys – function keys ii) use the following keys: – cursor movement and editing keys – special PC operation keys – numeric keypads keys iii) explain the practical keyboard skills
|
– Discussion
method – Question and answer method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 23-26 Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 32-37 Aid Computer System |
|
| 6 | 1 | Mouse skills | By the end of the lesson, the learner should be able to:
i) give the purpose of mouse pointer ii) outline the rules observed when using a mouse iii) explain the terminologies associated with the use of mouse – clicking – double clicking – right clicking – drag and drop |
– Brief lecture method
– Question and answer method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 27-28 Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 39-40
|
TERM : 1 TEACHER:
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 6 | 2 & 3 | COMPUTER SYSTEM
Description of a computer system |
By the end of the lesson, the learner should be able to:
i) name three main components of a computer system ii) list data capture devices iii) explain the pointing devices
|
– Discussion
method – Question and answer method |
Longhorn Bk 1 Computer Studies
By S. Mburu and G. Chemwa P 30-33
Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 43-50 |
|
| 7 | 1,2
& 3 |
MID TERM EXAMS,
THEN SCHOOL BREAK FOR HALF TERM/ RECESS |
||||
| 8 | 1 | Scanning devices | By the end of the lesson, the learner should be able to:
i) name two types of scanners ii) explain the scanning devices such as: – optical scanners – optical mark recognition (OMR) -optical bar recognition (OBR) – optical character recognition (OCR) -magnetic scanners Assignment The learner to read and write notes on the central processing unit Ref: Longhorn Bk 1 Computer studies P 37-39 |
– Question and
answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 33-34 Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 58-63 Aid – Scanner – Bar cords |
FORM : ONE
TERM : 1 TEACHER:
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 8 | 2 & 3 | Speech recognition or voice input | By the end of the lesson, the learner should be able to:
i) define the term speech recognition or voice input ii) outline the use of speech input iii) explain the limitation of speech input |
– Question and
answer method – Discussion method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 35-37 Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 56-57 |
|
| 9 | 1 | The central processing unit (CPU) | By the end of the lesson, the learner should be able to:
i) define the term central processing unit ii) explain the components of central processing unit iii) classify the computer memory and explain read only memory (ROM)
|
– Question and answer method – Discussion method |
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 37-40
Ref Computer Studies Bk 1 By Dr. John Onunga &Renu Shah P 69-73
|
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 9 | 2 & 3 | Main memory (primary storage or working storage) | By the end of the lesson, the learner should be able to:
i) explain the main memory namely: -random access memory -special purpose memory ii) give the characteristics and types of: – random access memory iii) explain the memory capacity
Assignment The learner to read and write notes on output devices Ref: Longhorn Bk 1 Computer Studies P 45-47 |
– Question and
answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 40-41
Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 73-75 |
|
| 10 | 1 | Speech recognition or voice input | By the end of the lesson, the learner should be able to:
i) define the term speech recognition or voice input ii) outline the use of speech input iii) explain the limitation of speech input |
– Discussion
method – Question and answer method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 35-37 Ref Computer Studies Bk 1 By Dr. John Onunga &Renu Shah P 56-57 |
YEAR : 2011 HEAD OF DEPT:
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 10 | 1 | Overall functional organizational of the CPU | By the end of the lesson, the learner should be able to:
i) explain three types of buses -control bus – address bus – data bus ii) describe types of processors and their clock speeds |
– Question and
answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 42-43
Ref Computer Studies Bk 1 By Dr.John Onunga &Renu Shah P 49-51 |
|
| 2 & 3 | Output devices | By the end of the lesson, the learner should be able to:
i) define the term output devices ii) name two types of output e.g. – softcopy – hardcopy iii) explain two types of output devices namely: – softcopy output devices – hardcopy output devices iii) describe types of graphical adapters e.g. – hercules graphics card – color graphics adapter – enhanced graphics adapter – super video graphics array – liquid crystal display
|
– Question and answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 45-47
Ref Computer Studies Bk 1 By Dr. john Onunga & Renu Shah P 79-83
Aid Computer CRT monitor and Flat panel screen |
YEAR : 2011 HEAD OF DEPT:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 11 | 1 | Sound output and hard copy output devices | By the end of the lesson, the learner should be able to:
i) explain the examples of: -sound output devices -light-emitting diodes ii) describe examples of hard copy devices namely: – impact printers – non-impact printers – plotters |
– Question and
answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 47-50
Ref Computer Studies Bk 1 By Dr.John Onunga & Renu Shah P 85-94
Aid Computer system and projector on plotters
|
|
| 2 & 3 | Secondary (auxiliary) storage devices and media | By the end of the lesson, the learner should be able to:
i) define the terms auxiliary media ii) name disadvantages of using magnetic tape, and care taken to magnetic storage media iii) explain various removable storage devices such as: – magnetic tape – magnetic disk -zip disk – jaz disk |
– Question and answer method – Discussion method
|
Ref
Longhorn Bk 1 Computer Studies By S. Mburu and G. Chemwa P 51-54
Ref Computer Studies Bk 1 By Dr. John Onunga & Renu Shah P 96-104 |
Computer Studies Schemes of Work
FORM : ONE
TERM : 1 TEACHER:
YEAR : 2011 HEAD OF DEPT:
|
WK
|
LESSON |
TOPIC/ SUB-TOPIC |
OBJECTIVES
|
TEACHING METHOD
|
TEACHING/ LEARNING RESOURCES
|
REMARKS
|
| 12
and 13 |
|
End of Term Examination
|
By the end of the test lesson, the learner should be able to:
i. answer all question in the test
ii. score at least 75% of the questions correctly
|
Pre-test activities – learners are introduced to the purpose of the test and encouraged to put forth their test efforts – teacher ensure proper seating arrangements, adequate lighting and a quiet environment – test papers are distributed to learners – instructions are clearly given to learners and corrections made on possible errors such as typographical errors – total time is announce – attempt all the questions in the test sheet
Post test activities – the teacher invigilate the exam – time briefing is done at interval of 30 minutes – answer sheets are collected – getting feedback from students about the exam |
Learner to:
i. spread out in an orderly manner before getting the question paper ii. maintain total silence iii. receive the question paper from the invigilator (teacher) and write their name. iv. read the instructions and attempt all the questions in the test sheet as instructed v. hand their scripts (answer sheet) after the session is over vi. give the feedback after they have collected the scripts |
Computer Studies Schemes of Work
Naturally, there are basically three states of matter: Solid, Liquid and gas:
(i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density.
(ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space.
(iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container.
2.Gases are affected by physical conditions. There are two physical conditions:
(i)Temperature
(ii)Pressure
Degrees Celsius/Centigrade(oC) are also used.
The two units can be interconverted from the relationship:
oC + 273= K
K -273 = oC
Practice examples
(i) O oC
oC + 273 = K substituting : O oC + 273 = 273 K
(ii) -273 oC
oC + 273 = K substituting : -273oC + 273 = 0 K
(iii) 25 oC
oC + 273 = K substituting : 25 oC + 273 = 298 K
(iv) 100 oC
oC + 273 = K substituting : 100 oC + 273 = 373 K
(i) 10 K
K -273 = oC substituting: 10 – 273 = -263 oC
(ii) (i) 1 K
K -273 = oC substituting: 1 – 273 = -272 oC
(iii) 110 K
K -273 = oC substituting: 110 – 273 = -163 oC
(iv) -24 K
K -273 = oC substituting: -24 – 273 = -297 oC
The standard temperature is 273K = 0 oC.
The room temperature is assumed to be 298K = 25oC
The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2).
The standard pressure is the atmospheric pressure.
Atmospheric pressure is equal to about:
(i)101325 Pa
(ii)101325 Nm-2
(iii)760 mmHg
(iv)76 cmHg
(v)one atmosphere.
Physical conditions change the volume occupied by gases in a closed system.
The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws.
“the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ”
Mathematically:
Volume α 1 (Fixed /constant Temperature)
Pressure
V α 1 (Fixed /constant T) ie PV = Constant(k)
P
From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved.
Graphically therefore a plot of volume(V) against pressure (P) produces a curve.
V
P
Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line
V
1/P
For two gases then P1 V1 = P2 V2
P1 = Pressure of gas 1
V1 = Volume of gas 1
P2 = Pressure of gas 2
V2 = Volume of gas 2
Practice examples:
Working
P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2
V2 = 102300 x 25 = 12.5cm3
(102300 x 2)
P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3)
V2 = 100000 x 100 = 33333.3333 Nm-2
(100 x 3)
3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?
P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2
V2 = 101325 x 60 = 67.55 cm3
90000
The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination.
7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ”
Mathematically:
Volume α Pressure (Fixed /constant pressure)
V α T (Fixed /constant P) ie V = Constant(k)
T
From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled.
Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure.
The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas.
This temperature is called absolute zero. It is the lowest temperature at which a gas can exist.
Graphically therefore a plot of volume(V) against Temperature(T) in:
(i)oC produces a straight line that is extrapolated to the absolute zero of -273oC .
V
-273oC 0oC
T(oC)
(ii)Kelvin/K produces a straight line from absolute zero of O Kelvin
V
0 T(Kelvin)
For two gases then V1 = V2
T1 T2
T1 = Temperature in Kelvin of gas 1
V1 = Volume of gas 1
T2 = Temperature in Kelvin of gas 2
V2 = Volume of gas 2
Practice examples:
V1 = V2 substituting 500 = V2
T1 T2 (0 +273) (-4 +273)
= 500 x (-4 x 273) = 492.674cm3
(0 + 273)
The capacity of cylinder (500cm3) is less than new volume(492.674cm3).
7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder.
V1 = V2 substituting 40000 = V2
T1 T2 (25 +273) (30 +273)
= 40000 x (30 x 273) = 40671.1409cm3
(25 + 273)
The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3).
The tyre thus bursts.
V1 = V2 substituting 80 = V2
T1 T2 (25 +273) (-30 +273)
= 80 x (-30 x 273) = 65.2349cm3
(25 + 273)
The capacity of balloon (80cm3) is more than new volume (65.2349cm3).
The balloon thus remained intact.
The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas.
The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion.
Examples
1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2.
2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2.
The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that:
“the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure”
Mathematically
R α 1 and since density is proportional to mass then R α 1
√ p √ m
For two gases then:
R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas.
√M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas.
Since rate is inverse of time. i.e. the higher the rate the less the time:
For two gases then:
T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse.
√M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas.
Practice examples:
Molar mass CO2=44.0 Molar mass NO2=46.0
Method 1
100cm3 CO2 takes 30seconds
150cm3 takes 150 x30 = 45seconds
100
T CO2 = √ molar mass CO2 => 45seconds = √ 44.0
T NO2 √ molar mass NO2 T NO2 √ 46.0
T NO2 =45seconds x √ 46.0 = 46.0114 seconds
√ 44.0
Method 2
100cm3 CO2 takes 30seconds
1cm3 takes 100 x1 = 3.3333cm3sec-1
30
R CO2 = √ molar mass NO2 => 3.3333cm3sec-1 = √ 46.0
R NO2 √ molar mass CO2 R NO2 √ 44.0
R NO2 = 3.3333cm3sec-1 x √ 44.0 = 3.2601cm3sec-1
√ 46.0
3.2601cm3 takes 1seconds
150cm3 take 150cm3 = 46.0109seconds
3.2601cm3
Molar mass CO2 = 44g Molar mass HCl = 36.5g
T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0
T HCl √ molar mass HCl T HCl √ 36.5
T HCl = 200seconds x √ 36.5 = 182.1588 seconds
√ 44.0
Molar mass O2 = 32g Molar mass Z = x g
T O2 = √ molar mass O2 => 250 seconds = √ 32.0
T Z √ molar mass Z 227seconds √ x
√ x = 227seconds x √ 32 = 26.3828 grams
250
Molar mass CO2 = 44.0 Molar mass CO = 28.0
Method 1
25cm3 CO takes 25seconds
75cm3 takes 75 x25 = 75seconds
25
T CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0
T CO √ molar mass CO 75 √ 28.0
T CO2 =75seconds x √ 44.0 = 94.0175 seconds
√ 28.0
Method 2
25cm3 CO2 takes 25seconds
1cm3 takes 25 x1 = 1.0cm3sec-1
25
R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0
R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0
R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1
√ 44.0
0.7977cm3 takes 1 seconds
75cm3 takes 75cm3 = 94.0203seconds
0.7977cm3
(b)Introduction to the mole, molar masses and Relative atomic masses
The Avogadros Constant contain 6.023 x10 23 particles. i.e.
1mole = 6.023 x10 23 particles = 6.023 x10 23
2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24
0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22
0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21
(i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g.
Molar mass of carbon(C)= relative atomic mass = 12.0g
6.023 x10 23 particles of carbon = 1 mole =12.0 g
Molar mass of sodium(Na) = relative atomic mass = 23.0g
6.023 x10 23 particles of sodium = 1 mole =23.0 g
Molar mass of Iron (Fe) = relative atomic mass = 56.0g
6.023 x10 23 particles of iron = 1 mole =56.0 g
(ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule.
The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.
Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g
6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g
6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g
Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g
6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g
(ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
(i)Molar mass Water(H2O) = relative formular mass =[(1.0 x 2 ) + 16.0]g =18.0g
6.023 x10 23 particles of Water molecule = 1 mole = 18.0 g
6.023 x10 23 particles of Water molecule has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-1 x 6.023 x10 23 particles of Oxygen atoms
(ii)Molar mass sulphuric(VI)acid(H2SO4) = relative formular mass
=[(1.0 x 2 ) + 32.0 + (16.0 x 4)]g =98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) = 1 mole = 98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
–1 x 6.023 x10 23 particles of Sulphur atoms
-4 x 6.023 x10 23 particles of Oxygen atoms
(iii)Molar mass sodium carbonate(IV)(Na2CO3) = relative formular mass
=[(23.0 x 2 ) + 12.0 + (16.0 x 3)]g =106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) = 1 mole = 106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) has:
– 2 x 6.023 x10 23 particles of Sodium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(iv)Molar mass Calcium carbonate(IV)(CaCO3) = relative formular mass
=[(40.0+ 12.0 + (16.0 x 3)]g =100.0g.
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) = 1 mole = 100.0g
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) has:
– 1 x 6.023 x10 23 particles of Calcium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(v)Molar mass Water(H2O) = relative formular mass
=[(2 x 1.0 )+ 16.0 ]g =18.0g
6.023 x10 23 particles of Water(H2O) = 1 mole = 18.0g
6.023 x10 23 particles of Water(H2O) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-2 x 6.023 x10 23 particles of Oxygen atoms
Practice
(i)0.23 g of Sodium atoms
Molar mass of Sodium atoms = 23g
Moles = mass in grams = > 0.23g = 0.01moles
Molar mass 23
(ii) 0.23 g of Chlorine atoms
Molar mass of Chlorine atoms = 35.5 g
Moles = mass in grams = > 0.23g = 0.0065moles /6.5 x 10-3 moles
Molar mass 35.5
(iii) 0.23 g of Chlorine molecules
Molar mass of Chlorine molecules =( 35.5 x 2) = 71.0 g
Moles = mass in grams = > 0.23g = 0.0032moles /3.2 x 10-3 moles
Molar mass 71
(iv) 0.23 g of dilute sulphuric(VI)acid
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
(i) 0.23 g of dilute sulphuric (VI)acid
Method I
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
1 mole has 6.0 x 10 23 atoms
2.3 x 10-3 moles has (2.3 x 10-3 x 6.0 x 10 23) = 1.38 x 10 21 atoms
1
Method II
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
98.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 1.38 x 10 21 atoms
98
(ii)0.23 g of sodium carbonate(IV)decahydrate
Molar mass of Na2CO3.10H2 O=
[(2 x 23) + 12 + (3 x16) + (10 x 1.0) + (10 x 16)] = 276.0g
Method I
Moles = mass in grams = > 0.23g = 0.00083moles /
Molar mass 276 8.3 x 10-4 moles
1 mole has 6.0 x 10 23 atoms
8.3 x 10-4 moles has (8.3 x 10-4 moles x 6.0 x 10 23) = 4.98 x 10 20 atoms
1
Method II
276.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 4.98 x 10 20 atoms
276.0
(iii)0.23 g of Oxygen gas
Molar mass of O2 = (2 x16) = 32.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00718moles /
Molar mass 32 7.18 x 10-3 moles
1 mole has 2 x 6.0 x 10 23 atoms in O2
7.18 x 10-3moles has (7.18 x 10-3moles x 2 x 6.0 x 10 23) =8.616 x 10 21atoms
1
Method II
32.0g = 1 mole has 2 x 6.0 x 10 23 atoms in O2
0.23 g therefore has (0.23 g x 2 x 6.0 x 10 23 ) = 8.616 x 10 21atoms
32.0
(iv)0.23 g of Carbon(IV)oxide gas
Molar mass of CO2 = [12 + (2 x16)] = 44.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00522moles /
Molar mass 44 5.22 x 10-3 moles
1 mole has 3 x 6.0 x 10 23 atoms in CO2
7.18 x 10-3moles has (5.22 x 10-3moles x 3 x 6.0 x 10 23) =9.396 x 10 21atoms
1
Method II
44.0g = 1 mole has 3 x 6.0 x 10 23 atoms in CO2
0.23 g therefore has (0.23 g x 3 x 6.0 x 10 23 ) = 9.409 x 10 21atoms
44.0
(c)Empirical and molecular formula
1.The empirical formula of a compound is its simplest formula. It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound
3.Practically the empirical formula of a compound can be determined as in the following examples.
To determine the empirical formula of copper oxide
(a)Method 1:From copper to copper(II)oxide
Procedure.
Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3).
Sample results
| Mass of crucible(M1) | 15.6g |
| Mass of crucible + copper before heating(M2) | 18.4 |
| Mass of crucible + copper after heating(M3) | 19.1 |
Sample questions
Mass of crucible + copper before heating(M2) = 18.4
Less Mass of crucible(M1) = – 15.6g
Mass of copper 2.8 g
Method I
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible + copper before heating(M2) = – 18.4g
Mass of Oxygen = 0.7 g
Method II
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible = – 15.6g
Mass of copper(II)Oxide = 3.5 g
Mass of copper(II)Oxide = 3.5 g
Mass of copper = – 2.8 g
Mass of Oxygen = 0.7 g
(i) copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
Observation
Colour change from brown to black
Explanation
Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide
Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesium oxide and magnesium nitride. This causes experimental mass errors.
(b)Method 2:From copper(II)oxide to copper
Procedure.
Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below;
Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator.
When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating.
Continue passing the gases until the glass tube is cool.
Turn off the gas generator.
Carefully remove the porcelain boat form the combustion tube.
Reweigh (M3).
Sample results
| Mass of boat(M1) | 15.6g |
| Mass of boat before heating(M2) | 19.1 |
| Mass of boat after heating(M3) | 18.4 |
Sample questions
Mass of boat before heating(M2) = 19.1
Mass of empty boat(M1) = – 15.6g
Mass of copper(II)Oxide 3.5 g
(i) Oxygen.
Mass of boat before heating(M2) = 19.1
Mass of boat after heating (M3) = – 18.4g
Mass of oxygen = 0.7 g
(ii)Copper
Mass of copper(II)Oxide = 3.5 g
Mass of oxygen = 0.7 g
Mass of oxygen = 2.8 g
(i) Copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
Observation
Colour change from black to brown
Explanation
Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal.
Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal.
(i) Hydrogen
CuO(s) + H2(g) -> Cu(s) + H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )
(ii)Carbon(II)oxide
CuO(s) + CO (g) -> Cu(s) + CO2(g)
(Black) (brown) (colourless gas, form
white ppt with lime water )
(iii)Ammonia
3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )
(i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide.
Air combine with hydrogen in presence of heat causing an explosion
(ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped.
Hot metallic copper can be re-oxidized back to copper(II)oxide
(iii) A stream of excess carbon (II)oxide gas should be ignited to burn
Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas.
(i)All copper(II)oxide may not be reduced to copper.
(ii)Some copper(II)oxide may be blown out the boat by the reducing agent.
4.Theoreticaly the empirical formula of a compound can be determined as in the following examples.
(a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
% of Oxygen = 100% – % of Copper => 100- 80 = 20% of Oxygen
| Element | Copper | Oxygen |
| Symbol | Cu | O |
| Moles present = % composition
Molar mass |
80
63.5 |
20
16 |
| Divide by the smallest value | 1.25
1.25 |
1.25
1.25 |
| Mole ratios | 1 | 1 |
Empirical formula is CuO
(b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0)
Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen
| Element | Magnesium | Oxygen |
| Symbol | Mg | O |
| Moles present = % composition
Molar mass |
0.84
24 |
0.56
16 |
| Divide by the smallest value | 0.35
0.35 |
0.35
0.35 |
| Mole ratios | 1 | 1 |
Empirical formula is MgO
(c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0)
Mass of Oxygen = 100 – 47 => 53% of Oxygen
| Element | Silicon | Oxygen |
| Symbol | Si | O |
| Moles present = % composition
Molar mass |
47
28 |
53
16 |
| Divide by the smallest value | 1.68
1.68 |
3.31
1.68 |
| Mole ratios | 1 | 1.94 = 2 |
Empirical formula is SiO2
(d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0)
Mass of Oxygen = 100 – 47 => 53% of Oxygen
| Element | Silicon | Oxygen |
| Symbol | Si | O |
| Moles present = % composition
Molar mass |
47
28 |
53
16 |
| Divide by the smallest value | 1.68
1.68 |
3.31
1.68 |
| Mole ratios | 1 | 1.94 = 2 |
Empirical formula is SiO2
2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained:
Mass of evaporating dish =300.0g
Mass of evaporating dish + hydrated salt = 305.0g
Mass of evaporating dish + anhydrous salt = 303.2g
Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI)
(Cu =64.5, S = 32.0,O=16.0, H = 1.0)
Working
Mass of Hydrated salt = 305.0g -300.0g = 5.0g
Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g
Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g
Molar mass of water(H2O) = 18.0g
Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g
| Element/compound | anhydrous copper (II)
sulphate(VI) |
Oxygen |
| Symbol | CuSO4 | O |
| Moles present = composition by mass
Molar mass |
3,2
160.5 |
1.8
18 |
| Divide by the smallest value | 0.0199
0.0199 |
0.1
18 |
| Mole ratios | 1 | 5 |
The empirical formula of hydrated salt = CuSO4.5H2O
Hydrated salt has five/5 molecules of water of crystallizations
The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.
The molecular formula is a multiple of empirical formula .It is determined from the relationship:
(i) n = Relative formular mass
Relative empirical formula
where n is a whole number.
(ii) Relative empirical formula x n = Relative formular mass
where n is a whole number.
Practice sample examples
If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0)
Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = % composition
Molar mass |
92.3
12 |
7.7
1 |
| Divide by the smallest value | 7.7
7.7 |
7.7
7.7 |
| Mole ratios | 1 | 1 |
Empirical formula is CH
The molecular formular is thus determined :
n = Relative formular mass = 78 = 6
Relative empirical formula 13
The molecular formula is (C H ) x 6 = C6H6
If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0)
| Element | Carbon | Hydrogen | Oxygen |
| Symbol | C | H | O |
| Moles present = % composition
Molar mass |
54.55
12 |
9.09
1 |
36.36
16 |
| Divide by the smallest value | 4.5458
2.2725 |
9.09
2.2725 |
2.2725
2.2725 |
| Mole ratios | 2 | 4 | 1 |
Empirical formula is C2H4O
The molecular formula is thus determined :
n = Relative formular mass = 88 = 2
Relative empirical formula 44
The molecular formula is (C2H4O ) x 2 = C4H8O2.
4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water.
If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure.
Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:
Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 =>
Molar mass of CO2
12 x 5.28 = 1.44g√
44
Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O =>
Molar mass of H2O
2 x 2.16 = 0.24g√
18
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = mass
Molar mass |
1.44g
12 |
0.24g√
1 |
| Divide by the smallest value | 0.12
0.12 |
0.24
0.12
|
| Mole ratios | 1 | 2√ |
Empirical formula is CH2√
The molecular formular is thus determined :
n = Relative formular mass = 84 = 6√
Relative empirical formula 14
The molecular formula is (CH2 ) x 6 = C6H12. √
molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene)
Molecular structure
H H H H H H
H C C C C C C H√
H H H H
Mass of N in A = 5.2% x 0.085 = 0.00442 g
Mass of C in A = 12 x 0.224 = 0.0611g
44
Mass of H in A = 2 x 0.0372 = 0.0041g
18
Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O)
=> 0.0611g + 0.0041g = 0.0652g (Mass of C,H)
0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g
| Element | Nitrogen | Carbon | Hydrogen | Oxygen |
| Symbol | N | C | H | O |
| Moles present = mass
Molar mass |
0.00442 g
14 |
0.0611g
12 |
0.0041g
1 |
0.0154 g 16 |
| Divide by the smallest value | 0.00032
0.00032 |
0.00509
0.00032 |
0.0041g
0.00032 |
0.00096
0.00032 |
| Mole ratios | 1 | 16 | 13 | 3 |
Empirical formula = C16H13NO3
(d)Molar gas volume
The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is:
(i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p).
i.e. 1mole of all gases =24dm3/24litres/24000cm3 at r.t.p
Examples
1mole of O2 = 32g =6.0 x1023 particles= 24dm3/24litres/24000cm3 at r.t.p
1mole of H2 = 2g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
1mole of CO2 = 44g = 6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
1mole of NH3 = 17g =6.0 x1023 particles = 24dm3/24litres/24000cm3 at r.t.p
1mole of CH4 = 16g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
(ii)22.4dm3/22.4litres/22400cm3 at standard temperature(0oC/273K) and pressure(s.t.p)
i.e. 1mole of all gases =22.4dm3/22.4litres/22400cm3 at s.t.p
Examples
1mole of O2 = 32g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of H2 = 2g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of CO2 = 44g = 6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of NH3 = 17g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of CH4 = 16g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
The volume occupied by one mole of a gas at r.t.p or s.t.p is commonly called the molar gas volume. Whether the molar gas volume is at r.t.p or s.t.p must always be specified.
From the above therefore a less or more volume can be determined as in the examples below.
Practice examples
(Avogadros constant =6.0 x1023mole-1 )
(i) 2.24dm3 of Oxygen.
22.4dm3 -> 6.0 x1023
2.24dm3 -> 2.24 x 6.0 x1023
22.4
=6.0 x1022 molecules = 2 x 6.0 x1022. = 1.2 x1023 atoms
(ii) 2.24dm3 of Carbon(IV)oxide.
22.4dm3 -> 6.0 x1023
2.24dm3 -> 2.24 x 6.0 x1023
22.4
=6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms
Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√
Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√
120cm3
Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:
Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 =>
Molar mass of CO2
12 x 0.41 = 0.1118g√
44
Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O =>
Molar mass of H2O
2 x 0.209 = 0.0232g√
18
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = % composition
Molar mass |
0.g118
12 |
0.0232g√
1 |
| Divide by the smallest value | 0.0093
0.0093 |
0.0232
0.0093√
|
| Mole ratios | 1 x2 | 2.5×2 |
| 2 | 5√ |
Empirical formula is C2H5√
The molecular formular is thus determined :
n = Relative formular mass = 58 = 2√
Relative empirical formula 29
The molecular formula is (C2H5 ) x 2 = C4H10.√
Molecule name Butane
Molecula structure
H H H H
H C C C C H√
H H H H
(e)Gravimetric analysis
Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples:
Chemical equation
CaCO3(s) -> CaO(s) + CO2(g)
Mole ratios 1: 1: 1
Molar Mass CaCO3 =100g
Method 1
100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p
5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3
100g
Method 2
Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles
100 g
Mole ratio 1:1
Moles of CO2(g) = 0.05moles
Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3
Chemical equation
Copper does not react with hydrochloric acid
2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)
Method 1
3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al
840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium
3 x 22.4 x 1000
Total mass of alloy – mass of aluminium = mass of copper
=> 1.0g – 0.675g =0.325g of copper
% copper = mass of copper x100% = 32.5%
Mass of alloy
Method 2
Mole ratio 2Al: 3H2 = 2:3
Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles
Molar gas volume 22400cm3
Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles
Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g
Total mass of alloy – mass of aluminium = mass of copper
=> 1.0g – 0.675g = 0.325 g of copper
% copper = mass of copper x100% = 32.5%
Mass of alloy
(f)Gay Lussac’s law
Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure”
Gay Lussacs law thus only apply to gases
Given the volume of one gas reactant, the other gaseous reactants can be deduced thus:
Examples
Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
Volume ratios 2 : 1 : 0
Reacting volumes 50cm3 : 25cm3
50cm3 of Oxygen is used
Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
Volume ratios 2 : 1 : 0
Reacting volumes 50cm3 : 25cm3
50cm3 of Oxygen is used
21% = 25cm3
100% = 100 x 25 =
21
3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.
Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g)
Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3
Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5
Reacting volume ratios 1volume 3 volume 2 volume 2 volume
Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g)
If “4H” are in 2H2O(g) the y=4
If “2C” are in 2CO2 (g) the x=2
Thus(i) chemical formula of hydrocarbon = C2H4
(ii) chemical name of hydrocarbon = Ethene
4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.
Chemical equation: NO (g) + O2 (g) -> NOx
Volumes 100cm3 : 50cm3 : 100
Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50
Reacting volume ratios 2volume 1 volume 2 volume
Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g)
Thus(i) chemical formula of the nitrogen compound = 2 NO2
(ii) chemical name of compound = Nitrogen(IV)oxide
5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.
(a)What volume of Oxygen was used during the reaction.(1mk)
Volume of Oxygen used =100-25 =75cm3√
(P was completely burnt)
(b)Determine the molecular formula of the hydrocarbon(2mk)
CxHy + O2 -> xCO2 + yH2O
15cm3 : 75cm3
15 15
1 : 3√
=> 1 atom of C react with 6 (3×2)atoms of Oxygen
Thus x = 1 and y = 2 => P has molecula formula CH4√
(g) Ionic equations
An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products.
Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions)
An ionic equation is usually derived from a stoichiometric equation by using the following guidelines
Guidelines for writing ionic equations
1.Write the balanced stoichiometric equation
2.Indicate the state symbols of the reactants and products
3.Split into cations and anions all the reactants and products that exist in aqueous state.
4.Cancel out any cation and anion that appear on both the product and reactant side.
Practice
(a)Precipitation of an insoluble salt
All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt
Balanced stoichiometric equation
AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
Ag+(aq) + NO3– (aq) + Na+(aq) + Cl–(aq) -> AgCl(s) + Na+(aq)+ NO3– (aq)
Cancel out ions appearing on reactant and product side
Ag+(aq) + NO3– (aq) + Na+(aq) + Cl–(aq) -> AgCl(s) + Na+(aq)+ NO3– (aq)
Rewrite the equation
Ag+(aq) + Cl–(aq) -> AgCl(s) (ionic equation)
Balanced stoichiometric equation
Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq)
Split reactants product existing in aqueous state as cation/anion
Ba2+(aq) + 2NO3– (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3– (aq)+ Cu2+(aq)
Cancel out ions appearing on reactant and product side
Ba2+(aq) + 2NO3– (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3– (aq) + Cu2+(aq)
Rewrite the equation
Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation)
3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.
Balanced stoichiometric equation
Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
Pb2+(aq) + 2NO3– (aq) + 2K +(aq) + 2I – (aq) -> PbI2 (s) + 2NO3– (aq)+ 2K +(aq)
Cancel out ions appearing on reactant and product side
Pb2+(aq) + 2NO3– (aq) + 2K +(aq) + 2I – (aq) -> PbI2 (s) + 2NO3– (aq)+ 2K +(aq)
Rewrite the equation
Pb2+(aq) + 2I – (aq) -> PbI2 (s) (ionic equation)
(b)Neutralization
Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.
(i)Reaction of alkalis with acids
1.Reaction of nitric(V)acid with potassium hydroxide
Balanced stoichiometric equation
HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
H+(aq) + NO3– (aq) + K +(aq) + OH – (aq) -> H2O (l) + NO3– (aq)+ K +(aq)
Cancel out ions appearing on reactant and product side
H+(aq) + NO3– (aq) + K +(aq) + OH – (aq) -> H2O (l) + NO3– (aq)+ K +(aq)
Rewrite the equation
H+ (aq) + OH – (aq) -> H2O (l) (ionic equation)
2.Reaction of sulphuric(VI)acid with ammonia solution
Balanced stoichiometric equation
H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq)
Split reactants product existing in aqueous state as cation/anion
2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH – (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
Cancel out ions appearing on reactant and product side
2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH – (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
Rewrite the equation
2H+ (aq) + 2OH – (aq) -> 2H2O (l)
H+ (aq) + OH – (aq) -> H2O (l) (ionic equation)
3.Reaction of hydrochloric acid with Zinc hydroxide
Balanced stoichiometric equation
2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq)
Split reactants product existing in aqueous state as cation/anion
2H+(aq) + 2Cl– (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl– (aq)+ Zn 2+ (aq)
Cancel out ions appearing on reactant and product side
2H+(aq) + 2Cl– (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl– (aq)+ Zn 2+ (aq)
Rewrite the equation
2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation)
(h)Molar solutions
A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.
Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.
One cubic decimeter is equal to one litre and also equal to 1000cm3.
The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution.
Examples
2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.
0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.
“2M” is more concentrated than“0.02M”.
Preparation of molar solution
Procedure
Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask.
Using a wash bottle add about 200cm3 of distilled water.
Stopper the flask.
Shake vigorously for three minutes.
Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved.
Add more water slowly upto exactly the 250 cm3 mark.
Sample questions
1.Calculate the number of moles of sodium hydroxide pellets present in:
(i) 4.0 g.
Molar mass of NaOH = (23 + 16 + 1) = 40g
Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles
Molar mass 40
(ii) 250 cm3 solution in the volumetric flask.
Moles in 250 cm3 = 0.1 / 1.0 x 10 -1 moles
(iii) one decimeter of solution
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 x 10 -1 moles x 1000cm3 =
250cm3
= 0.4 M / 0.4 molesdm-3
Method 2
250cm3 solution contain 1.0 x 10 -1 moles
1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles
250 cm3
= 0.4 M / 0.4 molesdm-3
Theoretical sample practice
(i) 4.0 g sodium hydroxide dissolved in 500cm3 solution
Molar mass of NaOH = (23 + 16 + 1) = 40g
Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles
Molar mass 40
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 x 10 -1 moles x 1000cm3
500cm3
= 0.2 M / 0.2 molesdm-3
Method 2
500 cm3 solution contain 1.0 x 10 -1 moles
1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles
500 cm3
= 0.2 M / 0.2 molesdm-3
(ii) 5.3 g anhydrous sodium carbonate dissolved in 50cm3 solution
Molar mass of Na2CO3 = (23 x 2 + 12 + 16 x 3) = 106 g
Moles = Mass => 5.3 = 0.05 / 5. 0 x 10-2 moles
Molar mass 106
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 moles x 1000cm3 =
50cm3
=1.0 M
Method 2
50 cm3 solution contain 5.0 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles
50 cm3
= 1.0M / 1.0 molesdm-3
(iii) 5.3 g hydrated sodium carbonate decahydrate dissolved in 50cm3 solution
Molar mass of Na2CO3.10H2O = (23 x 2 + 12 + 16 x 3 + 20 x 1 + 10 x 16) =286g
Moles = Mass => 5.3 = 0.0185 / 1.85 x 10 -2 moles
Molar mass 286
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.85 x 10 -2 moles x 1000cm3 =
50cm3
= 0.37 M/0.37 molesdm-3
Method 2
50 cm3 solution contain 1.85 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 1.85 x 10 -2 moles
50 cm3
= 3.7 x 10-1 M / 3.7 x 10-1 molesdm-3
(iv) 7.1 g of anhydrous sodium sulphate(VI)was dissolved in 20.0 cm3 solution. Calculate the molarity of the solution.
Method 1
20.0cm3 solution ->7.1 g
1000cm3 solution -> 1000 x 71 = 3550 g dm-3
20
Molar mass Na2SO4 = 142 g
Moles dm-3 = Molarity = Mass 3550 = 2.5 M/ molesdm-3
Molar mass 142
Method 2
Molar mass Na2SO4 = 142 g
Moles = Mass => 7.1 = 0.05 / 5.0 x 10 -2 moles
Molar mass 142
Method 2(a)
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 5.0 x 10 -2 moles x 1000cm3
20cm3
= 2.5 M/2.5 molesdm-3
Method 2(b)
20 cm3 solution contain 5.0 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles
20 cm3
= 2.5 M/2.5 molesdm-3
(iv) The density of sulphuric(VI) is 1.84gcm-3 Calculate the molarity of the acid.
Method 1
1.0cm3 solution ->1.84 g
1000cm3 solution -> 1000 x 1.84 = 1840 g dm-3
1
Molar mass H2SO4 = 98 g
Moles dm-3 = Molarity = Mass = 1840
Molar mass 98
= 18.7755 M/ molesdm-3
Method 2
Molar mass H2SO4 = 98 g
Moles = Mass => 1.84 = 0.0188 / 1.88 x 10 -2 moles
Molar mass 98
Method 2(a)
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.88 x 10 -2 moles x 1000cm3
1.0cm3
= 18.8M/18.8 molesdm-3
Method 2(b)
20 cm3 solution contain 1.88 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 1.88 x 10 -2 moles
1.0 cm3
= 18.8M/18.8 molesdm-3
(i) 25 cm3 of 0.2M sodium hydroxide solution(Na =23.0.O =16.0, H=1.0)
Molar mass NaOH = 40g
Moles in 25 cm3 = Molarity x volume => 0.2 x 25 = 0.005/5.0 x 10-3moles
1000 1000
Mass of NaOH =Moles x molar mass = 5.0 x 10-3 x 40 = 0.2 g
(ii) 20 cm3 of 0.625 M sulphuric(VI)acid (S =32.0.O =16.0, H=1.0)
Molar mass H2SO4 = 98g
Moles in 20 cm3 = Molarity x volume=> 0.625 x 20 = 0.0125/1.25.0 x 10-3moles
1000 1000
Mass of H2SO4 =Moles x molar mass => 5.0 x 10-3 x 40 = 0.2 g
(iii) 1.0 cm3 of 2.5 M Nitric(V)acid (N =14.0.O =16.0, H=1.0)
Molar mass HNO3 = 63 g
Moles in 1 cm3 = Molarity x volume => 2.5 x 1 = 0.0025 / 2.5. x 10-3moles
1000 1000
Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g
(a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution
Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3
Molarity 0.8
(ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution.
C1 x V1 = C2 x V2 where:
C1 = molarity/concentration before diluting/adding water
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water
=> 0.8M x 312.5cm3 = C2 x (312.5 + 100)
C2 = 0.8M x 312.5cm3 = 0.6061M
412.5
(b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution .
Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3
Molarity 0.01
(ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution.
C1 x V1 = C2 x V2 where:
C1 = molarity/concentration before diluting/adding water
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water
=> 0.01M x 1000 cm3 = 0.008 x V2
V2 = 0.01M x 1000cm3 = 1250cm3
0.008
Volume added = 1250 – 1000 = 250cm3
(c)Volumetric analysis/Titration
Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another.
Reactions take place in simple mole ratio of reactants and products.
Knowing the concentration/ volume of one reactant, the other can be determined from the relationship:
M1V1 = M2V2 where:
n1 n2
M1 = Molarity of 1st reactant
M2 = Molarity of 2nd reactant
V1 = Volume of 1st reactant
V1 = Volume of 2nd reactant
n1 = number of moles of 1st reactant from stoichiometric equation
n2 = number of moles of 2nd reactant from stoichiometric equation
Examples
1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0)
Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l)
Method 1
M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3
n1 n2 1 2
=> M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3
5.0 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio MCO3 : HCl = 1:2
Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
Stoichiometric equation: M2CO3 (aq) + 2HCl(aq) -> 2MCl (aq) + CO2(g) + H2O(l)
Method 1
M1V1 = M2V2 -> 0.5 x 2.0cm3 = 0.1M x V2 cm3
n1 n2 2 1
=> V2 = 0.5 x 2.0 x1 = 1.25M / 1.25 moledm-3
0.1 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.5 x 2.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio M2CO3 : HCl = 1:2
Moles M2CO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity M2CO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
(ii)the mass of Lead(II)Iodide formed
(Pb=207.0, I =127.0)
Stoichiometric equation: 2NaI(aq) + Pb(NO3)2(aq) -> 2NaNO3(aq) + PbI2(s)
(i)Volume of Lead(II)nitrate(V) used
Method 1
M1V1 = M2V2 -> 5 x 0.1cm3 = 0.1M x V2 cm3
n1 n2 2 1
=> V2 = 0.1 x 5.0 x 1 = 1.25M / 1.25 moledm-3
0.1 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.1 x 5.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio M2CO3 : HCl = 1:2
Moles M2CO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity M2CO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
Moles of NaOH used = molarity x volume
1000
=> 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles
1000
Mole ratio B : NaOH = 1:1
Moles B = 0.0044175 /4.4175 x 10-3moles
Molar mass B = mass => 0.388
moles 0.0044175 /4.4175 x 10-3moles
= 87.8324 gmole-1
X-COOH = 87.8324 where X is an alkyl group
X =87.8324- 42 = 42.8324=43
By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43
Molecula formula : CH3CH2 CH2COOH
Molecule name : Butan-1-oic acid
Molecular structure
H H H O
H C C C C O H
H H H H
Equation for neutralization
NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l)
Mole ratio NaOH(aq):HCl(aq)= 1:1
Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles
1000 1000
Excess moles of NaOH(aq)= 0.0425 moles
Equation for reaction with ammonium salt
2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l)
Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1
Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles
1000 1000
Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 – 0.0425 = 0.1575moles
Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles
Molar mass (NH4) 2SO4= 132 gmole-1
Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g
Mass of impurities = 10.5 -10.395 = 0.105 g
% impurities = 0.105 x 100 = 1.0 %
10.5
Practically volumetric analysis involves titration.
Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point.
Set up of titration apparatus
The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings.
To reduce errors, titration process should be repeated at least once more.
The results of titration are recorded in a titration table as below
Sample titration table
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution used(cm3) | 20.0 | 20.0 | 20.0 |
As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration.
For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council.
As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.
Calculate the average volume of solution used
24.0 + 24.0 + 24.0 = 24.0 cm3
3
As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point.
As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value .
The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates.
Bonus mark is awarded for averaged reading within 0.1 school value as Final answer.
Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the:
(i)relationship among the mole, molar mass, mole ratios, concentration, molarity.
(ii) mathematical application of 1st principles.
Very useful information which candidates forget appears usually in the beginning of the question paper as:
“You are provided with…”
All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point.
Candidates are expected to use a non programmable scientific calculator.
(a)Sample Titration Practice 1 (Simple Titration)
You are provided with:
0.1M sodium hydroxide solution A
Hydrochloric acid solution B
You are required to determine the concentration of solution B in moles per litre.
Procedure
Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1
Sample results Titration table 1
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution B used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
(i)solution A were present in 25cm3 solution.
Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles
1000 1000
(ii)solution B were present in the average volume.
Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles
(iii) solution B in moles per litre.
Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M
Volume 20
(b)Sample Titration Practice 2 (Redox Titration)
You are provided with:
Acidified Potassium manganate(VII) solution A
0.1M of an iron (II)salt solution B
8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C
You are required to
(i)standardize acidified potassium manganate(VII)
(ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O.
Procedure 1
Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears.
Record your results to complete table 1.
Table 1:Sample results
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution A used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
(i)solution B were present in 25cm3 solution.
Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles
1000 1000
(ii)solution A were present in the average volume. Assume one mole of B react with five moles of B
Mole ratio A : B = 1:5
=> Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles
5 5
(iii) solution B in moles per litre.
Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000
Volume 20
= 0.025 M /moles per litre /moles l-1
Procedure 2
Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2.
Table 2:Sample results
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution A used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
(i)solution A were present inin the average titre.
Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles
1000 1000
(ii)solution C in 25cm3 solution given the equation for the reaction:
MnO4– (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Mole ratio MnO4– (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4– (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles
5 5
(iii) solution B in 250cm3.
Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25
(N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0)
Molar mass = mass perlitre = 8.5 = 8500 g
Moles per litre 1.0 x 10 -3 moles
NH4)2SO4FeSO4.xH2O = 8500
284 + 18x =8500
8500 – 284 = 8216 = 18x = 454.4444
18 18
x = 454 (whole number)
(c)Sample Titration Practice 3 (Back titration)
You are provided with:
(i)an impure calcium carbonate labeled M
(ii)Hydrochloric acid labeled solution N
(iii)solution L containing 20g per litre sodium hydroxide.
You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M.
Procedure 1
Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark)
Sample Table 1
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 6.5 | 6.5 | 6.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 6.5 | 6.5 | 6.5 |
Sample questions
(a) Calculate the average volume of solution N used
6.5 + 6.5 + 6.5 = 6.5 cm3
3
(b) How many moles of sodium hydroxide are contained in 25cm3of solution L
Molar mass NaOH =40g
Molarity of L = mass per litre => 20 = 0.5M
Molar mass NaOH 40
Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles
1000 1000
(c)Calculate:
(i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above.
Mole ratio NaOH : HCl from stoichiometric equation= 1:1
Moles HCl =Moles NaOH => 0.0125 moles
(ii)the molarity of hydrochloric acid solution N.
Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3
6.5 6.5
Procedure 2
Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.5 | 24.5 | 24.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.5 | 24.5 | 24.5 |
Sample calculations
(a)Calculate the average volume of solution L used(1mk)
24.5 + 24.5 + 24.5 = 24.5cm3
3
(b)How many moles of sodium hydroxide are present in the average volume of solution L used?
Moles = molarity x average burette volume => 0.5 x 24.5
1000 1000
= 0.01225 /1.225 x 10-2 moles
(c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?
Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles
Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles
25cm3(volume pipetted)
=0.49 /4.9 x 10-1moles
(d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used
Original moles = Original molarity x pipetted volume =>
1000cm3
1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles
1000
(e)How many moles of hydrochloric acid were used to react with calcium carbonate present?
Moles that reacted = original moles –moles in average titre =>
0.04807/4.807 x 10-2moles – 0.01225 /1.225 x 10-2moles
= 0.03582/3.582 x 10 -2 moles
(f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
(g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.
From the equation CaCO3(s):2HCl(aq) = 1:2
=> Moles CaCO3(s) = 1/2moles HCl
= 1/2 x 0.03582/3.582 x 10 -2 moles
= 0.01791 /1.791 x 10-2moles
(h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0)
Molar mass CaCO3 = 100g
Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g
= 1.791g
(i)Determine the % of calcium carbonate present in the mixture
% CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775%
Mass of impure 4.0
(d)Sample titration practice 4 (Multiple titration)
You are provided with:
(i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.
(ii)solution M which is acidified potassium manganate(VII)
(iii)solution N a mixture of sodium ethanedioate and ethanedioic acid
(iv)0.1M sodium hydroxide solution P
(v)1.0M sulphuric(VI)
You are required to:
(i)standardize solution M using solution L
(ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture.
Procedure 1
Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.
Sample Table 1
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.0 | 24.0 | 24.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.0 | 24.0 | 24.0 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
24.0 + 24.0 + 24.0 = 24.0cm3
3
(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0)
Molar mass H2X.2H2O= mass per litre => 5.0g/litre = 100g
Moles/litre 0.05molesdm-3
H2X.2H2O =100
X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 66
(c) Calculate the number of moles of the dibasic acid H2X.2H2O.
Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles
1000 1000
(d)Given the mole ratio manganate(VII)(MnO4–): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4–) in the average titre.
Moles H2X = 2/5 moles of MnO4–
=> 2/5 x 0.0125/1.25 x10 -2 moles
= 0.005/5.0 x 10 -3moles
(e)Calculate the concentration of the manganate(VII)(MnO4–) in moles per litre.
Moles per litre/molarity = moles x 1000
average burette volume
=>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M
24.0
Procedure 2
With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 12.5 | 12.5 | 12.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 12.5 | 12.5 | 12.5 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
12.5 + 12.5 + 12.5 =12.5cm3
3
(b)Calculations:
(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?
Moles = molarity of solution M x average burette volume
1000
=> 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles
1000
(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:
2MnO4– (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.
From the stoichiometric equation,mole ratio MnO4– (aq): C2O42- (aq) = 2:5
=> moles C2O42- = 5/2 moles MnO4– => 5/2 x 0.0026 / 2.5 x 10-3 moles
= 0.0065 /6.5 x10-3 moles
(iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.
25cm3 pipette volume -> 0.0065 /6.5 x10-3 moles
250cm3 ->
0.0065 /6.5 x10-3 moles x 250 = 0.065 / 6.5 x10-2 moles
25
Procedure 3
Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.9 | 24.9 | 24.9 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.9 | 24.9 | 24.9 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
24.9 + 24.9 + 24.9 = 24.9 cm3
3
(b)Calculations:
(i)How many moles of sodium hydroxide solution P were contained in the average volume?
Moles = molarity of solution P x average burette volume
1000
=> 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles
1000
(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:
2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l)
Calculate the number of moles of ethanedioic acid that were used in the reaction
From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1
=> moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles
= 0.001245/1.245 x10-3 moles.
(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?
25cm3 pipette volume -> 0.001245/1.245 x10-3 moles
250cm3 ->
0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles
25
(iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution)
Molar mass H2C2O4 = 90.0g
Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4
=>0.01245/1.245 x10-2 moles x 90.0
= 1.1205g
% by mass of sodium ethanedioate
=(Mass of mixture – mass of H2C2O4) x 100%
Mass of mixture
=> 2.0 – 1.1205 g = 43.975%
2.0
Note
(i) L is 0.05M Oxalic acid
(ii) M is 0.01M KMnO4
(iii) N is 0.03M oxalic acid(without sodium oxalate)
Practice example 5.(Determining equation for a reaction)
You are provided with
-0.1M hydrochloric acid solution A
-0.5M sodium hydroxide solution B
You are to determine the equation for thereaction between solution A and B
Procedure
Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1
Table 1(Sample results)
| Titration | 1 | 2 | 3 |
| Final volume(cm3) | 12.5 | 25.0 | 37.5 |
| Initial volume(cm3) | 0.0 | 12.5 | 25.0 |
| Volume of solution A used(cm3) | 12.5 | 12.5 | 12.5 |
Sample questions
Calculate the average volume of solution A used.
12.5+12.5+12.5 = 12.5cm3
3
Theoretical Practice examples
Chemical equation
2NaOH(aq) + H2X(aq) -> Na2X (aq) + 2H2O(aq)
Mole ratio NaOH(aq) :H2X(aq) = 2:1
Method 1
Ma Va = na => Ma x 25.0 = 1 => Ma =0.06 x 30.0 x1
Mb Vb = nb 0.06 x 30.0 2 25.0 x 2
Molarity of acid = 0.036M/Mole l-1
Mass of acid per lite = 1.0 x1000 = 4.0 g/l
250
0.036M/ Mole l-1 -> 4.0 g /l
1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1 = 111.1111 g
0.036
Molar mass (CH2)x = 111.1111 – (HOOCCOOH = 90.0) = 21.1111
(CH2)x = 14x = 21.1111 = 1.5 = 1 (whole number)
14
Method 2
Moles of sodium hydroxide = Molarity x volume = 0.06 x 30 = 1.8 x 10 -3moles
1000
Moles of Hydrochloric acid = 1/2 x 1.8 x 10 -3moles = 9.0 x10 -4moles
Molarity of Hydrochloric acid = moles x 1000 = 9.0 x10 -4moles x1000
Volume 25
Molarity of acid = 0.036M/Mole l-1
Mass of acid per lite = 1.0 x1000 = 4.0 g/l
250
0.036M/ Mole l-1 -> 4.0 g /l
1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1 = 111.1111 g
0.036
Molar mass (CH2)x = 111.1111 – (HOOCCOOH = 90.0) = 21.1111
(CH2)x = 14x = 21.1111 = 1.5 = 1 (whole number)
14
MnO4– (aq) + 8H+(aq)+ 5Fe2+(aq)-> 5Fe3+(aq) + Mn2+(aq) + 4H2O(aq)
Fe=56.0,S= 32.0, O=16.0).
Moles of MnO4– (aq) = Molarity x volume = 0.05 x 20.0 = 0.001 Moles
1000 1000
Mole ratio MnO4– (aq): 5Fe2+(aq)= 1:5
Moles 5Fe2+(aq) = 5 x0.001 = 0.005 Moles
Moles of 5Fe2+(aq) per litre/molarity = Moles x 1000 = 0005 x 1000
Volume 25.0
= 0.2 M/ Moles/litre
Molar mass =FeSO4=152 g
Mass of in the mixture = Moles x molar mass => 0.2 x 152 = 30.4 g
Mass of impurity = 40.0 – 30.4 =9.6 g
% impurity = 9.6 g x100 = 24.0 % impurity
40.0
3.9.7 g of a mixture of Potassium hydroxide and Potassium chloride was dissolved to make one litre solution.20.0cm3 of this solution required 25.0cm3 of 0.12M hydrochloric acid for completed neutralization. Calculate the percentage by mass of Potassium chloride.(K=39.0,Cl= 35.5)
Chemical equation
KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)
Moles of HCl = Molarity x volume => 0.12 x 25.0 = 0.003/3.0 x 10 -3 moles
1000 1000
Mole ratio KOH(aq) : HCl(aq) -= 1:1
Moles KOH =0.003/3.0 x 10 -3 moles
Method 1
Molar mass KOH =56.0g
Mass KOH in 25cm3 =0.003/3.0 x 10 -3 moles x56.0 = 0.168g
Mass KOH in 1000cm3/1 litre = 0.168 x1000= 8.4 g/l
20
Mass of KCl = 9.7g – 8.4g = 1.3 g
% of KCl = 1.3 x 100 = 13.4021%
9.7
Method 2
Moles KOH in 1000cm3 /1 litre = Moles in 20cm3 x 1000 =>0.003 x 1000
20 20
=0.15M/Moles /litre
Molar mass KOH =56.0g
Mass KOH in 1000/1 litre = 0.15M/Moles /litre x 56.0 = 8.4g/l
Mass of KCl = 9.7g – 8.4g = 1.3 g
% of KCl = 1.3 x 100 = 13.4021%
9.7
4.A certain carbonate, GCO3, reacts with dilute hydrochloric acid according to the equation given below:
GCO3(s) + 2HCl(aq) -> GCl2 (aq) + CO2 (g) + H2O(l)
If 1 g of the carbonate reacts completely with 20 cm3 of 1 M hydrochloric acid ,calculate the relative atomic mass of G (C = 12.0 = 16.0)
Moles of HCl = Molarity x volume=> 1 x20 = 0.02 moles
1000 1000
Mole ratio HCl; GCO3 = 2:1
Moles of GCO3= 0.02 moles = 0.01moles
2
Molar mass of GCO3 = mass => 1 = 100 g
moles 0.01moles
G= GCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40(no units)
Equation
Chemical equation
NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)
Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles
1000 1000
Mole ratio HCl; NaOH = 1:1
Excess moles of HCl = 0.002 moles
25cm3 -> 0.002 moles
1000cm3 -> 1000 x 0.002 = 0.08moles
25cm3
Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles
Moles of HCl reacted with MCO3 = 1.0 – 0.08 moles = 0.92moles
Chemical equation
MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l)
Mole ratio MCO3(s) : HCl(aq) =1:2
Moles of MCO3 = 0.92moles => 0.46moles
2
Molar mass of MCO3= mass => 46g = 100 g
moles 0.46moles
M= MCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40
A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.
Mole ratio Fe2+ :Mn04– = 5:1
Moles Mn04– used = 0.02 x 15 = 3.0 x 10-4 moles
1000
Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles
5
Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1
25
Since Zinc reduces Fe3+ to Fe2+ in the mixture:
Moles Mn04– that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles
1000
Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles
5
Moles of Fe3+ = 3.8 x 10-4 – 6.0 x 10-5 = 1.6 x 10-5 moles
Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1
25
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UPGRADE
CHEMISTRY
FORM 3
ORGANIC Chemistry 1&2
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
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Introduction to Organic chemistry
Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made.
Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.
It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because:
(i) carbon uses all the four valence electrons to form four strong covalent bond.
(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself.
(iii)carbon atoms can covalently bond to form a very long chain or ring.
When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons
A.HYDROCARBONS (HCs)
Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.
Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as:
(i) Alkanes
(ii) Alkenes
(iii) Alkynes
(i) Alkanes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule.
The carbon atoms are linked by single bond to each other and to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name
|
| 1 | CH4 | H
H C H
H
|
Methane |
| 2 | C2H6 | H H
H C C H
H H
|
Ethane
|
| 3 | C3H8 | H H H
H C C C H
H H H
|
Propane |
| 4 | C4H10 | H H H H
H C C C C H
H H H H
|
Butane |
| 5 | C5H12 | H H H H H
H C C C C C H CH3 (CH2) 6CH3
H H H H H
|
Pentane |
| 6 | C6H14 | H H H H H H
H C C C C C C H CH3 (CH2) 6CH3
H H H H H H
|
Hexane |
| 7 | C7H16 | H H H H H H H
H C C C C C C C H
H H H H H H H
|
Heptane |
| 8 | C8H18 | H H H H H H H H
H C C C C C C C C H
H H H H H H H H
|
Octane |
| 9 | C9H20 | H H H H H H H H H
H C C C C C C C C C H
H H H H H H H H H
|
Nonane |
| 10 | C10H22 | H H H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H H H
|
decane |
Note
1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g.
Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.
2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g
Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom.
The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms.
3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.
4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons.
5.One member of the alkane differ from the next/previous by a CH2 group.
e.g
Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms
6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series.
7.A homologous series:
(i) differ by a CH2 group from the next /previous consecutively
(ii)have similar chemical properties
(iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula CnH2n+2.
(iv)the physical properties (e.g.melting/boiling points)show steady gradual change)
8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms.
9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g
| Alkane name | molecular structure
CnH2n+2 |
Alkyl name | Molecula structure
CnH2n+1 |
| methane | CH4 | methyl | CH3 |
| ethane | CH3CH3 | ethyl | CH3 CH2 |
| propane | CH3 CH2 CH3 | propyl | CH3 CH2 CH2 |
| butane | CH3 CH2 CH2 CH3 | butyl | CH3 CH2 CH2 CH2 |
(b)Isomers of alkanes
Isomers are compounds with the same molecular general formula but different molecular structural formula.
Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula.
The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature uses the following basic rules/guidelines:
1.Identify the longest continuous carbon chain to get/determine the parent alkane.
2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible
4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane.
Practice on IUPAC nomenclature of alkanes
(a)Draw the structure of:
(i)2-methylpentane
Procedure
Butane is the parent name CH3 CH2 CH2 CH3
The methyl group is attached to Carbon “2”
Position of the branch at carbon “2”
Number of branches at carbon “1”
Type of the branch “methyl” hence
Molecular formula
CH3
CH3 CH CH2 CH3 // CH3 CH (CH3 ) CH2CH3
Structural formula
H H H H
H C C C C H
H H H
H C H
H
(ii)2,2-dimethylpentane
Procedure
Butane is the parent name CH3 CH2 CH2 CH3
The methyl group is attached to Carbon “2”
Position of the branch at carbon “2”
Number of branches at carbon “2”
Type of the branch two“methyl” hence
Molecular formular
CH3
CH3 C CH2 CH3 // CH3 C (CH3 )2 CH2CH3
CH3
Structural formula
H
H C H
H H H
H C C C C H
H H H
H C H
H
(iii) 2,2,3-trimethylbutane
Procedure
Butane is the parent name CH3 CH2 CH2 CH3
The methyl group is attached to Carbon “2 and 3”
Position of the branch at carbon “2 and 3”
Number of branches at carbon “3”
Type of the branch three “methyl” hence
Molecular formular
CH3
CH3 C CH CH3 // CH3 C (CH3 )3 CH2CH3
CH3 CH3
Structural formula
H
H C H
H H
H C C C H
H H
H
H C C H
H
H C H
H
(iv) 1,1,1,2,2,2-hexabromoethane
Molecular formula
CBr3 CBr3
Structural formula
Br Br
Br C C Br
Br Br
(v) 1,1,1-tetrachloro-2,2-dimethylbutane
CH3
CCl 3 C CH3 // C Cl 3 C (CH3 )2 CH3
CH3
Structural formula
Cl
Cl C Cl
H H
H C C C H
H H
H C H
H
(c)Occurrence and extraction
Crude oil ,natural gas and biogas are the main sources of alkanes:
(i)Natural gas is found on top of crude oil deposits and consists mainly of methane.
(ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced.
(iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first.
As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.
Uses of different crude oil fractions
| Carbon atoms in a molecule | Common name of fraction | Uses of fraction |
| 1-4 | Gas | L.P.G gas for domestic use |
| 5-12 | Petrol | Fuel for petrol engines |
| 9-16 | Kerosene/Paraffin | Jet fuel and domestic lighting/cooking |
| 15-18 | Light diesel | Heavy diesel engine fuel |
| 18-25 | Diesel oil | Light diesel engine fuel |
| 20-70 | Lubricating oil | Lubricating oil to reduce friction. |
| Over 70 | Bitumen/Asphalt | Tarmacking roads |
(d)School laboratory preparation of alkanes
In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime.
Chemical equation:
Sodium alkanoate + soda lime -> alkane + Sodium carbonate
CnH2n+1COONa(s) + NaOH(s) -> C n H2n+2 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2.
Examples
Sodium ethanoate + soda lime -> methane + Sodium carbonate
CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4.
Sodium propanoate + soda lime -> ethane + Sodium carbonate
CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3
Sodium butanoate + soda lime -> propane + Sodium carbonate
CH3 CH2CH2COONa(s) + NaOH(s) -> CH3 CH2CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3
Sodium pentanoate + soda lime -> butane + Sodium carbonate
CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3
Laboratory set up for the preparation of alkanes
(d)Properties of alkanes
Alkanes are colourless gases, solids and liquids that are not poisonous.
They are slightly soluble in water.
The solubility decrease as the carbon chain and thus the molar mass increase
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids .
The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases.
This reduces the volume occupied by a given mass of the compound.
Summary of physical properties of alkanes
| Alkane | General formula | Melting point(K) | Boiling point(K) | Density gcm-3 | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Methane | CH4 | 90 | 112 | 0.424 | gas |
| Ethane | CH3CH3 | 91 | 184 | 0.546 | gas |
| Propane | CH3CH2CH3 | 105 | 231 | 0.501 | gas |
| Butane | CH3(CH2)2CH3 | 138 | 275 | 0.579 | gas |
| Pentane | CH3(CH2)3CH3 | 143 | 309 | 0.626 | liquid |
| Hexane | CH3(CH2)4CH3 | 178 | 342 | 0.657 | liquid |
| Heptane | CH3(CH2)5CH3 | 182 | 372 | 0.684 | liquid |
| Octane | CH3(CH2)6CH3 | 216 | 399 | 0.703 | liquid |
| Nonane | CH3(CH2)7CH3 | 219 | 424 | 0.708 | liquid |
| Octane | CH3(CH2)8CH3 | 243 | 447 | 0.730 | liquid |
II.Chemical properties
(i)Burning.
Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water.
Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.
Alkane + Air -> carbon(II) oxide + water (limited air)
Examples
1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Methane + Air -> carbon(IV) oxide + water (excess air/oxygen)
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g)
(b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Methane + Air -> carbon(II) oxide + water (excess air/oxygen)
2CH4(g) + 3O2(g) -> 2CO(g) + 4H2O(l/g)
2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g)
(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Ethane + Air -> carbon(II) oxide + water (excess air/oxygen)
2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g)
3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Propane + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g)
(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Ethane + Air -> carbon(II) oxide + water (excess air/oxygen)
2C3H8(g) + 7O2(g) -> 6CO(g) + 8H2O(l/g)
ii)Substitution
Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light.
Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes.
During substitution:
(i)the halogen molecule is split into free atom/radicals.
(ii)one free halogen radical/atoms knock /remove one hydrogen from the alkane leaving an alkyl radical.
(iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane.
(iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide.
(v)substitution stops when all the hydrogen in alkanes are replaced with halogens.
Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst.
Examples of substitution reactions
Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane explode to form colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades.
Chemical equation
1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride
CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g)
H H
H C H + Cl Cl -> H C Cl + H Cl
H H
(b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride
CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g)
H H
H C Cl + Cl Cl -> H C Cl + H Cl
H Cl
(c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride
CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g)
Cl H
H C Cl + Cl Cl -> Cl C Cl + H Cl
H Cl
(c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride
CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g)
H Cl
Cl C Cl + Cl Cl -> Cl C Cl + H Cl
Cl Cl
Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades.
Chemical equation
(a)Ethane + chlorine -> Chloroethane + Hydrogen chloride
CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g)
H H H H
H C C H + Br Br -> H C C H + H Br
H H H Br
Bromoethane
H H H Br
H C C H + Br Br -> H C C H + H Br
H Br H Br
1,1-dibromoethane
H Br H Br
H C C H + Br Br -> H C C Br + H Br
H Br H Br
1,1,1-tribromoethane
H Br H Br
H C C Br + Br Br -> H C C Br + H Br
H Br Br Br
1,1,1,2-tetrabromoethane
H Br H Br
H C C Br + Br Br -> Br C C Br + H Br
Br Br Br Br
1,1,1,2,2-pentabromoethane
H Br Br Br
Br C C Br + Br Br -> Br C C Br + H Br
Br Br Br Br
1,1,1,2,2,2-hexabromoethane
Uses of alkanes
1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes.Butane is used as the Laboratory gas.
2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process.
3.In manufacture of Carbon black which is a component in printers ink.
4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane.
(ii) Alkenes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule.
The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name
|
| 1 | Does not exist
|
||
| 2 | C2H6
|
H H
H C C H
CH2 CH2
|
Ethene |
| 3 | C3H8 | H H H
H C C C H
H
CH2 CH CH3
|
Propene |
| 4 | C4H10 | H H H H
H C C C C H
H H
CH2 CH CH2CH3
|
Butene |
| 5 | C5H12 | H H H H H
H C C C C C H
H H H CH2 CH (CH2)2CH3
|
Pentene |
| 6 | C6H14 | H H H H H H
H C C C C C C H
H H H H
CH2 CH (CH2)3CH3
|
Hexene |
| 7 | C7H16 | H H H H H H H
H C C C C C C C H
H H H H H H H
CH2 CH (CH2)4CH3
|
Heptene |
| 8 | C8H18 | H H H H H H H H
H C C C C C C C C H
H H H H H H
CH2 CH (CH2)5CH3
|
Octene |
| 9 | C9H20 | H H H H H H H H H
H C C C C C C C C C H
H H H H H H H
CH2 CH (CH2)6CH3
|
Nonene |
| 10 | C10H22 | H H H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H
CH2 CH (CH2)7CH3 |
decene |
Note
1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond.
2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons.
3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series.
e.g
Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 4.A homologous series of alkenes like that of alkanes:
(i) differ by a CH2 group from the next /previous consecutively
(ii)have similar chemical properties
(iii)have similar chemical formula represented by the general formula CnH2n
(iv)the physical properties also show steady gradual change
5.The = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound.
8.A saturated hydrocarbon is one without a double =C=C= or triple – C C – carbon bonds in their molecular structure.
Most of the reactions of alkenes take place at the = C = C =bond.
(b)Isomers of alkenes
Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula.
Ethene and propene do not form isomers. Isomers of alkenes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature of naming alkenes uses the following basic rules/guidelines:
1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene.
2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible.
3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3-diene.
4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e
But-2-ene means the double =C = C= is between Carbon “2”and “3”
Pent-1,3-diene means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4”
6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of double C = C bonds and branches attached to the alkene.
7.Position isomers can be formed when the=C = C= double bond is shifted between carbon atoms e.g.
But-2-ene means the double =C = C= is between Carbon “2”and “3”
But-1-ene means the double =C = C= is between Carbon “1”and “2”
Both But-1-ene and But-2-ene are position isomers of Butene
8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e.
Butene has the molecular/general formular C4H8 position but can form both But-1-ene and But-2-ene as position isomers.
10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g
Butene and 2-methyl propene both have the same general formualr but different branching chain.
Practice on IUPAC nomenclature of alkenes
Name the following isomers of alkene
H H H H
H C C C C H But-1-ene
H H
H H H H
H C C C C H But-2-ene
H H
H H H H H H
H C C C C C C H 4-methylhex-1-ene
H H H
H C H
H
H
H C H
H H H H H
H C C C C C C H 4,4-dimethylhex-1-ene
H H H
H C H
H
H C H
H H H H
H C C C C C H 4,4-dimethylpent -1- ene
H H
H C H
H
H C H
H H H H
H C C C C C H 5,5-dimethylhex-2- ene
H C H H H
H C H
H
H
H C H
H H H
H C C C C H 2,2-dimethylbut -2- ene
H H
H C H
H
8.H2C CHCH2 CH2 CH3 pent -1- ene
9.H2C C(CH3)CH2 CH2 CH3 2-methylpent -1- ene
10.H2C C(CH3)C(CH3)2 CH2 CH3 2,3,3-trimethylpent -1- ene
11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3 2,3,3,4,4-pentamethylpent -1- ene
12.H3C C(CH3)C(CH3) C(CH3)2 CH3 2,3,4,4-tetramethylpent -2- ene
(c)Occurrence and extraction
At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures.
Cracking is a major source of useful hydrogen gas for manufacture of ammonia/nitric(V)acid/HCl i.e.
Long chain alkane -> smaller/shorter alkane + Alkene + Hydrogen gas
Examples
1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas.
Chemical equation
CH3CH2CH3 (g) -> CH4(g) + CH2=CH2(g) + H2(g)
2.Octane undergo cracking to form hydrogen gas, butene and butane gases
Chemical equation
CH3(CH2) 6 CH3 (g) -> CH3CH2CH2CH3(g) + CH3 CH2CH=CH2(g) + H2(g)
(d)School laboratory preparation of alkenes
In a school laboratory, alkenes may be prepared from dehydration of alkanols using:
(i) concentrated sulphuric(VI)acid(H2SO4).
(a) aluminium(III)oxide(Al2O3) i.e
Alkanol –Conc. H2SO4 –> Alkene + Water
Alkanol –Al2O3 –> Alkene + Water e.g.
1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene.
The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities.
It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.
Chemical equation
CH3CH2OH (l) –conc H2SO4/180oC–> CH2=CH2(g) + H2O(l)
(b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene.
Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration.
Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene.
Chemical equation
CH3CH2OH (l) –(Al2O3/strong heat–> CH2=CH2(g) + H2O(l)
2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers).
Chemical equation
CH3CH2 CH2OH (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)
Propan-1-ol Prop-1-ene
CH3CHOH CH3 (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)
Propan-2-ol Prop-1-ene
(b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene
Chemical equation
CH3CH2 CH2OH (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)
Propan-1-ol Prop-1-ene
CH3CHOH CH3 (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)
Propan-2-ol Prop-1-ene
3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively
Chemical equation
CH3CH2 CH2 CH2OH (l) — conc H2SO4/180oC –>CH3 CH2CH2=CH2(g) + H2O(l)
Butan-1-ol But-1-ene
CH3CHOH CH2CH3 (l)– conc H2SO4/180oC –>CH3CH=CH CH2(g) + H2O(l)
Butan-2-ol But-2-ene
(b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.
Chemical equation
CH3CH2 CH2 CH2OH (l) — Heat/Al2O3 –> CH3 CH2CH2=CH2(g) + H2O(l)
Butan-1-ol But-1-ene
CH3CHOH CH2CH3 (l) — Heat/Al2O3 –> CH3CH=CH CH2(g) + H2O(l)
Butan-2-ol But-2-ene
Laboratory set up for the preparation of alkenes/ethene
Caution
(i)Ethanol is highly inflammable
(ii)Conc H2SO4 is highly corrosive on skin contact.
(iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.
(i)Using conentrated sulphuric(VI)acid
Some broken porcelain or sand should be put in the flask when heating to:
(i)prevent bumping which may break the flask.
(ii)ensure uniform and smooth boiling of the mixture
The temperatures should be maintained at above160oC.
At lower temperatures another compound –ether is predominantly formed instead of ethene gas.
