Chemistry 题库

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全英文对标 HKEAA。Paper 1 (75%, MC + structured) · Paper 2 Elective (25%) · SBA。所有定义须按官方英文表述背诵。 完成客观题自查 + 主观题交 英文答题批改指令；错题录入 错题档案。

Part 1 Structure, Bonding & Calculations（Q1–Q40）

Q1 · Basic · Atomic

The atomic number equals the number of: A. Neutrons B. Protons C. Nucleons D. Electrons only Answer: BMarking: Z = proton number.

Q2 · Basic · Mass number

Mass number = protons + ___. A. Electrons B. Neutrons C. Isotopes D. Orbitals Answer: BMarking: A = p + n.

Q3 · Drill · Isotopes

Isotopes of an element have the same ___ but different ___. A. protons, neutrons B. neutrons, protons C. electrons, protons D. mass, charge Answer: AMarking: same Z, different A (neutrons).

Q4 · Basic · Electron config

Electron arrangement of oxygen (Z=8): A. 2,6 B. 2,8 C. 8 D. 2,4 Answer: AMarking: fill shells 2,6.

Q5 · Drill · Group

Noble gases are in Group ___ and have ___ valence electrons (except He). A. 1, 1 B. 18 (0), 8 C. 17, 7 D. 2, 2 Answer: BMarking: Group 18, full outer shell (8; He 2).

Q6 · Basic · Ionic bonding

Sodium chloride is formed by ___ between Na and Cl. A. Sharing electrons B. Transfer of electrons C. Metallic bonding D. Van der Waals Answer: BMarking: ionic = electron transfer.

Q7 · Drill · Covalent

A covalent bond involves: A. Transfer of electrons B. Sharing a pair of electrons C. Free electrons D. Ionic lattice Answer: BMarking: covalent = shared pair.

Q8 · Basic · Metallic bonding

Metallic bonding consists of positive ions in a sea of ___. A. Protons B. Delocalised electrons C. Neutrons D. Molecules Answer: BMarking: sea of delocalised electrons.

Q9 · Basic · Mole

Number of moles in 18 g of H₂O (Mr=18): A. 0.5 B. 1 C. 2 D. 18 Answer: BMarking: n = m/Mr = 18/18 = 1.

Q10 · Drill · Mole mass

Mass of 2 mol of CO₂ (Mr=44): A. 22 g B. 44 g C. 88 g D. 88 kg Answer: CMarking: m = nM = 2×44 = 88 g.

Q11 · Basic · Avogadro

One mole of any substance contains ___ particles. A. 6.02×10²³ B. 3.01×10²³ C. 12 D. 1 Answer: AMarking: Avogadro constant.

Q12 · Drill · Concentration

Concentration of 0.5 mol dissolved in 2 dm³? A. 0.25 mol/dm³ B. 1 mol/dm³ C. 4 mol/dm³ D. 0.1 mol/dm³ Answer: AMarking: c = n/V = 0.5/2 = 0.25.

Q13 · Basic · Empirical formula

Compound: C=12g, H=2g, O=16g. Emp formula? (C=12,H=1,O=16) A. CH₂O B. CHO C. C₂H₄O₂ D. COH₂ Answer: AMarking: mol C=1,H=2,O=1 → CH₂O.

Q14 · Drill · Molar volume

At room temp & pressure (r.t.p.), molar volume ≈ ___ dm³/mol. A. 22.4 B. 24 C. 12 D. 240 Answer: BMarking: rtp ≈ 24 dm³/mol (s.t.p. = 22.4).

Q15 · Basic · Gas volume

Volume of 0.5 mol gas at r.t.p.? A. 12 dm³ B. 24 dm³ C. 11.2 dm³ D. 48 dm³ Answer: AMarking: 0.5×24 = 12 dm³.

Q16 · Drill · Percentage yield

Theoretical yield 10 g, actual 8 g. % yield? A. 80% B. 125% C. 20% D. 8% Answer: AMarking: actual/theoretical ×100 = 80%.

