2025 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

First, write the balanced chemical equation: \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). Calculate the moles of sulfuric acid: \(\text{moles} = \text{volume in dm}^3 \times \text{concentration} = 0.0200\text{ dm}^3 \times 0.150\text{ mol/dm}^3 = 0.00300\text{ mol}\). From the stoichiometry of the equation, 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{NaOH}\). Therefore, the moles of \(\text{NaOH}\) required = \(2 \times 0.00300\text{ mol} = 0.00600\text{ mol}\). Calculate the volume of \(\text{NaOH}\) needed: \(\text{volume} = \text{moles} / \text{concentration} = 0.00600\text{ mol} / 0.100\text{ mol/dm}^3 = 0.0600\text{ dm}^3 = 60.0\text{ cm}^3\).

1 mark for the correct volume of 60.0 cm3.

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\) (whereas chromium(III) also forms a green precipitate but it dissolves in excess sodium hydroxide to form a green solution). A white precipitate with dilute nitric acid and aqueous barium nitrate indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the solid is iron(II) sulfate, \(\text{FeSO}_4\).

1 mark for identifying the compound as iron(II) sulfate.

The reaction between calcium carbonate and hydrochloric acid produces calcium chloride (which is highly soluble), water, and carbon dioxide gas: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). Since no precipitate is formed, the solution remains clear and there is no formation of a solid to obscure a cross. Thus, measuring the time taken for a cross to become obscured is not a suitable method. Methods A, B, and D are all standard ways to measure rate by monitoring mass loss or gas volume over time.

1 mark for choosing the incorrect/unsuitable method.

A reducing agent undergoes oxidation (its oxidation state increases or it loses hydrogen / gains oxygen). In reaction D, hydrogen sulfide (\(\text{H}_2\text{S}\)) reacts with chlorine to form hydrogen chloride and sulfur. The sulfur in \(\text{H}_2\text{S}\) has an oxidation state of -2 and is oxidized to 0 in elemental sulfur (\(\text{S}\)), while also losing hydrogen. Therefore, \(\text{H}_2\text{S}\) acts as a reducing agent. In the other reactions, the underlined substances undergo reduction and act as oxidizing agents.

1 mark for identifying the reaction where H2S is the reducing agent.

Calcium oxide (\(\text{CaO}\)) is a basic metal oxide, and silicon(IV) oxide (\(\text{SiO}_2\)) is an acidic non-metal oxide. Their reaction to form calcium silicate (\(\text{CaSiO}_3\), also known as slag) is an acid-base neutralization reaction that removes acidic impurities from the furnace.

1 mark for identifying the formation of slag as a neutralization reaction.

The reaction of methane with chlorine in the presence of UV light is a photochemical substitution reaction. One hydrogen atom in methane is replaced by a chlorine atom, producing chloromethane (\(\text{CH}_3\text{Cl}\)) and hydrogen chloride (\(\text{HCl}\)) gas. If excess chlorine is present, further substitution can occur to form dichloromethane, trichloromethane, and tetrachloromethane.

1 mark for the correct statement regarding the substitution reaction and products.

Sulfur dioxide is produced by the combustion of fossil fuels (such as coal) that contain sulfur impurities. Once released into the atmosphere, it reacts with water and oxygen to form sulfuric acid, which causes acid rain. Carbon monoxide is toxic but does not cause acid rain. Nitrogen dioxide forms in car engines but is not responsible for destroying the ozone layer (CFCs are). Methane is a greenhouse gas but is not a major component of car exhausts (incomplete combustion of petrol produces carbon monoxide and unburnt hydrocarbons, but methane's main sources are agriculture and landfills).

1 mark for identifying the correct row for sulfur dioxide, its source, and its effect.

Calculate the mass of each part: 1) Ammonium ions: \(2 \times [14 + (4 \times 1)] = 2 \times 18 = 36\). 2) Iron: \(56\). 3) Sulfate ions: \(2 \times [32 + (4 \times 16)] = 2 \times 96 = 192\). 4) Water of crystallization: \(6 \times [(2 \times 1) + 16] = 6 \times 18 = 108\). Sum them together: \(36 + 56 + 192 + 108 = 392\).

1 mark for the correct calculation of relative formula mass.

The balanced chemical equation is: \(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). First, calculate the moles of NaOH: \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.150\text{ mol/dm}^3 = 0.00375\text{ mol}\). From the stoichiometry, the moles of \(\text{H}_2\text{SO}_4\) needed is \(0.00375 \div 2 = 0.001875\text{ mol}\). Finally, calculate the volume of sulfuric acid required: \(V = n \div C = 0.001875\text{ mol} \div 0.100\text{ mol/dm}^3 = 0.01875\text{ dm}^3 = 18.75\text{ cm}^3\).

1 mark for the correct option B. Award 1 mark for correct calculation showing 18.75 cm3.

The green precipitate that is insoluble in excess aqueous sodium hydroxide indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). Chromium(III) ions also form a green precipitate with sodium hydroxide, but it dissolves in excess to form a green solution. The evolution of a gas that turns damp red litmus paper blue (ammonia gas) when heated with aqueous sodium hydroxide confirms the presence of ammonium ions, \(\text{NH}_4^+\). Thus, the cations present are \(\text{Fe}^{2+}\) and \(\text{NH}_4^+\).

1 mark for the correct option B. Award 1 mark for identifying both Fe2+ and NH4+ based on the qualitative analysis tests.