(ii)Using aluminium(III)oxide
(e)Properties of alkenes
Like alkanes, alkenes are colourles gases, solids and liquids that are not poisonous.
They are slightly soluble in water.
The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene.
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure.
The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene.
Summary of physical properties of the 1st five alkenes
| Alkene | General formula | Melting point(oC) | Boiling point(K) | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Ethene | CH2CH2 | -169 | -104 | gas |
| Propene | CH3 CHCH2 | -145 | -47 | gas |
| Butene | CH3CH2 CHCH2 | -141 | -26 | gas |
| Pent-1-ene | CH3(CH2 CHCH2 | -138 | 30 | liquid |
| Hex-1-ene | CH3(CH2) CHCH2 | -98 | 64 | liquid |
(a)Burning/combustion
Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water.
Alkene + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water.
Alkene + Air -> carbon(II) oxide + water (limited air)
Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the =C=C= double bond because they have higher C:H ratio.
A homologous series with C = C double or C C triple bond is said to be unsaturated.
A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a
C C single bond .
Examples of burning alkenes
1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Ethene + Air -> carbon(IV) oxide + water (excess air/oxygen)
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g)
(b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Ethene + Air -> carbon(II) oxide + water (limited air )
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g)
2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C3H6(g) + 9O2(g) -> 6CO2(g) + 6H2O(l/g)
(a) Propene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g)
(b)Addition reactions
An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond.
(i)Hydrogenation
Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes.
Examples
1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.
Hydrogenation is thus used to harden oils to solid fat especially margarine.
During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.
Chemical equation
H2C=CH2 + H2 -Ni/Pa-> H3C – CH3
H H H H
C = C + H – H – Ni/Pa -> H – C – C – H
H H H H
2.Propene undergo hydrogenation to form Propane
Chemical equation
H3C CH=CH2 + H2 -Ni/Pa-> H3C CH – CH3
H H H H H H
H C C = C + H – H – Ni/Pa-> H – C – C – C- H
H H H H H
3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane
Chemical equation
But-1-ene + Hydrogen –Ni/Pa-> Butane
H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C – C = C + H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H H
But-2-ene + Hydrogen –Ni/Pa-> Butane
H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C = C – C -H + H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H
But-1,3-diene + Hydrogen –Ni/Pa-> Butane
H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C – C = C -H + 2(H – H) – Ni/Pa-> H – C- C – C – C- H
H H H H
(ii) Halogenation.
Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane.
The double bond in the alkene break and form a single bond.
The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces.
One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon.
Examples
1Ethene reacts with bromine to form 1,2-dibromoethane.
Chemical equation
H2C=CH2 + Br2 H2 Br C – CH2 Br
H H H H
C = C + Br – Br Br – C – C – Br
H H H H
Ethene + Bromine 1,2-dibromoethane
2.Propene reacts with chlorine to form 1,2-dichloropropane.
Chemical equation
H3C CH=CH2 + Cl2 H3C CHCl – CH2Cl
Propene + Chlorine 1,2-dichloropropane
H H H H H H
H C C = C + Cl – Cl H – C – C – C- Cl
H H H Cl H
H H H H H H H H
H C C – C = C + I – I H – C- C – C – C- I
H H H H H H H H
3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2-diiodobutane and 2,3-diiodobutane
Chemical equation
But-1-ene + iodine 1,2 diiodobutane
H3C CH2 CH=CH2 + I2 H3C CH2CH I – CH2I
But-2-ene + Iodine 2,3-diiodobutane
H3C CH= CH-CH2 + F2 H3C CHICHI – CH3
H H H H H H H H
H C C = C – C -H + I – I H – C- C – C – C- H
H H H I I H
But-1,3-diene + iodine 1,2,3,4-tetraiodobutane
H2C= CH CH=CH2 + 2I2 H2CI CHICHI – CHI
H H H H H H H H
H C C – C = C -H + 2(I – I) H – C- C – C – C- H
I I I I
(iii) Reaction with hydrogen halides.
Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond.
The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .
Examples
Chemical equation
H2C=CH2 + HBr H3 C – CH2 Br
H H H H
C = C + H – Br H – C – C – Br
H H H H
Ethene + Bromine bromoethane
Chemical equation
H3C CH=CH2 + HI H3C CHI – CH3
| Carbon atom with more Hydrogen atoms gets extra hydrogen |
Propene + Chlorine 2-chloropropane
H H H H H H
H C C = C + H – Cl H – C – C – C– H
H H H Cl H
Chemical equation
But-1-ene + hydrogen bromide 2-bromobutane
H3C CH2 CH=CH2 + HBr H3C CH2CHBr -CH3
H H H H H H H H
H C C – C = C + H – Br H – C- C – C – C- H
H H H H H Br H
But-2-ene + Hydrogen bromide 2-bromobutane
H3C CH= CH-CH2 + HBr H3C CHBrCH2 – CH3
H H H H H H H H
H C C = C – C -H + Br – H H – C- C – C – C- H
H H H Br H H
But-1,3-diene + iodine 2,3-diiodobutane
H2C= CH CH=CH2 + 2HI2 H3CCHICHI – CH3
H H H H H H H H
H C C – C = C -H + 2(H – I) H – C- C – C – C- H
H I I H
(iv) Reaction with bromine/chlorine water.
Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid).
During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol.
Bromine water + Alkene -> bromoalkanol
Chlorine water + Alkene -> bromoalkanol
Examples
1Ethene reacts with bromine water to form bromoethanol.
Chemical equation
H2C=CH2 + HOBr H2 Br C – CH2 OH
H H H H
C = C + Br – OH Br – C – C – OH
H H H H
Ethene + Bromine water bromoethanol
2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1-ol.
Chemical equation
I.H3C CH=CH2 + HOCl H3C CHCl – CH2OH
Propene + Chlorine water 2-chloropropane
H H H H H H
H C C = C + HO – Cl H – C – C – C- OH
H H H Cl H
II.H3C CH=CH2 + HOCl H3C CHOH – CH2Cl
Propene + Chlorine chloropropan-2-ol
H H H H H H
H C C = C + HO – Cl H – C – C – C- Cl
H H H OH H
3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1-ol /3-bromobutan-2-ol respectively
Chemical equation
I.But-1-ene + bromine water 2-bromobutan-1-ol
H3C CH2 CH=CH2 + HOBr H3C CH2CH Br – CH2OH
H H H H H H H H
H C C – C = C + HO– Br H – C- C – C – C- OH
H H H H H Br H
II.But-2-ene + bromine water 3-bromobutan-2-ol
H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3
H H H H H H H H
H C C – C = C + HO– Br H – C- C – C – C- OH
H H H H H Br H
The reaction uses two moles of bromine water molecules to break the two double bonds.
But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol
H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr
H H H H H H H H
H C C – C = C -H + 2(HO – Br) H – C- C – C – C- H
HO Br HO Br
(v) Oxidation.
Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.
When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green.
When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized.
Examples
1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution.
The purple acidified potassium/sodium manganate(VII) solution is decolorized.
The orange acidified potassium/sodium dichromate(VI) solution turns to green.
Chemical equation
H2C=CH2 [O] in H+/K2Cr2O7 HO CH2 – CH2 OH
H H H H
C = C+ [O] in H+/KMnO4 H – C – C – H
H H OH OH
Ethene + [O] in H+/KMnO4 ethan-1,2-diol
The purple acidified potassium/sodium manganate(VII) solution is decolorized.
The orange acidified potassium/sodium dichromate(VI) solution turns to green.
Chemical equation
H3C CH=CH2 [O] in H+/KMnO4 H3C CHOH – CH2OH
Propene [O] in H+/KMnO4 propan-1,2-diol
H H H H H H
H C C = C [O] in H+/KMnO4 H – C – C – C- OH
H H H OH H
3.Both But-1-ene and But-2-ene react with bromine water to form butan-1,2-diol and butan-2,3-diol
Chemical equation
I.But-1-ene + [O] in H+/KMnO4 butan-1,2-diol
H3C CH2 CH=CH2 + [O] H3C CH2CHOH – CH2OH
H H H H H H H H
H C C – C = C + [O] H – C- C – C – C- OH
H H H H H OH H
(v) Hydrolysis.
Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound.
Alkenes undergo hydrolysis to form alkanols .
This takes place in two steps:
(i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI).
Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI)
(ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed.
alkylhydrogen sulphate(VI) + water -warm-> Alkanol.
Examples
(i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII)
Chemical equation
H2C=CH2 + H2SO4 CH3 – CH2OSO3H
H H H O-SO3H
C = C + H2SO4 H – C – C – H
H H H H
Ethene + H2SO4 ethylhydrogen sulphate(VI)
(ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol
Chemical equation
CH3 – CH2OSO3H + H2O CH3 – CH2OH + H2SO4
H OSO3H H OH
H – C – C – H + H2O H – C – C – H + H2SO4
H H H H
ethylhydrogen sulphate(VI) + H2O Ethanol
Chemical equation
CH3H2C=CH2 + H2SO4 CH3CH2 – CH2OSO3H
H H H H H O-SO3H
C = C – C – H + H2SO4 H – C – C – C – H
H H H H H H
Propene + H2SO4 propylhydrogen sulphate(VI)
(ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol
Chemical equation
CH3 – CH2OSO3H + H2O CH3 – CH2OH + H2SO4
H H OSO3H H H OH
H – C – C – C – H + H2O H – C – C – C – H + H2SO4
H H H H H H
propylhydrogen sulphate(VI) + H2O propanol
(vi) Polymerization/self addition
Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule.
Only alkenes undergo addition polymerization.
Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene
During addition polymerization
(i)the double bond in alkenes break
(ii)free radicals are formed
(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.
Examples of addition polymerization
1.Formation of Polyethene
Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H H H H H H H H
Ethene + Ethene + Ethene + Ethene + …
(ii)the double bond joining the ethane molecule break to free readicals
H H H H H H H H
H H H H H H H H
Ethene radical + Ethene radical + Ethene radical + Ethene radical + …
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H H H H H H H H
Lone pair of electrons can be used to join more monomers to form longer polyethene.
Polyethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C-
H H H H H H H H
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H H
Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
Examples
Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )
Number of monomers/repeating units in polyomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760
Substituting 4760 = 170 ethene molecules
28
The commercial name of polyethene is polythene.
It is an elastic, tough, transparent and durable plastic.
Polythene is used:
(i)in making plastic bag
(ii)bowls and plastic bags
(iii)packaging materials
2.Formation of Polychlorethene
Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H Cl H Cl H Cl H Cl
chloroethene + chloroethene + chloroethene + chloroethene + …
(ii)the double bond joining the chloroethene molecule break to free radicals
H H H H H H H H
H Cl H Cl H Cl H Cl
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H Cl H Cl H Cl H Cl
Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
Polychloroethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H Cl H Cl H Cl H Cl
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H Cl
Examples
Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
3.Formation of Polyphenylethene
Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H C6H5 H C6H5 H C6H5 H C6H5
phenylethene + phenylethene + phenylethene + phenylethene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
H C6H5 H C6H5 H C6H5 H C6H5
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H C6H5 H C6H5 H C6H5 H C6H5
Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.
Polyphenylethene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H C6H5 H C6H5 H C6H5 H C6H5
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H C6H5
Examples
Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760
Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
104
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
4.Formation of Polypropene
Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H CH3 H CH3 H CH3 H CH3
propene + propene + propene + propene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
H CH3 H CH3 H CH3 H CH3
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H CH3 H CH3 H CH3 H CH3
Lone pair of electrons can be used to join more monomers to form longer propene.
propene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H CH3 H CH3 H CH3 H CH3
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H CH3
Examples
Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760
Substituting 4760 = 108.1818 =>108 propene molecules(whole number)
44
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
5.Formation of Polytetrafluorothene
Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
F F F F F F F F
C = C + C = C + C = C + C = C + …
F F F F F F F F
tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + …
(ii)the double bond joining the tetrafluoroethene molecule break to free radicals
F F F F F F F F
F F F F F F F F
(iii)the free radicals collide with each other and join to form a larger molecule
F F F F F F F F lone pair of electrons
F F F F F F F F
Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.
polytetrafluoroethene molecule can be represented as:
F F F F F F F F extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
F F F F F F F F
Since the molecule is a repetition of one monomer, then the polymer is:
F F
( C – C )n
F F
Examples
Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
6.Formation of rubber from Latex
Natural rubber is obtained from rubber trees.
During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.
Latex is a mixture of rubber and lots of water.
The latex is then added an acid to coagulate the rubber.
Natural rubber is a polymer of 2-methylbut-1,3-diene ;
H CH3 H H
CH2=C (CH3) CH = CH2 H – C = C – C = C – H
During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
H CH3 H H H CH3 H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H CH3 H H
-(- C – C = C – C -)n–
H H
Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.
During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.
| Sulphur atoms make cross link between polymers |
H CH3 H H H CH3 H H
– C – C – C – C – C – C – C – C –
H S H H S H
H CH3 S H H CH3 S H
– C – C – C – C – C – C – C – C –
H H H H H H
Vulcanized rubber is used to make tyres, shoes and valves.
7.Formation of synthetic rubber
Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot.
Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;
H Cl H H
CH2=C (Cl CH = CH2 H – C = C – C = C – H
During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;
H Cl H H H Cl H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H Cl H H
-(- C – C = C – C -)n–
H H
Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.
(c)Test for the presence of – C = C – double bond.
(i)Burning/combustion
All unsaturated hydrocarbons with a – C = C – or – C = C – bond burn with a yellow sooty flame.
Experiment
Scoop a sample of the substance provided in a clean metallic spatula. Introduce it on a Bunsen burner.
| Observation | Inference |
| Solid melt then burns with a yellow sooty flame
|
– C = C –,
– C = C – bond |
(ii)Oxidation by acidified KMnO4/K2Cr2O7
Bromine water ,Chlorine water and Oxidizing agents acidified KMnO4/K2Cr2O7 change to unique colour in presence of – C = C –
or – C = C – bond.
Experiment
Scoop a sample of the substance provided into a clean test tube. Add 10cm3 of distilled water. Shake. Take a portion of the solution mixture. Add three drops of acidified KMnO4/K2Cr2O7 .
| Observation | Inference |
| Acidified KMnO4 decolorized
Orange colour of acidified K2Cr2O7 turns green
Bromine water is decolorized
Chlorine water is decolorized |
– C = C –
– C = C – bond
|
(d)Some uses of Alkenes
(iii) Alkynes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n–2 and C C double bond as the functional group . n is the number of Carbon atoms in the molecule.
The carbon atoms are linked by at least one triple bond to each other and single bonds to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name |
| 1 | Does not exist
|
– | |
| 2 | C2H2
|
H C C H CH CH
|
Ethyne |
| 3 | C3H4 | H
H C C C H
H CH C CH3
|
Propyne |
| 4 | C4H6 | H H
H C C C C H
H H CH C CH2CH3
|
Butyne |
| 5 | C5H8 | H H H
H C C C C C H
H H H CH C (CH2)2CH3
|
Pentyne |
| 6 | C6H10 | H H H H
H C C C C C C H
H H H H CH C (CH2)3CH3
|
Hexyne |
| 7 | C7H12 | H H H H H
H C C C C C C C H
H H H H H H H CH C (CH2)4CH3 |
Heptyne |
| 8 | C8H14 | H H H H H H
H C C C C C C C C H
H H H H H H CH C (CH2)5CH3 |
Octyne |
| 9 | C9H16 | H H H H H H H
H C C C C C C C C C H
H H H H H H H CH C (CH2)6CH3 |
Nonyne |
| 10 | C10H18 | H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H CH C (CH2)7CH3 |
Decyne |
Note
e.g
Propyne differ from ethyne by (14 a.m.u) one carbon and two Hydrogen atoms from ethyne.
4.A homologous series of alkenes like that of alkanes:
(i) differ by a CH2 group from the next /previous consecutively
(ii) have similar chemical properties
(iii)have similar chemical formula with general formula CnH2n-2
(iv)the physical properties also show steady gradual change
5.The – C = C – triple bond in alkyne is the functional group. The functional group is the reacting site of the alkynes.
(b)Isomers of alkynes
Isomers of alkynes have the same molecular general formula but different molecular structural formula.
Isomers of alkynes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature of naming alkynes uses the following basic rules/guidelines:
1.Identify the longest continuous/straight carbon chain which contains the – C = C- triple bond to get/determine the parent alkene.
3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3-diyne.
4.The position indicated must be for the carbon atom at the lower position in the
-C = C- triple bond. i.e
But-2-yne means the triple -C = C- is between Carbon “2”and “3”
Pent-1,3-diyne means there are two triple bonds; one between carbon “1” and “2”and another between carbon “3” and “4”
6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of triple – C = C- bonds and branches attached to the alkyne.
7.Position isomers can be formed when the – C = C- triple bond is shifted between carbon atoms e.g.
But-2-yne means the double – C = C- is between Carbon “2”and “3”
But-1-yne means the double – C = C- is between Carbon “1”and “2”
Both But-1-yne and But-2-yne are position isomers of Butyne.
Butyne and 2-methyl propyne both have the same general formular but different branching chain.
(More on powerpoint)
(c)Preparation of Alkynes.
Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction
Chemical equation
CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g)
(d)Properties of alkynes
Like alkanes and alkenes,alkynes are colourles gases, solids and liquids that are not poisonous.
They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. Ethyne has a pleasant taste when pure.
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st three straight chain alkynes (ethyne,propyne and but-1-yne)are gases at room temperature and pressure.
The density of straight chain alkynes increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkyne.
Summary of physical properties of the 1st five alkenes
| Alkyne | General formula | Melting point(oC) | Boiling point(oC) | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Ethyne | CH CH | -82 | -84 | gas |
| Propyne | CH3 C CH | -103 | -23 | gas |
| Butyne | CH3CH2 CCH | -122 | 8 | gas |
| Pent-1-yne | CH3(CH2) 2 CCH | -119 | 39 | liquid |
| Hex-1-yne | CH3(CH2) 3C CH | -132 | 71 | liquid |
(a)Burning/combustion
Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water.
Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water.
Alkyne + Air -> carbon(II) oxide /carbon + water (limited air)
Burning of alkynes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the – C = C – triple bond because they have very high C:H ratio.
Examples of burning alkynes
1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water.
Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g)
(b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water.
Ethyne + Air -> carbon(II) oxide + water (limited air )
C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g)
2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Propyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H4(g) + 4O2(g) -> 3CO2(g) + 2H2O(l/g)
(a) Propyne when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C3H4(g) + 5O2(g) -> 6CO(g) + 4H2O(l/g)
(b)Addition reactions
An addition reaction is one which an unsaturated compound reacts to form a saturated compound. Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple – C = C- to single C- C bond.
(i)Hydrogenation
Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes.
Examples
1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single.
Chemical equation
HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C – CH2
H H H H H H
C = C + H – H – Ni/Pa -> H – C = C – H + H – H – Ni/Pa -> H – C – C – H
H H H H H H
2.Propyne undergo hydrogenation to form Propane
Chemical equation
H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH – CH3
H H H H H H
H C C = C + 2H – H – Ni/Pa-> H – C – C – C- H
H H H H H
3(a) But-1-yne undergo hydrogenation to form Butane
Chemical equation
But-1-yne + Hydrogen –Ni/Pa-> Butane
H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H
H C C – C = C + 2H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H
(b) But-2-yne undergo hydrogenation to form Butane
Chemical equation
But-2-yne + Hydrogen –Ni/Pa-> Butane
H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H
H C C = C – C H + 2H – H- Ni/Pa-> H – C- C – C – C- H
H H H H H H
(ii) Halogenation.
Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane.
The reaction of alkynes with halogens with alkynes is faster than with alkenes. The triple bond in the alkyne break and form a double then single bond.
The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases.
Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon.
Examples
1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane.
Chemical equation
HC = CH + 2Br2 H Br2 C – CH Br2
H H H H
C = C + 2Br – Br Br – C – C – Br
Br Br
Ethyne + Bromine 1,1,2,1-tetrabromoethane
2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane.
Chemical equation
H3C C = CH + 2Cl2 H3C CHCl2 – CHCl2
Propyne + Chlorine 1,1,2,2-tetrachloropropane
H H Cl H
H C C = C + 2Cl – Cl H – C – C – C- Cl
H H H Cl Cl
Propyne + Iodine 1,1,2,2-tetraiodopropane
H3C C = CH + 2I2 H3C CHI2 – CHI2
H H H H H I H
H C C – C = C + 2I – I H – C- C – C – C- I
H H H H I I
3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine
Chemical equation
But-1-yne + iodine 1,1,2,2-tetrabromobutane
H3C CH2 C = CH + 2I2 H3C CH2C I2 – CHI2
H H H H I I
H C C – C = C -H + 2I – I H – C- C – C – C- H
H H H H H I I
(b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine
But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane
H3C C = C -CH2 + 2F2 H3C CF2CF2 – CH3
H H H H H H H H
H C C = C – C -H + F – F H – C- C – C – C- H
H H H H H H
But-1,3-diene + iodine 1,2,3,4-tetraiodobutane
H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2
I I I I
H C C – C = C -H + 4(I – I) H – C- C – C – C- H
I I I I
(iii) Reaction with hydrogen halides.
Hydrogen halides reacts with alkyne to form a halogenoalkene then halogenoalkane. The triple bond in the alkyne break and form a double then single bond.
The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .
Examples
Chemical equation
H C = C H + 2HBr H3 C – CH Br2
H H H H
C = C + 2H – Br H – C – C – Br
H Br
Ethyne + Bromine 1,1-dibromoethane
Chemical equation
H3C C = CH + 2HI H3C CHI2 – CH3
| Carbon atom with more Hydrogen atoms gets extra hydrogen |
Propene + Chlorine 2,2-dichloropropane
H H I H
H C C = C + 2H – I H – C – C – C– H
H H H I H
Chemical equation
But-1-ene + hydrogen bromide 2,2-dibromobutane
H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3
H H H H Br H
H C C – C = C + 2H – Br H – C- C – C – C- H
H H H H H Br H
But-2-yne + Hydrogen bromide 2,2-dibromobutane
H3C C = C -CH3 + 2HBr H3C CBr2CH2 – CH3
H H H Br H H
H C C = C – C -H + 2Br – H H – C- C – C – C- H
H H H Br H H
But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane
H C = C C = C H + 4HI H3C C I2 C I2 CH3
H H H I I H
H C C – C = C -H + 4(H – I) H – C- C – C – C- H
H I I H
B.ALKANOLS(Alcohols)
(A) INTRODUCTION.
Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include
| n | General / molecular formular | Structural formula | IUPAC name |
| 1 | CH3OH |
H – C –O – H │ H
|
Methanol |
| 2 | CH3 CH2OH
C2H5 OH |
H H
H C – C –O – H │ H H
|
Ethanol |
| 3 | CH3 (CH2)2OH
C3H7 OH |
H H H
H C – C – C –O – H │ H H H
|
Propanol |
| 4 | CH3 (CH2)3OH
C4H9 OH |
H H H H
H C – C – C – C –O – H │ H H H H |
Butanol |
| 5 | CH3(CH2)4OH
C5H11 OH |
H H H H H
H C – C – C- C- C –O – H │ H H H H H
|
Pentanol |
| 6 | CH3(CH2)5OH
C6H13 OH |
H H H H H H
H C – C – C- C- C– C – O – H │ H H H H H H
|
Hexanol |
| 7 | CH3(CH2)6OH
C7H15 OH |
H H H H H H H
H C – C – C- C- C– C –C- O – H │ H H H H H H H
|
Heptanol |
| 8 | CH3(CH2)7OH
C8H17 OH |
H H H H H H H H
H C – C – C- C- C– C –C- C -O – H │ H H H H H H H H
|
Octanol |
| 9 | CH3(CH2)8OH
C9H19 OH |
H H H H H H H H H
H C – C – C- C- C– C –C- C –C- O – H │ H H H H H H H H H
|
Nonanol |
| 10 | CH3(CH2)9OH
C10H21 OH |
H H H H H H H H H H
H C – C – C- C- C– C –C- C –C- C-O – H │ H H H H H H H H H H
|
Decanol |
Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where:
(i)general name is derived from the alkane name then ending with “-ol”
(ii)the members have –OH as the fuctional group
(iii)they have the same general formula represented by R-OH where R is an alkyl group.
(iv) each member differ by –CH2 group from the next/previous.
(v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.
(vi)they show similar and gradual change in their chemical properties.
Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines:
(i)Like alkanes , identify the longest carbon chain to be the parent name.
(ii)Identify the position of the -OH functional group to give it the smallest /lowest position.
(iii) Identify the type and position of the side branches.
Practice examples of isomers of alkanols
(i)Isomers of propanol C3H7OH
CH3CH2CH2OH – Propan-1-ol
OH
CH3CHCH3 – Propan-2-ol
Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes.
(ii)Isomers of Butanol C4H9OH
CH3 CH2 CH3 CH2 OH Butan-1-ol
CH3 CH2 CH CH3
OH Butan-2-ol
CH3
CH3 CH3 CH3
OH 2-methylpropan-2-ol
Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes.
2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.
(iii)Isomers of Pentanol C5H11OH
CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)
CH3 CH2 CH CH3
OH Pentan-2-ol (Position isomer)
CH3 CH2 CH CH2 CH3
OH Pentan-3-ol (Position isomer)
CH3
CH3 CH2 CH2 C CH3
OH 2-methylbutan-2-ol (Position /structural isomer)
CH3
CH3 CH2 CH2 C CHOH
CH3 2,2-dimethylbutan-1-ol (Position /structural isomer)
CH3
CH3 CH2 CH C CH3
CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer)
(iv)1,2-dichloropropan-2-ol
CClH2 CCl CH3
OH
(v)1,2-dichloropropan-1-ol
CClH2 CHCl CH2
OH
(vi) Ethan1,2-diol
H H
HOCH2CH2OH H-O – C – C – O-H
H H
(vii) Propan1,2,3-triol H OH H
HOCH2CHOHCH2OH H-O – C- C – C – O-H
H H H
For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation.
In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination.
Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver.
Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast.
It involves three processes:
(i)Conversion of starch to maltose using the enzyme diastase.
(C6H10O5)n (s) + H2O(l) –diastase enzyme –> C12H22O11(aq)
(Starch) (Maltose)
(ii)Hydrolysis of Maltose to glucose using the enzyme maltase.
C12H22O11(aq)+ H2O(l) — maltase enzyme –>2 C6H12O6(aq)
(Maltose) (glucose)
(iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase.
C6H12O6(aq) — zymase enzyme –> 2 C2H5OH(aq) + 2CO2(g)
(glucose) (Ethanol)
At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction.
To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol).
Methanol is much more poisonous /toxic than ethanol.
Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours.
School laboratory preparation of ethanol from fermentation of glucose
Measure 100cm3 of pure water into a conical flask.
Add about five spatula end full of glucose.
Stir the mixture to dissolve.
Add about one spatula end full of yeast.
Set up the apparatus as below.
Preserve the mixture for about three days.
D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS
Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols
Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process.
A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.
More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate.
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s)
H2O(l) + CO2 (g) + CaCO3(s) -> Ca(HCO3) 2 (aq)
(c)Effects on litmus paper
Experiment
Take the prepared sample and test with both blue and red litmus papers.
Repeat the same with pure ethanol and methylated spirit.
Sample Observation table
| Substance/alkanol | Effect on litmus paper |
| Prepared sample | Blue litmus paper remain blue
Red litmus paper remain red |
| Absolute ethanol | Blue litmus paper remain blue
Red litmus paper remain red |
| Methylated spirit | Blue litmus paper remain blue
Red litmus paper remain red |
Explanation
Alkanols are neutral compounds/solution that have characteristic sweet smell and taste.
They have no effect on both blue and red litmus papers.
(d)Solubility in water.
Experiment
Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water.
Repeat the same with pure ethanol and methylated spirit.
Observation
No layers formed between the two liquids.
Explanation
Ethanol is miscible in water.Both ethanol and water are polar compounds .
The solubility of alkanols decrease with increase in the alkyl chain/molecular mass.
The alkyl group is insoluble in water while –OH functional group is soluble in water.
As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases.
e)Melting/boiling point.
Experiment
Place pure ethanol in a long boiling tube .Determine its boiling point.
Observation
Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure.
Explanation
The melting and boiling point of alkanols increase with increase in molecular chain/mass .
This is because the intermolecular/van-der-waals forces of attraction between the molecules increase.
More heat energy is thus required to weaken the longer chain during melting and break during boiling.
f)Density
Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other.
This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density).
Summary table showing the trend in physical properties of alkanols
| Alkanol | Melting point
(oC) |
Boiling point
(oC) |
Density
gcm-3 |
Solubility in water |
| Methanol | -98 | 65 | 0.791 | soluble |
| Ethanol | -117 | 78 | 0.789 | soluble |
| Propanol | -103 | 97 | 0.803 | soluble |
| Butanol | -89 | 117 | 0.810 | Slightly soluble |
| Pentanol | -78 | 138 | 0.814 | Slightly soluble |
| Hexanol | -52 | 157 | 0.815 | Slightly soluble |
| Heptanol | -34 | 176 | 0.822 | Slightly soluble |
| Octanol | -15 | 195 | 0.824 | Slightly soluble |
| Nonanol | -7 | 212 | 0.827 | Slightly soluble |
| Decanol | 6 | 228 | 0.827 | Slightly soluble |
g)Burning
Experiment
Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit.
Observation/Explanation
Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting.
Pure ethanol and methylated spirit easily catch fire / highly flammable.
They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water.
Ethanol is thus a saturated compound like alkanes.
Chemica equation
C2 H5OH(l) + 3O2 (g) -> 3H2O(l) + 2CO2 (g) ( excess air)
C2 H5OH(l) + 2O2 (g) -> 3H2O(l) + 2CO (g) ( limited air)
2CH3OH(l) + 3O2 (g) -> 4H2O(l) + 2CO2 (g) ( excess air)
2 CH3OH(l) + 2O2 (g) -> 4H2O(l) + 2CO (g) ( limited air)
2C3 H7OH(l) + 9O2 (g) -> 8H2O(l) + 6CO2 (g) ( excess air)
C3 H7OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air)
2C4 H9OH(l) + 13O2 (g) -> 20H2O(l) + 8CO2 (g) ( excess air)
C4 H9OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air)
Due to its flammability, ethanol is used;
(h)Formation of alkoxides
Experiment
Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker.
Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.
Sample observations
| Substance/alkanol | Effect of adding sodium |
| Fermentation prepared sample | (i)effervescence/fizzing/bubbles
(ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue |
| Pure/absolute ethanol/methylated spirit | (i)slow effervescence/fizzing/bubbles
(ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue |
Explanations
Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas.
If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.
Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas
Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas
Sodium + Water -> Sodium hydroxides + Hydrogen gas
Potassium + Water -> Potassium hydroxides + Hydrogen gas
Examples
1.Sodium metal reacts with ethanol to form sodium ethoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
2.Potassium metal reacts with ethanol to form Potassium ethoxide
Potassium metal reacts with water to form Potassium Hydroxide
2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s)
2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s)
3.Sodium metal reacts with propanol to form sodium propoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
4.Potassium metal reacts with propanol to form Potassium propoxide
Potassium metal reacts with water to form Potassium Hydroxide
2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s)
2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s)
5.Sodium metal reacts with butanol to form sodium butoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
6.Sodium metal reacts with pentanol to form sodium pentoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
(i)Formation of Esters/Esterification
Experiment
Place 2cm3 of ethanol in a boiling tube.
Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.
Warm/Heat gently.
Pour the mixture into a beaker containing about 50cm3 of cold water.
Smell the products.
Repeat with methanol
Sample observations
| Substance/alkanol | Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid |
| Absolute ethanol | Sweet fruity smell |
| Methanol | Sweet fruity smell |
Explanation
Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory.
Alkanol + Alkanoic acid –Conc. H2SO4-> Ester + water
Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters .
Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.
Ethanol + Ethanoic acid -> Ethylethanoate + Water
Ethanol + Propanoic acid -> Ethylpropanoate + Water
Ethanol + Methanoic acid -> Ethylmethanoate + Water
Ethanol + butanoic acid -> Ethylbutanoate + Water
Propanol + Ethanoic acid -> Propylethanoate + Water
Methanol + Ethanoic acid -> Methyethanoate + Water
Methanol + Decanoic acid -> Methyldecanoate + Water
Decanol + Methanoic acid -> Decylmethanoate + Water
During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.
R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O
e.g.
Ethanol + Ethanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –> CH3COOCH2CH3(aq) +H2O(l)
Ethanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2CH3(aq) +H2O(l)
Methanol + Ethanoic acid –Conc. H2SO4 –>Methylethanoate + Water
CH3OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO CH3(aq) +H2O(l)
Methanol + propanoic acid –Conc. H2SO4 –>Methylpropanoate + Water
CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq) +H2O(l)
Propanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq) +H2O(l)
CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)
(j)Oxidation
Experiment
Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII).
Sample observation table
| Substance/alkanol | Adding acidified KMnO4/K2Cr2O7 | pH of resulting solution/mixture | Nature of resulting solution/mixture |
| Pure ethanol | (i)Purple colour of KMnO4decolorized
(ii) Orange colour of K2Cr2O7turns green. |
pH= 4/5/6
pH = 4/5/6 |
Weakly acidic
Weakly acidic
|
Explanation
Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour:
(i) Purple KMnO4 is reduced to colourless Mn2+
(ii)Orange K2Cr2O7is reduced to green Cr3+
The pH of alkanoic acids show they have few H+ because they are weak acids i.e
Alkanol + [O] -> Alkanal + [O] -> alkanoic acid
NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7
Examples
1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4
Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid
CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH
2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
methanol + [O] -> methanal + [O] -> methanoic acid
CH3OH + [O] -> CH3O + [O] -> HCOOH
3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
Propanol + [O] -> Propanal + [O] -> Propanoic acid
CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH
4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
Butanol + [O] -> Butanal + [O] -> Butanoic acid
CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH
Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.
(k)Hydrolysis /Hydration and Dehydration
Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e.
Alkenes + Water – H3PO4 catalyst-> Alkanol
Examples
(i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol
Ethene + water —60 atm/300oC/ H3PO4 –> Ethanol
H2C =CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2OH(l)
This is the main method of producing large quantities of ethanol instead of fermentation
(ii) Propene + water —60 atm/300oC/ H3PO4 –> Propanol
CH3C =CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2 CH2OH(l)
(iii) Butene + water —60 atm/300oC/ H3PO4 –> Butanol
CH3 CH2 C=CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2 CH2 CH2OH(l)
Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e
Alkanol –Conc. H2 SO4/180oC–> Alkene + Water
Examples
Ethanol —180oC/ H2SO4 –> Ethene + Water
CH3 CH2OH(l) –180oC/ H2SO4 –> H2C =CH2 (g) + H2O(l)
Propanol —180oC/ H2SO4 –> Propene + Water
CH3 CH2 CH2OH(l) –180oC/ H2SO4 –> CH3CH =CH2 (g) + H2O(l)
Butanol —180oC/ H2SO4 –> Butene + Water
CH3 CH2 CH2CH2OH(l) –180oC/ H2SO4 –> CH3 CH2C =CH2 (g) + H2O(l)
Pentanol —180oC/ H2SO4 –> Pentene + Water
CH3 CH2 CH2 CH2 CH2OH(l)–180oC/ H2SO4–>CH3 CH2 CH2C =CH2 (g)+H2O(l)
(l)Similarities of alkanols with Hydrocarbons
Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g.
Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.
CH2 CH2OH(l) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l)
CH2 CH2OH(l) + 2O2(g) -Limited air-> 2CO (g) + 3H2 O(l)
CH3 CH3(g) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l)
2CH3 CH3(g) + 5O2(g) -Limited air-> 4CO (g) + 6H2 O(l)
Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond:
(i)decolorize acidified KMnO4
(ii)turns Orange acidified K2Cr2O7 to green.
Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH).
Alkenes are oxidized to alkanols with duo/double functional groups.
Examples
1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid.
Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid
CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH
2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol.
Ethene + [O] -> Ethan-1,2-diol.
H2C=CH2 + [O] -> HOCH2 -CH2OH
III. Differences with alkenes/alkynes
Alkanols do not decolorize bromine and chlorine water.
Alkenes decolorizes bromine and chlorine water to form halogenoalkanols
Example
When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol.
Ethene + Bromine water -> Bromoethanol.
H2C=CH2 + HOBr -> BrCH2 -CH2OH
Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne)
This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding.
Two alkanol molecules form a dimer joined by hydrogen bonding.
Example
In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen.
This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen.
Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer.
| Hydrogen bonds |
H H H
| Covalent bonds |
H C C O H H
H H H O C C H
H H
Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting.
E.USES OF SOME ALKANOLS
(a)Methanol is used as industrial alcohol and making methylated spirit
(b)Ethanol is used:
2.as antiseptic to wash woulds
3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate
4.as a fuel when blended with petrol to make gasohol.
B.ALKANOIC ACIDS (Carboxylic acids)
(A) INTRODUCTION.
Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:
| n | General /molecular
formular |
Structural formula | IUPAC name |
| 0 | HCOOH |
H – C –O – H │ O |
Methanoic acid
|
| 1 | CH3 COOH
|
H
H – C – C – O – H │ H O |
Ethanoic acid |
| 2 | CH3 CH2 COOH
C2 H5 COOH |
H H
H-C – C – C – O – H
H H O |
Propanoic acid |
| 3 | CH3 CH2 CH2 COOH
C3 H7 COOH |
H H H
H- C – C – C – C – O – H
H H H O |
Butanoic acid |
| 4 | CH3CH2CH2CH2 COOH
C4 H9 COOH |
H H H H
H – C – C – C – C – C – O – H
H H H H O |
Pentanoic acid |
| 5 | CH3CH2 CH2CH2CH2 COOH
C5 H11 COOH |
H H H H H
H C – C – C – C – C – C – O – H
H H H H H O
|
Hexanoic acid |
| 6 | CH3CH2 CH2 CH2CH2CH2 COOH
C6 H13 COOH |
H H H H H H
H C C – C – C – C – C – C – O – H
H H H H H H O
|
Pentanoic acid
|
Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where:
(i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows.
(ii) the members have R-COOH/R C-O-H as the functional group.
O
(iii)they have the same general formula represented by R-COOH where R is an alkyl group.
(iv)each member differ by –CH2– group from the next/previous.
(v)they show a similar and gradual change in their physical properties e.g. boiling and melting point.
(vi)they show similar and gradual change in their chemical properties.
(vii) since they are acids they show similar properties with mineral acids.
(B) ISOMERS OF ALKANOIC ACIDS.
Alkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines
(i)Like alkanes. identify the longest carbon chain to be the parent name.
(ii)Identify the position of the -C-O-H functional group to give it the smallest
O
/lowest position.
(iii)Identify the type and position of the side group branches.
Practice examples on isomers of alkanoic acids
1.Isomers of butanoic acid C3H7COOH
CH3 CH2 CH2 COOH
Butan-1-oic acid
CH3
H2C C COOH 2-methylpropan-1-oic acid
2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does.
2.Isomers of pentanoic acid C4H9COOH
CH3CH2CH2CH2 COOH pentan-1-oic acid
CH3
CH3CH2CH COOH 2-methylbutan-1-oic acid
CH3
H3C C COOH 2,2-dimethylpropan-1-oic acid
CH3
3.Ethan-1,2-dioic acid
O O
HOOC- COOH // H – O – C – C – O – H
4.Propan-1,3-dioic acid
O H O
HOOC- CH2COOH // H – O – C – C – C – O – H
H
5.Butan-1,4-dioic acid
O H H O
HOOC CH2 CH2 COOH H- O – C – C – C – C –O – H
H H
6.2,2-dichloroethan-1,2-dioic acid
HOOCCHCl2 Cl
H – O – C – C – Cl
O H
(C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.
In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming.
The oxidation converts the alkanol first to an alkanal the alkanoic acid.
NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7
General equation:
R- CH2 – OH + [O] –H+/KMnO4–> R- CH –O + H2O(l)
(alkanol) (alkanal)
R- CH – O + [O] –H+/KMnO4–> R- C –OOH
(alkanal) (alkanoic acid)
Examples
1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid .
CH3– CH2 – OH + [O] –H+/KMnO4–> CH3– CH –O + H2O(l)
(ethanol) (ethanal)
CH3– CH – O + [O] –H+/KMnO4–> CH3– C –OOH
(ethanal) (ethanoic acid)
2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid
CH3– CH2 CH2 – OH + [O] –H+/KMnO4–> CH3– CH2 CH –O + H2O(l)
(propanol) (propanal)
CH3– CH – O + [O] –H+/KMnO4–> CH3– C –OOH
(propanal) (propanoic acid)
Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from:
(a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e.
Alkenes + Steam/water — H2PO4 Catalyst–> Alkanol
The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.
Alkanol + Air — MnSO4 Catalyst/5 atm pressure–> Alkanoic acid
Example
Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol.
CH2=CH2 + H2O -> CH3 CH2OH
(Ethene) (Ethanol)
This is the industrial large scale method of manufacturing ethanol
Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.
CH3 CH2OH + [O] — MnSO4 Catalyst/5 atm pressure–> CH3 COOH
(Ethanol) (Ethanoic acid)
(b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.
Alkyne + Water — Mercury(II)sulphate(VI)catalyst–> Alkanal
The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid.
Alkanal + air/oxygen — Manganese(II)sulphate(VI)catalyst–> Alkanoic acid
Example
Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal.
CH = CH + H2O –HgSO4–> CH3 CH2O
(Ethyne) (Ethanal)
This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas.
Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.
CH3 CH2O + [O] — MnSO4 Catalyst/5 atm pressure–> CH3 COOH
(Ethanal) (Oxygen from air) (Ethanoic acid)
(D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.
I.Physical properties of alkanoic acids
The table below shows some physical properties of alkanoic acids
| Alkanol | Melting point(oC) | Boiling point(oC) | Density(gcm-3) | Solubility in water |
| Methanoic acid | 18.4 | 101 | 1.22 | soluble |
| Ethanoic acid | 16.6 | 118 | 1.05 | soluble |
| Propanoic acid | -2.8 | 141 | 0.992 | soluble |
| Butanoic acid | -8.0 | 164 | 0.964 | soluble |
| Pentanoic acid | -9.0 | 187 | 0.939 | Slightly soluble |
| Hexanoic acid | -11 | 205 | 0.927 | Slightly soluble |
| Heptanoic acid | -3 | 223 | 0.920 | Slightly soluble |
| Octanoic acid | 11 | 239 | 0.910 | Slightly soluble |
| Nonanoic acid | 16 | 253 | 0.907 | Slightly soluble |
| Decanoic acid | 31 | 269 | 0.905 | Slightly soluble |
From the table note the following:
II Chemical properties of alkanoic acids
The following experiments shows the main chemical properties of ethanoic (alkanoic) acid.
(a)Effect on litmus papers
Experiment
Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid.
Sample observations
| Solution/acid | Observations/effect on litmus papers | Inference |
| Ethanoic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Succinic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Citric acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Oxalic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Tartaric acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Nitric(V)acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
Explanation
All acidic solutions contains H+/H3O+(aq) ions. The H+ /H3O+ (aq) ions is responsible for turning blue litmus paper/solution to red
(b)pH
Experiment
Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid.
Sample observations
| Solution/acid | pH | Inference |
| Ethanoic acid | 4/5/6 | Weakly acidic |
| Succinic acid | 4/5/6 | Weakly acidic |
| Citric acid | 4/5/6 | Weakly acidic |
| Oxalic acid | 4/5/6 | Weakly acidic |
| Tartaric acid | 4/5/6 | Weakly acidic |
| Sulphuric(VI)acid | 1/2/3 | Strongly acidic |
Explanations
Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.
All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO–
Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H+ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e
Examples
(ethanoic acid) (ethanoate ion) (few H+ ion)
(propanoic acid) (propanoate ion) (few H+ ion)
(Butanoic acid) (butanoate ion) (few H+ ion)
(methanoic acid) (methanoate ion) (few H+ ion)
(sulphuric(VI) acid) (sulphate(VI) ion) (many H+ ion)
(nitric(V) acid) (nitrate(V) ion) (many H+ ion)
(c)Reaction with metals
Experiment
Place about 4cm3 of ethanoic acid in a test tube. Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations | Inference |
| Ethanoic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Succinic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Citric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Oxalic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Tartaric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Nitric(V)acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
Explanation
Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group -COOH is /are displaced and not in the alkyl hydrocarbon chain.
Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e.
Examples
2R – COOH + 2M -> 2R- COOM + 2H2(g)
2.For a divalent metal with monobasic acid
2R – COOH + M -> (R- COO) 2M + H2(g)
3.For a divalent metal with dibasic acid
HOOC-R-COOH+ M -> MOOC-R-COOM + H2(g)
4.For a monovalent metal with dibasic acid
HOOC-R-COOH+ 2M -> MOOC-R-COOM + H2(g)
5 For mineral acids
(i)Sulphuric(VI)acid is a dibasic acid
H2 SO4 (aq) + 2M -> M2 SO4 (aq) + H2(g)
H2 SO4 (aq) + M -> MSO4 (aq) + H2(g)
(ii)Nitric(V) and hydrochloric acid are monobasic acid
HNO3 (aq) + 2M -> 2MNO3 (aq) + H2(g)
HNO3 (aq) + M -> M(NO3 ) 2 (aq) + H2(g)
Examples
1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas.
Caution: This reaction is explosive.
CH3COOH (aq) + Na(s) -> CH3COONa (aq) + H2(g)
(Ethanoic acid) (Sodium ethanoate)
2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas.
2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)
(Ethanoic acid) (Calcium ethanoate)
3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.
HOOC-COOH+ 2Na -> NaOOC – COONa + H2(g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate.
4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas.
HOOC-R-COOH+ Mg -> ( OOC – COO) Mg + H2(g)
(ethan-1,2-dioic acid) (magnesium ethan-1,2-dioate)
5.Magnesium reacts with
(i)Sulphuric(VI)acid to form Magnesium sulphate(VI)
H2 SO4 (aq) + Mg -> MgSO4 (aq) + H2(g)
(ii)Nitric(V) and hydrochloric acid are monobasic acid
2HNO3 (aq) + Mg -> M(NO3 ) 2 (aq) + H2(g)
(d)Reaction with hydrogen carbonates and carbonates
Experiment
Place about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations | Inference |
| Ethanoic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Succinic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Citric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Oxalic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Tartaric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Nitric(V)acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water.
Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint.
Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water.
Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide
Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide
Examples
CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + H2O(l) + CO2 (g)
(Ethanoic acid) (Sodium ethanoate)
2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
2CH3COOH (aq) + Na2CO3 (s) -> 2CH3COONa (aq) + H2O(l) + CO2 (g)
(Ethanoic acid) (Sodium ethanoate)
3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
HOOC-COOH+ Na2CO3 (s) -> NaOOC – COONa + H2O(l) + CO2 (g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
HOOC-COOH+ 2NaHCO3 (s) -> NaOOC – COONa + H2O(l) + 2CO2 (g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
(e)Esterification
Experiment
Place 4cm3 of ethanol acid in a boiling tube.
Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame.
Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations |
| Ethanoic acid | Sweet fruity smell |
| Succinic acid | Sweet fruity smell |
| Citric acid | Sweet fruity smell |
| Oxalic acid | Sweet fruity smell |
| Tartaric acid | Sweet fruity smell |
| Dilute sulphuric(VI)acid | No sweet fruity smell |
Explanation
Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids.
Alkanol + Alkanoic acids -> Ester + water
Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.
Ethanol + Ethanoic acid -> Ethylethanoate + Water
Ethanol + Propanoic acid -> Ethylpropanoate + Water
Ethanol + Methanoic acid -> Ethylmethanoate + Water
Ethanol + butanoic acid -> Ethylbutanoate + Water
Propanol + Ethanoic acid -> Propylethanoate + Water
Methanol + Ethanoic acid -> Methyethanoate + Water
Methanol + Decanoic acid -> Methyldecanoate + Water
Decanol + Methanoic acid -> Decylmethanoate + Water
During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.
R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O
Examples
Ethanol + Ethanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –> CH3COOCH2CH3(aq) +H2O(l)
Ethanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2CH3(aq) +H2O(l)
Methanol + Ethanoic acid –Conc. H2SO4 –>Methylethanoate + Water
CH3OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO CH3(aq) +H2O(l)
Methanol + propanoic acid –Conc. H2SO4 –>Methylpropanoate + Water
CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq) +H2O(l)
Propanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq) +H2O(l)
CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)
Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to:
(i)dissolve substances which water can not e.g grease ,oil, fat
(ii)be washed away after cleaning.
There are two types of detergents:
(a)Soapy detergents
(b)Soapless detergents
Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e.
Sodium hydroxide + octadecanoic acid -> Sodium octadecanoate + water
NaOH(aq) + CH3 (CH2) 16 COOH(aq) -> CH3 (CH2) 16 COO – Na+ (aq) +H2 O(l)
Commonly ,soap can thus be represented ;
R- COO – Na+ where;
R is a long chain alkyl group and -COO – Na+ is the alkanoate ion.
In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.
Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol
Fats/Oils are esters with fatty acids and glycerol parts in their structure;
C17H35COOCH2
C17H35COOCH
C17H35COOCH2
When boiled with concentrated sodium hydroxide solution NaOH;
(i)NaOH ionizes/dissociates into Na+ and OH– ions
(ii)fat/oil split into three C17H35COO– and one CH2 CH CH2
(iii) the three Na+ combine with the three C17H35COO– to form the salt C17H35COO– Na+
(iv)the three OH–ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol)
C17H35COOCH2 CH2OH
C17H35COOCH +NaOH -> 3 C17H35COO– Na+ + CHOH
C17H35COOCH2 CH2OH
Ester Alkali Soap glycerol
Generally:
CnH2n+1COOCH2 CH2OH
CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO– Na+ + CHOH
CnH2n+1COOCH2 CH2OH
Ester Alkali Soap glycerol
R – COOCH2 CH2OH
R – COOCH +NaOH -> 3R-COO– Na+ + CHOH
R- COOCH2 CH2OH
Ester Alkali Soap glycerol
During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out.
The soap is then added colouring agents ,perfumes and herbs of choice.
School laboratory preparation of soap
Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve.
The action of soap
Soapy detergents:
(i)act by reducing the surface tension of water by forming a thin layer on top of the water.
(ii)is made of a non-polar alkyl /hydrocarbon tail and a polar -COO–Na+ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar -COO–Na+ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water.
Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is removed from the garment by rinsing with fresh water.The repulsion of the soap head prevent /ensure the droplets do not mix.Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules.
Advantages and disadvantages of using soapy detergents
Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution.
Soapy detergents have the diadvatage in that:
(i)they are made from fat and oils which are better eaten as food than make soap.
(ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate formed when soap reacts with Ca2+ and Mg2+ present in hard water.
Chemical equation
2C17H35COO– Na+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2Na+(aq)
(insoluble Calcium octadecanote/scum)
2C17H35COO– Na+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2Na+(aq)
(insoluble Magnesium octadecanote/scum)
This causes wastage of soap.
Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive.
(b)SOAPLESS DETERGENTS
Soapless detergent usually called detergent is a long chain salt fromed from by-products of fractional distillation of crude oil.Commonly used soaps include:
(i)washing agents
(ii)toothpaste
(iii)emulsifiers/wetting agents/shampoo
Soapless detergents are derived from reacting:
(i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI)
Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water
R –OH + H2SO4 -> R –O-SO3H + H2O
(ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI)
Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.
alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water
sulphate(VI) hydroxide alkyl hydrogen sulphate(VI)
R –O-SO3H + NaOH -> R –O-SO3– Na+ + H2O
Example
Step I : Reaction of Octadecanol with Conc.H2SO4
C17H35CH2OH (aq) + H2SO4 -> C17H35CH2–O- SO3– H+ (aq) + H2O (l)
octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water
Step II: Neutralization by an alkali
C17H35CH2–O- SO3– H+ (aq) + NaOH -> C17H35CH2–O- SO3– Na+ (aq) + H2O (l)
Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water
sulphate(VI) hydroxide hydrogen sulphate(VI)
School laboratory preparation of soapless detergent
Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water.
Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent.
The action of soapless detergents
The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COO–Na+
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO3– Na+
(long hydrophobic /non-polar alkyl tail) (hydrophilic/polar/ionic head)
The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water.
The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching.
The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment.
The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away.
Advantages and disadvantages of using soapless detergents
Soapless detergents are non-biodegradable unlike soapy detergents.
They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death.
Soapless detergents have the advantage in that they:
(i)do not form scum with hard water.
(ii)are cheap to manufacture/buying
(iii)are made from petroleum products but soapis made from fats/oil for human consumption.
Sample revision questions
| KOH |
| Fat/oil |
| Residue X |
| Filtrate Y |
| Filtration |
| Sodium Chloride |
| Boiling |
(a)Identify the process
Saponification
(b)Fats and oils are esters. Write the formula of the a common structure of ester
C17H35COOCH2
C17H35COOCH
C17H35COOCH2
(c)Write a balanced equation for the reaction taking place during boiling
C17H35COOCH2 CH2OH
C17H35COOCH +3NaOH -> 3 C17H35COO– Na+ + CHOH
C17H35COOCH2 CH2OH
Ester Alkali Soap glycerol
(d)Give the IUPAC name of:
(i)Residue X
Potassium octadecanoate
(ii)Filtrate Y
Propan-1,2,3-triol
(e)Give one use of fitrate Y
Making paint
(f)What is the function of sodium chloride
To reduce the solubility of the soap hence helping in precipitating it out
(g)Explain how residue X helps in washing.
Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat
Has a polar /ionic hydrophilic head that dissolves in water.
From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment.
(h)State one:
(i)advantage of continued use of residue X on the environment
Is biodegradable and thus do not pollute the environment
(ii)disadvantage of using residue X
Uses fat/oil during preparation/manufacture which are better used for human consumption.
(i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below.
| Water sample
|
|||
|
A |
B |
C |
|
| Drops of residue X | 15 | 2 | 15 |
| Drops of residue X in boiled water | 2 | 2 | 15 |
(i)State and explain which sample of water is:
Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating.
Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water.
III. Temporary hard
Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water.
(ii)Write the equation for the reaction at water sample C.
Chemical equation
2C17H35COO– K+ (aq) + CaSO4(aq) -> (C17H35COO– )Ca2+ (s) + K2SO4(aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2K+(aq)
(insoluble Calcium octadecanote/scum)
Chemical equation
2C17H35COO– K+ (aq) + MgSO4(aq) -> (C17H35COO– )Mg2+ (s) + K2SO4(aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2K+(aq)
(insoluble Magnesium octadecanote/scum)
(iii)Write the equation for the reaction at water sample A before boiling.
Chemical equation
2C17H35COO– K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO– )Ca2+ (s) + 2KHCO3 (aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2K+(aq)
(insoluble Calcium octadecanote/scum)
Chemical equation
2C17H35COO– K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO– )Mg2+ (s) + 2KHCO3 (aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2K+(aq)
(insoluble Magnesium octadecanote/scum)
(iv)Explain how water becomes hard
Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness.
(v)State two useful benefits of hard water
-Used in bone and teeth formation
-Coral polyps use hard water to form coral reefs
-Snails use hard water to make their shells
2.Study the scheme below and use it to answer the questions that follow.
| Conc. H2SO4 |
| Substance B |
| 6M sodium hydroxide |
| Brown solid A |
| Ice cold water |
| Olive oil |
(a)Identify :
(i)brown solid A
Alkyl hydrogen sulphate(VI)
(ii)substance B
Sodium alkyl hydrogen sulphate(VI)
(b)Write a general formula of:
(i)Substance A.
O
R-O-S O3 H // R- O – S – O – H
O
(ii)Substance B O
R-O-S O3 – Na+ R- O – S – O – Na+
O
(c)State one
(i) advantage of continued use of substance B
–Does not form scum with hard water
-Is cheap to make
-Does not use food for human as a raw material.
(ii)disadvantage of continued use of substance B.
Is non-biodegradable therefore do not pollute the environment
(d)Explain the action of B during washing.
Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat.
Has a polar/ionic hydrophilic head that dissolves in water
Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment.
(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B.
Product A
Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI)
H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H
Product B
Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water
hydrogen sulphate(VI)
H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3–Na+ + H2O
(f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.
Product A
Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water
H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O
Product B
Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water
hydrogen sulphate(VI)
H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3–Na+ + H2O
3.Below is part of a detergent
H3C – (CH2 )16 – O – SO3 – K +
(a)Write the formular of the polar and non-polar end
Polar end
H3C – (CH2 )16 –
Non-polar end
– O – SO3 – K +
(b)Is the molecule a soapy or saopless detergent?
Soapless detergent
(c)State one advantage of using the above detergent
-does not form scum with hard water
-is cheap to manufacture
4.The structure of a detergent is
H H H H H H H H H H H H H
H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO–Na+
H H H H H H H H H H H H H
CH3(CH2)12COO–Na+
Soapy detergent
Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.
Polymers and fibres are either:
(a)Natural polymers and fibres
(b)Synthetic polymers and fibres
Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include:
-proteins/polypeptides making amino acids in animals
-cellulose that make cotton,wool,paper and silk
-Starch that come from glucose
-Fats and oils
-Rubber from latex in rubber trees.
Synthetic polymers and fibres are man-made. They include:
-polyethene
-polychloroethene
-polyphenylethene(polystyrene)
-Terylene(Dacron)
-Nylon-6,6
-Perspex(artificial glass)
Synthetic polymers and fibres have the following characteristic advantages over natural polymers
Synthetic polymers and fibres however have the following disadvantages over natural polymers
To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used:
1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment.
2.Production of biodegradable synthetic polymers and fibres that rot away.
There are two types of polymerization:
(a)addition polymerization
(b)condensation polymerization
(a)addition polymerization
Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.
Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene
During addition polymerization
(i)the double bond in alkenes break
(ii)free radicals are formed
(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.
Examples of addition polymerization
1.Formation of Polyethene
Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H H H H H H H H
Ethene + Ethene + Ethene + Ethene + …
(ii)the double bond joining the ethane molecule break to free readicals
H H H H H H H H
H H H H H H H H
Ethene radical + Ethene radical + Ethene radical + Ethene radical + …
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H H H H H H H H
Lone pair of electrons can be used to join more monomers to form longer polyethene.
Polyethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H H H H H H H H
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H H
Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
Examples
Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )
Number of monomers/repeating units in polyomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760
Substituting 4760 = 170 ethene molecules
28
The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used:
(i)in making plastic bag
(ii)bowls and plastic bags
(iii)packaging materials
2.Formation of Polychlorethene
Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H Cl H Cl H Cl H Cl
chloroethene + chloroethene + chloroethene + chloroethene + …
(ii)the double bond joining the chloroethene molecule break to free radicals
H H H H H H H H
H Cl H Cl H Cl H Cl
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H Cl H Cl H Cl H Cl
Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
Polychloroethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H Cl H Cl H Cl H Cl
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H Cl
Examples
Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
3.Formation of Polyphenylethene
Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H C6H5 H C6H5 H C6H5 H C6H5
phenylethene + phenylethene + phenylethene + phenylethene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
H C6H5 H C6H5 H C6H5 H C6H5
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H C6H5 H C6H5 H C6H5 H C6H5
Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.
Polyphenylethene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H C6H5 H C6H5 H C6H5 H C6H5
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H C6H5
Examples
Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760
Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
104
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
4.Formation of Polypropene
Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H CH3 H CH3 H CH3 H CH3
propene + propene + propene + propene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
H CH3 H CH3 H CH3 H CH3
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
H CH3 H CH3 H CH3 H CH3
Lone pair of electrons can be used to join more monomers to form longer propene.
propene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H CH3 H CH3 H CH3 H CH3
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H CH3
Examples
Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760
Substituting 4760 = 108.1818 =>108 propene molecules(whole number)
44
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
5.Formation of Polytetrafluorothene
Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
F F F F F F F F
C = C + C = C + C = C + C = C + …
F F F F F F F F
tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + …
(ii)the double bond joining the tetrafluoroethene molecule break to free radicals
F F F F F F F F
F F F F F F F F
(iii)the free radicals collide with each other and join to form a larger molecule
F F F F F F F F lone pair of electrons
F F F F F F F F
Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.
polytetrafluoroethene molecule can be represented as:
F F F F F F F F extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
F F F F F F F F
Since the molecule is a repetition of one monomer, then the polymer is:
F F
( C – C )n
F F
Examples
Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
5.Formation of rubber from Latex
Natural rubber is obtained from rubber trees.
During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.
Latex is a mixture of rubber and lots of water.
The latex is then added an acid to coagulate the rubber.
Natural rubber is a polymer of 2-methylbut-1,3-diene ;
H CH3 H H
CH2=C (CH3) CH = CH2 H – C = C – C = C – H
During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
H CH3 H H H CH3 H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H CH3 H H
-(- C – C = C – C -)n–
H H
Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.
During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.
| Sulphur atoms make cross link between polymers |
H CH3 H H H CH3 H H
– C – C – C – C – C – C – C – C –
H S H H S H
H CH3 S H H CH3 S H
– C – C – C – C – C – C – C – C –
H H H H H H
Vulcanized rubber is used to make tyres, shoes and valves.
6.Formation of synthetic rubber
Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot.
Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;
H Cl H H
CH2=C (Cl CH = CH2 H – C = C – C = C – H
During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;
H Cl H H H Cl H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H Cl H H
-(- C – C = C – C -)n–
H H
Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.
(b)Condensation polymerization
Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water).
Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds
During condensation polymerization:
(i)the two monomers are brought together by high pressure to reduce distance between them.
(ii)monomers realign themselves at the functional group.
(iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl)
(iv)the two monomers join without the simple molecule of H2O/HCl
Examples of condensation polymerization
1.Formation of Nylon-6,6
Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group.
During the formation of Nylon-6,6:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O H H
H- O – C – (CH2 ) 4 – C – O – H + H –N – (CH2) 6 – N – H
(iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage .
O O H H
H- O – C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + H 2O
.
Polymer bond linkage
Nylon-6,6 derive its name from the two monomers each with six carbon chain
Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioyl dichloride with hexan-1,6-diamine.
Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group.
The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen
During the formation of Nylon-6,6:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O H H
Cl – C – (CH2 ) 4 – C – Cl + H –N – (CH2) 6 – N – H
(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .
O O H H
Cl – C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + HCl
.
Polymer bond linkage
The two monomers each has six carbon chain hence the name “nylon-6,6”
The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets.
2.Formation of Terylene
Method 1: Terylene can be made from the condensation polymerization of ethan-1,2-diol with benzene-1,4-dicarboxylic acid.
Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH.
During the formation of Terylene:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O
H- O – C – C6H5 – C – O – H + H –O – CH2 CH2 – O – H
(iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage .
O O
H- O – C – C6H5 – C – O – (CH2) 6 – N – H + H 2O
.
Polymer bond linkage of terylene
Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol.
Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring.
The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen
During the formation of Terylene
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O
Cl – C – C5H5 – C – Cl + H –O – CH2 CH2 – O – H
(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .
O O
Cl – C – C5H5 – C – O – CH2 CH2 – O – H + HCl
.
Polymer bond linkage of terylene
The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits.
Practice questions Organic chemistry
(i) Name and write the formula of the main products
Name………………………………….
Formula……………………………………..
(ii) Which homologous series does the product named in (i) above belong?
| HC = CH2 |
| O |
Explain why a lot of soap is needed when washing with hard water
| O
CH3CH2CH C OH
NH2 |
(a) Write the structural formula of the repeat unit of the polymer
(b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer
were obtained.
Determine the number of the monomer molecules in this polymer.
(C = 12; H = 1; N = 14; O =16)
Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
| H H H H
C C C C
O H O H n
|
(a) Give the name of the monomer and draw its structures
(b) Identify the type of polymerization that takes place
(c) State one advantage of synthetic polymers
of ethanol and pentane
| GLUCOSE SOLUTION |
| CRUDE ETHANOL |
| 95% ETHANOL |
| ABSOLUTE ETHANOL |
| G |
| H |
(a) What is absolute ethanol?
(b) State two conditions required for process G to take place efficiently
peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer
the questions that follow:-
| Time in seconds | Volume of Oxygen evolved (cm3) |
| 0
30 60 90 120 150 180 210 240 270 300 |
0
10 19 27 34 38 43 45 45 45 45 |
(i) Plot a graph of volume of oxygen gas against time
(ii) Determine the rate of reaction at time 156 seconds
(iii) From the graph, find the time taken for 18cm3 of oxygen to be produced
(iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence
of manganese (IV) Oxide
(b) The diagram below shows how a Le’clanche (Dry cell) appears:-
(i) What is the function of MnO2 in the cell above?
(ii) Write the equation of a reaction that occurs at the cathode
(iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows
through the above cell for 30minutes (1F =96500c Zn =65)
(i) CH3COOCH2CH3 *
| CH2 = C – CHCH3
Br |
(ii)
(b) The structure below shows some reactions starting with ethanol. Study it and answer
| CH3COOH |
| CH3COONa |
| CH3CH2OH |
| CH2=CH2 |
| CH3CH3 |
| CH2 CH2 |
| n |
| S |
| P |
| CH4 |
| T |
| Na Metal |
| Compound U |
| Step II |
| Step I |
| Step III |
| CH3COOH
Drops of Conc. H2SO4 |
| Reagent R |
| NaOH(aq) |
| Heat |
| Excess Cl2 /U.V |
the questions that follow:
(i) Write the formula of the organic compounds P and S *
(ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :-
(I) Step I *
(II) Step II *
(III) Step III *
(iii) Name reagent R …………………………………………………………… *
(iv) Draw the structural formula of T and give its name *
(v) (I) Name compound U………………………………………………………..
(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1)
(c) State why C2H4 burns with a more smoky flame than C2H6 *
CH3CH2CH2ONa
| CH3CH2CH3 |
| P |
| Step
V |
| CH3CH = CH2 |
| Step
W |
| CH3CH2CH2OH |
| CH3CH2COOH |
| Step R |
| Gas F |
| CH3CH2COONa |
| NaOH
Step J |
| K2CO3 |
| Step X |
| NaOH
Heat |
| CH3CH2COOCH3
G |
| Product T
+ Na2CO3 |
| H H
C – C
CH3 H |
| n |
| K |
iii) State the type of reaction that occurred in step J
vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12)
| H2 (g) Ni
High temp |
| Polymer Q |
| Polymerization |
| Compound P |
| CH3CH2CH3 |
| CH3CH2CH2ONa + H2
|
| Na(s) |
| Propan-l-ol |
| Step I |
| Propylethanoate |
| CH3CH2COOH
|
| Solution T + CO2 (g)
|
| Step III |
| Na2CO3(aq) |
| Conc. H2SO4 180oC |
| Step II |
(a) (i) Name compound P ……………………………………………………………………
(ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3
(b) State one use of polymer Q
(c) Name one oxidising agent that can be used in step II …………………………………..
(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of
monomers in the polymer (H = 1, C = 12)
(e) Name the type of reaction in step I …………………………………………………………..
(f) State one industrial application of step III
(g)State how burning can be used to distinguish between propane and propyne. Explain your
answer
(h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour.
Calculate the minimum volume of air needed for the complete combustion of ethene
(Air contains 20% by volume of oxygen)
| CH3CH2COOCH2CH2CH3 |
| CH3CHCH2 |
| CH3CH2CH2ONa + Gas P |
| CH3CH2CH2OH |
| X |
| V |
| HCl Step 5 |
| Step 1 R |
| Na
Step2 |
| H+ |
| Step 3 |
| Q + H2O |
| MnO–4
Step 4 |
| Ni H2 |
(i) Identify the following:
Substance Q ………………………………………………………………………………………………..
Substance R…………………………………………………………………………………………………
Gas P…………………………………………………………………………………………………………..
(ii) Name:
Step 1…………………………………………………………………………………….
Step 4…………………………………………………………………………………….
(iii) Draw the structural formula of the major product of step 5
(iv) State the condition and reagent in step 3
| M |
| KMnO4/H+ |
| CH2CH2 |
| Ethyl Ethanoate |
| CH2CH2OH |
| L |
| J |
| K |
| CO2 (g) |
| STEP 2 |
| Reagent Q
Step 3 |
| KMnO4/H+(aq) |
| Ni/H2(g)
Step 4 |
| Reagent P |
(a) (i) Name the following organic compounds:
M……………………………………………………………..……..
L…………………………………………………………………..
(ii) Name the process in step:
Step 2 ………………………………………………………….….
Step 4 ………………………………………………………….…
(iii) Identify the reagent P and Q
(iv) Write an equation for the reaction between CH3CH2CH2OH and sodium
iii) CH3C – O- CH2CH3 ……………………………………………………………………
| Step I |
| Step V
Complete combustion |
| Products |
| CH ºCH |
| C2H5COONa |
| Step II |
| Step IV + Heat |
| CH2 = CH2 |
| C2H6 |
| Step III |
|
CH2 = CHCl |
| n |
Step I: ………………………………………………………………………
Step II ……………………………………………………………………
Step III ………………………………………………………………………
iii) Explain one disadvantage of continued use of items made form the compound formed
in step III
sulphur 11.5%, water 45.3%
volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution
in moles per litre. (Given that the molecula mass of the salt is 278)
iii) Explain one disadvantage of continued use of items made form the compound formed
in step III
OH
ii)CH3 – CH – CH2 – CH2 – CH3
C2H5
iii)CH3COOCH2CH2CH3
O
R – S – O–Na+
O
CH3 CH3 CH3
ï ï ï
― CH – CH2 – CH- CH2 – CH – CH2 ―
| Ethanol |
| B
process I |
| Ethanoic acid
|
| Na2CO3 |
| Salt A + CO2 + H2O |
(a) Identify substances A and B
(b) Name the process I
ammonia gas
(b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of
ammonia (H=1, N=14, O=16)
(b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate
(b) Draw the structure and give the name of the organic compound formed when ethanol
reacts with pentanoic acid in presence of concentrated sulphuric acid
| Q |
| P |
| CH3COONa |
| Ethanol |
| CH3CH2ONa |
| C |
| H2SO4(l)
170oC step 1 |
| Cr2O7(aq) / H+(aq) |
| Na(s) |
| Step 2 |
| Step 4 |
| CH3CH2OH/H2SO4 |
| Step 3 |
| 2- |
that follow:-
(i) Name and draw the structure of substance Q
(ii) Give the names of the reactions that take place in steps 2 and 4
(iii) What reagent is necessary for reaction that takes place in step 3
They belong to two different homologous series of organic compounds. If both A and B
react with potassium metal:
(a) Name the common product produced by both
(b) State the observation made when each of the samples A and B are reacted with sodium
hydrogen carbonate
(i) A
(ii) B
electrons in the outermost energy level are shown
| W |
| U |
| V |
| 19P
20N |
| Z |
| Y |
key
P = Proton
N = Neutron
X = Electron
(a) Identify the particle which is an anion
(a) What term is used to describe plastic and rubbers used in this way?
(b) Explain why plastics and rubbers are used this way
| R |
| Conc.
H2SO4 |
| R |
|
SO3H |
| Cleaning agent X |
(a) Draw the structure of X and state the type of cleaning agent to which X belong
(b) State one disadvantage of using X as a cleaning agent
of the isotope is 20days
(a) Find the initial mass of the isotope
(b) Give one application of radioactivity in agriculture
| H H
-C – C –
CH3 H |
| n
n |
(i) Name the polymer above……………………………………………………………………….
(ii) Determine the value of n if giant molecule had relative molecular mass of 4956
from the Indian ocean. Explain
iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference
in their bleaching properties
| H H H H
C C C C
O H O H |
| n |
(a) Give the name of the polymer
(b) Draw the structure of the monomer used to manufacture the polymer
|
UPGRADE
CHEMISTRY
FORM 3
The Chemistry of NITROGEN
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
|
Contents
A NITROGEN …………………………………………………………………….3
Occurrence………………………………………………………..……..…..3
Isolation from air…………………………………………………………….3
Fractional distillation …………………………………………………….….4
Preparation of Nitrogen…………………………………………………..….6
Properties of Nitrogen…………………………………………………….…..6
B.OXIDES OFNITROGEN……………………………………………….……..7
Nitrogen(I)Oxide………………………..……………………………………7
Nitrogen(II) oxide……………………….…………………………………..9
Nitrogen(IV) oxide …………………………………………………………11
C.AMMONIA…..………………………….…………………………….……..13
Occurrence…………………………………………………………………15
Preparation………………………………………………………………….15
Properties……………………………………………………………………16
Haber process……………………………….………………………………12
Preparation…………………………………….……………………………13
Properties…………………………………………………………………..33
Ostwalds process…………………………………………………………..12
E.NO3– and NO2– salts ……………………….……………………………….……27
A.NITROGEN
Nitrogen is found in the atmosphere occupying about 78% by volume of air.