Q17 · Basic · Separation

To separate a solid from a liquid: use ___. A. Distillation B. Filtration C. Chromatography D. Evaporation only Answer: BMarking: filtration separates insoluble solid.

Q18 · Drill · Distillation

Distillation separates liquids by differences in ___. A. Colour B. Boiling point C. Density D. Mass Answer: BMarking: boiling point difference.

Q19 · Basic · Periodic trend

Across a period left→right, atomic radius generally ___. A. Increases B. Decreases C. Stays same D. Doubles Answer: BMarking: increased nuclear charge pulls in.

Q20 · Drill · Reactivity

Group 1 metal reactivity ___ down the group. A. Decreases B. Increases C. Constant D. Disappears Answer: BMarking: easier electron loss down group.

Q21 · Basic · Valency

Chlorine (Group 17) has valency ___. A. 1 B. 2 C. 3 D. 7 Answer: AMarking: needs 1 electron → valency 1.

Q22 · Drill · Formula writing

Formula of calcium chloride (Ca²⁺, Cl⁻): A. CaCl B. CaCl₂ C. Ca₂Cl D. Ca₂Cl₂ Answer: BMarking: balance charges: CaCl₂.

Q23 · Basic · Ionic compound property

Ionic compounds conduct electricity when ___. A. Solid B. Molten or in solution C. Cold D. Dry Answer: BMarking: ions mobile only when molten/dissolved.

Q24 · Drill · Covalent property

Simple molecular substances have ___ melting points because of ___. A. high, strong covalent bonds B. low, weak intermolecular forces C. high, ionic lattice D. low, metallic bonds Answer: BMarking: weak IM forces between molecules.

Q25 · Basic · Giant covalent

Diamond is hard because of ___. A. Weak forces B. A giant covalent network C. Ionic bonds D. Delocalised electrons Answer: BMarking: 3D covalent network.

Q26 · Drill · Alloy

An alloy is a mixture of metals (often with non-metal) that is usually ___ than pure metal. A. Softer B. Harder C. Lighter D. More reactive Answer: BMarking: disrupted layers → harder.

Q27 · Basic · Oxidation number

In H₂O, oxidation number of O is: A. −2 B. −1 C. +2 D. 0 Answer: AMarking: O usually −2.

Q28 · Drill · Oxidation number

In NaCl, Na is ___ (oxidation state). A. 0 B. +1 C. −1 D. +2 Answer: BMarking: Group 1 = +1.

Q29 · Basic · Acids

An acid releases ___ ions in aqueous solution. A. OH⁻ B. H⁺ (H₃O⁺) C. Na⁺ D. Cl⁻ Answer: BMarking: Arrhenius: H⁺ donor.

Q30 · Drill · Base

A base ___ H⁺ ions or releases ___ ions. A. donates, H⁺ B. accepts, OH⁻ C. releases, H⁺ D. ignores, OH⁻ Answer: BMarking: base accepts H⁺ / gives OH⁻.

Q31 · Basic · Neutralisation

Acid + base → salt + ___. A. CO₂ B. Water C. H₂ D. O₂ Answer: BMarking: H⁺ + OH⁻ → H₂O.

Q32 · Drill · Salt naming

HCl + NaOH → salt is ___. A. Sodium chloride B. Sodium hydroxide C. Hydrochloric acid D. Water Answer: AMarking: NaCl.

Q33 · Basic · pH

A solution of pH 3 is ___ and pH 13 is ___. A. alkaline, acidic B. acidic, alkaline C. neutral, acidic D. acidic, neutral Answer: BMarking: pH<7 acid, >7 alkali.

Q34 · Drill · Indicators

Litmus turns red in ___ and blue in ___. A. alkali, acid B. acid, alkali C. neutral, acid D. base, base Answer: BMarking: red in acid, blue in alkali.

