2026 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

At the cathode (negative electrode), \(\text{Cu}^{2+}\) ions gain electrons in preference to \(\text{H}^+\) ions because copper is less reactive than hydrogen, forming solid copper (a brown/pink solid). At the anode (positive electrode), hydroxide ions (\(\text{OH}^-\)) from water are discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)), producing oxygen gas (\(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\)). As \(\text{Cu}^{2+}\) ions are removed from the solution, the blue colour of the electrolyte gradually fades.

B is correct [1]. A is incorrect as hydrogen is not formed at the cathode. C is incorrect as sulfate is not discharged to sulfur dioxide. D is incorrect as the colour fades rather than remains unchanged.

From the stoichiometric ratio \(1\text{ Mg} : 2\text{ HCl}\), \(0.15\text{ mol}\) of \(\text{HCl}\) reacts with \(0.15 / 2 = 0.075\text{ mol}\) of \(\text{Mg}\). Therefore, \(\text{HCl}\) is the limiting reactant and \(\text{Mg}\) is in excess by \(0.10 - 0.075 = 0.025\text{ mol}\). The amount of \(\text{H}_2\) formed is determined by the limiting reactant: \(0.15 / 2 = 0.075\text{ mol}\). The volume of \(\text{H}_2\) gas at r.t.p. is \(0.075\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.80\text{ dm}^3\).

C is correct [1]. A is incorrect because Mg is in excess, not HCl. B is incorrect as 0.025 mol of Mg remains unreacted. D is incorrect because the volume of hydrogen produced is 1.80 dm^3, not 2.40 dm^3.

Increasing the temperature increases the average kinetic energy of the particles. As a result, particles move faster, which increases the collision frequency. More significantly, a much greater proportion of the colliding particles have kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)), leading to a higher frequency of successful collisions.

D is correct [1]. A is incorrect because activation energy is not altered by temperature (only catalysts change activation energy). B is incorrect because temperature increases kinetic energy, not decreases it. C is incorrect as the proportion of particles with sufficient energy increases, not decreases.

A dicarboxylic acid reacting with a diamine eliminates water molecules to form amide linkages (\(-\text{CO}-\text{NH}-\)). The resulting synthetic condensation polymer is a polyamide (such as nylon).

A is correct [1]. B is incorrect as polyesters contain ester linkages and are made from dicarboxylic acids and diols. C is incorrect because addition polymers do not involve condensation or amide linkages. D is incorrect because ester linkages join acids and alcohols.

Structural isomers are compounds with the same molecular formula but different structural formulae. But-1-ene has the molecular formula \(\text{C}_4\text{H}_8\). Methylpropene (2-methylpropene) also has the molecular formula \(\text{C}_4\text{H}_8\) with a branched chain, making it a structural isomer. Butane is \(\text{C}_4\text{H}_{10}\), cyclobutane is not in the unbranched/branched alkene series listed, and propene is \(\text{C}_3\text{H}_6\).

B is correct [1]. A is incorrect as butane has formula C4H10. C is incorrect as propene has formula C3H6. D is incorrect as butan-1-ol has formula C4H10O.

The forward reaction is exothermic (\(\Delta H < 0\)), so decreasing the temperature shifts the equilibrium to the right to produce more heat. The forward reaction goes from 3 moles of gas to 2 moles of gas, so increasing the pressure shifts the equilibrium to the side with fewer gas moles (the right).

C is correct [1]. A is incorrect because increasing temperature shifts equilibrium to the left. B is incorrect because decreasing pressure shifts equilibrium to the left. D is incorrect because increasing temperature and decreasing pressure both shift equilibrium to the left.

A green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). A white precipitate with dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, the salt is iron(II) sulfate.

A is correct [1]. B is incorrect as iron(III) gives a red-brown precipitate. C is incorrect as chloride gives a white precipitate with silver nitrate, not barium nitrate. D is incorrect as iron(III) chloride gives red-brown precipitate and no reaction with barium nitrate.

Energy required for bond breaking = \(1 \times (\text{H}-\text{H}) + 1 \times (\text{Cl}-\text{Cl}) = 436 + 242 = +678\text{ kJ/mol}\). Energy released from bond making = \(2 \times (\text{H}-\text{Cl}) = 2 \times 431 = 862\text{ kJ/mol}\). Enthalpy change \(\Delta H = \text{energy broken} - \text{energy formed} = 678 - 862 = -184\text{ kJ/mol}\).

A is correct [1]. B is incorrect because it misses the negative sign. C is incorrect from failing to multiply H-Cl by 2 (678 - 431 = +247 kJ/mol). D is incorrect from incorrect sign and stoichiometry.

According to Avogadro's law, equal volumes of gases under the same conditions contain equal numbers of moles. Therefore, the volume ratio gives the molar ratio:

\(1\text{ volume of }\text{C}_x\text{H}_y : 5\text{ volumes of }\text{O}_2 \rightarrow 3\text{ volumes of }\text{CO}_2\)

From the carbon balance:
\(1 \times x = 3 \implies x = 3\)

From the oxygen balance:
Total oxygen atoms on the reactant side = \(5 \times 2 = 10\text{ atoms}\).
Oxygen atoms in \(3\text{CO}_2 = 3 \times 2 = 6\text{ atoms}\).
Oxygen atoms in \(\text{H}_2\text{O} = 10 - 6 = 4\text{ atoms}\), which means \(4\text{H}_2\text{O}\) are formed.

From the hydrogen balance:
\(y = 4 \times 2 = 8\)

Therefore, the formula is \(\text{C}_3\text{H}_8\).

[1 mark] B is correct.
A is incorrect: corresponds to \(\text{C}_3\text{H}_6\) which would require \(90\text{ cm}^3\) of \(\text{O}_2\).
C is incorrect: corresponds to \(\text{C}_4\text{H}_{10}\).
D is incorrect: corresponds to \(\text{C}_2\text{H}_6\).