Proteins, amino acids, polypeptides in living things contain nitrogen.
Nitrogen can be isolated from other gases present in air like oxygen, water (vapour), carbon (IV) oxide and noble gases as in the school laboratory as in the flow chart below:
Water is added slowly into an “empty flask” which forces the air out into another flask containing concentrated sulphuric (VI) acid. Concentrated sulphuric (VI) acid is hygroscopic. It therefore absorb/remove water present in the air sample.
More water forces the air into the flask containing either concentrated sodium hydroxide or potassium hydroxide solution. These alkalis react with carbon IV) oxide to form the carbonates and thus absorbs/remove carbon IV) oxide present in the air sample.
Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (aq) + H2O(l)
Chemical equation 2KOH (aq) + CO2 (g) -> K2CO3 (aq) + H2O(l)
More water forces the air through a glass tube packed with copper turnings. Heated brown copper turnings react with oxygen to form black copper (II) oxide.
Chemical equation 2Cu (s) + O2 (g) -> CuO (s)
(brown) (black)
The remaining gas mixture is collected by upward delivery/downward displacement of water/over water. It contains about 99% nitrogen and 1% noble gases.
On a large scale for industrial purposes, nitrogen is got from fractional distillation of air.
c)Nitrogen from fractional distillation of air.
For commercial purposes nitrogen is got from the fractional of air.
Air is first passed through a dust precipitator/filter to remove dust particles.
The air is then bubbled through either concentrated sodium hydroxide or potassium hydroxide solution to remove/absorb Carbon(IV) oxide gas.
Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (aq) + H2O(l)
Chemical equation 2KOH (aq) + CO2 (g) -> K2CO3 (aq) + H2O(l)
Air mixture is the cooled to -25oC.At this temperature, water (vapour ) liquidifies and then solidify to ice and thus removed.
The air is further cooled to -200oC during which it forms a blue liquid.
The liquid is then heated. Nitrogen with a boiling point of -196oC distils first then Argon at-186oC and then finally Oxygen at -183oC boils last.
The diagram below shows the set up of the school laboratory preparation of nitrogen gas.
d.Properties of Nitrogen gas(Questions)
1.Write the equation for the reaction for the school laboratory preparation of nitrogen gas.
Chemical equation NH4Cl (s) + NaNO2(s)->NaCl (g)+ NH4NO2 (s)
Chemical equation NH4NO2 (s) -> N2 (g) + H2O (l)
– colourless, odourless, less dense than air ,neutral and slightly soluble in water
Observation; It continues burning with a blight blindening flame forming white ash.
ExplanationMagnesium burns to produce enough heat /energy to reacts with nitrogen to form white magnesium nitride.
Chemical equation3Mg (s) + N2 (g) -> Mg3N2 (s)
(white ash/solid)
-manufacture of ammonia from Haber process
– As a refrigerant in storage of semen for Artificial insemination.
Nitrogen forms three main oxides:
i)Nitrogen(I) oxide(N2O)
iii) Nitrogen (IV) oxide( NO2)
Nitrogen (I) oxide does not occur naturally but prepared in a laboratory.
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (I) oxide in a school laboratory.
Chemical equation NH4NO2(s) -> H2O (l) + N2O (g)
2.a) State and explain three errors made in the above set up
–Oxygen is being generated instead of Nitrogen (I) oxide.
Ammonium Nitrate(V) should be used instead of potassium manganate(VI) and manganese(IV)oxide.
-slightly soluble in water.
-colourless
-odourless
-less dense than air
-slightly sweet smell
Observation – Continues to burn with a bright flame
-White solid/residue is formed
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation
Mg(s) + N2O (g)-> MgO (s) + N2(g)
Observation – Continues to burn with an orange glow
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas. Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation C(s) + 2N2O (g)-> CO2 (g) + 2N2(g)
Observation – Continues to burn with a blue flame
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation S(s) + 2N2O (g)-> SO2 (g) + 2N2(g)
-As laughing gas because as anesthesia the patient regain consciousness laughing hysterically after surgery.
-improves engine efficiency.
–Oxygen is odourless while nitrogen (I) oxide has faint sweet smell
-Both relight/rekindle a glowing wooden splint but Oxygen can relight a feeble glowing splint while nitrogen (I) oxide relights well lit splint.
-Both are slightly soluble in water but nitrogen (I) oxide is more soluble.
Nitrogen (II) oxide does not occur naturally but prepared in a laboratory.
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (II) oxide in a school laboratory.
Chemical equation 3Cu(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Cu(NO3)2(aq)
Chemical equation 3Zn(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Zn(NO3)2(aq)
Chemical equation 3Mg(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g)+2Mg(NO3)2(aq)
-insoluble in water.
-colourless
-odourless
-denser dense than air
-has no effect on both blue and red litmus papers
Observation – Continues to burn with a bright flame
-White solid/residue is formed
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation 2Mg(s) + 2NO (g)-> 2MgO (s) + N2(g)
Observation – Continues to burn with an orange glow
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation C(s) + 2NO (g)-> CO2 (g) + N2(g)
Observation – Continues to burn with a blue flame
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation S(s) + N2O (g)-> SO2 (g) + N2(g)
Observation – Continues to produce dense white fumes
Explanation-Phosphorus burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.
Chemical equation 4P(s) + 10NO (g)-> 2P2O5(g) + 5N2(g)
As an intermediate gas in the Ostwalds process for manufacture of nitric(V) gas.
Observation–brown fumes produced/evolved that turn blue litmus paper red.
Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.
Chemical equation 2NO (g)+ O2(g)-> 2NO2 (g)
(colorless) (brown)
Nitrogen (IV) oxide occurs -naturally from active volcanic areas.
-formed from incomplete combustion of the internal combustion engine of motor vehicle exhaust fumes.
-from lightening
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (IV) oxide in a school laboratory.
Chemical equation Cu(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Cu(NO3)2(aq)
Chemical equation Zn(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Zn(NO3)2(aq)
Chemical equation Fe(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Fe(NO3)2(aq)
-soluble/dissolves in water.
-brown in colour
-has pungent irritating poisonous odour/smell
-denser dense than air
-turns blue litmus papers to red
Observation–The gas dissolves and thus brown colour of the gas fades
-A colourless solution is formed
-solution formed turns blue litmus papers to red
-solution formed has no effect on red
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (IV) oxide gas dissolves then react with water to form an acidic mixture of nitric(V) acid andnitric(III) acid.
Chemical equationH2O (l) + 2NO 2(g)->HNO3(aq) + HNO2(aq)
(nitric(V) acid) (nitric(III) acid)
Observation on cooling
-Brown colour fades
-Yellow liquid formed
Observation on gentle heating
–Brown colour reappears
–Yellow liquid formed changes to brown fumes/gas
Observation on gentle heating
–Brown colour fades
–brown fumes/gas changes to a colourless gas
Explanation-Brown nitrogen (IV) oxide gas easily liquefies to yellow dinitrogen tetraoxide liquid.When the yellow dinitrogen tetraoxide liquid is gently heated it changes back to the brown nitrogen (IV) oxidegas.When the brown nitrogen (IV) oxide gas is strongly heated it decomposes to colourless mixture of Nitrogen (II) oxide gas and Oxygen.
Chemical equation O2(s) + 2NO (g) ===== 2NO2 (g) ===== N2O4(l)
(colourless gases) (brown gas) (yellow liquid)
Observation – Continues to burn with a bright flame
-White solid/residue is formed
-Brown fumes/colour fades
Explanation-Magnesium burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation 4Mg(s) + 2NO 2(g)-> 4MgO (s) + N2(g)
Observation – Continues to burn with an orange glow
-Brown fumes/colour fades
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation2C(s) + 2NO 2(g)-> 2CO2 (g) + N2(g)
Observation – Continues to burn with a blue flame
-Brown fumes/colour fades
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation2S(s) + 2NO2 (g)-> 2SO2 (g) + N2(g)
Observation– Continues to produce dense white fumes
-Brown fumes/colour fades
Explanation-Phosphorus burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.
Chemical equation 8P(s) + 10NO2 (g)-> 4P2O5(g) + 5N2(g)
-In theOstwald process for industrial manufacture of nitric (V) gas.
-In the manufacture of T.N.T explosives
Observation–brown fumes produced/evolved that turn blue litmus paper red.
Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.
Chemical equation 2NO (g) + O2(g) -> 2NO2 (g)
(colourless) (brown)
Ammonia is a compound of nitrogen and hydrogen only. It is therefore a hydride of nitrogen.
Ammonia gas occurs -naturally from urine of mammals and excretion of birds
-formed in the kidney of human beings
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare dry Ammonia gas in a school laboratory.
Set up method 1
Chemical equation
Ca (OH)2(s) + NH4 Cl(s)->CaCl2 (aq) + H2O(l) + 2NH3(g)
b)Method 2
Chemical equation
NaOH (aq) + NH4 Cl(aq) -> NaCl (aq) + H2O(l) + NH3(g)
-has a pungent choking smell of urine
-Colourless
-Less dense than air hence collected by upward delivery
-Turns blue litmus paper blue thus is the only naturally occurring basic gas (at this level)
-Calcium chloride reacts with ammonia forming the complex compound CaCl2.8H2O.
Chemical equation CaCl2 (s) + 8NH3(g) -> CaCl2 .8NH3(g)
-Concentrated sulphuric(VI) acid reacts with ammonia forming ammonium sulphate(VI) salt compound
Chemical equation 2NH3(g) +H2SO4(l) ->(NH4)2SO4(aq)
4.Describe the test for the presence of ammonia gas.
Using litmus paper:
Dip moist/damp/wet blue and red litmus papers in a gas jar containing a gas suspected to be ammonia.The blue litmus paper remain blue and the red litmus paper turns blue.Ammonia is the only basic gas.(At this level)
Using hydrogen chloride gas
Dip a glass rod in concentrated hydrochloric acid. Bring the glass rod near the mouth of a gas jar suspected to be ammonia. White fumes (of ammonium chloride)are produced/evolved.
Ammonia is very soluble in water.
When a drop of water is introduced into flask containing ammonia, it dissolves all the ammonia in the flask. If water is subsequently allowed into the flask through a small inlet, atmospheric pressure forces it very fast to occupy the vacuum forming a fountain. If the water contains three/few drops of litmus solution,the litmus solution turns blue because ammonia is an alkaline/basic gas. If the water contains three/few drops of phenolphthalein indicator,the indicator turns pink because ammonia is an alkaline/basic gas. Sulphur(IV)oxide and hydrogen chloride gas are also capable of the fountain experiment . If the water contains three/few drops of phenolphthalein indicator, the indicator turns colourless because both Sulphur(IV) oxide and hydrogen chloride gas are acidic gases.
6.State and explain the observation made when hot platinum /nichrome wire is placed over concentrated ammonia solution with Oxygen gas bubbled into the mixture.
Observations
Hot platinum /nichrome wire continues to glow red hot.
Brown fumes of a gas are produced.
Explanation
Ammonia reacts with Oxygen on the surface of the wire .This reaction is exothermic producing a lot of heat/energy that enables platinum wire to glow red hot. Ammonia is oxidized to Nitrogen(II)oxide gas and water. Hot platinum /nichrome wire acts as catalyst to speed up the reaction. Nitrogen(II)oxide gas is further oxidized to brown Nitrogen(IV)oxide gas on exposure to air.
Chemical equation
(i)4NH3(g) + 5O2(g) -Pt-> 4NO(g) + 6H2O(l)
(ii)2NO(g) + O2(g) -> 2NO2(g)
Observations
– Ammonia gas burns with a green flame
-Colourless gas produced
Explanation
Ammonia gas burns with a green flame in air enriched with Oxygen to from Nitrogen gas and water.
Chemical equation
2NH3(g) + O2(g) -> N2(g) + 3H2O(l)
(a)State the observations made in tube K
-Colour changes from black to brown
-Colourless liquid droplet form on the cooler parts of tube K
(b)(i)Identify liquid L.
-Water/ H2O(l)
(ii)Explain a chemical and physical test that can be used to identify liquid L.
Chemical test
(i) Add three/few drops of liquid L into anhydrous copper(II)sulphate(VI).
Colour changes from white to blue.
Explanation-Water changes white anhydrous copper(II)sulphate(VI) to blue hydrated copper(II)sulphate(VI)
(ii) Add three/few drops of liquid L into anhydrous cobalt(II)Chloride.
Colour changes from blue to pink.
Explanation-Water changes blue anhydrous cobalt(II)Chloride to pink hydrated cobalt(II)Chloride.
Physical test
(i)Heat the liquid. It boils at 100oC at sea level (1atmosphere pressure/760mmHg pressure, 101300Pa,101300Nm-2).
(ii)Cool the liquid. It freezes at 0.0oC .
(iii)Determine the density. It is 1.0gcm-3
(c)Write the equation for the reaction that take place.
2NH3(g) + 3CuO(s) -> N2(g) + 3H2O(l) + 3Cu(s)
(black) (brown)
2NH3(g) + 3PbO(s) -> N2(g) + 3H2O(l) + 3Pb(s)
(brown when hot) (grey)
8.(a)What is aqueous ammonia
Aqueous ammonia is formed when ammonia gas is dissolved in water.
NH3(g) + (aq) -> NH3(aq)
A little NH3(aq) reacts with ammonia water to form ammonia solution(NH4OH)
NH3 (aq) + H2O(l) OH–(aq) + NH4+(aq)
This makes a solution of aqueous ammonia is a weak base /alkali unlike other two alkalis.
9.Using dot and cross to represent outer electrons show the bonding in:
(a) NH3
(b) NH4+
(c)NH4Cl
10.Name four uses of ammonia
(i)In the manufacture of nitrogenous fertilizers.
(ii) In the manufacture of nitric(V)acid from Ostwalds process.
(iii)As a refrigerant in ships and warehouses.
(iv)In softening hard water.
(v)In the solvay process for the manufacture of sodium carbonate.
(vi)In the removal of grease and stains.
11.(a)Calculate the percentage of Nitrogen in the following fertilizers:
(i) (NH4)2SO4
Molar mass of (NH4)2SO4 = 132g
Mass of N in (NH4)2SO4= 28g
% of N => 28 x 100 = 21.2121%
132
(ii) (NH4)3PO4
Molar mass of (NH4)3PO4 = 149g
Mass of N in (NH4)3PO4= 42g
% of N => 42 x 100 = 28.1879%
149
(b)State two advantages of fertilizer a (i) over a (ii) above.
(i)Has higher % of Nitrogen
(ii)Has phosphorus which is necessary for plant growth.
(c) Calculate the mass of Nitrogen in a 50kg bag of:
(i) (NH4)2SO4
% of N in (NH4)2SO4 = 21.2121%
Mass of N in 50 kg (NH4)2SO4= 21.2121 x 50 = 10.6 kg
100
(ii) NH4NO3
Molar mass of NH4NO3 = 80g
Mass of N in (NH4)3PO4= 28g
% of N => 28 x 100 = 35%
80
% of N in NH4NO3 = 35%
Mass of N in 50 kg (NH4)2SO4= 35 x 50 = 17.5 kg
100
NH4NO3 therefore has a higher mass of Nitrogen than (NH4)2SO4
d).Manufacture of Ammonia /Haber process
Most of the Ammonia produced for industrial purposes uses the Haber process developed by the German Scientist Fitz Haber.
(i)Raw materials
The raw materials include:
(i)Nitrogen from Fractional distillation of air from the atmosphere.
(ii)Hydrogen from:
C(s) + H2O(l) -> CO(g) + H2 (g)
II .Passing natural gas /methane through steam.
CH4(g)+ H2O(l) -> CO(g) + 3H2 (g)
(ii)Chemical process
Hydrogen and Nitrogen are passed through a purifier to remove unwanted gases like Carbon(IV)oxide,Oxygen,sulphur(IV)oxide, dust, smoke which would poison the catalyst.
Hydrogen and Nitrogen are then mixed in the ratio of 3:1 respectively. The mixture is compressed to 200-250atmoshere pressure to liquidify. The liquid mixture is then heated to 400- 450oC.The hot compressed gases are then passed over finely divided Iron catalyst promoted/impregnated with Al2O3 /K2O .Promoters increase the efficiency of the catalyst.
Optimum conditions in Haber processs
Chemical equation
N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ
Equilibrium/Reaction rate considerations
(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.
(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 200atmospheres is normally used.
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 450oC is normally used.
(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.
a)Introduction.
Nitric(V)acid is one of the mineral acids .There are three mineral acids; Nitric(V)acid, sulphuric(VI)acid and hydrochloric acid.Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.
Nitric(V)acid is prepared in a school laboratory from the reaction of Concentrated sulphuric(VI)acid and potassium nitrate(V) below.
(c)Properties of Concentrated Nitric (V)acid(Questions)
1.Write an equation for the school laboratory preparation of nitric(V)acid.
KNO3(s) + H2SO4(l) -> KHSO4(s) + HNO3(l)
2.Sodium nitrate(V)can also be used to prepare nitric(V)acid. State two reasons why potassium nitrate(V) is preferred over Sodium nitrate(V).
(i) Potassium nitrate(V) is more volatile than sodium nitrate(V) and therefore readily displaced from the less volatile concentrated sulphuric(VI)acid
(ii) Sodium nitrate(V) is hygroscopic and thus absorb water . Concentrated sulphuric(VI)acid dissolves in water. The dissolution is a highly exothermic process.
Hot concentrated nitric(V) acid vapour is highly corrosive and attacks rubber cork apparatus if used.
Hot concentrated nitric(V) acid decomposes to brown nitrogen(IV)oxide and Oxygen gases.
4HNO3(l/g) -> 4NO2(g) + H2O (l) +O2(g)
Once formed the brown nitrogen(IV)oxide dissolves in the acid forming a yellow solution .
Observation
Brown fumes are produced.
Colourless gas that relights/rekindles glowing splint
Explanation
Hot concentrated nitric(V) acid decomposes to water, brown nitrogen(IV)oxide and Oxygen gases. Oxygen gas is not visible in the brown fumes of nitrogen (IV) oxide.
4HNO3(g) -> 4NO2(g) + H2O (l) +O2(g)
(a) About 2cm3 of Iron(II)sulphate(VI) solution is added about 5 drops of concentrated nitric(V) acid and the mixture then heated/warmed in a test tube.
Observation
(i)Colour changes from green to brown.
(ii)brown fumes /gas produced on the upper parts of the test tube.
Explanation
Concentrated nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes green Fe2+ ions in FeSO4 to brown/yellow Fe3+ .The acid is reduced to colourless Nitrogen(II)oxide.
Chemical equation:
6FeSO4(aq) + 3H2SO4 (aq) + 2HNO3(aq) -> 3Fe2(SO4) 3 (aq)+ 4H2O + 2NO(g)
Colourless Nitrogen(II)oxide is rapidly further oxidized to brown Nitrogen(IV)oxide by atmospheric oxygen.
Chemical equation:
2NO(g) + O(g) -> 2NO2 (g)
(colourless) (brown)
(b) A spatula full of sulphur powder in a claen dry beaker was added to 10cm3 concentrated nitric (V) acid and then heated gently/warmed.
Observation
(i)Yellow colour of sulphur fades .
(ii)brown fumes /gas produced.
Explanation
Concentrated nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes yellow sulphur to colourless concentrated sulphuric(VI)acid. The acid is reduced to brown Nitrogen(IV)oxide gas.
Chemical equation:
S(s) + 6HNO3(l) -> 4NO2(g) + H2O (l) +H2SO4(l)
(c) A few/about 1.0g pieces of copper turnings/Zinc granules/ Magnesium ribbon are added 10cm3 of concentrated nitric(V) acid in a beaker.
Observation
(i) brown fumes /gas produced.
(ii) blue solution formed with copper turnings
(iii) colourless solution formed with Zinc granules/Magnesium ribbon
Explanation
Concentrated nitric (V) acid is a powerful/strong oxidizing agent. It oxidizes metals to their metal nitrate (VI) salts. The acid is reduced to brown Nitrogen (IV) oxide gas.
Chemical equation:
Cu(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Cu(NO3) 2 (aq)
Zn(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Zn(NO3) 2 (aq)
Mg(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Mg(NO3) 2 (aq)
Pb(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Pb(NO3) 2 (aq)
Ag(s) + 2HNO3(l) -> NO2(g) + H2O (l) + AgNO3 (aq)
(d)Properties of Dilute Nitric (V)acid(Questions)
(i)What is dilute nitric(v)acid
When concentrated nitric(v)acid is added to over half portion of water ,it is relatively said to be dilute. A dilute solution is one which has more solvent/water than solute/acid. The number of moles of the acid are present in a large amount/volume of the solvent.This makes the molarity /number of moles present in one cubic decimeter of the solution to be low e.g. 0.02M.
If more water is added to the acid until the acid is too dilute to be diluted further then an infinite dilute solution if formed.
(ii))1cm length of polished Magnesium ribbon was put is a test tube containing 0.2M dilute nitric(v)acid. State and explain the observation made.
Observation
-Effervescence/bubbling/fizzing
-Colourless gas produced that extinguish burning splint with an explosion/pop sound
-Colourless solution formed
-Magnesium ribbon dissolves/decrease in size
Explanation
Dilute dilute nitric(v)acid reacts with Magnesium to form hydrogen gas.
Mg(s) + 2HNO3(aq) -> H2 (g) + Mg(NO3) 2 (aq)
With other reactive heavy metals, the hydrogen gas produced is rapidly oxidized to water.
Chemical equation 3Pb(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Pb(NO3)2(aq)
Chemical equation 3Zn(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Zn(NO3)2(aq)
Chemical equation 3Fe(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Fe(NO3)2(aq)
Hydrogen gas therefore is usually not prepared in a school laboratory using dilute nitric (v)acid.
(iii)A half spatula full of sodium hydrogen carbonate and Copper(II) carbonate were separately to separate test tubes containing 10cm3 of 0.2M dilute nitric (V) acid.
Observation
-Effervescence/bubbling/fizzing
-Colourless gas produced that forms a white precipitate with lime water.
-Colourless solution formed with sodium hydrogen carbonate.
– Blue solution formed with Copper(II) carbonate.
Explanation
Dilute dilute nitric (v)acid reacts with Carbonates and hydrogen carbonates to form Carbon(IV)oxide, water and nitrate(V)salt
CuCO3 (s) + 2HNO3(aq) -> H2O (l) + Cu(NO3) 2 (aq) + CO2 (g)
ZnCO3 (s) + 2HNO3(aq) -> H2O (l) + Zn(NO3) 2 (aq) + CO2 (g)
CaCO3 (s) + 2HNO3(aq) -> H2O (l) + Ca(NO3) 2 (aq) + CO2 (g)
PbCO3 (s) + 2HNO3(aq) -> H2O (l) + Pb(NO3) 2 (aq) + CO2 (g)
FeCO3 (s) + 2HNO3(aq) -> H2O (l) + Fe(NO3) 2 (aq) + CO2 (g)
NaHCO3 (s) + HNO3(aq) -> H2O (l) + NaNO3 (aq) + CO2 (g)
KHCO3 (s) + HNO3(aq) -> H2O (l) + KNO3 (aq) + CO2 (g)
NH4HCO3 (aq) + HNO3(aq) -> H2O (l) + NH4NO3 (aq) + CO2 (g)
Ca(HCO3) 2 (aq) + 2HNO3(aq) -> 2H2O (l) + Ca(NO3) 2 (aq) + 2CO2 (g)
Mg(HCO3) 2 (aq) + 2HNO3(aq) -> 2H2O (l) + Mg(NO3) 2 (aq) + 2CO2 (g)
(iii) 25.0cm3 of 0.1M Nitric(V) acid was titrated with excess 0.2M sodium hydroxide solution using phenolphthalein indicator.
Colourless
pH 1/2/3
A little of the acid when added to the base changes the colour of the indicator to show the end point. The end point therefore is acidic with low pH of Nitric(V) acid. Nitric(V) acid is a strong acid with pH 1/2/3.
III. Calculate the number of moles of acid used.
Number of moles = molarity x volume => 0.1 x 25 = 2.5 x 10-3moles
1000 1000
Volume of sodium hydroxide in cm3
= 1000 x Number of moles => 1000x 2.5 x 10-3 = 12.5cm3
Molarity 0.2
(e)Industrial large scale manufacture of Nitric (V)acid
(i)Raw materials
Oxygen is got from fractional distillation of air
Ammonia from Haber process.
Air from the atmosphere is passes through electrostatic precipitators/filters to remove unwanted gases like Nitrogen,Carbon(IV)oxide,dust,smoke which may poison the catalyst.The ammonia -air mixture is compressed to 9 atmospheres to reduce the distance between reacting gases.
The mixture is passed through the heat exchangers where a temperature of 850oC-900oC is maintained.
The first reaction take place in the catalytic chamber where Ammonia reacts with the air to form Nitrogen(II)Oxide and water.
Optimum condition in Ostwalds process
Chemical equation
4NH3(g) + 5O2(g) === Pt/Rh === 4NO(g) + 6H2O(g) ΔH = –950kJ
The reaction is reversible and exist in dynamic equilibrium where the products reform back the reactants. The following factors are used to increase the yield/amount of Notrogen(II)oxide:
(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II) oxide.
More/higher yield of Nitrogen(II) oxide is attained as reactants try to return the equilibrium balance.
(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules .
Less/lower yield of Nitrogen(II)oxide is attained.
Very low pressures increases the distance between reacting NH3 and O2 molecules.
An optimum pressure of about 9 atmospheres is normally used.
Cooling the mixture condenses the water vapour to liquid water
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ).
Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained.
Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 900oC is normally used.
(iv)Platinum can be used as catalyst.
Platinum is very expensive. It is:
-promoted with Rhodium to increase the surface area/area of contact.
-added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used.
The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.
Nitrogen(II)oxide formed is passed through an oxidation reaction chamber where more air oxidizes the Nitrogen(II)Oxide to Nitrogen(IV)Oxide gas.
Chemical equation
2NO(g) + O2(g) -> 2NO2(g)
Nitrogen(IV)Oxide gas is passed up to meet a downward flow of water in the absorption chamber. The gas react with water to form a mixture of Nitric(V) and Nitric(III)acids
Chemical equation.
2NO2(g) + H2O (l) -> HNO2(aq) + HNO3(aq)
Excess air is bubbled through the mixture to oxidize Nitric(III)/ HNO2(aq) to Nitric(V)/HNO3(aq)
Chemical equation.
O2(g) + 2HNO2(aq) -> 2HNO3(aq)
Overall chemical equation in the absorption chamber.
O2(g) + 4NO2(g) + 2H2O (l) -> 4HNO3(aq)
The acid is 65% concentrated.It is made 100% concentrated by either:
(i)fractional distillation or
(ii)added to concentrated sulphuric(VI) acid to remove the 35% of water.
A factory uses 63.0 kg of 68% pure nitric(V)acid per day to produce an ammonium fertilizer for an agricultural county. If the density of the acid is 1.42 gcm-3 , calculate :
(i)the concentration of the acid used in moles per litre.
Molar mass HNO3 = 63
Method 1
Moles of HNO3 in 1cm3 = Mass in 1cm3 1.42 => 1.42 = 0.0225 moles
Molar mass HNO3 63
Molarity = Moles x 1000=>0.0225 moles x10000 = 22.5molesdm-3/M
1 cm3
100% = 22.5molesdm-3/M
68% = 68 x 22.5 = 15.3M/ molesdm-3
100
Method 2
Moles of HNO3 in 1000cm3 = Mass in 1000cm3 =>1.42 x1000
Molar mass HNO3 63
=22.5397 molesdm-3/M
100% = 22.5397 molesdm-3/M
68% = 68 x 22.5397 = 15.327 molesdm-3
100
(ii)the volume of ammonia gas at r.t.p used. (H=1.0,N=14.0,O=16.0,one mole of gas = 24 dm-3 at r.t.p)
Chemical equation
HNO3 (aq) + NH3 (g) -> NH4NO3(aq)
Mole ratio HNO3 (aq) : NH3 (g) = 1 : 1
1 mole HNO3 (aq) -> 24dm3 NH3 (g)
15.327 mole HNO3 (aq) ->15.327 mole x 24 dm3 = 367.848dm3
1dm3
(iii)the number of crops which can be applied the fertilizer if each crop require 4.0g.
HNO3 (aq) + NH3 (g) -> NH4NO3(aq)
Molar mass NH4NO3 =80 g
Mole ratio HNO3 : NH4NO3 = 1 : 1
Mass of HNO3 in 63.0 kg = 68% x 63 =42.84kg
1 mole HNO3 (aq)=63g -> 80g NH4NO3
(42.84×1000)g HNO3 (aq) -> (42.84×1000)g x 80
63
= 54400g
Mass of fertilizer = 54400g = 13600 crops
Mass per crop 4.0
Nitrate(V) /NO3– and Nitrate(III) /NO2– are salts derived from Nitric(V)/HNO3 and Nitric(III)/HNO2 acids. Both HNO3 and HNO2 are monobasic acids with only one ionizable hydrogen in a molecule.
Only KNO2 ,NaNO2 and NH4NO2 exist. All metallic nitrate(V)salts exist.
All Nitrate(V) /NO3– and Nitrate(III) /NO2– are soluble/dissolve in water.
(a)Effect of heat on Nitrate(V) /NO3– and Nitrate(III) /NO2– salts(Test for presence of Nitrate(V) /NO3– ions in solid state)
Ammonium nitrate(III) NH4NO2 is a colourless solid that decompose to form Nitrogen gas and water.
Chemical equation
NH4NO2 (s) -> H2O(l) + N2(g)
This reaction is used to prepare small amounts of Nitrogen in a school laboratory.
Experiment
Put ½ spatula full of sodium nitrate(V) into a a test tube. Place moist blue/red litmus papers on the mouth of the test tube. Heat strongly when test tube is slanted.
Test the gases produced using glowing splint
Caution (i) Wear safety gas mask and hand gloves
(ii)Lead(II)nitrate(V)decomposes to Lead(II)oxide that react and fuses with the test tube permanently.
Repeat with potassium nitrate(V), copper(II) nitrate(V), Lead(II)nitrate(V), silver nitrate(V), Zinc nitrate(V), Magnesium nitrate(V) and Ammonium nitrate(V).
Observations
Cracking sound
Brown fumes/gas produced except in potassium nitrate(V) and Sodium nitrate(V)
Glowing splint relights/rekindles but feebly in Ammonium nitrate(V).
Black solid residue with copper(II) nitrate(V)
White residue/solid with sodium nitrate(V), potassium nitrate(V),silver nitrate(V), Magnesium nitrate(V)
Yellow residue/solid when hot but white on cooling with Zinc nitrate(V)
Brown residue/solid when hot but yellow on cooling with Lead(II)nitrate(V)
Explanation
Chemical equation.
2KNO3(s) -> 2KNO2(s) + O2 (g)
2NaNO3(s) -> 2NaNO2(s) + O2 (g)
2.Heavy metal nitrate(V)salts decomposes to form the oxide, brown nitrogen (IV) oxide and Oxygen gas.
Copper(II)oxide is black.Zinc oxide is yellow when hot and white when cool/cold. Lead(II)oxide is yellow when cold/cool and brown when hot/heated.
Hydrated copper(II)nitrate is blue. On heating it melts and dissolves in its water of crystallization to form a green solution. When all the water of crystallization has evaporated,the nitrate(V)salt decomposes to black Copper(II)oxide and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.
Chemical equation
2Cu(NO3)2(s) -> 2CuO (s) + 4NO2(g) + O2(s)
2Ca(NO3)2(s) -> 2CaO (s) + 4NO2(g) + O2(s)
2Zn(NO3)2(s) -> 2ZnO (s) + 4NO2(g) + O2(s)
2Mg(NO3)2(s) -> 2MgO (s) + 4NO2(g) + O2(s)
2Pb(NO3)2(s) -> 2PbO (s) + 4NO2(g) + O2(s)
2Fe(NO3)2(s) -> 2FeO (s) + 4NO2(g) + O2(s)
Silver nitrate(V)and Mercury(II)nitrate decomposes to the corresponding metal and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.
Chemical equation
2AgNO3 (s) -> 2Ag (s) + 2NO2(g) + O2(s)
Hg(NO3)2(s) -> Hg(l) + 2NO2(g) + O2(s)
The production/evolution of brown fumes of Nitrogen(IV)oxide gas on heating a salt is a confirmatory test for presence of NO3– ions of heavy metals
(b)Brown ring test (Test for presence of Nitrate(V) /NO3– ions in aqueous/ solution state)
Experiment
Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 8 drops of freshly prepared Iron(II)sulphate(VI)solution. Swirl/ shake.
Using a test tube holder to firmly slant and hold the test tube, carefully add 5cm3 of Concentrated sulphuric (VI) acid down along the side of test tube.Do not shake the test tube contents.
Caution: Concentrated sulphuric (VI) acid is highly corrosive.
Observation.
Both Potassium nitrate(V)and freshly prepared Iron(II)sulphate (VI)do not form layers
On adding Concentrated Sulphuric(VI)acid,two layers are formed.
A brown ring is formed between the layers.
Explanation
All nitrate(V)salts are soluble. They form a miscible mixture when added freshly prepared Iron(II)sulphate(VI)solution. Concentrated sulphuric(VI)acid is denser than the miscible mixture thus settle at the bottom.
At the junction of the layers, the acid reacts with nitrate(V)salts to form Nitric(V)acid/HNO3. Nitric(V)acid/HNO3 is reduced to Nitrogen (II)oxide by the Iron(II)sulphate(VI) salt to form the complex compound Nitroso-iron(II)sulphate(VI)/FeSO4.NO . Nitroso-iron(II)sulphate(VI) is brown in colour.It forms a thin layer at the junction between concentrated sulphuric (VI)acid and the miscible mixture of freshly prepared Iron(II) sulphate(VI) and the nitrate(V)salts as a brown ring.
Chemical equation
FeSO4(aq) + NO(g) -> FeSO4.NO(aq)
(Nitroso-iron(II)sulphate(VI)complex)
The brown ring disappear if shaken because concentrated sulphuric (VI)acid mixes with the aqueous solution generating a lot of heat which decomposes Nitroso-iron(II)sulphate(VI)/FeSO4.NO to iron(II)sulphate(VI) and Nitrogen(II)oxide.
Chemical equation
FeSO4.NO(aq) ->FeSO4(aq) + NO(g) ->
Iron(II)sulphate(VI) solution is easily/readily oxidized to iron(III)sulphate(VI) on exposure to air/oxygen. The brown ring test thus require freshly prepared Iron(II) sulphate(VI) solution
(c)Devardas alloy test (Test for presence of Nitrate(V) /NO3– ions in aqueous/ solution state)
Experiment
Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 5 drops of sodium hydroxide solution. Swirl/ shake. Add a piece of aluminium foil to the mixture.Heat.Test any gases produced using both blue and red litmus papers.
Observation. Inference
Effervescence/bubbles/fizzing
colourless gas that has a pungent smell of urine NO3–
Blue limus paper remain blue
Red litmus paper turn red.
Explanation
The Devardas alloy test for NO3– ions in solution was developed by the Italian scientist Artulo Devarda(1859-1944)
When a NO3–salt is added sodium hydroxide and aluminium foil, effervescence of ammonia gas is a confirmatory test for NO3– ions.
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UPGRADE CHEMISTRY
FORM 3
Chemistry of SULPHUR
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
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A.SULPHUR (S)
Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S2-
Sulphur mainly occurs :
(i) as free element in Texas and Louisiana in USA and Sicily in Italy.
(ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near Naivasha in Kenya
(iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of the world.
Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in.
Digging of tunnels is thus discouraged in trying to extract the mineral.
Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters.
Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe.
The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas.
The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC.
A compressed air at 15 atmospheres is forced /pumped through the innermost pipe.
The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank.
It is about 99% pure.
Diagram showing extraction of Sulphur from Fraschs Process
(i)Rhombic sulphur
(ii)Monoclinic sulphur
| Rhombic sulphur | Monoclinic sulphur |
| Bright yellow crystalline solid
Has a melting point of 113oC Has a density of 2.06gcm-3 Stable below 96oC Has octahedral structure |
Pale yellow crystalline solid
Has a melting point of 119oC Has a density of 1.96gcm-3 Stable above 96oC Has a needle-like structure |
Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96oC.This is the temperature at which one allotrope changes to the other.
(i)Plastic sulphur-
Plastic sulphur is prepared from heating powdered sulphur to boil then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur.
(ii)Colloidal sulphur-
Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is added hydrochloric acid to form a yellow precipitate.
A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8.
On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily.
On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily.
This is because the S8 rings break into S8 chain that join together to form very long chains made of over 100000 atoms of Sulphur.
The long chains entangle each other reducing their mobility /flow and hence increases their viscosity.
On continued further heating to above 160oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous.
This is because the long chains break to smaller/shorter chains.
At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring of “flowers of sulphur” on the cooler parts.
Summary of changes on heating sulphur
| Observation on heating | Explanation/structure of Sulphur |
| Solid sulphur
Heat to 113oC Amber yellow liquid
Heat to 160oC Liquid darkens
Heat to 444oC Liquid boils to brown vapour
Cool to room temperature Yellow sublimate (Flowers of Sulphur) |
Puckered S8 ring
Puckered S8 ring in liquid form (low viscosity/flow easily)
Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow)
Mixture of S8 ,S6 ,S2 vapour
Puckered S8 ring |
Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC.
Observations
-Sulphur melts then burns with a blue flame
Colourless gas produced that has a pungent smell
Red litmus paper remains red. Blue litmus paper turns red.
Explanation
Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e
(i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid
(ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid
Chemical equation
S(s) + O2(g) -> SO2(g) (Sulphur(IV)Oxide gas)
2S(s) + 3O2(g) -> 2SO3(g) (Sulphur(VI)Oxide gas traces)
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid) SO3(g) + H2O(l) -> H2 SO4 (aq) ( sulphuric(VI)acid).
Observations
Before heating, the magnet attracts iron filings leaving sulphur
After heating, the magnet does not attract the mixture.
After heating, a red glow is observed that continues even when heating is stopped..
Black solid is formed.
Explanation Iron is attracted to a magnet because it is ferromagnetic.
When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating.
Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic.
Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ –∆H)
Heated powdered heavy metals combine with sulphur to form black sulphides.
Cu(s) + S(s) -> CuS(s)
Zn(s) + S(s) -> ZnS(s)
Pb(s) + S(s) -> PbS(s)
4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid.
(i)State and explain the observation made.
Observation
Yellow colour of sulphur fades
Orange colour of potassium dichromate(VI)paper turns to green.
Explanation
Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas. The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed.
Chemical equation
S(s) + 3H2 SO4 (l) -> 3SO2(g) + 3H2O(l) +SO3(g)
Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green.
(ii)State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated Nitric (V) acid in the above set up.
Observation
Yellow colour of sulphur fades
Colurless solution formed
Brown fumes/gas produced.
Explanation
Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The Nitric (V) acid is reduced to brown nitrogen(IV)oxide gas.
Chemical equation
S(s) + 6HNO3 (l) -> 6NO2(g) + 2H2O(l) +H2SO4 (l)
NB:
Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids.
5.State three main uses of sulphur .
Sulphur is mainly used in:
(i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid.
(ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.
(iii)Making gun powder and match stick heads
(iv) As ointments to treat fungal infections
The diagram below represents the extraction of sulphur by Fraschs process. Use it to answer the questions that follow.
| N |
| M |
| L |
(a)Name the substances that passes through:
M Superheated water at 170oC and 10 atmosphere pressure
L Hot compressed air
N Molten sulphur
(b)What is the purpose of the substance that passes through L and M?
M- Superheated water at 170oC and 10 atmosphere pressure is used to melt the sulphur
L- Hot compressed air is used to force up the molten sulphur.
(c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow.
| A | B | |
| Appearance | Bright yellow | Pale yellow |
| Density(gcm-3) | 1.93 | 2.08 |
| Melting point(oC) | 119 | 113 |
| Stability | Above 96oC | Below 96oC |
I.What are allotropes?
Different forms of the same element existing at the same temperature and pressure without change of state.
B . Rhombic sulphur
III. State two main uses of sulphur.
-Manufacture of sulphuric(VI)acid
-as fungicide
-in vulcanization of rubber to make it harder/tougher/ stronger
-manufacture of dyes /fibres
(d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies 24 dm3 at room temperature)
Chemical equation
S(s) + O2(g) -> SO2(g)
Mole ratio S: SO2 = 1:1
Method 1
32.0 g of sulphur -> 24 dm3 of SO2(g)
0.4 g of sulphur -> 0.4 g x 24 dm3 = 0.3 dm3
32.0 g
Method 2
Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles
Molar mass of sulphur 32
Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles
Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3
B.COMPOUNDS OF SULPHUR
The following are the main compounds of sulphur:
(i) Sulphur(IV)oxide
(ii) Sulphur(VI)oxide .
(iii) Sulphuric(VI)acid
(iv) Hydrogen Sulphide
(v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts
(i) Sulphur(IV)oxide(SO2)
(a) Occurrence
Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.
(b) School laboratory preparation
In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of
Method 1:Using Copper and Sulphuric(VI)acid.
Method 2:Using Sodium Sulphate(IV) and hydrochloric acid.
(c)Properties of Sulphur(IV)oxide(Questions)
(i)Method 1
Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l)
Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l)
Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas.
(ii)Method 2
Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g) + 2H2O(l)
K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g) + 2H2O(l)
BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g) + H2O(l)
CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g) + H2O(l)
PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g) + H2O(l)
Lead(II)chloride is soluble on heating thus reactants should be heated to prevent it coating/covering unreacted PbSO3(s)
2.State the physical properties unique to sulphur (IV)oxide gas.
Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and choking smell which liquidifies easily. It is about two times denser than air.
Sulphur(IV) oxide is very soluble in water.
One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a vacuum.
If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum.
An acidic solution of sulphuric (IV)acid is formed which turns litmus solution red.
Chemical equation
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)
4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made.
Observations
(i)Dry Blue litmus paper remains blue.
Dry red litmus paper remains red.
(ii) Wet/damp/moist blue litmus paper turns red.
Moist/damp/wet red litmus paper remains red.
Both litmus papers are then bleached /decolorized.
Explanation
Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red litmus papers.
Wet/damp/moist litmus papers contain water that dissolves /react with dry sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3 (aq)).
Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions:
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic.
The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas.
It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples:
(a)Bleaching agent
Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas.
This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/ material to form sulphuric(VI)acid.
Chemical equations
(i)Formation of sulphuric(IV)acid
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Decolorization/bleaching of the dye/removal of atomic oxygen.
Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye
(coloured) (colourless)
Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (aq) + (dye – O)
(coloured) (colourless)
Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a dye unlike chlorine that bleaches by oxidation /adding oxygen.
The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary.
This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g.
Old newspapers turn brown on exposure to air on regaining the atomic oxygen.
The bleaching through adding oxygen by chlorine gas is permanent.
(b)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas.
Observation:
Orange acidified potassium dichromate(VI) turns to green.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
This is a confirmatory test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue.
(c)Decolorizes acidified potassium manganate(VII)
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4– ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of MnO4– (aq)
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
This is another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving a yellow residue.
(d)Decolorizes bromine water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or;
(ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of MnO4– (aq)
SO32-(aq) + Br2 (aq) + H2O(l) -> SO42-(aq) + 2HBr(aq)
(yellow) (colourless)
This can also be used as another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue.
(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green
Explanation:
Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq)
SO32-(aq) + 2Fe3+ (aq) +3H2O(l) -> SO42-(aq) + 2Fe2+(aq) + 2H+(aq)
(yellow) (green)
(f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Sulphur(IV) oxide gas. Swirl.
Observation:
Brown fumes of a gas evolved/produced.
Explanation:
Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g)
(brown fumes/gas)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V) acid.
Observation:
A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid.
Explanation:
Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either SO32- ,SO42- ,CO32- ions. i.e.
Chemical/ionic equation:
SO32-(aq) + Ba2+ (aq) -> BaSO3(s)
white precipitate
SO42-(aq) + Ba2+ (aq) -> BaSO4(s)
white precipitate
CO32-(aq) + Ba2+ (aq) -> BaCO3(s)
white precipitate
If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate:
(i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present.
(ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because:
Chemical equation
BaSO3(s) +2H+(aq) -> Ba2+ (aq) + SO2(g) + H2O(l)
Chemical equation
BaCO3(s) +2H+(aq) -> Ba2+ (aq) + CO2(g) + H2O(l)
5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples.
(a)Reduction by burning Magnesium
Experiment
Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas
Observation
Magnesium ribbon continues to burn with difficulty.
White ash and yellow powder/speck
Explanation
Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen.
The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash.
Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.
During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent)
Chemical equation
SO2(g) + 2Mg(s) -> 2MgO(s) + S(s)
(white ash/solid) (yellow speck/powder)
(b)Reduction by Hydrogen sulphide gas
Experiment
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas.
Or
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl
Observation
Yellow powder/speck
Explanation
Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar.
A little moisture/water act as catalyst /speeds up the reaction.
Chemical equation
SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s)
(yellow speck/powder)
6.Sulphur(IV)oxide has many industrial uses. State three.
(i)In the contact process for the manufacture of Sulphuric(VI)acid
(ii)As a bleaching agent of pulp and paper.
(iii)As a fungicide to kill microbes’
(iv)As a preservative of jam, juices to prevent fermentation
(ii) Sulphur(VI)oxide(SO3)
(a) Occurrence
Sulphur (VI)oxide is does not occur free in nature/atmosphere
(b) Preparation
In a Chemistry school laboratory Sulphur (VI)oxide may prepared from:
Method 1;Catalytic oxidation of sulphur(IV)oxide gas.
Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .
The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas.
Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water.
The solid fumes out on heating to a highly acidic poisonous gas.
Chemical equation
2SO2(g) + O2(g) –platinised asbestos–> 2SO3 (g)
Method 2; Heating Iron(II)sulphate(VI) heptahydrate
When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white.
Chemical equation
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
(green solid) (white solid)
On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas.
Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water.
Iron(III)oxide is left as a brown residue/solid.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g)
(green solid) (brown solid)
Caution
On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky.
(c) Uses of sulphur(VI)oxide
One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid.
(iii) Sulphuric(VI)acid(H2SO4)
(a) Occurrence
Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids;
Nitric(V)acid
Sulphuric(VI)acid
Hydrochloric acid.
Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.
(b)The Contact process for industrial manufacture of H2SO4 .
The main raw materials for industrial preparation of Sulphuric(VI)acid include:
(i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS)
(ii)Oxygen from fractional distillation of air
(iii)Water from rivers/lakes
The contact process involves four main chemical processes:
(i)Production of Sulphur (IV)oxide
As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources;
Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners
Chemical equation
S(s) + O2(g) –> SO2 (g)
Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like:
– Lead from Lead(II)sulphide/Galena,(PbS)
– Zinc from zinc(II)sulphide/Zinc blende, (ZnS)
– Copper from Copper iron sulphide/Copper pyrites, (CuFeS2)
On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced.
Chemical equation
(i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g)
(ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g)
(ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s)
Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely.
(ii)Purification of Sulphur(IV)oxide
Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface.
The impurities are removed by electrostatic precipitation method .
In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is :
(i) cheaper/less expensive
(ii) less easily poisoned by impurities
(iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide
Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger.
The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst.
The close “contact” between the reacting gases and catalyst give the process its name.
Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas.
Chemical equation
2SO2 (g) + O2(g) — V2O5 –> 2SO2 (g)
This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450oC to ensure that:
(i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC
(ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC.
(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid
Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic.
To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower.
The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid.
Chemical equation
H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)
Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid .
Chemical equation
H2S2O7 (l) + H2O (l) -> 2H2SO4 (l)
The acid is stored ready for market/sale.
III. Environmental effects of contact process
Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere.
In the Contact process, about 2% of these gases do not form sulphuric (VI) acid.
The following precautions prevent/minimize pollution from Contact process:
(i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers.
(ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water.
This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid.
(iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid.
This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3
Chemical equation
Ca(OH)2 (aq) + SO2(g) –> CaSO3 (aq) + H2O (g)
Sulphuric (VI) acid is used:
(i) in making dyes and paint
(ii)as acid in Lead-acid accumulator/battery
(iii) for making soapless detergents
(iv) for making sulphate agricultural fertilizers
(c) Properties of Concentrated sulphuric(VI)acid (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC.
(ii) Concentrated sulphuric(VI)acid is very soluble in water.
The solubility /dissolution of the acid very highly exothermic.
The concentrated acid should thus be diluted slowly in excess water.
Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container.
(iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions.
Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red.
(iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution.
This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases.
(v)The following are some chemical properties of concentrated sulphuric (VI) acid:
Experiment I;
Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker.
Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.
Observation;
Colour( in brown sugar )change from brown to black.
Colour (in glucose) change from white to black.
10cm3 beaker becomes very hot.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.
When added to sugar /glucose a vigorous reaction that is highly exothermic take place.
The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon.
Caution
This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi.
Chemical equation
Glucose: C6H12O6(s) –conc.H2SO4–> 6C (s) + 6H2O(l)
(white) (black)
Sugar: C12H22O11(s) –conc.H2SO4–> 12C (s) +11H2O(l)
(brown) (black)
Experiment II;
Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm .
Observation;
Colour change from blue to white.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds.
The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate .
Chemical equation
CuSO4.5H2O(s) –conc.H2SO4–> CuSO4 (s) + 5H2O(l)
(blue) (white)
Experiment III;
Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.
Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI)solution on the mouth of the boiling tube. Heat strongly.
Caution:
Absolute ethanol is highly flammable.
Observation;
Colourless gas produced.
Orange acidified potassium dichromate (VI) paper turns to green.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates ethanol to ethene gas at about 170oC.
Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green.
Chemical equation
C2H5OH(l) –conc.H2SO4/170oC –> C2H4 (g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of ethene gas
Experiment IV;
Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Heat gently
Caution:
This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.
Chemical equation
HCOOH(l) –conc.H2SO4 –> CO(g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas
Experiment V;
Put about 4cm3 of ethan-1,2-dioic/oxalic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water.Heat gently
Caution:
This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Gas produced forms a white precipitate with lime water.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases.
Chemical equation
HOOCCOOH(l) –conc.H2SO4 –> CO(g) + CO2(g) + H2O(l)
NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas.
Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution.
Experiment I
Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately.
Observation;
Effervescence/fizzing/bubbles
Blue solution formed with copper,
Green solution formed with Iron
Colourless solution formed with Zinc
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent.
It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas.
Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a green solution.
Chemical equation
Cu(s) + 2H2SO4(aq) –> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(aq) –> ZnSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(aq) –> FeSO4(aq) + SO2(g) + 2H2O(l)
Experiment II
Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube.
Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately.
Warm.
Observation;
Black solid charcoal dissolves/decrease
Yellow solid sulphur dissolves/decrease
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes non-metals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to Sulphur(IV)oxide .
Chemical equation
C(s) + 2H2SO4(aq) –> CO2(aq) + 2SO2(g) + 2H2O(l)
S(s) + 2H2SO4(aq) –> 3SO2(g) + 2H2O(l)
III. As the least volatile acid
Study the table below showing a comparison in boiling points of the three mineral acids
| Mineral acid | Relative molecula mass | Boiling point(oC) |
| Hydrochloric acid(HCl) | 36.5 | 35.0 |
| Nitric(V)acid(HNO3) | 63.0 | 83.0 |
| Sulphuric(VI)acid(H2SO4) | 98.0 | 333 |
1.Which is the least volatile acid? Explain
Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by Hydrogen bonds making it to have the highest boiling point/least volatile.
(i)Chemical equation
KNO3(s) + H2SO4(aq) –> KHSO4(l) + HNO3(g)
NaNO3(s) + H2SO4(aq) –> NaHSO4(l) + HNO3(g)
This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3).
(ii)Chemical equation
KCl(s) + H2SO4(aq) –> KHSO4(s) + HCl(g)
NaCl(s) + H2SO4(aq) –> NaHSO4(s) + HCl(g)
This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl).
(d) Properties of dilute sulphuric(VI)acid.
Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated sulphuric
(VI) acid is carefully added to about 90cm3 of distilled water.
Diluting concentrated sulphuric (VI) acid should be done carefully because the reaction is highly exothermic.
Diluting concentrated sulphuric (VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive.
On diluting concentrated sulphuric(VI) acid, water ionizes /dissociates the acid fully/wholly into two(dibasic)free H+(aq) and SO42-(aq)ions:
H2SO4 (aq) -> 2H+(aq) + SO42-(aq)
The presence of free H+(aq)ions is responsible for ;
(i)turn litmus red because of the presence of free H+(aq)ions
(ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence a strongly acidic solution.
(iii)Reaction with metals
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon.
Observation:
No effervescence/ bubbles/ fizzing with copper
Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium
Colourless gas produced that extinguishes burning splint with a “pop” sound.
Colourless solution formed with Zinc and Magnesium.
Green solution formed with Iron
Explanation:
When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen gas.
Impure hydrogen gas extinguishes burning splint with a “pop” sound.
A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher than hydrogen in the reactivity/electrochemical series.
They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc sulphate(VI).
. When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place.
Copper thus do not react with dilute sulphuric(VI)acid.
Chemical/ionic equation
Mg(s) + H2SO4(aq) –> MgSO4(aq) + H2(g)
Mg(s) + 2H+(aq) –> Mg2+ (aq) + H2(g)
Zn(s) + H2SO4(aq) –> ZnSO4(aq) + H2(g)
Zn(s) + 2H+(aq) –> Zn2+ (aq) + H2(g)
Fe(s) + H2SO4(aq) –> FeSO4(aq) + H2(g)
Fe(s) + H+(aq) –> Fe2+ (aq) + H2(g)
NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
(ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid
(iv)Reaction with metal carbonates and hydrogen carbonates
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium hydrogen carbonate.
Observation:
Effervescence/ bubbles/ fizzing.
Colourless gas produced that extinguishes burning splint.
Colourless solution formed with Zinc carbonate, sodium hydrogen carbonate and sodium carbonate.
Green solution formed with Iron(II)Carbonate
Blue solution formed with Copper(II)Carbonate
Explanation:
When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas extinguishes a burning splint and forms a white precipitate when bubbled in lime water.
A sulphate (VI) salts is formed.
Chemical/ionic equation
ZnCO3(s) + H2SO4(aq) –> ZnSO4(aq) + H2O(l) + CO2(g)
ZnCO3(s) + 2H+(aq) –> Zn2+ (aq) + H2O(l) + CO2(g)
CuCO3(s) + H2SO4(aq) –> CuSO4(aq) + H2O(l) + CO2(g)
CuCO3(s) + 2H+(aq) –> Cu2+ (aq) + H2O(l) + CO2(g)
FeCO3(s) + H2SO4(aq) –> FeSO4(aq) + H2O(l) + CO2(g)
FeCO3(s) + 2H+(aq) –> Fe2+ (aq) + H2O(l) + CO2(g)
2NaHCO3(s) + H2SO4(aq) –> Na2SO4(aq) + 2H2O(l) + 2CO2(g)
NaHCO3(s) + H+(aq) –> Na+ (aq) + H2O(l) + CO2(g)
Na2CO3(s) + H2SO4(aq) –> Na2SO4(aq) + H2O(l) + CO2(g)
NaHCO3(s) + H+(aq) –> Na+ (aq) + H2O(l) + CO2(g)
(NH4)2CO3(s) + H2SO4(aq) –> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
(NH4)2CO3 (s) + H+(aq) –> NH4+ (aq) + H2O(l) + CO2(g)
2NH4HCO3(aq) + H2SO4(aq) –> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
NH4HCO3(aq) + H+(aq) –> NH4+ (aq) + H2O(l) + CO2(g)
NB:
Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
(v)Neutralization-reaction of metal oxides and alkalis/bases
Experiment I:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir.
Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place of the Copper(II)Oxide.
Observation:
Blue solution formed with Copper(II)Oxide
Colourless solution formed with other oxides
Explanation:
When a metal oxide is put in a test tube containing dilute sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt.
Chemical/ionic equation
ZnO(s) + H2SO4(aq) –> ZnSO4(aq) + H2O(l)
ZnO(s) + 2H+(aq) –> Zn2+ (aq) + H2O(l)
CuO(s) + H2SO4(aq) –> CuSO4(aq) + H2O(l)
CuO(s) + 2H+(aq) –> Cu2+ (aq) + H2O(l)
MgO(s) + H2SO4(aq) –> MgSO4(aq) + H2O(l)
MgO(s) + 2H+(aq) –> Mg2+ (aq) + H2O(l)
Na2O(s) + H2SO4(aq) –> Na2SO4(aq) + H2O(l)
Na2O(s) + 2H+(aq) –> 2Na+ (aq) + H2O(l)
K2CO3(s) + H2SO4(aq) –> K2SO4(aq) + H2O(l)
K2O(s) + H+(aq) –> 2K+ (aq) + H2O(l)
NB:
Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides.
Experiment II:
Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution
Observation:
Colour of phenolphthalein changes from pink to colourless at the end point.
Explanation
Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only.
Colour of the indicator used changes when a slight excess of acid is added to the base at the end point
Chemical equation:
2NaOH(aq) + H2SO4(aq) –> Na2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
2KOH(aq) + H2SO4(aq) –> K2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
2NH4OH(aq) + H2SO4(aq) –> (NH4)2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
(iv) Hydrogen sulphide(H2S)
(a) Occurrence
Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.
It is present in rotten eggs and human excreta.
(b) Preparation
Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid.
(c) Properties of Hydrogen sulphide(Questions)
Chemical equation: FeS (s) + 2HCl (aq) -> H2S (g) FeCl2 (aq)
Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red.
H2S(aq) -> H+(aq) + HS–(aq)
H2S(aq) -> 2H+(aq) + S2- (aq)
Hydrogen sulphide therefore can form both normal and acid salts e.g
Sodium hydrogen sulphide and sodium sulphide both exist
Hydrogen/ H2
Iron(II)sulphide contains Iron as impurity .The iron will react with dilute hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes with hydrogen sulphide gas.
Observations
Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black.
Explanation
When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed.
All sulphides are insoluble black salts except sodium sulphide, potassium sulphide and ammonium sulphides.
Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper .
The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide.
This is the chemical test for the presence of H2S other than the physical smell of rotten eggs.
Chemical equations
Pb2+(aq) + H2S -> PbS + 2H+(aq)
(black)
Fe2+(aq) + H2S -> FeS + 2H+(aq)
(black)
Zn2+(aq) + H2S -> ZnS + 2H+(aq)
(black)
Cu2+(aq) + H2S -> CuS + 2H+(aq)
(black)
2Cu+(aq) + H2S -> Cu2S + 2H+(aq)
(black)
| Dry Hydrogen sulphide gas |
| Flame A |
(i) State the observations made in flame A
Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g)
Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water.
Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s)
(a)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing Hydrogen sulphide gas.
Observation:
Orange acidified potassium dichromate (VI) turns to green.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur.
Chemical/ionic equation:
4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l)
This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide gas.
Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a yellow residue.
(b)Decolorizes acidified potassium manganate(VII)
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Hydrogen Sulphide gas.
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO4– ions to green Mn2+ ions leaving a residue as the gas itself is oxidized to sulphur.
Chemical/ionic equation:
5H2S(g) + 2MnO4– (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l)
(purple) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving no yellow residue.
(c)Decolorizes bromine water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or;
(ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to sulphur.
Chemical/ionic equation:
H2 S(g) + Br2 (aq) -> S (s) + 2HBr(aq)
(yellow solution) (yellow solid) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving no yellow residue.
(d)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green.
Yellow solid solid
Explanation:
Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur.
Chemical/ionic equation:
H2S(aq) + 2Fe3+ (aq) -> S (s) + Fe2+(aq) + 2H+(aq)
(yellow solution) (yellow residue) (green)
(e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Brown fumes of a gas evolved/produced.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur.
Chemical/ionic equation:
H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)
(yellow residue) (brown fumes)
(f)Reduces sulphuric(VI)acid to Sulphur
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or;
(ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur.
Chemical/ionic equation:
3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s)
(yellow residue)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur
Chemical/ionic equation:
H2S(g) + H2O2 (l) -> 2H2O(l) + S (s)
(yellow residue)
8.Name the salt formed when:
(i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution:
Sodium hydrogen sulphide
Chemical/ionic equation:
H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq)
(ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution:
Sodium sulphide
Chemical/ionic equation:
H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq)
Practice
Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below
| Hydrogen sulphide |
| Blue solution |
| Brown solution |
| Black solid |
| Green solution |
(a)Name the black solid Copper(II)sulphide
(b)Identify the cation responsible for the formation of:
III. Brown solution Fe3+(aq)
(c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide
-Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution.
-Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.
-Yellow solid residue/deposit is formed with Hydrogen sulphide
Chemical/ionic equation:
4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator.
ObservationGas continues burning with a blue flame
Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g)
(v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4–) salts.
i.e.
H2SO4 (aq) -> 2H+(aq) + SO42-(aq)
H2SO4 (aq) -> H+(aq) + HSO4 –(aq)
All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4)
All Hydrogen sulphate (VI) (HSO3–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids.
Other Hydrogen sulphate (VI) (HSO4–) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2).
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4–) salts.
i.e.
H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
H2SO3 (aq) -> H+(aq) + HSO3 –(aq)
All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3)
All Hydrogen sulphate (IV) (HSO3–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids.
Other Hydrogen sulphate (IV) (HSO3–) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2).
5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts:
Experiment:
In a clean dry test tube place separately about 1.0g of :
Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers.
Observations:
-Colourless droplets of liquid forms on the cooler parts of the test tube in all cases.
-White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).
-Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI)
-Colour changes from blue to white then black in case of Copper (II) sulphate (VI)
-Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV)
-Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI).
Explanation
(i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g.
K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l)
Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l)
MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l)
CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l)
ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l)
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l)
All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).
(i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.
Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water.
Sulphur (IV) oxide does not.
(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas.
Chemical equation
2CuSO4 (s) -> CuO(s) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.
Experiments/Observations:
(a)Using Lead(II)nitrate(V)
| Observation | Inference |
| White precipitate/ppt | SO42- , SO32- , CO32- , Cl– ions |
Observation 1
| Observation | Inference |
| White precipitate/ppt persists | SO42- , Cl– ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves | SO32- , CO32- , ions |
III.(a)To the preserved sample observation 1 in (II) above, Heat to boil.
Observation 1
| Observation | Inference |
| White precipitate/ppt persists on boiling | SO42- ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves on boiling | Cl – ions |
.(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
Observation 1
| Observation | Inference |
| (i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium dichromate(VI) turns to green |
SO32- ions
|
Observation 2
| Observation | Inference |
| (i)acidified potassium manganate(VII) not decolorized
(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green
|
CO32- ions
|
Experiments/Observations:
(b)Using Barium(II)nitrate(V)/ Barium(II)chloride
| Observation | Inference |
| White precipitate/ppt | SO42- , SO32- , CO32- ions |
Observation 1
| Observation | Inference |
| White precipitate/ppt persists | SO42- , ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves | SO32- , CO32- , ions |
III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
Observation 1
| Observation | Inference |
| (i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium dichromate(VI) turns to green |
SO32- ions
|
Observation 2
| Observation | Inference |
| (i)acidified potassium manganate(VII) not decolorized
(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green
|
CO32- ions
|
Explanations
Using Lead(II)nitrate(V)
(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl–), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV).
Chemical/ionic equation:
Pb2+(aq) + Cl– (aq) -> PbCl2(s)
Pb2+(aq) + SO42+ (aq) -> PbSO4 (s)
Pb2+(aq) + SO32+ (aq) -> PbSO3 (s)
Pb2+(aq) + CO32+ (aq) -> PbCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
– Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists.
– Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g)
PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g)
(iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed;
– Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling)
– Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling.
(iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
– Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l)
These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.
Using Barium(II)nitrate(V)/ Barium(II)Chloride
(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV).
Chemical/ionic equation:
Ba2+(aq) + SO42+ (aq) -> BaSO4 (s)
Ba2+(aq) + SO32+ (aq) -> BaSO3 (s)
Ba2+(aq) + CO32+ (aq) -> BaCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
– Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists.
– Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g)
BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g)
(iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
– Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l)
These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.
Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts
| Heat to boil |
| White ppt persist
on heating in SO42- in CO32-
|
| White ppt dissolves on heating in Cl–
|
| Acidified KMnO4 decolorized in SO32- |
| White ppt with lime water in CO32- |
| Acidified KMnO4
K2Cr2O7 / Lime water |
| white ppt persist /remains in SO32- and CO32-
|
| white ppt dissolves in SO32- and CO32- |
| Dilute nitric(V) acid |
| Lead(II)nitrate(V) |
| White precipitates of Cl–, SO42- , SO32- and CO32- |
| Unknown salt |
Practice revision question
| Sodium salt solution |
| Acidified K2Cr2O7 |
| Gas G and colour of solution
changes orange to green |
| Colourless solution B |
| Barium nitrate(VI) (VI)(aq) |
| White precipitate |
| Dilute HCl |
(a)Identify the:
I: Sodium salt solution
Sodium sulphate(IV)/Na2SO3
II: White precipitate
Barium sulphate(IV)/BaSO3
III: Gas G
Sulphur (IV)Oxide /SO2
IV: Colourless solution H
Barium chloride /BaCl2
(b)Write an ionic equation for the formation of: I.White precipitate
Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s)
II.Gas G
Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq)
III. Green solution from the orange solution
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
(i)Write equation for the reaction taking place at:
I.The roasting furnace (1mk)
2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g)
II.The absorption tower (1mk)
H2SO4 (l) + SO3 (g) -> H2S2O7(l)
III.The diluter (1mk)
H2S2O7(l) + H2 O(l) -> 2H2SO4 (l)
(ii)The reaction taking place in chamber K is
SO2 (g) + 1/2O2 (g) SO3 (g)
To ensure all the SO2 reacts
II.Name another substance used in chamber K
Vanadium(V)oxide
3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI).
Add acidified Barium nitrate(V)/chloride.
White precipitate formed with sodium sulphate (VI)
No white precipitate formed with sodium sulphate (IV)
(b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3)
Chemical equation
Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g)
Mole ratio Cu(s: SO2 (g) = 1:1
Method 1
1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3
(120 x 1000) g -> (120 x 1000) g x 22.4.dm3)
63.5 g
= 42330.7087
Method 2
Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles
63.5
Moles SO2 = Moles of Cu = 1889.7639 moles
Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4)
= 42330.7114
(a)Identify the:
(i)cation responsible for the green solution T
Cr3+
(ii)possible anions present in white precipitate R
CO32-, SO32-, SO42-
(b)Name gas V
Sulphur (IV)oxide
(c)Write a possible ionic equation for the formation of white precipitate R.
Ba2+ (aq) + CO32- (aq) -> BaCO3(s)
Ba2+ (aq) + SO32- (aq) -> BaSO3(s)
Ba2+ (aq) + SO42- (aq) -> BaSO4(s)
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UPGRADE CHEMISTRY
FORM 3
Chemistry of CHLORINE
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
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CONTENTS
A.CHLORINE
Chlorine is a non-metallic element in group VII (Group 17) of the periodic table. It has electronic configuration 2:8:7. It gains one valence election to form stable Cl–ion, it belongs to the chemical family of halogens.
Occurrence
-As Brine-concentration sodium chloride solution dissolved in salty seas water, oceans and lakes e.g. Lake Magadi in Kenya is very salty.
-As rock-salt solid sodium chloride crystals in the earths crust all over the world.
Chlorine gas may be prepared in the school laboratory from the following:
a)Heating solid Manganese (iv) Oxide and Concentrated Hydrochloric acid.
c)Reacting Potassium Manganate (VII) with concentrated Hydrochloric acid
d)Reacting Potassium /sodium Dichromate (VI) Acid with Concentrated Hydrochloric acid.
Set up of school laboratory preparation of chlorine.
Pale green.
Pungent irritating smell.
-Downward delivery.
-Chlorine is 11/2 denser than air.
4.(i) What is the purpose of concentrated sulphuric (VI) acid.
-To dry the gas.
(ii) Name two other substances that can be used in place of concentrated sulphuric (VI) acid.
-Calcium chloride
-Silica gel
(iii) Name a substance that cannot be used in place of concentrated sulphuric (VI) acid explain.
-Calcium oxide reacts with chlorine.
5.(a)Write three possible reactions between concentrated hydrochloric acid and the oxidizing agents.
2.K2Cr2O7(s) +14HCl(aq) → 2KCl(aq) + 2CrCl3(aq) + 7H2O(l) + 3Cl2(g)
3.Na2Cr2O7(s) + 14HCl(aq) → 2NaCl(aq) + CrCl3(aq) + 7H2O(l) + 3Cl2(g)
4.PbO2(s) + 4HCl(aq) → PbCl2(aq) + Cl2(g) + 2H2O(l)
5.MnO2(s)+ 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O(l)
(b) Why is Hydrochloric acid used in all the above cases?
Oxidizing agents KMnO4/PbO2/MnO2/K2Cr2O/Na2Cr2O7 readily oxidize hydrochloric acid to chlorine themselves reduced to their chlorides.
Generally:
2HCl (aq) + [O] → Cl2 (g) + H2O (l)
(From oxidizing agent)
Observation
-Pale yellow colour of chlorine fades.
-yellow solution formed.
Explanation
Chlorine dissolves then reacts with water to form yellow chlorine water. Chlorine water is chemically a mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid).
A mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid) is commonly called Chlorine water
Chemical equation:
Cl2(g) + H2O(l) → HCl(aq) + HClO(aq)
Chlorine (I) acid is an unstable compound.
After two hours the chloric (I) acid in chlorine water decomposes to hydrochloric acid and releases oxygen gas. This reaction takes place in sunlight.
Chemical equation
2HOCl(aq) → 2HCl(aq) + O2 (g)
Observation
The blue litmus turns red then both the red/blue litmus papers are bleached/decolourized.
Explanation
Chlorine reacts with water in the litmus papers to form acidic hydrochloric acid and chloric (l) acid that turns blue litmus papers red.
Chemical Equation
Cl2(g) + H2O(l) → HCl(aq) + HClO(aq)
Explanation
Unstable chloric (I) acid oxidizes the dye/colured litmus paper to colourless material
Chemical Equation
HClO(aq) + dye → HCl(aq) + (dye + O)
(coloured) (colourless)
Or:
HClO(aq) + dye-O → HCl(aq) + dye
(coloured) (colourless)
NB Chlorine does not therefore bleach/decolourize dry litmus paper/dye because chloric(I) acid cannot be formed in absence of water.
Observation
blue litmus papers were bleached /decolorized.
Pale green colour of chlorine fades.
Explanation
-Sodium hydroxide reacts with chlorine to form sodium chloride and sodium hypochlorite. Sodium hypochlorite bleaches dyes by oxidation.
Chemical Equation
Cl2 + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O
NaClO(aq) + dye → NaCl(aq) + (dye + O)
(coloured) (Colourless)
NaClO(aq) + (dye-O) → NaCl(aq) + dye
(Coloured) (Colourless)
10.Blue litmus papers were put in flask containing hot concentrated sodium hydroxide. Chlorine gas was bubbled into the solution. State and explain the observations made.
Observation.
blue litmus papers were bleached.
Pale green colour of chlorine fades.
Explanation
Hot concentrated sodium hydroxide reacts with chlorine to form sodium chloride and sodium chloride (V).Sodium chlorate (V) bleaches by oxidation.