Q35 · Basic · Metal reactions

A metal above hydrogen in the reactivity series ___ displace H from acid. A. can B. cannot C. never D. only if noble Answer: AMarking: more reactive than H → H₂ evolved.

Q36 · Drill · Reactivity series

Which reacts with cold water violently? A. Copper B. Potassium C. Iron D. Silver Answer: BMarking: K, Na reactive with water.

Q37 · Basic · Redox

Oxidation is ___ of electrons; reduction is ___ of electrons. A. gain, loss B. loss, gain C. sharing, transfer D. gain, gain Answer: BMarking: OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Q38 · Drill · Redox identification

In 2Mg + O₂ → 2MgO, Mg is ___. A. Reduced B. Oxidised C. Unchanged D. Catalyst Answer: BMarking: Mg loses e⁻ → oxidised.

Q39 · Basic · Electrolysis

During electrolysis, cations move to the ___ electrode. A. Negative (cathode) B. Positive (anode) C. Neutral D. Earth Answer: AMarking: positive ions → negative cathode.

Q40 · Drill · Electrolysis products

Electrolysis of brine (NaCl aq) gives H₂ at cathode, Cl₂ at anode, and ___ in solution. A. NaOH B. HCl C. NaCl D. O₂ Answer: AMarking: H₂ + Cl₂ + NaOH (chlor-alkali).

Part 2 Energetics, Rates, Equilibrium & Tests（Q41–Q75）

Q41 · Basic · Exothermic

An exothermic reaction ___ heat (ΔH ___ 0). A. absorbs, > B. releases, < C. releases, > D. absorbs, < Answer: BMarking: exo = releases, ΔH negative.

Q42 · Drill · Endothermic

An endothermic reaction has ΔH ___ 0 and feels ___. A. <, warm B. >, cold C. <, cold D. =, same Answer: BMarking: endo absorbs, ΔH>0, feels cold.

Q43 · Basic · Activation energy

Activation energy is the ___ energy needed to start a reaction. A. Minimum B. Maximum C. Average D. Zero Answer: AMarking: minimum E to react.

Q44 · Drill · Catalyst

A catalyst ___ the rate by providing an alternative pathway with ___ activation energy. A. decreases, higher B. increases, lower C. stops, zero D. increases, higher Answer: BMarking: lowers Ea → faster.

Q45 · Basic · Rate factor

Increasing temperature generally ___ the reaction rate. A. Decreases B. Increases C. Stops D. No effect Answer: BMarking: more collisions, more energetic.

Q46 · Drill · Surface area

Smaller particles (greater surface area) ___ the rate. A. decrease B. increase C. stop D. no effect Answer: BMarking: more exposed area.

Q47 · Basic · Equilibrium

At dynamic equilibrium, forward and reverse rates are ___, concentrations ___. A. equal, constant B. zero, changing C. equal, zero D. different, constant Answer: AMarking: rates equal, [ ] constant.

Q48 · Drill · Le Chatelier

For N₂ + 3H₂ ⇌ 2NH₃ (+heat), increasing pressure shifts equilibrium ___ (fewer moles side). A. left B. right C. no shift D. stops Answer: BMarking: 4 mol → 2 mol, high P favours right.

Q49 · Drill · Le Chatelier temperature

For an exothermic forward reaction, raising temperature shifts equilibrium ___. A. right B. left (favours endothermic reverse) C. no shift D. stops Answer: BMarking: heat = reactant for exo; add heat → shift left.

Q50 · Basic · Reversible

In a closed system, a reversible reaction reaches ___. A. Completion B. Equilibrium C. Explosion D. No change Answer: BMarking: equilibrium in closed system.

Q51 · Basic · Gas test

A gas that relights a glowing splint is ___. A. CO₂ B. O₂ C. H₂ D. NH₃ Answer: BMarking: oxygen relights splint.