In concentrated aqueous copper(II) chloride:
- At the cathode (negative electrode): \(\text{Cu}^{2+}\) ions are discharged in preference to \(\text{H}^+\) ions because copper is less reactive than hydrogen (\(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)), forming a reddish-brown solid deposit.
- At the anode (positive electrode): Chloride ions (\(\text{Cl}^-\)) are present in high concentration, so they are discharged in preference to hydroxide ions (\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\)), producing bubbles of a pale yellow-green gas (chlorine).

[1 mark] C is correct.
A is incorrect: hydrogen gas is not discharged at the cathode.
B is incorrect: oxygen is not the main product at the anode for concentrated chloride solutions.
D is incorrect: electrodes are reversed.

Increasing temperature increases the kinetic energy of particles, which makes them move faster and collide more frequently (Statement 1 is correct). It also significantly increases the proportion of particles possessing energy \(E \ge E_a\), leading to a higher frequency of successful collisions (Statement 2 is correct). Increasing temperature does not change the activation energy \(E_a\); only adding a catalyst lowers \(E_a\) (Statement 3 is incorrect).

[1 mark] A (1 and 2 only) is correct.
B, C, and D are incorrect because statement 3 is false (temperature does not alter activation energy).

In silicon(IV) oxide (silica, \(\text{SiO}_2\)), each silicon atom is covalently bonded tetrahedrally to four oxygen atoms, and each oxygen atom forms covalent bonds to two silicon atoms. It is a giant covalent macromolecule with a very high melting point, does not contain mobile ions or delocalised electrons, and is insoluble in water.

[1 mark] A is correct.
B is incorrect: \(\text{SiO}_2\) contains no free ions or delocalised electrons.
C is incorrect: \(\text{SiO}_2\) has strong covalent bonds throughout its giant lattice, leading to a high melting point.
D is incorrect: \(\text{SiO}_2\) is insoluble in water.

An oxide that reacts with both acids (such as dilute sulfuric acid) and bases/alkalis (such as aqueous potassium hydroxide) to produce a salt and water is classified as an amphoteric oxide. According to the Cambridge IGCSE syllabus, amphoteric oxides are aluminium oxide (\(\text{Al}_2\text{O}_3\)) and zinc oxide (\(\text{ZnO}\)). Therefore, element \(\text{E}\) is aluminium.

[1 mark] B (Aluminium) is correct.
A (Calcium) forms a basic oxide only.
C (Magnesium) forms a basic oxide only.
D (Sulfur) forms an acidic oxide only.

Energy required to break bonds in reactants:
\(\text{Energy in} = (1 \times \text{H}-\text{H}) + (1 \times \text{Cl}-\text{Cl}) = 436 + 242 = 678\text{ kJ/mol}\)

Energy released when bonds form in products:
\(\text{Energy out} = 2 \times (\text{H}-\text{Cl}) = 2 \times 431 = 862\text{ kJ/mol}\)

\(\Delta H = \text{Energy in} - \text{Energy out} = 678 - 862 = -184\text{ kJ/mol}\).

[1 mark] D is correct.
A is incorrect: sign inverted (\(+184\text{ kJ/mol}\)).
B is incorrect: failed to multiply \(\text{H}-\text{Cl}\) by 2 (\(678 - 431 = +247\text{ kJ/mol}\)).
C is incorrect: inverted subtraction with one bond only (\(-247\text{ kJ/mol}\)).

- With \(\text{NaOH}\), \(\text{Al}^{3+}\) and \(\text{Zn}^{2+}\) give a white precipitate soluble in excess.
- With aqueous \(\text{NH}_3\), \(\text{Al}^{3+}\) gives a white precipitate insoluble in excess, whereas \(\text{Zn}^{2+}\) gives a white precipitate soluble in excess. Thus, the cation is \(\text{Al}^{3+}\).
- With dilute nitric acid and aqueous barium nitrate, a white precipitate of \(\text{BaSO}_4\) confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)).
Therefore, compound \(\text{Q}\) is aluminium sulfate.

[1 mark] A (Aluminium sulfate) is correct.
B (Zinc sulfate) is incorrect because zinc hydroxide dissolves in excess aqueous ammonia.
C (Aluminium chloride) is incorrect because chloride ions give a white precipitate with silver nitrate, not barium nitrate.
D (Calcium nitrate) is incorrect because calcium ions do not dissolve in excess sodium hydroxide.

In the condensation polymerisation to form a polyester:
- The diol loses \(-\text{H}\) from its \(-\text{OH}\) groups to become \(-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\).
- The dicarboxylic acid loses \(-\text{OH}\) from its \(-\text{COOH}\) groups to become \(-\text{C}(=\text{O})-\text{C}(=\text{O})-\).
- These link via ester linkages (\(-\text{COO}-\)) with the elimination of water molecules.
Joining these gives the repeat unit \(-[\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{CO}-\text{CO}]_n-\).

[1 mark] C is correct.
A is incorrect: represents an addition polymer or polyamide linkage.
B is incorrect: contains incorrect alternating ether and carbonyl arrangements.
D is incorrect: has missing oxygen atoms in the backbone.

During the electrolysis of aqueous copper(II) sulfate with inert graphite electrodes: at the cathode (negative electrode), \(\text{Cu}^{2+}\) ions are discharged in preference to \(\text{H}^+\) ions because copper is less reactive than hydrogen, forming a pink-brown/red-brown solid coating of copper metal: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). At the anode (positive electrode), \(\text{OH}^-\) ions are discharged in preference to \(\text{SO}_4^{2-}\) ions, producing colourless bubbles of oxygen gas: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\).

