MITIHANI POPOTE EXAMINATIONS SERIES FORM FOUR CHEMISTRY EXAMINATION SERIES 06 (With Marking Guide)

FORM FOUR CHEMISTRY EXAMINATION SERIES 06

Chemistry Examination Answers
Series 06

Section A (15 Marks)

Question 1

(i) If the results you obtain from an experiment do not support your hypothesis C. Give ideas for further testing to find a solution

Scientific method requires analyzing unexpected results to refine understanding. Discarding data (B) or changing variables (E) would compromise scientific integrity. The proper approach is to design new experiments to investigate the discrepancy.

(ii) Which of the following sets of processes uses a gas that ignites with a pop sound when a lighted splint is passed through it? B. Hardening oil, balloon filling and welding

Hydrogen gas (H₂) produces the characteristic "pop" sound when ignited. It's used in:

• Hydrogenation of oils (hardening)
• Filling balloons (though largely replaced by helium for safety)
• Oxy-hydrogen welding (H₂ + O₂ produces very high temperatures)

(iii) One advantage of hard water is that it D. Contains minerals which are useful to the body

Hard water contains calcium and magnesium ions that contribute to dietary mineral intake. While options A, C, and E describe disadvantages, the mineral content can benefit bone health and cardiovascular function.

(iv) A and B are elements in the same period of the periodic Table. A is in group II and B is in group III. Which of the following statements is not true about the two elements? D. A has one electron more than B in its outermost shell

In the same period:

• Group II elements have 2 valence electrons
• Group III elements have 3 valence electrons
• Thus A has one fewer valence electron than B (not more)
• All other statements are correct about adjacent group elements

(v) An example of a salt which is insoluble in cold water but can dissolve in hot water is B. Lead chloride

PbCl₂ has solubility that increases significantly with temperature (endothermic dissolution). Other options:

• NaCl (A) is soluble in both cold and hot water
• CaCO₃ (C) and AgCl (D) remain insoluble in hot water
• CuCO₃ (E) decomposes in hot water

(vi) Which state is involved when drying wet clothes? D. Liquid to gas

The process involves evaporation of liquid water (from clothes) into water vapor (gas). This phase change requires energy (latent heat of vaporization) which is why clothes dry faster in warm, dry conditions.

(vii) The formula below represents some chemical substances. Which formula represents a substance that does not contribute to global warming? A. H₂

Greenhouse gases trap infrared radiation:

• CH₄ (B): Methane (25× more potent than CO₂)
• CO₂ (C): Primary greenhouse gas
• N₂O (D): Nitrous oxide (300× CO₂ potency)
• SO₂ (E): Contributes to acid rain but minimal greenhouse effect
• H₂ (A): Not a greenhouse gas

(viii) An alkyl group has a general formula D. CₙH₂ₙ₊₁

Alkyl groups are alkane derivatives with one hydrogen removed:

• Example: Methyl (CH₃-) from methane (CH₄)
• CₙH₂ₙ₊₂ (C) is alkane formula
• CₙH₂ₙ (B) is alkene formula
• CₙH₂ₙ₋₁ (A) and CₙH₂ₙ₋₂ (E) are incorrect

(ix) The empirical formula of a certain compound is CH₃. Its molar mass is 30g what will be its molecular formula? C. C₂H₆

Calculation:

• Empirical formula mass = 12 + (3×1) = 15
• Multiplier = 30 ÷ 15 = 2
• Molecular formula = (CH₃)₂ = C₂H₆ (ethane)

(x) What type of fire is associated with electrical equipment? A. Class E

Fire classifications:

• Class A: Ordinary combustibles (wood, paper)
• Class B: Flammable liquids
• Class C: Flammable gases
• Class D: Metal fires
• Class E: Electrical fires (note: some systems use Class C)
• Class F: Cooking oils/fats

Question 2 - Matching

List A	List B
(i) A solvent which dissolves most substances to form solutions	C. Water
(ii) A substance that has no definite shape or size	F. Gas
(iii) A substance that has a fixed shape and volume	A. Solid
(iv) A substance whose components can be separated by physical means	E. Milk
(v) Homogeneous mixture of two or more substance	B. Solution

Key concepts:

• Water is the "universal solvent" (i-C)
• Gases expand to fill containers (ii-F)
• Solids maintain shape/volume (iii-A)
• Milk can be separated by centrifugation (iv-E)
• Solutions are homogeneous mixtures (v-B)

Section B (70 Marks)

Question 3 7 marks

(a) Characteristics of good fuel:

1. High energy value: More energy per unit mass means less fuel is needed for the same output, improving efficiency and reducing storage/transport costs.
2. Affordable: Must be economically viable for widespread use; expensive fuels limit accessibility and practical applications.
3. Low non-combustible content: Impurities reduce energy output, cause pollution (ash, SO₂), and may damage equipment (slag formation).

