JAMB Chemistry 1997

B: 31%

The reaction is 2H₂ + O₂ → 2H₂O. The 12 cm³ of oxygen consumes 24 cm³ of hydrogen and forms 24 cm³ of steam. Hydrogen remaining is 35 − 24 = 11 cm³. At 110 °C the total gas volume is 11 + 24 = 35 cm³, so hydrogen accounts for 11/35 × 100 = 31% to the nearest whole percent.

D: A pure compound

The horizontal section represents freezing at a constant temperature. In the usual cooling-curve interpretation this is characteristic of a pure substance, making a pure compound the best listed description. The graph alone does not identify its bonding or whether it is a hydrated salt.

B: 31.0

Vapour density relative to hydrogen is half the relative molecular mass. Therefore Mᵣ(XH₃) = 2 × 17.0 = 34.0. Subtracting the three hydrogen atoms gives Aᵣ(X) = 34.0 − 3 = 31.0.

C: 21 s

By Graham’s law, rate is inversely proportional to the square root of molar mass. Time also increases with the volume required. Thus t = 7 × (60/30) × √(71/32) = 20.85 s, or 21 s to the nearest second.

C: The heat of vaporisation is drawn from the body

Evaporation requires energy to overcome attractions between liquid molecules. When that energy is supplied by the body, the body loses thermal energy and cools.

C: 1s²2s²2p⁶3s¹ and 1s²2s¹

The configurations in C describe sodium and lithium. Each has one electron in its outermost shell and belongs to Group 1, so they have similar chemical properties. The other pairs have different outer-shell electron counts.

D: Atomic radius

Atomic radius generally decreases from left to right across a period because the increasing effective nuclear charge attracts electrons more strongly. It increases down a group as additional occupied electron shells are added.

C: 2 and 1

P is sodium, which forms P⁺, and Q is oxygen, which forms Q²⁻ in the oxide. Two P⁺ ions balance one Q²⁻ ion, giving P₂Q. Thus x = 2 and y = 1.

B: 16.2

The relative atomic mass is the abundance-weighted average: (16 × 0.90) + (18 × 0.10) = 16.2. The percentages describe the specified sample.

C: 21%

Heated copper removes oxygen as copper(II) oxide. The volume removed is 200 − 158 = 42 cm³. Comparing volumes at the same temperature and pressure gives oxygen percentage = 42/200 × 100 = 21%.

C: 10⁻⁴ mol dm⁻³

At 25 °C, pH + pOH = 14.00. Thus pOH = 4.00 and [OH⁻] = 10⁻⁴ mol dm⁻³.

C: 0.104 mol dm-3

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The NaOH amount is 0.02500 × 0.125 = 0.003125 mol, so acid amount is half this. Acid concentration = 0.0015625/0.01500 = 0.104 mol dm⁻³.

A: Acidic

Copper(II) ions are reduced to copper at the cathode. At the platinum anode, water is oxidised: 2H₂O → O₂ + 4H⁺ + 4e⁻. The increasing H⁺ concentration makes the solution more acidic.

B: +3

Potassium contributes +1 per atom and chloride −1. For neutral K₃ZCl₆, 3(+1) + Z + 6(−1) = 0, so Z has oxidation number +3.

B: −198 kJ

Reaction enthalpy is the sum of product formation enthalpies minus the reactant sum. O₂ in its standard state contributes zero. Therefore ΔH = 2(−396) − 2(−297) = −198 kJ for the reaction as written.

B: 85.48 kJ mol⁻¹

Use ΔG = ΔH − TΔS with consistent units. At 25 °C, T ≈ 298 K and ΔS = 0.0118 kJ mol⁻¹ K⁻¹. Therefore ΔG = 89 − 298(0.0118) = 85.4836 kJ mol⁻¹, or 85.48 kJ mol⁻¹.

C: n + m

The overall reaction order is the sum of the exponents of concentration in the experimentally determined rate law. For rate = k[X]ⁿ[Y]ᵐ, the order is n + m.

A: increase temperature at constant pressure

Increasing temperature favours the endothermic direction, so it increases the equilibrium proportion of products for an endothermic forward reaction. A pressure change depends on the numbers of gas particles on each side.

D: Disinfecting water

The reaction produces chlorine: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O. Chlorine is used in water disinfection because it destroys many disease-causing microorganisms.

C: [G]ᵖ[H]ᑫ / ([E]ᵐ[F]ⁿ)

The concentration equilibrium expression places products above reactants, with each concentration raised to its stoichiometric coefficient. Thus Kc = [G]ᵖ[H]ᑫ / ([E]ᵐ[F]ⁿ).

A: SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s), in dilute HNO₃

Sulfate ions form a white precipitate of barium sulfate with Ba²⁺: Ba²⁺ + SO₄²⁻ → BaSO₄(s). Acidifying with dilute nitric acid helps exclude interfering carbonate precipitates.

B: Reaction of an acid with a base

Rust contains iron oxides and hydroxides. These basic materials react with acid to form iron salts and water, so their removal by acid is based on an acid–base reaction.

A: Electrolysis using mercury as the cathode

In the historical mercury-cell process, sodium forms an amalgam at the mercury cathode. The amalgam then reacts with water in a separate decomposer to produce NaOH and hydrogen, while mercury is recycled. This identifies the mercury-cathode electrolysis process.

B: Alkanoate

An alcohol and a carboxylic acid undergo acid-catalysed esterification to form an ester, also called an alkanoate, and water. Concentrated sulfuric acid provides the acid catalyst.

A: CH₃–CHI₂

The first addition gives CH₂=CHI. Further addition of HI produces CH₃CHI₂, with both iodine atoms on the same carbon. The overall reaction is HC≡CH + 2HI → CH₃CHI₂.

B: The calcium salts of detergents are water-soluble

Common anionic detergents form calcium and magnesium salts that remain soluble. Soap instead forms insoluble salts with these ions, producing scum and reducing its cleaning action in hard water.

C: CH₃COOH

KOH provides 0.025 × 0.02 = 0.00050 mol OH⁻. A monobasic acid reacts with it in a 1:1 ratio, giving molar mass 0.030/0.00050 = 60 g mol⁻¹. CH₃COOH has this molar mass: 2(12) + 4(1) + 2(16) = 60.