WASSCE Chemistry past questions and answers

A carbon atom is classified by how many OTHER CARBON atoms it is directly bonded to (count carbons only — bonds to H, O or other atoms do not count):

• Primary (1°) — bonded to 1 carbon • Secondary (2°) — bonded to 2 carbons • Tertiary (3°) — bonded to 3 carbons • Quaternary (4°) — bonded to 4 carbons

Reading the structure (CH₃–CH(OH)–CH(CH₃)–CH₃): • Atom 1 (CH₃, right end) — bonded to 1 carbon → primary • Atom 2 (the CH–OH carbon) — bonded to atoms 1 and 3 = 2 carbons → SECONDARY ✓ • Atom 3 — bonded to atoms 2, 4 and the CH₃ branch = 3 carbons → tertiary • Atom 4 (CH₃, left end) — bonded to 1 carbon → primary

So the secondary carbon is atom 2. The answer is B. (Note: the –OH on atom 2 does not affect the classification, which counts carbon neighbours only.)

In fractional distillation of crude oil, the fractionating column is hottest at the bottom and coolest at the top. Fractions with the lowest boiling points do not condense easily, so they rise to the top of the column before being collected. Refinery gas (small molecules, C₁–C₄) has the lowest boiling point and is collected at the very top — so the answer is C. The other fractions have higher boiling points and are collected lower down: kerosene and diesel in the middle, and bitumen — with the highest boiling point — left as the residue at the bottom.

POWDERED marble has a much larger TOTAL SURFACE AREA than the same mass of lumps. More surface is exposed to the acid, so there are more collisions per second and the reaction is FASTER. On a volume-of-gas vs time graph this shows as a STEEPER initial gradient that reaches the final volume sooner.

Curve X rises more steeply and reaches the plateau first, so Curve X represents the POWDER. The answer is C.

BOTH curves level off at the SAME final volume, because the same MASS of marble (CaCO₃) is used in each — surface area changes the RATE, not the total amount of gas produced.

Why the others are wrong: (A) Curve Y is the slower one (lumps); (D) the curves are NOT the same — they have different gradients; (B) one curve clearly is the powder, so 'neither' is wrong.

In molten lead(II) bromide the ions are free to move. Lead ions (Pb2+) migrate to the cathode (−) and gain electrons: Pb2+ + 2e⁻ → Pb, forming silvery lead metal. Bromide ions (Br⁻) migrate to the anode (+) and lose electrons: 2Br⁻ → Br₂ + 2e⁻, forming a brown bromine vapour. So lead forms at the cathode and bromine at the anode — option C. There is no water present, so no hydrogen (D) or oxygen (B) can form; A is wrong because it reverses the two electrodes.

At a HIGHER temperature the particles have more kinetic energy, collide more often and with enough energy to react, so the reaction is FASTER. On a volume-of-gas vs time graph this shows as a STEEPER initial gradient that levels off (reaches the final volume) sooner.

Curve P rises more steeply and reaches the plateau first, so Curve P is the reaction at the higher temperature. The answer is D.

Key point: BOTH curves level off at the SAME final volume, because the same mass of marble chips (CaCO₃) is used, producing the same total amount of CO₂ — temperature changes the RATE, not the total amount of gas.

Why the others are wrong: (A) and (B) claim one curve produces 'more gas' — false, the final volumes are equal; (C) says the higher-temperature curve 'starts slower' — the opposite is true, it starts faster.

Across a period the atomic radius DECREASES from left to right. Going Li → F, each step adds one more proton to the nucleus (nuclear charge rises from +3 to +9) AND one more electron — but the new electrons all enter the SAME outer shell (n = 2). They do not add a new shell, so they provide little extra shielding. The growing positive nuclear charge therefore pulls the same-shell electrons in more strongly, shrinking the atom. The answer is B.

Why the others are wrong: (A) neutron number does not control atomic size; (C) the number of shells stays the SAME across a period (it only increases DOWN a group), so it cannot explain the decrease; (D) electron–electron repulsion actually rises slightly as electrons are added, which would tend to expand the atom — the opposite of what is seen.

To read a solubility from the curve: find 60°C on the Temperature axis, go straight UP to the curve, then straight ACROSS to the Solubility axis. At 60°C the curve is at 60 g per 100 g of water, so the answer is B.

(The curve rises from left to right, showing that — like most solids — salt X becomes MORE soluble as the temperature increases.)

BF₃ has 3 bonding pairs and no lone pairs on the central boron atom, giving a trigonal planar shape with 120° bond angles. CH₄ is tetrahedral, H₂O is bent/V-shaped, and NH₃ is trigonal pyramidal.