A reducing agent reduces another substance and is itself oxidised. In the reaction \(\text{Fe}_2\text{O}_3 + 3\underline{\text{CO}} \rightarrow 2\text{Fe} + 3\text{CO}_2\), carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)), meaning it is oxidised. It reduces the iron(III) oxide to iron metal. Therefore, \(\text{CO}\) is the reducing agent.

1 mark for the correct option C. Award 1 mark for identifying the substance that undergoes oxidation as the reducing agent.

Copper(II) sulfate is a soluble salt prepared from an insoluble base (copper(II) oxide) and an acid (sulfuric acid). Copper metal does not react with dilute sulfuric acid, making option A incorrect. Copper(II) sulfate is highly soluble, so it cannot be prepared by precipitation (options B and D are incorrect because no precipitate of copper(II) sulfate forms). Adding excess copper(II) oxide ensures all the acid is neutralised, filtration removes the excess solid, and crystallisation yields pure copper(II) sulfate crystals.

1 mark for the correct option C. Award 1 mark for selecting the correct method of preparing a soluble salt from an insoluble base.

The extraction of zinc involves first roasting the zinc blende (ZnS) in air to convert it to zinc oxide (ZnO) and sulfur dioxide (\(\text{SO}_2\)). The zinc oxide is then reduced in a furnace using carbon (or carbon monoxide). Because zinc has a relatively low boiling point (907 °C), it forms a vapour inside the furnace, which is distilled and condensed, rather than being tapped off as a liquid from the bottom like iron.

1 mark for the correct option B. Award 1 mark for identifying the correct chemical steps involved in the pyrometallurgical extraction of zinc.

The general formula \(\text{C}_n\text{H}_{2n}\text{O}_2\) represents compounds with one double bond equivalent and two oxygen atoms. Both carboxylic acids (such as butanoic acid, \(\text{C}_3\text{H}_7\text{COOH}\)) and esters (such as ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) share this general formula and are functional group isomers of each other.

1 mark for the correct option C. Award 1 mark for identifying that both carboxylic acids and esters share the general formula CnH2nO2.

Alkanes like methane react with halogens in the presence of ultraviolet (UV) light. This is a photochemical substitution reaction where a hydrogen atom in methane is replaced by a chlorine atom to form chloromethane (\(\text{CH}_3\text{Cl}\)) and hydrogen chloride (\(\text{HCl}\)).

1 mark for the correct option C. Award 1 mark for correctly identifying the product as chloromethane and the reaction type as substitution.

Since X is soluble in water while Y and Z are insoluble, adding water dissolves X. Filtering separates the solution of X (filtrate) from the residue (Y and Z). Evaporating the filtrate yields pure solid X. Drying the residue ensures no water remains. Adding ethanol to the dried residue dissolves Y but not Z. Filtering separates the solution of Y (filtrate) from the insoluble Z (residue). Evaporating the second filtrate yields pure solid Y, and drying the final residue yields pure solid Z.

1 mark for the correct option A. Award 1 mark for identifying the correct, logically sound sequence of dissolution, filtration, and evaporation/drying.

1. Write the balanced chemical equation for the reaction:
\(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\).

2. Calculate the moles of \(\text{H}_2\text{SO}_4\) used:
\(\text{moles} = \text{concentration} \times \text{volume} = 0.0500\text{ mol/dm}^3 \times \frac{20.00}{1000}\text{ dm}^3 = 0.00100\text{ mol}\).

3. Determine the moles of \(\text{NaOH}\) reacted using the mole ratio (2:1):
\(\text{moles of NaOH} = 2 \times 0.00100\text{ mol} = 0.00200\text{ mol}\).

4. Calculate the concentration of \(\text{NaOH}\) in \(25.0\text{ cm}^3\):
\(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00200\text{ mol}}{0.0250\text{ dm}^3} = 0.0800\text{ mol/dm}^3\).

Award 1 mark for the correct option B.
- Option A: Incorrect ratio of 1:1 used instead of 2:1.
- Option D: Incorrectly multiplied by 4 instead of 2.

1. The reaction of the cation with aqueous ammonia produces a white precipitate that is soluble in excess. This is characteristic of zinc ions (\(\text{Zn}^{2+}\)). Aluminium ions (\(\text{Al}^{3+}\)) form a white precipitate that is insoluble in excess ammonia, and calcium ions (\(\text{Ca}^{2+}\)) do not form a precipitate with ammonia.
2. The reaction of the anion with nitric acid followed by silver nitrate produces a cream precipitate, which confirms the presence of bromide ions (\(\text{Br}^-\)). Chloride ions would produce a white precipitate.
Therefore, the compound is zinc bromide.

Award 1 mark for the correct option B.
- Option A: Aluminium precipitate is insoluble in excess ammonia.
- Option C: Calcium does not form a white precipitate with ammonia.
- Option D: Chloride produces a white precipitate, not a cream precipitate.

To measure the rate of a gas-producing reaction by gas volume:
- A conical flask is needed to hold the reactants.
- A gas syringe is required to collect and measure the volume of gas produced.
- A stopwatch is essential to measure the time intervals.
- A balance is needed to measure the initial mass of the calcium carbonate.
Beakers are open containers and cannot be used with a gas syringe, while condensers and filter funnels are irrelevant for this rate measurement.

Award 1 mark for the correct option A.
- Option B is incorrect because a beaker cannot collect gas.
- Option C contains a filter funnel which is for filtration.
- Option D contains a condenser which is for distillation.