B [1 mark] - Correctly identifies colourless bubbles of oxygen gas at the anode and a pink-brown solid at the cathode.

1. Calculate moles of magnesium: \(\text{moles of Mg} = \frac{0.24\text{ g}}{24\text{ g/mol}} = 0.010\text{ mol}\).
2. According to the balanced equation, \(1\text{ mol of Mg}\) produces \(1\text{ mol of H}_2\), so \(\text{moles of H}_2 = 0.010\text{ mol}\).
3. Calculate volume of \(\text{H}_2\): \(\text{Volume} = 0.010\text{ mol} \times 24\text{ dm}^3\text{/mol} = 0.24\text{ dm}^3 = 240\text{ cm}^3\).

C [1 mark] - Accurate calculation of molar quantity and conversion to gas volume in \(\text{cm}^3\).

Increasing the temperature increases the average kinetic energy of the reacting particles. This shifts the energy distribution so that a significantly greater fraction of particles have kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)). While increasing concentration, pressure, or surface area increases collision frequency, only increasing temperature (and adding a catalyst by lowering \(E_a\)) alters the proportion of particles with \(E \ge E_a\).

D [1 mark] - Correctly identifies temperature as the factor changing the proportion of particles with energy \(\ge E_a\).

The polymer contains the amide linkage \(-\text{CO}-\text{NH}-\), which classifies it as a polyamide (such as nylon). Condensation polymerisation between a dicarboxylic acid and a diamine eliminates water (\(\text{H}_2\text{O}\)) as the small molecule byproduct.

A [1 mark] - Correct identification of polyamide and water elimination.

Test 1: Aqueous sodium hydroxide giving a green precipitate insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\).
Test 2: Dilute nitric acid followed by aqueous barium nitrate giving a white precipitate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, \(\text{X}\) is iron(II) sulfate, \(\text{FeSO}_4\).

B [1 mark] - Correct deduction of cation \(\text{Fe}^{2+}\) and anion \(\text{SO}_4^{2-}\).

The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle:
1. Lower temperature shifts the equilibrium in the exothermic direction (to the right), increasing the yield of \(\text{SO}_3\).
2. Higher pressure shifts the equilibrium towards the side with fewer moles of gas (3 moles of gas on the left produce 2 moles of gas on the right), shifting the equilibrium to the right and increasing the yield of \(\text{SO}_3\).
Therefore, high pressure and low temperature give the highest equilibrium yield.

C [1 mark] - Correct application of Le Chatelier's principle to temperature and pressure.

Amphoteric oxides (such as aluminium oxide, \(\text{Al}_2\text{O}_3\), and zinc oxide, \(\text{ZnO}\)) react with both acids and bases/alkalis to form a salt and water.

C [1 mark] - Correct definition/behaviour of amphoteric oxides.

1. Energy required to break bonds (endothermic): \(\text{Energy in} = 1(\text{H}-\text{H}) + 1(\text{Cl}-\text{Cl}) = 436 + 242 = +678\text{ kJ/mol}\).
2. Energy released by forming bonds (exothermic): \(\text{Energy out} = 2(\text{H}-\text{Cl}) = 2 \times 431 = 862\text{ kJ/mol}\).
3. Enthalpy change: \(\Delta H = \text{Energy in} - \text{Energy out} = 678 - 862 = -184\text{ kJ/mol}\).

A [1 mark] - Correct calculation of energy in, energy out, and final \(\Delta H\) including sign.

When copper electrodes are used in the electrolysis of aqueous copper(II) sulfate, the copper anode dissolves ( ext{Cu(s)} ightarrow ext{Cu}^{2+} ext{(aq)} + 2 ext{e}^-) resulting in a loss of mass at the anode. At the cathode, copper ions are discharged and deposited ( ext{Cu}^{2+} ext{(aq)} + 2 ext{e}^- ightarrow ext{Cu(s)}). Therefore, the concentration of ext{Cu}^{2+} in the solution remains constant and the blue colour does not fade.

C [1 mark]: The mass of the anode decreases as copper dissolves.

Increasing the temperature increases the average kinetic energy of the particles. Consequently, a significantly greater proportion of the colliding particles possess kinetic energy greater than or equal to the activation energy (\( E \ge E_a \)), leading to a higher frequency of successful collisions. The activation energy itself remains unchanged without a catalyst.

B [1 mark]: A greater proportion of colliding particles possess energy greater than or equal to the activation energy.

The alkene structural isomers with the molecular formula \( \text{C}_4\text{H}_8 \) are:
1. but-1-ene (\( \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3 \))
2. but-2-ene (\( \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \))
3. 2-methylprop-1-ene (\( \text{CH}_2=\text{C}(\text{CH}_3)_2 \))

Therefore, there are 3 structural alkene isomers.

B [1 mark]: 3

The polymer shown contains the \( -\text{CO}-\text{NH}- \) linkage, which is an amide linkage (as in nylon). It is formed by the condensation polymerisation of a dicarboxylic acid and a diamine, releasing water (\( \text{H}_2\text{O} \)) as the small molecule eliminated.

A [1 mark]: Amide linkage; \( \text{H}_2\text{O} \).

According to Avogadro's law, gas volumes at the same temperature and pressure are directly proportional to the number of moles.
Ratio of volumes:
\( \text{C}_x\text{H}_y : \text{O}_2 : \text{CO}_2 = 20 : 100 : 60 = 1 : 5 : 3 \)

The equation is:
\( 1\,\text{C}_x\text{H}_y + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + \frac{y}{2}\,\text{H}_2\text{O} \)

From carbon balance: \( x = 3 \).
From oxygen balance: \( 5 \times 2 = (3 \times 2) + \frac{y}{2} \implies 10 = 6 + \frac{y}{2} \implies \frac{y}{2} = 4 \implies y = 8 \).

