JAMB Chemistry 1989

C: It melts over the range 300°C to 375°C

A broad melting interval supports the presence of more than one component. A pure crystalline substance normally melts sharply at a fixed pressure. Grinding, a density value and moisture absorption can also be properties of pure substances, so they do not by themselves establish a mixture.

D: C₆H₁₂

The empirical formula from C:H = 1:2 is CH₂, with mass 14. The vapour amount is 0.032/22.4 mol, giving molar mass 0.12/(0.032/22.4) = 84 g mol⁻¹. Since 84/14 = 6, the molecular formula is C₆H₁₂.

A: 737 mmHg

The collected gas contains hydrogen and water vapour. Dalton’s law gives total pressure = hydrogen pressure + water-vapour pressure. Therefore the hydrogen pressure is 760 − 23 = 737 mmHg.

C: The number of occupied electron shells increases from Li to Na to K

Lithium, sodium and potassium occupy successive periods in Group 1. Their occupied shells increase from two to three to four. The extra shells and shielding make the outer electrons farther from the nucleus, explaining the increase in atomic radius.

C: Y

For one mole of an ideal gas, PV = RT. Therefore PV/RT = 1 at every pressure, so the horizontal curve Y represents ideal-gas behaviour.

D: Y₂X₃

X has 8 electrons and is oxygen, which commonly forms X²⁻. Y has 13 electrons and is aluminium, which commonly forms Y³⁺. Two Y ions give +6 and three X ions give −6, so the neutral formula is Y₂X₃.

B: 55 protons and 78 neutrons

Atomic number counts protons, so the nucleus contains 55 protons. Mass number counts protons plus neutrons: neutrons = 133 − 55 = 78. Electrons occupy regions outside the nucleus.

B: Q

Atomic number 10 is neon. Its electron arrangement is 2,8, with a filled outer shell, so it is a noble gas. The other numbers identify beryllium, magnesium and silicon.

C: Smelting copper sulfide ores

Copper sulfide ores contain sulfur. During roasting and smelting, sulfur is oxidised and sulfur dioxide enters the process gases. Without effective capture, copper smelting is therefore a significant SO₂ emission source.

B: An amphoteric oxide

An amphoteric oxide reacts with both acids and bases. ZnO behaves as a base with acid, forming a zinc salt and water, and reacts with strong alkali to form soluble zincate species. This dual behaviour identifies it as amphoteric.

B: Differ by one proton

A Brønsted–Lowry acid forms its conjugate base by losing one proton, H⁺. For example, NH₄⁺ and NH₃ differ by one proton. Their charges differ by one unit, but they need not have opposite charges.

C: −4

Each cyanide ligand, CN⁻, contributes −1. With six cyanides and Z in oxidation state +2, the total charge is y = +2 + 6(−1) = −4. The anion is therefore [Z(CN)₆]⁴⁻.

D: +64 kJ mol⁻¹

Hess’s law gives ΔH(combustion) = 2ΔHf(CO₂) + 2ΔHf(H₂O) − ΔHf(C₂H₄), since O₂ has zero standard enthalpy of formation. Thus −1428 = 2(−396) + 2(−286) − x, giving x = +64 kJ mol⁻¹ with the supplied data.

D: IV only

There are two moles of gas on each side, so changing pressure does not favour either side in the ideal-gas model. Because the forward reaction is exothermic, higher temperature favours the reverse direction. Adding excess steam shifts equilibrium toward CO₂ and H₂, so only IV favours hydrogen formation.

D: Y

During melting, added heat changes solid into liquid while the temperature stays constant. Both phases coexist on the first horizontal section, labeled Y. The later horizontal section is the liquid-to-gas change.

B: Cu + 4HNO₃ → Cu(NO₃)₂ + 2H₂O + 2NO₂

Concentrated nitric acid oxidises copper to Cu²⁺ while nitrate is reduced mainly to NO₂. The balanced equation is Cu + 4HNO₃ → Cu(NO₃)₂ + 2H₂O + 2NO₂. Equation C describes the familiar dilute-acid reaction producing NO instead.

C: Vanadium(V) oxide

Vanadium(V) oxide, V₂O₅, catalyses the oxidation of SO₂ to SO₃ in the contact process. The SO₃ is then used to produce sulfuric acid. The catalyst increases the reaction rate without changing the equilibrium position.

D: Coke and ammoniacal liquor

Heating coal without air produces coke as the solid residue. Volatile products include ammonia, which dissolves in condensed water to form ammoniacal liquor. Coal tar and coal gas are other products of this process.

A: An oxidant

Potassium nitrate acts as the oxidising agent. It supports oxidation of the combustible components rather than acting as a solvent or as a catalyst. An oxidant is reduced while causing another substance to be oxidised.

C: Atmospheric carbon dioxide reacts with it and releases chlorine

Carbon dioxide in moist air reacts with bleaching powder, leading to loss of available chlorine. This lowers its bleaching strength during exposure. Atmospheric nitrogen does not displace chlorine in this way.

D: N₂O and water

Under controlled, gentle heating, ammonium nitrate decomposes to nitrous oxide and water: NH₄NO₃ → N₂O + 2H₂O. The nitrogen atoms total two on each side, and hydrogen and oxygen are also conserved.

A: 1,2-Dichloroethane

Chlorine adds across the carbon–carbon double bond of ethene. Each carbon receives one chlorine atom, forming CH₂Cl–CH₂Cl, or 1,2-dichloroethane. The carbon chain stays two atoms long.

C: Sulfur is chemically combined with the rubber

In sulfur vulcanisation, sulfur forms cross-links between rubber polymer chains. These links limit chain slippage and improve elastic recovery and mechanical strength. Joining isoprene monomers is polymerisation, which occurs before this treatment.

B: Neutralisation

Ethanoic acid reacts with sodium hydroxide to form sodium ethanoate and water: CH₃COOH + NaOH → CH₃COONa + H₂O. An acid reacting with a base in this way is neutralisation.

D: A hydrogen bond

The O–H hydrogen of one ethanoic acid molecule can be attracted to an oxygen lone pair on another molecule. This intermolecular attraction is a hydrogen bond. The covalent bonds are within each acid molecule.

D: Propane-1,2,3-triol

Fats and oils are esters of glycerol. Alkaline hydrolysis breaks their ester links to give fatty-acid salts, which are soaps, and glycerol. Glycerol has an OH group on each of its three carbons, so its name is propane-1,2,3-triol.

C: 3-Chloro-2-methylprop-1-ene

The longest chain containing the double bond has three carbons and is numbered to give that bond position 1. The methyl group is on carbon 2 and chlorine is on carbon 3. The name is therefore 3-chloro-2-methylprop-1-ene.

D: Methane

Methane can escape from coal seams and accumulate in mine air. A methane–air mixture can ignite and explode if an ignition source is present. Methane is therefore a major gas-explosion hazard in coal mines.

C: X

X burns with a sooty flame but does not readily decolourise bromine water, behaviour characteristic of an aromatic hydrocarbon such as benzene. Y decolourises bromine water, indicating a reactive multiple bond, while Z has the cleaner flame expected of a saturated hydrocarbon.