Q52 · Drill · Gas test

A gas that "pops" with a lit splint is ___. A. O₂ B. H₂ C. CO₂ D. Cl₂ Answer: BMarking: hydrogen pops.

Q53 · Drill · Gas test

A gas that turns limewater milky is ___. A. O₂ B. CO₂ C. H₂ D. N₂ Answer: BMarking: CO₂ + Ca(OH)₂ → CaCO₃ (milky).

Q54 · Basic · Flame test

A crimson (red) flame indicates ___ ion. A. Sodium B. Lithium C. Potassium D. Copper Answer: BMarking: Li = crimson; Na = yellow; K = lilac; Cu = green.

Q55 · Drill · Flame test

A yellow flame indicates ___ ion. A. Potassium B. Sodium C. Calcium D. Copper Answer: BMarking: Na = yellow (persistent).

Q56 · Basic · Anion test

To test for carbonate, add acid and check for ___ gas. A. O₂ B. CO₂ (limewater turns milky) C. H₂ D. NH₃ Answer: BMarking: CO₂ confirms carbonate.

Q57 · Drill · Cation test

Adding NaOH to a solution giving a blue precipitate indicates ___ ion. A. Fe²⁺ B. Cu²⁺ C. Ca²⁺ D. Mg²⁺ Answer: BMarking: Cu(OH)₂ = blue.

Q58 · Basic · Precipitate colour

Fe³⁺ + NaOH → ___ precipitate. A. Green B. Brown C. Blue D. White Answer: BMarking: Fe(OH)₃ brown.

Q59 · Drill · Ammonium test

Add NaOH and warm; gas turns damp red litmus blue → ___ ion present. A. Carbonate B. Ammonium C. Sulphate D. Chloride Answer: BMarking: NH₃ from NH₄⁺ + OH⁻.

Q60 · Basic · Sulphate test

Add BaCl₂ (after HCl) → white precipitate of ___ confirms sulphate. A. BaSO₄ B. BaCO₃ C. BaCl₂ D. NaCl Answer: AMarking: BaSO₄ insoluble white.

Q61 · Basic · Alkane

General formula of alkanes is ___. A. CₙH₂ₙ B. CₙH₂ₙ₊₂ C. CₙH₂ₙ₋₂ D. CₙHₙ Answer: BMarking: saturated CₙH₂ₙ₊₂.

Q62 · Drill · Alkene

Alkenes contain a ___ bond (general CₙH₂ₙ). A. Single B. Double (C=C) C. Triple D. Ionic Answer: BMarking: unsaturated, C=C.

Q63 · Basic · Saturated/Unsaturated

Bromine water is decolourised by ___ compounds (e.g. alkenes). A. Saturated B. Unsaturated C. Alkane D. Inert Answer: BMarking: C=C adds Br₂ → decolourise.

Q64 · Drill · Combustion

Complete combustion of a hydrocarbon produces ___ and ___. A. CO, H₂O B. CO₂, H₂O C. C, H₂ D. Soot, O₂ Answer: BMarking: plenty O₂ → CO₂ + H₂O.

Q65 · Basic · Alcohol

Ethanol formula is ___. A. CH₃OH B. C₂H₅OH C. CH₃COOH D. C₂H₆ Answer: BMarking: C₂H₅OH.

Q66 · Drill · Carboxylic acid

Vinegar contains ___ acid (CH₃COOH). A. Ethanoic B. Citric C. Sulphuric D. Hydrochloric Answer: AMarking: ethanoic acid in vinegar.

Q67 · Basic · Esterification

Alcohol + carboxylic acid (with H₂SO₄ catalyst) → ___ + water. A. Ester B. Salt C. Alkene D. Amine Answer: AMarking: ester + H₂O.

Q68 · Drill · Ester

Esters typically have a ___ smell. A. Fruity B. Rotten C. None D. Bitter Answer: AMarking: esters = fruity odour.