The hydrocarbon is \( \text{C}_3\text{H}_8 \).

C [1 mark]: \( \text{C}_3\text{H}_8 \)

- Zinc ions (\( \text{Zn}^{2+} \)) form a white precipitate soluble in both excess aqueous sodium hydroxide and excess aqueous ammonia.
- Aluminium ions (\( \text{Al}^{3+} \)) form a white precipitate soluble in excess NaOH, but insoluble in excess ammonia.
- Iodide ions (\( \text{I}^- \)) form a yellow precipitate with aqueous silver nitrate acidified with dilute nitric acid.
Therefore, the compound is zinc iodide.

D [1 mark]: zinc iodide

- The forward reaction is exothermic (\( \Delta H < 0 \)), so decreasing temperature shifts the equilibrium to the right to produce more heat.
- There are 3 moles of gas on the left-hand side and 2 moles of gas on the right-hand side. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas (to the right).
Therefore, decreasing temperature and increasing pressure will both increase the equilibrium yield of \( \text{SO}_3 \).

A [1 mark]: decrease temperature and increase pressure

- Basic oxides react only with acids (e.g. dilute \( \text{HCl} \)), so \( \text{W} \) is basic.
- Acidic oxides react only with bases (e.g. aqueous \( \text{NaOH} \)), so \( \text{X} \) is acidic.
- Amphoteric oxides react with both acids and bases, so \( \text{Y} \) is amphoteric.
- Neutral oxides react with neither acids nor bases, so \( \text{Z} \) is neutral.

A [1 mark]: W is basic, X is acidic, Y is amphoteric, Z is neutral

According to kinetic particle theory, the rate of diffusion of a gas is inversely related to its relative molecular mass, \(M_{\text{r}}\). The gas with the lowest \(M_{\text{r}}\) has the highest average particle speed at a given temperature and thus diffuses the fastest.

- Carbon dioxide, \(\text{CO}_2\): \(M_{\text{r}} = 12 + (2 \times 16) = 44\)
- Methane, \(\text{CH}_4\): \(M_{\text{r}} = 12 + (4 \times 1) = 16\)
- Nitrogen, \(\text{N}_2\): \(M_{\text{r}} = 2 \times 14 = 28\)
- Sulfur dioxide, \(\text{SO}_2\): \(M_{\text{r}} = 32 + (2 \times 16) = 64\)

Methane has the smallest \(M_{\text{r}}\) (16) and therefore diffuses most rapidly.

B [1 mark] — identifies methane as having the lowest relative molecular mass (\(M_{\text{r}} = 16\)).

During the electrolysis of aqueous copper(II) sulfate with inert electrodes:

- At the cathode (negative electrode), copper ions (\(\text{Cu}^{2+}\)) are preferentially discharged over hydrogen ions because copper is less reactive than hydrogen: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\) (pink/brown solid formed).
- At the anode (positive electrode), hydroxide ions (\(\text{OH}^-\)) from water are discharged rather than sulfate ions, releasing oxygen gas: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\).
- As \(\text{Cu}^{2+}\) ions are removed from solution and not replenished by inert electrodes, the blue colour fades and the solution eventually turns colourless.

C [1 mark] — correctly identifies anode product as oxygen, cathode product as copper, and electrolyte turning colourless.

Under identical conditions of temperature and pressure, equal volumes of gases contain equal numbers of moles (Avogadro's law).

Volume ratio of reacting gases and products:
\(\text{C}_x\text{H}_y : \text{O}_2 : \text{CO}_2 = 20 : 100 : 60 = 1 : 5 : 3\)

General combustion equation:
\(\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\)

From the ratio:
1. Moles of \(\text{CO}_2\) per mole of hydrocarbon: \(x = 3\)
2. Moles of \(\text{O}_2\) per mole of hydrocarbon: \(x + \frac{y}{4} = 5 \implies 3 + \frac{y}{4} = 5 \implies y = 8\)

Therefore, the formula is \(\text{C}_3\text{H}_8\).

C [1 mark] — determines molar ratio 1 : 5 : 3 to find x = 3 and y = 8 giving \(\text{C}_3\text{H}_8\).

Enthalpy change \(\Delta H = \Sigma(\text{bond energies broken}) - \Sigma(\text{bond energies formed})\)

1. Energy absorbed to break bonds in reactants:
\(1 \times (\text{H–H}) + 1 \times (\text{Cl–Cl}) = 436 + 242 = +678\text{ kJ/mol}\)

2. Energy released by forming bonds in products:
\(2 \times (\text{H–Cl}) = 2 \times 431 = 862\text{ kJ/mol}\)

3. Overall enthalpy change:
\(\Delta H = +678 - 862 = -184\text{ kJ/mol}\)

A [1 mark] — calculates \(\Delta H = 678 - 862 = -184\text{ kJ/mol}\).

- Effect of temperature: The forward reaction is exothermic (\(\Delta H < 0\)). Lowering the temperature shifts the position of equilibrium in the exothermic direction (to the right), increasing the yield of \(\text{SO}_3\).
- Effect of pressure: The reaction involves 3 moles of gaseous reactants producing 2 moles of gaseous products. Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (to the right), increasing the yield of \(\text{SO}_3\).

Therefore, decreasing temperature and increasing pressure yields the most \(\text{SO}_3\).

B [1 mark] — recognises that low temperature favours exothermic direction and high pressure favours fewer moles of gas.

- Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is an amphoteric oxide because it reacts with both acids and alkalis to form salts.
- Calcium oxide (\(\text{CaO}\)) is a basic oxide because it is a metal oxide that reacts with acids to form salt and water.
- Sulfur dioxide (\(\text{SO}_2\)) is an acidic oxide because it is a non-metal oxide that dissolves in water to form an acid and reacts with bases.
- Carbon monoxide (\(\text{CO}\)) is a neutral oxide because it does not react with either acids or bases.

