Question 1 · MCQ · Analyse · 1 mark
A nail changes a blue solution
Types of chemical reactions
A clean iron nail is placed in blue copper sulphate solution. A reddish-brown coating forms and the solution becomes pale green. Which explanation fits both observations?
(A) Iron displaces copper and forms iron(II) sulphate.
(B) Copper displaces iron and forms copper oxide.
(C) Only the water evaporates.
(D) Iron and copper sulphate undergo no chemical change.
Show answer and solution
Answer: (A) Iron displaces copper and forms iron(II) sulphate.
Worked solution: Iron is more reactive than copper. Fe + CuSO₄ → FeSO₄ + Cu explains both deposited copper and the pale-green iron(II) sulphate solution.
Question 2 · MCQ · Analyse · 1 mark
Silver salt in sunlight
Decomposition reactions
White silver chloride turns grey in sunlight. What does this show?
(A) A combination reaction caused by cooling.
(B) A photochemical decomposition producing silver.
(C) A displacement reaction caused by iron.
(D) A physical change caused only by drying.
Show answer and solution
Answer: (B) A photochemical decomposition producing silver.
Worked solution: Light decomposes silver chloride: 2AgCl → 2Ag + Cl₂. Finely divided silver gives the grey colour.
Question 3 · MCQ · Analyse · 1 mark
Counting atoms in a reaction
Balanced equations
A student balances aluminium reacting with oxygen to form aluminium oxide. Which coefficient set conserves both elements?
(A) Al + O₂ → Al₂O₃
(B) 2Al + O₂ → Al₂O₃
(C) 4Al + 3O₂ → 2Al₂O₃
(D) 4Al + 2O₂ → 2Al₂O₃
Show answer and solution
Answer: (C) 4Al + 3O₂ → 2Al₂O₃
Worked solution: The balanced equation has 4 aluminium atoms and 6 oxygen atoms on each side: 4Al + 3O₂ → 2Al₂O₃.
Question 4 · MCQ · Analyse · 1 mark
A solid appears on mixing
Precipitation reactions
Clear aqueous barium chloride and sodium sulphate solutions are mixed. A white insoluble solid forms. Identify it.
(A) Sodium chloride
(B) Barium metal
(C) Sodium sulphate
(D) Barium sulphate
Show answer and solution
Answer: (D) Barium sulphate
Worked solution: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). The insoluble barium sulphate is the white precipitate.
Question 5 · A-R · Analyse · 1 mark
Black copper oxide changes colour
Oxidation and reduction
Assertion (A): Hydrogen can reduce hot black copper(II) oxide to reddish-brown copper.
Reason (R): Hydrogen removes oxygen from copper(II) oxide and forms water.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (A) Both A and R are true, and R correctly explains A.
Worked solution: CuO + H₂ → Cu + H₂O. Removal of oxygen from CuO is reduction; hydrogen is oxidised as it gains oxygen.
Question 6 · A-R · Analyse · 1 mark
A balanced equation and energy
Exothermic and endothermic reactions
Assertion (A): Every balanced chemical equation represents an exothermic reaction.
Reason (R): Balancing an equation ensures equal numbers of each kind of atom on both sides.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (D) A is false, but R is true.
Worked solution: Atom conservation does not determine whether heat is absorbed or released. Balanced equations can describe either exothermic or endothermic reactions.
Question 7 · VSA · Apply · 2 marks
Protecting a steel gate
Oxidation and corrosion
Two identical steel gates face the same moist air. One has an intact paint coating. Predict which rusts faster and explain why.
Show answer and solution
Answer: The unpainted gate rusts faster.
Worked solution: Rusting requires oxygen and water. Intact paint forms a barrier that reduces their contact with the steel surface.
Question 8 · VSA · Apply · 2 marks
A warm beaker
Exothermic and endothermic reactions
A teacher adds water to quicklime and the beaker becomes hot. Name the reaction type with respect to heat and write the equation.
Show answer and solution
Answer: Exothermic; CaO + H₂O → Ca(OH)₂ + heat.
Worked solution: Calcium oxide combines with water to form calcium hydroxide. The temperature rises because energy is released to the surroundings.
Question 9 · SA · Analyse · 3 marks
A mass appears to disappear
Balanced equations
A student heats calcium carbonate in an open test tube and finds the remaining solid has less mass. Explain why this does not violate conservation of mass, and give the equation.
Show answer and solution
Answer: Carbon dioxide escapes; total mass is conserved when the gas is included. CaCO₃ → CaO + CO₂.
Worked solution: Heating decomposes calcium carbonate into calcium oxide and carbon dioxide. The open tube loses gaseous CO₂, so weighing only the solid omits a product. In a suitable closed system, total reactant and product masses are equal.
Question 10 · SA · Analyse · 3 marks
Testing a food-packaging claim
Oxidation and reduction
A snack manufacturer fills an opaque packet with nitrogen instead of air. Explain the purpose of the gas, the role of the opaque packet, and the chemical change they help reduce.
Show answer and solution
Answer: Nitrogen reduces oxygen exposure, opacity reduces light exposure, and both help slow rancidity.
Worked solution: Oils and fats can oxidise and develop an unpleasant smell and taste. Nitrogen provides a less reactive atmosphere; blocking light also slows oxidation. These measures delay rancidity rather than reverse it.
Question 11 · LA · Analyse · 5 marks
Classifying laboratory observations
Types of chemical reactions
Classify and write balanced equations for: (i) burning magnesium in oxygen; (ii) heating calcium carbonate; (iii) iron placed in copper sulphate; (iv) mixing barium chloride and sodium sulphate. (v) Explain why the last reaction is also called precipitation.
Show answer and solution
Answer: (i) Combination; (ii) decomposition; (iii) displacement; (iv) double displacement; (v) insoluble BaSO₄ separates.
Worked solution: (i) 2Mg + O₂ → 2MgO. (ii) CaCO₃ → CaO + CO₂. (iii) Fe + CuSO₄ → FeSO₄ + Cu. (iv) BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl. (v) The ions exchange partners and insoluble barium sulphate forms a solid precipitate in the aqueous mixture.
Question 12 · Case Study · Analyse · 4 marks
Three rusting conditions
Oxidation and corrosion
Three identical clean iron nails are kept for several days: P in dry air with a drying agent, Q in boiled water covered by oil, and R in ordinary water exposed to air. Assume the oil excludes air from Q.
Answer the following:
(i) Which nail should show the most rust? [1 mark]
(ii) Which requirement for rusting is absent from P? [1 mark]
(iii) Explain why Q rusts much less than R. [2 marks]
Show answer and solution
Answer: (i) R. (ii) Moisture. (iii) Boiling removes much dissolved oxygen and oil prevents its replenishment.
Worked solution: Rusting needs both oxygen and moisture. P has oxygen but lacks moisture. Q has water but little available oxygen under the stated conditions. R has access to both, so it shows the greatest rusting.