Chemical equation
2Cl2(g) + 4NaOH(aq) → 3NaCl(aq) + NaClO3(aq) + H2O(l)
NaClO3(aq) + 3(dyes) → NaCl(aq) + 3(dye + O)
NaClO3(aq) + 3(dyes-O) → NaCl(aq) + 3 dyes
NaClO3 is also a weed killer
-Manufacture of polyvinyl chloride (P.V.C) // polychloroethene pipes.
-Manufacture of hydrochloric acid used in “Pickling” of metals.
-Manufacture of bleaching agents
-Chlorination of water to kill germs.
Method I
Method II
| Suction pump |
| HEAT |
| Chlorine gas |
| Concentrated sodium/potassium hydroxide. |
| Iron/Aluminium |
| Aluminium(III)oxide/iron(III)oxide |
-Conc. H2SO4 / Concentrated sulphuric (VI) acid.
-Anhydrous Calcium (II) Chloride.
-Silica gel
Observation
Iron glows red hot
Brown crystals are formed
Explanation
Iron reacts with chlorine to form dark brown crystals of iron (III) Chloride.
This reaction is exothermic and requires no farther heating once started.
Iron (III) Chloride sublimes away ensuring the unreacted Iron completely reacts with chlorine gas.
Chemical equation
2Fe(s) + 3Cl2 (g) → 2FeCl3(g)
-Heated iron (III) Chloride crystals sublime to gas and solidify on the cooler parts.
(ii) Name another metal that can be used in place of iron to react with chlorine and collected at similar point on heating explain.
Metal Aluminum
Explanation
Aluminum reacts with chlorine to form a while sublimate of aluminum (III) chloride at the cooler parts
Chemical equation
2Al(s) + 3Cl2(g) → 2AlCl3(s/g)
To pull the gaseous products into the set up.
(i) Sodium hydroxide in method II. Explain.
To absorb poisonous/toxic excess unreacted chlorine gas.
Sodium hydroxide reacts with chlorine to form sodium chloride, Sodium hypochlorite and water.
Chemical equation:
2NaOH(aq) + Cl2(g) → NaCl(aq) + NaClO(aq) + H2O(l)
2KOH(aq) + Cl2(g) → KCl(aq) + KClO(aq) + H2O(l)
(ii) Anhydrous calcium chloride/calcium oxide in method I. Explain.
To absorb moisture/water in the set up to prevent it from hydrolyzing iron (III) chloride/aluminium oxide.
Explanation
Iron (III) chloride and Aluminium chloride fumes and reacts with small traces of water to form a solution of iron (III) hydroxide/aluminium hydroxide and hydrogen chloride gas.
Chemical equation
FeCl3(s) + 3HCl(aq) → Fe(OH)3(aq) + 3HCl(g)
AlCl3(s) + 3HCl(aq) → Al(OH)3(aq) + 3HCl(g)
(i) method II to ensure correct results.
-Tube B should be completely dry to prevent hydrolysis of iron (III) Chloride to iron (III) hydroxide.
(ii) Carrying out method I
-Should be done in a fume chamber or in the open because chlorine gas is poisonous/toxic.
(g) Name another substance that can be used place of Sodium hydroxide in method I
Potassium hydroxide
(h) Calcium oxide cannot be used in place of calcium chloride during preparation of chlorine. Explain.
Calcium oxide is a base. It reacts /absorbs water to form calcium hydroxide solution.
Calcium hydroxide reacts with chlorine to form a mixture of calcium chloride and calcium hypochlorite.
Chemical equation
2Ca (OH)2(aq) + 2Cl2(g) → CaCl2(aq) + CaOCl2(aq) + H2O(l)
-Magnesium ribbon continues burning with a bright flame.
-White solid formed.
-Pale yellow colour of chlorine fades
Explanation:
Magnesium reacts with chlorine forming a white solid of magnesium chloride.
Chemical equation
Mg(s) + Cl2(g) → MgCl2(s)
(b) Write the equation for the reaction that takes place if zinc is used.
Zn(s) + Cl2(g) → ZnCl2(s)
-Phosphorus continues to burn.
-Dense white fumes formed.
-Pale green colour of chlorine fades.
P4(s) + 6Cl2(g) → 4 PCl3(s)
P4(s) + 10Cl2(g) → 4 PCl3(s)
(c) State two reasons why the deflagrating spoon with rid/cover should be used.
-Chlorine in the gas jar is poisonous/toxic.
-Burning phosphorus produces poisonous/toxic phosphorus (III) chloride // phosphorus (V) chloride.
-Ensure the reaction is not affected by air/oxygen from the atmosphere.
(d) After the reaction is complete, 2cm3 of distilled water were added. The solution formed was tested with both blue and red litmus papers.
(i) State the observations made.
-Blue litmus paper turns red
-Red litmus paper remain red
(ii) Explain the observation made in d(i) above
-Phosphoric (V) Chloride hydrolyze in water to phosphoric (V) acid and produce hydrogen chloride gas. Both hydrogen chloride and phosphoric (V) acid are acidic.
Chemical equation
PCl5 (l) + 4H2O(l) → H3PO4 (aq) + 5HCl(g)
Observation
Yellow solid formed.
Pale colour of chlorine fades
Explanation
Chlorine oxidizes hydrogen sulphide to sulphur itself reduced to hydrogen chloride gas. A little water catalyzes the reaction.
Chemical equation
H2S(g) + Cl2(g) → S(s) + HCl(g)
(yellow solid) (White Fume)
Observation:
White fumes evolved.
Pale green colour of chlorine fades.
Explanation
Chlorine reacts with ammonia gas to form a dense white fume of ammonia chloride and Nitrogen gas is produced.
Chemical equation
8NH3(g) + 3Cl2(g) → 6NH4Cl(s) + N2(g)
Observation
Pale green colour of chlorine fades.
Pale yellow solution is formed.
Explanation
Chlorine reacts withhot concentrated sodiumsodium hydroxide / Potassium hydroxide solution to form pale yellow solution of metal chlorate (V) and chlorides of the metal
Chemical equation
Cl2(g) + 2NaOH → NaClO(aq) + NaCl(aq) + H2O(l)
(sodium hydroxide) (Sodium Chlorate (I))
Cl2(g) + 2KOH → KClO(aq) + NaCl(aq) + H2O(l)
(Potassium hydroxide) (Potassium Chlorate (I))
(b)The experiment in 17(a) was repeated with hot concentrated sodium hydroxide solution. Explain the observation made.
Observation
Pale green colour of chlorine fades.
Pale yellow solution is formed.
Explanation
Chlorine reacts with hot concentrated Sodium hydroxide/Potassium hydroxide solution to form pale yellow solution of metal chlorate (v) and chlorides of metals.
Chemical equation
3Cl2(g) + 6NaOH(aq) → NaClO3 (aq) + 5NaCl(aq) + 3H2O(l)
(Sodium hydroxide) (Sodium Chlorate (V))
3Cl2(g) + 6KOH(aq) → KClO3 (aq) + 5KCl(aq) + 3H2O(l)
(Potassium hydroxide) (Potassium Chlorate (V))
The products formed when chlorine reacts with alkalis depend thus on temperature and the concentration of alkalis.
(c) (i) Write the equation for the formation of calcium chlorite (I) and calcium chlorate (V).
2Ca (OH)2(aq) + 2Cl2(g) → CaCl2(aq) + CaOCl2(aq) + H2O(l)
(Calcium hydroxide) (Calcium Chlorate(I))
(Cold/dilute)
Ca (OH)2(aq) + Cl2(g) → CaCl2(aq) + Ca(ClO3)2(aq) + H2O(l)
(Calcium Chlorate(V))
B: THE HALOGENS
These are elements in group VII of the periodic table. They include:
| Element | Symbol | Atomic number | Electric configuration | Charge of ion | Valency | State at Room Temperature |
| Fluorine
Chlorine
Bromine
Iodine
Astatine |
F
Cl
Br
I
At |
9
17
35
53
85 |
2:7
2:8:7
2:8:18:7
2:8:18:18:7
2:8:18:32:18:7 |
F–
Cl–
Br–
I–
At– |
1
1
1
1
1 |
Pale yellow gas
Pale green gas Red liquid
Grey Solid
Radioactive |
The radius of chlorine is smaller than the ionic radius o the chloride ion.
Effective nucleus attraction on outer energy level in chloride ion is less than chlorine atom because of extra gained electron gained electron that repelled thus causes the outer energy level to expand/increase.
Atomic radius of Fluorine is smaller than that of chlorine.
Chlorine has more energy levels than fluorine occupied by more electrons.
–Bromine, Chlorine and iodine exists as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces.
-The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius.Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid.
Electronegativity is the tendency/ease of acquiring /gaining electrons by an element during chemical reaction.
It is measured using Pauling’s scale.
Fluorine with Pauling scale 4.0 is the most electronegative element in the periodic table and thus the highest tendency to acquire/gain extra electron.
(ii) The table below shows the electronegativity of the halogens.
| Halogen | F | Cl | Br | I | At |
| Electronegativity (Pauling’s scale) | 4.0 | 3.0 | 2.8 | 2.5 | 2.2 |
Explain the trend in electronegativity of the halogens.
Decrease down the group from fluorine to Astatine
Atomic radius increase down the group decreasing electron – attracting power down the group from fluorine to astatine.
(f) (i)What is electron affinity
Electron affinity is the energy required to gain an electron in an atom of an element in its gaseous state.
(ii) Study the table below showing the election affinity of halogens for the process x + e → x–
| Halogen | F | Cl | Br | I |
| Electron affinity kJmole-1 | -333 | -364 | -342 | -295 |
(iii) Explain the trend in electron affinity of the halogens.
-Decrease down the group
-Atomic radius of halogens increase down the group thus incoming/gained electron is attracted less strongly by the progressively larger atoms with a decreasing effective nuclear charge on outer energy level
(iv) Which is a move stable ion Cl– or Br – explain?
-Cl– ion.
-Has a more negative/exothermic electron affinity than Br–
(v) Differentiate between electron affinity and:
Ionization energy is the energy required to lose /donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state.
Both are measured in kilojoules per mole.
-Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process:
X(g) + e → X–(g)
Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.
It does not involve use of energy but theoretical arbitrary Pauling’s scale of measurements.
(g) (i) 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions were put separately in test tubes. 5 drops of chlorine water was added to each test tube: state and explain the observation made.
Observation
Yellow colour of chlorine water fades in all test tubes expect with sodium chloride.
-Coloured Solution formed.
Explanation
Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine.
(ii) The experiment in g (i) was repeated with 5 drops of bromine water instead of chlorine water .explain the observation made.
Observation
Yellow colour of bromine water fades in test tube containing sodium iodide.
Explanation
Bromine is more electronegative than iodide but less 6than chlorine.
On adding Bromine water, iodide displaced from its solution but not chlorine.
(iii) Using the knowledge in g(i) and (ii) above,
| Halide ion Halogen
ion in solution Halogen |
F– | Cl– | Br– | I– |
| F2 | X | √ | √ | √
|
| Cl2 | X | X | √ | √
|
| Br2 | X | X | X | √
|
| I2 | X | X | X | √ |
Write an ionic equation for the reaction where there is (V)
F2 (g)+ 2Cl– (aq) -> 2F–(aq) + Cl2(g)
F2 (g)+ 2Br– (aq) -> 2F–(aq) + Br2(aq)
F2 (g)+ 2I– (aq) -> 2F–(aq) + I2(aq)
Cl2 (g) + 2Br– (aq) -> 2Cl–(aq) + Br2(aq)
Cl2 (g)+ 2I– (aq) -> 2Cl–(aq) + I2(aq)
Br2 (aq)+ 2I– (aq)- -> 2Br–(aq) + I2(aq)
(h) State one uses of:
Manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber.
Reduce tooth decay when added in small amounts/equations in tooth paste.
Note: large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis.
Hydrogen fluoride is used to engrave word pictures in glass.
Silver bromide is used to make light sensitive photographic paper/films.
Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine.
| Element
Halogen |
H | Na | Mg | Al | Si | C | P |
| F | HF | NaF | MgH2 | AlF3 | SiF4 | CF4 | PF3
|
| Cl | HCl | NaCl | MgCl | AlCl3 | SiCl3 | CCl4 | PCl3
|
| Br | HBr | NaBr | MgBr2 | AlBo3 | SiBr4 | CBr4 | PBr3
|
| I | Hl | Nal | Mgl2 | All3 | SiL4 | Cl2 | Pb3
|
(j) (i) Using dot (.) and Cross (x) to represent electrons, show the bonding in chlorine molecule.
Covalent.
Bond Bond energy k J mole-1
Cl-Cl 242
Br-Br 193
I-I 151
Bond energy is the energy required to break/ form one mole of chemical bond
–Decrease down the group from chlorine to Iodine
-Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear charge an outer energy level that take part in bonding.
(k) Some compounds of chlorine are in the table below the oxidation state of chlorine in each compound.
Compound Oxidation state Name of compound
NaClO3 +5 Sodium chlorate (V)
ClO2 +4 Chloric (IV) oxide
KClO2 +3 Potassium chlorate (III)
NaClO +1 Sodium Chlorite (I)
Cl2 0 Chlorine Molecule
NaCl -1 Sodium Chloride (I)
MgCl2 -1 Magnesium Chloride (I)
Hydrogen Chloride does not occur free in the atmosphere or in nature
Hydrogen chloride may be prepared in the school laboratory by reacting solid sodium/potassium chloride crystals with concentrated sulphuric (Vi) acid as in the set up below.
-Wear protective clothing/gloves to avoid accidental contact with skin.
-Concentrated sulphuric (VI) acid is highly corrosive-it causes painful wounds when in contact with skin.
-Downward delivery// upward displacement of water
-Hydrogen chloride is denser than air.
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
NaCl is commonly used because it is cheaper than KCl
-is the least volatile mineral acid, thus displace the more volatile hydrogen chloride from its salt (KCl/NaCl)
-Drying agent / to dry the gas.
-Is hygroscopic – absorbs water but do not form solution.
iii) Name another substance which can be used to dry chlorine gas.
-anhydrous Calcium chloride
– silica gel
iv)Using a chemical equation, explain why anhydrous calcium oxide cannot be used in flask B
-Calcium oxide reacts with water /moisture to form calcium hydroxide. The calcium hydroxide formed reacts with chlorine to form calcium hypochlorite.
Chemical equations:
CaO(s) + 2H2O(l) -> Ca(OH)2(aq) + H2O(l)
Ca(OH)2(aq) + Cl2 (g) -> CaOCl2(aq) + H2O(l)
This reduces the amount of Chlorine produced.
d)Blue and red litmus papers were dipped in the hydrogen chloride prepared above. The Procedure was repeated with damp/wet/moist litmus papers. Explain the differences in observations made.
–Dry blue litmus papers remain blue
-Dry red litmus papers remain red
-Damp/moist/wet blue litmus papers turn red
-Damp/moist/wet red litmus paper turns red.
-Dry hydrogen chloride is a molecular compound that is joined by covalent bonds between the atoms. The gas is polar thus dissolves in water and ionize completely to free H+ that are responsible to turning blue litmus paper red.
Water – polar
Methylbenzene – non-polar
(ii) State and explain the observations made in the beaker containing:
(i)Methylbenzene
Colour of litmus solution remain.
Hydrogen chloride is a molecular substance. When dissolved in non-polar solvent, it does not dissociate / ionize to release H+ ions that changes the colour of litmus solution.
(ii)Water
Colour of litmus solution change to red.
Hydrogen chloride is a molecular substance. When dissolved in polar solvent like water, it dissociate/ionize to release H+ ions that changes litmus solution to red.
(iii)Why should an inverted filter funnel be used to dissolve hydrogen chloride.
– The filter funnel is dipped just below the water surface to increase the surface area of dissolving the gas and prevent suck back.
(iv)Name the solution formed when hydrogen chloride dissolves in water.
Hydrochloric acid
(f) Describe the test for presence of hydrogen chloride gas.
-Dip a glass rod in ammonia. Bring it to the mouth of a gas jar containing a gas suspected to be hydrogen chloride
-White fumes of ammonia chloride are formed.
(g) Place 5cm3 of dilute hydrochloric acid into a four separate test tubes. To separate test tube add zinc, magnesium iron and copper metals. State and explain the observations made.
Observation
– Effervescence/bubbles/fizzing in all cases except copper
Explanation
Metals above hydrogen in reactivity series react with hydrochloric and liberating hydrogen gas.
Chemical Equation:
Concentrated hydrochloric acid is a weak oxidizing agent than other concentrated acids i.eSulphuric (VI) acid and nitric (V) acid that react with all metals even those lower in the reactivity series.
(h) Place 5cm3 of dilute hydrochloric acid into five separate test tubes. To separate test tubes, add calcium carbonate, silver carbonate, copper carbonate, iron (II) carbonate and Sodium hydrogen carbonate. Explain the observations made.
Observation
Effervescence/bubbles/fizzing vigorously except in silver carbonate and lead (II) carbonate that stop later.
Explanation.
Carbonates and hydrogen carbonate react with dilute hydrochloric acid to produce carbon (IV) oxide, water and form chlorides.
All chlorides formed are soluble Except Lead (II) Chloride (soluble on heating/warming) and silver chloride.
Chemical equation:
CaCO3 (s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
(Colourless solution)
Chemical equation:
Ag2CO3 (s) + 2HCl(aq) → AgCl(s) + H2O(l) + CO2(g)
(Coats/Cover Ag2CO3)
Chemical equation:
CuCO3 (s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)
(Blue Solution)
Chemical equation:
FeCO3 (s) + 2HCl(aq) → FeCl2(aq) + H2O(l) + CO2(g)
Chemical equation:
NaHCO3 (s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Observation
Colour of Phenophthalein indicator change from pink to colourless.
Explanation
Hydrochloric acid neutralizes alkalis to salt and water
When all the alkali has reacted with the acid, An extra slight excess acid turns the indicator used to colourless.
Chemical equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Chemical equation:
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
Chemical equation:
NH4OH(aq) + HCl(aq) → NHaCl(aq) + H2O(l)
(j) Place 5cm3 of hydrochloric acid into four separate test tube tubes Separately add about 1g of each of copper (II) Oxide, Zinc (II) Oxide, Lead (II) Oxide< Calcium (II) Oxide. What happens to each test tube? Explain.
Observation:
All Solid dissolves except Lead (II) Oxide
Colourless solution formed with zinc Oxide and calcium (II) Oxide blue solution formed with copper (II) Oxide.
Explanation:
Metal oxides dissolves in dilute hydrochloric acid to form water and chloride salt Insoluble Lead (II) chloride and silver chloride once formed cover/coat unreacted oxides stopping further reaction.
Chemical equation: CuO(s) + HCl (aq) → CaCl2(aq) + H2O(l)
Chemical equation: CaO(s) + HCl (aq) → CaCl2 (aq) + H2O (l)
Chemical equation: PbO(s) + 2HCl (aq) → PbCl2 (aq) + H2O (l)
Chemical equation: ZnO(s) + HCl (aq) → ZCl2 (aq) + H2O (l)
(k) Manufacture of Hydrochloric acid.
(i) Raw Materials
(i) During electrolysis of Brine from the flowing mercury-cathode cell during the manufacture of sodium hydroxide solution.
(ii)From water gas by passing steam in heated charcoal.
C(s) + H2O → CO(g) + H2(g)
(iii)From partial oxidation of natural gas/methane
CH4(g) + O2(g) → CO(g) + 3H2(g)
2.Chlorine
(i)From electrolysis of fused/solid sodium chloride in the downs process during extraction of sodium
(ii)From electrolysis of brine/concentrated sodium chloride solution in the flowing mercury-cathode during the manufacture of sodium hydroxide solution.
(ii)Chemical processes.
Large amount of hydrogen explodes.
Chemical Equation
H2(g) + Cl(g) → 2HCl(g)
Chemical Equation
HCl(g) + (aq) → HCl(aq)
The absorption chamber is shelved and packed with broken glass beads to
(i)Slow down the downward flow of water.
(ii)Increase surface area over which the water dissolves
The hydrochloric acid is then transported in steel tanks lined with rubber for market
(iii)Uses of Hydrochloric Acid
| Cold water |
(iv)Diagram Showing Industrial manufacture.
| Dry Hydrogen gas |
| Dry chlorine gas |
| Burning Hydrogen gas |
| Absorbtion
chamber |
| 35% Conc. HCl |
(ii)Environmental effects of manufacturing HCl.
D: CHLORIDE (Cl–) SALTS
| Metal | K | Na | Li | Mg | Ca | Al | Zn | Fe | Pb | H. | Cu | Ag | Hg |
| Formula of chloride | KCl | NaCl | LiCl | MgCl2 | CaCl2 | AlCl3 | ZnCl2 | FeCl2
FeCl3 |
PbCl
PbCl4 |
HCl | CuCl
CuCl2 |
AgCl | Hg2Cl2
HgCl2 |
(i)Both FeCl2 and FeCl3 exists but FeCl2 is readily oxidized to FeCl3 because it is more stable.
(ii)PbCl2 and PbCl4 exist but PbCl4 is only oxidized to form PbCl2 by using excess chlorine. It is less stable.
(iii)CuCl and CuCl2 exists but CuCl2 is (thermodynamically) more stable than CuCl. CuCl disproportionate to Cu and CuCl2..
(iv)HgCl and HgCl2exists as molecular compounds.
Place about 2g of solid ammonium chloride crystals in a clean dry boiling tube.Heat gently then strongly.
Observation
–red litmus paper turn blue
-blue litmus paper remains blue
Then later:
-both blue litmus papers turn red
Explanation:
Ammonium chloride on heating decomposes through chemical sublimation to ammonia and hydrogen chloride gas. Ammonia gas is less dense than hydrogen chloride. It is a basic gas and diffuses out faster to turn red litmus paper to blue. Hydrogen chloride is an acidic gas .It is denser than ammonia gas and thus diffuses slower than ammonia gas to turn the already both blue litmus paper to red.
Chemical equation
NH4Cl(s) -> HCl(g) + NH3 (g)
(acidic gas) (basic/alkaline gas)
Place about 1g of sodium chloride, Zinc chloride and copper (II) chloride in separate boiling tubes. Place moist blue and red litmus papers on the mouth of the test tube. Carefully, add three drops of concentrated sulphuric (VI) acid.
Dip a glass rod in aqueous ammonia solution then bring it to the mouth of the boiling tube.
| observation | inference |
| -red litmus paper remain red
-blue litmus paper turn red
-vigorous effervescence/fizzing /bubbling
-white fumes produced on |
H+ ions
Cl– ions
HCl gas suspected |
(b)Explanation:
Concentrated sulphuric (VI) acid is the less volatile mineral acid.
It vigorously displaces chlorine in metal chlorides to evolve acidic hydrogen chloride gas fumes.
Chemical equation
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
CuCl2(s) + H2SO4(l) -> CuSO4(aq) + 2HCl(g)
ZnCl2(s) + H2SO4(l) -> ZnSO4(aq) + 2HCl(g)
Hydrogen chloride and ammonia gases react and form white fumes of ammonium chloride that confirms presence of Cl– ions in the solid substance.
Chemical equation
NH3(g) + HCl(g) -> NH4Cl(s)
(i)Using aqueous Lead (II) nitrate(V)
(a)Procedure:
I.Place about 5cm3 of sodium chloride, Iron (III) chloride and copper (II) chloride in separate boiling tubes. Add four drops of Lead (II) nitrate(V) solution to each. Preserve.
| Observation | Inference |
| White precipitate/ppt | SO42-, SO32-, Cl–,CO32-
|
II.To the preserved sample, add six drops of nitric (V) acid. Preserve.
| Observation | Inference |
| White precipitate/ppt persist | SO42-, Cl–
|
III. To the preserved sample, heat the mixture to boil
| Observation | Inference |
| White precipitate/ppt dissolves on boiling/warming | Cl–
|
Explanation:
I.When Lead(II) nitrate(V) solution is added to an unknown salt , a white precipitate/ppt of Lead(II) sulphate(VI) Lead(II) carbonate(IV) Lead(II) sulphate(IV) Lead(II) chloride(I) are formed.
Ionic equation:
Pb2+ (aq) + SO42-(aq) -> PbSO4(s)
Pb2+ (aq) + SO32-(aq) -> PbSO3(s)
Pb2+ (aq) + CO32-(aq) -> PbCO3(s)
Pb2+ (aq) + Cl–(aq) -> PbCl2(s)
II.When the white precipitate/ppt formed is acidified with dilute nitric(V) acid, the white precipitate of Lead(II) sulphate(VI) and Lead(II) chloride(I) persist/remain while that of Lead(II) carbonate(IV) and Lead(II) sulphate(IV) dissolves.
III.On heating /warming Lead (II) chloride (I) dissolves but on cooling it recrystallizes.This shows the presence of Cl– ions in aqueous solutions
.
(ii)Using aqueous silver (I) nitrate(V)
Procedure
| Observation | Inference |
| White precipitate/ppt | Cl–, CO32-
|
| Observation | Inference |
| White precipitate/ppt persist | Cl–
|
Explanation:
I.When silver(I) nitrate(V) solution is added to an unknown salt , a white precipitate /ppt of silver(I) carbonate(IV) and silver(I) chloride(I) are formed.
Ionic equation:
2Ag+ (aq) + CO32-(aq) -> Ag2CO3(s)
Ag+ (aq) + Cl–(aq) -> AgCl(s)
Silver (I) carbonate (IV) dissolves when reacted with nitric (V) acid.
COMPREHENSIVE REVISION QUESTIONS
(i)For the school laboratory preparation of hydrogen chloride gas.
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii)in tube S
Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
b)State and explain the observation made in tube V.
Observations-colour of solid changes from black to brown
-colourless liquid forms on the cooler parts of tube V
Explanation-Hydrogen produced in tube S reduces black copper(II) oxide to brown copper metal and the gas oxidized to water vapour that condense on cooler parts..
Chemical equation.
CuO(s) +H2(g) ->Cu(s) + H2O(l)
(c)How would the total mass of tube S and tube V and their contents compare before and after the experiment.
Tube S- Mass increase/rise because Zinc combine with chlorine to form heavier Zinc Chloride.
Tube V- Mass decrease/falls/lowers because copper (II) oxide is reduced to lighter copper and oxygen combine with hydrogen to form water vapour that escape.
a)What is the role of the following in the reaction;
(i) concentrated sulphuric(VI)
To produce hydrogen chloride gas by reacting with the solid sodium chloride.
(ii)potassium manganate(VII)
To oxidize hydrogen chloride gas to chlorine
3.Use the flow chart below to answer the questions that follow.
a)(i) Name:
gas X Hydrogen chloride
solution W hydrochloric acid
gas Q chlorine
bleaching agent Z sodium chlorate(V)
b)Write the chemical equation for the formation of :
(i) gas X
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii)solution W
HCl(g) + (aq) -> HCl(aq)
(iii)gas Q
2KMnO4 + 16HCl(aq) -> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
(iv)bleaching agent Z
6NaOH(aq) + 3Cl2(g) ->NaCl(aq) + NaClO3(aq) + 3H2O(l)
c)State and explain the following observations;
(i) a glass rod dipped in aqueous ammonia is brought near gas X
Observation: Dense white fumes
Explanation:Ammonia gas reacts with hydrogen chloride gas to form dense white fumes of ammonium chloride.
Chemical equation: NH3(g) +HCl(g) -> NH4Cl(s)
(ii)Wet blue and red litmus papers were dipped into gas Q
Observations: Blue litmus paper turned red the both are bleached
/decolorized.
Explanations: chlorine reacts with water to form both acidic hydrochloric and chloric (I) acids that turn blue litmus paper red. Unstable chloric (I) acid oxidizes the dye in the papers to colourless.
Chemical equations
Cl2(g) + HCl(aq) ->HCl(aq) + HClO(aq)
Coloured dye +HClO(aq) ->HCl(aq) + (Colourless dye +O)//
(Coloured dye-O) + HClO(aq) ->HCl(aq) + Colourless dye
4.Use the flow chart below to answer the questions that follow
Liquid A Concentrated sulphuric(VI) acid
Process Z Neutralization
White solid X Ammonium chloride
b)Write the equation for the formation of:
(i) Hydrogen chloride
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii) B
HCl(g) + (aq) -> HCl(aq)
(iii)process Z (using ionic equation)
H+ (aq) + OH–(aq) -> H2O(l)
(iv)C (using ionic equation)
Ag+ (aq) + Cl–(aq) -> AgCl(s)
c)Describe how solution B is obtained.
Bubbling hydrogen chloride gas through inverted funnel into distilled water until no more dissolve.
5 The results obtained when halogens are bubbled into test tubes containing solutions of halide A,B and C is as in the table below. Tick(v) means a reaction took place.Cross(x) means no reaction took place.
|
Halogens |
Halide ions in solution | ||
| A | B | C | |
| I2 | x | – | x |
| Br2 | x | v | – |
| Cl2 | – | v | v |
a)Identify the halide ions represented by letter
A Cl–
B I–
C Br–
b)Write the ionic equation for the reaction that take place with halide:
(i) C
Cl2(g) + 2Br–(aq) -> 2Cl–(aq) + Br2(aq)
(ii) B
Cl2(g) + 2Br–(aq) -> 2Cl–(aq) + Br2(aq)
Cl2(g) + 2I–(aq) -> 2Cl–(aq) + I2(aq)
6.The diagram below shows a set up of apparatus for the school laboratory collection of dry chlorine gas.
a)Name:
(i) substance Q
Concentrated hydrochloric acid
(ii)suitable drying agent L
-Concentrated sulphuric(VI) acid
-anhydrous calcium chloride
-silica gel
-Heat/Heating
c)Red and blue litmus papers were dipped into the chlorine gas from the above set up .State and explain the observations made.
Observation: Blue litmus paper remain blue. Red litmus paper remain red.
Explanation: Dry chlorine has no effect on dry litmus papers.
d)Write the equation for the reaction taking place in the conical flask
MnO4 (s) + 4HCl(aq) -> MnCl2(aq) + 2H2O(l) + Cl2(g)
e)Name two other substances that can be used in place of MnO2
Lead(IV) oxide (PbO2)
Potassium manganate(VI)(KMnO4)
Potassium dichromate(K2Cr2O4)
Bleaching powder(CaOCl2)
7.The set up below shows the apparatus used to prepare and collect anhydrous iron(III) chloride.
a)Name salt K
Iron(III)cchloride
2Fe(s) + 3Cl2 (g) -> 2FeCl3 (s/g)
(i)Small amount of water is added to iron (II) chloride in a test tube then shaken
Solid dissolves to form a green solution. Iron(II) chloride is soluble in water
(ii)I.Three drops of aqueous sodium hydroxide is added to aqueous iron(II) chloride and then added excess of the alkali.
Observation:
Green precipitate is formed that persist/remain /insoluble in excess akali.
Explanation:
Iron(II) chloride reacts with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide.
Ionic equation:
Fe2+(aq) + OH–(aq) -> Fe(OH)2(s)
II.Six drops of hydrogen peroxide is added to the mixture in d(ii) above.
Observation:
Effervescence/bubbling/fizzing take place and the green precipitate dissolve to form a yellow/brown solution.
Explanation:
hydrogen peroxide oxidizes green Fe2+to yellow/ brown Fe3+solution.
9.Use the flow chart below to answer the questions that follow.
a)Write the chemical equation for the formation of gas A
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
b)Identify:
(i) four possible ions that can produce white precipitate B
SO42-,SO32-, CO32-, Cl–
(ii)two possible ions that can produce;
I.White precipitate C
SO42-,Cl–
II.colourless solution D
SO32-, CO32-
(iii)possible ions present in
I.White precipitate E
SO42-
II.colourless solution F
Cl–
State the observation made when aluminium wool is heated.
Glows red hot.
b)(i) Identify salt A
aluminium(III) chloride// AlCl3
(ii)Write the equation for the formation of salt A
2Al(s) + 3Cl2(g) -> 2AlCl3(s/g)
(iii)What property of salt A is exhibited as shown in the experiment.
It sublimes//sublimation.
(iv)Calculate the minimum volume of chlorine required to form 700kg of iron(III) chloride at room temperature.(Fe= 56.0, Cl=35.5, 1 mole of a gas =24000cm3, 1000g = 1kg)
Mole ratio Fe : Cl2 = 2: 3 molar mass FeCl3 = 162.5g
Method 1
2 x 162.5 g FeCl3 -> 3x 22400 cm3 Cl2
700 x1000 gFeCl3 -> (700 x1000 x3 x22400)/(2 x 162.5)
=1.4474 x 10-8 cm3
Method 2
Moles of FeCl3= mass/ molar mass
=> (700 x 1000) / 162.5 = 4307.6923 moles
Moles of Cl2= 3/2 moles of FeCl3
=>3/2 x 4307.6923 = 6461.5385 moles
Volume of chlorine= moles x molar gas volume
=>6461.5385 x 24000 = 1.5508 x 10-8 cm3
Iron
d)(i) What is the purpose of anhydrous calcium chloride.
-ensure the apparatus are water free.
-prevent water from the atmosphere from entering and altering//hydrolysing salt A
(ii) Write the equation for the reaction that take place if anhydrous calcium chloride is not used in the above set up.
AlCl3(s) + 3H2O(l) -> Al(OH)3(aq) + 3HCl(g)
(iii) Write the equation for the reaction that take place when Iron metal is reacted with dry hydrogen chloride gas.
Fe(s) + 2HCl(g) -> FeCl2(s) + H2(g)
(iv)Calculate the mass of Iron(II)chloride formed when 60cm3 of hydrogen chloride at r.t.p is completely reacted. (1 mole of a gas =24dm3 at r.t.p, Fe = 56.O, Cl= 35.5)
Chemical equation Fe(s) + 2HCl(g) -> FeCl2(s) + Cl2(g)
Mole ratio HCl: FeCl2 = 1:1
Molar mass FeCl2 = 127g
Moles of HCl used = 60cm3 /24000cm3 = 2.5 x 10 -3 moles
Moles of FeCl2 = Moles of HCl => 2.5 x 10 -3 moles
Mass of FeCl2 = moles x molar mass => 2.5 x 10 -3 x 127 =0.3175g
12.Study the flow chart below and use it to answer the questions that follow
a)Identify substance:
P Iron(II) chloride//FeCl2
Q Chlorine // Cl2
R Iron(III) chloride//FeCl3
b)Write the equation for the reaction for the formation of:
(i) gas Q
2KMnO4 (s) + 16HCl(aq) -> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
(ii) the green precipitate (using ionic equation)
Ionic equation:
Fe2+(aq) + 2OH–(aq) -> Fe(OH)2(s)
(ii) the brown precipitate (using ionic equation)
Ionic equation:
Fe3+(aq) + 3OH–(aq) -> Fe(OH)3(s)
c)A glass rod was dipped in aqueous ammonia. The rod was then brought near hydrogen chloride. State and explain the observation made.
Observation:
White fumes
Explanation:
Ammonia gas reacts with hydrogen chloride gas to form white fumes of ammonium chloride.
Chemical equation:
NH3(g) + HCl(g) -> NH4Cl(s)
(b)How many electrons in :
(i)aluminium atoms are used in bonding.
Six electrons(three valence electrons in each aluminium atom)
(ii)chlorine atoms atoms are used in dativebonding.
four electrons(two lone pairs of valence electrons in two chlorine atoms)
(iii)the molecule are used in bonding.
Sixteen electrons
-six valence electrons from aluminium atom through covalent bond
-six valence electrons from chlorine atoms through covalent bond.
– four valence electrons from chlorine atoms through dative bond
(c)How many lone pair of electrons do not take part in bonding within the molecule.
Sixteen(16) lone pairs from six chlorine atoms(32 electrons)
(d)Aluminium chloride does not conduct electricity in molten state but Magnesium chloride conduct.
Aluminium chloride is a molecular compound that has no free mobile Al3+ and Cl– ions which are responsible for conducting electricity. Magnesium chloride has free mobile Mg2+ and Cl– ions because it is an ionic compound.
a)Write an equation for the school laboratory formation of hydrogen chloride gas
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
b)Name:
II green precipitateIron (II) hydroxide (Fe (OH)2
III Gas Y Chlorine (Cl2)
c)Blue and red litmus papers were dipped into bleaching agent A. Write the equation for the reaction that takes place.
Coloured dye +NaOCl(aq) ->NaCl(aq) + (Colourless dye + O)//
(Coloured dye-O) + NaOCl(aq) ->NaCl(aq) + Colourless dye
d)State four uses of gas Z
curriculum.
(b) In what five ways can teaching of C.R.E enhance national unity in Kenya?
(b) Give six reasons for studying Christian Religious Education in Secondary schools.
in Kenya
local languages
(b) State seven problems which church leaders encounter in their work of evangelization
(b) Give four ways in which the Bible is used to spread the Gospel today
(b) Explain four ways in which the translation of the Bible into African languages
led to African mass evangelism.
(c) Give reasons why Kenyans are attracted to the Church.
with mankind.
(b) Explain the comparison between the Biblical concept of sin and the Traditional African
concept of evil
(b) Mention six causes of evil in Traditional African Society
(b) What are the consequences of breaking taboo in Traditional African Communities?
(b) Give six lessons that Christians learn about work from the Genesis stories of creation
(c) State four ways on how Christians can care for God’s creation today
(b) What is the importance of the decalogue to Christians today
(c)What can Christians learn about God on the call of Moses in the wilderness?
the Israelites
(b) Identify six similarities between the Jewish Passover and Christian Easter
(c)State four reasons why Christians should live by laws of God
the Exodus
the covenant
(b) Explain the conditions given to the Israelites during the renewal of the Sinai covenant
10 (a) Explain four ways in which Moses early life prepared him for his future role.
(b) List seven plagues Moses had to perform in Egypt before the Israelites could be released.
(c) State five lessons Moses learnt about God from his call.
covenant
(b) Describe how the Sinai covenant was sealed
(c) State seven leadership qualities a modern Christian can learn from Moses
(b) Show how God cared for the Israelites during the Exodus
(c) Give six reasons why some Christians have lost faith in God today
(c) Why do Christians find it difficult to follow the laws of God?
18 a) Identify the commandments given to the Israelites that teach on how to relate to one another
(c) Why do Christians find it difficult to follow the laws of God?
(b) How did the Kings of Israel lead people back to God?
(b) How do Christians demonstrate their faith in God today?
(b) How did the Kings of Israel lead people back to God?
(c) Give five ways in which Christians show their trust in God.
(b) How do Christians demonstrate their faith in God today?
as a nation
7 a) State six promises that God gave to King David
of King David
(b) Give four ways in which King David promoted the worship of Yahweh in Israel
(c) Identify reasons why political leaders in Kenya have failed to perform their duties effectively
15 a) Outline six promises that God gave to king David through prophet Nathan.
(b) What problems were faced by Prophet Elijah in Israel?
(c) State six reasons why Christians should fight against the spread of devil worship in the society.
(b) Show ways in which prophet Elijah tried to restore the true worship of Yahweh in Israel.
(c) What can Christians learn from the happenings at the Mt. Carmel contest about the, nature
of God.
(b) Explain the failures of King Ahab
(c) Give reasons why political leaders in Kenya failed to perform their duties effectively
(b) List five forms of corruption in Kenya today
(c) Explain the relevance of Elijah’s prophetic mission to Christians today
(b) Give five reasons why some Christians have lost faith in God
(b) Give five qualities in Elijah that may influence the life of a true Christian
8 (a) Explain any four characteristics of the Canaanite religion.
(b) Give five reasons why it was difficult for Prophet Elijah to stop idolatry in Israel.
(c) Mention any seven factors that lead people away from the worship of God today
(b) Give five signs used by God to show that Elijah is the true prophet of God
can help reduce corrupt in the society
1 Kings 21.
(b) State what is revealed about the nature of God during the contest at Mount Carmel
(b) What problems were faced by Prophet Elijah in Israel?
(c) State six reasons why Christians should fight against the spread of devil worship in
the society.
(a) Prophets
(b) What problems did the Israelites experience during the Babylonian exile
(c) State five difficulties Christian go through as they serve God
messages to the people
(c) Amos
(b) What were the similarities in the Prophetic vocations of Amos and Jeremiah?
(c) What lessons can Christians learn from Amos teachings on judgement
people of the Northern Kingdom.
today.
(b) Outline seven teachings of Prophet Amos about the day of the Lord
(c) Identify seven ways in which Christians can help the church leaders to perform their duties
effectively
(b) Give ways through which Christians promote mutual responsibility in the society today
c ) Identify the prophecy of Amos on Yahweh’s judgement.
African society.
15 a) Explain the purpose of bride wealth in the traditional African community.
(d) Jeremiah
(b) Give reasons for Jeremiah temple sermon.
(c) What is the relevance of the suffering and lamentation of Jeremiah to Christians?
punishment to the Israelites
(b) Explain Jeremiah’s teaching on the new covenant
(c) Give five lessons that Christians learn from Prophet Jeremiah’s teaching on the new
covenant
(b) State any seven challenges that modern Christians face in their struggle to condemn social
injustices in the society today
Sinaic covenant
9 (a) Explain four reforms King Josiah made in Judah during the time of prophet Jeremiah.
(b) Give five reasons why Jeremiah condemned necromancy in Judah.
(c) What can modern leaders learn from Jeremiah’s sermon in the temple?
(b) What are the evils that Prophet Jeremiah would condemn in Kenya today
Judgements and punishment to the Israelites
false prophets
(b) Explain eight reasons why the temple of Jerusalem was considered as an important place
for the Israelites
(c) In what ways do Christians use the print media to spread the Gospel
(b) Give reasons why child labour is morally wrong
(c) Identify ways in which Christians spend their leisure time to glorify God
(e) Nehemiah
(b) What were the socio-economic problems that Nehemiah faced as the Governor of Judah
after completing the wall of Jerusalem
(c) Mention six good qualities in Nehemiah which can help modern leaders in Church and
society to Execute good leadership today
today
Nehemiah.
(a). The African concept of God, Spirits and Ancestors.
(b) Give the occasions when sacrifices were offered in Traditional African communities.
(c) State six aspects of traditional religion which have been integrated into the Christian faith
them.
(b) Explain four reasons which made traditional African communities to offer sacrifices.
(c) Give six ways the Traditional Africans used to maintain their relationship with the
ancestors
(b) Identify seven traditional African practices which demonstrated their belief in God.
(c) Outline five African understanding of evil.
(b) African moral and cultural values
Community.
(c) Show how Christians can contribute towards the resolution of conflicts in society today.
communities
(b) List down the traditional African practices which show that life is sacred
(c) State the challenges facing rite of initiation today
Communities.
Traditional Society
society today
(b) Discuss factors that have contributed to harmony and mutual responsibility in the Traditional
African Communities
the traditional African communities
(b) What changes have taken place in the traditional African attitude to orphans?
African society.
of the messiah
4 Explain the concept of the Messiah in the New Testament.
according to Isaiah 53
Old Testament prophecy about the messiah
(b) Outline activities which the youth may involve in the church today
Messiah.
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FORM 2
Soil Fertility II
(Inorganic Fertilizers)
Introduction
Essential Elements
Macro-nutrients
They include;
Role of Macronutrients in Plants
Nitrogen (NO3,NH4++)
Sources:
Role of Nitrogen in Plants
Deficiency Symptoms
Effect of Excess Nitrogen
Loss of Nitrogen From the Soil:
Phosphorus (H2 P04, HPO2-4 P2O5)
Sources:
Role of Phosphorus
Deficiency symptoms
Loss of Phosphorus From the Soil
Potasium (K+, K2O)
Sources;
Role of Potassium in Plants
Deficiency Symptoms
Loss of Potassium From the Soil
Calcium (Ca2+)
Source:
Role of Calcium in Plants
Deficiency symptoms
Loss of Calcium
Magnesium (Mg2+)
Sources:
Role of Magnesium in Plants
Deficiency symptoms
Sulphur (S042-,SO2)
Sources:
Role of Sulphur in Plants
Deficiency Symptoms
Micro-nutrients
They include;
Role of Micronutrients and Their Deficiency Symptoms
Inorganic Fertilizers
Classification According to:
Properties and Identification of Fertilizers
Nitrogenous Fertilizers
Characteristics
Examples:
Physical appearance:
Phosphate Fertilizers
Examples;
Potassic Fertilizers
Characteristics:
Examples;
Compound or Mixed Fertilizers
Examples;
Advantages of application of compound fertilizers
Disadvantages of compound fertilizers application
Methods of fertilizer application
Determination of Fertilizer Rates
Contents of fertilizers are expressed as fertilizer grade or fertilizer analysis.
Example 10:20:0 means for every 10kg of the mixture there are 10kg of nitrogen, 20kg of P2O 5 and 0kg of K2O.
Example
A farmer was asked to apply fertilizers as follows:
How much sulphate of ammonia (20%) would be required per hectare?
How much double super-phosphate (40%) P2O5would be required per hectare?
How much muriate of potash (50% K2O) would be required per hectare?
Answer/Solution
= 60
20x 100 =300kg SA
60
= 40x 100 =150kg DSP
= 60x 100=100kg muriate of potash
60
Example
A farmer was asked to apply fertilizers as follows:
How much P2O5 did the farmer apply per acre?
How much K2O did the farmer apply per hectare?
How much N did the farmer apply per hectare?
Solution/Answer
40
= 100x 200= 80kg/ha P2O 5
60
= 100×150=90kg/ha K2O
20
= 100 x 150= 30kg/ha N
Soil Sampling
Soil Sampling Procedures
Sites to Avoid
Methods Of Soil Sampling:
Soil Testing
Importance of Soil testing:
How Soil pH affects Crop Production
Methods of pH Testing
Labour Records– show labour utilization and labour costs. Crop production II (Planting)
Types of planting materials
Seeds
Advantages of using seeds as planting materials.
Disantivantages of using seeds as planting materials.
Advantages of using vegetative materials for planting.
Disadvantages.
Plant parts used for vegetative propagation.
Factors affecting rooting of cuttings.
Selection of planting materials
When selecting materials for planting the following factors must be considered:
In Kenya certified seeds are produced by the Kenya seed company (KSC) and distributed by Kenya Farmers Association (KFA) and other agents.
PREPARATION OF PLANTING MATERIALS.
After the planting materials are selected they are prepared in different ways before they are planted. Some of the methods used to prepare planting materials include the following:
(a)Breaking the seed dormancy.
Some seeds undergo a dormancy period between maturity and the time they sprout. The dormancy period is the stage whereby a seed cannot germinate, the stage of inhibited growth of seed. It should be broken before the seed is planted.
Methods of breaking seed dormancy.
The following methods are used to break seed dormancy:
(I) Mechanical method: This is a method which aims at scratching the seed coat to make it permeable to water. Scarification is done by rubbing small sized seeds against hard surface such as sand paper, while filling or nicking the seed coat with a knife is done to large sized seeds such as croton seeds.
(ii) Heat treatment: this involves the use of hot water or burning the seeds lightly. It softens the seed coat making it permeable to water and thus is able to germinate. The seeds are soaked in hot water about 80’c for 3-4 minutes after which the water is allowed to drain off. Example of seeds treated in this way include: leucean calliadra and acacia.
Light burning also serves the same purpose as hot water treatment. In this case trash is spread over the seeds which are already covered with a thin layer of soil. The trash is burned, after which the seeds are retrieved and planted. Examples include acacia and wattle tree seeds. Overheating should be avoided as this will cook the seeds.
(iii)Chemical treatment: seeds are dipped in specific chemicals such as concentrated sulphuric acid, for two minutes and then removed. The chemical wears off the seed coat making it permeable to water. Care should be taken not to leave the seeds in the chemicals for too long as this will kill the embryo. Cotton seeds are normally treated with chemicals to remove the lint or fibres.
iv)Soaking in water: seeds are soaked in water for a period of between 24 – 48 hours until they swell. They are then removed and planted immediately. The seeds treated thus germinate very fast. Pre-germinated seeds are used when raising rice in the nurseries.
This is the coating of seeds with fungicides or an insecticide or a combination of the two chemicals. This is particularly common with cereals, sugar-cane and legumes.
The chemicals protect the seedlings from soil-borne diseases and pests. Certified seeds which are sold by seed merchants in Kenya have been dressed with these chemicals. Farmers can also buy the chemicals and dress their own seeds.
In areas where soils are deficient in nitrogen, legumes such as beans, clovers and peas should be coated with an inoculant. An innoculant is a preparation which contains the right strain of Rhizobium depending on the type of legume and encourages nodulation, hence nitrogen fixation. Below is a table showing different legume crops and their right strain of Rhizobium.
| Crop inoculation group | Rhizobium Species |
| Lucerne | R. melioti |
| Clover | R. trifoli |
| Pea | R. leguminosarum |
| Bean | R. phaseoli |
| Lupin | R. lupini |
| soyabean | R. japonicum |
When handling inoculated seeds, care should be taken to prevent them from coming in contact with chemicals. This means that inoculated seeds should not be dressed with chemicals as these will kill the bacterium. They should also be planted when the soil is moist to avoid dehydration which kills the bacterium.
This practice is also referred to as sprouting. The selected seed potatoes ‘setts’ which are used as planting materials are sprouted before planting to break their dormancy. The setts of about 3-6 cm in diameter are arranged in layers of 2 or 3 tubers deep in a partially
darkened room. The setts should be arranged with the rose- end facing upwards and the heel-end downwards. Diffused light encourages the production of short, green and healthy sprouts. If Chitting is done in complete darkness, long, pale thin sprouts develop which break easily during planting. During Chitting potato aphids and tuber months should be controlled by dusting or spraying the sett with dimethoate. Sometimes a chemical known as Rendite is used to break dormancy, thus inducing sprouting. Chitting is done mainly to make sure that growth commences immediately the seed is planted so as to make maximum use of rains for high yields.
Time of planting
The timing of planting or sowing is influenced by the type of crop to be planted and the environmental conditions of the area.
Factors to consider in timing planting.
Timely planting is necessary and should be done at the onset of rains. In some areas where rainfall is scare dry planting is recommended.
Advantages of timely planting.
Methods of planting.
There are two main methods of planting :-
Broadcasting.
This method involves scattering the seeds all over the field in a random manner. It is commonly adapted for light tiny seeds such as those of pasture grasses. It is easier, quicker and cheaper than row planting. However, it uses more seeds than row planting and the seeds are spread unevenly leading to crowding of plants in some places. This results in poor performance due to competition. Broadcasting gives a good ground cover, but weeding cannot be mechanized. For good results, the seedbed should be weed-free, firm and have a fine tilth.
Row planting.
The seeds or other planting materials are placed in holes, drills or furrows in rows. The distance between one row to the other and from one hole to the other is known. In Kenya, both large and small – scale farmers practice row planting. It is practiced when planting many types of crops, especially perennial, annual and root crops.
Advantages of row planting.
Disadvantages of row planting.
Seeds can also be planted by dibbling where the planting holes are dug by use of pangas or jembe, or by a dibbling stick (dibbler). Most of the dibbling is done randomly although rows can also be used when using a planting line. Random dibbling is not popular in commercial farming due to low levels of production. It is only common among conservative farmers in planting of legumes such as beans, pigeon peas and cow peas.
Over-sowing.
This is the introduction of a pasture legume such as desmodium in an existing grass pasture. Some form of growth suppression of existing grass such as burning, slashing or hard grazing plus slight soil disturbance is recommended before over sowing. A heavy dose of superphosphate, preferably single supers at a rate of 200-400 kg/ha is applied. The grass must be kept short until the legume is fully established. Regardless of the method of establishment, the pastures and fodder stands should be ready for light grazing 4-5 months after planting if rainfall and soil fertility are not limiting.
Under-sowing.
This refers to the establishment of pasture under a cover crop, usually maize. Maize is planted as recommended and weeded 2-3 weeks after the onset of rains. Pasture seeds are then broadcasted with half the recommended basal fertilizer. No further weeding should be done and maize should be harvested early to expose the young pasture seedlings to sunlight. The benefits of under sowing include facilitating more intensive land utilization and encouraging an early establishment of pastures.
Fodder crops and vegetetively propagated pasture species may also be under sown as long as rainfall is adequate for their establishment. Timing is not very crucial in this case and planting can be done as late as 6-8 weeks after the onset of rains.
Plant population
This refers to the ideal number of plants that can be comfortably accommodated in any given area, without overcrowding or too few to waste space. Agricultural research has arrived at the optimum number of various crop plants to be recommended to farmers. Plant population is determined by dividing the planting area by spacing of the crop. This may be simplified thus:
Area of land
Plant population =
Pacing of crop
Example
Given that maize is planted at a spacing of 75 x25 cm, calculate the plant population in a plot of land measuring 4×3 m.
Working
Area of land
Plant population =
Pacing of crop
Area of land = 400cm x 300 cm
Spacing of maize = 75 cm x 25 cm
Therefore, plant population = 400 cm x 300 cm
75 cm x 25 cm
= 64 plants.
Spacing
It is the distance of plants between and within the rows. Correct spacing for each crop has been established as shown in table below.
| crop | spacing |
| Maize
(Kitale) hybrids |
75 – 90 cm x 23 – 30 cm |
| Coffee
(Arabica) tall varieties |
2.75 cm x 2.75m |
| Tea | 1.5 m by 0.75 m |
| Beans (erect type) | 45 -60 m by 25 cm |
| Bananas | 3.6 – 6.0 m by 3.6 – 4.5 m |
| Coconut | 9 m x 9 m |
| Tomatoes (Money maker) | 100 x 50 cm |
| kales | 60 x 60 cm |
Spacing determines plant population and the main aim of correct spacing is to obtain maximum number of plants per unit area which will make maximum use of environmental factors. Wider spacing leads to a reduced plant population which means lower yields, whereas closer spacing could lead to overcrowding of plants and competition for nutrients and other resources would occur. Correctly spaced crops produce yield of high quality that are acceptable in the market.
Spacing is determined by the following factors:
The space between the rows should allow free passage of the machinery which can be used in the field. For example, the spacing between rows of coffee is supposed to allow movement of tractor drawn implements.
A fertile soil can support high plant population. Therefore closer spacing is possible.
Tall crop varieties require wider spacing while short varieties require closer spacing, for example, Kitale hybrid maize is widely spaced than Katumani maize.
Areas with higher rainfall are capable of supporting a large number of plants hence closer spacing than areas of low rainfall.
Crop grown for the supply of forage or silage material is planted at a closer spacing than for grain production.
When crops are properly spaced, pests might find it difficult to move from one place to the other, for example, aphids in groundnuts.
Spreading and tillering crop varieties require wider spacing than erect type.
Seed rate.
Seed rate is the amount of seeds to be planted in a given unit area governed by ultimate crop stand which is desired. The objective of correct spacing of crop is to obtain the maximum yields from a unit area without sacrificing quality. Most crops are seeded at lighter rates under drier conditions than under wet or irrigated conditions. Seeds with low germination percentage are planted at higher rates than those which have about 100% germination percentage. There is an optimal seed rate for various crops. For example, the seed rate for maize is 22 kg per hectare, wheat is 110 kg per hectare and cotton is between 17 to 45 kg per hectare.
Factors to consider in choosing seed rates.
When planting seed which is pure or with a high germination percentage, less seed is required. On the contrary, more seeds are required when using impure or mixed seeds.
Less seed is used when its germination percentage is higher. Seed of lower germination percentage is required in large amounts.
At closer spacing, more seeds are used than in a wider spacing.
When two or more seeds are planted per hole, higher seed rate is required than when only one seed is planted per hole.
A crop to be used for silage making is spaced more closely than one meant for grain production. This would require use of more seeds. Maize to be used for silage making, for example, requires more seeds than that meant for production of grain.
Depth of planting.
This is the distance from the soil surface to where the seed is placed. The correct depth of planting is determined by:
Suggested Activities.
Crop Production III
(Nursery Practices)
Introduction
Importance of Nursery Bed in Crop Production
Selection of a Nursery Site
Factors to consider;
Types of Nurseries
Categories of nurseries:
Establishment of Vegetable Nurseries.
– Clear bushes from the site.
– Dig deeply to remove all perennial weeds.
– Map out/measure out the dimension of the bed : 1 m wide and any convenient length.
– Harrow to obtain a fine tilth.
– Broadcast phosphatic fertilizer or well rotten organic manure.
– Level the beds using garden rake to mix fertilizer or organic manure with soil and remove any trash.
– Make shallow drills 10 – 20 cm apart .
– Drill seeds uniformly.
– Cover the seeds lightly with soil.
– Water the nursery bed thoroughly.
– Apply light mulch /thin layer of mulch.
Vegetable Propagation Nurseries:
Procedure of preparing a vegetative nursery bed.
– Select a suitable site.
– Clear and level the area.
– Mark out/measure the bed, measuring 3.66 m by 1.22 m.
– Fill polythene sleeves with a mixture of rooting medium.
– Arrange the polythene sleeves on the site to form a vegetative propagation unit.
– Water the polythene sleeves.
– Push well-prepared cuttings (tea) into each of the polythene sleeves and at the centre.
– Make sure that cutting leaves do not touch the soil.
– Construct a shade over the cuttings, using wooden loops which are covered with a polythene sheet.
Preparation of a tree seedling nursery.
– Choose the site.
– Clear the site.
– Mark the bed.
– Dig deeper than in the vegetable nursery.
– Plant directly or in polythene sleeves containing rooting medium.
Importance of polythene sleeves.
– Alleviates/ control root disturbances during transplanting.
– Seedlings can be stored awaiting conducive environmental conditions for planting.
– It is easy to transport the seedlings.
– The mixture used to fill the polythene sleeves is free from soil borne pests and diseases.
– Seedlings establish faster in the field.
Nursery Management Practices:
They include:
Preparation of vegetative materials for planting:
Methods of Grafting
of the scion.
Other types of grafting include ;
Budding:
Methods of Budding:
Importance of Budding and Grafting:
Layering
Types of layering;
Tissue Culture for Crop Propagation
The Right Conditions for tissue culture:
Importance of Tissue Culture in Crop Propagation
Disadvantages of tissue culture.
– high level of technical know-how is needed.
– requires a special working environment/controlled conditions are required.
– it is expensive.
– takes time before desired results are realized.
Transplanting Seedlings
1) Vegetable seedlings.
It should be done when:
PROCEDURE
– water the nursery before uprooting seedlings to avoid root damage.
-uproot seedlings using a garden trowel.
– ensure a lump of soil is attached to the roots.
– place the seedlings gently in a planting hole.
– mix soil thoroughly with either well rotten organic manure or phosphatic fertilizer in the planting hole.
– cover the seedlings to the same depth as they were in the nursery.
– firm the soil around the seedling base .
– water the seedling adequately.
– apply light mulch around the seedling base.
2) Transplanting of tree seedlings.
– transplant after 6 – 12 months.
– open the planting holes so that they are 60 cm by 60 cm in size.
– separate the topsoil from the subsoil.
– ensure that the planting holes are dug
Crop production IV (Field Practices I)
Introduction
They include the following:
Crop Rotation
Importance of Crop Rotation
Factors to consider when designing a crop rotational programme.
Mulching
Importance of Mulching
Types of Mulching Materials
Advantages of Mulching (organic mulches).
Disadvantages of Mulching
Importance of inorganic mulch.
– moderates soil temperature.
– conserves soil moisture.
– smothers weeds / controls weed growth.
– prevents splash erosion.
3) Thinning
Importance of thinning.
– helps to avoid unnecessary competition.
– helps to maintain the appropriate plant population.
– encourages good crop growth and high yields.
– high light intensity to remaining plants.
Precautions to observe when thinning.
– should be done when the crop is young.
– ensure that there is appropriate moisture in the soil.
– uproot carefully to avoid root damage.
4)Gapping
– provides efficient ground cover hence controlling erosion.
– ensures that the appropriate plant population is achieved.
– efficient ground cover controls weed growth.
Rogueing
Pruning
Reasons for pruning are:
Note: Tools used are secateur, pruning saw and pruning knife.
Earthing-up
Crop Protection
Weed Control
Pest Control
Control of Crop Diseases
Harvesting
Time of harvesting depends on:
Methods of harvesting is determined by:
Post-Harvest Practices
Storage
Purpose of storage is to;
Requirements for proper store are:
Types of Storage
Preparation of the Store
Crop Production V: (Vegetables)
Introduction
Vegetables are grouped into the following categories:
Tomatoes (Lycopersicon esculentum)
Ecological Requirements
Varieties
Nursery Practices
Seedbed Preparation
Transplanting
Field Maintenance
Pests Controls
Disease Control
Caused by;
Harvesting
Cabbage
Ecological Requirements
Varieties
Nursery Practices
Seedbed Preparation
Transplanting
Field Maintenance
Pest Control
Disease Control
Harvesting
Carrots (Daucuscarota)
Ecological Requirements
Varieties
Land Preparation
Planting
Field Practice
Pest Control
Disease Control
Harvesting and Marketing
Onions (Allium cepa)
Ecological Requirements
Varieties
Land Preparation
Planting
Field Management Practices
Thinning
Topdressing
Pest Control
Onion Thrips:
Disease Control
Purple Blotch and Downey Mildew
Harvesting and Marketing
Livestock Health I
(Introduction to Livestock Health)
Introduction
Importance of Keeping Livestock Healthy:
Predisposing Factors to Livestock Diseases
They include:
Signs of ILL-Healthin Livestock
Causes of Diseases
Categories of Diseases
General Methods of Disease Control
Appropriate Methods of Handling Livestock
Animals are handled for the following reasons:
When carrying out these activities animals should be restrained in a crush.
Other methods of restraining animals include the use of;
Livestock Health II (Parasites)
Introduction
The effects of parasite on the host animal are:
General Symptoms of Parasites Infestation:
Types of Parasites
There are two types of parasites:
External parasites are;
Life Cycle of ticks
One–Host Tick
Two-Host Tick
Three-Host Tick
Control of Ticks
Endo-parasites (internal Parasites)
They can be divided into:
General Symptoms of Helminthiasis
Trematodes (Liver Fluke)
Life Cycle of the Liver Fluke
Control of Liver Fluke
Tapeworms
Eexample;
The adults live in the small intestines of man (the primary host).
Life Cycle of Tapeworm
Control of Tapeworms
Nematodes (Roundworms)
Common ones are;
Nature of Damage
Control of Roundworms
Livestock Production II (Nutrition)
Introduction
Components of Food material
Water
Sources
Functions
Factors Determining the Requirements of Water by Livestock
Protein
Sources:
Functions:
Digestion of Proteins
In non-ruminants, protein digestion takes placed in the stomach.
In ruminants, protein digestion initially takes place in the rumen.
Carbohydrates
Sources:
Functions:
Digestion of Carbohydrates
Fats and Oils
Sources:
Functions:
Digestion of lipids in Ruminants
Vitamins
Sources:
Functions:
Examples:
Minerals
Sources:
Functions:
Examples:
Classification of Animal Feeds
This is based on nutrient composition:
Roughages
Examples:
Characteristics
Concentrates
Examples:
Characteristics
Feed Additives
These are substances added to the feed to increase;
There are two types:
Functions
Compounded Feeds
Poultry feeds can be categorized as:
Meaning of terms used to express feed values
Computation of Livestock Rations
Steps in ration formulation
Methods used in ration formulation
Examples;
Mix a Pigs ration 22% protein using soya bean meal 40% DCP and maize meal containing 8%DCP.
Soya bean meal (14 *100)=43.75kg
32
Maize meal (18*100=56.25kg
32
Digestion and digestive systems
Digestion of food in livestock takes place in three stages;
Rumen-
Reticulum:
Omasum:
Abomasum:
Comparison Between Digestion in Ruminant and Non–ruminants
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Functions of the Parts of Poultry
Comparison Between Digestion In Ruminants and Non–Ruminants
SimilaritiesBetween Digestion In Ruminants and Non–Ruminants
Water absorption takes place in the colon in both ruminants and nonruminants
UNIT 1: INTRODUCTION TO CHEMISTRY.
Unit Checklist.
Meaning of chemistry
Role of chemistry in society.
Laboratory rules.
Common apparatus used in a chemistry laboratory.
The Bunsen burner.
Methods of gas collection.
Drying of gases
Drugs and drug abuse.
Meaning of chemistry.
– It is a branch of science which deals with the composition and properties of matter.
Matter
– Matter is anything that has mass and occupies space.
States of matter.
– Matter exists in three different states: solids, liquids and gases.
Properties of the different states of matter.
| Solid | Liquids | Gases | |
| Mass | Definite | Definite | Definite |
| Shape | Definite | Indefinite: they take the volume of the container in which they are in. | Indefinite: they take the shape of the container in which they are. |
| Volume | Definite | Definite | Indefinite: volume may increase with increase in temperature; and decrease with decrease in pressure |
Note: Conductors and non-conductors:
– The flow of electric current through materials is called electrical conductivity.
– Solid substances which allow electric current to flow through them are called conductors.
– Solid substances that do not allow electric current to flow through them are called non-conductors.
Role of chemistry in a society.
– Chemistry has enabled extraction of chemicals from plants.
– It is used in the manufacture of substances such as soap, glass, plastics, medicine, rubber, textiles etc from naturally occurring substances.
– Purification of substances from natural raw materials.
– It forms a basis for entry into careers e.g. teaching, medicine, chemical engineering etc.
Laboratory rules and safety symbols.
Necessity of laboratory safety rules.
The laboratory safety rules and regulations.
– Never run while in the laboratory;
Reason: You may injure others or yourself in the laboratory.
– Never taste or eat something in the laboratory;
Reason: to avoid poisoning.
– Always consult your teacher before trying out any experiment; so as to avoid accidents.
– Label all chemicals in use so as to avoid confusion.
– Always use a clean spatula for scooping a substance from a container to minimize contamination.
– Always hold test-tubes and boiling tubes using test tube holder when heating; to avoid being burned.
– When heating a substance never let the open end of the tube face yourself or anybody else, because the liquid may spurt out and cause injury.
– Never look directly into flasks and test tubes where reactions are taking place, because the chemicals may spurt into your eyes and cause injury.
– Never smell gases directly. Instead, waft the gaseous fumes near your nose with your hand.
– Experiments in which poisonous gases and vapours are produced must be carried out in a fume cupboard or an open space outdoors.
– Always keep flammable substances away from flames because they easily catch fire.
– Always report any accidents to the teacher or the laboratory technician immediately for necessary action.
– In case of an accident do not scramble for the same exit, because it may hinder easy escape.
– Always put off flames that are not in use in order to avoid accidents and minimize fuel wastage.
– If a chemical gets on your skin or mouth rinse it immediately with a lot of water.
– Always dispose off the chemicals already used safely to avoid explosions and contaminations.
– Always work on a clean bench. After completing your experiment, clean all the pieces of apparatus you have used and return them to their correct storage places.
– Always read the label of the reagents before using them.
Safety symbols.
– These are signs found on the labels of bottles or cartons containing dangerous chemicals.
– The common safety symbols are as follows:
| Symbols | Meaning. |
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Toxic: are very poisonous and can easily kill if swallowed, inhaled or on contact with the skin.
Examples: Chlorine and mercury; |
| Harmful: Less harmful (dangerous) than the toxic substances; are only likely to cause pain and discomfort.
Examples: copper (II) sulphate, lead (II) oxide |
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| Highly flammable: are substances that catch fire easily and must not be handled near open fire.
Examples: ethanol, hydrogen |
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| Oxidizing: rapidly provide oxygen and can cause fire to burn more fiercely.
Examples: potassium manganate (VII), hydrogen peroxide |
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| Corrosive: are substances that cause burns to skin and fabric; and can also react with other substances such as metals
Examples: nitric (V) acid, conc. sulphuric acid, bromine. |
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| Irritant: can cause blisters or reddening of the skin; usually irritate the respiratory tract, skin, eyes etc.
Examples: calcium chloride and zinc sulphate |
Common Chemistry laboratory apparatus and their uses.
| Name of apparatus | Diagram. | Use. |
| Test tubes |
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– General laboratory experiments; like heating solids; qualitative analysis etc |
| Boiling tubes
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– Mainly used for heating small amounts of solids and liquids. | |
| Test tube holder
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Used for holding test tubes and boiling tubes during heating experiments. | |
| Measuring cylinder
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– Measuring accurate volumes of liquids in the laboratory | |
| Beaker
– Lipped glass or plastic vessels of various capacities.
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– Glass beakers are used for boiling liquid substances;
– Holding solutions during chemistry experiments. |
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| Filtering funnel
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– Directing liquids into containers with small narrow mouths;
– Holding filter papers during filtration; |
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| Name of apparatus | Diagram | Use |
| Stirring rod
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| Watch glass
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| Thermometer
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– Measuring temperatures during experiments. | |
| Conical flask
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– Normal laboratory experiments like titration.
– May be used for measuring volumes if graduated. |
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| Round-bottomed flask
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– Used when heating liquid substances because heat is supplied uniformly. | |
| Flat-bottomed flask
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– Used for general laboratory experiments. |
| Evaporating dish
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– Used when evaporating liquids. |
| Name of apparatus | Diagram | Use |
| Crucible
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– Use when heating solid substances that require strong heating. | |
| Pestle and mortar
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– Crushing substances while the mortar holds the substances being crushed. | |
| Pie clay (ceramic) triangle
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– Supporting crucibles during heating. | |
| Tripod stand
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– Supporting beakers and flasks in which liquids are being heated.
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| Wire gauze
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– Used when glass apparatus are being heated; to facilitate even distribution of heat when heating substances in beakers or flasks | |
| Clamp
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– Supporting and holding pieces of apparatus during experiments. |
| Name of apparatus | Diagram | Use |
| Deflagrating spoon
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– Holding burning substances. | |
| Spatula
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– Scooping solid substances from containers | |
| Crucible tongs
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– Holding solid chemicals. | |
| Condenser
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| Separating funnel
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– Separating immiscible liquids. | |
| Thistle funnel
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– Delivering liquid substances into other containers like flasks during reactions. |
| Name of apparatus | Diagram | Use |
| Wash bottle
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– Holding water for rinsing apparatus |
| Dropping funnel
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| Test tube rack
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– Holding boiling tubes and test tubes. |
| Teat pipette (dropper)
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– Sucking liquid chemicals and placing them in another container dropwise. |
| Burette
– It consists of a long narrow tube with a tap and a jet at the bottom.
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– Delivering accurate volumes of liquids |
| Name of apparatus | Diagram | Use |
| Pipette
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– Delivering a specified volume of liquid accurately. | |
| Gas jar
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– Gas collection. |
| Trough
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– Holding some amount of water for some experiments e.g. gas preparation. | |
| Reagent bottles
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– Storing chemicals in liquid state. |
| Desiccator
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– Drying substances or keeping substances free from moisture. |
| Name of apparatus | Diagram | Use |
| Spirit lamp
Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments
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– Heating substances in the laboratory.
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| Kerosene stove
Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments
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– Heating substances in the laboratory. |
| Electric heater
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– Heating substances in the laboratory. |
| Candles
Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments
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– Heating substances in the laboratory. |
| Bunsen burner
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– Heating substances in the laboratory. |
| Name of apparatus | Diagram | Use |
| Stop watch (clock)
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– Measuring time particularly in determination of reaction rates.
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| Beam balance
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– To take accurate weight measurements | |
| Electronic balance |
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– Take accurate weight measurements; and can take extremely low weight measurements. |
| Volumetric flask
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Note: most chemistry apparatus are made of glass:
Advantages of glassware:
– It does not react with most chemicals
– Glass is transparent and hence reactions can easily be observed as they progress.
– Glass materials are easy to wash and rinse after experiments.
– They can be used comfortable in heating experiments.
Disadvantages:
– Have higher chances of breakages in case they fall during experiments.
– They are comparatively expensive to plastics
– Some materials like beakers may however be made of plastics.
Advantages of plastic apparatus.
– Have lower chances of breaking.
– They are relatively cheaper to buy.
Disadvantages:
– Plastics tend to react with some laboratory chemicals
– may not be transparent and hence reactions cannot easily be observed as they progress.
– Glass materials are difficult to wash and rinse after experiments.
– They cannot be used in heating experiments.
The Bunsen burner.
– is the most common heating apparatus in the laboratory.
– Was invented by a German scientist known as Wilhelm Bunsen hence the name.
– It uses natural (cooking) gas for heating.
