Chemistry (9701) Past Papers & Revision Resources

Fix

Check each C=C, and any ring with two substituted carbons, for cis/trans.

Check each carbon for four different groups.

ClCH=CHCHClCH₃: 2 cis/trans forms × 2 optical isomers = 4 stereoisomers, not 2.

Fix

Only temperature changes Kc or Kp, even when heating raises pressure.

2 mol butane, 3 mol methylpropane at 90 kPa: p(butane) = 2 ÷ 5 × 90 = 36 kPa, not 2 ÷ 3 × 90.

Fix

Acidified K₂Cr₂O₇ oxidises every primary and secondary OH group, never a tertiary one.

Hexane-2,5-diol gives hexane-2,5-dione: both OH groups are oxidised.

3-ethylpentan-3-ol does not react: the dichromate stays orange.

Fix

Look for a CH₃CO- ketone or aldehyde, or a CH₃CH(OH)- alcohol.

Hexan-2-one and hexan-2-ol give the yellow precipitate.

Cyclohexanone, a ketone, and hexan-1-ol, an alcohol, do not.

Fix

Multiply each ΔHf or ΔHc by its coefficient.

2CH₃CHO(l) (ΔHf −192) + O₂(g) (ΔHf 0) → 2CH₃COOH(l) (ΔHf −484), in kJ mol⁻¹.

ΔH = 2(−484) − 2(−192) = −584 kJ mol⁻¹, not −292.

Fix

Start arrows at a lone pair or bond.

Hexan-3-one + HCN: lone pair on C of CN⁻ → C=O carbon.

C=O π bond → O.

O⁻ lone pair → H⁺, not H⁺ → O⁻.

Fix

Compare the carbocations, not the products.

But-1-ene + HBr: CH₃CH₂CH⁺CH₃ has two alkyl groups on C⁺, CH₃CH₂CH₂CH₂⁺ only one.

The greater inductive effect stabilises the secondary cation, so 2-bromobutane is major.

Fix

Single bond: one σ.

Double: one σ + one π.

Triple: one σ + two π.

But-2-yne, CH₃C≡CCH₃: 6 C-H + 2 C-C + 1 in C≡C = 9 σ and 2 π.

A π bond: sideways overlap of p orbitals, not head-on.

Fix

K(g) → K⁺(g) + e⁻: (g) on both species.

K's first IE is lower than Na's.

Its 4s electron, further out and more shielded, is less attracted despite a larger nuclear charge.

Fix

Boiling simple molecular CCl₄ overcomes id-id forces between molecules; no C-Cl bond breaks.

Melting KBr breaks ionic bonds; melting diamond breaks covalent bonds.

Neither has intermolecular forces to overcome.

Fix

Write burette readings to two decimal places, ending in 0 or 5.

Initial 2.15, final 21.50 (not 21.5), titre 19.35 cm³.

Rough titration: write both readings, not just the titre.

Fix

Average only titres within 0.10 cm³ of each other, never the rough.

Accurate titres 16.70, 16.35, 16.35 cm³: mean 16.35 cm³, not 16.47 cm³.

Tick the titres you used.

Fix

Describe exactly what you see.

Lithium carbonate in dilute HCl: 'effervescence, colourless solution', not 'gas given off, clear'.

Record 'gas gives a white ppt. with limewater', not 'goes milky'.

Fix

After adding NaOH(aq), warm the mixture and hold damp red litmus in the gas at the tube's mouth.

Record 'gas turns damp red litmus blue', not 'litmus turns blue'.

Fix

m = 1.00 g per cm³ of all the liquid.

1.73 g solid + 45.0 cm³ acid: m = 45.0 g, not 1.73 g.

35.0 cm³ acid + 27.5 cm³ alkali: m = 62.5 g.

Fix

${E^\ominus}$ = +0.34 V exceeds ${E^\ominus}$ = −0.13 V, so Cu²⁺ is reduced.

Pb + Cu²⁺ → Pb²⁺ + Cu.

${E^\ominus_\mathrm{cell}}$ = 0.34 − (−0.13) = +0.47 V.

The reverse, Cu + Pb²⁺, gives −0.47 V: not feasible.

Fix

In cold HCl(aq), H₂N(CH₂)₄CONH₂ becomes H₃N⁺(CH₂)₄CONH₂.

The amine N lone pair forms a dative bond to H⁺.

The amide N lone pair is delocalised into C=O and far less available.

Fix

One repeat unit holds one residue of each monomer.

Pentanedioic acid with propane-1,3-diol gives ‑CO(CH₂)₃CO-O(CH₂)₃O‑.

Draw a continuation bond at each end; 2H₂O are lost per unit.

Fix

Bromination of benzene: arrow 1 runs from inside the ring to Br⁺.

In C₆H₆Br⁺ the horseshoe spans five carbons other than CHBr, + inside.

Arrow 2 starts on the C‑H bond.

Fix

A transition element is a d-block element forming one or more stable ions with incomplete d orbitals.

Scandium (atom [Ar]3d¹4s²) is excluded: Sc³⁺ is [Ar].

Mn³⁺ is [Ar]3d⁴, not [Ar]3d²4s².

Fix

Take [Zn(H₂O)₆]²⁺ + 4NH₃ ⇌ [Zn(NH₃)₄]²⁺ + 6H₂O.

${K_\mathrm{stab} }={ \dfrac{[[\mathrm{Zn(NH_3)_4}]^{2+}]}{[[\mathrm{Zn(H_2O)_6}]^{2+}][\mathrm{NH_3}]^4}}$ with no [H₂O].

Coordination number: [Cr(en)₂(H₂O)₂]³⁺ has four ligands but six dative bonds, so 6.

Fix

MgO's lattice energy is more exothermic than Na₂O's.

Mg²⁺ is smaller and doubly charged, so it attracts O²⁻ more strongly.

For MgBr₂, count both Br: 2 × (−325) = −650 kJ mol⁻¹.

Fix

Convert ΔS to kJ K⁻¹ mol⁻¹ and T to kelvin.

Example: ΔH = −120 kJ mol⁻¹, ΔS = −150 J K⁻¹ mol⁻¹, 25 °C.

ΔG = −120 − 298 × (−0.150) = −75.3 kJ mol⁻¹.

Fix

The dependent variable is the reading you take.

Flask on a balance losing gas: 'mass of flask and contents', not 'rate of reaction'.

Rate comes later, from the mass-time graph.

Fix

250 cm³ beaker, 1000 cm³ flask: dissolve in about 150 cm³ distilled water, not 1000 cm³.

Transfer to the flask; rinse the beaker into it.

Fill to the mark; invert.

Fix

Two readings: show 2 × half the smallest division.

Titre 16.80 cm³: (2 × 0.05) ÷ 16.80 × 100 = 0.595%, not 0.298%.

Mass by difference on a 0.01 g balance: 2 × 0.005 g.

Fix

Read two points off your line, at least half its length apart.

Axis marked × 10⁻³: quote (1.2 × 10⁻³, 0.50) and (4.6 × 10⁻³, 1.85).

Gradient = 1.35 ÷ (3.4 × 10⁻³) = 397, not 0.397.

Fix

Leave the anomaly out of your best-fit line.

Give a cause that fits its direction.

Titre above the line: 'water left in the burette diluted the titrant', not 'wrong concentration'.