- From 0 to 4 minutes: Substance \(Y\) is a solid warming up. At 2 minutes, it is solid (not a mixture).
- From 4 to 10 minutes: The temperature is constant at \(80^\circ\text{C}\) because melting is occurring. This is its melting point.
- From 10 to 15 minutes: The substance has completely melted and is warming up as a liquid.
- From 15 to 22 minutes: The temperature is constant at \(120^\circ\text{C}\) because boiling is occurring. It is a mixture of liquid and gas.

Award 1 mark for the correct option C.
- Option A is incorrect because melting only starts at 4 minutes.
- Option B is incorrect because \(80^\circ\text{C}\) is the melting point.
- Option D is incorrect because at 18 minutes the substance is boiling and is a liquid-gas mixture.

Reduction is defined as the gain of electrons.
- In equation A, iron(II) ions lose an electron to form iron(III) (oxidation).
- In equation B, chloride ions lose electrons to form chlorine molecules (oxidation).
- In equation C, the manganese in the manganate(VII) ion gains electrons to form manganese(II) ions (reduction).
- In equation D, zinc atoms lose electrons to form zinc ions (oxidation).

Award 1 mark for the correct option C.
- Options A, B, and D all represent oxidation since electrons are on the product side (lost).

Barium sulfate is an insoluble salt and is best prepared by precipitation:
1. Mix two soluble solutions (aqueous barium chloride and dilute sulfuric acid) to form the insoluble precipitate.
2. Filter the mixture to collect the precipitate as the residue.
3. Wash the residue with distilled water to remove soluble impurities.
4. Dry the precipitate to obtain the pure, dry salt.
Option A is incorrect because barium carbonate is insoluble and the final salt cannot be recovered by crystallization. Option C forms barium sulfide. Option D is used for soluble salts.

Award 1 mark for the correct option B.
- Option A is incorrect because barium carbonate is insoluble, making the reaction incomplete.
- Option C is incorrect because heating elements produces barium sulfide, not sulfate.
- Option D is incorrect because titration is not suitable for insoluble salts.

- Option A is incorrect because carbon dioxide is reduced, not oxidized, by carbon to form carbon monoxide.
- Option B is incorrect because iron(III) oxide is reduced by carbon monoxide, not oxidized.
- Option C is correct because limestone (calcium carbonate) undergoes thermal decomposition to calcium oxide, which then acts as a basic oxide to neutralize acidic silicon dioxide impurities, forming slag (calcium silicate).
- Option D is incorrect because the combustion of coke is highly exothermic, releasing energy to heat the furnace.

Award 1 mark for the correct option C.
- Option A confuses reduction and oxidation of carbon compounds.
- Option B confuses reduction and oxidation of iron oxide.

- Alkanes have the general formula \(\text{C}_n\text{H}_{2n+2}\) (no oxygen).
- Carboxylic acids have the general formula \(\text{C}_n\text{H}_{2n}\text{O}_2\) (e.g., butanoic acid is \(\text{C}_3\text{H}_7\text{COOH}\) or \(\text{C}_4\text{H}_8\text{O}_2\)).
- Esters have the general formula \(\text{C}_n\text{H}_{2n}\text{O}_2\) (e.g., ethyl ethanoate is \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) or \(\text{C}_4\text{H}_8\text{O}_2\)).
- Monohydric alcohols have the general formula \(\text{C}_n\text{H}_{2n+1}\text{OH}\) (only one oxygen atom).
Therefore, the compound could be a carboxylic acid or an ester.

Award 1 mark for the correct option B.
- Options A, C, and D are incorrect because they include alkanes (no oxygen) or alcohols (one oxygen under standard homologous series definitions).

First, calculate the number of moles of \(\text{NaOH}\): \(\text{moles} = 0.150\text{ mol/dm}^3 \times (25.0 / 1000)\text{ dm}^3 = 0.00375\text{ mol}\). From the stoichiometry of the balanced equation, \(2\text{ moles}\) of \(\text{NaOH}\) react with \(1\text{ mole}\) of \(\text{H}_2\text{SO}_4\). Therefore, the number of moles of \(\text{H}_2\text{SO}_4\) needed is \(0.00375 / 2 = 0.001875\text{ mol}\). Finally, calculate the volume of sulfuric acid needed: \(\text{volume} = \text{moles} / \text{concentration} = 0.001875\text{ mol} / 0.100\text{ mol/dm}^3 = 0.01875\text{ dm}^3 = 18.75\text{ cm}^3\).

1 mark for the correct option A. (Method: calculate moles of NaOH, use 2:1 ratio to find moles of acid, and convert moles of acid to volume in cm3).

The reaction of the solution with aqueous sodium hydroxide produces a green precipitate insoluble in excess, which is characteristic of iron(II) ions, \(\text{Fe}^{2+}\). (Note that chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide). The reaction with aqueous barium nitrate in acidic conditions produces a white precipitate of barium sulfate, which indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Thus, the unknown solid is iron(II) sulfate.

1 mark for the correct option C.

In this reaction, \(\text{CO}\) reduces the iron(III) oxide to iron by removing oxygen from it, meaning \(\text{CO}\) acts as the reducing agent. The oxidation state of iron in \(\text{Fe}_2\text{O}_3\) is \(+3\), and it is reduced to \(0\) in iron metal. Carbon's oxidation state increases from \(+2\) in \(\text{CO}\) to \(+4\) in \(\text{CO}_2\), so carbon is oxidized, not reduced.