Parts of a bunsen burner
– Chimney
– The air hole
– Collar
– The jet
– Base
– Gas inlet
Diagram: structure of a bunsen burner. Diagram: Bunsen burner-components detached
Functions of the various parts:
– It is a hollow metallic cylinder with an air hole near its lower end.
Function:
– allows air and the laboratory gas from the jet mix before they start burning at the upper end of the chimney.
– Its upper opened end provided a site where the gas burns.
– Is a small aperture found at the lower end of the chimney and smaller than the collar.
– Its diameter (size) is regulated by the collar.
Function:
– Allows air (oxygen) to enter and mix with the laboratory gas in the chimney.
– Is a metallic ring with an air hole whose diameter is the same size as that of the air hole in the chimney.
– It fits into the lower part of the chimney; and can rotate around the chimney opening or closing the air hole.
Function:
– Regulates the amount of air entering the chimney.
– It is a very tiny opening just below the air hole, that connects the gas inlet to the chimney.
Function:
– allows the laboratory gas (methane) into the chimney at high pressure.
– It is a hollow metal connected to the base and extending into the jet.
– Its external opening is usually connected to a rubber tubing that is connected to a gas tap.
Function:
– Allows laboratory gas from the cylinders (reservoirs) in the lab; through the gas taps into the Bunsen burner.
– A thick heavy metal, that is usually circular or oval.
Function:
– It supports the Bunsen burner on the bench.
The Bunsen burner flames.
– A flame is a mass of burning gases.
– A bunsen burner can produce two types of flames depending on the size of the air hole and hence amount of air entering the chimney.
Types of flames.
(a). The luminous flame.
– It is a large bright yellow flame produced when the air hole is closed and hence no air enters the chimney.
Characteristics of a luminous flame.
– It is large, quiet and bright yellow.
– Colour is not uniform and it ahs four zones.
– It produces less heat.
– It gives a lot of light to the glow of unburnt hot carbon particles
– It produces soot.
Diagram: the luminous flame.
Parts of a luminous flame.
– Is a fairly visible, narrow zone on the outer surface of the flame.
– At this point methane (lab gas) mixes with sufficient air from the outside and burns completely to carbon (IV) oxide and water.
– It is a large bright yellow zone that lies beneath the thin outer zone.
– Here, air supply is insufficient resulting to incomplete combustion of the gas.
– Consequently the gas burns producing tiny carbon particles instead of carbon (IV) oxide.
– The white hot carbon particles glow brightly and are responsible for the yellow colour and the emission of light.
– On cooling the carbon particles form soot, which blackens the bottom of the apparatus being heated.
– Is found below the yellow inner zone; and consists mainly of unburnt gases.
– Is found on the outer side of the base of the flame.
– Here, air near the flame rises rapidly due to convection currents and mixes with the burning gas.
– This makes burning more complete than in the two upper parts above it.
Advantages of the luminous flame:
– Can be used for lighting purposes; because it produces more light.
Disadvantages.
– Produces less heat hence inefficient in heating.
– Due to production of soot it blackens apparatus thus preventing better observations of experiments.
(b). The non-luminous flame.
– It is a small blue flame produced when the air hole is completely open and hence a lot of air enters the chimney.
Characteristics of a non-luminous flame.
– It is small, noisy and blue.
– Colour is uniform and it ahs three regions.
– It produces comparatively more (a lot of) heat.
– It does not produce soot, due to complete combustion hence no carbon particles remain.
– It produces less light due to lack of white-hot carbon particles.
Diagram: the luminous flame.
Parts of a non-luminous flame.
– It is a large light blue zone.
– Here, there is a lot of air coming up the chimney from the air hole and from the outside.
– The air gas mixture thus burns completely to carbon (IV) oxide and water.
– No soot formation because there are no carbon particles.
– consists of partially burnt gas-air mixture, due to insufficient air supply.
– However as the mixture rises up the pale blue region, it undergoes complete combustion due to plenty of air (from outside)
– Is located at the base of the flame.
– It consists of unburnt gas-air mixture.
Advantages of a non-luminous flame.
– Gives out a lot of heat hence very efficient in hating.
– It does not form soot hence will leave apparatus clean even after experiment (heating).
Disadvantages:
– It uses a lot of laboratory gas in burning.
– cannot be used for lighting purposes since it produces very little light.
Differences between a luminous and a non-luminous flame.
| Luminous flam | Non-luminous flame |
| Bright yellow in colour | Blue in colour |
| Produces a lot of light | Produces a lot of light. |
| Large and unsteady | Small and steady |
| Produces soot | Does not produce soot |
| Has four zones | Has three zones |
| Burns quietly | Burns noisily |
| Moderately hot | Very hot |
Experiments on Bunsen burner flames.
Apparatus:
– Bunsen burners, 250 ml beakers, lighter, stopwatch, tripod stand, wire gauze.
Procedure
– 100 cm3 of water is put into ach of the two 250 ml beakers.
– One beaker is put over a luminous flame while the other is simultaneously put over a luminous flame
– Time taken for water to boil is noted for each set up.
– The bottom of ach beaker is observed for any changes.
Apparatus
Observations.
– Water heated over the non-luminous flame boiled ion a shorter time than the same amount of water heated over a non-luminous flame.
– The bottom of the beaker heated over the non-luminous flame remained clear but the one heated over the luminous flame was covered with black deposits of soot.
Explanations.
– The non-luminous flame is hotter than the luminous flame; hence boils the water faster
– The hottest part of the luminous flam is the outer blue zone.
– Incomplete combustion in the luminous flame leads to production of carbon particles, which when hot glow yellow and on cooling forms black soot on the beaker;
– Incomplete combustion in a non-luminous flame leads to production of carbon (IV) oxide and steam only, hence no soot formation.
Conclusions.
– The non-luminous flame is hotter than the luminous flame.
– The non-luminous flame is cleaner than the luminous flame.
Requirements
– Bunsen burner, stiff white paper (cardboard), wooden splint.
Procedure
– A bunsen burner is ignited with the air hole open to get anon-luminous flame.
– A piece of white paper (cardboard) is slipped into the flame in region marked X as shown below.
– The piece of paper is removed quickly before it catches fire.
– A fresh piece of paper is then slipped into region marked Y as shown below; then again quickly removed before it catches fire.
– The experiment for each of the regions marked X and Y is then repeated using wooden splints.
– The splints should be held long enough for some of their parts to get charred
Apparatus
Observations.
– In region X, the part of the paper that was in contact with the flame was charred uniformly as shown below.
– In region Y, the part of the paper in contact with the flame had a charred ring with an unburnt part in the middle of the ring as shown below
Diagrams
– In region X, the part of the splint in contact with the flame was charred uniformly as shown below.
– In region Y, the part of the splint in contact with the flame had an unburnt part in between two charred regions as shown below.
Diagrams:
Explanations.
– Regions which become charred indicate that they are the hottest part of the flame.
– Region X corresponds to the outermost blue region of a non-luminous flame.
– Region Y is the almost colourless region of the non-luminous flame, which is however surrounded by the middle greenish blue and the outer pale blue zones.
– Thus in region X, the uniform charring of the paper and splint indicate that the outer pale blue zone is the hottest pat of the flame.
– Similarly the charred ring for experiment in region Y show that the parts in contact with the outer pale blue zone gets burnt faster before the parts in contact with the almost colourless or the greenish blue zones.
Conclusions.
– The hottest part of the non-luminous flame is the outermost pale blue zone.
– During heating the object being heated should not be placed nearer the chimney; these parts are less hot.
– For efficient heating the object being heated should be placed at the outermost region of the flame.
Apparatus:
– Bunsen burner, tongs, narrow hard glass tubing.
Procedure
– A bunsen burner is lit and adjusted to get a non-luminous flame.
– A narrow hard glass tubing is held with a pair of tongs and one of its end s is placed in the colourless zone of the flame.
– A match is lit and placed at the free end of the glass tubing.
Apparatus
Observations.
– A flame is obtained at the free end of the glass tubing.
Explanations.
– The tubing trapped unburnt gases at the almost colourless zone of the flame.
– The trapped gases combined with atmospheric air (oxygen) at the other (free) end of the tubing hence the flame.
Conclusions.
– The almost colourless region contains unburnt gases.
Apparatus:
– Bunsen burner, match stick
Procedure
– A matchstick is placed at the top of the bunsen burner chimney using a pin.
– A bunsen burner is lit and adjusted to get a non-luminous flame.
– The match stick is observed fro sometime for any changes.
– If no observable changes are made, the matchstick is then slowly raised towards the blue zone and observed keenly.
Apparatus
Observations.
– The matchstick did not ignite while it was at the bottom of the flame (resting on top of the chimney).
– It ignited as it was being raised towards the outer pale blue zone.
Explanations.
– The bottom of the flame (just on top of chimney) corresponds to the almost colourless zone.
– This zone contains unburnt gases, hence no burning occurs and is thus least hot to cause ignition of the matchstick.
– As the matchstick is raised upwards it moves past the greenish blue zone (where there is partial combustion) then to the outer pale blue zone where there is complete combustion and hence most heat.
– The heat in this region is adequate to cause ignition of the matchstick.
Conclusions.
– The outer pale blue zone is the hottest part of the non-luminous flame, and is thus the correct position to place an object during heating.
Methods of gas collection.
– Various chemical reactions produce gases; some of which are colourless while others are coloured.
– Additionally some gases are poisonous to the human body, while others are major causes of environmental pollution.
Examples:
Chlorine (green-yellow); nitrogen (IV) oxide (brown); bromine (red-brown)
Oxygen; carbon (II) oxide; carbon (IV) oxide; sulphur (IV) oxide; hydrogen; ammonia etc.
Factors affecting method used in collecting a gas.
– Density
– Solubility in water.
– Colour
– Toxicity
Summary on collection methods.
| Method | Apparatus | Characteristic of gas |
| Upward delivery
– Also called downward displacement of air |
|
– Must be less dense than air.
Examples: Hydrogen, ammonia gas. Note: being lighter the gas is supported by the denser air from below; – When used for colourless gases, it is not possible to know when the gas jar is full; |
| Downward delivery (upward displacement of air) | – Must be denser than air.
Examples: carbon (IV) oxide; nitrogen (IV) oxide; chlorine gas; Note: – The gas displaces air and settle at the bottom of the collecting vessel – Unless the gas is coloured, it is difficult to know when the container is full |
|
| Over water | – insoluble or only slightly soluble in water;
– does not react with water Examples: carbon (IV) oxide; hydrogen; carbon (II) oxide; Note: – with this method it is easy to tell when the gas jar or collecting tube is full of gas; – This method cannot be used when the gas is required dry;
|
|
| Collecting syringe
– the gas produced is collected in a syringe; |
– Mainly for poisonous gases; since the gases are confined and leakages are limited;
Note: this method allows collection of small volumes of gases; – It also allows direct measurement of volume of gas produced; |
Drying of gases.
– Is the process by which the moisture in a gas being prepared is removed prior to collection.
– This is done by passing the gas through chemicals that absorb moisture.
– Such chemicals are called drying agents.
– The drying agents should not react with the gases being dried.
Examples of drying agents.
– Anhydrous calcium chloride
– Concentrated sulphuric acid.
– Calcium oxide.
Apparatus and drying agents
Collection and drying of some gases
| Gas | Collection method | Drying agent |
| Oxygen | Over water | Concentrated sulphuric (VI) acid; anhydrous calcium chloride. |
| Hydrogen | Over water, upward delivery | Concentrated sulphuric (VI) acid; anhydrous calcium chloride |
| Nitrogen | Over water | Concentrated sulphuric (VI) acid; anhydrous calcium chloride |
| Carbon (IV) oxide | Over water, downward delivery | Concentrated sulphuric (VI) acid; anhydrous calcium chloride |
| Ammonia | Upward delivery | Calcium oxide |
Drugs and drug abuse.
Drug: is a chemical substance that alters the functioning of the body.
Types of drugs
(i). Medicinal drugs (medicines):
– Are drugs mainly used for treatment and prevention of diseases.
– Are also classified into two: over the counter drugs and prescription drugs.
– Are medicinal drugs that can be bought at a pharmacy or retail shop without written instructions from a doctor.
Examples: Mild painkillers like aspirin, panadol, paracetamol, drugs for flu etc.
– Are strong medicines which should only be taken upon a doctors instruction (prescription).
– In this prescription, the doctors give a dosage, which indicates the amount and the rate at which it should be taken.
(ii). Leisure drugs.
– Are drugs that are usually taken for pleasure.
– Are classified into two:
Drug abuse:
– Is the indiscriminate use of a drug for purposes which it is meant for; or administration of an overdose or underdose of a drug; as well as use of drugs for leisure purposes.
Note:
– The worst form of drug abuse is the taking of drugs for leisure purposes; and the most commonly abused drugs are the leisure drugs.
Effects of commonly abused drugs.
– Affects the brain and the nervous system
– Damages the liver, and is a common cause of liver cirrhosis.
– Poor health due to loss of appetite.
– Time for working is wasted in drinking and hence less productivity and even lose of jobs; which results to poverty and family disintegrations.
– Bad breath, discoloured fingers and teeth
– Cause diseases such as bronchitis and tuberculosis.
– damages the lungs and is a common cause of lung cancer due to chemicals found in the cigarettes.
– Smoking during pregnancy is a common cause of miscarriages or still births.
– It is expensive: money used for other better uses is wasted in cigarette smoking.
– Interferes with the functioning of the brain.
– Results to addiction and drug dependency.
– Some are administered directly into the blood through syringes and hence common routes of transmission of HIV/AIDS.
General effects of drug abuse on the society.
– Drug abuses spend most of their money on drugs and hence neglect their family leading to misery and societal breakdown.
– Drunk drivers cause accidents.
– People who are drunk with a drug are unreasonable and cannot make logical decisions; and hence cannot be productive at that time.
– Drug abuse has resulted into loss of morals leading to higher rates of rapes, violent crimes, murders, prostitution etc.
– Drug abuse has fueled the spread of sexually transmitted diseases and HIV/AIDS.
Note:
Drug addiction:
– Is a situation in which an individual becomes dependent on a particular drug such that he cannot function normally without it; and lack of it result to some discomfort.
UNIT 2: SIMPLE CLASSIFICATION OF SUBSTANCES
Unit Checklist:
Elements compound and mixtures.
(a). Element:
– Is a pure substance that cannot be split up into simpler substances by chemical means.
Examples: copper, hydrogen, carbon.
(b). Compound:
– A pure substance that consists of two or more elements that are chemically combined.
Examples:
| Compound | Elements in the compound |
| Calcium carbonate | Calcium, carbon and oxygen |
| Sodium chloride | Sodium and chlorine |
| Ammonium nitrate | Nitrogen, hydrogen, oxygen |
| Iron (II) sulphate | Iron, sulphur, oxygen |
(c). Mixture:
– A substance that consists of two or more elements or compounds that are not chemically combined
– Some mixtures can be naturally occurring while some are artificial.
Examples
Naturally occurring mixtures.
| Mixture | Components |
| Air | Nitrogen, oxygen, carbon (IV) oxide, water vapour, noble gases etc |
| Sea water | Water and various salts like chlorides of sodium, potassium and magnesium |
| Crude oil | A mixture of hydrocarbons like methane, petrol, bitumen, etc |
| Magadi soda | Sodium carbonate, sodium hydrogen carbonate and sodium chloride |
Artificial mixtures.
| Mixture | Components |
| Soft drinks | Water, citric acid, sugar, carbon (IV) oxide, stabilizers, sodium benzoate |
| Black ink | Blue, black, yellow dyes and solvent |
| Cement | Oxides of aluminium, iron, silicon, calcium and calcium carbonate. |
Types of mixtures:
– There are two types of mixtures;
(i). Homogenous mixtures.
– Is a mixture with a uniform composition and properties throughout its mass.
– The parts (components) of the mixture are uniformly distributed throughout the mixture
Examples:
Tea with sugar solution.
(ii). Heterogenous mixture:
Is a mixture without uniform composition throughout its mass.
Examples:
– Soil, rocks and sand mixture.
Separating mixtures.
A mixture can be separated into its various components (constituents) by appropriate physical means, depending on type of mixture.
Basic concepts:
Residue: solid that remains on the filter paper during filtration
Filtrate: liquid that passes past the filter paper during filtration
Solute: a solid that dissolves in a particular liquid
Solvent: the liquid in which a solute dissolves.
Saturated solution: a solution in which no more solute can dissolve at a particular temperature
Unsaturated solution: a solution that can take more of the solute (solute) at a particular temperature.
Miscible liquids: liquids that can mix together completely.
Immiscible liquids: liquids that cannot mix together completely.
There are various methods that can be used to separate mixtures.
These include:
– Is a method used to separate insoluble solids from liquids; a heterogenous mixture.
Procedure:
– The solid-liquid mixture is allowed to stand in a container.
– The insoluble solid settles at the bottom and the upper liquid portion poured out with care.
Apparatus.
Examples:
– Separation of sand-water mixture
– Separation of maize flour-water mixture.
Limitations (disadvantages) of decantation.
– It is not efficient as some fine suspended solids may come long with the liquid during pouring.
– Is the separation of an insoluble solid from a heterogenous mixture (liquid) using a porous filter that does not allow the solids to pass through.
– Upon filtration the undissolved solid is left on the filter paper and is called the residue.
– The liquid that passes the filter paper is called filtrate.
Examples: separation of sand from water.
(i). Procedure.
– The filter paper is folded into ¼ and opened to from a cone.
Diagram: folding a filter paper.
– It is carefully placed inside a filter funnel.
– The apparatus are then arranged as shown blow.
– The sand-water mixture is then poured into the filter paper in the filter funnel.
– The collecting liquid is directed into a conical flask.
(ii). Apparatus.
Applications of filtration.
– Filtration of domestic water.
– Extraction of medicinal substances from plants.
– Extraction of sugar from sugarcane.
– Operation of a vacuum cleaner.
– Fuel filters in automobile engines.
– Is used to separate a soluble solid from its solution.
– Such solutions are usually homogenous mixtures.
– The solid is called a solute while the liquid is called a solvent.
Example: separation of salt from salt solution.
(i). Procedure:
– The salt solution is poured in an evaporating dish.
– The set up is then arranged as in the apparatus shown below.
– The solution is boiled under steam or sand bath until all the water in the salt solution evaporates and salt crystals remain in the dish.
(ii). Apparatus.
(iii). Observations and explanations.
– Upon heating the solution, water evaporates because it has lower boiling point than the salt.
– The solution is boiled until salt crystals start appearing on a glass rid dipped into the solution.
– This shall indicate that the solution is saturated.
– The saturated solution is allowed to cool and crystallize.
– The mother liquor (liquid that remains with the crystals) is poured and the salt (solid) dried between absorbent papers.
Note:
A crystal: is a solid that consists of particles arranged in an orderly repetitive manner.
– It is advantageous to boil the solution under a steam or sand bath rather than directly.
Reason:
– The steam or sand bath prevents the mixture from splashing out (spitting) of the evaporating dish.
– It also reduces chances of the evaporating dish cracking.
Applications of evaporation:
– Extraction of soda ash from Lake Magadi.
(a). Crystallization:
– is the process of formation of crystals from a solution.
– It involves evaporation of the solution to form a concentrated solution.
Example: crystallization of potassium nitrate from its solution.
(i). Procedure:
– About 5g of powdered potassium nitrate is added to 10cm3 of water in a boiling tube.
– The solution is heated until all the solid dissolves and then allowed to cool and crystallize.
Note:
– More potassium nitrate dissolves in hot water than in cold water.
– The resultant solution is then heated until crystals start appearing; and this can be confirmed by dipping a glass rod into the solution and feeling for crystals.
– This is called a saturated solution i.e. a solution that cannot take in any more of the solute at a given temperature.
– The saturated solution is then allowed to cool and crystallize.
(ii). Observations:
– The resultant solid particles have definite shapes.
– Some are needle-like while others are flat and sharp-edged.
– These are the potassium nitrate crystals.
(b). Recrystallization:
– Is used in obtaining pure crystals from a soluble solid containing impurities.
– Involves filtration and evaporation.
Examples:
Obtaining pure copper (II) sulphate crystals from impure copper (II) sulphate.
Purification of rock salt.
Note: The process can be enhanced by suspending a small piece of pure crystal into the saturated solution.
Diagram: recrystallization of copper (II) sulphate.
Applications of crystallization.
– Separation of Trona from sodium chloride in Lake Magadi.
– Is the vapourisation of a liquid from a mixture and then condensing the vapour.
– Is used in the purification of liquids and separation of liquids from a mixture.
– It utilizes the differences in boiling points of the components of the mixture.
– Are of two types:
(i). Simple distillation.
– Is mainly used for purification of liquids containing dissolved substances.
– It is also useful in separating two miscible liquids with widely differing boiling points
Note:
Miscible liquids: Liquids that mix to from a uniform a uniform homogenous solution
– The liquid with the lower boiling point usually distills over first, and is collected.
Example: To obtain pure water from sea water.
(i). Procedure:
– Salty sea water is poured into a distillation flask.
– A few pieces of pumice or porcelain is added to the solution.
Reason:
– To increase the surface area fro condensation and evaporation.
– The solution is heated until it starts boiling, then the burner removed so that the liquid boils gently.
– The boiling goes on until the liquid (distillate) starts collecting in the beaker.
(ii). Apparatus.
(iii). Observations and explanations:
– The water boils and the resultant steam is passed through the Liebig condenser.
– As the vapour passes through the condenser, it is cooled by circulating cold water through the jacket of the condenser.
– The cold water enters through the lower bottom and leaves through the top upper part.
Reason:
– To provide more time for the cold water to condense the vapours.
– The distillate is collected in the beaker while the residue remains in the distillation flask.
Applications of simple distillation.
– Manufacture of wins and spirits.
– Desalinization of sea water to obtain fresh water.
(ii). Fractional distillation:
– Is a method used fro the separation of miscible liquids with very close boiling points.
Examples:
Ethanol and water.
– It is a modification of simple distillation in which the fractionating column is inserted on top of the distillation flask.
– All the components must be volatile at different extents in order for separation to be possible.
The fractionating column.
– Is usually an elongated (glass) tube, packed with pieces of glass beads or pieces of broken glass.
Role of glass beads.
– To increase the surface area for vapourisation of the various components of the mixture and allow the separation of the vapours to occur.
– Thus the more the glass beads in the fractionating column, the higher the efficiency of separation.
Note:
– The efficiency of the fractional distillation so s to get more pure components can also be done by:
Volatile liquids:
– Are liquids with the ability to change into vapour.
– More volatile liquids vapourize and condense faster than the less volatile liquids.
Note:
– During fractional distillation, the components of a mixture are collected at intervals, one at a time with the most volatile (lowest boiling point) coming out first.
– Each component collected in the receiver is called a fraction.
Example: separation of ethanol and water.
(i). Apparatus.
(ii). Procedure:
– Water-ethanol mixture is poured into a round-bottomed flask.
– The apparatus is then connected and set up as shown below.
Note:
– The thermometer bulb must be at the vapour outlet to the condenser.
Reason:
– For accurate determination of the vapourisation temperature for each fraction.
– The mixture is then strongly heated until the first fraction comes out of the distillation flask into the conical flask.
– Collection of the fractions should be done in a conical flask other than in a beaker.
Reason:
– To reduce the rates of evaporation of the fractions, especially the highly volatile ones (in this case ethanol)
– For this particular separation the first temperatures recorded by the thermometer should not exceed 80oC; to ensure that the first fraction is only ethanol.
(iii). Discussion.
– Ethanol boils at 78oC and water boils at 100oC.
– When the mixture is heated, ethanol and water evaporate and pass through the fractionating column which is filled with glass beads to offer a large surface area
– The large surface area encourages evaporation of ethanol and condensation of water vapour.
– Water can be seen dropping back into the distillation flask.
-Ethanol vapour passes through the condenser and warm liquid ethanol is collected in the conical flask.
Note:
– The first portion is almost pure ethanol (about 97%) and burns quietly with a blue flame.
– It also has the characteristic smell of alcohols.
Industrial applications of fractional distillation.
– Separation of air into various components in BOC gases Kenya limited.
– Separation of crude oil into paraffin, petrol, kerosene diesel and other components in the Kenya oil refinery.
– Distillation of ethanol from molasses at Muhoroni Agro-chemicals company.
– Is the process by which a solid changes directly to gaseous state upon heating.
– It is used to separate a mixture in which one of the components sublimes on heating.
Note:
Solid Gas
Solids that sublime have very weak forces of attraction between the atoms and hence are easily broken on slight heating.
Examples of solids that sublime on heating.
– Iodine; sublimes to from a purple vapour.
– Ammonium chloride; sublimes to from dense white fumes;
– Solid carbon (IV) oxide (dry ice);
– Anhydrous iron (III) chloride; sublimes to give red brown fumes.
Examples: separation of iodine from sodium chloride.
(i). apparatus:
(ii). Procedure:
– The iodine-common salt mixture is poured into a beaker and placed in a tripod stand.
– A watch glass full of cold water is placed on the beaker.
– The beaker is heated gently until some dense purple fumes are observed.
(iii). Observations:
– A purple vapour appears in the beaker.
– A dark-grey shiny solid collects on the bottom of the watch-glass.
– White solid remains in the beaker.
(iv). Explanations:
– Upon heating the mixture iodine sublimes and condenses on the cold watch glass to form a sublimate of pure iodine.
Note:
– Solid carbon (IV) oxide (dry ice) is used a s refrigerant by ice cream and soft drink vendors.
Reason:
– It sublimes on heating; as it sublimes it takes latent heat from ice cream (soft drinks) thus leaving it cold.
– It is also advantageous as it does not turn into liquid, which could be cumbersome to carry and would mess up the ice cream.
– Is the separation of coloured substances using an eluting solvent.
– It is also used to identify the components of a coloured substance.
– It involves the use of a moving liquid (eluting solvent) on a material that absorbs the solvent.
– It involves two major processes:
The tendency of a substance to dissolve in a solvent.
The tendency of a substance to stick on an adsorbent material.
Examples:
(i). Procedure:
– A filter paper is placed on the rim of an evaporating dish or a small beaker.
– A drop of the black ink is placed at the centre of the filter paper; allowed to spread out and dry.
– A drop of water (ethanol) is then added to the ink and allowed to spread.
– After complete spread of the drop, a second drop is added.
– Water drops are added continuously until the disc of coloured substances almost reaches the edge.
(ii). Observations:
Note: The dry filter paper showing the separated components of a mixture is called a chromatogram.
(iii). Explanations:
– Water is the eluting solvent since ink is soluble in it.
– The various dyes in the black ink move at different distances from the black spot hence the bands.
Reasons:
– The dyes have different solubilities in the solvent; the more soluble the dye, the further the distance it travels on the absorbent paper
– They have different rates of adsorption i.e. the tendency of the dyes to stick on the absorbent material; dyes with low rates of absorption travel far from the original spot.
Note:
Solvent front.
– Is the furthest distance reached by the eluting solvent on the filter paper.
Baseline:
– The point at which the dye to be separated is placed; i.e. it is the starting point of separation.
(i). apparatus:
(ii). Observations:
(iii). Explanations:
– The mixtures A to D have various components with varying solubilities in the solvent (ethanol)
– Mixture D is the most pure because it has only one spot.
– Mixture C is the least pure (most impure), as it has the highest number of spots indicating it is composed of so many dyes (four)
– Mixture C has the most soluble dye; its last component is the one nearest to the solvent front.
– Mixtures with similar dyes in their composition have spots at same levels; in this case A, B and D.
Applications of chromatography.
– Purification of natural products such as hormones, vitamins and natural pigments.
– Detection of food poisons e.g. in canned foods and soft drinks.
– Is the extraction of a solute from its original solvent by using a second solvent in which it has a higher solubility
Example: extraction of oil from nuts.
(i). Apparatus.
(ii). Procedure:
– Some nuts are crushed in a mortar using a pestle; to increase the surface area for solubility.
– A suitable solvent such as hexane or propanone (acetone) is added.
– The nuts are further crushed in the solvent.
– The resultant solution is decanted in an evaporating dish, and left in the sun to evaporate.
– The liquid remaining in the evaporating dish is smeared onto a clean filter paper.
(iv). Observations;
– A permanent translucent mark appears on the filter paper.
(v). Explanations:
– The nuts are crushed when in contact with the solvent to bring more of the oil in the nuts closer to the solvent.
– Upon evaporation oil is left behind because it has a higher boiling point than the solvent.
– A permanent translucent mark verifies the presence of oils.
Applications of solvent extraction.
– Used by dry-cleaners to remove dirt (grease) and stains from dry-clean-only clothes such as sweaters, suits, dresses etc.
– Is used fro separating a mixture with two or more immiscible liquids.
– Such liquids do not mix but instead form layers based on their densities.
– The heaviest liquid layer is found at the bottom of the separating funnel; while the lightest liquid is found at the top of the separating funnel.
– The liquids are drained one after the other by opening and closing the tap of the separating funnel.
Example: Separation of oil from water
Apparatus:
Beakers, separating funnel, paraffin oil, distilled water, rubber stopper.
Apparatus.
Procedure.
– The tap of the separating funnel is closed.
– Equal volumes of water and paraffin are put in a separating funnel until it is half full.
– The mouth of the funnel is closed with a stopper and the mixture shaken.
– The mixture is allowed to stand until two distinct layers are formed.
– The stopper is removed and the tap opened to allow the bottom layer to drain into the beaker.
– The tap is closed after most of the bottom layer has drained off.
– The beaker is removed and the rest of the bottom layer is drained into a separate container and discarded; to ensure that no part of the top layer (paraffin) gets into the beaker containing the bottom (water) layer.
– The other (top) layer is then drained into another beaker.
Observations:
– After the mixture has settled oil and paraffin separate into two layers.
– The first beaker contains only water; while the second beaker contains only paraffin.
Conclusion.
– Paraffin and water are immiscible.
– The top layer contains water which is denser while the top layer contains oil (paraffin) which is lighter.
Practical application:
– Extraction of useful substances from complex mixtures.
Example: Separation of iodine from sodium chloride.
******************************
– Is used to separate solid mixtures, one of which is magnetic (usually iron).
– The iron is picked with a magnet leaving the other components of the mixture behind.
Examples:
– Separation of powdered iron from iron powder-sulphur mixture.
Practical applications:
-In the extraction (mining of iron); where magnetic iron ore is separated from other materials in the crushed ore.
– Separation of scrap iron from non-magnetic materials like glass and plastics in recycling plants.
Criteria for purity.
– Is a substance that contains only one type of compound or element.
Determination of purity
(a). Solids:
– Purity of solids is determined by measuring the melting point.
– Pure solids melt sharply over a narrow temperature range.
Examples:
– Naphthalene melts at 80oC 81oC.
– Water melts at 0oC.
Effects of impurity on melting point.
– Impurities lower the melting point of a substance making it melt over a wide range of temperatures i.e. the melting point is not sharp.
Applications:
– Impurities are added to purified metal ores to lower their melting points, hence save energy and extraction costs.
(b). Liquids:
– Purity of a liquid is determined by measuring its boiling point.
– A pure liquid has a sharp boiling point.
Examples:
Pure water boils at 100oC at 1 atmospheric pressure.
Pure ethanol boils at 78oC.
Effect of impurity on boiling point.
– Impurities raise the boiling point of a liquid.
Example:
– Sea water boils at a higher temperature than pure water due to the presence of dissolved salts.
States of matter.
– Matter is anything that occupies space and has mass.
– It is composed of either pure substances or a mixture of substances.
States of matter:
Matter exists in three states:
Kinetic theory of matter:
– States that matter is made up of small particles which are in continuous random motion.
– The continuous random motion of particles in matter is called Brownian motion.
– The rate of movement of particles in matter depends on the state
(a). Solid state:
– Have closely packed particles held by strong forces of attraction.
– Particles do not move from one point to another but vibrate about a fixed position.
– Upon heating, they start to vibrate vigorously.
Reason:
– Due to increase in their kinetic energy.
(b). Liquid state:
– Particles are further apart from one another than those in the solid state
– Forces of attraction between the particles are weaker than those in the solid.
– The particles remain close, but are free to move from one position to another.
– On heating they gain energy and move rapidly.
(c). Gaseous state:
– The particles are far apart and free to move randomly in all directions.
– Consequently they lack definite shape and volume, but occupy the whole space within a container.
– Forces of attraction between the particles are very weak.
The effect of heat on substances.
– A substance can change from one state to another upon heating or cooling.
– These are physical changes and can be reversed.
– There are five processes involved in changes of state:
The processes involved in change of states of matter.
(i). Melting:
– Is a change of state from solid to liquid.
– Is the constant temperature at which the melting takes place.
Examples:
– Ice melts at 0oC.
– Sodium chloride melts at 800oC.
– During melting the energy supplied to the particles is used to weaken the forces of attraction so that particles can move about.
(ii). Vapourisation (evaporation)
– Is change of state from liquid to gas.
– Is the constant temperature at which a liquid changes from solid to a gas.
– During boiling, the energy supplied is used to break the forces of attraction in the liquid thus moving the particles far away from each other.
Examples:
Water boils at 100oC, while ethanol boils at 78oC at one atmospheric pressure.
Reason:
The forces of attraction between the water particles are stronger than those of ethanol.
(c). Condensation:
– Is the change of state from a gas to a liquid.
– Is a change due to decrease in temperature.
– When the temperature of gas is decreased, the particles lose kinetic energy to the surroundings to move slowly.
– The attractive forces become stronger, and the sample changes to a liquid.
Note:
– The temperature at which condensation occurs is the same as the boiling point.
(d). Freezing:
– Is the change from a liquid to a solid.
– It is also due to decrease in temperature.
– When a liquid is cooled, the particles lose energy and move very slowly.
– They attract one another strongly, and ultimately remain in fixed positions.
Note:
– The freezing point is the same as the melting point.
(e). Sublimation.
– Is the process whereby a solid does not melt when heated, but changes directly to the gaseous state.
Example:
– Iodine solid changes to purple vapour when heated to 70oC.
– Dry ice (solid CO2), used to cool ice cream, evaporates without leaving a liquid.
Note:
– The reverse of sublimation, whereby a gas changes directly to solid is called deposition.
Summary on changes of state.
E
Solid Liquid Gas
F
Key: A: melting; B: vapourisation; C: Freezing; D: Condensation; E: Sublimation; F: Deposition;
Experiment: Investigating changes in temperature when ice is heated.
(i). Procedure:
– A 250ml beaker is half-filled with dry ice, and the initial temperature recorded.
– The ice is heated, while stirring with a thermometer and the temperature recorded every 30 seconds.
– Heating and recording is done until the resultant water starts to boil.
– A graph of temperature against time is plotted.
(ii). Results:
| Temperatures (oC) | -10 | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 |
| Time (seconds) |
(iii). Graph:
Effect of heat on pure ice
(iv). Explanations:
– As the ice is heated the temperature rises steadily from -10oC to 0oC.
Reason:
– The heat supplied increases the kinetic energy of the ice (solid water) molecules; collisions between them hence increased temperature.
-The temperature of the ice remains constant even as heat is applied.
Reason:
– Heat supplied is used to break the forces of attraction between the water molecules in ice.
– This is the melting point hence at B-C the ice melts.
– At C, all the ice has already melted (turned to water).
– Between C and D, the temperature of the water increases as heating continues.
Reason:
– The heat supplied increases the kinetic energy of the water molecules; their rate of collision increases hence increased temperatures.
– Temperature of the water remains constant even as heat is being supplied.
– Point D-E is the boiling point i.e. 100oC.
Reason:
– The heat supplied is used to break the forces of attraction between water molecules in the liquid.
– At point E, all the liquid water has turned into vapour.
– Thus between E and F, the temperature of the vapour rises as heat is applied.
Note: Heating curve for an impure solid.
Effect of heat on impure ice
The cooling curve.
– Is a curve that shows how the temperature of a substance changes with time as it is cooled from a gas into a solid.
– It is the opposite of a heating curve.
Example: the cooling curve of water
Explanations:
– Gaseous state; temperature is declining.
– Heat loss results into decrease in kinetic energy of the gaseous molecules.
– This is the condensation point;
– The water vapour condenses to the liquid state; bonds are formed as the hest is lost.
– Temperature of the liquid water is declining;
– Heat loss results into decrease in kinetic energy of the liquid molecules.
– This is the freezing point;
– The temperature remains constant as the heat is being lost;
– The heat being lost results into bond formation; as the liquid forms a solid.
– The water is now in solid state.
– The temperature of the solid declines as heat is being lost.
Permanent and temporary changes.
– Heat causes matter to change.
– Changes due to heat can either be permanent or non-permanent (temporary).
(a). Permanent changes.
– Are also called chemical changes.
– Involves substances that are relatively less stable to heat.
Characteristics of permanent (chemical) changes.
(i). New substances are formed.
(ii). Involves considerable heat changes; energy is either given out or absorbed.
(iii). The mass of the substance changes.
Types of permanent (chemical) changes.
– Chemical changes are of two main types:
(i). Reversible permanent changes.
– Are chemical changes in which the final new products can recombine to form the original substance, under certain conditions.
Generally:
A + B C + D
Reactants Products
Examples:
(i). Apparatus:
– Test tubes, Bunsen burner, test tube holder
– Hydrated copper (II) sulphate
(ii). Apparatus:
(ii). Procedure:
Dry crystals of hydrated blue copper (II) sulphate are put in a clean dry test tube.
– The apparatus are arranged as above.
– The copper (II) sulphate is heated until no further change.
– The delivery tube is removed from the collected liquid while heating continues.
Reason:
– To avoid sucking back of the condensing liquid which would otherwise rehydrate the anhydrous copper (II) sulphate
– The test tube is allowed to cool and the remaining solid is divided into two portions.
– To one portion of the powder, add distilled water, while to the other potion add the condensed liquid.
(iii). Observations:
– A white solid/ powder remains in the test tube after heating.
– A colourless liquid condenses in the test tube dipped into the ice cold water.
– The colourless liquid turns the white solid into blue.
(iv). Explanations:
– Hydrated blue copper (II) sulphate have water of crystallization, giving it the characteristic blue colour.
– During heating, the heat energy supplied is used to drive out the water molecules (particles) out of the crystals;
– Without water, the copper (II) sulphate turns white and thus called anhydrous copper (II) sulphate;
– The water driven out of the crystals condenses in the test tube immersed in the ice cold water.
|
Equation
|
Hydrated copper (II) sulphate anhydrous copper (II) sulphate + Water
Blue White
|
Chemically:
CuSO4.5H2O(s) CuSO4(s) + 5H2O(g);
Confirmatory test for water.
Confirmatory test that the water is pure:
Note:
– The delivery tube is removed from the collecting liquid while heating is continued; to ensure that no water condenses back into the copper (II) sulphate as this would cause rehydration;
– Addition of water to the anhydrous copper (III) sulphate changes its colour from white to blue;
Conclusion:
The effect of heat on copper (II) sulphate is a reversible chemical change;
– When ammonium chloride is heated, it produces ammonia gas and hydrogen chloride gas.
– These are seen as dense white fumes.
– Reversible, when ammonia and hydrogen chloride are gases are reacted or allowed to cool, they produce ammonium chloride;
|
In summary:
|
Ammonium chloride solid ammonium chloride + hydrogen chloride;
White Dense white fumes
|
Chemically:
NH4Cl(s) NH3(g) + HCl(g);
Ammonium chloride Ammonia gas hydrogen chloride gas
|
In summary:
|
Calcium carbonate solid Calcium oxide + Carbon (IV) oxide;
(ii). Irreversible chemical change.
– Are chemical changes in which the resultant products cannot recombine to form the original substance (reactants);
– Majority of the chemical changes are irreversible;
Generally:
A + B C + D
Reactants Products
Examples:
1. Action of heat on potassium manganate (VII)
(i). Apparatus and chemicals.
– Bunsen burner, test tube, trough, wooden splint;
– Potassium managnate (VII);
(ii). Apparatus set up.
|
|
(iii). Procedure:
– 2 end-fulls of a spatula of potassium manganate (VII) are put in a hard glass test tube;
– The set up is assembled as shown above;
– The solid potassium manganate (VII) is heated, and the resultant gas collected over water;
– The resultant gas(es) is tested with a glowing splint;
(iv). Observations:
– The purple solid turns black;
– A colourless gas collects over water;
– The colourless relights a glowing splint;
(v). Explanations:
– Potassium manganate (VII), a purple solid was decomposed (splint up) on heating to yield (give potassium manganate (III) and oxygen.
– The potassium manganate (III) is the black residue;
– The colourless gas is oxygen; and relighting a glowing splint is the confirmatory test;
In summary:
Potassium manganate (VII) → Potassium manganate (III) + oxygen
Purple solid Black solid Colourless gas
Note: It is not possible for oxygen and potassium manganate (III) to recombine back to potassium manganate (VII); hence the change is irreversible;
– The blue solid decomposes to form a black solid; copper (II) oxide, red-brown fumes of nitrogen (IV) oxide and a colourless gas, oxygen;
In summary:
Copper (II) nitrate → Copper (II) oxide + Nitrogen (IV) oxide + oxygen
Blue solid Black solid Brown (red-brown) fumes Colourless gas
Note: Further examples of chemical changes
– The burning of any substance (except platinum);
– The rusting of iron;
– Addition of water to calcium oxide;
– Explosion of natural gas or hydrogen with air;
– Reacting of sodium in water;
Note: Exothermic and endothermic reactions:
(i). Exothermic reactions.
– Are reactions in which heat is released // given out to the surrounding;
– Usually the final temperature of the reaction vessel // mixture (e.g. beaker is higher than initially;
Examples:
– Freezing;
– Condensation;
– Deposition // sublimation of fumes to solid;
(ii). Endothermic reactions.
– Are chemical reactions in which heat is absorbed from the surrounding;
– The final temperature of the reaction vessel or reaction mixture is usually lower than the initial i.e. they are accompanied by a drop in temperature;
Examples.
– Melting;
– Vapourization;
– Sublimation (of solid to gas)
Summary on chemical changes.
| Reaction | Appearance of substance | Changes during reaction | New substance(s) | Type of change |
| Heating hydrated copper (II) sulphate | Blue | Blue crystals turn into a white powder; colourless liquid condenses on cooling;
|
Anhydrous copper (II) sulphate and water | Chemical |
| Heating potassium manganate (VII) | Shiny purple crystals | The purple solid turns black; evolution of a colourless gas;
|
Potassium manganate (III) and oxygen; | Chemical; |
| Heating ammonium chloride | White solid // powder; | Dense white fumed that cools to a white solid;
|
Ammonia gas and hydrogen chloride gas; | Chemical; |
| Heating lead (II) nitrate; | White solid | The white solid turns into a red solid during heating which on cooling turns yellow;
– Decrepitating sound; – Brown fumes; – colourless gas;
|
– Lead (II) oxide; nitrogen (IV) oxide and oxygen gas; | Chemical; |
| Heating lead (II) nitrate; | White solid | The white solid turns into a yellow solid during heating which on cooling turns white;
– Decrepitating sound; – Brown fumes; colourless gas; |
– Zinc (II) oxide; nitrogen (IV) oxide and oxygen gas; | Chemical; |
| Heating copper turnings | Brown turnings;
|
– Brown turnings // solid turn black; | – Copper (II) oxide; | Chemical; |
| Rusting of iron; | Grey solid | – Grey solid turns into a red brown solid;
|
– Hydrated iron (III) oxide; | Chemical; |
| Heating Copper (II) nitrate; | Blue solid | The blue solid turns into a black solid;
– Brown fumes; – colourless gas;
|
– Copper (II) oxide; nitrogen (IV) oxide and oxygen gas; | Chemical; |
| Heating copper (II) carbonate | Green solid | The green solid turns into a black solid;
– colourless gas; |
– copper (II) oxide and carbon (IV) oxide; | Chemical; |
(b). Temporary (non-permanent) changes.
– Are also called physical changes;
– They are changes that involve substances that are more stable to heat;
– On heating they do not decompose hence no new substances are formed;
Characteristics of permanent changes.
Examples:
| Solid | Original appearance | Observations during heating then cooling |
| 1. Candle wax | White sticky solid; | – The solid melts into a colourless viscous liquid; and on cooling solidifies to the original solid wax again; |
| 2. Iodine solid. | Shiny dark-grey crystals; | – The solid turns directly to purple vapour (sublimation);
– On cooling the purple iodine vapour (gas) changes directly to solid iodine (deposition);
ie. Iodine solid ═ iodine vapour; Dark grey Purple
|
| 3. Zinc oxide | White solid; | – The white solid turns yellow on heating and upon cooling changes back to the original white colour;
ie. Zinc oxide ═ Zinc oxide; White (cold) Yellow (hot)
|
| 4. Ice | White | – The solid water melts into liquid and on further heating the liquid vapourizes and turns into gas;
– On cooling the gas condenses to liquid which then freezes back into solid;
i.e Ice ═ Water ═ Gas
|
| 5. Platinum wire; | – A white glow of the metal is seen on heating, but on cooling the metal changes back to its original grey colour;
|
|
| 6. Lead (II) oxide; | Yellow | – The yellow solid turns red on heating and upon cooling changes back to the original yellow colour;
ie. Lead (II) oxide ═ Lead (II) oxide; White (cold) Yellow (hot)
|
Differences between Physical and Chemical changes.
| Physical change | Chemical change |
| 1. Produces no new kind of substance; | – Always produces a new kind of substance; |
| 2. UIs usually (generally) irreversible; | Are generally irreversible; with only few exceptions (i.e. most are irreversible); |
| 3. The mass of the substance does not change; | – The mass of the substance changes; |
| 4. No energy is given out or absorbed i.e. are not accompanied by great heat changes; | – Energy is usually given out or absorbed i.e are usually accompanied by great heat changes; |
Constituents of matter
– A detailed examination of matter reveals that it is built of very tiny units called toms;
– Presently about 115 atoms have been identified;
– The arrangement and number of atoms in a substance will result into other much larger constituents of matter;
– These are:
– Is the smallest particle of matter that can take part in a chemical reaction;
– It is the smallest particle into which an element can be divided without losing the properties of the element;
– Atoms of various elements all differ from one another;
Examples:
– Copper is made up of many copper atoms;
– Sodium element is made up of many sodium atoms;
– An element is a substance that cannot be split into anything simple by any known chemical means;
– An element consists of a single type of atom;
– There are about 155 known elements, 90 of which occur naturally.
– Elements are classified into two main groups;
– All are solids at room temperature (except mercury); and are good conductors of electricity;
– Exists as solids and gases;
– All are poor electric conductors except graphite;
Examples of elements.
(i). Metals:
– Sodium, magnesium, potassium, aluminium, lead, iron, zinc, silver, gold, tin, platinum, uranium, calcium, manganese etc.
(ii). Non-elements.
– Carbon, nitrogen, sulphur, oxygen, chlorine, fluorine, argon, neon, bromine, iodine, silicon, boron, xenon, krypton.
– Is the smallest particle of a substance that can exist independently;
– It is made when 2 or more atoms (similar or dissimilar) are chemically combined together;
– However atoms of noble/ inert gases exist as single atoms;
Note:
– Depending on number of atoms molecules can be categorized into:
(i). Monoatomic molecule;
– made up of only one atom;
Examples:
– Argon;
→
Argon atom Argon molecule;
– Neon;
– Helium;
(ii). Diatomic molecules.
– Made up of 2 similar atoms; chemically combines;
Examples
– Oxygen gas;
→
Oxygen atom Oxygen molecule;
– Nitrogen gas;
– Hydrogen gas;
– Chlorine gas;
(ii). Triatomic molecules.
– Made up of 3 similar atoms; chemically combines;
Examples
– Ozone molecule;
→
Oxygen atom Ozone molecule;
Note:
– Other molecules are also made from atoms of different elements chemically combined together;
Examples:
(i). Hydrogen chloride;
+ →
Hydrogen Chlorine Hydrogen chloride molecule;
Atom. Atom;
(ii). Water molecule;
+ + →
Hydrogen atoms oxygen atom; Water molecule;
– A compound is a pure substance consisting of two or more elements that are chemically combined.
– Compounds usually have different properties from those of its constituent elements;
– Properties of a compound are uniform throughout any given sample and from one sample to another;
Examples of compounds and their constituent elements.
| Compound | Constituent elements |
| Colourless water liquid; | Oxygen and hydrogen; |
| Green copper (II) carbonate; | Copper, carbon and oxygen; |
| White sodium nitrate; | Sodium, nitrogen and oxygen; |
| Black copper (II) oxide; | Copper and oxygen; |
| Blue copper (II) nitrate; | Copper, nitrogen and oxygen; |
| Blue copper (II) sulphate; | Copper, sulphur, oxygen and hydrogen; |
| Ammonium chloride; | Nitrogen, hydrogen and chlorine; |
Note:
– Carbonates are derivatives of (derived from or made of) carbon and oxygen;
– Nitrates are derivatives of nitrogen and oxygen;
– Sulphates are derivatives of sulphur and oxygen;
– Hydrogen carbonates are derivatives of hydrogen, carbon and oxygen;
– A substance that consists of two or more elements or compounds that are not chemically combined.
Characteristics of mixtures.
– It properties are the average of the properties of its elements;
– Its components can be separated by physical means e.g. filtration, magnetism, distillation etc.
– Its components are not necessarily if fixed positions;
– Are formed by physical means; i.e. there is usually no heat change during its formation.
Examples of mixtures:
1. Air:
– A mixture of oxygen, nitrogen, carbon (IV) oxide, water vapour, and noble gases.
– A mixture of sugar and water.
– Water, dissolved salts;
Experiment: To distinguish between an element and a compound.
(i). Apparatus.
– Watch glass, test tube, wooden splint, magnet, iron fillings, sulphur powder, dilute hydrochloric acid.
(ii). Procedure.
– Approximately 7g of iron fillings and 4g of sulphur are mixed in a test tube and the mixture strongly heated;
Observations:
– A red glow starts and spreads throughout the mixture forming a black solid.
– The black solid is iron (II) sulphide.
– The two products // substances in steps 1 and 2 are subjected to the following tests;
| Test // Analysis | Observations | |
| Iron-sulphur mixture. | Iron (II) sulphide | |
| 1. Colour: The colour of the substance is noted; | – The resultant substance is yellow-grey due to the yellow sulphur and the grey iron powder; | – The yellow-grey mixture changes to a black solid; iron (II) sulphide on heating; |
| 2. Separation: A magnet is passed over the substances separately;
– Alternatively, water was added to each substance; |
– Before heating the iron could be separated from sulphur by use of a magnet or sedimentation;
Note: These are physical methods; |
– Magnetism and sedimentation have no effect on iron (II) sulphide; |
| 3. Reaction with dilute hydrochloric acid: To each of the substances, a few drops of hydrochloric acid is added; | – Iron reacted with dilute hydrochloric acid to form a colourless gas that burns with a pop sound.
– This is hydrogen gas; – Sulphur is not affected; |
– Iron (II) sulphide reacted with hydrochloric acid to produce a colourless gas with a characteristic pungent (rotten egg) smell;
– The gas is hydrogen sulphide; |
| 4. Heat change. | No heat was produced or applied in mixing iron and sulphur; | – After heating the mixture, the formation of the new substance, iron (II) sulphide produced enough heat hence the bright red glow; |
Explanations:
– These four experiments summarize the four main differences between compounds and mixtures.
– From the results, iron and sulphur powder is a mixture; while iron (II) sulphide is a compound.
Differences between a mixture and a compound.
| Compounds | Mixtures |
| 1. Components are in fixed positions; | – Components are in any positions; |
| 2. Components can only be separated by chemical means; which require large amounts of energy; | – Components ca be separated by physical means; |
| 3. The properties are different from those of the constituent substances; | – The properties are the average of the properties of the constituent elements; |
| 4. Are formed by chemical means // methods; i.e. a new substance is formed and there is evolution of heat; | – Are formed by physical mans; no new substance is formed and there is no // negligible heat change; |
| 5. Formation involves heat changes; either liberation or absorption; | – No heat change in the formation of a mixture; |
Names and symbols of common elements.
Chemical symbols.
– Are chemical short hands, written to refer to elements.
– They are usually based on the letters of the element;
– Chemical symbols consist of one or two letters which are usually derivatives of the Latin or English name of the element;
Rules in writing chemical symbols.
Note: The abbreviations of the chemical symbols are mainly derivatives of English, Latin or German names.
Examples:
Copper is ymbolised as Cu; derived from Cuprum which is ltin;
– The symbol of each element represents one atom of that element.
Example:
– Ag represents one atom of silver;
– 2Ag represents 2 atoms of silver;
Some common elements and their symbols.
| Element | Latin // Greek // German name | Symbol |
| Carbon
Fluorine Hydrogen Iodine Nitrogen Oxygen Phosphorus Sulphur Aluminium Argon Barium Calcium Chlorine Helium Magnesium Neon Silicon Zinc Copper Iron Lead Mercury Potassium Silver Sodium Gold Tin Manganese |
–
– – – – – – – – – – – – – – – – – Cuprum Ferrum Plumbum Hydagyrum Kalium Argentum Natrium Aurum – – |
C
F N O P S Ar Ba Ca He Mg Ne Si Zn Cu Fe Pb Hg K Ag Na Au Sn Mn |
Importance of chemical names and symbols over common names.
Examples:
– Compounds whose names end in -ide; contains only two elements;
– Compounds whose names end in ate contain three elements and oe of them is oxygen;
Simple word equations.
Equations:
– Is a linear summary of a chemical reaction, showing the reactants and products.
Examples:
Copper (II) oxide + hydrogen → copper + water;
Explanations:
– Substances on the left hand side are called reactants;
– Substances on the right hand side are called products;
– The addition sign (+) on the left hand side means “reacts with”;
– The arrow (→) means to form;
– The addition sign (+) on the right hand side (products side) means “and”.
Conclusion:
Copper (II) oxide reacts with hydrogen to form copper and water;
Note:
– Some chemical reactions are reversible and hence have two opposite arrows ( ) between reactants and products
– The arrows ( ) in chemistry means a reversible chemical reaction;
Further examples:
1. Copper (II) carbonate → Copper (II) oxide + carbon (IV) oxide;
UNIT 3: ACIDS BASES AND INDICATORS.
Unit checklist
– Are substances that dissolve in water to release hydrogen ions.
– Acids can either be organic acids or mineral acids;
(i). Organic acids:
– Are acids found in plants and animals;
Examples:
– lactic acid in sour milk;
– Citric acid in citrus fruits like oranges;
– Ethanoic acid in vinegar;
– Tartaric acid in baking powder;
– Methanoic acid in bee and ant stings;
– Tannic acid in tea;
(ii). Mineral acids.
– Are acids made from minerals containing elements such as sulphur, chlorine, nitrogen etc.
– Are formed from reactions of chemicals;
– Main examples include:
Note:
– Mineral acids are more powerful than organic acids; because they yield // release more hydrogen ions in water
– They are thus more corrosive.
– Are substances that dissolve in water to yield // release hydroxyl ions;
– Just like acids they are bitter to taste;
Examples:
– Sodium hydroxide;
– Ammonium hydroxide;
– Calcium hydroxide;
Note:
– Some bases insoluble in water while some are soluble in water;
– Soluble bases are called alkalis;
– Are substances which give definite colours in acidic or basic solutions;
– Are substances which can be used to determine whether a substance is an acid or a base;
– Consequently they are called acid-base indicators;
– The determination is based on colour changes, where each indicator have particular colourations in acids and bases.
– Indicators can be commercially or locally prepared in the laboratory;
Indicators:
Experiment: preparation of simple acid-base indicators from flower extracts.
(i). Apparatus and chemicals.
– Test tubes;
– Pestle and mortar;
– Flower petals;
– Ethanol // propanone;
– Water;
– Various test solutions: sulphuric (VI) acid, hydrochloric acid, Ethanoic acid, sodium hydroxide, magadi soda, ammonia solution.
(ii). Procedure:
– Flowers from selected plants are collected and assembled e.g. bougainvillea, hibiscus etc;
– They are crushed in a mortar using a pestle and some ethanol added with continued crushing;
– The resultant liquid is decanted into a small beaker; and its colour recorded.
– Using a dropper, two to three drops of the resultant indicator are added to the test solutions.
(iii). Observations:
(a). Colour of extract in acids and bases
| Plant extract | Colour in hydrochloric acid | Colour in dilute sodium hydroxide |
| 1 | ||
| 2 |
(b). Result with various test solutions:
| Test substance | Colour (change) | Type of substance (acid/base) |
| Lemon juice
Wood ash Ammonia Sour milk Vinegar Nitric (V) acid Toothpaste Lime water Baking powder; Sugar Potassium hydroxide |
Note:
– Plant extracts acid-base indicators are not normally preferred in Chemistry experiments.
Reason:
– They dont give consistent (reproducible results because they are impure.
– Commercial indicators give more distinctive and reproducible results.
Commercial indicators.
– Are commercially prepared indicators which are sold in already purified forms.
Advantages of commercial indicators.
– They are relatively pure hence give consistent and reproducible results;
– They are readily available and easy to store in a Chemistry laboratory;
Main examples:
– Phenolpthalein;
– Methyl orange;
– Bromothymol blue;
– Litmus paper;
Note:
– Litmus is a blue vegetable compound which is extracted from plants called lichens;
– Litmus paper is an adsorbent paper which has been dipped in litmus indicator solution then dried;
Colours of various commercial indicators in acids and bases.
| Indicator | Colour in. | ||
| Neutral | Base | Acid | |
| 1. Litmus;
2. Phenolphthalein; 3. Methyl orange; 4. Bromothymol blue |
Purple
Colourless; Orange Blue |
Blue;
Pink; Yellow Blue |
Red;
Colourless; Pink; Yellow; |
Classification of various substances as acids or bases using indicators.
| Substance | Colour in | Classification | |||
| Litmus | Phenolphthalein | Methyl orange | Bromothymol blue | ||
| Hydrochloric acid
Sodium hydroxide Omo (detergent) Soda Actal tablets Lemon juice Sour milk Bleach (jik) Fresh milk Wood ash |
|||||
The universal indicator.
– Is a full range indicator which gives range of colours depending on the strength of the acid or alkali.
– It is prepared by suitable mixing certain indicators;
– It gives a range of colour depending on the strength of acids and bases;
– Each universal indicator is supplied with a chart, to facilitate this classification.
The pH scale.
– Is a scale of numbers which shows the strength of acids or bases.
– It refers to the potential (power) of hydrogen;
– It ranges from 0 14;
– To determine the strength of an acid or base, the colour it gives in universal indicators solution is compared to the shades on the pH chart of the indicator;
Diagram: The pH scale.
Note:
– The strongest acid has a pH of 1;
– The strongest alkali has a pH of 14;
– Neutral substances have a pH of 7;
– Any pH less than 7 is acidic solution; while any pH above 7 is for a alkaline / basic solution;
Colour and pH of various solutions in universal indicator;
| Substance | Colour | pH on chart | Classification |
| Hydrochloric acid
Sodium hydroxide Omo (detergent) Soda Actal tablets Lemon juice Sour milk Bleach (jik) Fresh milk Wood ash |
Properties of acids.
(a). Physical properties.
Examples:
– The sour taste of citric fruits is due to the citric acid in them.
– The sour taste in sour milk is due to lactic acid;
– Red litmus will remain red in acidic solution; blue litmus will turn red;
– Acids conduct electric current when dissolved in water;
– This is because they dissolve in water to release hydrogen ions; which are the ones that conduct electric current;
Chemical properties.
– Acids react with alkalis to form salt and water only;
– These types of reactions are called neutralization reactions;
– The hydrogen ions of the acid react with the hydroxyl ion of the alkali to form water;
– The name of the salt is usually derived from the acid;
Examples:
| Acid | Derivative salt |
| Sulphuric (VI) acid
Hydrochloric acid Nitric (V) acid Phosphoric acid |
Sulphates;
Chlorides Nitrates; Phosphates; |
Summary:
Acid + base (alkali) → salt + water; (a neutralization reaction);
Examples:
1. Sodium hydroxide + Hydrochloric acid → Sodium chloride + water;
– Acids react with some metals to produce hydrogen;
Examples: Reaction with dilute hydrochloric acid and zinc metal;
Procedure:
– 2 cm3 of hydrochloric acid is put in a test tube;
– A spatula end-full of zinc powder is added.
– A burning splint is lowered in the test tube.
Observations.
– Effervescence of a colourless gas;
– The colourless gas burns with a pop sound;
Explanations.
– Zinc metal displaces the hydrogen ions in the acid which form the hydrogen gas;
– When a glowing splint is introduced into the hydrogen gas; it burns with a pop sound;
– This is the chemical test to confirm that a gas is hydrogen;
Conclusion;
– The gas produced is hydrogen gas;
– Thus, acids react with some metals to produce hydrogen gas, and a salt;
General equation:
Metal + Dilute acid → salt + Hydrogen gas;
Reaction equation:
Zinc + Hydrochloric acid → Zinc chloride + Hydrogen chloride;
Further examples:
i. Magnesium + Dilute sulphuric (VI) acid → magnesium sulphate + hydrogen gas;
– Metal carbonates and hydrogen carbonates react with acids to form carbon (IV) oxide, water and a salt;
General equation:
Metal carbonate + Dilute acid → A salt + water + carbon (IV) oxide;
Metal hydrogen carbonate + Dilute acid → A salt + water + carbon (IV) oxide;
Examples: Reaction of sodium carbonate with dilute hydrochloric acid.
Procedure:
– About 2 cm3 of dilute hydrochloric acid is put in a test tube;
– A spatula end-full of sodium carbonate powder is then added;
– A burning splint is carefully lowered into the test tube.
Apparatus.
Observations.
– An effervescence occurs (bubbles); and a colourless gas is produced;
– The colourless gas does not relight a glowing splint; showing that it is carbon (IV) oxide;
Properties of bases.
Note:
– Bases are substances that release hydroxyl ions when added to water;
– Soluble bases are called alkalis;
Examples:
Sodium hydroxide + Water → Sodium ions + hydroxyl ions;
(a). Physical properties of bases
(b). Chemical properties.
– They react with acids to form a salt and water as the only products;
– This is a neutralization reaction; and is used to cure indigestion;
Example:
– Actal tablets contain a base that neutralizes the stomach acid.
– Addition of some alkalis to salt solutions results in formation of solids;
– Most of these are normally hydroxides;
– A solid that is formed when two solutions are mixed is called a precipitate;
Example:
– Copper (II) sulphate + Sodium hydroxide → Copper (II) hydroxide + Sodium sulphate solution;
Blue solid;
– Most metal hydroxides are decomposed by heat to form their oxides and water;
|
General equation:
Metal hydroxide Metal oxide + Water;
Example:
Zinc hydroxide → Zinc oxide + water;
Applications of acids and bases.
Application of acids
Examples:
– Wasp and bee stings can be treated by applying vinegar (Ethanoic acid) or lemon juice;
– These acidic substances neutralize the alkaline insect stings;
Uses of bases.
UNIT 4: AIR AND COMBUSTION.
Checklist.
Introduction:
– Air is a gaseous mixture constituted of several gases, water vapour and pollutants.
Combustion:
– Is the burning of substances, usually in presence of air // oxygen;
– During combustion only the oxygen component of air is used; .e the active part of air.
Percentage composition of air.
| Component | Percentage volume. |
| Nitrogen | 78% |
| Oxygen | 21% |
| Carbon (IV) oxide | 0.03% |
| Noble gases (argon) | About 1% |
| Water vapour | Variable |
| Smoke/dust particles | Variable; |
| Others | Trace |
Note:
– From the noble gases argon is the most abundant, constituting about 0.93% of the entire 1%
Oxygen and combustion.
– When substances burn in air they consume oxygen.
– Thus the process of combustion utilizes mainly oxygen;
– The reactions in combustion are normally exothermic (give out heat) and often involve flame.
Note:
Combustion in which a flame is used is called burning;
– In combustion if all the oxygen in a given volume of air is used, the final volume of air reduces by about 21.0%;
– Since oxygen is the only constituent of air participating in combustion its termed the active part of air.
Experiments: Determination of the active part of air.
Apparatus and requirements.
– Candle;
– Cork / evaporating dish;
– Sodium hydroxide solution;
Procedure:
– A candle about 3cm long is put on a wide cork/ evaporating dish;
– It is then floated in a dilute solution of sodium hydroxide solution just above the beehive shelf;
– It is carefully covered with a dry 100cm3 measuring cylinder, during which the level of solution in the cylinder is noted and marked;
– The measuring cylinder is removed and the candle lit;
– The lighting candle is then covered with a measuring cylinder;
– The experiment is allowed to proceed until the candle goes off;
Observations:
– The candle went off after sometime;
– The sodium hydroxide level inside the gas jar rises;
– The sodium hydroxide level in the trough goes down;
Diagrams:
Explanations:
– The candle wax is made up of hydrogen and carbon, hence called a hydrocarbon;
– During burning it melts in air consuming oxygen and producing carbon (IV) oxide and water vapour;
– The water vapour condenses giving a negligible volume of water;
– The resultant carbon (IV) oxide is absorbed by the sodium hydroxide;
– Absorption of carbon (IV) oxide in the gas jar creates a partial vacuum within it;
– The sodium hydroxide in the trough rises to fill the resultant space; and hence a drop in the sodium hydroxide level in the tough;
Conclusion.
– Oxygen is the active part of air that is utilized during burning;
– Air is basically made up of 2 parts; an active part that supports burning and an inactive part that does not support burning;
Apparatus and chemicals.
– Tough;
– Beehive;
– Candle and gas jar;
– A 30ml ruler;
– Sodium hydroxide solution;
Procedure:
– The entire apparatus is arranged as shown below;
– An empty gas jar is inverted over the candle before lighting it;
– The initial height A, is measured and recorded;
– The gas jar is then removed; the candle lit and covered with the gas jar again;
– The set up is allowed to run till the candle extinguishes (goes off); and the final height (B) of the air column measured.
Diagrams
Calculations:
– Amount of air in the gas jar initially = A cm3;
– Final amount of air remaining after burning; = Bcm3;
– Amount of oxygen used; = (A B)
Thus;
Percentage of oxygen in air: = (A B) x 10
A
= C%
Sample data:
Volume of air in the gas jar before burning =
Volume of air in the gas jar after burning =
Volume of air used during burning =
Percentage of air (by volume) used up = Volume used in burning x 100
Original (initial) volume
Substituting:
Conclusion:
– When candle burns in air, about 20% of air, which is oxygen used up;
Apparatus:
– Gas jar;
– Trough;
– Beehive;
– Iron fillings//powder
Procedure:
– The gas jar is divided into five equal portions by marking around it using a waterproof marker;
– The gas jar is wet near the bottom and some iron fillings sprinkled on it;
– Some water is put in a trough and the jar with iron fillings // wool // powder inverted over it;
– The initial colour of iron fillings is noted;
– The set up is left undisturbed for a few days until the water shows no further change in rising;
Diagrams of apparatus set up:
Observations:
– The iron fillings change colour from a grey to form a brown solid;
– Water level in the gas jar rises // increases until the first mark in the gas jar;
– The water level in the trough decreases;
Explanations:
– The gas jar is moistened to make the iron fillings stick onto its surface so that the fillings do not fall in the water when the gas jar is inverted;
– The brown substance formed is called rust and its chemical name is hydrated iron (III) oxide;
– During rusting, oxygen is utilized, thus creating a partial vacuum in the gas jar;
– This causes the water level in the gas jar to rise up and the water level in the trough to go down;
– The rise in water level is equivalent to about 1/5 of the original air volume, which translates to about 20%;
Conclusion:
– When rusting occurs about 20% of air, which is oxygen, is used up;
Apparatus and chemicals.
– Two 100cm3 syringes labeled Y and Y;
– Hard glass test tube;
– Glass wool;
– Bunsen burner;
– Copper turnings;
Procedure:
– A small amount of copper turnings is put in a hard glass tube and glass wool put at both ends of the tube;
– All the air in the syringe Y is removed by pushing the plunger inside, upon which the syringe is tightly fixed at one end of the tube// glass tube;
– The plunger of syringe Z is pulled out to the 100cm3 mark; to fill it with air.
The apparatus is arranged as below.
Diagram:
– The tube containing copper turnings is strongly heated;
– Air is then passed over the hot copper turnings by slowly pushing the plunger Z to and fro for several times.
Reason:
– Ensure complete reaction between the hot copper turnings and oxygen (air);
– When no further air change in volume of air in the syringe occurs, the apparatus is allowed to cool;
– The volume of air left in syringe Z is recorded;
Observations:
– The brown solid (copper) turns into a black solid (copper (II) oxide);
– The plunger of syringe Z moves inwards to approximately 80 cm3 mark;
Explanations:
– The heated copper reacted with oxygen in air to form black copper (II) oxide;
– The percentage of oxygen that was in the air is approximately 20%, causing the plunger to move inwards to the 80 cm3 mark;
Equation:
Copper + Oxygen → Copper (II) oxide;
(Brown) (Colourless) (Black)
2Cu(s) + O2(g) → 2CuO(s)
Brown (Colourless) (Black)
Conclusion:
– Burning of copper in air utilizes oxygen and produces black copper (II) oxide.
Sample results and calculations:
– Initial air volume in syringe Z = cm3;
– Final air volume in syringe Z = cm3;
– Volume of air used = (100 80) = 20 cm3;
Percentage of oxygen in air = 20 x 100 = 20%;
100
– When the same set up is used to investigate the percentage of air used up in combustion of magnesium the volume of air used up is relatively higher than the 20%.
Reason:
Magnesium produces a lot of heat during combustion and thus reacts with both oxygen and nitrogen to form two products; magnesium oxide and magnesium nitride respectively;
Observations:
– Magnesium glows giving a bright blinding flame;
– Formation of a mixture of two white powders.
Equations:
Reaction with oxygen:
Magnesium + Oxygen → Magnesium oxide;
(Grey) (Colourless) (White)
2Mg(s) + O2(g) → 2MgO(s)
(Grey) (Colourless) (White)
Reaction with nitrogen:
Magnesium + Nitrogen → Magnesium nitride;
(Grey) (Colourless) (White)
3Mg(s) + N2(g) → Mg3N2(s)
(Grey) (Colourless) (White)
Note:
– Sodium metal will also react with both oxygen and nitrogen during combustion; forming sodium oxide and sodium nitride respectively;
Apparatus and requirements.
– Graduated measuring cylinder;
– Water tough // pneumatic trough;
– Copper wire;
– White phosphorus;
Procedure:
– An empty measuring cylinder is inverted in a water trough and the water level noted;
– A small piece of white phosphorus is attached to the end of a piece of copper wire then put // inserted into the inverted measuring cylinder ensuring it is above the water;
– The set up is left undisturbed for 24 hours;
Precaution:
– Avoid contact with the phosphorus;
– Avoid inhalation of the fumes;
Observations:
– White fumes inside the cylinder at the start of the experiment;
– After 24 hours:
– water level inside the measuring cylinder rises;
– Water level in the trough drops;
Explanations:
– Yellow or white phosphorus smoulders in air; due to the fact that phosphorus reacts with oxygen to form phosphorus oxides;
-The phosphorus oxides are the white fumes;
– The phosphorus oxides then dissolves in water; forming acidic solutions of phosphoric acids;
– The water level rises inside the cylinder to occupy the volume of oxygen used up in reaction with phosphorus;
Equations:
Phosphorus + Oxygen → Phosphorus (V) oxide
White // yellow Colourless White fumes
P4(s) + 5O2(g) → 2P2O5(g);
White Colourless White fumes
Yellow
Phosphorus + Oxygen → Phosphorus (III) oxide
White // yellow Colourless White fumes
P4(s) + 3O2(g) → 2P2O3(g);
White Colourless White fumes
Yellow
Conclusion:
– Phosphorus smolders easily in air, reacting with oxygen (active part of air) to form phosphorus (III) or phosphorus (V) oxide;
– For this reason phosphorus is stored under water; to prevent it from reacting with atmospheric oxygen;
Note:
– This reaction can be made much faster by heating the copper wire; which will transmit heat to the piece of phosphorus at the tip, causing rapid burning of phosphorus to give dense white fumes of phosphorus (V) oxide // phosphorus (III) oxides;
Test for evidence of some components of air.