A [1 mark] — correctly assigns amphoteric (\(\text{Al}_2\text{O}_3\)), basic (\(\text{CaO}\)), acidic (\(\text{SO}_2\)), and neutral (\(\text{CO}\)).

The repeating unit contains the ester linkage \(-\text{COO}-\), identifying it as a polyester (polyethylene terephthalate, PET). Polyesters are synthetic condensation polymers manufactured by reacting a diol (such as ethane-1,2-diol) with a dicarboxylic acid (such as benzene-1,4-dicarboxylic acid), with the elimination of small molecules (water).

C [1 mark] — identifies the polymer as a polyester formed from a diol and a dicarboxylic acid.

- With \(\text{NaOH}\text{(aq)}\), \(\text{Al}^{3+}\) and \(\text{Zn}^{2+}\) both give a white precipitate soluble in excess.
- With \(\text{NH}_3\text{(aq)}\), \(\text{Zn}^{2+}\) gives a white precipitate soluble in excess, whereas \(\text{Al}^{3+}\) gives a white precipitate insoluble in excess. This confirms the cation is \(\text{Zn}^{2+}\).
- Acidified silver nitrate forming a yellow precipitate confirms the presence of iodide ions (\(\text{I}^-\)).

Hence, solid \(X\) is zinc iodide (\(\text{ZnI}_2\)).

D [1 mark] — identifies cation as \(\text{Zn}^{2+}\) (soluble in excess \(\text{NH}_3\)) and anion as \(\text{I}^-\).

An amphoteric oxide is an oxide that reacts with both acids and bases/alkalis. According to the syllabus, amphoteric oxides are aluminium oxide (\(\text{Al}_2\text{O}_3\)) and zinc oxide (\(\text{ZnO}\)).

aluminium oxide / zinc oxide [1]
ALLOW: \(\text{Al}_2\text{O}_3\) / \(\text{ZnO}\)
REJECT: basic oxides (e.g. magnesium oxide) / acidic oxides (e.g. sulfur dioxide)

In the Contact process, sulfur dioxide is converted to sulfur trioxide using vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) as a catalyst at approximately \(450\,^\circ\text{C}\).

vanadium(V) oxide [1]
ALLOW: vanadium pentoxide / \(\text{V}_2\text{O}_5\)
REJECT: vanadium / iron / nickel

Sulfur dioxide is a reducing agent that reduces purple manganate(VII) ions (\(\text{MnO}_4^-\)) to colourless manganese(II) ions (\(\text{Mn}^{2+}\)).

sulfur dioxide [1]
ALLOW: \(\text{SO}_2\)
REJECT: sulfate / sulfite

Both synthetic polyamides like nylon and natural condensation polymers like proteins contain the amide linkage (also known as a peptide link in proteins), represented as \(-\text{CONH}-\).

amide (link / linkage) [1]
ALLOW: peptide (link / linkage) / \(-\text{CONH}-\) / \(-\text{C}(=\text{O})\text{NH}-\)
REJECT: ester link

In concentrated aqueous sodium chloride (brine), chloride ions (\(\text{Cl}^-\)) are oxidised in preference to hydroxide ions at the anode, releasing chlorine gas (\(\text{Cl}_2\)).

chlorine [1]
ALLOW: \(\text{Cl}_2\)
REJECT: chloride / \(\text{Cl}^-\)

Natural gas consists predominantly of methane, which is an alkane with the chemical formula \(\text{CH}_4\).

methane [1]
ALLOW: \(\text{CH}_4\)
REJECT: propane / butane / petroleum

The reaction of ethanol (alcohol) with propanoic acid (carboxylic acid) produces the ester ethyl propanoate along with water: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\).

ethyl propanoate [1]
ALLOW: ethyl propionate
REJECT: propyl ethanoate

In Period 3, sodium, magnesium, and aluminium have metallic structures; silicon has a giant covalent macromolecular structure (like diamond); phosphorus, sulfur, and chlorine exist as simple molecules; and argon is monatomic.

silicon [1]
ALLOW: \(\text{Si}\)
REJECT: silicon dioxide / silica (question asks for an element)

Ammonia (\(\text{NH}_3\)) is an alkaline gas that dissolves in moisture on damp red litmus paper to produce hydroxide ions (\(\text{OH}^-\)), turning the indicator blue.

ammonia / \(\text{NH}_3\) [1]
IGNORE alkaline gas
REJECT ammonium / \(\text{NH}_4^+\)

Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is an amphoteric oxide because it reacts with both acids and bases to produce a salt and water. \(\text{CO}_2\) and \(\text{SO}_2\) are acidic oxides, while \(\text{Na}_2\text{O}\) and \(\text{CaO}\) are basic oxides.

aluminium oxide / \(\text{Al}_2\text{O}_3\) [1]

During the electrolysis of molten lead(II) bromide, bromide ions (\(\text{Br}^-\)) migrate to the positive anode, where they lose electrons (are oxidised) to form diatomic bromine (\(\text{Br}_2\)).

bromine / \(\text{Br}_2\) [1]
REJECT bromide / \(\text{Br}^-\)

Iron (\(\text{Fe}\)) is used as a finely divided solid catalyst in the Haber process to increase the rate of the reversible reaction between nitrogen and hydrogen at \(450\,^\circ\text{C}\) and \(200\text{ atm}\).

iron / \(\text{Fe}\) [1]
IGNORE finely divided / powdered
REJECT iron(III) oxide / \(\text{Fe}_2\text{O}_3\) / vanadium(V) oxide

Structural isomers are compounds that share identical molecular formulae (the same numbers and types of atoms) but differ in the spatial arrangement and connectivity of their covalent bonds.

structural isomers / isomers / structural isomerism [1]