1 mark for the correct option C.

The reaction of an alkane like ethane with a halogen like chlorine is a photochemical substitution reaction. Chlorine atoms replace hydrogen atoms in the alkane to yield hydrogen chloride (\(\text{HCl}\)) and chlorinated alkanes such as chloroethane (\(\text{C}_2\text{H}_5\text{Cl}\)) or dichloroethane (\(\text{C}_2\text{H}_4\text{Cl}_2\)). Free hydrogen gas (\(\text{H}_2\)) is not a product of this reaction.

1 mark for the correct option C.

Because copper(II) sulfate is soluble in water, it cannot be filtered out at the beginning. Ethanol (boiling point \(78^\circ\text{C}\)) and water (boiling point \(100^\circ\text{C}\)) are miscible liquids, so they must be separated using fractional distillation, which allows the collection of a pure ethanol distillate. The remaining aqueous copper(II) sulfate solution is then heated to saturation, cooled to allow crystallisation to occur, filtered, and the crystals are dried to obtain pure copper(II) sulfate.

1 mark for the correct option B.

At high temperatures and pressures inside car engines, nitrogen gas and oxygen gas from the air react together to form nitrogen oxides (mainly nitrogen monoxide, \(\text{NO}\), and nitrogen dioxide, \(\text{NO}_2\)). Catalytic converters decrease air pollution by reducing these oxides back to harmless nitrogen gas. The primary greenhouse gases are carbon dioxide and methane, not nitrogen oxides.

1 mark for the correct option B.

To prepare an insoluble salt, precipitation is used. This involves mixing two soluble reactants. Lead(II) nitrate is a soluble salt, and dilute sulfuric acid is a soluble acid. Mixing them produces a precipitate of insoluble lead(II) sulfate: \(\text{Pb}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{PbSO}_4(\text{s})\). The precipitate is filtered out as the residue, washed with distilled water to remove spectator ions, and then dried.

1 mark for the correct option C.

The rate of diffusion of a gas is inversely proportional to its relative molecular mass, \(M_r\). The gas with the lowest \(M_r\) will diffuse the fastest. Computing the \(M_r\) values: Carbon dioxide (\(\text{CO}_2\)) = \(12 + 2(16) = 44\); Ammonia (\(\text{NH}_3\)) = \(14 + 3(1) = 17\); Hydrogen chloride (\(\text{HCl}\)) = \(1 + 35.5 = 36.5\); Oxygen (\(\text{O}_2\)) = \(2(16) = 32\). Ammonia has the smallest \(M_r\) and therefore diffuses the fastest.

1 mark for the correct option B.

1. Write the balanced equation for the reaction:
\(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)

2. Calculate the moles of \(\text{NaOH}\) reacted:
\(\text{moles of NaOH} = \frac{25.0}{1000} \times 0.150 = 0.00375\text{ mol}\)

3. Determine the moles of \(\text{H}_2\text{SO}_4\) reacted using the mole ratio (2:1):
\(\text{moles of H}_2\text{SO}_4 = \frac{0.00375}{2} = 0.001875\text{ mol}\)

4. Calculate the concentration of \(\text{H}_2\text{SO}_4\):
\(\text{concentration} = \frac{0.001875}{18.75 / 1000} = 0.100\text{ mol/dm}^3\)

Award 1 mark for selecting correct option A.
- Reject incorrect calculations.

Standard qualitative analysis tests indicate:
- A green precipitate insoluble in excess sodium hydroxide confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note that chromium(III) ions also form a green precipitate, but it is soluble in excess sodium hydroxide).
- The addition of dilute nitric acid followed by barium nitrate forms a white precipitate of barium sulfate, confirming the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, the compound \(X\) is iron(II) sulfate.

Award 1 mark for the correct option B.

In this reaction:
- The oxidation state of iron changes from \(+3\) in \(\text{FeCl}_3\) to \(+2\) in \(\text{FeCl}_2\) (reduction).
- The oxidation state of sulfur changes from \(-2\) in \(\text{H}_2\text{S}\) to \(0\) in \(\text{S}\) (oxidation).
- Therefore, sulfur is oxidised because its oxidation state increases from \(-2\) to \(0\), making statement C correct.

Award 1 mark for correct option C.

To measure the rate of a gas-producing reaction by mass loss over time, we need:
- A vessel to hold the reactants (conical flask),
- A device to measure the mass (electronic balance),
- A device to measure time elapsed (stop-watch),
- Cotton wool in the neck of the flask to prevent acid spray from escaping (which would cause an inaccurate mass loss reading) while allowing carbon dioxide gas to escape freely.

Award 1 mark for correct option B.

Alkanes are saturated hydrocarbons and react with halogens via substitution reactions in the presence of UV light. In the first step of the reaction between ethane (\(\text{C}_2\text{H}_6\)) and chlorine (\(\text{Cl}_2\)), one hydrogen atom is replaced by one chlorine atom, yielding chloroethane, \(\text{C}_2\text{H}_5\text{Cl}\).

Award 1 mark for correct option B.

Barium sulfate is insoluble and must be prepared by a precipitation reaction between two soluble solutions, such as barium nitrate and sodium sulfate. The steps are:
1. Mix the two solutions together to form the precipitate.
2. Filter the mixture to collect the precipitate as the residue on the filter paper.
3. Wash the residue with distilled water to remove any soluble impurities (like sodium nitrate).
4. Dry the residue (precipitate) in a warm oven or with filter paper.