(i). Formation of dew;
(ii). When white anhydrous copper (II) sulphate is left in the open overnight; it forms a blue solid of hydrated copper (II) sulphate;
Reason:
– The white anhydrous copper (II) sulphate absorbs atmospheric water vapour;
– Upon hydration the copper (II) sulphate turns blue;
(iii). Sodium hydroxide pellets form a colourless solution when left in the open air overnight.
Reason:
– The sodium hydroxide pellets absorbs atmospheric water vapour and dissolves in it forming sodium hydroxide solution;
(iv). When air is passed through anhydrous calcium hydroxide solid in a U-tube for sometime; there is formation of a colourless solution in the U-tube.
Apparatus:
Reason:
– The anhydrous calcium chloride absorbs atmospheric water vapour forming a colourless solution of calcium chloride;
Equation:
Calcium chloride + water → calcium chloride solution;
White Colourless solution;
Note:
– Substances that absorb moisture from the air to form a colourless solution are called deliquescent substances.
– Other examples of deliquescent substances include: iron (III) chloride, magnesium chloride and zinc chloride;
(i). Glass stoppers of reagent bottles containing sodium hydroxide solution tend to stick when left on for sometime in the laboratory;
Reason:
– The sodium hydroxide solution at the edges of the stopper is exposed to air; and thus reacts with atmospheric carbon (IV) oxide forming white sodium carbonate solid.
Equation:
Sodium hydroxide + Carbon (IV) oxide → Sodium carbonate + Water;
2NaOH(aq) + CO2(g) → Na2CO3(s) + H2O(l);
(ii). Bubbling atmospheric air through lime water (calcium hydroxide) to form a white insoluble salt of calcium carbonate.
Apparatus:
Reason:
– Atmospheric carbon (IV) oxide reacts with limewater (calcium hydroxide) to form a white insoluble precipitate of calcium carbonate salt;
Equation:
Calcium hydroxide + Carbon (IV) oxide → Calcium carbonate + Water;
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l);
Note:
– When the air is bubbled on even after the formation of the white precipitate; the white precipitate dissolves after sometime to form a colourless solution;
Reason:
– Excess carbon (IV) oxide reacts with the calcium carbonate to form soluble calcium hydrogen carbonate.
Equation:
Calcium carbonate + Water + Carbon (IV) oxide → Calcium hydroxide solution;
CaCO3(s) + H2O(g) + CO2(g) → Ca(HCO3)2(aq);
Fractional distillation of Liquefied air.
– Air is a mixture of gases;
– It can be separated into its constituents by fractional distillation of liquid air.
– During the process air is passed through a series of steps during which it is purified, some components eliminated then it is compressed into liquid prior to fractional distillation.
– The process can be divided into two main stages;
Purification and liquefaction;
Fractional distillation of air;
(a). Purification and liquefaction.
Step 1: Purification:
– The air is purified by removal of dust particles ;
– This is done through the following ways:
Step 2: Removal of carbon (IV) oxide.
– The dust-free air is passed through a chamber containing calcium hydroxide solution;
– The sodium hydroxide solution dissolves the carbon (IV) oxide present in the air;
– During the reaction, sodium carbonate and water are formed;
– Over a prolonged time; the sodium carbonate absorbs more (excess) carbon (IV) oxide forming sodium hydrogen carbonate;
Equations:
Sodium hydroxide + carbon (IV) oxide → Sodium carbonate + Water;
2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l);
In excess;
Sodium carbonate + Water + Carbon (IV) oxide → Sodium hydrogen carbonate;
Na2CO3(aq) + H2O(l) + CO2(g) → 2NaHCO3(aq);
Step 3: Removal of water vapour;
– The dust-free, CO2 – free air is then cooled to -25oC;
– This process solidifies the water vapour out as ice;
– This cooling process may be done at temperatures a s low as -80; so as to solidify any carbon (IV) oxide (freezing point -78oC) that may have escaped absorption by the sodium hydroxide;
– The removal of water vapour and carbon (IV) oxide are important because it prevents blockage of the pipes in the rest of the system;
Step 4: Liquefaction of air;
– The dry, dust-free and carbon (IV) oxide-free air is compressed to about 100 atmospheres of pressure; causing it to warm;
– The compressed air is cooled by refrigeration;
– The cold compressed air is made to expand rapidly by passage through a nozzle which cools it further;
– The repeated compression, cooling and expansion of air causes it to liquefy at about -200oC:
Note:
At this temperature only neon and helium whose boiling points re lower than -200oC remain in gaseous states;
(b). Fractional distillation;
– The liquid air now consists only of nitrogen, oxygen and noble gases (especially argon);
– The liquid air is fed at the bottom of a fractionating column;
– It is warmed to a temperature of -192oC;
-Nitrogen distils over fast at -196oC because it has a lower boiling point; and is collected at the top of the fractionating column;
Note:
– Any vapours of oxygen and argon which rise together with nitrogen vapour condense in the column and fall back as liquids;
– The nitrogen collected is 99% pure;
The small amounts of impurities include neon and helium;
– The liquids remaining at the bottom of the fractionating column after vaporization of all nitrogen is mainly oxygen and argon; with traces of krypton and xenon;
– The liquid is again warmed further to a temperature of -185oC; causing the vapourization of argon whose boiling point is -186oC;
– This is collected as a gas at the top of the fractionating column;
– The residue liquid is mainly oxygen with minute quantities of krypton and xenon which have even high boiling points;
– The oxygen is drained off and stored as pressurized oxygen in steel cylinders;
Uses of the products:
Oxygen;
– Used in hospitals with patients with breathing difficulties;
– It is used by mountain climbers and deep-sea divers for breathing;
– It is used to burn fuels;
– It is combined with acetylene to form oxy-acetylene flame which is used in welding;
– During steel making oxygen is used to remove carbon impurities;
Nitrogen:
– Manufacture of ammonia;
– Used in light bulbs; because of its inert nature it dies not react with the filament;
– As a refrigerant e.g. storage of semen for artificial insemination;
Rusting.
– Is the corrosion of iron in presence of oxygen and moisture to form brown hydrated iron (III) oxide;
– The chemical name rust is therefore hydrated iron (III) oxide with the formula Fe2O3.2H2O;
– Rust itself is a brown porous substance;
Disadvantage of rusting:
– It weakens the structure of the metal (iron) and hence eventually destroys them.
Experiment: To show the conditions necessary for rusting.
| Experiment | Procedure | Observation | Explanation |
| 1 | – Two clean iron nails are put inside the test tube;
– 10 cm3 of tap water are then added; – Examine for two days;
|
– Iron nails turn brown implying there is rusting; | – There is presence of both oxygen and water; |
| 2 | – Two clean iron nails are added to the test tube;
– 10 cm3 of boiled hot water is added followed by about 3 cm3 of oil; – Examine for two days;
|
– No rusting occurs; | – There is water but no oxygen so no rusting occur;
– Boiling the water removes any dissolved oxygen; – Addition of the oil on top prevents dissolution // entry of any air containing oxygen into the water; |
| 3 | – Two clean iron nails are added to the test tube;
– Push a piece of cotton wool half way the test tube; – Place some anhydrous calcium chloride on it and cork the tube tightly; – Examine for two days;
|
– No rusting occurs; | – There is no air // oxygen but no water;
– Anhydrous calcium chloride absorbs any moisture form the air in the test tube; – Corking the tube tightly prevents more moisture from the atmosphere from getting into the tube as the calcium chloride may get saturated and allow moisture into the nails; |
| 4 | – Two clean iron nails are added to the test tube;
– Examine for two days;
|
– Some little rusting occurs; | – Air contains and oxygen and some moisture that will facilitate rusting; |
| 5 | – Two clean iron nails are added to the test tube;
– Add salty water; – Examine for two days; |
Rusting occurs; and at a faster rate than the rest;
|
– Rusting occurs due to presence of both water and oxygen in the salty water;
– Rusting is faster because the salty water contains ions which gain electrons hence facilitate faster oxidation of iron; |
Summary diagrams
Further explanations:
– During rusting the first step is the oxidation of iron b7y xygen (in the air) to form anhydrous iron (III) oxide;
Equation:
Iron + Oxygen → Iron (III) oxide;
4Fe(s) + 3O2(g) → 2Fe2O3(s);
– The anhydrous ion (III) oxide then undergoes hydration with water to form brown hydrated iron (III) oxide;
Equation:
Anhydrous iron (III) oxide + Water → Hydrated iron (III) oxide.
Black Brown;
– Rusting occurs faster in salty conditions;
Reason:
– The initial step is the oxidation of iron, from iron (II) ions (Fe2+) to iron (III) ions (Fe3+);
– During oxidation iron (II) ions give out electrons to undergo oxidation and form iron (III) ions;
– Salty water contains several dissolved salts whose ions easily accept electrons from the iron (II) ions and thus accelerating the oxidation of iron and hence rusting;
Prevention of rusting.
Note:
– Rusting destroys materials; equipment and roofs made of iron;
– Rust is porous and thus allows air and water to reach the iron beneath.
– Thus if not removed iron will continue corroding until it is all eaten up.
Methods of preventing rusting.
– Is the coating of iron with a small layer of zinc;
– Can be done by either dipping the iron object in molten zinc, spraying with a spray of molten zinc, or by electroplating (electrolytic deposition);
– On exposure to air the zinc acquires an inert layer of zinc oxide that is impervious to both air and water;
– The iron beneath is thus prevented from air and water and thus rusting.
Note:
– The iron is protected even if the zinc coating is scratched.
Reason:
– Upon scratching both the iron and zinc get into contact with air and water;
– Since zinc is more reactive than iron, air and water reacts with zinc at the expense (instead) of iron;
– This is done through the process of electrolysis where floe of electric current causes the less reactive metal to coat the metal being protected from rusting;
Example:
– Most tin cans are in fact made yup of steel coated with a thin layer of tin.
– Other than being non-toxic tin is unreactive and rarely reacts with the contents of the can or air;
Note:
– Unlike in galvanizing, when an electroplated material gets scratched, the metal underneath (iron) rusts, and very fast;
Reason:
– Both iron and the electroplating metal (tin) are exposed to air and water;
– Since iron is more reactive than tin (the less reactive electroplating material) it reacts with oxygen and water in preference to tin;
– This explain why galvanization is more durable than electroplating;
– Other less reactive metals that are used to coat iron objects include chromium, silver and gold;
– Some electroplating metals such as gold and silver also increase the aesthetic value of the electroplated object;
– Blocks of a more reactive metal such as zinc or magnesium are attached to the iron structure;
– The more reactive metal will be corroded in preference to iron;
– To keep the iron structure from rusting, the block of reactive metal has to be replaced regularly;
– This metal is used for the protection of underground water pipes as well as ship hulls;
– The blocks of reactive metal are either attached directly to the iron structure or connected to it by a wire.
Diagrams:
– The paint coats the metal surface and thus prevents contact with air and water hence no rusting;
– However if the paint is scratched, rusting occurs quickly;
– It is used mainly in ion railings, gates, bridges, roofs, ships ad cars;
– Alloys are mixtures of two or more metals;
– Thus to prevent iron from rusting it may be mixed with one o more metals resulting into a substance that does not rust;
Example:
– Stainless steel is an alloy of iron with chromium, nickel and manganese and it resistant to rusting.
– Oil is used in moving engine parts while grease is used I other movable metal joints;
– The oil // grease forms a barrier that prevents water and air from coming into contact with the metal surface and hence preventing rusting;
– Oiling and greasing are unique in the sense that they are the only methods that can be used to prevent rusting in movable car parts;
Oxygen.
– A very important constituent of air;
– Lavoisier (1743 1794), A French Chemist showed that it is the component of air used in respiration and also in burning fuels;
– It is the most abundant of all elements; occurring both freely as well as in combination with other elements;
– Freely it constitutes about 21% by volume of atmospheric air;
Laboratory preparation of oxygen gas.
Apparatus.
– Zinc; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, hydrogen peroxide (20% by volume), manganese (IV) oxide.
Diagram;
Procedure.
– Some manganese (IV) oxide is placed into a flat-bottomed flask;
– The apparatus is set up as shown in the diagram above;
– Add hydrogen peroxide from a thistle funnel into the flask dropwise;
– The gas is collected as shown;
Observations:
– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;
– The colourless gas collects on top of the water;
Explanations:
– Hydrogen peroxide decomposes slowly to oxygen and water under normal conditions;
– This process is however slow to collect enough volumes of oxygen;
– On addition of manganese (IV) oxide the decomposition is speeded up;
– Thus manganese (IV) oxide speeds up the decomposition of hydrogen peroxide and thus acts a s a catalyst;
Equation:
Hydrogen peroxide → Water + oxygen;
2H2O2(l) → 2H2O(l) + O2(g) (slow process)
|
Hydrogen peroxide Water + oxygen;
2H2O2(l) → 2H2O(l) + O2(g) (faster process)
Note:
The first few bubbles of oxygen gas are not collected.
Reason: The gas is mixed with air which was originally in air and hence impure.
Method of collection;
– Over water collection.
Reason:
– It is insoluble in water and less dense than water
Physical properties of oxygen gas.
– It is colourless;
– It is odourless;
– Has a low boiling point of about -183oC;
– Almost insoluble in water (hence collected over water);
Chemical test for oxygen gas.
– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;
Drying of oxygen gas.
– The resultant oxygen is usually moist due to the fact that it is collected over water;
– If required dry the gas ca be died using either of the two methods:
(i). Using sulphuric (VI) acid.
– Bubbling the gas through a wash bottle containing concentrated sulphuric (VI) acid;
– The concentrated sulphuric (VI) acid absorbs moisture from the gas leaving it dry;
– The dry gas is then draw into collection syringe;
Diagram;
(ii). Using anhydrous calcium chloride.
– From the flask the gas is passed through a U-tube containing anhydrous calcium chloride;
– The anhydrous calcium chloride also absorbs moisture from the gas leaving it dry;
– The dry gas is then drawn into a collection syringe;
Diagram:
Alternative methods of oxygen preparation.
Apparatus:
– Sodium peroxide; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, water;
Diagram of apparatus.
Procedure.
– Some sodium peroxide is placed into a flat-bottomed flask;
– The apparatus is set up as shown in the diagram above;
– Add water from a thistle funnel into the flask dropwise;
– The gas is collected as shown;
Observations:
– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;
– The colourless gas collects on top of the water;
Explanations:
– Sodium peroxide reacts with water to liberate oxygen;
– A solution of sodium hydroxide remains in the flask;
– This solution will turn litmus paper blue showing it is alkaline.
Equation:
Sodium peroxide + water → Sodium hydroxide + oxygen;
2Na2O2(l) + 2H2O(l) → 4NaOH(aq) + O2(g)
Note:
The first few bubbles of oxygen gas are not collected.
Reason: The gas is mixed with air which was originally in air and hence impure.
Method of collection;
– Over water collection;
Reason:
– It is insoluble in water and less dense than water;
Chemical test for oxygen gas.
– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;
Apparatus:
– Ignition tube // boiling tube; means of heating; solid potassium manganate (VII); rubber stopper, deliver tubes, beehive shelf, trough, gas jars, wooden splint, water;
Diagram of apparatus.
Procedure:
– The apparatus is set up as shown above.
– Some solid potassium manganate (VII) is put in a hard ignition// combustion tube and strongly heated as shown above.
– The resultant gas is collected over water as shown above.
Observations;
– The purple solid forms a black solid (potassium manganate (II) solid);
– Bubbles of a colourless gas are evolved and collect over water;
Explanations:
– Upon heating potassium manganate (VII) decompose to manganese (VI) oxide; potassium
Equation:
Potassium manganate (VII) → Potassium manganate (II) + Oxygen gas
KMnO4(s) → KMnO2(s) + O2(g);
Uses of oxygen.
Examples:
– It combines with hydrogen to form a very hot oxy-hydrogen flame that is used in welding and cutting metals;
– It combines with acetylene to form oxy-acetylene flame which is also used in welding and cutting metals;
– During this process oxygen is blown over hot impure iron.
– The oxygen react with carbon impurities forming carbon (IV) oxide which escapes laving pure iron which is steel due to its higher purity;
Burning substances in air.
– When substances burn in air they mainly react mainly with oxygen (the active part of air);
– Some metals however also react with nitrogen;
– During burning there is usually change in mass;
Experiment: To investigate burning substances in air.
Requirements:
– Metal (magnesium ribbon); crucible; tripod stand; pipe clay triangle; means of heating;
Apparatus.
Procedure:
– About 1g of magnesium is put in the crucible;
– The crucible (with the magnesium is then weighed)
– The apparatus is set up as above;
– The crucible is heated with the lid lifted occasionally; so as to allow in air;
– No content of the crucible is allowed to escape; to ensure all products of the burning are retained;
– After all the magnesium has burned the crucible is allowed to cool;
– The crucible and its contents are weighed again;
Observations
Mass of crucible + magnesium before burning = xg
Mass of crucible + contents after burning = yg
Change in mass = (x y) g;
Mass of product before burning is lower // less than the mass of the product after burning;
Explanations:
– When the magnesium is burned in a closed crucible in a closed container, most of the air is consumed;
– It is therefore necessary to allow in air so that the burning can continue;
– During burning the magnesium combines with air to form a new product;
– Magnesium combines with both oxygen and nitrogen in air to form magnesium nitride and magnesium oxide;
Equations
With oxygen:
Magnesium + oxygen → Magnesium oxide;
Mg(s) + O2(g) → MgO(s);
With nitrogen:
Magnesium + Nitrogen →Magnesium nitride;
3Mg(s) + N2(g) → Mg3N2(s);
Conclusion:
– Generally when metals burn in air, there is increase in mass;
– All metals react with oxygen to form metal oxides;
– Only more reactive metals react with nitrogen in air;
Note:
– During burning if the product(s) of the burning is gaseous, then there would be decrease in mass.
Examples:
Phosphorus → Phosphorus (V) oxide;
Lead (II) nitrate → Phosphorus (V) oxide + Nitrogen (IV) oxide + Oxygen gas;
Calcium carbonate → Calcium oxide + carbon (IV) oxide;
Burning metals in air and in oxygen.
Requirements:
– Metals; deflagrating spoon; gas jar; source of heat;
Diagram of apparatus;
Procedure:
– A piece of sodium is warmed on a deflagrating spoon until it begins to burn;
– It is then lowered into a gas jar of air as shown above;
– The flame colour is noted;
– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;
– Any gases produced are tested by smell and also with litmus papers;
– The experiment is then repeated with pure oxygen;
– The whole procedure is repeated with other metals;
Observations;
– When substances burn in oxygen they form only oxides; as opposed to burning substances in air where some react with both air and nitrogen;
– Different substances produce different flame colours;
– Many metals burn in air and in oxygen at different speeds; with more reactive metals burning more vigorously than the less reactive metals;
– Burning is faster in oxygen than in air;
Reason:
– Oxygen is pure but in air there are other constituents such as nitrogen, carbon (IV) oxide and noble gases which slow down the burning;
– In air products are generally oxides and in some few cases (magnesium and sodium) nitrides as well;
– Metals that tend to be more reactive are the ones that react with both oxygen and nitrogen;
– In oxygen products are strictly oxides;
– Some of then products are soluble in water while others are not.
Sample equations:
Magnesium:
With oxygen:
Magnesium + oxygen → Magnesium oxide;
Mg(s) + O2(g) → MgO(s);
With nitrogen:
Magnesium + Nitrogen →Magnesium nitride;
3Mg(s) + N2(g) → Mg3N2(s);
Sodium:
With oxygen:
Sodium + oxygen → Sodium oxide;
4Na(s) + O2(g) → 2Na2O(s);
With nitrogen:
Sodium + Nitrogen →Magnesium nitride;
6Na(s) + N2(g) → 2Na3N(s);
Summary: burning metals in air.
| Metal | How it burns | Appearance of product | Name of products | Solubility of product in water | Effect of solution on litmus paper |
| Magnesium | Burns with a bright white flame; | White powder | Magnesium oxide and magnesium nitride; | Slightly soluble; alkaline gas (ammonia) is produced during the process; | Turns blue; |
| Copper | Burns with a blue flame; surface turns black; | Black solid; | Copper (II) oxide; | Insoluble; | No effect; |
| Iron. | Glows to red hot; produces sparks; | Brown black (dark brown) solid; | Iron (II) oxide | Insoluble; | No effect; |
| Sodium | Buns very vigorously with a golden yellow flame; | White solid; | Sodium oxide and sodium nitride | Soluble; alkaline gas (ammonia) is produced in the process; | Turns litmus blue; |
| Calcium | Vigorous with a red flame; | White solid; | Calcium oxide and calcium nitride; | Slightly soluble; alkaline gas evolved in the process; | Turns blue; |
| Zinc | Yellow solid which cools to white; | Zinc oxide | Insoluble; | No effect | |
| Lead | Red solid which cools to yellow; | Lead (II) oxide | Insoluble; | No effect; | |
| Potassium | Very vigorously with a lilac flame; | White solid; | Potassium oxide and potassium nitride; | Soluble; alkaline gas evolved in the process; | Turns blue; |
Note:
– When metals combine with oxygen, it forms metal oxides. In these reactions oxygen is added to the metals; hence the reaction is called oxidation.
– Oxidation refers to the addition of oxygen to a substance;
– The reactivity of various metals with oxygen differs.
– The arrangement of the metals in order of their activity forms the reactivity series;
– Metallic oxides generally turn litmus paper blue and are thus said to be basic oxides;
– Some metallic oxides however have both acidic and basic properties and are thus termed amphoteric oxides e.g. aluminium oxides;
The Reactivity series of metals;
Potassium; Most reactive;
Sodium;
Calcium;
Aluminium;
Zinc; increasing reactivity;
Iron;
Lead;
Copper;
Mercury;
Silver;
Gold; Least reactive
Burning non-metals in oxygen
Requirements:
– Non-metals; deflagrating spoon; gas jar; source of heat;
Diagram of apparatus;
Procedure:
– A piece of sulphur is heated on a deflagrating spoon until it begins to burn;
– It is then lowered into a gas jar of oxygen as shown above;
– The flame colour is noted;
– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;
– Any resultant solution is tested with litmus papers;
– Any gases produced are tested by smell and also with litmus papers;
– The experiment is then repeated with pure oxygen;
– The whole procedure is repeated with other non-metals such as carbon and phosphorus;
Explanations:
1. Sulphur.
– Burns in oxygen with a blue flame to give a colourless gas with a choking irritating smell;
– The gas is sulphur (IV) oxide;
Equation:
Sulphur + Oxygen → Sulphur (IV) oxide;
S + O2(g) → SO2(g);
– The sulphur (IV) oxide dissolves in water to form sulphurous acid, which turns litmus rd;
SO2(g) + H2O(l) → H2SO3(l);
– Glows red to give a colourless gas that forms a white precipitate in lime water;
– The gas is Carbon (IV) oxide;
Equation:
Carbon + Oxygen → Carbon (IV) oxide;
C + O2(g) → CO2(g);
– The Carbon (IV) oxide dissolves in water to form weak carbonic acid, which turns litmus rd;
CO2(g) + H2O(l) → H2CO3(l);
Note:
– In limited oxygen the carbon undergoes partial oxidation forming carbon (II) oxide;
Equation:
2C(s) + O2(g) → 2CO(g)
– Burns in oxygen with a white flame to give dense white fumes;
– The white fumes are either phosphorus (V) oxide or phosphorus (III) oxide;
– Both solids // fumes dissolve in water to form phosphoric acid;
Equations:
With limited supply of oxygen:
Phosphorus + Oxygen → Phosphorus (III) oxide;
P4(s) + 3O2(g) →2P2O3(g);
With excess oxygen.
Phosphorus + Oxygen → Phosphorus (V) oxide;
P4(s) + 5O2(g) →P2O10(g);
Summary: effects of burning non-metals in air.
| Non-metal | How it burns in oxygen | Name of products formed | Appearance of the product | Effect of solution on litmus paper |
| Sulphur | Burns with a blue flame; | Sulphur (IV) oxide | White fumes; | Turns ed |
| Carbon | Glows red | Carbon (IV) oxide | Colourless gas; | Turns red; |
| Phosphorus | Burns with a white flame | Phosphorus (V) oxide and phosphorus (III) oxide | White fumes | Turns red; |
Precautions:
The experiment should be done in a fume cupboard.
Reason:
Products of burning sulphur and phosphorus in air are poisonous.
Note:
– Most non-metallic oxides are acidic in nature and therefore turn litmus red and are thus referred to as acidic oxides;
– Some non-metallic oxides form oxides which are neither acidic nor basic and are thus termed neutral oxides; e.g. carbon (II) oxide and water (hydrogen oxide);
Competition for combined oxygen among elements.
Apparatus:
– Metal oxides, source of heat, metals.
Procedure:
– A spatula end full of copper (II) oxide in a bottle top.
– Magnesium powder and mixed well;
– Record the observations;
– The experiment is repeated using other metal oxides with various other metals like zinc, iron etc.
Observations:
| Metal
Metal oxide |
Magnesium | Zinc | Iron | Lead | Copper |
| Magnesium oxide (white) | No reaction | No reaction | No reaction | No reaction | No reaction |
| Zinc oxide (white) | White magnesium oxide and grey zinc metal | No reaction | No reaction | No reaction | No reaction |
| Iron (III) oxide | White magnesium oxide and grey iron metal;
|
White zinc oxide and iron; | No reaction | No reaction | No reaction |
| Lead (II) oxide (yellow when | White magnesium oxide and | White zinc oxide and lead; | Iron (III) oxide and lead; | No reaction | No reaction |
| Copper (II) oxide (Black | White magnesium oxide and brown copper metal; | White zinc oxide and brown copper metal; | Brown iron (III) oxide and brown copper metal | Yellow lead (II) oxide and brown copper metal | No reaction |
Explanations:
– A more reactive metal takes away oxygen from a less reactive metal;
– This is because a more reactive metal reacts more readily with a less reactive metal;
– These reactions are called displacement reactions;
– Some metals can displace other metals from their oxides upon heating;
– Metals which are higher in the reactivity series can displace metals which are lower in the reactivity series from their oxides;
– From the table none of the metals can displace magnesium from its oxide, while copper can be displaced from its oxides by all the metals.
– Thus from the list magnesium is the most reactive while copper is the least reactive.
– Such results of displacement reactions can also be used to develop a reactivity series of the metals (elements) concerned.
Selected equations:
CuO(s) + Mg(s) → MgO(s) + Cu(s);
Black Grey White Brown.
White when cold Grey White Grey.
Yellow when hot
ZnO(s) + Mg(s) → MgO(s) + Cu(s);
Black Grey White when cold Brown.
Yellow when hot
CuO(s) + Mg(s) → MgO(s) + Cu(s);
Yellow when cold Grey White Grey.
Redwhen hot
CuO(s) + Mg(s) → MgO(s) + Cu(s);
CuO(s) + Zn(s) → MgO(s) + Zn(s);
Black Grey White Brown.
Typical reactivity series from the results above:
Magnesium ↑ Most reactive
Zinc
Iron
Lead
Copper ↓ Least reactive
Note:
– Removal of oxygen is called reduction;
– Addition of oxygen is called oxidation;
– A substance that loses oxygen during a reaction is said to be reduced while a substance that removes oxygen from another is called reducing agent;
– A substance that gains oxygen during a reaction is said to be oxidized while a substance that loses / donates oxygen to another is called an oxidizing agent;
Examples:
CuO(s) + Mg(s) → MgO(s) + Cu(s);
Black Grey White Brown.
Reducing agent: magnesium
Oxidizing agent: Copper (II) oxide
Oxidized species: Magnesium
Reduced species: Copper
CuO(s) + Zn(s) → MgO(s) + Zn(s);
Black Grey White Brown.
Reducing agent: Zinc
Oxidizing agent: Copper (II) oxide
Oxidized species: Zinc
Reduced species: Copper
– In the above reactions both reduction and oxidation take place at the same time;
– A reaction in which both reduction and oxidation occur at the same time is called a redox reaction; red from reduction and ox from oxidation;
Application of Redox reactions:
– Ores of metals such as zinc, iron lead etc are roasted in air to form corresponding metal oxides;
– The metal oxides are then reduced to corresponding metals using common reducing agents like carbon and carbon (II) oxide.
Examples:
Zinc (II) oxide + Carbon (II) oxide → Zinc + carbon (IV) oxide;
Atmospheric pollution and percentage composition of air.
– Human activities have changes the normal composition of air in some places;
– This has not only altered the percentage composition of the main components but also added other components into the air.
Examples:
– Mining increases the amount of dust particles in the air;
– Geothermal power drilling may result into emission of gases like hydrogen sulphide, sulphur (IV) oxide into the air;
– Industrial processes like manufacture of nitric (V) acid, contact process etc may add gases ilke sulphur (IV) oxide, nitrogen oxides into the air;
These gases and emissions cause atmospheric pollution:
– Gases like sulphur (IV) oxide and nitrogen (IV) oxide dissolve I rain water to form acidic rain that causes corrosion of buildings, iron sheet roofing, bleaching of plants; irritation in bodies and respiratory surfaces in animals etc;
– Dust particles may block stomata in plants; cause smog formation hence reducing visibility (leading to more cases of road accidents).
Uses of oxygen:
1. Used for breathing in hospitals fro patients with breathing difficulties.
UNIT 5: WATER AND HYDROGEN.
Checklist:
1. Introduction
Introduction:
– Water is the most abundant substances on earth;
– It covers about 71% of the earths surface;
– Main sources of water include seas, lakes, rivers, oceans.
Burning candle wax in air.
Apparatus:
Procedure.
– The candle is lit under the funnel and the suction pump turned on.
– The set up is left undisturbed for about 15 minutes.
Observations;
– The candle continues to burn.
– Droplets of a colourless liquid in the tube A;
– The colourless liquid turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt (II) chloride into pink;
– A white precipitate forms in the calcium hydroxide in tube B;
– Deposits of a black solid on the inner sides of the funnel;
Explanations;
– The suction pump ensures continuous supply of air hence the candle continues to burn;
– Candle wax buns in oxygen to form carbon (IV) oxide and steam;
– The carbon (IV) oxide is sucked out through the apparatus by the suction pump;
– Carbon (IV) oxide forms a white precipitate of calcium carbonate when bubbled through lime water (calcium hydroxide)
– Incomplete combustion of the carbon in the candle wax produces carbon particles which cools and deposits as black solids;
Equations:
As the candle burns:
Carbon + Oxygen → Carbon (IV) oxide;
C(s) + O2(g) → CO2(g);
Hydrogen + Oxygen → Carbon (IV) oxide;
2H2(s) + O2(g) → 2H2O(g);
For the formation of the black deposits (soot)
Carbon + Oxygen → Carbon + Carbon (IV) oxide;
2C(s) + O2(g) → C(s) + CO2(g);
In the calcium hydroxide:
Carbon (IV) oxide + calcium hydroxide → Calcium carbonate + Water
Colourless Colourless White precipitate Colourless
CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l);
– The steam condenses into water in the boiling tube;
– Water turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt chloride paper into pink;
General equation:
Hydrocarbon + Oxygen → Water + Carbon (IV) oxide;
Conclusion:
– Candle wax is a compound of carbon and hydrogen only; and such compounds are defined as hydrocarbons;
– When burned in air (oxygen) hydrocarbons produce carbon (IV) oxide ad steam (water);
– Other examples of hydrocarbons include: petrol; diesel; kerosene etc;
Note: Effects of repeating the same experiment without a suction pump.
Apparatus:
Observations:
– The candle went off;
– Deposition of black solid on the inner sides of the funnel;
– No colourless liquid in tube A;
– No white precipitate in tube B;
Explanations:
– The carbon (IV) oxide and steam produced would accumulate in the filter funnel hence making the flame to go off;
– Incomplete combustion of the candle would produce carbon particles which cool as soot;
– Only negligible amount of water and carbon (IV) oxide would pass through the apparatus;
Reactions of water with metals:
1. Potassium.
Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It explosively melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.
– It darts on the surface; due to propulsion by hydrogen;
– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;
– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;
(b). Reaction equations.
Equation I
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);
Equation II
4K(s) + O2 (g) → 2K2O(s);
Equation III:
K2O(s) + H2O(l) → 2KOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.
– It darts on the surface; due to propulsion by hydrogen;
– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;
– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;
(b). Reaction equations.
Equation I
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);
Equation II
4Na(s) + O2 (g) → 2Na2O(s);
Equation III:
Na2O(s) + H2O(l) → 2NaOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Procedure:
– A small piece of calcium metal is cut and dropped into a trough containing water;
– A filter funnel is inverted over it;
– A test tube filled with water is then inverted over the funnel;
– The gas given out is collected as shown in the apparatus below.
– The resultant gas is then tested with a burning splint;
– The resultant solution in the trough is tested with litmus paper.
Diagram of apparatus:
Observations and explanations:
– Calcium sinks to the bottom of the beaker; because it is denser than water;
– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;
– Soapy solution formed; due to formation of alkaline calcium hydroxide;
– A white suspension is formed; because calcium hydroxide is slightly soluble in water;
Reaction equation:
Ca(s) + H2O (l) → Ca (OH) 2(aq) + H2 (g);
Effect of resultant solution on litmus paper;
– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;
– Magnesium reacts with atmospheric oxygen to form magnesium oxide that coast the metal surface;
– Thus before reacting it with water this oxide layer has to be removed e.g. by polishing metal surface using sand paper;
– Reaction between magnesium and cold water is generally very slow; with very slow evolution of hydrogen gas;
– Zinc and iron metals do not react with cold water;
Reaction of metals with steam.
Note:
– Metals that react with cold water would react very explosively with steam and thus their reactions with steam should not be attempted in the laboratory;
– However some metals which react only sparingly with cold water or do not react with cold water at all react with steam to produce respective metal oxide and hydrogen gas;
Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;
– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;
– The magnesium ribbon is heated strongly then the wet sand is warmed gently;
– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;
Diagram of apparatus:
Observations and explanations.
– Magnesium burns with a bright blinding flame;
– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;
– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;
Reaction equation.
Magnesium + Steam → Magnesium oxide + Hydrogen gas;
Mg(s) + H2O(g) → MgO(s) + H2(g);
Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;
– A small piece of zinc put in the middle of the combustion tube;
– The zinc is heated strongly then the wet sand is warmed gently;
– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;
Diagram of apparatus:
Observations and explanations.
– Zinc metal does not burn but rather glows;
– Grey solid (zinc) forms a yellow solid which cools to a white solid (zinc oxide);
– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;
Reaction equation.
Zinc + Steam → Zinc oxide + Hydrogen gas;
Grey Colourless Yellow when hot Colourless
White on cooling
Zn(s) + H2O(g) → ZnO(s) + H2(g);
Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;
– A small piece of iron put in the middle of the combustion tube;
– The iron is heated strongly then the wet sand is warmed gently;
– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;
Diagram of apparatus:
Observations and explanations.
– Iron metal does not burn but rather glows;
– Grey solid (zinc) forms a black solid; due to formation of tri-iron tetra-oxide;
– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;
Reaction equation.
Iron + Steam → Tri-iron tetra-oxide + Hydrogen gas;
Grey Colourless Black Colourless
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g);
Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;
– A small piece of aluminium put in the middle of the combustion tube;
– The aluminium is heated strongly then the wet sand is warmed gently;
– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;
Diagram of apparatus:
Observations and explanations.
– Aluminium burns in steam but the reaction quickly stops; because the reaction forms a layer of aluminium oxide that coats the metal surface preventing further reaction;
– Grey solid (aluminium) forms a white solid of aluminium oxide;
– Slight evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;
– The production of the gas however stops soon after the reaction starts because the oxide layer stops further reaction;
Reaction equation.
Aluminium + Steam → Zinc oxide + Hydrogen gas;
Grey Colourless White Colourless
2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g);
– Lead and copper do not react with steam;
Summary of the reaction between metals with cold water and steam
| Metal | Action of metal on water | Action of metal on steam |
| Potassium
Sodium Calcium Magnesium Aluminium Zinc Iron Lead Copper |
Violent
Violent Moderate Very slow No reaction No reaction No reaction No reaction No reaction |
Explosive
Explosive Violent Rapid Slow Slow Slow No reaction No reaction |
Note:
– Metals can thus be arranged in order of their reactivities with water; resulting to a reactivity series similar to that obtained form reaction between metals with oxygen;
Reactivity series of metals:
Potassium; Most reactive;
Sodium;
Calcium;
Magnesium
Aluminium;
Zinc; increasing reactivity;
Iron;
Lead;
Copper; Least reactive;
Hydrogen
– An element that does not exist freely in nature;
– Generally exists in compounds such as water, sugars, fuels etc;
Laboratory preparation of hydrogen gas.
Note:
– Hydrogen gas is generally prepared by the reaction between dilute acids and metals;
– Most suitable acids are dilute hydrochloric acid and dilute sulphuric (VI) acid;
– Most suitable metal is zinc metal;
Apparatus:
Procedure:
– Zinc granules are added to dilute sulphuric (VI) acid;
– Small amounts of copper (II) sulphate are added to the zinc – acid mixture;
Reason: To act as a catalyst hence speed up the reaction;
– The resultant colourless gas is collected over water;
Reason: The gas is insoluble in water;
– If the gas is required dry, the gas is passed through concentrated sulphuric (VI) acid or a U-tube containing calcium chloride;
Diagrams for drying the gas:
Zinc granules
– The dry gas is collected by upward delivery (downward displacement of air);
Reason: It is less dense than air (note that hydrogen is the lightest gas known);
Note:
– Nitric (V) acid is not used in preparation of hydrogen gas; except very dilute nitric (V) acid and magnesium
Reason: Nitric acid is a strong oxidizing agent hence the hydrogen formed is immediately oxidized to water
– Potassium, sodium, lithium and calcium are not used in laboratory preparation of hydrogen gas;
Reason: They react explosively with acids;
– Magnesium is not usually used for laboratory preparation of hydrogen;
Reason: It is expensive;
– Prior to using it for preparation of hydrogen; aluminium should be washed with concentrated hydrochloric acid;
Reason: To remove the protective oxide layer that usually forms on the aluminium surface on its exposure to air;
– Hydrogen gas produced from iron metal tends to have a foul smell;
Reason: Iron gives a mixture of gases due to impurities in the iron; the foul smell is usually due to production of hydrogen sulphide that results from sulphide impurities in the iron metal;
Reaction equations:
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
FeS(s) + H2SO4(aq) → FeSO4(aq) + H2S(g);
Other sources of hydrogen.
– Cracking of alkanes;
– Fractional distillation of petroleum;
Properties of Hydrogen gas;
Physical properties.
Chemical properties.
Chemical test for hydrogen;
– When a burning splint is introduced into a gas jar full of hydrogen gas; the gas buns with a “pop sound”
Note:
– The intensity of the “pop sound” diminishes as the purity of hydrogen increases;
Reactions of hydrogen:
Requirements:
Copper (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;
Diagram of apparatus.
Procedure:
– Apparatus are arranged as shown above;
– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;
Reason:
– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;
– The gas is then lit at the jet and the copper oxide heated;
– This is done until no further change;
– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;
Reason:
– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;
Reason:
Observations and explanations
– The black solid turns into a brown solid;
– Droplets of a colourless liquid on the cooler parts of the combustion tube;
Reason:
The hot black copper (II) oxide is reduced by hydrogen gas into brown copper metal while hydrogen gas is oxidized to water;
– The colourless liquid is confirmed to be water by:
Reaction equations:
In the combustion tube:
Copper (II) oxide + Hydrogen → Copper + Water;
Black solid Colourless gas Brown solid Colourless liquid
CuO(s) + H2(g) → Cu(s) + H2O(l);
At the jet:
– Hydrogen burns with a blue flame producing water;
Equation:
2H2(g) + O2(g) → 2H2O(l);
Conclusion:
– Hydrogen is a reducing agent and thus reduces the copper (II) oxide to copper metals while hydrogen itself undergoes oxidation to form water;
– Hydrogen is the reducing agent;
– Copper (II) oxide is the oxidizing agent;
Requirements:
Lead (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;
Diagram of apparatus.
Procedure:
– Apparatus are arranged as shown above;
– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;
Reason:
– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;
– The gas is then lit at the jet and the copper oxide heated;
– This is done until no further change;
– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;
Reason:
– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;
Reason:
Observations and explanations
– The yellow solid turns red on heating then finally into a grey solid;
– Droplets of a colourless liquid on the cooler parts of the combustion tube;
Reason:
The hot red lead (II) oxide is reduced by hydrogen gas into grey lead metal while hydrogen gas is oxidized to water;
– The colourless liquid is confirmed to be water by:
Reaction equations:
In the combustion tube:
Lead (II) oxide + Hydrogen → Lead + Water;
Yellow-cold; red – hot Colourless gas Grey solid Colourless liquid
PbO(s) + H2(g) → Pb(s) + H2O(l);
At the jet:
– Hydrogen burns with a blue flame producing water;
Equation:
2H2(g) + O2(g) → 2H2O(l);
Conclusion:
– Hydrogen is a reducing agent and thus reduces the lead (II) oxide to lead metals while hydrogen itself undergoes oxidation to form water;
– Hydrogen is the reducing agent;
– Lead (II) oxide is the oxidizing agent;
Requirements:
Iron (III) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;
Diagram of apparatus.
Procedure:
– Apparatus are arranged as shown above;
– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;
Reason:
– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;
– The gas is then lit at the jet and the copper oxide heated;
– This is done until no further change;
– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;
Reason:
– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;
Reason:
Observations and explanations
– The Brown solid turns into a grey solid;
– Droplets of a colourless liquid on the cooler parts of the combustion tube;
Reason:
The hot brown iron (III) oxide is reduced by hydrogen gas into grey iron metal while hydrogen gas is oxidized to water;
– The colourless liquid is confirmed to be water by:
Reaction equations:
In the combustion tube:
Iron (III) oxide + Hydrogen → Iron + Water;
Brown solid Colourless gas Grey solid Colourless liquid
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l);
At the jet:
– Hydrogen burns with a blue flame producing water;
Equation:
2H2(g) + O2(g) → 2H2O(l);
Conclusion:
– Hydrogen is a reducing agent and thus reduces the iron (III) oxide to iron metal while hydrogen itself undergoes oxidation to form water;
– Hydrogen is the reducing agent;
– Iron (III) oxide is the oxidizing agent;
Note:
– Hydrogen does not reduce (remove oxygen) from oxides of metals above it in the reactivity series;
Products of burning hydrogen gas in air.
Apparatus and requirements:
– The apparatus is arranged as shown below.
Procedure:
– Apparatus is arranged as shown below.
– A stream of hydrogen is passed through anhydrous calcium chloride;
– The gas is tested for purity by collecting samples over the jet and testing with a burning splint.
Note: pure hydrogen gas should burn smoothly without the typical “pop sound”;
– The gas is then lit and the pump tuned on;
– The products of burning hydrogen is drown in through the apparatus using the pump for about 15 minutes;
– The product condensing in the test tube in cold water is tested with white anhydrous copper (II) sulphate and blue cobalt chloride paper;
Observations:
– Pure hydrogen burns with a blue flame;
– A colourless liquid condenses in the test tube immersed in cold water;
– The liquid turns white anhydrous copper (II) sulphate blue;
– The colourless liquid turns blue anhydrous cobalt chloride pink;
Explanations:
– The calcium chloride in the U-tube is used to dry the gas;
– The pure dry hydrogen gas burns with a blue flame to form steam which condenses into liquid water;
– Water turns anhydrous copper (II) chloride from white to blue; and turns blue anhydrous cobalt chloride paper into pink;
Reaction equations:
At the jet:
2H2(g) + O2(g) → 2H2O(g);
Uses of Hydrogen
– This refers to hardening of oils into fats.
– In this reaction Hydrogen gas is bubbled into liquid oil in presence o0f nickel catalyst;
– The oil takes up hydrogen and is converted into fat;
– Usually a radio transmitter is connected to a weather balloon filled with air; as the balloon floats in air the transmitter collects information which is conveyed to weather stations for interpretation by meteorologists;
MATHEMATICS I
PART I
SECTION I (50 MARKS)
0.01712 X 3
855 X 0.531 (2 Mks)
5 hours a day will be required to pack 2500 parcels in 2 days (3 Mks)
AB = 3OC. cm: mB = 2:3. Find in terms of i and j
C vector Om (3 Mks)
O A
5x + 5y = 1
4y + 3x = 5 (3 Mks)
Calculate ÐOAC (3 Mks)
A C
B
2(3x – 1)2 – 9 (3x – 1) + 7 = 0 (4 Mks)
Cts and for every two shillings Kamau gets, Omondi gets three shillings. By how much does Maina’s
share exceed Omondi’s (3 Mks)
(i) You will fail in two attempts (1 Mk)
(ii) In three attempts, you will at least fail once (3 Mks)
equation of the line (2 Mks)
3 5 6
–25. Find the sum of the first 21 terms (5 Mks)
the cone. (4 Mks)
O
20cm 1250 20 cm
z2 – w2
a bearing of 055 from Q. Calculate the distance between P and R. (3 Mks)
SECTION II (50 MARKS)
U T P
Q
S 390
(a) ÐRST (1 Mk)
(b) ÐSUT (3 Mks)
(c) Obtuse angle ROT (2 Mks)
(d) ÐPST (2 Mks)
1 US$ = Kshs.74.75
1 French Franc (Fr) = Kshs.11.04
(a) How much in dollars did he receive from his Kshs.350,000 in 4 s.f? (2 Mks)
(b) The tourist spend ¼ of the amount in America and proceeded to France where he spend Fr
16,200. Calculate his balance in French Francs to 4 s.f (3 Mks)
(c) When he flies back to Kenya, the exchange rate for 1 Fr = Kshs.12.80. How much more in
Kshs. does he receive for his balance than he would have got the day he left? (3 Mks)
(a) Draw a line through points (0,2) and (1,0) and extend it to intersect with curve y = 5 + 2x – 3 x 2
read the values of x where the curve intersects with the line (2 Mks)
(b) Find the equation whose solution is the values of x in (a) above (2 Mks)
and angle PQR = 300 (2 Mks)
(b) Construct a circle passing through points P, Q and R (2 Mks)
(c) Calculate the difference between area of the circle formed and triangle PQR (4 Mks)
Y
4
2 R (3,3)
X
-3 5
| Mass (Kg) | 60 – 64 | 65 – 69 | 70 – 74 | 75 – 79 | 80 – 84 | 85 – 89 |
| No. of women | 8 | 14 | 18 | 15 | 3 | 2 |
(a) State (i) The modal class (1 Mk)
(ii) The median class (1 Mk)
(b) Estimate the mean mark (4 Mks)
(c) Draw a histogram for the data (2 Mks)
at points A, B, C respectively. XY = 10 cm,
YZ = 8 cm and XZ = 12 cm. (2 MKS)
Z
C
.. B
X
A Y
(a) Calculate, length XA (2 Mks)
(b) The shaded area (6 Mks)
(a) After how long to the nearest 1 decimal place will the value of the car be Kshs.130,000 (4 Mks)
(b) Calculate the rate of depreciation to the nearest one decimal place which would make the value of
the car be half of its original value in 5 years (4 Mks)
MATHEMATICS I
PART II
SECTION 1 (50 MARKS)
b15 4a6 (2 Mks)
Ö0.375 cos 75
tan 85.6 (4 Mks)
A 5cm D
C O
B
5 – 2Ö3 5 + 2Ö3
ABC. AC = 8m, BC = 8m, BD = 10m and (ACB = 1200. 8m
If a scout needs 2.5 m3 of air, how many scouts can fit 120o C E
in the tent. 8m (4 mks)
B D
10m
labelled net of the figure and show the dimensions of the net
angle of elevation of 200. After walking 10m towards the bottom of the flag post, the top is observed at angle of elevation of 400. Calculate the height of the flag post (4 mks)
larger Container is 120 cm2. Calculate base area of the smaller container. (3 mks)
30km per hour. Calculate the speed of the overtaking train (3 mks)
Determine the distance at the faster speed (3 mks)
(2 Mks)
SECTION B (50 MARKS)
1 – 39,600 10
39,601 – 79,200 15
79,201 – 118,800 25
118,801 – 158,400 35
158,401 – 198,000 45
is 3/5 if the first is correct and it is 4/5 if the first was wrong. The chance of the third correct is
2/5 if the second was correct and it is 1/5 if the second was wrong. Find the probability that
(a) All the three are correct (2 Mks)
(b) Two out of three are correct (3 Mks)
(c) At least two are correct (3 Mks)
purchase reduced his average cost per pen by Sh.1.50. Calculate the number of pens bought earlier and the difference in cost of the total purchase at the two prices (8 mks)
OPAQ is a parallelogram.
Determine in terms of a and b vectors
(a) OP (2 Mks)
(b) PQ (2 Mks)
(c) QN (2 Mks)
(d) PN (2 mks)
cm per second. If the water flows for 10 hours a day
(a) Calculate the volume in M3 added in 2 days (4 ms)
(b) If the tank has a height of 8 m and it takes 15 days to fill the tank, calculate the base radius
of the tank (4 mks)
School B needed 24 million to complete their projects. The sponsor raised Shs.16.9 million while other
guest raised Shs.13.5 million.
(a) If it was decided that the sponsor’s money be shared according to the needs of the school
with the rest equally, how much does each school get (5 mks)
(b) If the sponsor’s money was shared according to the schools needs while the rest was in the ratio of
students, how much does each school get if school A has 780 students and school B 220
students (3 mks)
V = KEn where k and n are constants. The table below shows values of V and E
V |
0.35 | 0.49 | 0.72 | 0.98 | 1.11 |
| E | 0.45 | 0.61 | 0.89 | 1.17 | 1.35 |
(a) Draw triangle PQR and its image PIQIRI of PQR under translation T = 3 on the provided grid 4 (2 Mks)
(b) Under transformation matrix m = 4 3 , PIQIRI is mapped on to PIIQIIRII. Find the
co-ordinates of PIIQIIRII and plot it 1 2 on the given grid (4 Mks)
(c) If area of D PIQIRI is 3.5 cm2, find area of the images PIIQIIRII (2 Mks)
MATHEMATICS I
PART 1
MARKING SCHEME
= 2 X 10-6 A1
2
= 75 A1
Extra men = 75 – 6 = 69 B1
3
= 2i + 7j + 2/5 (8i + 11j) M1
= 26 i + 57 j
5 5 A1
3
3 4 y 5 M1
x -1/7 5/7 1
y = 3/7 -2/7 5 M1
x = 3
y -1
x, 3, y = -1 A1
3
Obtuse ÐAOC = 360 – 252 = 1080 B1
= 1/2 (180 – 108)0
= 360 B1
3
6x2 – 13x + 6 = 0 B1Ö equation
6x2 – 9x – 4x + 6 = 0
3x(2x – 3) (3x – 2) = 0 M1
x = 2/3 or A1
x =1 ½ B1
4
= 80/- B1Ö Omondi’s
3
= 64 + 96x + 60x2 A1
2.46 = (2 + 1/2 (0.8))6
= 64 + 96 (0.8) + 60 (0.64) M1
= 179.2
@179 to 3 s.f A1
4
= 49/100 B1
P (at least one fail) = 1 – P (FI FI FI)
= 1- 13/20 3 M1
= 1 – 2197 M1
8000
= 5803
4
= -1 B1
y = mx + c
y = -x + 5 B1
2
1000 10 M1Ö conv. to litres
= 500,000 Lts A1
3
10a – 18a + 42 = 5a – 10
– 13a = -52 M1
a = 4 A1
3
2a + 4d = -10 M1
8d = 12
d = 11/2 A1
a = -8 B1
S21 = 21/2 (-16 + 20 X 3/2) M1
= 147 A1
5
360
r = 6.667 cm A1
h = Ö 400 – 44.44 M1
= 18.86 cm A1
4
(z – w) (z + w)
= (w – y) (x – z) (w + z) M1Ö grouping
(z – w) (z + w)
= (w – y) (x – z)
z – w A1
3
R
250 B1Ö sketch
Q 125 PR = 4 sin 125 M1
Sin 25
A1
30
P 3
(b) < RTQ = 900– 740 = 160 B1
< PTR = 460 + 160 = 620 B1
< SUT = 620 – 390 = 230 B1
(c) Reflex ÐRQT = 180 – 2 x 16
= 180 – 32 = 1480 B1
Obtuse ROT = 360 – 148 = 2120 B1
(d) < PTS = 46 + 180 – 129 = 970 B1
< PST = 180 – (97 + 39) = 440 B1
8
(a) Kshs.350,000 = $ 350,000 M1
74.75
= $ 4682 A1
(b) Balance = 3/4 x 4682
= $ 3511.5 B1
$3511.5 = Fr 3511.5 x 74.75 M1
11.04
= Fr 23780 A1
Expenditure = Fr 16 200
Balance = Fr 7580
(c) Value on arrival = Kshs.7580 X 12.80
= Kshs.97,024
Value on departure = Kshs.7580 X 11.04 B1 bothÖ
= Kshs.83 683.2
Difference = Kshs.97,024 – 83683.2 M1
= Kshs.13,340.80 A1
8
| X | -2 | -1 | 0 | 1 | 2 | 3 |
| Y | -11 | 0 | 5 | 4 | -3 | -16 |
B1Övalues
y
S1Ö scale
8 — P1Ö plotting
6 — C1 Ö curve
4 —
2
-2 — 1 2 3 x
-4 —
-6 —
-8 — y=2x=2
-10 —
-12 —
-14 — x =-0.53 + 0.1 BI
-16 — Nx = 1.87+ 0.1
y = 5+2x-3x2 =2-2x MI for equation
3x2-4x-4x-3=0 AI equation
8
x = -0.53 ± 0.1 B1
mx = 1.87 ± 0.1
y = 5 + 2x – 3x2 = 2 – 2x M1 Ö for equation
\ 3×2 – 4x – 3 = 0 MA1 Ö equation
8
20.
B1 Ö 300
R B1 Ö 2 ^ PQ, QR
B1 Ö 2 ^ bisectors
B1 Ö circle
9 Q
Radius = 4.2 ± 0.1 B1Ö radius
Area of circle = 22/7 x 4.22
= 55.44 ± 3 cm2
Area of D PQR = 1/2 x 3.5 x 7.5 sin 30 M1Ö D and circle
= 6.5625 cm2
= 48.88 cm2 A1
8
5y + 2x = 10 B1Öequation
5y + 2x = 10 B1Ö inequality
3y = 4x + 12 B1Ö equation
3y < 4x + 12 or 3y – 4x < 12 B1Ö inequality
3y + x = 12 or 3y = -x + 12 B1Ö equation
3y + x < 12 B1Ö inequality
x – 3 2
2y + 3x = 15 B1Ö equation
\ 2y + 3x £ 15 B1Ö equation
8
CLASS |
F | x | Fx | Cf |
| 60 – 64
65 – 69 70 – 79 75 – 79 80 – 84 85 – 89 |
8
14 18 15 3 2 |
62
67 72 77 82 87 |
496
938 1296 1155 246 174 |
8
28 40 55 58 60 |
| Sf = 60 | Sfx 3809 |
B1Ö x column
B1Ö f column
(a) (i) Modal class = 70 – 74 B1Ö model class
(ii) Median class = 70 – 74 B1Ö median
(b) Mean = 3809
= 63.48 A1
S1Ö scale
B1 Ö blocks
59.5 – 64.5
64.5 – 69.5 e.t.c.
8
(c)
Histogram
20 —
15 —
10 – –
5 —
55 60 65 70 75 80 85 90
YZ = 10 – a + 12 – a = 8 M1
2a = 14
a = 7 cm A1
Cos X = 100 + 144 – 64
240 M1Ö any angle of the D
= 0.75
X = 41.410
1/2 X = 20.700 A1Ö 1/2 of the angle
r = OA = 7tan 20.7 B1 Ö radius
= 2.645 cm
Shaded area = 1/2 X 10 X 12 sin 41.41 – 22/7 X 2.6452 M1 Ö D & circle
= 39.69 – 21.99
= 17.7 cm2 A1Ö
8
1.15n = 0.2
n = log 0.2 M1Ö
log 0.85
= 1.3010
1.9294
= – 0.6990 M1
– 0.0706
= 9.9 years A1
(b) 650,000 (1 – r/100) 5 = 325,000 M1
(1 – r/100) 5 = 0.5
1 – r/100 = 0.5 1/5 M1
= 0.8706
r/100 = 0.1294 A1
r = 12.9 % B1
8
MATHEMATICS I
PART II
MARKING SCHEME
SECTION I (50 MARKS)
32a10 9b4 M1Ö reciprocal
= 2a5 A1
27 2
No. Log.
0.375 1.5740 +
cos 75 1.4130
2.9870 _
tan 85.6 1.1138
3.8732 = 4 + 1.8732
2 2
2.9366
0.0864
100
= Shs.480 B1
Cost Price = x
96x = 480 M1
100
x = Shs.500 A1
3
r2 + hr + (h/2)2 = A/2A + h/4 M1
(r + h/2)2 = Ö 2A + h2
4p M1
r = -h/2 ± Ö2A + h2 A1
4p 4
4r2 – qr = 0
r(4r – q) = 0 M1
r = 0
or r = 2.25 A1
3
(5 – 2Ö3) (5 + 2Ö3)
= 10 + 4Ö3 – 5 +2Ö3 M1
13
= 5 + 6Ö3 A1
13 3
6 = 4k + c
16 = 9k + c M1 Ö subtraction
5k = 10
k = 2
c = -2 A1 Ö k and c
q = Ö33
= ± 5.745 A1
4
2.5 M1
= 110.8
= 110 A1
3
2
= 2 B1
% error = 2/26 X 100 M1
= 7.692% A1
G 3
H D G H B1 Ö dimen. FE must be 10cm
4cm 4cm
B1 Ö labelling
E 4cm A 12cm F 10cm E 3
4cm 12cm
E
F
= 1 + 12x + 60x2 A1
(1.04)6 = (1 + 2(0.02))6
= 1 + 12 (0.02) + 60(0.02)2 M1
= 1.264 A1
4
CT = 10 sin 40 M1
= 6.428 m A1
A1 10cm B C h = 6.428 + 1.5
1-5 = 7.928 B1
4
960 64 10
smaller area = 29 X 120 M1
164
= 67.5 cm2 A1
3
2 km = 2 hrs
(x – 30)km/h 60 M1
2x – 60 = 120
x = 90 km/h A1
3
x Km (22 – x) Km
x + 22 – x = 11
16 5 4 M1
5x + 352 – 16x = 220 M1Ö x-multiplication
11x = 132
x = 12 km A1
3
\ PQ = 62/3 – 22/3 = 4 cm B1 Ö C.A.O
2
= Shs.9200 p. m
= Shs.110,400 p.a A1
Tax dues = 10/100 x 39600 + 15/100 x 39600 + 25/100 x 31200 M1 Ö first 2 slabs
= 3960 + 5940 + 7800 M1 Ö last slab
= Shs.17,700 p.a
= 1475 p.m A1
net tax = 1475 – 400
= Shs.1075 B1 Ö net tax
Total deductions = 1075 + 130
= Shs.1205
net income = 8000 – 1205 M1
= Shs.6795 A1
8
(a) P (all correct) = 2/3 x 3/5 x 2/5 M1
= 12/125 A1
(b) P (2 correct) = 2/5 x 3/5 x 3/5 + 2/5 x 2/5 X 1/5 + 3/5 x 4/5 x 2/5
= 18/125 + 4/125 + 24/125 M1
= 46/125 A1
(c) P (at least 2 correct)
= P(2 correct or 3 correct)
= 46/125 + 12/125 M1
= 46 + 12 M1
125
= 58
8
x
New price/pen = 494 B1Öboth expressions
x + 3
440 – 494 = 1.50
x x + 3 M1 Ö expression
440(x + 3) – 494x = 1.5x2 + 4.5x M1Ö x-multiplication
x2 + 39x – 880 = 0 A1 Ö solvable quad. Eqn
x2 + 55x – 16x – 880 = 0 M1 Ö factors or equivalent
(x – 16) (x + 55) = 0
x = -55
or x = 16 A1 Ö both values
\ x = 16
difference in purchase = 19 X 1.50 M1
= Shs.28.50 A1
8
= 3/4 a + 1/4 b A1
(b) PQ = PO + OQ
= –3/4 a – 1/4 b + 1/4 (a – b) M1
= –1/2 a – 1/2 b A1
(c) QN = QO + ON
= 1/4 (b – a) + 5/3 b M1
= 23/12 b – 1/4 a A1
(d) PN = PB + BN
= 3/4 (b – a) + 2/3 b M1
= 17/12 b – 3/4 a A1
8
7 2 2 1,000,000 M1 Ö volume in cm3
= 103.95 m3 M1 Ö volume in m3
(b) 22 X r2 x 8 = 103.95 x 15 x 7 M1
7 2
r2 = 103.95 x 15 x 7 M1
= 31.01 M1
r = 5.568 m A1
8
A’s share = 5/13 x 16.9 + 1/2 x 13.5 M1
= 13.25 Million A1
B’s share = (13.5 + 16.9) – 13.25 M1
= 13.25 M1
6.5 + 10.53
= 17.03 m A1
B’s share = 30.4 – 17.03 M1
= 13.37 Million A1
8
| Log V | -0.46 | -0.13 | -0.14 | -0.01 | 0.05 |
| Log E | -0.35 | -0.21 | -0.05 | 0.07 | 0.13 |
B1Ö log V all points
B1Ö log E all points
S1 Ö scale
P1Ö plotting
Log V = n log E + log K L1 Ö line
K = 1.2 ± 0.01 B1 Ö K
N = 0.06/0.06 B1 Ö n
= 1 ± 0.1
\ v = 1.2E B1Ö v
when E = 0.75, V = 0.9 ± 0.1 8
4 PI (0,5), QI (2,2) RI (1,0)
PI QI RI PII QII RII
(b) 4 3 0 2 1 = 15 14 4 M1 Ö
1 2 5 2 0 10 6 1 A1 Ö
PII (15,10), QII (14,6), RII (4,1) B1Ö
(c) Area s.f = det M
= 5
area of PII QII RII = 5 (area PIQIRI)
= 5 X 3.5 M1Ö
= 16.5 cm2 A1
8
MATHEMATICS 2
PART I
SECTION A:
0.0368 x 43.92
361.8
2x2 – 5x + 1 = 0
2
hence find P when Q = 4 a (3mks)
½ of 5/6
3 x 1 = 2
2 1 -1 y
Find its centre and radius
(i) Gradient of the line
(ii) The magnitude of AB
(iii) The equation of the perpendicular bisector of AB.
SECTION B:
BCA = 450
transformation by the matrix -2 0 to obtain its image AI B I CI DI. under a second transformation
0 – 2
which has a rotation centre (0,0) through –900 , the image AII BII CII DII of AI BI CI DI is
obtained. Plot the three figures on a cartesian plane (6mks)
(b) Find the matrix of transformation that maps the triangle ABC where A (2,2) B (3,4) C (5,2)
onto A B C where A( 6,10) B (10,19 ) C ( 12, 13). ( 2mks)
19.
In the triangle OAB, OA = 3a , OB = 4b and OC = 5/3 OA. M divides OB in the ratio 5:3
Calculate:
The area of the intersection of the two circles
The area of the quadrilateral S P Q R
The area of the shaded region
| X | 0 | 4 | 8 | 12 | 16 | 20 |
| Y | 1.0 | 0.64 | 0.5 | 0.42 | 0.34 | 0.28 |
Y = q
P + x
where p and q are constants. (3mks)
(b) Use your graph to determine p and q (3mks)
(c ) Estimate the value of (i) y when x = 14
(ii) x when y = 0.46 (2mks)
Hence
Area of AXB
Area of DXC
(scale 1cm rep 10 km)
From this find,
(a) The distance of AD and DC in km (2mks)
(b) The distance and bearing of B from D (2mks)
(c) The bearing of C from A (1mk)
MATHEMATICS 2
PART 1
MARKING SCHEME (100MKS)
= 3.6502
0.3681 2.5660
0.3682 1.6427 + -4 = 1.6502 = 2.8251
0.2087 Logs 2
361.8 2.5585 + – v ans (4) 6.6850 x 10 -2
3.6502 = 0.06685
x2 – 5 x + ½ = 0
2
x2 – 5 x = – ½
2
x – 5x + –5 2 = –½ + –5 2 (m)
2 4 4
= x – 5 = – ½ + 25 = 17 (3)
4 16 16
= x – 5/4 = 17/16 = 1.0625
x – 5/4 ± 1.031
X1 = -1.031 = 1.25 = 0.2192
X2 = 1.031 + 1.25 = 1.281
A2 = P + PRT/100 = 6000 + 15 X 2 = 6000 + 1800
100
= Shs. 7800
Amount by simple interest is more by Shs. (7800 – 7661. 40)
Shs. 138.60
3p + 4c = 380 x 5 15p + 20c = 1900
4p + 5c = 490 x 4 16p + 20c = 1960
-p -60 (m)
p = Shs 60
3(60) + 4 c = 380
4c = 380 –180 = 2000 (3)
c= Shs. 50
Plate = Shs. 60 , Cup = Shs. 50 (A both)
In 1 glass = 1/8 x 200 = Sh 25
3 litres = 300 ml (undiluted concentrate) (3)
No. of glasses =v 3000 = 120 glasses
25
Cos 30 , 330, 390, 690, 750 ….
2 θ = 210 , = 1050 (3)
2 θ = 390 – 120 = 2700 , θ2 1350
2 θ = 690 – 120 = 5700 , θ3 2850 (for 4 ans)
θ4= 315o ( for >2)
2 θ = 750 – 120 = 6300 ,
Q2 9
K = 4 X 4 = 16
9 9
P = 16 v when Q = 4
9Q2
9x4x4
Max perimeter = (3.65 + 2.85) x 2 = 23 cm Expressions
% error = 13 –12.8 x 100 m = 0.2 x 100 (3)
12.8 12.8
= 1.5620% (A)
½ of 5/6 ½ of 5/6 5/12 5/12
= -1/24 = -1 x 12 = -1
5/12 24 5 10 (3)
Volume of each cube = 2x2x2 = 8 cm3 A
No. of cubes = 704 /8 = 88 cm3 A
8
Tan 67.5 = h
4
h = 4 x 2.414 A
= 9.650cm
Area of 1 triangle = ½ x 8 x 9.656 x 8 cm = 38.628 x 8 vm
Octagon area = 38.628 x 8 m
= 309.0 cm2 (A)
=
2 1 -1 y
3 – x = 2 (1) x = 1 (2)
2 – 1 = y y = 1 (A)
x2 – 6x + (-3)2 + y2 + 8y + (4)2 = 11 + (-3)2 + (4)2 (completing the square)
(x – 3)2 + (y+4)2 = 11 + 9 + 16 = 36
(x – 3)2 + (y + 4)2 = 62
Centre is (3, -4)
Radius = 6 units As (3)
14.
Figs A C B and D C E are similar
AB = AC = and AB = BC
10 = 6 + x
3 6
= 10 = 15 + y, m
3 y 60 = 18 + 3x
10y = 15 + 3y 3x = 42
7y = 15 x = 14
y = 15/7 (A) (3)
A (4 , 3) B(8,13)
change in x 8-4 4 2
(ii) Mag AB = 8 -4 4 =
13 -3 10
Length = Ö42 + 102 = Ö116 = 10.77 units
(iii) Mid point = 4 +8 , 3 + 3
2 2
= (6, 8) (mid point) (5 mks)
gdt of perpendicular to AB = -ve rec. of 5/2
-2/5
Eqn is y = -2/5 x + c
8 = -2/5 x 6 + c = 40 = -12 + 5c
= c = 52/5
y = -2/5 x + 52/5 (A)
Time
= (34 – 10) m/s = 24 m/s
60 x 10 600
= 0.04m/s2- (2)
17.
Triangle (8)
AC = 9cm
Circumference Centre
Circle
Perpendicular bisector of AB
P
Q
c d 2 4 2 10 19 13
2a +2b = 6 x 2 = 49 + 4b = 12
3a + 4b = 10 3a + 4b = 10
a = 2 4 + 2b = b
2c + 2d = 10×2 = 4c + 4d = 20 2 b = 2 b = 1
3c + 4d = 19 3c + 4d = 19
c = 1
2 (1) + 2d = 10
2d = 8 Matrix is 2 1 (A)
d = 4 1 4
OC = 5/3 (31) = 5A
19.
(a) = AO + OB MC = MO + OC
= -3a = 4b = -5/8 (4b) + 5
= 5A – 5/2 b
(b) MN = 5 Mc = 3(5a – 5/2 b)
= 5 s a – 5/2 s b
MN = BN + BN
= 3/8 (4 b) + (1 – t) (-BA)
= 3/8 (4 b) + (1 – t)(3a – 4 b)
= 3/2 b + 3 ta –4b + 4tb
= (3-3t) a (4t – 5/2)b
MN = MN
= 5 s a – 5/2 sb = (3-3t)a + (4t – 5/2 )b
= 5 a = 3 – 3t = 5s + 3t =3
= -5/2 s = 4t –5/2 v 5s + 8t = 5
-5t = -2 t = 2/5
5 s = 3 – 3(2/5)
= 3 – 6/5 = 9/5
= 3 – 6/5 = 9/5
s = 9/25
(i) AN : NB = 2 : 3
(ii) MN : 9 : 16
20.
θ x pr2
360
= 121.6
Area of triangle SPR ½ x 13.2 x 13.2 x sin 80
= 85.8 cm2
(m of area of ) A (at least one)
(m of area) A(at least one)
Area of segment = 121.6 – 85.8
= 35.8 cm2
Area of sector QPR = 90/360 x 3.142 x 12 x12
Area of PQR = ½ x 12 x 12 = 722
Area of segment = 113.1 – 72
= 41.1cm2
Area of intersection = (35.8 + 41.1) = 76.9 cm2
b). Area of quadrilateral = Area of PQR + SPR
= 85.8 + 72 = 157.8cm2
Area of shaded region = Area of Quadrilateral – Area of sector SPR
= 157.8 – 121.6
= 36.2 cm2
p + x y y q q
Gradient = 1/q at (0, 0.95) (8,2.0) (8,2.0) gradient = 2.0 – 0.95 = 1.05
8 8
1 = 0.1312
q
q = 1 = 7.619
0.1312
q = 7.62.
y(1/y) Intercept p = 0.95 P = 0.95
q 7.62
p = 7.62 x 095 = 7.27
at x = 14, y = 2.7
at y = 0.46, 1/y = 2.174
x = 9.6.
uphill Time = 75/v hrs
Downhill speed = ( + 20) km/h
Downhill Time = 75 hrs.
Takes larger uphill
75 – 75 = 1
v v+20
75 (v+20) – 75v = 1
v(v + 20) 1
75v + 1500 – 75v = v(v + 20) = v2 + 20v.
v2 + 20v – 1500 = 0
v = – 20 + 202 – 4(1) (-1500)
2(1)
v = –20 + 400 + 6000 = –20 + v6400
2 2.
V1 = –20 + 80 = 30km/hr
2
V2 = – 20 – 80 X impossible
2
speed uphill = 30 km /hr, T = 75 time = 2 ½ hrs
30
speed downhill = 50 km /hr Time = 75 Time = 2 ½ hr
50
Average speed = Total distance = 150km = 37.5 km/ hr
Total time 4hrs
| X | 0 | 4 | 8 | 12 | 16 | 20 |
| Y | 1.0 | 0.64 | 0.5 | 0.42 | 0.34 | 0.28 |
| 1/y | 1.0 | 1.56 | 2.0 | 2.38 | 2.94 | 3.57 |
D C
A x X x C = BX . XD
8 x 4 = 6BX
BX = 8 x 142 = 16
6 3
X AD = XBC
XA = 8 = 24 = 3
XB 16 16 2
XD = 6 = 3
XC 4 2
< AXD = BXC (vertically opposite <s))
SAS holds : they are similar.
LSF = 3/2 ASF = (3/2)2 = 9/4
Area A x A = 6cm2 Area B x C = 6 x 9 = 27 = 13.5cm2
4
24.