Q69 · Basic · Addition polymer

Ethene polymerises to form ___. A. Poly(ethene) B. Poly(propene) C. PVC D. Nylon Answer: AMarking: (C₂H₄)ₙ polythene.

Q70 · Drill · Polymer

In addition polymerisation, monomers join without forming ___. A. A by-product B. A chain C. Double bonds remain D. Heat Answer: AMarking: no small molecule lost (vs condensation).

Q71 · Basic · Hydrocarbon pollution

Incomplete combustion produces toxic ___. A. CO₂ B. CO C. H₂O D. O₂ Answer: BMarking: CO toxic from incomplete.

Q72 · Drill · Crude oil

Crude oil is separated by ___ into fractions. A. Filtration B. Fractional distillation C. Evaporation D. Chromatography Answer: BMarking: fractions by boiling point.

Q73 · Basic · Air composition

The most abundant gas in dry air is ___. A. Oxygen B. Nitrogen C. CO₂ D. Argon Answer: BMarking: ~78% N₂.

Q74 · Drill · Rusting

Rusting of iron requires ___ and ___. A. Oxygen, water B. Acid only C. Heat only D. Salt only Answer: AMarking: Fe + O₂ + H₂O → hydrated Fe₂O₃.

Q75 · Basic · Sacrificial protection

Zinc coating protects iron by ___ (sacrificial). A. Painting B. Being more reactive (oxidises first) C. Insulating D. Cooling Answer: BMarking: Zn more reactive → sacrifices itself.

Part 3 Structured & Applied（Q76–Q110）

Q76 · Structured · Mole calculation

Calculate the mass of 0.25 mol of H₂SO₄ (Mr=98). Answer: m = nM = 0.25×98 = 24.5 g. Marking: formula (1); compute (1).

Q77 · Structured · Empirical → molecular

A compound has empirical formula CH₂O and Mr=60. Find molecular formula. Answer: EF mass = 30; 60/30=2 → molecular = C₂H₄O₂. Marking: ratio (1); multiply (1).

Q78 · Structured · Gas volume

What volume (r.t.p.) does 4 g of O₂ (Mr=32) occupy? Answer: n = 4/32 = 0.125 mol; V = 0.125×24 = 3 dm³. Marking: moles (1); volume (1).

Q79 · Structured · Concentration dilution

250 cm³ of 1.0 mol/dm³ HCl is diluted to 1.0 dm³. New concentration? Answer: n = 1.0×0.25 = 0.25 mol; c = 0.25/1.0 = 0.25 mol/dm³. Marking: moles conserved (1); new c (1).

Q80 · Structured · Neutralisation stoichiometry

25.0 cm³ of 0.100 mol/dm³ NaOH neutralises 20.0 cm³ HCl. Find HCl concentration (1:1). Answer: n(NaOH)=0.100×0.025=0.0025 mol → n(HCl)=0.0025; c=0.0025/0.020=0.125 mol/dm³. Marking: mol NaOH (1); mol HCl (1); conc (1).

Q81 · Structured · Redox half-equations

Write half-equations for Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu. Overall? Answer: Overall: Zn + Cu²⁺ → Zn²⁺ + Cu. Marking: two halves (1); combined (1).

Q82 · Structured · Electrolysis

During electrolysis of CuSO₄(aq) with inert electrodes: (a) product at cathode, (b) at anode. Answer: (a) Cu deposited at cathode; (b) O₂ evolved at anode (OH⁻ oxidised). Marking: Cu²⁺ reduced (1); O₂ from water (1).

Q83 · Structured · Energy profile

Draw (describe) an energy profile for an exothermic reaction. Answer: Products lower than reactants; activation energy hump; ΔH = E(reactants)−E(products) < 0. Marking: relative levels (1); Ea label (1); ΔH negative (1).