Carbon monoxide (\(\text{CO}\)) is formed when carbon-containing fuels burn with an insufficient supply of oxygen. It binds irreversibly with haemoglobin, preventing normal oxygen transport in the bloodstream.

carbon monoxide / \(\text{CO}\) [1]
REJECT carbon dioxide / \(\text{CO}_2\)

Bauxite is the primary commercial ore of aluminium, containing impure hydrated aluminium oxide.

bauxite [1]
ALLOW phonetic spellings (e.g. baucsite)

Poly(ethene) is formed by addition polymerisation, where the \(\text{C}=\text{C}\) double bonds in unsaturated alkene monomers break to form long saturated polymer chains without eliminating small molecules.

addition (polymerisation) [1]
REJECT condensation (polymerisation)

1. Identify the reacting ions in solution that form the insoluble precipitate: aqueous iron(III) ions, \(\text{Fe}^{3+}(\text{aq})\), and aqueous hydroxide ions, \(\text{OH}^-(\text{aq})\).
2. Write the formula of the insoluble solid product: \(\text{Fe(OH)}_3(\text{s})\).
3. Balance the ionic charges and atoms: one \(\text{Fe}^{3+}\) ion requires three \(\text{OH}^-\prec\) ions.
4. Include state symbols for all species: \(\text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_3(\text{s})\).

M1: \(\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3\) [1]
(ALLOW stoichiometric multiples)

M2: correct state symbols \((\text{aq})\), \((\text{aq})\) and \((\text{s})\) [1]
(M2 is dependent on correct or near-miss formulae of species)

1. Carbon (Group IV / 14) has 4 valence electrons. It forms 4 single covalent bonds:
- Three \(\text{C–H}\) single bonds: each contains 1 shared pair of electrons (e.g. 1 dot from \(\text{C}\) and 1 cross from \(\text{H}\)).
- One \(\text{C–O}\) single bond: contains 1 shared pair of electrons (1 dot from \(\text{C}\) and 1 cross from \(\text{O}\)).

2. Oxygen (Group VI / 16) has 6 valence electrons. It forms 2 single covalent bonds and retains 2 lone pairs:
- One \(\text{O–H}\) single bond: contains 1 shared pair of electrons (1 cross from \(\text{O}\) and 1 dot/cross from \(\text{H}\)).
- Two lone pairs on the oxygen atom: represented as 4 unshared electrons (two pairs).

3. Each hydrogen atom has a complete duet (2 shared electrons) and carbon and oxygen each have a complete octet (8 outer-shell electrons).

• M1: Five single covalent bonding pairs shown correctly (3 × C–H, 1 × C–O, 1 × O–H) with no extra electrons on any H atom.
• M2: Carbon atom with complete octet and no lone pairs.
• M3 (0.5 marks): Oxygen atom with 4 non-bonding electrons (2 lone pairs) and a complete octet.
[ALLOW different symbols such as all dots, all crosses, or circles/crosses as long as the sharing is clear]
[REJECT any double bonds or extra lone pairs on carbon or hydrogen]

1. A magnesium atom (Group II, electronic configuration 2,8,2) loses its 2 outer-shell electrons to form a magnesium cation, \(\text{Mg}^{2+}\).
2. Two fluorine atoms (Group VII, electronic configuration 2,7) each gain 1 electron into their outer shell to form two fluoride anions, \(\text{F}^-\).
3. In the diagram:
- The magnesium ion is drawn with a charge of \(2+\) (with either an empty valence shell shown or 8 outer electrons, in square brackets).
- Two separate fluoride ions (or a single \(\text{F}^-\label{}\) ion preceded by the coefficient '2') are drawn with 8 outer electrons (7 of one symbol, 1 transferred electron of the other symbol) enclosed in square brackets with a \(1-\label{}\) (or \(-\)) charge.

• M1: Correct charge and outer shell for the magnesium ion (\(\text{Mg}^{2+}\)) with 0 or 8 outer electrons.
• M2: Correct charge and 8 outer-shell electrons on the fluoride ion (\(\text{F}^-\)) showing 1 transferred electron distinctly.
• M3 (0.5 marks): Correct ratio of ions shown (one \(\text{Mg}^{2+}\) to two \(\text{F}^-\), either drawn separately or with a coefficient of 2 in front of \(\text{F}^-\)).
[ALLOW all dots or all crosses if charges and numbers of ions are correct]
[REJECT covalent shared pairs or incorrect ion charges]

(a) A displayed formula must show every single and double bond and all atoms explicitly:
- Ethanoate part: \(\text{H}_3\text{C–C}(=\text{O})–\) with three \(\text{C–H}\) bonds on the terminal carbon, a \(\text{C–C}\) single bond to the carbonyl carbon, and a \(\text{C=O}\) double bond.
- Ester bridge: \(\text{–C–O–C–}\) single bond linking the carbonyl carbon to the alkyl oxygen.
- Propyl part: \(\text{–CH}_2\text{–CH}_2\text{–CH}_3\) with fully drawn \(\text{C–C}\) and \(\text{C–H}\) bonds on all three carbons.

(b) Esterification reaction:
\(\text{Ethanoic acid} + \text{propan-1-ol} \rightleftharpoons \text{propyl ethanoate} + \text{water}\)
- Carboxylic acid: ethanoic acid
- Alcohol: propan-1-ol (or propanol)

• (a) M1: Correct ester linkage drawn showing \(\text{C=O}\) and \(\text{C–O}\) bonds explicitly.
• (a) M2 (0.5 marks): Entire structure fully displayed showing all atoms and all bonds without grouping (e.g. no \(\text{CH}_3\) or \(\text{C}_3\text{H}_7\) abbreviations).
• (b) M1 (0.5 marks): Ethanoic acid.
• (b) M2 (0.5 marks): Propan-1-ol / propanol.
[REJECT (a) if any bond is omitted or grouped as a condensed formula]

1. During condensation polymerisation of a dicarboxylic acid and a diamine:
- Each \(\text{–COOH}\) group loses an \(\text{–OH}\) group.
- Each \(\text{–NH}_2\) group loses an \(\text{–H}\) atom.
- A molecule of water (\(\text{H}_2\text{O}\)) is eliminated for each amide (peptide) bond formed.