DC = 35km
BD = 90km
Bearing is 0200
Bearing is 134o (8mks)
MATHEMATICS I
PART II
SECTION (52 MARKS)
1.43 x 0.091 x 5.04
2.86 x 2.8 x 11.7 (3mks)
y = a/x + bx (3mks)
2x + 1< 10 – x < 6x – 1 (3mks)
(i) The distance of A to C (2mks)
(ii) The bearing of A from C (3mks)
Ö 3
1 – v3
| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No. of | 52 | 41 | 43 | 48 | 40 | 38 | 36 | 40 | 44 | 45 |
Calculate the 4 day moving averages for the data (3mks)
3mks
2 1
-1 2
4 3
2x – 3y = 7
4x + 3y +5 (3mks)
1/7 + 3/12 + 7/0.103
Given that O is the centre of the circle and OA is parallel to CB, and that angle
ABC = 1070, find
(i) Angles AOC, (ii) OCB (iii) OAB (3mks)
Log x 3
VY (2mks)
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 |
| No. of Students |
2 |
4 |
7 |
19 |
26 |
15 |
12 |
5 |
Using an assumed mean of 44.5, calculate
State the minimum co-ordinates
iii. The same number showing (4mks)
Two pulley wheels centers A and B are joined by a rubber band C D E F G H C round them. Given that larger wheel has radius of 12 cm and AB = 20 cm, CD and GF are tangents common to both wheels and that CBA = 60o), Find
iii. Arc length CHG and DEF, hence find the length of the rubber.
faces is inclined at 750 to the base. Calculate
Use your graphs to solve the equation 2 sin x = cos 2x
end to 3.5m at the deep end. The pool is 25m long and 12m wide. Calculate the volume of the pool
in cubic meters.
The pool is emptied by a cylindrical pipe of internal radius 9cm. The water flows through the pipe at speed of 3 metres per second. Calculate the number of litres emptied from the pool in two minutes to the nearest 10 litres. (Take II = 3.142)
reflection in the line y = x are B1 (-1, -3) and D1 (1,3) respectively.
State its co-ordinates, and these of A1 and C1.
PART II
2.86 X 2.8 X 11.7 105 2 x 28 x 117 x 103
(3)
= 0.007 (A)
Either
bx2 – yx + a = 0
x = y ± v y2 – 4ab
2b (3)
2x + 1 £ 10 – x 10 –x £ 6x –1
3x £ 9 11£ 7x
x £ 3 x £ 11/7 (3)
11/7 £ x £ 3
(a and r)
Sn = a(r n – 1)
X 5
0.20695
0.1 (4)
= 4000 (1.15 –1) (any)
1.1 –1 4000 (1.6 – 1)
0.1
A = 4000 ( 0.6105)
0.1
= Sh. 2442 = Sh. 24,420 (A) (4)
0.1
= 642 + 562– 2(64) (56) cos 78
= 4096 + 3136 – 7168 (0.2079)
= 7232 – km 1490.3
b2 = 5741.7 = 5.77 km (5)
(ii) b a
75.77 = 64
Sin 78 sin A Sin A = 64 x 0.9781
75.77
Sin A = 0.08262
A = 55.70 (or B = 46.30)
Bearing = 90 – 28 – 55.7
= 0.06.30 (A)
(1.02)6 = (1 +0.02)6 x = 1
y = 0.02
(1.02)6 = 1+6 (0.02) + 15 (0.02)2 + 15(0.02) + 20(0.02)3 + 15 (0.02)4
= 1 + 0.12 + 0.006 + 0.00016
= 1.12616
= 1.126 (to 3 d.p) (3)
(1- 3)(1+ 3) 1-3 2
M1 = 52 + 41 + 43 + 48 184 = 146
4 4
M2 = 184 – 52 + 40 = 172 = 43 for 7
4 4 for > 4
M3 = 172– 40 + 38 = 170 = 42.5
4 4
M4 = 170 – 38+36 = 168 = 42
4 4
M5 = 168 – 36 + 40 = 173 = 43 (3)
4 4
M6 = 172 – 40 + 44 = 176 = 44
4 4
M7 = 176 – 44 + 45 = 177 = 44.25
4 4
A(0,4) B (1,7) Object points
=
-1 2 4 7 8 13
Y = Mx + C
M = 13 – 8 = 5 = 1
9-4 5 1
y = x+c y = x + 4
8 = 4 + c c = 4
=
7 -8 -1 5 – 7 -2
AB = ½ BC and AB and BC share point B.
A,B,C are collinear. (3)
4 3 det. = 6 + 12 = 18
Inv.= 1 3 3
18
-4 2
1 3 3 2 -3 x 1 3 3 7
18 18
-4 2 4 2 y -4 2 5
x 36
1
y 18 -18 (3)
x = 2, y = -1 (A)
1/7 + 3/1.24 x 10-1 + 7/1.03 x 10-1
0.1429 + 3(0.8064) + 7 x 10 (0.9709)
10
= 0.1429 + 0.2419 + 67.96 (3)
=70.52 (A)
= 1460
(ii) OCB = x = 180 – 146 = 34
(iii) 360 – 107 – 146 – 34
= 73 0
Tan 600 = 1000 + y ; y = x tan 60 – 1000
X
X tan 300 = x tan 60 – 1000
0.5773 x = 1.732x – 1000
1.732x – 0.577 = 1000
1.155x = 1000
x = 1000
1.155 = 866.0 m (A) (4)
Time = 5/x
Faster = (x+20) k/h
Time = 5/x=20 T1 – T2 = 5/x – 5/x+20 = 30/3600
5 (x+20) –5x 1
x(x+20) 120
120 (5/x + 100 – 5x) = x2 + 20x (5)
x2 + 20x – 12000
x = –20 400 + 48000
2
x = -20 ± 220
2
Spd = 100 km/h
And x = 120 km/h (A)
Log x3 = Log x3 – log y ½
y
= 3 Log x – ½ Log y
= 8a – ½ ab
SECTION B
17.
| Marks | Mid point (x) | d = x-44.5 | F | E = d/10 | Ft | T2 | Ft2 v |
| 0-9 | 4.5 | -40 | 2 | -4 | -8 | 16 | 32 |
| 10-19 | 14.5 | -30 | 4 | -3 | -12 | 9 | 36 |
| 20-29 | 24.5 | -20 | 7 | -2 | -14 | 4 | 28 |
| 30-39 | 34.5 | -10 | 19 | -1 | -19 | 1 | 19 |
| 40-49 | 44.5 | -0 | 26 | 0 | 0 | 0 | 0 |
| 50-59 | 54.5 | -10 | 15 | 1 | 15 | 1 | 15 |
| 60-69 | 64.5 | 20 | 12 | 2 | 24 | 4 | 48 |
| 70-79 | 74.5 | 30 | 5 | 3 | 15 | 9 | 45 |
=90 =1 =223
(a) Mean = (1 / 90 x 10) + 44.5 = 44.5 + 0.111
= 44.610
(b) Standard deviation = 10 233/90 – (1/90)2
10 2.478 – 0.0001 (8)
10 2.478
10 x 1.574 = 15. 74 (A)
(c) Median 45.5th value = 39.5 + (13.5 x 10/ 26)
39.5 + 5.192 (A)
44.69
(a) The probability = Shaded area
Shaded area = ПR2 – П r2
= 22/7 (42 – 32) v = 22/7 x 7 = 22
Probability = 22 = 22 x 7 = 7
352/7 352 16
(b)
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6 |
| 2 | 2,1 | 2,2 | 2,3 | 2,4 | 2,5 | 2,6 |
| 3 | 3,1 | 3,2 | 3,3 | 3,4 | 3,5 | 3,6 |
| 4 | 4,1 | 4,2 | 4,3 | 4,4 | 4,5 | 4,6 |
| 5 | 5,1 | 5,2 | 5,3 | 5,4 | 5,5 | 5,6 |
| 6 | 6,1 | 6,2 | 6,3 | 6,4 | 6,5 | 6,6 |
(M)
(i) P(Product of 6) = P((1,6) or (2,3) or (3,2) or (6,1))
= 4/36 = 1/9
(4)
(ii) P (sum of 8) = P( (2,6) or (3,5) or (4,4) or (5,3) or (6,2) )
= 5/36 (A)
(iii) P (same number) = P (1,1) or (2,2) or (3,3) or (4,4) or (5,5) or (6,6)
6/36 = 1/6 (A)
(i) Cos 60 = x/20 x = 20 x 0.5 = 10 cm
BD = 12 – 10 = 2 cm
(ii) CD = y Sin 60 = y/20 y = 20x 0.8666
CD = 17.32 cm
(iii) CHG = 120 reflex = 2400
CHG = 240/360 x 2 x p x r
= 50.27
DBF = 1200/360 x 2 x П x r = 1/3 x 2 x 3.142 x 2
= 4.189 (A)
Length C D E f G H C = 50.27 + 2(17.32) + 4.189
= 89.189 (A)
Tan 750 = VO/2.5 v m
VO = 2.5 x 3.732
Perpendicular height = VO = 9.33 cm
2 (A)
VA2 = AO2 + VO2 (m)
3.5362 + 9.32
12.50 + 87.05
= 99.55 = 9.98 cm2 (A) (8)
(c ) = VAO Tan = 9.33 = 2.639
3.536
VAO = 69.240 (A)
(d) Cos VBA = = 2.5 /9.98 = 0.2505
VBA = 75.490
Area VBA = ½ x 5 x 4.99 x sin 75.45 m ( or other perimeter)
= 5 x 4.99 x 0.9681
= 24.15 cm2 (A)
X-sec Area = (1 x 25) + (½ x 25 x 2.5)
= 25 + 31.25 = 56. M
Volume = 56.25 x 12
= 675 m3
= П r2 x l = 3.14 x 9/100 x 9/100 x 3 (8)
= 0.07635 m3 /sec v (M)
Volume emptied in 2 minutes
= 0.07635 x 60 x 2
= 9.162 m2 (A)
1 m3 = 1000 l
= 9.162 litres
= 9160 litres (A)
24.
PART I
SECTION A (52 MARKS)
3Ö 0.09122 + Ö 3.152 (5mks)
0.1279 x 25.71
a2 – b2 (2mks)
Find the equation of the curve. (3mks)
has no inverse. (2mks) 3 k
log 16 – log 6 (3mks)
0.321 n2.2 (4mks)
point (3,-8) (3mks)
V = 3Ö k + x
sk – x
y
0.5 2 x
-1.5
-2
SECTION B (48 MARKS)
(i.) 3x + 2 – x2 = 0 (ii) –x2 – x = -2 (8mks)
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | |||||
| No. of students | 2 | 8 | 6 | 7 | 8 | 10 | |||||
| Marks | 60-69 | 70-79 | 80-89 | 90-99 | |||||||
| No. of Students | 9 | 6 | 3 | ||||||||
(i) The mean mark
(ii) The standard deviation (8mks)
AB such that the AK:KB = 1:1. The point R divides line OB in the ratio 3:2 and point S divides OK in
the ratio 3:1.
B
R
B K
0 a A
(a) Express in terms of a and b
(i) OK (iii) RS
(iii) OS (iv) RA
(b) Hence show that R,S and A are collinear. (8mks)
Calculate:
(i) The angle between planes BXC and ABCD.
(ii) The angle between planes ABXY and ABCD. (8mks)
(6mks)
From the graph, find:
(a) The value of x for which 2cosx = sin ½ x (1mk)
(b) The range of values of x for which –1.5 £ 2cos x £ 1.5 (1mk)
(a) Find the average speed of A. (6mks)
(b) How far was A from T when B reached S. (2mks)
(a) The distance between ports Q and R. (2mks)
(b) The distance between ports P and R. (3mks)
(c) The bearing of port R from port P. (3mks)
(a) Write down all the inequalities describing this information. (13mks)
(b) Graph the inequalities and find the maximum profit he makes from the crops in a year. (5mks)
PART II
3Ö 36.5 x 0.02573
1.938 (3mks)
4p
x + 1/x = 10/3 (3mks)
ÐXPY = 600. (2mks)
CDF = 680, BDC = 450 and BAE = 980.
Calculate the size of: (2mks)
Between the two places along the circle of Latitude:
x3
.
Find the area of the shaded segment. (3mks)
| Marks (x ) | Frequency (f) |
| 5
6 7 8 9 |
3
8 9 6 4 |
(4mks)
SECTION II (48MKS)
Calculate:
country.
| Mass(g) | 44 | 45 | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | ||||||||||
| Freq. | 1 | 2 | 2 | 1 | 6 | 11 | 9 | 7 | 10 | 12 | 16 | 16 | ||||||||||
| Mass(g) | 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 70 | |||||||||||
| Freq. | 10 | 11 | 9 | 7 | 5 | 3 | 4 | 3 | 3 | 1 | 1 | |||||||||||
Make a frequency Table with class-interval of 5g. Using 52g as a working mean, calculate the mean mass. Also calculate the median mass using ogive curve.
size. He wishes to order from supplies and find that he has room for 1000 cans. He knows that bitter
drinks has higher demand and so proposes to order at least twice as many cans of bitter as soft. He
wishes however to have at least 90cans of soft and not more than 720 cans of bitter. Taking x to be
the number of cans of soft and y to be the number of cans of bitter which he orders. Write down the
four inequalities involving x and y which satisfy these conditions. Construct and indicate clearly by
shading the unwanted regions.
iii) The bearing of A from B.
MATHEMATICS II
PART I
MARKING SCHEME:
Ö 3.152 = 1.776
3Ö 1.776 + 0.008317
0.1279 x 25.91
= 3Ö 1.784317 No. log
0.1279 x 25.91 1.784 0.2514
0.1279 -1.1069
25.71 1.4101 +
0.5170
-1.7344
x 1/3
10-1 x 8.155(6) 1-1.9115
Or 0.8155(6)
(a – b)(a + b) a + b
dx
y = x2 + 3x + c
-1 = 1 – 3 + c
c = 1 ; E.g y = x2 + 3x + 1
K = ± 3
18 9
log 16 log 8
6 3
= 2 log (8/3)
log (8/3)
= 2
2 2
Gradient of LN = 4/-14 = -2/7
Gradient of ^ bisector = 7/2
y – 2 = 7/2
x + 1
y = 7/2X + 11/2
100
0.5 = ( 85 )n
100
0.5 = 0.85n
log 0.5 = n log 0.85
log 0.5 = n
log 0.85
n = –1.6990 = -0.3010 = 4.264yrs
-1.9294 -0.0706
3.21 x 10-1 3.21
1 = 1 = 0.5807 = 0.005807
172.2 1.722 x 102 100
6.230 – 0.005807 = 6.224193
= 6. 224(3d.p)
| X | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 |
| y | 4 | 6.25 | 9 | 12.25 | 16 | 20.25 | 25 |
h = ½
Area= ½ x ½[29+2(6.25+9+12.25+16+20.25+25)]
= ¼ [29 + 127.5]
= ¼ x 156.5 = 39.125 sq. units.
cos q = 0 cos q = -0.5
q = 900, 2700 q = 1200, 2400
\ q = 900, 1200, 2400, 2700
9 -18
MP = 4 ( -9 ) = ( -4 )
9 -18 8
\ P is ( -1,0 )
300 = 30 + 3 (n – 1 )
300 = 30 + 3n – 3
300 – 27 = 3n
273 = 3n
91 = n
1 = 3m – 1
m = 2/3 = 0.6667
tan q = 0.6667 ; q = 33.690
FC = 25/3 = 8 1/3 cm
CX = 81/3 – 9/2 = 23/6 = 35/6 cm
CX x XK = XA x XN
33/6 x 3/2 = 3 x XN
\ XN = 111/12 cm
k – x
V3k – V3x = k + x
V3k – k = x + V3x
V3k – k = x( 1 + v3)
V3k – k = x
1 + V3
(ii) y = -2 Þ y > -2
(iii)pts. (0.5,0)
(0,-1.5)
m = -1.5 – 0 = 3
0 – 0.5
Eq. Y = 3x – 1.5 y < 3x – 1.5
SECTION B
| X | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| Y | -26 | -16 | -8 | -2 | 2 | 4 | 4 | 2 | -2 |
(i) Roots are x = -0.5 x = 3.6
(ii) y = -x2 + 3x + 2
0 = -x2 – x + 2
y = 4x (-2, -8) (1, 4)
Roots are x = -2, x = 1
0 – 9 4.5 2 – 70 – 140 4900 9800
10 – 19 14.5 8 – 60 – 480 3600 28,800
20 – 29 24.5 6 – 50 – 300 2500 15,000
30 – 39 34.5 7 – 40 – 280 1600 11,200
40 – 49 44.5 8 – 30 – 240 900 7,200
50 – 59 54.5 10 – 20 – 200 400 4,000
60 – 69 64.5 9 – 10 – 90 100 900
70 – 79 74.5 6 0 0 0 0
80 – 89 84.5 3 10 30 100 300
90 – 99 94.5 1 20 20 400 400
Sf = Sfd = Sfd2 = 77,600
60 -1680
(i) Mean = 74.5 + -1680
60
= 74.5 – 28 = 46.5
(ii) Standard deviation = Ö 77600 – ( –1680 )2
60 60
= Ö 1283.3 – 784
= Ö 499.3 = 22.35
(ii) OS = ¾ OK = 3/8 a + 3/8 b
(iii)RS = RO + OS = 3/8 a – 9/40 b
(iv) RA = RO + OA = – 3/5 b + a
= 3/8( a – 3/5 b)
\ RS = 3/8 RA
The vectors are parallel and they have a common
point R \ point R, S and A are collinear
KB = 3m NK = 1.5m XB = 5m
(i) XK = Ö 52 – 32 = Ö 16 = 4m
let ÐXKN = q
cos q = 1.5 = 0.375
4
q = 67.97(8)0
(ii) In DXNK
XN = Ö 42 – 1.52 = Ö 13.75 = 3.708
In D SMR; MR = KB = 3m
SM = XN = 3.708m
Let ÐSRM = a
tan a = 3.708 =1.236
3
a = 51.02(3)0
21.
21.
| 0 | 150 | 300 | 450 | 600 | 750 | 900 | 1050 | 1200 | 1350 | 1500 | 1650 | 1800 | |
| Y =2cosX | 2.00 | 1.93 | 1.73 | 1.41 | 1.00 | 0.52 | 0.00 | -0.52 | -1 | -1.41 | -1.73 | -1.93 | -2.00 |
| Y = sin ½ X | 0.00 | 0.13 | 0.26 | 0.38 | 0.50 | 0.61 | 0.71 | 0.79 | 0.87 | 0.92 | 0.97 | 0.99 | 1.00 |
(a) X = 730 ± 10
(b) Between 40.50 and 139.50
T S
Let the speed of A be X km/hr
Speed of B = (X + 10) km/hr
Time taken by A = 300 hrs
X
Time taken by B = 300 hrs
X + 10
300 – 300 = 5
x x + 10 4
300(x + 10) – 300x = 5
x(x + 10) 4
300x + 300 – 300x = 5
x2 + 10x
x2 + 10x – 2400 = 0.
x = 44.25
X = -54.25 N/A
(b) Distance covered by A in 1 ¼ hrs = 44.25 x 5/4 = 55.3 km
Distance of A from T is 300 – 55.3 = 244.7 km
2
(b)
PR2 = 2002 + 3002 –2x 200 x 300 cos700
= 130,000 – 41040 = 88,960
PR = 298.3 km
(c) 298.3 = 300
sin 700 sin a
sin a = 300 sin 700
298.3
= 0.9344
a = 69.10
Bearing of R from P is
40 + 69.1 = 109.10
(ii) 4,400X + 10,800Y £ 90,000
Simplifies to 11X + 27y £ 225
(iii) X + y £ 15
X > 0; y > 0
Boundaries
x = y pts (6,6) (12,12)
11x + 27y = 225 pts (13,3) (1,8)
X + y = 15 pts (0,15) (8,7)
Objective function
2400 x 3200y
(pt (2,1)
2400X + 3200y = 8000
Search line ® 3X + 4y = 10
Point that give maximum profit is (12,3)
\ maximum profit
= 2400 x 12 + 3200 x 3 = 38,400 shs.
MATHEMATICS II
PART II
MARKING SCHEME
36.5 1.5623
0.02573 –2.4104 +
-1.9727
1.938 0.2874 –
-1.6853
-3 + 2.6853
3 3
-1 + 0.8951
1.273(4) ¬ 0.1049
= 1.273(4)
let blouse be sh. y.
5x + 3y =1750 (i.)
3x + y = 850 (ii)
mult (ii) by 3
9x + 3y = 2550 (iii)
Subtract (iii) – (i.)
– 4x = -800
Subt for x
Shirt = sh 200 ; Blouse = sh 250
4p
y – c = 4pK2
y = 4pK2 + c
(b) y = 4 x 2 x 25 + 2 ; y = 202
3
3x2 – 10x + 3 = 0
3x (x – 3) – 1(x – 3) = 0
(3x – 1) (x – 3 ) = 0
x = 1/3 or x = 3
½ x 12.49 x 10 = 62.45cm2
Cos q = 10/16 = 0.625
q = 51.30 62.5
Sector 57.30 x 3.14 x 100 40.2 –
360 = 22.3
\ÐXC1Y = 1200
B1 \ÐC1XY = ÐC1YX
= 1800 – 1200 = 300
2
Construct 300 angles
at XY to get centres
B1 C1 and C2 mojar arcs drawn
2 on both sides with C1X and C2X
as centres.
ADB = 180 – (68 + 45 ) = 670
ABD = 180 – (67 + 82)
= 310
(a) 1800 – (67 + 82)0 = 310
ÐABD = 310 Opp = 1800
(b) (180 – 82)0 = 980 82 + 98 = 1800
1800 – (980 – 450) =
ÐCBD = 370 180 – (98 + 45)0
= 370
100 Discount = sh 32
Sold at sh 288
If no Discount = ( 320 x 20 ) % = 22.7%
288
Long diff x 60 x cos q nm
100 x 60 x Cos 500
100 x 60 x 0.866
5196nm = 100 x 2pR Cos 500
360
100 x 2 x 3.14 x 6371
360 = 5780Km
(b) 80 x 60 Cos 50 = 3895 Km
1m3 = 1000 L
20.16m3 = 20160 L
20160
3600
16560 L to fill
0.5 L – 1 sec
16560 L – ?
165600
5 x 3600
33120 hr
3600 @ 9.41 hrs ; @ 564.6 min.
a = -0.2
1 + 5(-0.2) + 10(-0.2)2 + 10(-0.2)3 + 5 (-0.2)4
1 – 1.0 + 0.4 – 0.08 + 0.008 = 0.3277 (4d.p)
p(R2 – r2)
3.14 (21 –19)
Vol 6.28cm2 x 3400cm
= 215.52m3
. 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
P(odd) = 3/6 = ½
P(Sum > 7 but < 10) = 9 /36
\ P(odd) and P(sum > 7 but < 10 )
= ½ x 9/36 = 9/72 = 1/8
ò( 8x2 – 3x-2) d4
16x3/3 + 6x-3/-3 + C
16x3/3 – 2/x3 + C
½ x 14 x 14 x 0.866 = 84.866cm2
Area of sector = 60 x3.14 x 14 x14 = 10.257
360
Shaded Area
84.666 – 10.257 = 74.409cm2
| Marks | F | Fx | fx2 |
| 5 | 3 | 15 | 75 |
| 6 | 8 | 48 | 288 |
| 7 | 9 | 63 | 441 |
| 8 | 6 | 48 | 384 |
| 9 | 4 | 36 | 324 |
åx = åf=30 åfx=210 1512
S.d = Ö åfx2 – ( åfx )2
åf åf
= Ö 1512 – (210)2
30 30
= Ö 50.4 – 49
= Ö 1.4 = 1,183
SECTION II .
(b) HB = Ö 52 + 7.072 = Ö 75 = 8.660
(c) q = Tan-1 5/5 = Tan-1 = 450
(d) b = Tan-1 5/7.071 = Tan-1 0.7071 = 35.30
(e) y = Tan-1 5/3.535 = Tan-1 = 54.70
(f) ÐAGX = 19.40
| x0 | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 |
| sin x0 | 0 | 0.50 | 0.66 | 1.00 | 0.866 | 0.500 | 0 |
y = 2 Sin (x0 + 100)
| X0 | 00 | 300 | 600 | 900 | 1200 | 1500 | 700 |
| 2 Sin(x +100) | 0.3472 | 1.286 | 1.8794 | 1.286 | 0.3472 | -0.3472 | -1.8794 |
Amplitudes for y = Sin x0 is 1
For
y = Sin(x+100) is 2.
| c.f | X | F |
| 61 | 53 | 12 |
| 16 | 54 | |
| 93 | 55 | 16 |
| 103 | 56 | 10 |
| 11 | 57 | |
| 123 | 58 | 9 |
| 130 | 59 | 7 |
| 135 | 60 | 5 |
| 138 | 61 | 3 |
| 142 | 62 | 4 |
| 145 | 63 | 3 |
| 148 | 64 | 3 |
| 149 | 65 | 1 |
| 150 | 70 | 1 |
Mean = x + 52 + -4
150
52 – 0.02
= 51.08
Median = 51.4g.
class interval 59
| Class interval | mid point | Freg. | c.f |
| 44-48 | 46 | 12 | 12 |
| 49-53 | 51 | 49 | 61 |
| 54-58 | 56 | 64 | 125 |
| 59-63 | 69 | 22 | 147 |
| 64-68 | 66 | 3 | 130 |
| 69-73 | 71 | 1 | 150 |
X £ 2Y
Y < 720
X > 90
21.(a) 1cm = 200Km/h
A = 200 x 7.5 = 1500 Km
B = 200 x 9 = 1800Km.
(b) (i.) 15.8cm x 200 (ii) Bearing 2240
= 3160 Km. (iii) Bearing 0490
= 0.5 x 0.6 0.5 x 0.4 P(R)’ x P(R)
= 0.3 = 0.2 0.5 x 0.6 = 0.3
0.5 x 0.4 = 0. 2= 0.5
= x3 – x2 – 2x
A1 = ò(x3 – x2 –2x) d4
-1[¼ x4 – 1/3 x2]-1
= 0 – ( ¼ + 1/3 – 1) = 5/12
A2 = 2ò(x3 – x2 –2x) d4
= 0ò ¼ x4 – 1/3 x3 – x2)-20
= ( ¼ .16 – 1/3 .8 – 8 )
= 4-0 – 8/3 – 4 = – 8/3
A1 = 5/12= A2 = 2 2/3
dy = 3x2 – 6x
At stationary
Points dy = 0
dx
i.e 3x2 – 6x = 0
3x(x – 2) = 0
x = 0 or 2
Distinguish
dy = 3x2 – 6x
dx
d2y = 6x – 6
dx2
(i) x = 0 dy2 = 6x – 6 = -6 (ii) x = 2
dx2 d2y = 6
-6 < 0 – maximum. dx2
\ (0,0) Max Pt. 6 > 0 hence
Minimum Pt.
x = 2, y = 8 – 12 = -4
(2, -4) minimum point.
MATHEMATICS II
PART I
SECTION 1 (52 Marks)
Ö7.5625 x 3Ö3.375
15 (5 mks)
y = 1 Ök + y
T2 k (3 mks)
-2 (3 mks)
rx4 – r
2xr – 2r
Log 3 (8-x) – log 3 (1+x) = 1
3,5,2,1,2,4,6,5 (4 mks)
½ tan x = sin x for -1800 £ x £ 3600. (3 mks).
|
|
y
SECTION 2 (48 Marks)
Use your graph to solve the equations.
(i) x2 + x – 6 = 0 (ii) x2 + 2x – 8 = 0 (8 mks)
Determine:-
Ksh = 1340 Italian lire. Answer the following questions:
|
|
Given that PM = kPA and BM = tBQ where k and t are scalars, express PM in two different ways and hence find the values of k and t.
Express PM in terms of r and h only. (8 mks)
| T | 6.56 | 17.7 | 47.8 | 129 | 349 | 941 | 2540 | 6860 |
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Draw a suitable straight line graph and use it to estimate the values of a and b. (8 mks)
MATHEMATICS III
PART II
Section I: (52 Marks)
8.67 (3 mks)
Ö 0.786 x (21.72)3
4 – 1
x2 – 4 x-2
ab
If B S (2 S 3) = 4 S 1, Find B. (3 mks)
M
850
1600
(b) By putting x = 0.01, find the approximate value of (1.02)6 correct to 4 S.F. (2 mks)
-5 2 5 3
(a) All pass: (1 mk)
(b) At least one fails: (2 mks)
C A
B P R
( 3x2 – 1 ) dx
4 x 2
1
(3,2)
(2,1) (4,1)
| Month | March | April | May | June | July |
| Depth (m) | 5.1 | 4.9 | 4.7 | 4.5 | 4.0 |
Calculate the three monthly moving averages. (3 mks)
n(n-7)
SECTION II: (48 Marks)
(i) Km (ii) nm (8 mks)
(take radius of earth to be 6400km, P = 3.14)
S = 2t3 + 4t3– 8t + 3.
Find:
(a) The velocity at : (i) t = 2 (ii) t = 3
| Time (t) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Velocity v(ms -1) | 0 | 12 | 24 | 35 | 41 | 45 | 47 |
Use trapezium rule to calculate the distance travelled between t = 1 and t = 6. (8 mks)
| Marks | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 |
| No. of students | 8 | 15 | 15 | 20 | 15 | 14 | 13 |
(a) Estimate the median mark. (2 mks)
(b) Using 44.5 as the assumed mean, calculate:-
(i) The mean mark: (2 mks)
(ii) The variance: (2 mks)
(iii) The standard deviation: (2 mks)
y = cosx + sin X for 00 Ð X Ð 3600 .
(b) Use your graph to deduce
(i) The amplitude
(ii) The period of the wave y = cos x + sin x.
(c) Use your graph to solve:
Cos x = – sin x for 00 Ð X Ð 3600 .
B
A E
C
D
(a) Write down the equation of the circle in the form ax2 + bx + cy2 + dy + e = 0 where a, b, c, d, e are constants. (2 mks)
(b) Calculate the length DE. (2 mks)
(c) Calculate the value of angle BED. (2 mks)
(d) Calculate the value of angle DCB. (2 mks)
(a) Represent all the information above as inequalities.
MATHEMATICS III
PART I
MARKING SCHEME
|
|
SOLUTION | MRK | AWARDING | |
| 1. | Ö7.5625 = 2.75
3Ö3.375 = 3Ö3375 X 3Ö10-3
= 3 Ö33 x 53 x 10-1 = 3 x 5 x 10-1 = 1.5
= 2.75 x 1.5 = 2.75 = 0.275 1.5 x 10 10
|
1
1
1 1 1
|
Method for Ö7.5625
Square root
Method for 3Ö 3Ö Answer |
|
| 5 | ||||
| 2. | T2y = Ö k+y
K T4y2k = k+y T4y2k – k = y K(T4y2-1) = y K = y T4y2 – 1
|
1
1
1
|
Removal of square root
Rearrangement of terms Answer |
|
| 3 | ||||
| 3. | (x 2) x = (8)
-2
x2 – 4 = 8
x = +Ö12 = + 2Ö3 = + 3.464
|
1
1
1
|
Matrix equation
Quadratic equation Answers in any form |
|
| 3 | ||||
| 4. | r(x2 – 1)
2r(x – 1)
r(x2 – 1)(x2 + 1) 2r (x – 1)
r(x – 1)(x + 1)( x2 + 1) 2r (x – 1)
= (x + 1)( x2 + 1) 2
|
1
1
1
|
Complete factorisation of numerator
Factorisation of denominator
Answer |
|
| 3 | ||||
| 5. | 1 = log3 3
8 – x = 3 1+x
-4x = -5
x = 5 4 |
1
1
1
|
Logarithic expression.
Equation
Answer
|
|
| 3 | ||||
| 6. | Let the centre be (a,b)
4-9 = -1 4-a -5-b 3-b
4-a = -4+9 -5-b = -3+b a = 4 b = -1 centre is (4,-1) |
1
1
1
|
Equation
Linear equations
Centre
|
|
| 3 | ||||
| 7. | Y = 4x + 5
Gradient = 4 Gradient of ^ line – ¼ y + 2 = – 1 x + 3 4 4y + x = -11
|
1 1
|
Gradient of ^ line. Equation.
|
|
| 2 | ||||
|
8. |
X = 28 = 3.5 8
standard deviation = Ö 22 = Ö2.75 = 1.658 8 |
1
1 1
1
|
Mean
d values d2 values
Answer |
|
| 4
|
||||
| 9. | a = 203 d = 7 L = 294
294 = 203 + 7(n-1) n = 14
S 14 = 14 (203 + 294) 2
= 7 x 497 = 3479
|
1
1
1
1
|
For both a and b
Equation
For n
Sum
|
|
| 4 | ||||
| 10. | Sin x = 2 sin x
Cos x
Sin x = 2 cosx Sin x
2 cos x = 1 cos x = 0.5
x = 600, 3000, -600 |
1
1
1
|
Simplification
Equation
All 3 values |
|
| 3 | ||||
| 11. | (1 +-2x)4 = 1-8x + 24x2 – 32x3 + 16x4
(0.82)4 = (1 + -2 x 0.09)4 x = 0.09 (0.82)4 = 1 – 0.72 + 0.1944 – 0.023328 + 0.00119376 = 0.35226576 @ 0.35227 (5 d..p) |
1
1 1
1
|
Expansion
Value of x All terms
Rounded |
|
| 4 | ||||
| 12. | 2 = 5m – 3
m = 1 tan q = 1 q = 450 |
1 1
|
Value of m. Angle |
|
| 2 | ||||
| 13. | Let the number be xy
3y = x + 14 10y + x = 10x + y + 36 = 9y – 9x Þ 36 3y – x = 14 9y – 9x = 36 y = 5 x = 1 the number is 15. |
1 1
1
1
|
1st equation 2nd equation
method of solving
Answer
|
|
|
S |
4 | |||
| 14. | Let ÐAOB = q
q x 2 x 22 x 7 = 11 360 7 q = 900
Area shaded = 90 x 22 x 7 x 7 – 1 x 7 x 7 360 7 2 = 77 – 49 2 2 = 28 = 14cm2 2 |
1
1
1
|
Value of q
Substitution
Answer |
|
| 3 | ||||
| 15. | P(WBb) = 6 x 15 x 15
36 35 34
= 15 476 |
1
1
|
Method
Answer |
|
| 2 | ||||
| 16. | Equation inequality
L1 y = x y £ x L2 y = -2 y ³ -2 L3 2y + 5x = 21 2y + 5x < 21 |
1
1 1 1
|
1 mark for each inequality.
Method for obtaining L3
|
|
|
(i) roots are x = -3 x = 2 (ii) y = x2 + x-6 0 = x2 + 2x-8 y = -x + 2 roots are x = -4 x = 2 |
4 | 2
1
1
1
1
1
|
For all correct points.
1 for atleast five correct points.
Correct plotting.
Scale
Smoothness of curve
Both roots
Linear equation
Both roots
|
|
|
|
8 | |||
| 18. | h = 15
h+50 40
h = 30cm H = 80cm
(a) Volume = 1/3 p x 40 x 40 x 80 – 1/3 p x 15 x 15 x 30
= 128000 p – 6750 p 3 3 3
(b) L2 = 802 + 402 L = 152 + 302 = 6400 + 1600 = 225 + 900 = 8000 = 1125 L = 89.44 cm L = 33.54 cm Curved surface area of bucket = p x 40 x 89.44 p x15x33.54 = 3577.6p – 503.1p = 3074.5cm2 |
1
1
1
1
1
1
1
1
|
Expression
Value of H
Substitution
Volume
L
L
Substitution
Area
|
|
| 8 | ||||
|
19. |
|
|||
|
20. |
(i) ÐRPQ = 130 ÐPQR = 320+900+240 = 1460 ÐPRQ = 1800 – (1460 + 130) = 210
(ii) P = 7 sin130 sin 210 P = 7 sin 130 Sin 210 = 4.394km
P T
(iii) Let PR = q
q = 7 sin 1460 sin 210
q = 7 sin 1460 sin 210 q = 10.92 km
sin 450 = RT 10.92
RT = 10.92 sin 450
= 7.72 km (2 d..p)
|
1
1
1
1
1
1
1
1
|
Fair sketch
ÐPRQ
Equation
Method
Equation
Distance PR
Equation
RT
|
|
| 8 | ||||
| 21. | (a) 4x = 9y
2(x+y) = y+44 Þ 2x + y = 44
4x – 9y = 0 4x + 2y = 88 11y = 88 y = 8
x = 18 (b) P(RR) = 18 x 18 = 81 26 26 169
|
1
1
2
1 1 1 1
|
Equation
Equation
Method of solving Value y Value x Method Answer |
|
| 8 | ||||
| 22. | (a) 67,000 Ksh = 67,000 US dollars
16.75 = 4,000 dollars
(b) 2 x 4,000 = 1600 US dollars 5 1600 US dollars = 1600 x 1340 = 2,144,000 Italian lire (c) Remainder = 2400 US dollars 5 x 2400 = 1500 US dollars 8 1500 US dollars = 1500 x 1.8 = 2700 Deutche marks (d) Remainder = 900 US Dollars 900 US Dollars = 900 x 16.75 Ksh. = 15,075 Ksh.
|
1
1
1
1 1
1 1 1
|
Method
Answer
Method
Answer
For 1500
Answer
Method Ksh. |
|
| 8 | ||||
| 23. | PM = kPA
= k(r + 1h) 2 = kr + 1kh 2 PM = PB + BM 3h + t BQ 4 = 3h + t(-3h + r) 4 4
= 3h – 3t h + tr 4 4 = 3 – 3t h + tr 4 4
t = k 3 – 3t = 1k 4 4 2 3 – 3t = 1 t 4 4 2 5t = 3 4 4 t = 3 + 4 4 5 = 3 5 \ k = 3 5 \ PM = 3r + 3h 5 10 |
1
1
1
1
1
1
1 1
|
PM
PM
PM simplified
Both equations
method
Value of k
Value k PM
|
|
| 8 | ||||
|
24. |
Y
LogT
Log T = log a + x log b Log T Þ 0.82, 1.25, 1.68, 2.11, 2.54, 2.97, 3.40, 3.84
y – intercept = log a = 0 a = 1 gradient = 3.84 – 0.82 = 3.02 9 – 2 7 = 0.4315
log b = 0.4315 = 0.4315 b = antilog 0.4315 b = 2.7 |
1
1 2
1
1
1 8 |
Plotting
Linear All correct logs
Value of a Method of gradient
Value of b |
MATHEMATICS III
PART II
MARKING SCHEME
| 1. | No log
8.69 0.9390 0.786 1.8954 21.72 1.3369 1.2323 1.7067 – 2
2 + 1.7067 2 2 – 1 + 0.8533 0.7134 x 10 -1 = 0.07134 |
M1
M1
A1
|
ü reading to 4 s.f
Rearranging |
| 3 | |||
| 2. |
4 – 1 (x-2)(x+2) (x-2)
– x+2 (x-2(x+2) – (x-2) (x-2(x+2)
-1 x+2 |
M1
M1
A1
|
|
| 3 | |||
| 3. |
Re6000 = Ksh. 75000 Spent 5000 Rem 2500 Rem 2500 1.25 Re 2000 |
M1
M1
A1
|
|
| 3 | |||
| 4. | 2x – 1 , 2z + 1 , 2x + 3
6x + 3 = 105 6x = 102 x = 17 |
M1
M1 A1 A1
|
Allow M1 for us of different variable. |
| 4 | |||
| 5. |
4 * 1 = 5 4 2 * 3 = 5 6 A * 5 = 5 6 4 A + 5 = 5 x 5A 6 4 6 A + 5 = 25 A 6 24 A = 20 |
M1
M1
A1 3 |
|
| 6. |
180 – M + 20 + 95 = 180 295 – M = 180 – M = – 115 M = 115
|
B1
B1
A1
|
|
| 3 | |||
|
7. |
1 + 2x + 60x2 + 160x3 + 1 + 0.2 + 0.006 + 0.00016 = 1.20616 = 1.206 |
M1 M1 M1 A1 4 |
Only upto term in x3. Correct substitution
Only 4 s.f.
|
| 8. |
3 -1 2 1 = I -5 2 5 3
6 -5 3 -3 -10 +10 -5 + 6
1 0 0 1
|
M1
M1
A1
|
Matrix multiplication gives :
I 1 0 0 1 |
| 3 | |||
| 9. |
(a) 2 x 3 x 4 = 2
3 4 5 5 (b) 2 x 3 x 1 + 2 x 1 x 4 + 1 x 3 x 4
1 + 4 + 1 10 15 5
= 17 10 |
M1
M1
A1
|
|
| 3 | |||
| 10. | ÐQCB = 300
180 – (27 + 30) = 1230 \ BAC = 570.
OBA = 370 OAB = 370
AOB = 1060 \ ACB = 530
|
M1
M1
A1
|
Isosceles triangle.
Angle at centre is twice angle at circumference. |
| 3 | |||
| 11. | V = 1 x 3.14 x r 2 x 10 = 270
L.S.F. 20 = 2 30 3 V.S.F = 2 3 = 8 3 27 Vol. of cone = 8 x 270 27 = 80cm3 \ Vol. Of frusturm = (270 – 80) = 190cm3 |
M1
M1
A1
|
|
| 2 | |||
| 12.
|
3x 3 – x -1 2 3 -1 1
x 3 + 1 2 x 1
8 + 1 – ( 1 – 1) 2 8 1 – 2 = 6 1 2 2 |
M1
A1 2 |
|
| 13. | (2 x 3 x 1.5) volume
9 m3 1L º 1000 cm3 1000 L = 1 m3 9000 L = 9 m3 1000 L = 1 tonne 9000 L = 9 tonnes. |
M1
A1
|
|
| 2 | |||
| 14. | y ³ 1 (i)
y < x – 1 (ii) y < 5 – x (iii) |
B1 B1
|
|
| 3 | |||
| 15. | M1 = 5.1 + 4.9 + 4.7 = 4.9
3 M2 = 4.9 + 4.7 + 4.5 = 4.7 3 M3 = 4.7 + 4.5 + 4.0 = 4.4 3 |
M1
M1 M1
|
|
| 3 | |||
| 16. | Original contribution per woman = 6300
N Contribution when 7 withdraw = 6300 (n-7) Increase – Diff. 6300 – 6300 n-7 n 6300n – 6300(n-7) n(n-7) 6300n – 6300 + 44100 n(n-7) 44100 n(n-7) |
M1
M1
1 3 |
|
| SECTION II (48 Marks)
|
|||
| 17. | (i)
1150
A B
Centre of circles of latitude 500 S. R Cos 500 AB = 115 x 2p R Cos 50o 115 x 40192 x 0.6428 360 = 8252.98 km
(ii) Arc AB 60 x 115 Cos 50 nm 60 x 115 x 0.6428 nm 4435 nm
|
M1 M1
M1
A1
M1 M1 M1 A1
|
No. log 60 1.7782 1+5 2.0607 0.6428 1.8080 4435nm 3.6469 |
| 8 | |||
| 18. |
(a) V = ds = 6t2 + 8t – 8
dt (i) t = 2 V = 6×4 + 8×2 – 8 = 32 ms-1 (ii) t = 3 V = 6×9 + 8×3 – 8 = 70ms-1
(b) Particle is at rest when V = 0 6t2 + 8t – 8 = 0 2(3t – 2) (t+2) = 0 t = 2 t = -2 3 particle is at rest at t = 2 seconds 3 |
Do not accept t = -2. Must be stated. |
|
| 8 | |||
| 19. | Area under velocity – time.
graph gives distance.
A = { h ½ (y1 + y6 ) + y2 + y3 + y4 + y5 )}
= 1 { ½ ( 12+47) + 24 + 35 + 41 + 45)} = 29.5 + 14.5 = 174.5m |
B1 B1 M1 M1 B1 B1 A1
|
Trapezium rule only accepted.
Formula.
Substitution into formular. |
| 8 | |||
| 20. | Drawing actual
Scale 1cm = 2cm
Radius 1cm = 2cm |
M1
M1
M1
M1
M1 M1
M1 M1
|
Bisect ÐA
Bisect Ð B
Intersection at centre of inscribed circle. Draw circle.
Measure radius. Arcs must be clearly shown. |
| 8 | |||
|
21. |
mean = 44.5 + 130 100 = 44.5 + 1.3 = 45.8
(b) Variance S (x – A) 2 = 2800 Sf 100 = 28 S.D. = Ö 28 = 5.292
|
M1
A1
M1
A1 M1 A1
|
|
| 8 | |||
|
22. |
y = sin x
x 0 60 120 180 240 30 360 sin x 0 0.866 0.866 0 -0.866 -0.866 0 y = cos x x q 60 120 180 240 300 360 cos x 1 0.5 -0.5 -1.0 -0.5 0.5 1.0 y = cosx + sinx x q 60 120 180 240 30 360 cosx + sinx 1 1.366 0.366 -1 -1.366 -0.366 1.0 (c) Cos x = – sin x x = 450 , 2250 |
||
|
23. |
(i) amplitude = 1.366 (ii) Period = 3000
(a) (x+1) 2 + (y-6)2 = 32 x2 + 2x + 1 + y2 – 12y + 36 = 9 x2 + 2x + y2 – 12y + 28 = 0
(b) cos 10 = OD DE = 3 DE 0.9848 DE = 3.046
(c) Twice ÐOED 100 x 2 = 200
(d) DAB = 800 \ DCB = 1000 |
M1
A1
M1 A1
M1 A1
M1 A1
|
Formular (x-a)2 + (y-b)2 = r2
Cyclic quad.
|
| 8 | |||
| 24. | Let number of trips by 12 tonne lorry be x.
Let number of trips by 7 tonne lorry be y.
(a) x > 0 ; y > 0 24x + 21y £ 150
12 x 25 x X + 15 x 7 x y £ 1200 300x + 105y £ 1200 x > 2y
(b) Ref. Graph paper. Minimising: 3 – 12 tonne lorry and 2 – 7 tonne lorries should be deployed. |
B1
B1 B1
|
|
MATHEMATICS IV
PART I
SECTION 1 (52MKS)
12.3 (4mks)
3 5 5
C D
B A
Express V in terms of S only. (3mks)
Find the value of K. (3mks)
alloy. He can only get metal P from a metallic ore which contains 20% of it. How many Kgs of the ore
does he need. (3mks)
-2 3
co-ordinate of point C. 3mks)
(4mks)
and AC to form a cone . Calculate the height of the cone formed.
(4mks)
5Ocm
50cm
sales above this. If in January he got shs 12,200 as commission, what were his total sales (4mks)
Ö 7 – 2
Leaving your answer in the form Öa + Öb
C
Where a ,b, and c are integers (3mks)
SECTION II (48 mks)
A F
Calculate (i) <CBE (3mks)
(ii) <BEA (2mks)
(iii) <EAB (3mks)
(a) Write down the first 4 terms of the sequence . (5mks)
(b) Find the sum of the first 5 terms using positive values of the common ratio (r)
(3mks)
constants. In an experiment , the following values of E and F were obtained .
| E | 2 | 4 | 6 | 8 |
| F | 16.1 | 127.8 | 431.9 | 1024 |
Use graphical method to determine the value of k and n (Graph paper provided) (8mks)
(b) If the density of the wire is 5g/cm3. Calculate the mass of the sphere in kg. (3mks)
9cm
(a) The co-ordinates of Q,R,S, (6mks)
(b) Find the equation of the diagonal SQ (2mks)
(8mks)
MATHEMATICS IV
PART II
SECTION I (52MKS)
Ö 0.0784 x 0.27 (leave your answer in standard form)
0.1875
old are they now? (3mks)
AB = 5cm. Calculate the perimeter of the trapezium ABCE (3mks)
E D C
A B
2b2 – 3a2c
Angle BAC = 600 . Calculate the total surface area of the bar in M2. (4mks)
School flag F is 150m away from C and on a bearing of 0200. Calculate the distance and
bearing of A from F. (5mks)
(i) Find the number of red balls (2mks)
(ii) If 2 balls are chosen at random without replacement, find the probability that they are of different colour. (2mks)
Determine the radius of the original circle. (3mks)
(i) Write down the first 3 terms of the G.P (1mk)
(ii) Calculate the sum of the first 5 terms (2mks)
SECTION II (48MKS)
(a) The three are white (2mks)
(b) At least two are blue (3mks)
(c) Two are white and one is blue (3mks)
(a) Calculate the number of tiles needed. (2mks)
(b) Tiles needed for 15 such rooms are packed in cartons containing 20 tiles. How many cartons are
there in total? (2mks)
(c) Each carton costs shs. 800. He spends shs. 100 to transport each 5 cartons. How much would one
sell each carton to make 20% profit ? (4mks)
Income in K£ P.a Rate (Ksh) £
1 – 2100 2
2101 – 4200 3
4201 – 6303 5
6301 – 8400 7
(a) Maina earns £ 1800 P.a. How much tax does he pay? (2mks)
(b) Okoth is housed by his employer and therefore 15% is added to salary to make taxable income. He
pays nominal rent of Sh.100 p.m His total tax relief is Shs.450. If he earns K£3600 P.a, how much
tax does he pay? (6mks)
Q
B
R
O A
(a) Write in terms of a and b vector PQ (2mks)
(b) Given that AR = hAB where h is a scalar, write OR in terms h, a. and b (2mks)
(c) PR = K PQ Where K is a scalar, write OR in terms of k, a and b (1mk)
(d) Calculate the value of k and h (3mks)
co-ordinates of A1B1C1 and plot ABC and A1B1C1 on the given grid.
Transformation Q maps A1B1C1 onto A11 (-6,2) B11(-2,3) and C11(-6,6). Find the matrix Q and plot
A11B11C11on the same grid. Describe Q fully. (8mks)
BC = 3.5cm and AC = 4.5cm. Escribe circle centre 0 on BC to touch AB and
AC produced at P and Q respectively. Calculate the area of the circle. (8mks)
330 334 354 348 337 349 343 335 344 355
392 341 358 375 353 369 353 355 352 362
340 384 316 386 361 323 362 350 390 334
338 355 326 379 349 328 347 321 354 367
(i) Make a frequency table with intervals of 10 with the lowest class starting at 31 (2mks)
(ii) State the modal and median class (2mks)
(iii) Calculate the mean mark using an assumed mean of 355.5 (4mks)
PART 1
MARKING SCHEME
| 1. |
Ö – 7.939 12.3
= No log 7.939 0.8998 12.3 1.0899 T.8099 1/3 = 3 + 2.8099 T.9363 3
= -0.8635 |
B1
B
M1
A14
|
Subtraction
Logs
Divide by 3
Ans |
| 2. | 5x – 3 (3x –7 ) = 3(x – 2 )
5x – 9x + 21 = 3x – 6 -7x = -27 x = 36/7
|
M1
M1
A13 |
Multiplication
Removal ( )
Ans |
| 3. | 3x +5y + x = 180
9x = 180 x = 20 y = 60 |
M1
A1 B13 |
Eqn
X B
|
| 4. |
. r = 3v 1/3 4P
. r = S ½ 4P
\ 3V 1/3 = S ½ 4P 4P
3V = S 3/2 4P 4P
V = 4P S 3/2 3 4P
|
B1
M1
A13
|
Value r
Equation
Expression |
|
5.
6. |
Grad line = ¼
y – 2 = ¼ x – 5 y = ¼ x + ¾ k = ¾ P in Alloy = 4/10 x 800 = 320g = 100 x 320 20 = 3.2 kg
|
M 1
A1 A 13
B1
M1
A 1
|
Equation
Equation K
P in alloy
Expression
Ans |
|
7. |
B (a,b) , C (x ,y) .a – 2 = 5 .b – 8 -2 .a = 8 b = 6 B(8, 6 ) x – 8 = 3 y – 6 4 x = 11, y = 10 c(11,10)
|
B1
M1
A13 |
B conduct
Formular
C |
| 8. |
80 – x
.h = x tan 70 h = (80 – x ) tan 60 \ x tan 70 = 80 tan 60-x tan 60 2.7475x + 1.732x = 138.6 4.4796 x = 138.6 .h = 138.6 x tan 60 4.4796
= 53.59 |
B1 M1
M1
A14 |
Expression for h both Equation
Expression for h
Ans |
| 9. | 2pr = 90 x 2p x 50
360 r = 12.5 h = Ö2500 – 156.25 = Ö2343.75 = 48.41 cm
|
M1
P A1 M1
A14 |
Equation
.r expression for h
ans
|
|
10. |
100 n = 302.323 n = 3.023 99n = 299.3 n = 2993 990 = 323/990 |
M1
A14
|
Equation
Ans
|
| 11. |
AB = 3-1
5-9 = 2 -4 BC = 4 -8 AB = ½ BC \ AB // BC But B is common \ A,B,C are collinear.
|
B1
B1
B13 |
A B & BC
Both
Both
|
| 12. | 4% of 200,000 = 8000/=
balance = 4200/= 6% of x = 4200/= x = 4200 x 100 6 = 70,000 sales = sh. 270,000 |
B1
M1 A1 B14
|
Both
Expression Extra sales Ans
|
|
13 . |
Time = 22/7 x 3.5/2x 3.5/2 x 200 hrs 22/7x 140x140x 140x 3600
= 8960 3600 = 2 hrs 29min |
M1 M1
M1
A14
|
Vol tank Vol tank
Div x 3600
Tank
|
| 14.
|
Ö3 = Ö3 Ö7 + Ö2
Ö7Ö2 Ö2Ö2 Ö7+ Ö2
= Ö3 Ö7 + Ö2 5
= Ö21 + Ö6 5 |
M1
M1
A13 |
Multi
Expression
Ans |
| 15. | 3 £ x 2 x2 £ 35
±1.732 £x x £ ± 5.916 1.732 £ x £ 5.916 integral x : 2, 3, 4, 5
|
B1
B1 B1 B14 |
Lower limit
Upper limit Range Integral values
|
| 16. | No of days = 8/6 x 5/8 x 12
= 10 days |
M1
A12 |
Expression
days |
| 17. | (i) ÐCED = ÐECD = 30
Ð CDE = 180 – 60 = 120 Ð CBE = 180-120 =60 (ii) Ð AEC = 90+30 = 120 Ð EAB = 180-(120+45) = 150 (iii) ÐBEO = 90-45 = 45 |
B1
B1 B1 B1
B1
B1 B1
B18 |
ÐA EB = 450
ÐBEO |
| 18. | .ar + ar2 = 9/4
3r + 3r2 = 9/4 12r2 + 12r – 9 = 0 4r2 + 3r – 3 = 0 4r2 + 6r – 2r –3 = 0 (2r – 1) (2r + 3) = 0 r = ½ or r = -11/2
Ss = 3(1- (1/2 )5) 1 – ½
= 3 (1-12/3 2) ½ = 6 ( 31/32) = 6 31/32
|
B1
B1
B1
M1 A1
M1
M1
A18
|
|
| 19. |
LOG E. 0.3010 0.6021 0.7782 0.9031LOG F 1.2068 2.1065 2.6354 3.0103
Log E =n log F + Log K
.n = gradient = 2 2.4 – 1.4 = 12 = 3 Log k. = 0.3 0.7 – 0.3 4 .k = 1.995 ¾ 2 E = 2F 3 |
B1
B1
S1
P1
L1
M1 A1
B18
|
Log E
Log F
Scale
Plotting
Line
Gradient
K
|
| 20 |
.x -2 -1 0 1 2 3 4 .y 17 6 1 6 9 22 41
.y = 3x 2 – 2x + 1 – 0 = 3x 2 – 3x – 2 y = x + 3
|
B2
B1
B1
S1 P1 C1
L1
B1
8
|
All values
At least 5
Line
Scale Plotting Smooth curve
Line drawn
Value of r
|
| 21. | .h = ¾ p x 18 x 18x 18
p x 0.04 x 0.04 = 24 x 18x 18x 18 0.04 x 0.04 x 100
= 48,600m
density = 4/3 x 22/7 x 18 x 18x 18x 15 kg 1000 = 122.2kg |
M1
M1 M1 M1
A1
M1 M1 A18 |
N of wire
¸ to length in cm ¸ for length conversing to metres
length
expression for density conversion to kg ans
|
| 22. |
H = Ö152 – 92 = Ö144 = 12
X/6 = 9/12 X = 4.5 Volume = 1/3 x 22/7x (81 x 12 –20.25×6 )
= 22/21 (972 – 121 -5)
= 891 cm3
|
M1
A1
M1 A1 M1 M1 M1
A18 |
Method
Method Radius Small vd Large vol Subtraction of vol.
Ans |
| 23. | R(-a , b) , Q (c,d), S(x , y) ,P (5,0)
PR is diagonal (a) Mid point PR (0,0) a + 5 = 0 2 .a = -5 b- 0 = 0 2 R (-5,0) Grad PQ = -2 Grad RS = -2 .d – 0 = -2 c –5 .d – 0 = ½ c+5 .d+ 2c = 10 2d – c = 5×2 – 4d – 2c = 10 5d = 20 d = 4 c = 3 Q (3, 4) x + 3 , y+4 = (0,0) 2 2 x = -3 , y = -4 \ s(-3 -4)
(b) y – 4 = 8 x – 3 6 3y = 8x – 12 |
B1
M1
M1
A1
M1 A1
M1
A18 |
Ans .
Expression both correct
Equation
Ans
Expression
Equation
|
| 1. | 784 X 27 =
187500 Ö 784 x 9 = 4 x 7x 3 62500 250 = 42 125 = 0.336
|
M1
M1
A1 |
Factors for Fraction or equivalent
C.A.O |
| 3 | |||
| 2. | Father 3x , r son = x
2(x +10) = 3x + 10 2x +20 = 3x + 10 x = 10 father = 30 |
M1
A1 B1
|
Expression
|
| 3 | |||
| 3. |
3 = sin 60
AE AE = 3 Sin 60 = 3.464 perimeter = 5×2 + 3.464 x 3 = 10+10.393 = 20.39 |
M1
A1
B1 |
Side of a triangle
Perimeter |
| 3 | |||
| 4. | .a3 – b-2c2 = (-2)3 – 3 –2(-1)2
2b2 – 3a2c 2(3)2 –3(-2)2(-1) = -8 –3-2 18 + 12 = -13 30 |
M1
M1
A1 |
Substitution
Signs
C.A.O |
| 3 | |||
| 5. | Ksh 189,000 = $ 189,000
75.6 = $ 2500 balance = $ 2500 = Kshs. 189,000 Kshs. 189,000 = 189,000 115.8 Uk ₤1632 |
M1
A1
M1 A1
A14
|
Conversion
Conversion
|
| 6. | Area of 2 triangles = 2 (½ x 8x 5 sin 60)
= 40 sin 60 = 40x 0.8660 = 34.64 cm2 Area of rectangle = 300 x 8 + 300 x 5 +300 x BC BC = Ö64 +25 – 2 x 40cos 60 = Ö89 – 80 x 0.5 = Ö89 – 40 = Ö49 = 7 Total S.A. = 300 (8+5+7) + 34.64 cm2 = 6000 + 34.64 = 6034.64 cm2 |
M1
M1
M1
A1 |
Areas of D
B.C. expression
Area
|
| 4 | |||
| 7. | AF2 = 32+42+-2+12x cos 50
= 25 – 24 x 0.6428 = 25-15.43 = 9.57 AF = 3.094 x 50 AF = 154.7m Sin Q = 200 sin 50o 154.7 = 0.9904 Q = 82.040 Bearing = 117.960 |
M1
A1 M1
A1 B1
|
Bearing |
| 5 | |||
| 8. | (i) No. of white = w
w = 2 w+9 3 3w = 2w + 18 w = 18 (ii) p(different colour ) = p(WB N BW) = 2 x 9 + 9 x 18 3 25 27 25 = 12/25 |
M1
A1 M1
A1 |
|
| 4 | |||
| 9. | A.sf = 1
49 smaller area = 1 x 441 p 49 = 9p pr2 = 9p r2 = 9 r = 3 |
M1
M1
A1
|
|
| 3 | |||
| 10. | Largest area = 22 x (14.5)2
7 = 660.8 cm 2 smallest area = 22/7 x (13.5)2 = 572.8 572.8 £ A £ 660.81 |
M1
M1
A1 |
|
| 3 | |||
| 11. | (1 +2 x)5 = 1 + 5 (2x) + 10 (2x)2 + 10 (2x)3
= 1 + 10x + 40x2 + 80x3 2.0455 = 1+2 (0.52)5 = 1+10 (0.52)+ 40(0.52)2+80(0.52)3 = 1+5.2 + 10.82 + 11.25 = 28.27 |
M1
A1
M1
A1 |
|
| 4 | |||
| 12. | Tn = 5x 2n –2
(i) T1 , T2, T3 = 2.5, 5, 10 (ii) S5 = 2.5(25-1) 2-1 = 2.5 (31) = 77.5 |
B1 M1
A1 |
All terms
|
| 3 | |||
| 13. | 12 = 22 x 3
18 = 2 x 32 30 = 2x3x5 Lcm = 22 x 32x 5 = 180 min = 3hrs time they ring together =11.55 +3 = 2.55 p.m |
M1
A1 B1 |
|
| 3 | |||
| 14. | Map area = 40cm 2
Actual area = 200x200x40m2 = 200x200x40ha 100×100 = 320ha |
M1
M1
A1 |
Area in m2
Area in ha
CAO |
| 3 | |||
| 15. | 3p + 2r = 13
p + 2r = 9 – 2p = 4 p = sh 2 r = 3.50 |
M1
A1 B1 |
|
| 3 | |||
| 16. | 110 + 100+130+2x +3x = 540
5x = 200 x = 400 2x , 3x = 80 and 1200 res |
M1
A A12 |
|
| 17. | Contribution / person = 180,000
X New contribution = 180,000 x – 2 180,000 – 180,000 = 24,000 x –2 x 180,000x – 180,000x +360,000 = 24,000(x-2)x 24,000x2 – 48,000x – 360,000 =0 x2 – 2x – 15 = 0 x2 – 5x + 3x – 15 = 0 x (x – 5)+ 3 (x – 5) = 0 (x + 3 )(x – 5) = 0 x = -3 or = 5 remaining members = 5-2 = 3 |
B1
B1
M1 M1
A1 M1
A1
B1
|
‘C’
eqn mult
eqn factor
both ans
remaining members |
| 8 | |||
| 18. | (a) P (3 white) = 8 x 7 x 6 = 28
13 12 11 143 (b) P(at least 2 blue)=p(WBBorBBWorBWB)orBBB = 8 x 5 x 4 + 5 x 4 x 8 13 12 11 13 12 11 + 5 x 8 x 4 + 8 x 7 x 6 13 12 11 13 12 11 = 204 429 = 68 143 (c) p(2 white and one blue )= p(WWB or WBW or BWW) = 8 x 7 x 5 + 8 x 5 x 7 + 5 x 8 x 7 13 12 11 13 12 11 13 12 11 = 3 x 8 x 7 x 5 13 x 12 x 11
= 70 143 |
M1
A1
M1
M1
A1
M1 M1
A1
|
|
| 8 | |||
| 19. | (a) recourt area = 10.5 x 6 m2
title area = 0.3 x 0.3 m2 No of tiles = 10.5 x 6 0.3 x 0.3 = 700 (b) No of cartons = 700 x 15 20 = 52.5
(c) Cost of 525 cartons = 525 x 100 + 800 x 525 + transport 5 = 10,500+420,000 = 430,500 sale price = 120 x 4.30,500 100 = sh 516,600 s.p of a carton = 516,600 525 = sh. 984 |
M1 A1
M1
A1
B1
M1
M1
A1
|
|
| 8 | |||
| 20. | (a) Maina`s tax dues = 1800 x 10
100 = 180 (b) Taxable income = 3600 x 115 – n rent 100 = 36 x 115 – 100 x 12 20 = 4140 – 60 = 4080 Tax dues = 10 x 2100 + 15 x 1980 100 100 = 210 + 297 = 507 Tax relief = 270- Tax paid = 237 |
M1
A1
M1
A1 M1 M1
A1
B1
|
1st slab 2nd slab |
| 8 | |||
| 21. | (a) PQ = –3/5 a + 3/1b
= 31/2 – 3/5 a (b) OR = h a + h b = a – ha + hb = (1-h) a + h b (c) OR = 3/5 a + k (31/2 b – 3/5a) = (3/5 – 3/5k)a +3k b (d) 1 – h = 3/5 – 3/5k (i) 3k = h (ii) Sub (i) 1 – 3k = 3/5 – 3/5k 5- 15k = 3-3k 12k = 2 k = 1/6 h = ½
|
B1
M1 A1 M1 A1
M1
A1 B1 |
|
|
8 |
|||
|
22. |
P(ABC) = 0 – 1 1 4 3 = -3 -1 -3 1 0 3 1 3 1 4 3 A1 (-3,1)B1 (-1,4)C1(-3,3) Q(A1B1C1) = a b -3 –1 -3 = -6 –2 –6 c d 1 4 3 2 8 6
=> -3a + b = -6 -3c + d = 2 -a + 4b = -2 x 3 -c + 4d = 8 x 3 – 3a + 12b = -6 – 3c + 12d = 24 11b = 0 -11d = -22 b = 0 d = 2 a = 2 d = 2 c = 0 Q = 2 0 0 2
|
M1 A1
M1
M1
A1
B1
B1
B1
|
A1 B1 C1
L Q
A1 B1 C1 drawn
All BII CII Ploted
Destruction
|
| 8 | |||
| 23.
24. |
R = 2.2CM ± 0.1
Area = 22 x 2.2 x 2-2 7 = 15.21cm2
Ef =40 efd = -80 (ii) model class = 351- 360 modern class = 341 – 350 (iii) mean = 355.5 – 80 40 = 355.5 – 2 = 353.5
|
B1
B1
B1
B1
B1
B1
M1
1 1
8
B1 B1
M1
A1
B1
B1 B1 B1
|
|
| A1 | |||
| 8 |
MATHEMATICS V
PART I
SECTION 1 (52 MARKS)
working in a day to pack 20 cartons in 18 days? (2mks)
SECTION II (48MARKS)
800 p.m. He pays service charge of Sh 150 and contributes Sh 730 to welfare. Calculate Mwangis net
salary per month. (8mks)
(b) Find the ratio in which Y divides AP (6mks)
| X | 0.5 | 1 | 2 | 3 | 10 | |
| Y | 2 | 8 | 32 | 200 | 800 |
By plotting a suitable straight line graph on the graph provided, determine the values of a and n.
20. Chalk box x has 2 red and 3 blue chalk pieces. Box Y has same number of red and blue
pieces. A teacher picks 2 pieces from each box. What is the probability that
(a) They are of the same colour. (4mks)
(b) At least one is blue (2mks)
(c) At most 2 are red (2mks)
21. Point P(50oN, 10oW) are on the earth’s surface. A plane flies from P due east on a parallel of
latitude for 6 hours at 300 knots to port Q.
(a) Determine the position of Q to the nearest degree. (3mks)
(b) If the time at Q when the plane lands is 11.20am what time is it in P. (2mks)
(c) The plane leaves Q at the same speed and flies due north for 9 hours along a longitude to
airport R. Determine the position of R. (3mks)
22. Using a ruler a pair of compasses only, construct :
(a) Triangle ABC in which AB = 6cm, AC = 4cm and Ð ABC = 37.5o. (3mks)
(b) Construct a circle which passes through C and has line AB as tangent to the circle at A. (3mks)
(c) One side of AB opposite to C, construct the locus of point P such that ÐAPB = 90o. (2mks)
23. A particle moves in a straight line and its distance is given by S = 10t2 – t3 + 8t where S is
distance in metres at time t in seconds.
Calculate:
(i) Maximum velocity of the motion. (4mks)
(ii) The acceleration when t = 3 sec. (2mks)
(iii) The time when acceleration is zero. (2mks)
matrix M = 2 2 ABCD is mapped onto A1B1C1D1
1 3
under transformation M = -1 0 A1B1C1D1 is mapped onto A11B11C11D11. Draw on the given grid
0 –2
(a) ABCD, A1B1C1D1 and A11B11C11D11 (4mks)
(b) If area of ABCD is 8 square units, find area of A11B11C11D11. (3mks)
(c) What single transformation matrix maps A11B11C11D11 onto A1B1C1D1 (1mk)
MATHEMATICS V
PART II
SECTION 1 (52 Marks)
| Year | 1994 | 1995 | 1996 | 1997 | 1998 |
| Index | 125 | 150 | 175 | 185 | 200 |
Calculate clothing index using 1995 as base year. (4mks)
B at a final speed of 40 knots. If to achieve his own aim, he has to steer his ship at a course of 350o.
Find the bearing of A from B. (3mks)
SECTION II (48 MARKS)
| Weight in Kg | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55-59 | 60-64 |
| No. of Girls | 4 | 10 | 8 | 11 | 8 | 6 | 3 |
(a) State the modal class. (1mk)
(b) Using an assumed mean of 47,
(i) Estimate the mean weight (3mks)
(ii) Calculate the standard deviation. (4mks)
| X | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
| a Cos(x-5) | 0.97 | 0.71 | 0.5 | -0.5 | -0.71 | ||||||
| b sin(x+3) | 1.00 | 2.00 | 1.00 | 0.00 |
(a) Determine the values of a and b (2mks)
(b) Complete the table (2mks)
(c) On the same axes draw the graphs of y = across(x – 15) and y = b sin(x + 30) (3mks)
(d) Use your graph to solve ½ cos (x – 15) = sin(x + 30) (1mk)
21. The diagram below is a clothing workshop. Ð ECJ = 30o AD, BC, HE, GF are vertical
walls. ABHG is horizontal floor. AB = 50m, BH = 20m, AD=3m
(a) Calculate DE (3mks)
(b) The angle line BF makes with plane ABHG (2mks)
(c) If one person requires minimum 6m3 of air, how many people can fit in the workshop (3mks)
than 16 vehicles all together.
(a) Write down 3 inequalities in A and B which are the number of vehicles used and plot them
in a graph. (3mks)
(b) What is the smallest number of vehicles he could use. (2mks)
(c) Hire charge for type A is Sh.1000 while hire for type B is Sh.1200 per vehicle. Find the cheapest
hire charge for the whole function (3mks)
A circle centre A has radius 8cm and circle centre B has radius 3cm. The two centres are
12cm apart. A thin tight string is tied all round the circles to form interior common tangent. The tangents CD and EF intersect at X.
(a) Calculate AX (2mks)
(b) Calculate the length of the string which goes all round the circles and forms the tangent.
(6mks)
40km/h from 200o. A pilot navigates his plane at an air speed of 200km/h from B to A.