Q84 · Structured · Le Chatelier applied

2SO₂ + O₂ ⇌ 2SO₃ (+heat). State effect of (i) raising temp, (ii) adding catalyst. Answer: (i) shift left (reverse endothermic favoured); (ii) no shift, faster equilibrium. Marking: temp (1); catalyst (1).

Q85 · Structured · Flame/emission

A sample gives a lilac flame. Identify the ion and one confirmation test. Answer: Potassium (K⁺); confirm with flame test (lilac) or Na-free; or flame spectroscope. Marking: K⁺ (1); method (1).

Q86 · Structured · Ion test sequence

A white solid may contain carbonate, sulphate, chloride. State the test order and observations. Answer: (1) Add dilute HNO₃ + observe CO₂ (effervescence, limewater milky) → carbonate; (2) add Ba(NO₃)₂ → white ppt BaSO₄ → sulphate; (3) add AgNO₃ → white ppt AgCl → chloride. Marking: 3 tests in correct order (3).

Q87 · Structured · Organic naming

Name CH₃CH₂CH₂OH and state its homologous series. Answer: Propan-1-ol; alcohol. Marking: name (1); family (1).

Q88 · Structured · Esterification equation

Write the equation: ethanol + ethanoic acid → ? Answer: CH₃CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₃ + H₂O (ethyl ethanoate + water; H₂SO₄ catalyst). Marking: ester (1); water (1); condenses (1).

Q89 · Structured · Polymerisation

Draw (state) the repeat unit of poly(ethene) from monomer CH₂=CH₂. Answer: −[CH₂−CH₂]− (n); addition polymer. Marking: repeat unit (1); type (1).

Q90 · Structured · Rates graph

A concentration-time graph for reactant levels off → what does it mean? Answer: Reaction reached equilibrium (or completed); rate → 0. Marking: equilibrium/completion (1-2).

Q91 · Structured · Reactivity series use

Why can iron be extracted from its ore by reduction with carbon but not aluminium? Answer: Fe is less reactive than C (below C in series) → C reduces Fe₂O₃; Al is more reactive than C → carbon cannot reduce Al ore (electrolysis needed). Marking: Fe reduced by C (1); Al more reactive (1); electrolysis (1).

Q92 · Structured · Neutralisation pH

25 cm³ HCl (pH 1) mixed with 25 cm³ NaOH (pH 13). Final pH? Explain. Answer: Equal moles H⁺ and OH⁻ (both 0.0025 mol if 0.1 M) → neutral, pH ≈ 7. Marking: equal neutralisation (2).

Q93 · Structured · Periodic trend explanation

Why does metallic character increase down Group 1? Answer: Atomic radius increases → outer electron further from nucleus, more shielded → easier to lose e⁻ → more metallic/reactive. Marking: radius + shielding + ease of loss (2-3).

Q94 · Structured · Bonding & properties

Explain why NaCl has high melting point but CO₂(s) (dry ice) sublimes easily. Answer: NaCl = giant ionic lattice, strong electrostatic forces → high mp. CO₂ = simple molecular, weak van der Waals between molecules → sublimes at low T. Marking: ionic lattice (1); molecular forces (1); contrast (1).

Q95 · Structured · Redox in combustion

Methane burns: CH₄ + 2O₂ → CO₂ + 2H₂O. Identify oxidation/reduction. Answer: C in CH₄ oxidised (C goes −4 → +4); O₂ reduced (0 → −2). Redox overall. Marking: C oxidised (1); O reduced (1).

Q96 · Structured · Acid-base titration

Describe how to titrate HCl with NaOH using phenolphthalein. Answer: Rinse burette with NaOH; pipette 25 cm³ HCl + indicator into conical flask; add NaOH until permanent pink endpoint; repeat for concordant titres; calculate. Marking: burette/pipette (1); indicator endpoint (1); calculate (1).

Q97 · Structured · Molar mass from gas

A gas has density 1.96 g/dm³ at s.t.p. (22.4 dm³/mol). Find Mr. Answer: Mr = density × molar volume = 1.96 × 22.4 = 43.9 ≈ 44 (CO₂). Marking: formula (1); compute (1).