2. The amide linkage has the structure \(\text{–C(=O)–NH–}\) (or \(\text{–CONH–}\)).

3. One repeat unit consists of one diamine residue and one dicarboxylic acid residue connected by an amide bond:
\(\text{–NH–(CH}_2\text{)}_6\text{–NH–C(=O)–(CH}_2\text{)}_4\text{–C(=O)–}\)
with open continuation bonds extending from the left-hand nitrogen and the right-hand carbonyl carbon.

• M1: Correct amide linkage shown with \(\text{C=O}\) and \(\text{N–H}\) (\(\text{–C(=O)–NH–}\) or \(\text{–CONH–}\)).
• M2: Correct carbon chain blocks: \(\text{–(CH}_2\text{)}_6\text{–}\) (or six carbons) attached to nitrogen atoms and \(\text{–(CH}_2\text{)}_4\text{–}\) (or four carbons) attached to carbonyl carbons.
• M3 (0.5 marks): Open continuation bonds extending from both ends (from N at one end and C=O at the other end).
[ALLOW displayed or structural representation of the hydrocarbon chains \(\text{–(CH}_2\text{)}_6\text{–}\) and \(\text{–(CH}_2\text{)}_4\text{–}\)]
[REJECT inclusion of terminal \(\text{–H}\) or \(\text{–OH}\) groups on the continuation bonds]

When temperature is increased, particles gain thermal energy which is converted into kinetic energy, meaning particles move faster. This causes collisions between reactant particles to occur more frequently (more collisions per unit time). Furthermore, a significantly greater proportion of the colliding particles possess energy equal to or greater than the activation energy (\(E \ge E_a\)), resulting in a higher rate of successful (effective) collisions.

M1: Particles gain kinetic energy / move faster [1]
M2: Higher frequency of collisions / more collisions per unit time [1] (REJECT: more collisions unqualified)
M3: Greater proportion / fraction of particles (or collisions) have energy greater than or equal to activation energy / \(E \ge E_a\) [1]

Increasing the concentration of nitric acid means that there are more hydrogen ions / acid particles in a given volume (greater number of particles per unit volume). As the particles are closer together, the frequency of collisions between acid particles and calcium carbonate increases, leading to more successful collisions per second.

M1: More particles per unit volume / more particles per \(\text{cm}^3\) / particles are crowded closer together [1] (REJECT: just 'more particles')
M2: More collisions per second / higher frequency of collisions [1]

Manganese(IV) oxide acts as a catalyst. It provides an alternative reaction pathway that has a lower activation energy (\(E_a\)). Because the activation energy barrier is reduced, a greater fraction of colliding particles have kinetic energy \(E \ge E_a\), which increases the number of successful (effective) collisions per unit time.

M1: Provides an alternative reaction pathway with a lower activation energy / lowers the activation energy [1]
M2: Greater proportion / percentage / fraction of particles / collisions have energy \(\ge E_a\) [1]
M3: Greater frequency of successful collisions / more successful collisions per unit time [1]

Zinc powder has a much larger surface area to volume ratio compared to large granules of zinc. This exposes more zinc atoms to the acid solution, resulting in an increased collision frequency (more collisions per unit time) between reactant particles, thereby increasing the rate of reaction.

M1: (Powder has a) greater / larger surface area (per unit mass) / more particles exposed [1]
M2: Higher collision frequency / more collisions per second / more collisions per unit time [1] (IGNORE: more successful collisions unqualified)

Increasing the pressure of a gaseous system decreases the volume available, packing the gas molecules closer together. This increases the concentration of gas particles (more particles per unit volume), causing reactant molecules to collide more frequently (higher rate of collisions per second).

M1: Particles are closer together / more particles per unit volume / more particles in a given space [1] (REJECT: particles move faster)
M2: Increased frequency of collisions / more collisions per unit time [1]

1. Calculate the moles of each element present in the sample:
- Moles of C = \(\frac{4.80}{12} = 0.40\text{ mol}\)
- Moles of H = \(\frac{0.80}{1} = 0.80\text{ mol}\)
- Moles of O = \(\frac{6.40}{16} = 0.40\text{ mol}\)

2. Determine the simplest whole-number molar ratio:
- \(\text{C} : \text{H} : \text{O} = \frac{0.40}{0.40} : \frac{0.80}{0.40} : \frac{0.40}{0.40} = 1 : 2 : 1\)
- Empirical formula = \(\text{CH}_2\text{O}\)

3. Determine the molecular formula:
- Empirical formula mass = \(12 + (2 \times 1) + 16 = 30\)
- Ratio = \(\frac{M_r}{\text{empirical mass}} = \frac{60}{30} = 2\)
- Molecular formula = \(2 \times \text{CH}_2\text{O} = \text{C}_2\text{H}_4\text{O}_2\)

M1: Moles of all three elements correctly evaluated [1]
(C = 0.40, H = 0.80, O = 0.40)

M2: Correct empirical formula: \(\text{CH}_2\text{O}\) [1]

M3: Correct molecular formula: \(\text{C}_2\text{H}_4\text{O}_2\) [1]
(ALLOW ECF for M3 based on an incorrect empirical formula mass compared against \(M_r = 60\))

1. Calculate the number of moles of \(\text{Li}_2\text{CO}_3\):
\(\text{moles of } \text{Li}_2\text{CO}_3 = \frac{\text{mass}}{M_r} = \frac{3.70}{74} = 0.050\text{ mol}\)