Award 1 mark for correct option D.

Silicon(IV) oxide is an acidic impurity. Limestone (calcium carbonate) added to the furnace decomposes to form calcium oxide (a basic oxide). Calcium oxide reacts with silicon(IV) oxide to form calcium silicate (slag, \(\text{CaSiO}_3\)), which is easily separated. The balanced equation for this acid-base neutralisation reaction is: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

Award 1 mark for correct option D.

Isomers are compounds that have the same molecular formula but different structural formulas. Because they have different structures (e.g., branched vs straight chains), they differ in physical properties such as boiling point. Branching reduces the surface area contact between molecules, leading to weaker intermolecular forces and a lower boiling point.

Award 1 mark for correct option D.

(a) Suitable indicator is methyl orange. The color change in the flask is from yellow (alkaline) to orange/red (neutral/acidic at endpoint) OR phenolphthalein which changes from pink to colorless.
(b)(i) Moles of \(\text{NaOH} = \frac{25.0}{1000} \times 0.150 = 0.00375\text{ mol}\).
(ii) From the chemical equation, 1 mole of \(\text{H}_2\text{X}\) reacts with 2 moles of \(\text{NaOH}\). Moles of \(\text{H}_2\text{X} = \frac{0.00375}{2} = 0.001875\text{ mol}\).
(iii) Concentration of \(\text{H}_2\text{X} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.001875}{18.75 / 1000} = 0.100\text{ mol/dm}^3\).
(c) Concentration in \(\text{g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r = 0.100 \times 126 = 12.6\text{ g/dm}^3\).
(d) A strong acid completely dissociates/ionises into ions in aqueous solution, whereas a weak acid only partially dissociates/ionises. To distinguish them: Add equal lengths of magnesium ribbon (or equal mass of calcium carbonate) to equal volumes of both acids of the same concentration. The strong acid will produce effervescence/bubbling at a much faster rate than the weak acid.

Part (a):
- Methyl orange (or phenolphthalein) [1]
- Initial color yellow (or pink) [1]
- Final color orange/red (or colorless) [1]

Part (b):
- (i) \(0.00375\text{ mol}\) [1]
- (ii) \(0.001875\text{ mol}\) (allow ecf = (i) / 2) [1]
- (iii) \(0.100\text{ mol/dm}^3\) (allow ecf = (ii) / 0.01875) [2]

Part (c):
- \(12.6\text{ g/dm}^3\) (allow ecf = (b)(iii) x 126) [2]

Part (d):
- Strong acid fully ionises/dissociates and weak acid partially ionises/dissociates [2]
- Add magnesium ribbon (or calcium carbonate) [1]
- Faster rate of bubbling / magnesium dissolves faster in strong acid [1]

(a)(i) The cation is iron(II) / \(\text{Fe}^{2+}\) because it forms a green precipitate with sodium hydroxide that is insoluble in excess.
(ii) The ionic equation is: \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\).
(b)(i) The anion is sulfate / \(\text{SO}_4^{2-}\) because it forms a white precipitate with barium nitrate in acidic conditions, and no precipitate with silver nitrate rules out halides.
(ii) Nitric acid is added to eliminate carbonate or sulfite ions that could react with barium ions to form a false-positive white precipitate.
(iii) The white precipitate is barium sulfate, \(\text{BaSO}_4\).
(c)(i) The gas that decolors acidified potassium manganate(VII) is sulfur dioxide, \(\text{SO}_2\).
(ii) It acts as a reducing agent because it reduces the purple manganate(VII) ions (where Mn has oxidation state +7) to colorless manganese(II) ions (where Mn is +2).
(d) To test for ammonium ions: add aqueous sodium hydroxide to the solution of \(\text{Z}\) and warm the mixture gently. Test the gas evolved with damp red litmus paper. The damp red litmus paper turns blue, confirming the gas is ammonia, which indicates the presence of ammonium ions.

Part (a):
- (i) Iron(II) / \(\text{Fe}^{2+}\) (reject iron / Fe) [1]
- (ii) \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\) [2: 1 mark for species, 1 mark for state symbols]

Part (b):
- (i) Sulfate / \(\text{SO}_4^{2-}\) [1]
- (ii) To remove carbonate / sulfite ions / prevent false-positive precipitate [1]
- (iii) Barium sulfate / \(\text{BaSO}_4\) [1]

Part (c):
- (i) Sulfur dioxide / \(\text{SO}_2\) [1]
- (ii) Reducing agent [1] because it reduces purple manganate(VII) to colorless manganese(II) / decreases the oxidation state of manganese [1]

Part (d):
- Add aqueous sodium hydroxide [1]
- Warm/heat [1]
- Effervescence / gas evolved / pungent gas [1]
- Gas turns damp red litmus paper blue [1]