Q98 · Structured · Equilibrium constant concept

For aA + bB ⇌ cC, Kc = [C]^c / ([A]^a[B]^b). If Kc large, equilibrium lies to the ___. Answer: right (products favoured). Marking: large K → products.

Q99 · Structured · Haber process

Give two conditions used in the Haber process and why. Answer: High pressure (favours fewer moles, more NH₃) + iron catalyst (faster) + moderate temp (compromise: exothermic so not too high). Marking: 2 conditions + reasons (2-3).

Q100 · Structured · Pollution

State one environmental problem from burning fossil fuels and one solution. Answer: CO₂ → global warming; solution: renewable energy / carbon capture / efficiency. Marking: problem (1); solution (1).

Q101 · Data Response · Fuel

A fuel releases 50 kJ/g. How much needed to heat 200 g water by 20°C? (c_water=4.2 J/g°C) Answer: Q = mcΔT = 200×4.2×20 = 16,800 J = 16.8 kJ. Mass fuel = 16.8/50 = 0.336 g (ideal, ignoring loss). Marking: heat needed (1); divide by energy density (1).

Q102 · Structured · Transition metals

Give one characteristic property of transition metals. Answer: Form coloured ions / multiple oxidation states / good catalysts / high mp & density. Marking: any one valid (1-2).

Q103 · Structured · Test for water

How to test whether a liquid is pure water? Answer: Anhydrous CuSO₄ turns blue; or measure boiling point 100°C / freezing 0°C. Marking: CuSO₄ white→blue (1); or bp (1).

Q104 · Structured · Redox titration concept

In redox titration, the equivalence point is when ___. Answer: exactly stoichiometric amounts of oxidising and reducing agents have reacted. Marking: stoichiometric equivalence (1-2).

Q105 · Structured · Halogens

Describe the trend in reactivity of Group 17 (halogens) down the group. Answer: Reactivity decreases down group (F most reactive); harder to gain an electron as atomic radius increases & shielding rises. Marking: decrease (1); reason (1-2).

Q106 · Structured · Displacement

Chlorine water added to KBr solution: what happens? (Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂) Answer: Br₂ formed (solution turns orange/brown); Cl₂ displaces Br⁻ (more reactive halogen displaces less reactive). Marking: displacement (1); observation (1).

Q107 · Structured · Empirical from combustion

4.4 g CO₂ and 1.8 g H₂O from combustion of hydrocarbon. Find C:H ratio. Answer: n(C)=4.4/44=0.1; n(H)=2×1.8/18=0.2 → C:H = 1:2 → empirical CH₂. Marking: C moles (1); H moles (1); ratio (1).

Q108 · Structured · Acid strength

HCl is a strong acid, CH₃COOH is weak. What differs? Answer: Strong acid fully ionises in water; weak acid only partially ionises (equilibrium favouring unionised). Marking: full vs partial ionisation (2).

Q109 · Structured · Electrochemical cell

In a simple cell Zn|Zn²⁺ || Cu²⁺|Cu, which is the negative terminal and why? Answer: Zn electrode is negative (anode); Zn more reactive, oxidises, releases e⁻ → supplies electrons. Marking: Zn negative (1); reason (1-2).

Q110 · Essay-style · Industrial importance

Evaluate the use of catalysts in the chemical industry (2 advantages, 1 limitation). Answer: Advantages: (1) lower Ea → less energy/cost & faster production; (2) allow lower T/P → safer, selective products. Limitation: catalysts can be poisoned/deactivated, need replacement; may favour side reactions if not selective. Marking: 2 advantages (2); 1 limitation (1); evaluation (1).

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有机/测试题是 DSE 高频：重点练 Q61–Q70、Q51–Q60。主观题交 英文答题批改指令 按评分标准阅卷。