(b) Chemistry in community:

1. Water treatment: Chemical processes (coagulation, chlorination) make water safe by removing pathogens and contaminants.
2. Agriculture: Fertilizers (NH₄NO₃), pesticides, and soil pH adjustment (liming) increase crop yields to support food security.

Question 4 7 marks

(a) First Aid uses:

Item	Use
(i) Soap	Cleaning wounds to prevent infection
(ii) Bandage	Securing dressings or supporting injured limbs
(iii) Sterile gauze	Covering wounds to absorb fluids and prevent contamination
(iv) Iodine tincture	Antiseptic for disinfecting skin around wounds
(v) Petroleum jelly	Protecting minor burns or chapped skin
(vi) Cotton wool	Cleaning wounds or applying antiseptics

(b) Mass of KClO₃ calculation:

Reactions:

4Al + 3O₂ → 2Al₂O₃

2KClO₃ → 2KCl + 3O₂

Moles of Al₂O₃ = 5.1g ÷ 102g/mol = 0.05 mol

From equation: 2 mol Al₂O₃ requires 3 mol O₂ ⇒ 0.05 mol needs 0.075 mol O₂

2 mol KClO₃ produces 3 mol O₂ ⇒ 0.05 mol KClO₃ needed

Mass of KClO₃ = 0.05 mol × 122.5 g/mol = 6.125 g

Question 5 7 marks

(a) Particle explanations:

1. Pouring liquids: Particles can slide past each other (weak intermolecular forces) while maintaining contact (no fixed shape but definite volume).
2. Gas filling containers: Particles move rapidly in random directions with negligible forces between them, expanding to occupy all available space.
3. Solid expansion: Heating increases particle vibration amplitude, causing the fixed lattice structure to occupy more space while maintaining orderly arrangement.

(b) Water hardness removal:

(i) Temporary hardness by boiling:

Ca(HCO₃)₂ → CaCO₃↓ + H₂O + CO₂↑

Heat decomposes soluble hydrogencarbonates into insoluble carbonates that precipitate out.

(ii) Permanent hardness by chemical means:

CaSO₄ + Na₂CO₃ → CaCO₃↓ + Na₂SO₄

Washing soda (Na₂CO₃) precipitates calcium as carbonate, removing sulfate-based hardness.

Question 6 7 marks

(a) Fluorine bonding:

(i) F₂ molecule:

F + F → F:F (or F-F)

Covalent bond - sharing one electron pair (single bond)

Lewis structure:
:F: + :F: → :F:::F: (each F has 3 lone pairs)

(ii) Other bond types:

Ionic bond - Fluorine gains one electron to form F⁻ (e.g., NaF - sodium fluoride)

(b) HNO₃ vs. H₃PO₄ differences:

Property	Dilute HNO₃	Dilute H₃PO₄
Acid strength	Strong acid (fully dissociates)	Weak acid (partial dissociation)
Reactivity with metals	Oxidizing - produces NO/NO₂	Non-oxidizing - produces H₂
Basicity	Monoprotic (1 H⁺)	Triprotic (3 H⁺)

Question 7 7 marks

(a) Bunsen burner flames:

(i) Why luminous flame is unsuitable:

1. Incomplete combustion produces soot (carbon particles) that contaminates apparatus
2. Lower temperature (∼300°C) compared to non-luminous flame (∼1500°C)

(ii) Non-luminous flame conditions:

Air hole fully open - sufficient oxygen for complete combustion (blue cone visible)

(b) Electrolysis time calculation:

Cu²⁺ + 2e⁻ → Cu (2 F deposits 1 mol Cu)

Moles of Cu = 1.184g ÷ 63.5g/mol = 0.01865 mol

Charge needed = 0.01865 × 2 × 96500 = 3600 C

Time = Charge ÷ Current = 3600 ÷ 2 = 1800 seconds = 30 minutes

Question 8 7 marks

(a) Water properties and oxygen test:

(i) Physical properties:

1. High specific heat capacity (4.18 J/g°C)
2. Maximum density at 4°C (anomalous expansion)

(ii) Oxygen identification:

Insert a glowing splint - oxygen supports combustion so the splint will relight brightly

(b) HCl concentration calculation:

NaOH + HCl → NaCl + H₂O

Moles NaOH = 0.5M × 0.0125dm³ = 0.00625 mol

Moles HCl in 25cm³ = 0.00625 mol

Moles in 250cm³ diluted = 0.0625 mol

This came from 10cm³ stock ⇒ Concentration = 0.0625 ÷ 0.01 = 6.25 M

Question 9 7 marks

(a) Chlorine diagrams:

(i) Chlorine atom (Cl):

Electron configuration: 2,8,7

○ Nucleus (17p⁺, 18n⁰)

Electron shells: K(2), L(8), M(7)

(ii) Chloride ion (Cl⁻):

Electron configuration: 2,8,8

Gains 1 electron → negative charge

(b) Carbon allotropes:

(i) Two crystalline forms:

1. Diamond - tetrahedral covalent network
2. Graphite - layered hexagonal structure

(ii) Conductor: Graphite

Reason: Delocalized π-electrons between layers can move freely, enabling electrical conductivity. Diamond has all electrons tightly bound in σ-bonds.

Question 10 7 marks

(a) Metal Z (Sodium):

(i) Extraction cell: Downs cell

(ii) Why not from aqueous solution:

Sodium is more reactive than hydrogen, so H⁺ would reduce instead: 2H₂O + 2e⁻ → H₂↑ + 2OH⁻

(b) Reaction types:

(i) Double displacement/precipitation - Ionic exchange forms insoluble Fe(OH)₃

(ii) Thermal decomposition - Compound breaks down into simpler substances when heated

Question 11 7 marks

(i) Reaction type: Exothermic

Reason: Negative ΔH (-55 kJ/mol) indicates heat is released to surroundings

(ii) Energy profile diagram description:

Y-axis: Energy (kJ)

X-axis: Reaction coordinate

Features:

• Reactants (A₂ + B₂) at higher energy than products (2AB)
• Energy difference = ΔH = -55 kJ
• Peak at activation energy (X kJ above reactants)

Question 12 7 marks

(a) Substance identification:

1. Lead(II) oxide (PbO)

   Question 15 15 marks

   Environmental Chemistry Applications:

   1. Air Pollution Control:

   • Catalytic converters in vehicles use platinum/palladium to convert harmful gases:

     2CO + 2NO → 2CO₂ + N₂

   • Flue gas desulfurization in power plants removes SO₂ using limestone:

     CaCO₃ + SO₂ → CaSO₃ + CO₂

   2. Water Treatment Processes:

   • Coagulation using alum (KAl(SO₄)₂) removes suspended particles:

     Al³⁺ + 3H₂O → Al(OH)₃ + 3H⁺

   • Chlorination disinfects water but requires careful dosage to avoid THM formation

   3. Waste Management:

   • Landfill gas (CH₄) capture and conversion to energy
   • Plastic pyrolysis to break down polymers into useful hydrocarbons

Question 16 15 marks

Industrial Chemistry Processes:

1. Haber Process (Ammonia Production):

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol

• Conditions: 450°C, 200 atm, iron catalyst
• Trade-off between rate (favored by high T) and yield (favored by low T)

2. Contact Process (Sulfuric Acid):

2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅ catalyst, 450°C)

SO₃ + H₂SO₄ → H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O → 2H₂SO₄

3. Electrolysis of Brine:

• Anode: 2Cl⁻ → Cl₂ + 2e⁻
• Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
• Products: Cl₂ gas, H₂ gas, and NaOH solution

Question 17 15 marks

Organic Chemistry Reactions:

1. Functional Group Transformations:

Reaction Type	Example	Conditions
Esterification	CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O	Conc. H₂SO₄, heat
Alkene Hydration	C₂H₄ + H₂O → C₂H₅OH	H₃PO₄ catalyst, 300°C

2. Polymerization:

• Addition: n(CH₂=CHCl) → [CH₂-CHCl]ₙ (PVC)
• Condensation: n(HOOC-C₆H₄-COOH) + n(HO-CH₂-CH₂-OH) → polyester + nH₂O

Practical Chemistry Tips

Laboratory Safety Reminders:

1. Always wear appropriate PPE (goggles, lab coat, gloves)
2. Never taste chemicals or directly smell vapors - waft gently
3. Dispose of waste according to local regulations
4. Know the locations of safety equipment (eyewash, shower, fire extinguishers)
5. Label all containers clearly with contents and hazards

Common Calculation Formulas:

Concept	Formula
Molarity	M = moles solute / dm³ solution
Dilution	M₁V₁ = M₂V₂
Electrolysis	Mass = (I × t × M) / (n × F)