(a) A suitable diagram must show a sealed conical flask containing the reaction mixture (calcium carbonate chips and hydrochloric acid) connected via a delivery tube to a gas syringe (or a graduated tube/measuring cylinder inverted over water in a trough). All key elements must be clearly labelled: 'conical flask', 'gas syringe' (or 'graduated cylinder'), 'calcium carbonate chips', 'hydrochloric acid'.
(b) Experimental Procedure:
- Controlled variables: Temperature of the mixture, total volume of liquid (acid + water mixture), and mass and surface area (size) of the calcium carbonate chips.
- Independent variable (concentration): Prepare different concentrations of acid by taking a fixed total volume (e.g., \(50\text{ cm}^3\)) of liquid but changing the ratio of stock hydrochloric acid to distilled water (e.g., \(50\text{ cm}^3\) acid; \(40\text{ cm}^3\) acid + \(10\text{ cm}^3\) water, etc.).
- Measurements: Start a stopwatch immediately upon adding the acid to the flask. Record the volume of gas collected in the syringe at regular intervals (e.g., every 20 seconds) until the reaction ceases (no more gas is evolved).
- Analysis: Plot a graph of volume of gas (y-axis) against time (x-axis) for each concentration. Compare the initial gradient/slope of the curves. The concentration that yields the steepest initial slope has the highest rate of reaction.
(c) According to collision theory, increasing the concentration of hydrochloric acid means there are more hydrogen ions per unit volume. This reduces the distance between reacting particles, leading to a higher frequency of collisions (more collisions per second) between the hydrogen ions and the calcium carbonate chips. Consequently, the frequency of successful collisions (those with kinetic energy equal to or exceeding the activation energy) increases, which increases the rate of the reaction.

Part (a):
- Flask containing reaction mixture with a delivery tube [1]
- Gas syringe / inverted measuring cylinder in water trough [1]
- Labels for reactants (calcium carbonate, acid) and gas syringe/cylinder, with airtight apparatus [1]

Part (b):
- Identifies at least two controlled variables: mass/size of carbonate, total volume, temperature [2]
- Explains how to vary concentration: dilution of acid with water [1]
- Explains measurements: volume of gas at specified regular time intervals [2]
- Explains comparison: plot graph and compare initial gradients / find time taken to collect a specific volume [1]

Part (c):
- More reactant particles / hydrogen ions per unit volume [1]
- Increased collision frequency / more collisions per second [1]
- Reactant particles must be specified (hydrogen ions and calcium carbonate) [1]
- Increased frequency of successful/effective collisions / more collisions with energy \(\ge\) activation energy [1]

(a)(i) During melting (segment B), the highly regular arrangement (lattice) of the solid breaks down into a random, closely-packed liquid arrangement. The movement of particles changes from vibrating about fixed positions to sliding past one another.
(ii) The temperature remains constant because the thermal energy supplied is absorbed to overcome/break the attractive intermolecular forces holding the particles together in the solid lattice, rather than increasing the kinetic energy of the particles.
(b)(i) At \(-20\ ^\circ\text{C}\), substance \(\text{W}\) is below its melting point, so it is a solid. At \(25\ ^\circ\text{C}\), it is between its melting and boiling points, so it is a liquid. At \(130\ ^\circ\text{C}\), it is above its boiling point, so it is a gas.
(ii) At \(25\ ^\circ\text{C}\) (liquid), the particles are close together/touching but randomly arranged, and they move slowly, sliding past each other. At \(130\ ^\circ\text{C}\) (gas), the particles are very far apart with large empty spaces between them, and they move extremely rapidly and randomly in all directions.
(c)(i) Evaporation occurs at any temperature below the boiling point and takes place only at the surface of the liquid, whereas boiling occurs only at a single temperature (the boiling point) and takes place throughout the entire bulk of the liquid (with bubbles forming inside).
(ii) Factors that increase evaporation rate include: increasing the temperature, increasing the surface area of the liquid, or increasing air movement (wind) across the surface.

Part (a):
- (i) Arrangement becomes random/less orderly [1]; movement changes from vibration to sliding/rolling over each other [1]
- (ii) Energy is used to overcome/break intermolecular forces/bonds [1]; kinetic energy of particles does not increase [1]

Part (b):
- (i) Solid, Liquid, Gas (1 mark each) [3]
- (ii) Arrangement difference: particles are close/touching at \(25\ ^\circ\text{C}\) but far apart at \(130\ ^\circ\text{C}\) [1]; Motion difference: particles slide past each other at \(25\ ^\circ\text{C}\) but move rapidly/randomly at \(130\ ^\circ\text{C}\) [1]; Correct comparison of density/randomness [1]

Part (c):
- (i) Evaporation occurs at any temperature / surface only [1]; boiling occurs only at boiling point / throughout the liquid [1]
- (ii) Increased temperature / increased surface area / wind [1]

(a)(i) Oxidation is the loss of electrons.
(ii) The species oxidized is \(\text{Fe}^{2+}\) because each \(\text{Fe}^{2+}\) ion loses one electron to form an \(\text{Fe}^{3+}\) ion: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\).
(b)(i) In the permanganate ion, \(\text{MnO}_4^-\): let \(x\) be the oxidation number of Mn. \(x + 4(-2) = -1 \Rightarrow x - 8 = -1 \Rightarrow x = +7\).
In the manganese(II) ion, \(\text{Mn}^{2+}\): the oxidation number is simply the charge on the monoatomic ion, which is \(+2\).
The change in oxidation number is from \(+7\) to \(+2\) (a decrease of 5).
(ii) A redox reaction involves both oxidation and reduction. Here, iron undergoes oxidation because its oxidation number increases from \(+2\) (in \(\text{Fe}^{2+}\)) to \(+3\) (in \(\text{Fe}^{3+}\)). Manganese undergoes reduction because its oxidation number decreases from \(+7\) (in \(\text{MnO}_4^-\)) to \(+2\) (in \(\text{Mn}^{2+}\)).
(c)(i) The salt bridge completes the electrical circuit, allowing the migration of ions between the two half-cells to maintain electrical neutrality.
(ii) Electrons flow through the wire in the external circuit from the iron(II) sulfate half-cell (where oxidation occurs and electrons are lost) to the potassium manganate(VII) half-cell (where reduction occurs and electrons are gained).
(d) When aqueous potassium iodide (a reducing agent) is added to an oxidising agent, it is oxidised to iodine. The color change is from colorless to brown. The substance responsible is iodine, \(\text{I}_2\).