(a) Calculate the actual speed of the plane. (3mks)
(b) What course does the pilot take to reach B? (3mks)
(c) How long does the whole journey take? (2mks)
PART I
| 1 | SOLUTION | MKS | AWARDING |
| No Log
13.6 1.1335 + Cos 40 1.8842 1.0177 – 63.4 1.8021 1. 2156 (4 + 3.2156) 1/4 1.8039 Antilog 0.6366 |
B1
M1
M1
A1 |
Log
+
divide by 4
C.A.O |
|
| 4 | |||
| 2. | (x + 3) (x + 3 – 5) = 0
(x +3)b (x – 2) = 0 x = -3 or x = 2 |
M1
A1
|
Factors
Both answers |
| 3 | BD = C Sin 30 = 0.05
CD = b Cos 25 = 0.9063b BC = 0.9063b + 0.5 C |
B1
B1 B1
|
BD in ratio from
CD in ratio form Addition |
| 3 | |||
| 4 | Dy = 3 – 3x2 dx x = 2, grad = 1 9 Point (2,3) y – 3 = 1 x – 2 9 9y – 27 = x – 2 |
B1
B1
M1
A1
|
Grad equ
Grad of normal
Eqn
Eqn
|
| 4 | |||
| 5 | 700 = 100 + n 2200 = 400 + n 1500 = 300m m = 5 n = 200 P = 5 + 200 |
M1
A1
B1 B2 |
Equan
Both ans
Eqn (law) Ans (P) |
| 4 | |||
|
6 |
4 Sin x + 2 cos y = 6 3 Sin x – 2 Cos y = 1 Sin x = 1 X = 90 Cos y = 1 Y = 0o |
M1 M1
A1
B1 |
Elim Sub
|
| 7 | 2(x +1) – 1(x + 2) + x + 2
(x+2) (x +1) (x +2) (x + 1) = 2x + 2 (x + 2) (x + 1) = 2 |
M1
M1
A1 |
Use of ccm
Substitution
Ans |
| 8 | (-2 – ½ x)5 = 25 – 5 (2)4 ( ½ x) + 10(2)3( ½ x)2
= 32 – 40x + 20x2 = 32 – 4 (0.08) + 20 (0.08)2 = 32 – 0.32 + 0.128 |
M1
A1
M1 A1 |
|
| 4 | |||
| 9. | Circle centre C = (3 +1, 0 + 4)
2 2 C( 2, 2) R =Ö (2 – 0)2 + (2 – 3)2 =Ö 5 (y – 2)2 + (x – 2)2 = Ö5 y2 + x2 – 4y – 4x = 8 + Ö5 |
B1
B1
M1
A1 |
Centre
Radius
|
| 4 | |||
| 10 | ar2 =2, ar5 = 16
a = 2 \ 2 r5 = 16 r2 r2 2r3 = 16 r3 = 8 r = 2, a = ½
S5= ½ (1 – ( ½ )5) ½ = 1 – 1/32 = 31/32 |
M1
A1
M1
A1 |
Both
Sub
CAO |
| 4 | |||
| 11 | NR – 3MT2 = 2RT2
T2(2R + 3M) = NR T2 = NR 2R + 3m T = ! Ö NR |
M1
M1
A1 |
X mult
72
ans |
| 3 | |||
| 12 | 2 = m 2 + n 6
2 0 4 2 = 2m + 6n 2 = 0 + 4n n = ½ m = – ½ \a = – ½ b + ½ c \a b c are linearly dep |
M1
A1
B1 |
|
| 3 | |||
| 13 | Volume = 22 x 2.1 x 2.1 x 2 x ¾ m3
7 Time = 11 x 0.3 x 2.1 x 3 x 1,000,000 500 x 3600 = 11.55 = 11.33 hrs time to fill = 8.03 pm |
M1
M1
A1 |
|
| 3 | |||
| 14 | Mass = 54 x 1.2 x 1,000,000
90 1000 = 720kg |
M1
A1 |
|
| 2 | |||
| 15 | V3 = P
P(0.9)3 = 200,000 P = 200,000 0.93 = 200,000 0.729 = Sh 274,348 |
M1
M1
A3 |
|
| 3 | |||
| 16 | No of hours = 8 x 12 x10 x 20
8 x 18 x 25 = 19200 3600 = 5hrs, 20 min |
M1
A1 |
|
| 2 | |||
| 17 | Taxable income = 8100 + 2400
= sh. 10,500 = ₤6300 Tax dues = Sh 1980 x 2 + 1980 x 3 + 1980 x 5 + 3670 x 7 12 = 22320 12 = Sh 1860 net tax = 1860 – 800 p.m. = Sh 1060 Total deduction = 1060 + 150 + 730 = 1940 Net salary = 10,500 – 1940 = Sh 8560 p.m. |
B1
M1 M1
A1
B1
B1
M1 A1 |
Tax inc
2 2
net tax
total dedu. |
| 8 | |||
| 18 | OR = 2/3 a + 1/3b or (1/3 (2a + b)
AP = 2/5 b – a OY = m OR = A + n (2/5b – a) 2/5m b + ma = (1 – n)a + 2/5 n b 2/5m = 2/5n \m = 1 – m 2m = 1 m = ½ = n ½ AP = Ay AY:AP = 1:1 |
B1
B1
B1 M1 M1 A1 A1
B1 |
EXP, OY Eqn M = n Sub CAO
Ratio |
| 8 | |||
|
19 |
Log y = n log x + log a Log a = 0.9031 A = 8 Grad = 1.75 – 0.5 0.4 + 0.2 = 1.25 = 2.08 n = 2 \y = 8x2 x = 3 y = 8 x 32 = 72 y = 200 x = 5
|
B1 B1
B1
B1 B1 S1 P1 L1 |
Log x Log y
A
N Missing x and y Scale Points Line |
| 8 | |||
|
20 |
P (same colour) = P (XRRrr orXBB or YXX or YBB) = ½ (2/5 x ¼ + 3/5 x 2/4) x 2 = 2 + 6 = 8 = 2/5 (b) P(at least 1B) = 1 – P(non blue) = 1 – P (XRR or YRR) = 1 – ½ (2/5 x ¼) x 2 = 1 – 1/10 = 9/10 (c) P(at most 2 Red) = 1 – P (BB) = 1 – ½ (3/5 x 2/4)2 = 1 – 6/20 = 14/20 or 7/10 |
M1 M1
M1
A1
M1
A1 M1
A1 |
Any 2 Any 2
Fraction
|
| 8 | |||
| 21 | (a) PQ = 1800nm
q = 1800 60 x 0.6428 = 46.67 = 47o Q (50oN, 37oE)
(b) Time diff = 47 x 4 = 3.08 Time at P = 9.12am (c) QR = 2700 nm x o = 2700 60 = 45o R (85oN, 133oW) |
M1
A1
M1
A1
M1
A1 B1 |
|
| 8 | |||
|
22 |
B1 B1
B1 B1 B1 B1 B1 B1
|
Bisector of 150 Bisector 75
AB AC ^ at A Bisector AC Circle Ð AB Locus P with A B excluded |
|
| 8 | |||
| 24 | A1B1 C1D1
2 2 1 3 3 1 = 4 8 10 6 1 3 1 1 2 2 4 6 9 7
A11 B11 C11 D11 -1 0 4 8 10 6 = -4 –8 -10 -6 0 –2 4 6 9 7 -8 -12 -18 -14
NM = -1 0 2 2 0 –2 1 3
= -2 -2 -2 -6
(b) det = Asf = 12 – 4 = 8 Area A11 B11 C11 D11 = 8 x 8 = 64 U2 (c) Single matrix = Inv N 0 –1
= -1 0 0 – ½
|
B1
B1
B1 M1 A1
B1
|
Product
Product
Det
Inverse |
| 6 | |||
| 23 |
Ds = 20t – 3t2 + 8 =0 Dt 3t2 – 20t – 8 = 0 T = 20 ! Ö400 + 4 x 3 x 8 6 t = 7.045 sec max vel = 148.9 – 140.9 – 8 = 0.9 m/s
dt2 when t = 3 a = -2m/s2 6t – 20 = 0 6t = 20 t = 3 2/3 sec |
M1
A1 M1 A1 M1
A1 M1
A1 |
|
| 8 | |||
PART II
| No | Solution | Mks | Awarding |
| 1 | 2744 x 125 1/3
1000 8
2744 1/3 x 53 1/3 1000 23
23 x 73 1/3 x 5 103 2
2 x 7 x 5 = 3.5 10 2 |
M1
M1 A1 |
Factor
Cube root
|
| 3 | |||
| 2 | (i) Highest – 10 x 7.5 = 75
Lowest – 6 x 4 = 24 – 51 (ii) Highest = 7.5 = 1.875 4 Lowest = 6 = 0.600 10 1.275 |
M1
A1
M1
A1 |
Highest
Fraction
|
| 4 | |||
| 3 | Cos q = 17 = 0.8095
21
q = Cos 0.8095 = 36.03o
Arc length = 72. 06 x 2 x 22 x 21 360 7 = 26.422cm |
M1
A1
M1
A1 |
q |
| 4 | |||
| 4 | x2 – 2x(x +3) = 0
x2 – 2x2 – 6x = 0 -x2 – 6x = 0 either x = 0 or x = 6 |
M1
M1
A1 |
Equ
Factor
Both A |
| 3 | |||
|
5 |
8 = x x 2 x 22 x 28 Cos 60o 360 7
8 = x x 44 x 28 x 0.5 360 7 x = 8 x 360 x 7 = 32.73o = 33o |
M1
M1
A1 B1 |
x exp |
|
6 |
ÐDMC = Ð AMB vert. Opp = q ÐMAB = Ð MDC = 180 – q BASE Ls of an isosc. < 2 2 <’s AMC and < CDM are equiangle
\ Similar proved
|
B1
B1
B1 |
|
| 3 | |||
| 7 | Tan x = 5/12
h = Ö b2 + 122 = Ö25 + 144 = Ö169 = 13
1 – Sinx = 1 – 5 sin x + 2 Cos x 5/13 + 2 x 12/13
12/13 = 12 x 13 = 12 29/13 13 29 29 |
M1 M1
A1 |
Hypo Sub
|
| 3 | |||
| 8 | Y = x 2 + 2
Area = h (y1, = y2 +……..yn) = 1(2.225 + 4.25 + 8.25 +14.25 + 22.25) = 51.25 sq units |
B1
M1
A1 |
Ordinals |
| 3 | |||
|
9 |
ÐCBA = 117o Ð ACD = 55 Ð BAC = 180 – (117 + 55) = 8o |
B1 B1 B1
3 |
|
| 10 |
|
B1
B1 B1 B1 |
1994
1996 1997 1998 |
| 4 | |||
| 11. | Xy = 35
y = 35/x 9x – 9y = -18 Sub x2 + 2x – 35 = 0 x2 + 7x – 5x – 35 = 0 x (x + 7) – 5(x + 7) = 0 (x – 5) (x + 7) = 0 x = -7 x = +5 y = 7 Smaller No. = 57 = 75 |
B1
M1
A1
B1
|
|
| 3 | |||
| 12 | Log5 (2x – 1 )4 = log552
20 4(2x – 1) = 52 20 2x – 1 = 25 5 2x – 1 = 125 2x = 126 x = 63 |
M1
M1
A1 |
|
| 3 | |||
| 13 |
C.P = 100 x 49.50
110 = 45/- 52x + 40y = 45 x + y 45x + 45y = 52x + 40 -7x = -54 x/y = 5/7 x : y = 5 : 7 |
B1 M1
M1
A1 |
|
| 4 | |||
| 14 |
2n – 4 it angle = 172 n (2n – 4) x 90 = 172n n 90 (2n – 4) x 90 = 172 n 180 n – 360 = 172n
180n – 172n = 360 8n = 360 n = 45 |
M1
A1
M1 |
|
| 2 | |||
| 15 | 2 x = 2. 1 + 3. 1
6.341 9.22 2x = 2 x 0. 1578 + 3 x 0.1085 = 0.3154 + 0.3254 = 0.6408 x = 0.3204 |
B1
A1 |
Tables |
| 2 | |||
| 16 | Bearing 140o
Sin q = 20 Sin 110 40 = 0.4698 = 228.02 Bearing of A from B = 198.42 |
M1
A1 B1 |
|
| 3 | |||
| 17 | Points that each tap fills in one hour
A = 1 B = 1 C – 1 In one hour all taps can fill = 1 + 1 + 1 = 11 50 25 20 100 In 6hrs all can fill = 11 x 6 = 33 parts 100 50 taps A and B can fill = = 1 + 1 = 3 part in 1 hr 50 25 50 In 4 1 hrs, A and B = 25 x 3 + 1 6 6 50 4 Parts remaining for B to fill = 1 – 33 + 1 = 1 – 91 = 9 parts 50 4 100 100 Time taken = 9 x 25 hrs = 2 ¼ hrs 100 1 7.30 am 6. hrs 13.30 4.10 5.40pm 2.15 7.55 pm
|
M1
B1
B1
B1
M1
A1
|
|
|
18 |
x2 + x – 8 = -2 – 2x y = x2 + 3x – 6 Points of intersection (-4, 1.4) y = x2 + x – 8 = 2x2 + 3x – 6 x2 + 2x + 2 y = x2 + x – 8 x 2 2y = 2x2 + 2x – 16 0 = 2x2 + 3x – 6 2y = -x – 10 y = – 2.6 Ny = 1.2
|
8
B1 B1
B1
B1 |
Eqn Point of inter
Line eqn
Both |
|
19 |
(a) Modal class = 45 – 49 (i) Mean = 47 + -55 50 = 47 – 1.1 = 45.9
(ii) Standard deviation = Ö 3575 – –55 2 = Ö71.5 – 1.21 =Ö 70.29 = 8.3839
|
4
B1 B1
M1
A1 B1
|
fd fd2 |
| 8 | |||
| 20 |
(a) a = 1 ½ cos (x – 15) = Sin (x + 30) has no solution in the domain |
B1 B1 B1
B1 |
All All A & b
|
| 8 | |||
| 21 | (a) O Cos 30 = 20
X X = 20 0.866 = 23.09
DE = Ö 502 + 23.092 = Ö 2500 + 533.36 = Ö 3033.36 = 55.076m
(b) GB = Ö 202 + 502 = 53.85 Tan q = 14.55 = 0.27019 q = 15.12o |
B1
M1
A1
M1
A1 |
|
| 8 | |||
| (c) Volume of air = 50 x 20 x 3 + ½ x 20 x 11.55 x 50
= 3000 + 5775 = 8775 No. of people = 8775 = 1462.5 j 1462 |
M1
M1
A1 |
||
| 8 | |||
| 22 | (a) A + B [ 16
5A + 3B ³ 50 2A + 3B [ 35
(b) 14 vehicles
(c) A – 6 vehicles B – 8 Cost = 6 x 1000 + 8 x 1200 = 6000 + 9600 = 15,600/= |
B1
B1
B1
M1
A1
|
In equation 3
Vehicles |
| 8 | |||
|
23 |
x = 8 12 – x 3
= 8.727 FBX = 3 = 0.9166 = 23.57
3FBX = 47.13
Reflex Ð FBD = 312.87
Reflex arc FD = 312.87 x 22 x 6
= 16.39cm Reflex Arc CE = 312.87 x 22 x 16
= 43.7cm
FE (tangent) = Ö144 – 121 = Ö 23 = 4.796cm 2 FE = 9.592
Total length = 9.592 + 4.796 + 43.7 + 16.39 = 74.48 cm2 |
M1
A1
M1
A1
M1
A1
M1 A1 |
|
| 8 | |||
|
24 |
(b) 200 = 40 Sin 50 Sin q
Sin q = 40Sin 50 q = 8.81o Ð ACB = 180 – (50 + 8.81)o = 121.19o x = 200
x = 200 x Sin 121.19 = 200 x 0.855645 = 223.36Km/h
(b) Course = 330o – 8.81o = 321.19o
(c) Time = 600
= 2.686 hrs
|
M1
A1
M1
M1
A1
B1
M1
A1
8
|
PHYSICS FORM THREE
CHAPTER ONE
LINEAR MOTION
Introduction
Study of motion is divided into two;
In kinematics forces causing motion are disregarded while dynamics deals with motion of objects and the forces causing them.
Distance moved by a body in a specified direction is called displacement. It is denoted by letter‘s’ and has both magnitude and direction. Distance is the movement from one point to another. The Si unit for displacement is the metre (m).
This is the distance covered per unit time.
Speed= distance covered/ time taken. Distance is a scalar quantity since it has magnitude only. The SI unit for speed is metres per second(m/s or ms-1)
Average speed= total distance covered/total time taken
Other units for speed used are Km/h.
Examples
Solution
Total distance covered=10+90=100m
Total time taken=4+10+6=20 seconds
Therefore average speed=100/20=5m/s
Solution
Distance covered=speed*time
=180*1000/60*60=50m/s
=50*30
=1,500m
Solution
Speed:360km/h=360*1000/60*60=100m/s
Time=distance/speed
3000*1000/100
=30,000 seconds.
This is the change of displacement per unit time. It is a vector quantity.
Velocity=change in displacement/total time taken
The SI units for velocity are m/s
Examples
Solution
=800+400/100+80
=1200/180
=6.67m/s
=800-400/180
=400/180
=2.22 m/s due North
= (800/100)-(400-80)
=8-5
=3m/s due North
Solution
Initial velocity(u)=-10m/s
Final velocity (v) = 10m/s
Therefore change in velocity= v-u
=10- (-10)
=20m/s
This is the change of velocity per unit time. It is a vector quantity symbolized by ‘a’.
Acceleration ‘a’=change in velocity/time taken= v-u/t
The SI units for acceleration are m/s2
Examples
Solution
Initial velocity= 72 km/h=20m/s
Final velocity= 144 km/h=40m/s
Therefore ‘a’ =v-u/t
= 40-20/10
2m/s2
Solution
Initial velocity=180km/h=50m/s
Final velocity= 0 m/s
A = v-u/t=0-50/20
= -2.5 m/s2
Hence retardation is 2.5 m/s2
Motion graphs
Distance-time graphs
| Stationary body |
b)
| A body moving with uniform speed |
c)
| A body moving with variable speed |
Area under velocity-time graph
Consider a body with uniform or constant acceleration for time‘t’ seconds;
Distance travelled= average velocity*t
=(0+v/2)*t
=1/2vt
This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under‘t’ seconds.
Example
A car starts from rest and attains a velocity of 72km/h in 10 seconds. It travels at this velocity for 5 seconds and then decelerates to stop after another 6 seconds. Draw a velocity-time graph for this motion. From the graph;
Solution
=(1/2×10×20)+(1/2×6×20)+(5×20)
=100+60+100
=260m
Also the area of the trapezium gives the same result.
Stage A gradient= 20-0/ 10-0 = 2 m/s2
Stage b gradient= 20-20/15-10 =0 m/s2
Stage c gradient= 0-20/21-15 =-3.33 m/s2
Using a ticker-timer to measure speed, velocity and acceleration.
It will be noted that the dots pulled at different velocities will be as follows;
Most ticker-timers operate at a frequency of 50Hzi.e. 50 cycles per second hence they make 50 dots per second. Time interval between two consecutive dots is given as,
1/50 seconds= 0.02 seconds. This time is called a tick.
The distance is measured in ten-tick intervals hence time becomes 10×0.02= 0.2 seconds.
Examples
| C |
| B |
| A |
| · |
| · |
| · |
Solution
Distance between two consecutive dots= 5cm
Frequency of the ticker-timer=50Hz
Time taken between two consecutive dots=1/50=0.02 seconds
Therefore, velocity of tape=5/0.02= 250 cm/s
Solution
Time between successive dots=1/100=0.01 seconds
Initial velocity (u) 0.5/0.01 50 cm/s
Final velocity (v) 2.5/0.01= 250 cm/s
Time taken= 4 ×0.01 = 0.04 seconds
Therefore, acceleration= v-u/t= 250-50/0.04=5,000 cm/s2
Equations of linear motion
The following equations are applied for uniformly accelerated motion;
v = u + at
s = ut + ½ at2
v2= u2 +2as
Examples
Solution
V2 = u2 +2as
= (60) +2×10×320
=3600+6400
= 10,000
Therefore v= (10,000)1/2
v= 100m/s
Solution
v = u+at
0= 30-3t
30=3t
t= 30 seconds.
Solution
s=ut+ ½ at2
=0×10+ ½ ×10×102
= 1000/2=500m
Motion under gravity.
The equations used for constant acceleration can be used to become,
v =u+gt
s =ut + ½ gt2
v2= u+2gs
Since the body goes against force of gravity then the following equations hold
v =u-gt ……………1
s =ut- ½ gt2 ……2
v2= u-2gs …………3
N.B time taken to reach maximum height is given by the following
t=u/g since v=0 (using equation 1)
Time of flight
The time taken by the projectile is the timetaken to fall back to its point ofprojection. Using eq. 2 then, displacement =0
0= ut- ½ gt2
0=2ut-gt2
t(2u-gt)=0
Hence, t=0 or t= 2u/g
t=o corresponds to the start of projection
t=2u/gcorresponds to the time of flight
The time of flight is twice the time taken to attain maximum height.
Maximum height reached.
Using equation 3 maximum height, Hmax is attained when v=0 (final velocity). Hence
v2= u2-2gs;- 0=u2-2gHmax, therefore
2gHmax=u2
Hmax=u2/2g
Velocity to return to point of projection.
At the instance of returning to the original point, total displacement equals to zero.
v2 =u2-2gs hence v2= u2
Thereforev=u or v=±u
Example
A stone is projected vertically upwards with a velocity of 30m/s from the ground. Calculate,
Solution
T=u/g=30/10=3 seconds
T=2t= 2×3=6 seconds
Or T=2u/g=2×30/10=6 seconds.
Hmax= u2/2g= 30×30/2×10= 45m
v2= u2-2gs, but s=0,
Hence v2=u2
Thereforev=(30×30)1/2=30m/s
The path followed by a body (projectile) is called trajectory. The maximum horizontal distance covered by the projectile is called range.
The horizontal displacement ‘R’ at a time‘t’ is given by s=ut+1/2at2
Taking u=u and a=0 hence R=ut, is the horizontal displacement and h=1/2gt2 is the vertical displacement.
NOTE
The time of flight is the same as the time of free fall.
Example
A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s. Calculate,
Solution
20= ½ ×10×t2
40=10t2
t2=40/10=4
t=2 seconds
=10×2
=20m
= 2×10=20m/s
CHAPTER TWO
REFRACTION OF LIGHT
Introduction
Refraction is the change of direction of light rays as they pass at an angle from one medium to another of different optical densities.
Exp. To investigate the path of light through rectangular glass block.
Apparatus: – soft-board, white sheet of paper, drawing pins (optical), rectangular glass block.
Procedure
Explanation of refraction.
Light travels at a velocity of 3.0×108in a vacuum. Light travels with different velocities in different media. When a ray of light travels from an optically less dense media to more dense media, it is refracted towards the normal. The glass block experiment gives rise to a very important law known as the law of reversibility which states that “if a ray of light is reversed, it always travels along its original path”. If the glass block is parallel-sided, the emergent ray will be parallel to the incident ray but displaced laterally as shown
‘e’ is called the angle of emergence. The direction of the light is not altered but displaced sideways. This displacement is called lateral displacement and is denoted by‘d’. Therefore
XY= t/Cos r YZ= Sin (i-r) ×xy
So, lateral displacement, d = t Sin (i-r)/Cos r
Laws of refraction
Sin i/sin r = constant (k)
Refractive index
Refractive index (n) is the constant of proportionality in Snell’s law; hence
Sin i/ sin r = n
Therefore sin i/sin r=n=1/sin r/sin i
Examples
Solution
gna= 1/ang = 1/1.5=0.67
Solution
Angle of incidence (i) = 900-600=300
1.5=sin 30o/sin r, sin r =sin 300/ 1.5=0.5/1.5
Sin r=0.3333, sin-10.3333= 19.50
R= 19.50
Refractive index in terms of velocity.
Refractive index can be given in terms of velocity by the use of the following equation;
1n2 = velocity of light in medium 1/velocity of light in medium 2
When a ray of light is travelling from vacuum to a medium the refractive index is referred to as absolute refractive index of the medium denoted by ‘n’
Refractive index of a material ‘n’=velocity of light in a vacuum/velocity of light in material ‘n’
The absolute refractive indices of some common materials is given below
| Material | Refractive index | |
| 1 | Air (ATP) | 1.00028 |
| 2 | Ice | 1.31 |
| 3 | Water | 1.33 |
| 4 | Ethanol | 1.36 |
| 5 | Kerosene | 1.44 |
| 6 | Glycerol | 1.47 |
| 7 | Perspex | 1.49 |
| 8 | Glass (crown) | 1.55 |
| 9 | Glass (flint) | 1.65 |
| 10 | Ruby | 1.76 |
| 11 | Diamond | 2.72 |
Examples
Solution
Since anw sin θw=ang sing
4/3 sin 300= 3/2 sin r
3/2 sin r= 4/3× 0.5
Sin r =4/6×2/3=4/9= 0.4444
r = 26.40
Solution
wng= gna×ang, but wna = 1/ anw=3/4
wng=3/4×3/2=9/8= 1.13
Real and apparent depth
Consider the following diagram
The depth of the water OM is the real depth, and the distance IM is known as the apparent depth. OI is the distance through which the coin has been displaced and is known as the vertical displacement. The relationship between refractive index and the apparent depth is given by;
Refractive index of a material=real depth/apparent depth
NB
This is true only if the object is viewed normally.
Example
A glass block of thickness 12 cm is placed on a mark drawn on a plain paper. The mark is viewed normally through the glass. Calculate the apparent depth of the mark and hence the vertical displacement. (Refractive index of glass =3/2)
Solution
ang= real depth/apparent depth
apparent depth= real depth/ ang=(12×2)/3= 8 cm
vertical displacement= 12-8=4 cm
Applications of refractive index
Total internal reflection
This occurs when light travels from a denser optical medium to a less dense medium. The refracted ray moves away from the normal until a critical angle is reached usually 900 where the refracted ray is parallel to the boundary between the two media. If this critical angle is exceeded total internal reflection occurs and at this point no refraction occurs but the ray is reflected internally within the denser medium.
Relationship between the critical angle and refractive index.
Consider the following diagram
From Snell’s law
gnw = sin C/sin 900,but ang = 1/gna since sin 900 = 1
Thereforeang= 1/sin C, hence sin C=1/n or n=1/sin C
Example
Calculate the critical angle of diamond given that its refractive index is 2.42
Solution
Sin C= 1/n=1/ 2.42= 0.4132= 24.40
Effects of total internal reflection
Applications of total internal reflection
CHAPTER THREE
NEWTON’S LAWS OF MOTION
Newton’s first law (law of inertia)
This law states that “A body continues in its state of rest or uniform motion unless an unbalanced force acts on it”. The mass of a body is a measure of its inertia. Inertia is the property that keeps an object in its state of motion and resists any efforts to change it.
Newton’s second law (law of momentum)
Momentum of a body is defined as the product of its mass and its velocity.
Momentum ‘p’=mv. The SI unit for momentum is kgm/s or Ns. The Newton’s second law states that “The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts”
Change in momentum= mv-mu
Rate of change of momentum= mv-mu/∆t
Generally the second law gives rise to the equation of force F=ma
Hence F=mv-mu/∆t and F∆t=mv-mu
The quantity F∆t is called impulse and is equal to the change of momentum of the body. The SI unit for impulse is Ns.
Examples
Solution
Momentum=mv=72km/h=(20m/s)×3×103 kg
=6.0×104kgm/s
Solution
Mass of the truck = (1.0×105)/10=6.0×104
Using F=ma
=1.5×10×104
=1.5×104 N
Solution
Since the car comes to rest, v=0, a=(v-u)/t =(0-45)/9=-5m/s (retardation)
F=ma =(1200×-5) N =-6,000 N (braking force)
Solution
Impulse = Ft=1,000×3= 3,000 Ns
Let v be the velocity after 3 seconds. Since the truck was initially at rest then u=0.
Change in momentum=mv-mu
= (2,000×v) – (2,000×0)
=2,000 v
But impulse=change in momentum
2,000 v = 3,000
v = 3/2=1.5 m/s.
Weight of a body in a lift or elevator
When a body is in a lift at rest then the weight
W=mg
When the lift moves upwards with acceleration ‘a’ then the weight becomes
W = m (a+g)
If the lift moves downwards with acceleration ‘a’ then the weight becomes
W = m (g-a)
Example
A girl of mass stands inside a lift which is accelerated upwards at a rate of 2 m/s2. Determine the reaction of the lift at the girls’ feet.
Solution
Let the reaction at the girls’ feet be ‘R’ and the weight ‘W’
The resultant force F= R-W
= (R-500) N
Using F = ma, then R-500= 50×2, R= 100+500 = 600 N.
Newton’s third law (law of interaction)
This law states that “For every action or force there is an equal and opposite force or reaction”
Example
A girl of mass 50 Kg stands on roller skates near a wall. She pushes herself against the wall with a force of 30N. If the ground is horizontal and the friction on the roller skates is negligible, determine her acceleration from the wall.
Solution
Action = reaction = 30 N
Force of acceleration from the wall = 30 N
F = ma
a = F/m = 30/50 = 0.6 m/s2
Linear collisions
Linear collision occurs when two bodies collide head-on and move along the same straight line. There are two types of collisions;
Collisions bring about a law derived from both Newton’s third law and conservation of momentum. This law is known as the law of conservation of linear momentum which states that “when no outside forces act on a system of moving objects, the total momentum of the system stays constant”.
Examples
Solution
Initial momentum of the bullet and the gun is zero since they are at rest.
Momentum of the bullet after firing = (0.005×350) = 1.75 kgm/s
But momentum before firing = momentum after firing hence
0 = 1.75 + 0.5 v where ‘v’ = recoil velocity
0.5 v = -1.75
v =-1.75/0.5 = – 3.5 m/s (recoil velocity)
Solution
Change in momentum = ∆P = mv – mu= Ft
= 12×10 = 12 Ns
Solution
Momentum before collision = momentum after collision
(1500×20) + (900×0) = (1500 +900)v
30,000 = 2,400v
v = 30,000/2,400 = 12.5 m/s (common velocity)
Distance = velocity × time
= 12.5×20
= 250m
= 1500 (20-12.5) for minibus or
=900 (12.5 – 0) for the car
= 11,250 Ns
Impulse force F = impulse/time = 11,250/2 = 5,625 N
K.E after collision = ½ × 2400 × 12.52 = 1.875×105 J
Therefore, change in K.E =(3.00 – 1.875) × 105 = 1.25× 105 J
Some of the applications of the law of conservation of momentum
Solid friction
Friction is a force which opposes or tends to oppose the relative motion of two surfaces in contact with each other.
Measuring frictional forces
We can relate weight of bodies in contact and the force between them. This relationship is called coefficient of friction. Coefficient of friction is defined as the ratio of the force needed to overcome friction Ff to the perpendicular force between the surfaces Fn. Hence
µ = Ff/ Fn
Examples
Solution
Ff = µFn
Fn= weight = 50×10 = 500 N
Ff = 0.30 × 500 = 150 N
Solution
Since motion is uniform, the applied force is equal to the frictional force
Fn = normal reaction = weight = 20 ×10 = 200 N
Therefore, µ =Ff/ Fn = 50/ 200 = 0.25.
Laws of friction
It is difficult to perform experiments involving friction and thus the following statements should therefore be taken merely as approximate descriptions: –
Applications of friction
Methods of reducing friction
Example
A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6. Calculate
Solution
= 0.6×30×10 = 180 N
From F =ma, then 20 = 30 a
a = 20 / 30 = 0.67 m/s2
Viscosity
This is the internal friction of a fluid. Viscosity of a liquid decreases as temperature increases. When a body is released in a viscous fluid it accelerates at first then soon attains a steady velocity called terminal velocity. Terminal velocity is attained when F + U = mg where F is viscous force, U is upthrust and mg is weight.
CHAPTER FOUR
ENERGY, WORK, POWER AND MACHINES
Energy
This is the ability to do work.
Forms of energy.
Transformation and conservation of energy
Any device that facilitates energy transformations is called transducer. Energy can be transformed from one form to another i.e. mechanical – electrical – heat energy. The law of conservation of energy states that “energy cannot be created or destroyed; it can only be transformed from one form to another”.
Work
Work is done when a force acts on a body and the body moves in the direction of the force.
Work done = force × distance moved by object
W = F × d
Work is measured in Nm. 1 Nm = 1 Joule (J)
Examples
Solution
Work done = force × distance
= (15× 10) × 2 = 300 Nm or 300 J
Solution
Work done = force × distance
= (50× 10) × (12 ×30) ÷ 100 = 500 × 3.6 = 1,800 J
Solution
A force of 7.5 produces an extension of 5.0 cm.
Hence 8.0 cm = (7.5 ×8)/ 5 = 12.0 N
Work done = ½ × force × extension
= ½ × 12.0 × 0.08 = 0.48 J
Solution
But 72 km/h = 20m/s
a = 0 -20/8 = – 2.5 m/s
Retardation = 2.5 m/s
Braking force F = 1,250 × 2.5
= 3,125 N
= ½ mv2 – ½ mu2
= ½ × 1250 × 02 – ½ × 1250 × 202
= – 2.5 × 105 J
Solution
Work = ½ ks2
= ½ × 100 × 0.22
= 2 J
Power
Poweris the time rate of doing work or the rate of energy conversion.
Power (P) = work done / time
P = W / t
The SI unit for power is the watt (W) or joules per second (J/s).
Examples
Solution
Power = work done / time = (force × distance) / time
= (500 ×3) / 4 = 375 W
Solution
Power = F v
= 2,000 × 12
= 24,000 W = 24 kW.
Machines
A machine is any device that uses a force applied at one point to overcome a force at another point. Force applied is called the effort while the resisting force overcome is called load. Machines makes work easier or convenient to be done. Three quantities dealing with machines are;-
M.A = load (L) / effort (E)
V.R = distance moved by effort/ distance moved by the load
Efficiency = (work output/work input) × 100
= (M.A / V.R) × 100
= (work done on load / work done on effort) × 100
Examples
Solution
Efficiency = (M.A / V.R) × 100 M.A = load/effort =60/20 = 3
V.R =DE/ DL = 8/2 = 4
Efficiency = ¾ × 100 = 75%
Some simple machines
M.A = Load/ Effort
V.R = no. of pulleys/ no. of strings supporting the load
Example
A block and tackle system has 3 pulleys in the upper fixed block and two in the lower moveable block. What load can be lifted by an effort of 200 N if the efficiency of the machine is 60%?
Solution
V.R = total number of pulleys = 5
Efficiency = (M.A /V.R) × 100 = 60%
0.6 = M.A/ 5 =3, but M.A = Load/Effort
Therefore, load = 3 ×200 = 600 N
V.R = R/r and M.A = R/r
Example
A wheel and axle is used to raise a load of 280 N by a force of 40 N applied to the rim of the wheel. If the radii of the wheel and axle are 70 cm and 5 cm respectively. Calculate the M.A, V.Rand efficiency.
Solution
M.A = 280 / 40 = 7
V.R = R/r = 70/5 = 14
Efficiency = (M.A/ V.R) × 100 = 7/14 × 100 = 50 %
V.R = 1/ sin θ M.A = Load/ Effort
Example
A man uses an inclined plane to lift a 50 kg load through a vertical height of 4.0 m. the inclined plane makes an angle of 300 with the horizontal. If the efficiency of the inclined plane is 72%, calculate;
Solution
Effort = load (mg) / effort (50×10)/ 1.44 = 347.2 N
Work output = mgh = 50×10×4 = 2,000 J
Work input = effort × distance moved by effort
347.2 × (4× sin 300) = 2,777.6 J
Therefore work done against friction = 2,777.6 – 2,000 = 777.6 J
V.R of screw = circumference of screw head / pitch P
= 2πr / P
Example
A car weighing 1,600 kg is lifted with a jack-screw of 11 mm pitch. If the handleis 28 cmfrom the screw, find the force applied.
Solution
Neglecting friction M.A = V.R
V.R = 2πr /P = M.A = L / E
1,600 / E = (2π× 0.28) / 0.011
E = (1,600 × 0.011 × 7) / 22×2×0.28 =10 N
V.R = revolutions of driver wheel / revolutions of driven wheel
Or
V.R = no.of teeth in the driven wheel/ no. of teeth in the driving wheel
Example
V.R = radius of the driven pulley / radius of the driving pulley
V.R = R2 / r2 where R- radius of the load piston and r- radius of the effort piston
Example
The radius of the effort piston of a hydraulic lift is 1.4 cm while that of the load piston is 7.0 cm. This machine is used to raise a load of 120 kg at a constant velocity through a height of 2.5 cm. given that the machine is 80% efficient, calculate;
Solution
Efficiency = M.A / V.R = (80 /100) × 25 = 20
But M.A = Load / Effort = (120×10) / 20 = 60 N
= (120 × 10× 2.5) / work input
80 / 100 = 3,000 / work input
Work input = (3,000 × 100) /80 = 3,750 J
Energy wasted = work input – work output
= 3,750 – 3,000 = 750 J
CHAPTER FIVE
CURRENT ELECTRICITY
Electric potential difference and electric current
Electric current
Electric potential difference (p. d) is defined as the work done per unit charge in moving charge from one point to another. It is measured in volts.
Electric current is the rate of flow of charge. P. d is measured using a voltmeter while current is measured using an ammeter. The SI units for charge is amperes (A).
Ammeters and voltmeters
In a circuit an ammeter is always connected in series with the battery while a voltmeter is always connected parallel to the device whose voltage is being measured.
Ohm’s law
This law gives the relationship between the voltage across a conductor and the current flowing through it. Ohm’s law states that “the current flowing through a metal conductor is directly proportional to the potential difference across the ends of the wire provided that temperature and other physical conditions remain constant”
Mathematically V α I
So V /I = constant, this constant of proportionality is called resistance
V / I = Resistance (R)
Resistance is measured in ohms and given the symbol Ω
Examples
Solution
V = IR = (2 × 10-3) × (2 × 103) = 4 V.
Solution
I = V/R = 5 / 20 = 0.25 A
Ohmic and non-ohmic conductors
Ohmic conductors are those that obey Ohms law(V α I) and a good example is nichrome wire i.e. the nichrome wire is not affected by temperature.
Non-ohmic conductors do not obey Ohms law i.e. bulb filament (tungsten), thermistor couple, semi-conductor diode etc. They are affected by temperature hence non-linear.
Factors affecting the resistance of a metallic conductor
Resistivity of a material is numerically equal to the resistance of a material of unit length and unit cross-sectional area. It is symbolized by ρ and the units are ohmmeter (Ωm). It is given by the following formula;
ρ = AR /lwhere A – cross-sectional area, R – resistance, l – length
Example
Given that the resistivity of nichrome is 1.1× 10-6Ωm, what length of nichrome wire of diameter 0.42 mm is needed to make a resistance of 20 Ω?
Solution
ρ = AR /l, hence l = RA/ ρ = 20 × 3.142 × (2.1×10-4) / 1.1 × 10-6 = 2.52 m
Resistors
Resistors are used to regulate or control the magnitude of current and voltage in a circuit according to Ohms law.
Types of resistors
| Carbon resistor |
| Wire-wound resistor |
Resistor combination
Consider the following loop
Since it is in series then,
VT = V1 + V2 + V3
The same current (I) flows through the circuit (resistors), hence
IRT = I (R1 + R2 + R3), dividing through by I, then
RT = R1 + R2 + R3
Therefore for resistors connected in series the equivalent resistance (Req) is equal to the total sum of their individual resistances.
Req = R1 + R2 + R3
Consider the following circuit
Total current is given by;
IT = I1 + I2 + I3. But IT = VT/RT = V1/R1 + V2/R2 + V3/R3
Since in parallel, VT = V1 = V2 = V3
Then 1/RT = 1/R1 + 1/R2 +1/R3, for ‘n’ resistors in parallel
1/RT = 1/R1 + 1/R2 +1/R3 ………… 1/Rn
If only two resistors are involved then the equivalent resistance becomes
1/Req = 1/R1 + 1/R2 = (R1 + R2)/ R1 R2
Examples
Solution
This reduces to
Combining the two in parallel;
1/Req = (R1 + R2)/R1 R2 = 20/96
1/Req = 20/96, therefore Req = 96/20 = 4.8 Ω
Lastly combining the two in series;
Then Req = 4 Ω + 4.8 Ω = 8.8 Ω
Solution
Combining those in series then this can be replaced by two resistors of 60 Ω and 40 Ω.
Current through 10 Ω = (p.d. between P and R)/ (30 + 10) Ω
p.d between P and R = 0.8 × Req. Req = (40 × 60)/ 40 + 60 = 2400/ 100 = 24 Ω
p.d across R and P = 0.8 × 24 (V=IR)
therefore, current through 10 Ω = 19.2 / 10 + 30 = 0.48 A
Electromotive force and internal resistance
Electromotive force (e.m.f.) is the p.d across a cell when no current is being drawn from the cell. The p.d across the cell when the circuit is closed is referred to as the terminal voltage of the cell. Internal resistance of a cell is therefore the resistance of flow of current that they generate. Consider the following diagram;
The current flowing through the circuit is given by the equation,
Current = e.m.f / total resistance
I = E / R + rwhere E – e.m.f of the cell
Therefore E = I (R + r) = IR + I r = V + I r
Examples
Solution
Let the internal resistance be ‘r’ and e.m.f be ‘E’.
Using E = V + I r = IR + I r
Substitute for the two sets of values for I and R
E = 0.6 × (2 + 0.6 r) = 1.2 + 0.36 r
E = 0.6 × (7 × 0.2 r) = 1.4 + 0.12 r
Solving the two simultaneously, we have,
E = 1.5 v and R = 0.5 Ω
Solution
When two identical cells are connected in series, the equivalent e.m.f is equal to that of only one cell. The equivalent internal resistance is equal to that of two such resistance connected in parallel. Hence Req = R1 R2 / R1 + R2 = (0.6 × 0.6) / 0.6 + 0.6 = 0.36 / 1.2 = 0.3 Ω
Equivalent e.m.f =1.5 / (0.7 + 0.3) = 1.5 A
Hence current flowing through 0.7 Ω resistor is 1.5 A
NAME…………………………………………………………ADM NO………………………
SIGN……………………………………… DATE…………………………….………
FORM 3
PAPER 1 – 313/1
CHRISTIAN RELIGIOUS EDUCATION
2½HRS
JOINT EVALUATION EXAMINATION TERM 3
INSTRUCTIONS
For Examiner’s use only
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5 |
6 |
TOTAL SCORE |
| SCORE |
|
b)Outline the contents of the Jeremiah’s letter to the exile. (7mks)
_________________________________________________________________
JOINT EXAMINATION
FORM 3CRE – PAPER 1
TERM 3-
MARKING SCHEME
(Accept first four correct) (4x2mks)
_________________________________________________________________________________________________
TERM 3
FORM 3
313/2
C.R.E
MARKING SCHEME
1a) Describe the concept of the Messiah in the New Testament.
2a) Relate the story of the healing of a man with a withered hand LK 6:6-11
3a) Give reasons why Jesus was arrested
4a) Mention St. Paul’s teachings on how the gifts of the Holy Spirit should be used in the Church.
in the old testament the Israelites were the true vine of God
their mission. (6 marks)
– They were not to carry money, food or clothing
– They were to preach peace in the house they entered.
– They were to heal the sick.
– They were to preach the kingdom of God.
– They were to shake off dust from their feet where they are rejected.
– They were not to salute anyone on the way.
– They were to remain in one house.
– They were to eat and drink what has provided to them.
6 x 1 = 6 marks
(8 marks)
– he used parable of the unfruitful fig tree i.e. he came to give sinners time to repent so as to
get to the kingdom of God.
– In parable of mustard seed, Jesus showed that the kingdom of God starts from a humble
beginning.
– In the parable of the yeast, he showed the kingdom grows secretly in the hearts of people.
– In the parable of the sower, he taught kingdom of God belongs to those who hear the word
of God and do according to it.
– Alternative religions/denominations
– too much wealth/riches
– it is too demanding.
– lack of absolute faith in God.
– Science and technology seems to provide solutions to man’s problems.
– lacking good role models
– discouragement from church leaders.
– permissiveness in the society.
– drug abuse makes people not to think about God.
6 x 1 = 6 marks
– He comforts the believers.
– Teaches the believers all things.
– He brings remembrance all that Jesus taught the disciples.
– He convicts the world of sin and leads to righteousness and judgment.
– Guides the believers to all truth.
– To glorify Jesus among all believers.
– Enables believers to witness about Christ
– To counsel & advice God’s people for he knows the mind of God.
– He reveals to God’s people what is yet to come.
– To enable believers to know the perfect will of God.
7 x 1 = 7 marks
– Sexual immorality.
– Abuse of the gifts of the holy spirit.
– Misconduct during the Lord’s supper
– Misunderstanding of Resurrection.
– Eating food/offered to idols.
– Fellow believers taking others to pagan courts.
– The position of women in the church.
– Dispute over marriage, divorce and celibacy.
7 x 1 = 7 marks
C What are the causes of disunity in the church today (6mrks)
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TERM ONE
OUR HOME
A _________is a place where a family lives (tree, home)
One or many _________will make a home (houses, people)
Our home and what is around us make up our _________ (compound, house)
Every home must have a ________ (car, house)
Our home is in ________ estate
Types of houses
A hut is a _________ house (traditional, modern)
A hut has ___________doors (one, two)
Many huts together make a ___________ (homestead)
A place where we store grains is called a __________ (granary, hut)
A__________ is house for the ____________ and ___________ (manyatta, bus, Luo, Maasai, Samburu)
A manyatta is a _______ house (traditional, bad)
__________are many houses in one tall building (flats, stairs)
Flats are mainly found in ________ (towns, villages)
A flat is a _______ house (traditional, modern)
A block of flats is also called_________ (apartment, semi – apartment)
This houses stay for a _________time (short, long)
A camp has many _______ together (tents, huts)
Semi-permanent houses include
GRADE ONE ENVIRONMENTAL ACTIVITIES
____
____
Makuti houses
Mabati houses
Timber houses
Small huts
Permanent houses
This houses stay for a _________ time (long, short)
It is a _________house (modern, traditional)
Permanent house include
Bungalow
Storied houses
Materials used in building
The roof of a hut is made of ___________
The wall of a hut is made of ___________
The floor of s hut is made of ___________
A manyatta is made of ___________and ____________
_________ are used to make a granary
A mabati house is made of ___________
Another name for mabati is ___________
A timber house is made of __________
The roof of a permanent house is made of ___________
The walls of a permanent house is made of ___________
Parts of a house
A house has different parts called __________ (rooms, toilet)
Name the rooms found in your home
Bedroom
____
Kitchen
Bedroom
Bathroom
Store
Toilet
The sitting room is also called____________
The _______is also called the washing area
A kitchen is also called the _________area
The bedroom is also called the _______area
Uses of different parts of the house
We ____________and rest in the sitting room
When it is time to sleep we go to ___________ (bedroom, store)
When it is time to eat we go to __________ (dining, store)
Food is cooked in the _____ (kitchen, store)
Farm tools are kept in the __________ (toilet, stores)
Tissue paper is found in the ________ (toilet, store)
People sleep in the ________
We wash our bodies in the ________
Importance of houses
Houses protect us from hot sun, heavy clouds and strong winds
Our house keeps us safe from dangerous _________________ (people, trees)
__________are bad people who steal from our houses (thieves, teachers)
A _______person is a person I do not know (mother, stranger)
Birds stay in a _________ (nest, house)
____
A dog lives in a _________ (kennel, nest)
SAFETY IN THE HOME
Things found in the home
Some things are found inside the house and others outside the house
Name of some things found in the house
Lamp
Broom
Chair
Jiko
Basin
Utensils
Beds
Cupboards
Some animals kept at home
Dog
Cat
Chicken
Pig
Donkey
Sheep
Camel
Uses of things found in the home
A panga is used for ___________
We use a lamp to ____________ (light) up a house
A __________is used to collect rubbish in the compound (jembe, rake)
Peter wants to cut grass; he will use a ______________ (panga, slasher)
We use a ________ to sweep
____
We listen to news from the _________
People use _______________to split firewood
Uses of some animals found at home
Donkeys and camels help us to carry ____________ things (heavy bad)
Name 3 animals that give us milk
_____________
_____________
______________
Sheep, rabbits and pigs give us___________ (meat, wool)
Match the animals and the meat
Cow mutton
Goat pork
Hen beef
Sheep goat meat
Pig chicken
A sheep give us_______and________
Dangerous objects at home
Some objects are dangerous they can ____________ us (hurt, love)
Do not play with ___________ (toys, fire)
Fire can ___________us or even our houses (burn, wash)
A hot jiko can __________us (burn,)
A dog can _________us and a cat can ___________us (scratch, bite)
Putting things like sticks into electrical sockets can cause_________ (shock, play)
____
Playing with matchsticks can cause __________ (fire, rain)
Medicine should be kept away from ___________ (children, teachers)
Broken bottles can __________us (hurt)
Name four sharp objects
______________
________________
________________
_____________
Keeping our homes clean
We sweep the floor to remove __________ (dirt, soap)
We clean utensils to keep away_________and _____________ (rats, dogs, cockroaches)
We should cut long _________around our home (grass, tree)
_________and__________ are some animals that can hide in long grass
We remove cobwebs to keep away __________ (spiders, cows)
Tools are kept safely in the ___________ (kitchen, store)
We should sleep under the _________ (bed, mosquito net)
A fence keeps away_________ from entering the home (friends, strangers)
Dirty toilets have ________ (flies, flowers)
A _________takes care of our home at night (teacher, watchman)
OUR FAMILY
____
MEANING OF A FAMILY
A group of people who are related is called ____________ (group, family)
A family has __________and _________
Father and mother are my _______________ (parents, enemies)
Father is a __________parent (male, female)
A ________ is a female parent (mother, father)
A male child is a _______ and a female child is a _____________
In a family we ________one another (love, hate)
Types of families
Nuclear family
It is made up of father, mother and children
__________is the head of the family (father, mother)
A nuclear family has _________parents (one, two)
A single parent family
It has __________parents (one, two)
It is made up of ___________and _____________
Father and _______make up a single family
An extended family
______are members of an extended family
____
My grandparents call me __________ (grandchild, sister)
Uncle, aunts, cousins are my _________ (relatives, friends)
A boy will be called ___by his grandparents (granddaughter, grandson)
How family members are related
My parents calls me a ________ or a ________ (son, daughter)
My mother’s sister is my ____________
My father’s mother is my __________
My brothers and sisters child is my ____________
The child of my aunt is my_________________ (uncle, aunt)
My father’s brother is my _________ (uncle, aunt)
Work done by family members
___cooks for us at home (mother, aunt)
Our ________pay our school fees
Her work is to tell us stories, she is my _____________
Children should help their parents by
___________
____________
______________
When work is shared it becomes _________________ (easier, good)
Sharing work makes us __________ (happy, sad)
Family celebrations
A celebration is a _________ (parade, ceremony)
Name three family celebrations
____
_______________
_______________
_______________
The day I was born is my _________
In a birthday party people are __________
We celebrate the birth of a child by giving___________ (gifts, stones)
When two people get __________there is a wedding ceremony (money, married)
Wedding celebrations are _________ ceremonies (sad, happy)
When someone dies we go for _________ (wedding, funeral)
A funeral is a ________ceremony (happy, sad)
Religious celebrations
Muslims worship in a ________
Hindus celebrate __________
Christians worship in a ___________
Iddul –fitr is celebrated by __________
Hindus worship in a _________
Christmas is celebrated by ___________
National celebrations
Match the celebrations
Celebration date
New year 1st may
Labour Day 20th October
Madaraka day 25th December
Jamhuri day 26th December
https://educationnewshub.co.ke/category/teachers-resources/
____
Mashujaa day 1st June
Christmas day 12th December
Boxing Day 1st January
TERM 2
OUR FAMILY NEEDS AND CHILD RIGHTS
FOOD
Food gives us __________to work and play
We eat food in order to grow _________
Food makes us look __________
We get food from plants and _______________
After eating food we should drink ___________
Water is ___________
We cover our food to keep away ______________
Match the following
Hen pork
Pig fillet
Cow chicken
Fish mutton
Sheep beef
Draw and name three foods we eat
__________
___________
___________
https://educationnewshub.co.ke/category/teachers-resources/
____
(Water, strong, animals, soil, energy, life, fat, healthy, water
Clothes
We wear ___________ to cover our bodies
When it is cold we wear_________clothes
In __________places people wear light clothes
Clothes we wear at school are called ______
We use ___________ on a rainy day
Shoes make us look __________
On a muddy day we wear ______________
Draw clothes we wear
(Heavy, light, umbrella, gumboots, smart, warm, clothes, uniform)
Shelter
People live in a _________
Another name for a house is _________
Houses protect us from _________-animals
Houses provide us with__________
We ___________i n houses at night
Every home must have a __________
Name three types of houses
___________
___________
____________
(Wild, shelter, domestic, sleep, food, toilet, manyatta, stoned house, hut)
https://educationnewshub.co.ke/category/teachers-resources/
____
How to meet family needs
To meet family needs we need to ____________
Our parent’s _________to get money
My mother is a __________
A__________sells fish to get money
A__________grows food
A cow gives us _________and___________
A sheep gives us ________and __________
We buy food from the ___________
A cobbler repairs our ________
Revision: basic needs
Name three basic needs
_____________
___________
___________
Name three sources of water
______________
______________
_____________
Shelter protect us from
__________________
__________________
__________________
Name three types of food that we eat
__________________
_________________
___________________
Name three types of clothes we wear
https://educationnewshub.co.ke/category/teachers-resources/
____
________________
________________
_______________
A person who makes clothes is a ______________
A carpenter uses_________to make a house
Maintaining a happy family
Good behaviour in the family
Good behaviour means _________manners
We should ____________other people
We should not ___________our parents and friends
We say __________when we do something wrong
We say _________when we wake up in the morning
Children should ________their parents
We say ________when people do good things
We say ________- when asking for something
(Good, bad, abuse, respect, love, sorry, good morning, beat, obey, thank you, please
Child rights
A right is a ________ to do something
Children’s right makes them to be treated ___________
All children have a right to __________life
All children need food to __________
Children should go to school
Child _________is when children work for money
www.arena.co.ke 0713779527 ____
____
When sick children should be taken to __________
Parents should__________from bad people
(Freedom, well, badly, good, die, labour, hospitals, protect)
OUR SCHOOL
Name of our school
Our school is called __________
People ________in School
A school is a place where we learn to ____________and _____________
Our school is a ____________school
Boys wear __________-and__________ while girls wear____________and _____________ to school
The pupils in our school are ________
Our school was started by _________
(Learn, fight, read, and write, public, private, shirts, skirts, dresses, blouses, boys and girls, boys alone)
Symbols of the school and their importance
School logo
A _____is a symbol of the school
Our school motto is __________
Name two thing found in the logo of our school
______________
_____________
Draw the school logo
Name two things that you can see on the school logo
(Logo, fearing God wisdom begins)
https://educationnewshub.co.ke/category/teachers-resources/
____
The flag
This is a __________
The flag of Kenya has ______colours
We stand at _________-while raising the flag
We rise the flag __________and _______________every week
Name the colours of our flag
_______________
________________
________________
________________
Red on our flag shows ___________
Green on our flag means ___________
_____________raises the flag at our school
The Colour on top of the flag is ___________
Flag, 4, 10, attention, ease, Monday, Friday, black, white, red, green, blood, Colour, natural resources,
scouts, black)
National anthem
______anthem is sang when raising the flag
National anthem is a ____________-for our country
Name two languages that we use when we sing national anthem
____________
_____________
There are __________verses in the national anthem
Complete the following sentence of the national anthem
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____
“Oh God of all ___________”
(Church, National, Prayer, English, Kiswahili, 4, 3, Creation)
Revision
Children should _________their parents
Patents and friends should not be __________by children
Good behaviour means good ___________
We should wash our __________after visiting the toilet
Children should wear___________
Children should live in a _________
Children should go to __________to learn
Good children say_____-when they do something wrong (obey, abused, good, hands, sorry, legs,
clothes, school, house)
School compound
Our school compound has ____________and____________
We help ourselves in the _____________
In ___________–we are given story books to read
Our school has __________where we go to play at breaktime
We plant ________and__________in our Shamba
We should keep our school compound _________
Draw and name two things found in the school compound
(Classroom, kitchen, dormitory, stores, library, toilet, bush, playground, maize, potatoes, cabbages,
coffee, small, clean)
Our school routine
https://educationnewshub.co.ke/category/teachers-resources/
____
We go to school from ___________to ______________
I go to school for ___________days in a week
Assembly is also called________
We go for assembly at ____________o clock in the morning
When the bell rings we change the subject
We go for lunch at ____________
After learning we leave school in the _________
(Monday to Sunday, Monday to Friday, 7, 8, school, circle, parade, 5, 9, 10.00am 12.45pm, every
morning, every afternoon)
People in the school and their roles
Some people in the school community are ____________
We have __________teachers in our school
There are _________pupils in our school
My class teacher is called________
The secretary types___________and______________
The __________guards the school during day and night
A __________drives our school bus
Food is cooked from the ____________
(Teachers, cooks, drivers, gateman, secretary, 2, 10, 60, tests, books, exams, 100, 700, kitchen)
Our classroom
Am in class _____________
We learn in the _________room
Name some things found in your classroom
We sit on ________
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____
In our classrooms desks are arranged into ____________rows
________sits behind in class
The ___________tells us what to do when the teacher is absent
We keep our books in _________after classes
We use _________to clean the blackboard
Name three things you do to keep the classroom clean
_____________
____________
____________
REVISION: OUR SCHOOL
The special song for our school is called__________
Our school motto says________
Write three needs in your school
____________
____________
____________
The school __________make we look different from other pupils
We should keep our school ____________
The colours of our school uniform are ________and ___________
The name of our head teacher is ___________
Class rules
__________guides us in class
Pupil’smust________the class rules
Name two rules in the class
We should _________our teachers
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____
The __________help us keep order in the classroom
We clean our school by____________
We ___________our class everyday
We use __________-for sweeping
Name three things you do to keep your classroom clean
____________
____________
___________
Taking care of things in our classroom/school
We write using _________in school
All children in school write on ____________
The teacher __________the cupboard so that the books cannot be stolen
Before going home we should ___________the windows
Our__________locks the classroom
Tearing books to make toys is ___________
It is bad to play with chairs and tables
A class timetable shows time and___________that we learn everyday
Safety on the way to and from school
Different ways of travelling to school
I go to school by _________
Kama lives far from school; he goes to school by __________
Some pupils ___________to school because they live near the school
Name three ways that we can use to go to school
https://educationnewshub.co.ke/category/teachers-resources/
____
_____________
______________
_____________
Draw two means of travelling to school
How to use the road safely
Road safety means free from___________on the road
Road_________helps us use the road safely
We should __________road signs
We should walk on the _________side of the road
Before crossing the road, look ____________, look_________look____–again and cross if the road is
clear
A__________is a person walking on foot
Another name for zebra crossing is ___________
The colours of a zebra crossing are __________-and _____________
Draw and Colour the road signs for
Bus stop
Stop
Dangers of talking and going with strangers
A ___________is a person I don’t know
We should not talk to _____________
Write two things that strangers might do to you
_____________
______________
Write three things we should not accept from strangers
https://educationnewshub.co.ke/category/teachers-resources/
____
_____________
_____________
_____________
Strangers can__________to us
We should walk with the other ____________
Gifts from strangers can be __________
Revision
We should cross the road only when it is _________
It is wrong to __________near the road
People we do not know are ____________
We should not get out of a ______________vehicle
Name two ways of travelling to school
Important features on the way to school
Features seen along the way
We see many things on our way to ___________
Things that we see are called __________features
Name three physical features that you see on your way to school
Many trees growing together make a __________
Animals that live in the forest are called __________animals
Name some of the wild animals
A_________has a lot of water
We buy vegetables and fruits from__________
https://educationnewshub.co.ke/category/teachers-resources/
____
Draw some things you see when coming to school
______________
_______________
_______________
Christians go to __________and Muslims pray in __________
Importance of the features to the people living near our school
We grow plants in a _________
Plants give us ____________
Wild animals are___________because they attract visitors
We get water and fish from a __________
Trees provide a ________for birds and monkeys
Tourists bring a lot of ___________when they visit us
We get ___________from trees in the forest
Visitors who come to see wild animals are called__________
We cross a river using a ____________
https://educationnewshub.co.ke/category/teachers-resources/
____
DATA SECURITY AND CONTROL
Introduction
Data & Information must be protected against unauthorized access, disclosure, modification or damage. This is because; it is a scarce & valuable resource for any business organization or government. It is mostly used in transactions, it can be shared, and has high value attached to it.
Data & Information security:
Data security is the protection of data & information from accidental or intentional disclosure to unauthorized persons.
Data & Information privacy:
Private data or information is that which belongs to an individual & must not be accessed by or disclosed to any other person, without direct permission from the owner.
Confidential data or information – this is data or information held by a government or organization about people. This data/information may be seen by authorized persons without the knowledge of the owner. However, it should not be used for commercial gain or any other unofficial purpose without the owner being informed.
Review Questions
(b) Recently, data and information security has become very important. Explain.
SECURITY THREATS TO DATA & INFORMATION
A virus is a computer code usually designed to carry out 2 tasks:
Types of computer viruses.
Types of destructions/damages caused by a virus attack
Sources of viruses.
If a diskette is used on a virus infected computer, it could become contaminated. If the same diskette is used on another computer, then the virus will spread.
Pirated software may be contaminated by a virus code or it may have been amended to perform some destructive functions which may affect your computer.
A virus could be introduced when the software is being developed in laboratories, and then copied onto diskettes containing the finished software product.
Some virus programs behave like games software. Since many people like playing games on computers, the virus can spread very fast.
Both freeware & shareware programs are commonly available in Bulletin board systems.
Such programs should first be used in controlled environment until it is clear that the program does not contain either a virus or a destructive code.
Viruses programs can be spread through software distributed via networks.
Symptoms of viruses in a computer system.
The following symptoms indicate the presence of a virus in your computer:
Control measures against viruses.
If they have to be used, they must be scanned for viruses.
Data & information is always under constant threat from people who may want to access it without permission. Such persons will usually have a bad intention, either to commit fraud, steal the information & destroy or corrupt the data.
Unauthorized access may take the following forms:
This is tapping into communication channels to get information, e.g., Hackers mainly use eavesdropping to obtain credit card numbers.
This is where a person may monitor all computer activities done by another person or people.
The information gathered may be used for different purposes, e.g., for spreading propaganda or sabotage.
Industrial espionage involves spying on a competitor so as to get or steal information that can be used to finish the competitor or for commercial gain.
The main aim of espionage is to get ideas on how to counter by developing similar approach or sabotage.
Control measures against unauthorized access.
Errors and accidental access to data & information may be as a result of:
Control measures against computer errors & accidents.
This is because; accidental access mistakes occur if the end-users have too much privilege that allows them to access or change sensitive files on the computer.
The threat of theft of data & information, hardware & software is real. Some information is so valuable such that business competitors or some governments can decide to pay somebody a fortune so as to steal the information for them to use.
Control measures against theft of information, hardware, & software.
Review Questions
(b) Give and explain two types of computer viruses.
(c) List three types of risks that computer viruses pose.
(d) List and explain five sources of computer viruses.
(e) Outline four symptoms of computer viruses.
(f) Explain the measures one would take to protect computers from virus attacks
COMPUTER CRIMES
Types of computer crimes.
The following are the major types of computer crimes:
Trespass.
Trespass is not allowed and should be discouraged.
Hacking.
Hacking is an attempt to invade the privacy of a system, either by tapping messages being transmitted along a public telephone line, or through breaking security codes & passwords to gain unauthorized entry to the system data and information files in a computer.
Reasons for hacking.
Hacking is done by skilled programmers referred to as Hackers. Hacker is a person who gains unauthorised access to a computer network for profit, criminal mischief, or personal gain.
Such people are able to break through passwords or find weak access points in software. They are involved in propagating computer viruses.
Tapping.
Tapping involves listening to a transmission line to gain a copy of the message being transmitted.
Tapping may take place through the following ways:
Cracking.
Cracking is the use of guesswork by a person trying to look for a weakness in the security codes of a software in order to get access to data & information.
These weak access points can only be sealed using sealed using special corrective programs called Patches, which are prepared by the manufacturing company.
A program patch is a software update that when incorporated in the current software makes it better.
NB: Cracking is usually done by people who have some idea of passwords or user names of the authorized staff.
Piracy.
Software, information & data are protected by copyright laws. Piracy means making illegal copies of copyrighted software, data, or information either for personal use or for re-sale.
Ways of reducing piracy:
Fraud.
Fraud is the use of computers to conceal information or cheat other people with the intention of gaining money or information.
Fraud may take the following forms:
Data input clerks can manipulate input transactions, e.g., they can create dummy (ghost) employees on the Salary file or a ghost supplier on the Purchases file.
E.g., a person created an intelligent program in the Tax department that could credit his account with cents from all the tax payers. He ended up becoming very rich before he was discovered.
Fraudsters can either be employees in the company or outsiders who are smart enough to defraud unsuspecting people.
Reasons that may lead to computer fraud.
Security measures to prevent fraud:
Sabotage.
Sabotage is the illegal or malicious destruction of the system, data or information by employees or other people with grudges with the aim of crippling service delivery or causing great loss to an organization.
Sabotage is usually carried out by discontented employees or those sent by competitors to cause harm to the organization.
The following are some acts of saboteurs which can result in great damage to the computer centres:
Alteration.
Alteration is the illegal changing of stored data & information without permission with the aim of gaining or misinforming the authorized users.
Alteration is usually done by those people who wish to hide the truth. It makes the data irrelevant and unreliable.
Alteration may take place through the following ways:
This is done by people with excellent programming skills. They do this out of malice or they may liaise with others for selfish gains.
This is normally done by authorized database users, e.g., one can adjust prices on Invoices, increase prices on selling products, etc, and then pocket the surplus amounts.
Security measures to prevent alteration:
Theft of computer time.
Employees may use the computers of an organization to do their own work, e.g., they may produce publications for selling using the computers of the company.
Theft of data (i.e., commercial espionage).
Employees steal sensitive information or copy packages and sell them to outsiders or competitors for profit.
This may lead to a leakage of important information, e.g., information on marketing strategies used by the organization, research information, or medical reports.
Review Questions
(b) State and explain various types of computer crimes.
DETECTION & PROTECTION AGAINST COMPUTER CRIMES
The following measures can be taken to detect & prevent computer crimes, and also seal security loopholes.
Audit trails
This is a careful study of an information system by experts in order to establish (or, find out) all the weaknesses in the system that could lead to security threats or act as weak access points for criminals.
An audit of the information system may seek to answer the following questions: –
Data encryption
Data being transmitted over a network faces the dangers of being tapped, listened to, or copied to unauthorized destinations.
To protect such data, it is mixed up into a form that only the sender & the receiver can be able to understand by reconstructing the original message from the mix. This is called Data encryption.
The flow diagram below shows how a message can be encrypted and decrypted to enhance security.
Cyphertext
Plain text Plain text
Encryption key Decryption key
The message to be encrypted is called the Plain text document. After encryption using a particular order (or, algorithm) called encryption key, it is sent as Cyphertext on the network.
After the recipient receives the message, he/she decrypts it using a reverse algorithm to the one used during encryption called decryption key to get the original plain text document.
This means that, without the decryption key, it is not possible to reconstruct the original message.
Log files
These are special system files that keep a record (log) of events on the use of the computers and resources of the information system.
Each user is usually assigned a username & password or account. The information system administrator can therefore easily track who accessed the system, when and what they did on the system. This information can help monitor & track people who are likely to violate system security policies.
Firewalls
A Firewall is a device or software system that filters the data & information exchanged between different networks by enforcing the access control policy of the host network.
A firewall monitors & controls access to or from protected networks. People (remote users) who do not have permission cannot access the network, and those within cannot access sites outside the network restricted by firewalls.
LAWS GOVERNING PROTECTION OF INFORMATION
Laws have been developed that govern the handling of data & information in order to ensure that there is ‘right of privacy’ for all people.
The following rules must be observed in order to keep within the law when working with data and information.
WEEKEND ASSIGNMENT 2015
FORM 3 AND 4
(b) Give and explain two types of computer viruses.
(c) List three types of risks that computer viruses pose.
(d) List and explain five sources of computer viruses.
(e) Outline four symptoms of computer viruses.
(f) Explain the measures one would take to protect computers from virus attacks
(b) Mention various threats to computer security.
GOOD LUCK
COMPUTER SECURITY
What is Computer security?
A computer system can only be claimed to be secure if precautions are taken to safeguard it against damage or threats such as accidents, errors & omissions.
The security measures to be undertaken by the organization should be able to protect:
Environmental threats to computers & Information systems.
Fire destroys data, information, software & hardware.
Security measures against fire:
This causes rusting of the metallic components of the computer.
Security measures against water, floods & moisture:
This causes power failure that can cause damage to data, which has not been transferred to permanent storage devices.
Security measures:
Excessive heat or temperature from the computer itself or from the surrounding environment can destroy computer storage media or devices.
Security measures:
A virus is a rogue software program that spreads rampantly through computer systems, destroying data or causing the system to break down.
Security measures against computer virus:
Dust and Smoke particles settle on storage devices and may scratch them during Read/write operation.
Security measures:
This includes activities such as:
Security measures:
People threats include:
Security measures against Carelessness & Clumsiness:
Security measures against Vandalism:
Review Questions
(b) Mention various threats to computer security.
CAUSES OF DATA LOSS IN COMPUTERS
Momentary interruptions or fluctuations of electrical power may cause:
The main cause of power disruptions are:
Precautions against data loss due to Power failure:
Frequent saving of documents ensures that minimum data is lost in case of any power failure.
Some application packages have an AutoSave feature, which should be activated to automatically save work after a specified time interval.
To eliminate any power quality defects or fluctuation, use power correction equipment such as a Stabilizer or Uninterruptible Power Supply (UPS). These equipments ensure a steady flow of input power to the computer system.
A computer virus destroys all the data files & programs in the computer memory by interfering with the normal processes of the operating system.
Precautions against computer viruses:
Use Antivirus software to detect & remove known viruses from infected files.
Some of the commonly used Antivirus software are: Dr. Solomon’s Toolkit, Norton Antivirus, AVG Antivirus, PC-Cillin, etc
NB: The best way to prevent virus is to have a memory-resident antivirus software, which will detect the virus before it can affect the system. This can be achieved by installing a GUARD program in the RAM every time the computer boots up. Once in the RAM, the antivirus software will automatically check diskettes inserted in the drives & warn the user immediately if a disk is found to have a virus.
Review Questions
(b) What are Anti-viruses? Explain how they detect and remove viruses.
Commands such as DELETE & FORMAT can be dangerous to the computer if used wrongly.
Both commands wipe out the information stored on the specified secondary storage media, e.g., formatting the Hard disk (drive C:) will destroy all the software on that system.
Precautions against Accidental erasure:
Use the Undelete facilities in case you accidentally delete your files.
There are two Undelete facilities depending on the operating system you are using.
To undelete at the DOS prompt, change to the drive & directory whose files were deleted, then type, e.g.,
C:\>UNDELETE <directory that contain the deleted file>
A list of all deleted files will be displayed with the first letter missing. Type in the first letter and the file will be recovered.
Norton utilities & PC Tools also have an undelete facility, which is similar to the DOS Undelete facility.
The Recycle Bin temporarily stores all deleted files & can be used to recover your files.
The Recycle Bin will restore all selected files to their original folders and disks.
NB: If you delete a file accidentally, don’t copy any files or install any applications to the disk that contains the deleted file. If you write anything to the disk, you might destroy parts of the deleted file, making it unrecoverable.
MS-DOS 6.0 has an Unformat facility which can be used to recover information stored on disks that have been accidentally formatted.
All data must be backed up periodically either on diskettes, tapes or CDs so that in case of any accidental loss, the backed up copy can be used to recover the data.
For small files, use the Copy command to make a copy of the data on a diskette. For larger amounts of data, use the Backup command to copy the data to several diskettes or to a tape drive.
Review Questions
When a hard disk crashes, the data or information on the disk cannot be accessed. The effect is the same as formatting the hard disk.
Crashing of a hard disk can occur due to the following reasons:
Precautions against crashing of Hard disks:
All data must be backed up regularly. In addition, all application programs & operating system software should also be kept safely so that in case of a complete system crash, everything can be re-installed/restored.
System tools such as Norton Utilities, PC Tools, QAPlus, etc can be used to revive a disk that has crashed.
Review Questions
Unauthorised access refers to access to data & information without permission.
Computer criminals can do the following harms:
Precautions against Unauthorised access:
Physical access to computer systems should be restricted to ensure that no unauthorised person gets access to the system.
Some of the ways of restricting physical access include:
Install a password to restrict access to the computer system.
A Password is a secret code that can be used to prevent unauthorised access of data in a computer.
Passwords can be put in at various levels:
When a valid password is entered, the user gets access to the computer system. Usually, the user is allowed three (3) attempts to get the password correct. If an invalid password is entered, access is denied after the 3 attempts.
Some computer security systems may generate an alarm if someone tries to use a fake password.
NB: You should never use passwords that can easily be linked to you, e.g., your name, birth date, or names of people close to you.
Review Questions
(b) What is a Password? Give its main importance.