Part (a):
- (i) Loss of electrons [1]
- (ii) \(\text{Fe}^{2+}\) [1] and explanation: it loses an electron / equation showing loss of electron [1]

Part (b):
- (i) \(+7\) for Mn in \(\text{MnO}_4^-\)[1]; \(+2\) for Mn in \(\text{Mn}^{2+}\) [1]; change is decrease of 5 (from \(+7\) to \(+2\)) [1]
- (ii) Oxidation number of Fe increases from \(+2\) to \(+3\) (oxidation) AND oxidation number of Mn decreases from \(+7\) to \(+2\) (reduction) [2]

Part (c):
- (i) To complete the circuit / allow ions to flow / maintain electrical neutrality [1]
- (ii) From \(\text{Fe}^{2+}\) half-cell to \(\text{MnO}_4^-\) half-cell [1] through the external wires [1]

Part (d):
- Colorless to brown / yellow-brown [1]
- Iodine / \(\text{I}_2\) [1]

(a) Excess zinc carbonate is used to ensure that all of the sulfuric acid is completely neutralised and reacted. This makes sure the final zinc sulfate solution is not contaminated with unreacted sulfuric acid, which would be difficult to separate.
(b) Step-by-step preparation:
1. Filtration: Filter the reaction mixture to remove the unreacted, excess zinc carbonate solid. The zinc sulfate solution is collected as the filtrate.
2. Heating: Transfer the filtrate to an evaporating basin and heat it to evaporate some of the water, concentrating the solution until the crystallization point is reached.
3. Testing for crystallization: Dip a cold glass rod into the hot solution; if small crystals form on the rod as it cools, the crystallization point has been reached and heating should be stopped.
4. Crystallization: Leave the hot, saturated solution to cool slowly. Crystals of hydrated zinc sulfate will form.
5. Isolation: Filter the mixture of crystals and liquid to isolate the zinc sulfate crystals.
6. Washing and drying: Wash the crystals with a tiny amount of cold distilled water to remove impurities, then gently pat them dry using pieces of filter paper or leave them in a warm oven (not too hot, to avoid dehydration/losing water of crystallization).
(c) The chemical equation is: \(\text{ZnCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\).
(d) Calculation:
1. \(\text{Moles of H}_2\text{SO}_4 = \text{volume in dm}^3 \times \text{concentration} = \frac{50.0}{1000} \times 1.50 = 0.0750\text{ mol}\).
2. From the equation, 1 mole of \(\text{H}_2\text{SO}_4\) yields 1 mole of \(\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}\).
So, \(\text{moles of ZnSO}_4 \cdot 7\text{H}_2\text{O} = 0.0750\text{ mol}\).
3. \(\text{Maximum mass of crystals} = \text{moles} \times M_r = 0.0750 \times 287 = 21.525\text{ g}\).
Rounding to 3 significant figures gives \(21.5\text{ g}\).

Part (a):
- To ensure all the sulfuric acid is completely reacted/neutralised [1]

Part (b):
- Filter to remove excess solid zinc carbonate [1]
- Heat the filtrate in an evaporating dish to evaporate water / concentrate the solution [1]
- Heat to crystallization point / saturation point [1]
- Test crystallization point with a cold glass rod (crystals form on it) [1]
- Allow to cool to crystallize [1]
- Wash crystals with a small amount of cold distilled water and dry with filter paper / in a warm oven (below \(60\ ^\circ\text{C}\)) [1]

Part (c):
- Correctly written equation: \(\text{ZnCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\text{O} + \text{CO}_2\) [1]
- State symbols correct: \(\text{ZnCO}_3(\text{s})\), \(\text{H}_2\text{SO}_4(\text{aq})\), \(\text{ZnSO}_4(\text{aq})\), \(\text{H}_2\text{O}(\text{l})\), \(\text{CO}_2(\text{g})\) [1]

Part (d):
- Number of moles of sulfuric acid calculated correctly: \(0.0750\text{ mol}\) [1]
- Number of moles of zinc sulfate crystals = \(0.0750\text{ mol}\) [1]
- Multiplying moles by 287: \(0.0750 \times 287\) [1]
- Correct final mass with units: \(21.5\text{ g}\) (accept \(21.525\text{ g}\) or \(21.53\text{ g}\)) [1]

To carry out the investigation: 1. Dissolve equal masses (e.g., 5.0 g) of each solid descaling agent in equal volumes (e.g., 100 cm3) of distilled water to make three solutions of equal concentration. 2. Set up a conical flask connected to a gas syringe using a delivery tube. 3. Place a fixed mass (e.g., 2.0 g) of uniform calcium carbonate chips into the conical flask. 4. Add a fixed volume (e.g., 25 cm3) of the solution of Brand A to the flask, quickly seal with a stopper, and start a stopwatch immediately. 5. Record the volume of carbon dioxide gas collected in the gas syringe at regular intervals (such as every 30 seconds) for 5 minutes. 6. Repeat the exact procedure using the same mass and size of calcium carbonate chips with equal volumes of the solutions of Brand B and Brand C. 7. Plot a graph of volume of gas against time for each brand. The brand that produces gas at the fastest rate (i.e., has the steepest initial gradient on the graph) is the most effective descaling agent.

Total 10 marks. - Prepare equal concentrations: weigh equal masses of the three descaling solids [1 mark] and dissolve each in the same volume of water [1 mark]. - Use a conical flask containing a fixed mass of calcium carbonate chips [1 mark]. - Specify that calcium carbonate chips must be of uniform size/surface area [1 mark]. - Add a fixed/controlled volume of the descaling solution [1 mark]. - Connect flask to a gas syringe/gas collection system [1 mark]. - Start stopwatch immediately on mixing and record gas volume at regular time intervals [1 mark]. - Control variable: keep temperature constant [1 mark]. - Analysis: Plot volume against time and determine the rate from the gradient of each curve [1 mark]. - Conclusion: The most effective brand is the one with the highest initial gradient / fastest rate of gas production [1 mark].

For (a), a burette is the standard laboratory apparatus used to deliver variable but highly precise volumes of liquid titrant. For (b), methyl orange is yellow in alkaline solution and turns orange or pink at the end-point. For (c), the titres are calculated by subtracting the initial reading from the final reading: Trial 1 = 18.4 - 0.0 = 18.4 cm3; Trial 2 = 36.6 - 18.4 = 18.2 cm3; Trial 3 = 18.8 - 0.5 = 18.3 cm3. Since all three are concordant (within 0.2 cm3 of each other), the average is (18.4 + 18.2 + 18.3) / 3 = 18.3 cm3. For (d), moles of H2SO4 = volume in dm3 * concentration = (18.3 / 1000) * 0.100 = 0.00183 mol. For (e), since the reaction ratio of NaOH : H2SO4 is 2 : 1, moles of NaOH = 2 * 0.00183 = 0.00366 mol. Concentration of NaOH = moles / volume in dm3 = 0.00366 / 0.0250 = 0.1464 mol/dm3 (or 0.146 mol/dm3).

Total 10 marks. - (a) Burette [1 mark]. - (b) Yellow [1 mark] to orange / pink [1 mark]. - (c) All three titres calculated correctly: 18.4 cm3, 18.2 cm3, 18.3 cm3 [1 mark]; average calculated correctly as 18.3 cm3 [1 mark]. - (d) Moles of acid calculation: (average titre / 1000) * 0.100 [1 mark] = 0.00183 mol [1 mark]. - (e) Moles of NaOH = 2 * moles of acid = 0.00366 mol [1 mark]; concentration of NaOH = 0.00366 / 0.025 = 0.146 (or 0.1464) mol/dm3 [1 mark].

For (a), the reaction is carried out in a conical flask closed with a rubber stopper. A delivery tube connects the flask to a graduated gas syringe. For (b), using a gas syringe avoids any loss of gas due to solubility in water and is easier to read accurately. For (c)(i), the rate is highest initially because the concentration of acid reactants is at its maximum, leading to the highest frequency of successful collisions. The rate eventually becomes zero because the limiting reactant (zinc) is completely used up. For (c)(ii), the initial rate of reaction is determined by drawing a tangent to the curve at time t = 0 and calculating the gradient of this tangent. For (d), zinc powder has a larger surface area than granules, so the reaction is faster, resulting in a steeper initial curve. Since the same mass of zinc is used, the total volume of hydrogen gas produced remains the same, so the curve levels off at the same height.

Total 10 marks: (a) Conical flask/vessel with rubber bung/stopper [1 mark], delivery tube connecting flask to gas syringe [1 mark], gas syringe correctly shown/described as graduated [1 mark]. (b) Gas syringe is more accurate/prevents loss of gas by dissolving [1 mark]. (c)(i) Maximum concentration of reactants at start leads to maximum rate [1 mark], concentration/amount of reactants decreases over time so rate decreases [1 mark], reaction stops when zinc is completely used up [1 mark]. (c)(ii) Draw a tangent to the curve at t = 0 [1 mark] and calculate the gradient [1 mark]. (d) Steeper curve [1 mark] leveling off at the same volume plateau [1 mark].

For (a), iron(II) ions react with hydroxide ions to form a green precipitate of iron(II) hydroxide, which is insoluble in excess sodium hydroxide. For (b), the reaction with aqueous ammonia also produces a green precipitate of iron(II) hydroxide, which remains insoluble in excess ammonia. For (c), the addition of hydrochloric acid and barium chloride is the standard test for sulfate ions, yielding a white precipitate of barium sulfate. For (d)(i), heating an ammonium salt with sodium hydroxide releases ammonia gas, which can be identified because it turns damp red litmus paper blue. For (d)(ii), the ion responsible is the ammonium ion. For (e), to test for the absence of carbonate ions, add dilute hydrochloric acid and observe that there is no bubbling/effervescence.

Total 10 marks: (a) Green precipitate [1 mark], insoluble in excess sodium hydroxide [1 mark]. (b) Green precipitate [1 mark], insoluble in excess ammonia [1 mark]. (c) White precipitate [1 mark] indicating the presence of sulfate ions [1 mark]. (d)(i) Ammonia gas [1 mark] which turns damp red litmus paper blue [1 mark]. (d)(ii) Ammonium ion [1 mark]. (e) Add dilute acid and observe no fizzing/effervescence [1 mark].