2. Determine the moles of \(\text{CO}_2\) produced using stoichiometry:
According to the equation, \(1\text{ mol } \text{Li}_2\text{CO}_3 \rightarrow 1\text{ mol } \text{CO}_2\).
\(\text{moles of } \text{CO}_2 = 0.050\text{ mol}\)

3. Calculate the volume of \(\text{CO}_2\) in \(\text{cm}^3\):
\(\text{volume} = 0.050\text{ mol} \times 24\,000\text{ cm}^3/\text{mol} = 1200\text{ cm}^3\)

M1: Moles of \(\text{Li}_2\text{CO}_3 = \frac{3.70}{74} = 0.050\text{ mol}\) [1]

M2: Moles of \(\text{CO}_2 = 0.050\text{ mol}\) (stated or implied by calculation) [1]

M3: Volume of \(\text{CO}_2 = 1200\text{ cm}^3\) [1]
(ALLOW \(1.2\text{ dm}^3\) only if the unit \(\text{dm}^3\) is explicitly written to replace \(\text{cm}^3\); ALLOW ECF from incorrect moles calculated in M1)

1. Calculate the theoretical maximum mass of \(\text{FeS}\) produced:
- \(M_r(\text{FeS}) = 56 + 32 = 88\)
- Moles of \(\text{Fe} = \frac{5.60}{56} = 0.100\text{ mol}\)
- Moles of \(\text{FeS}\) expected = \(0.100\text{ mol}\)
- Theoretical mass of \(\text{FeS} = 0.100 \times 88 = 8.80\text{ g}\)

2. Calculate the percentage yield:
- \(\text{Percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100\)
- \(\text{Percentage yield} = \frac{7.48}{8.80} \times 100 = 85.0\%\)

M1: Calculates theoretical mass of \(\text{FeS} = 8.80\text{ g}\) OR moles of \(\text{Fe} = 0.100\text{ mol}\) AND actual moles of \(\text{FeS} = \frac{7.48}{88} = 0.085\text{ mol}\) [1]

M2: Correct expression for percentage yield: \(\frac{7.48}{8.80} \times 100\) OR \(\frac{0.085}{0.100} \times 100\) [1]

M3: \(85.0\%\) or \(85\%\) [1]
(ALLOW ECF for M2 and M3 from arithmetic error in M1)

A volumetric pipette is specifically designed to measure and deliver a fixed, accurate volume of liquid (such as \(25.0\text{ cm}^3\)) during a titration procedure.

M1: (volumetric) pipette [1]
ALLOW: transfer pipette
IGNORE: burette / measuring cylinder
REJECT: dropper / beaker

When heating stops, the air and gases inside the boiling tube cool down and contract, creating a partial vacuum. If the delivery tube remains submerged, cold water will be sucked back into the hot glass tube, causing it to crack or shatter.

M1: to prevent suck-back (of cold water into the hot tube / prevent tube cracking) [1]
ALLOW: to prevent water being drawn back into the hot glassware
IGNORE: to keep water clean
REJECT: to prevent pressure build-up

A gas syringe is standard laboratory apparatus used to collect and measure volumes of evolving gases during rate investigations.

M1: gas syringe [1]
ALLOW: (inverted) measuring cylinder (filled with water) / inverted burette (filled with water) / graduated gas tube
IGNORE: delivery tube alone
REJECT: test-tube / beaker / conical flask

Nitrogen dioxide (\(\text{NO}_2\)) is a toxic and respiratory irritant gas. Experiments releasing harmful fumes must be carried out in a fume cupboard (fume hood) to prevent inhalation.

M1: use a fume cupboard / fume hood / work in a well-ventilated area [1]
ALLOW: perform the experiment outdoors / open all windows
IGNORE: wear gloves / tie back long hair / wear safety goggles (excluded by question stem)
REJECT: wear a dust mask

(a) Plot the volume of \(\text{KOH(aq)}\) on the \(x\)-axis and temperature on the \(y\)-axis. Points from \(0.0\text{ cm}^3\) to \(25.0\text{ cm}^3\) form an ascending linear trend (gradient \(= +0.48\,^\circ\text{C}/\text{cm}^3\)). Points from \(25.0\text{ cm}^3\) to \(40.0\text{ cm}^3\) form a descending linear trend (gradient \(= -0.24\,^\circ\text{C}/\text{cm}^3\)). Using a ruler, draw two straight lines of best fit.

(b)(i) The point of intersection of the two lines gives the equivalence point: \(\text{Volume} = 25.0\text{ cm}^3\).
(b)(ii) The temperature coordinate at the intersection point gives the maximum temperature reached: \(T = 32.0\,^\circ\text{C}\).
Tie lines / construction marks must be visible on the grid.

(c) Extending the cooling/dilution line (gradient \(= -0.24\,^\circ\text{C}/\text{cm}^3\)) to \(V = 50.0\text{ cm}^3\):
$$T = 32.0 - 0.24 \times (50.0 - 25.0) = 32.0 - 6.0 = 26.0\,^\circ\text{C}$$

(a) [1.5 marks]
- M1: All points plotted accurately to within half a small square [1]
- M2: Two straight lines of best fit drawn using a ruler, intersecting cleanly at a single point [0.5]
(REJECT if drawn freehand or single curved line drawn)

(b) [1.0 mark]
- (i) Volume read correctly from graph intersection \(= 25.0\text{ cm}^3\) [0.5]
- (ii) Temperature read correctly from graph intersection \(= 32.0\,^\circ\text{C}\) AND tie lines shown on graph [0.5]

(c) [1.0 mark]
- Extrapolation line shown extending to \(50.0\text{ cm}^3\) AND temperature read correctly as \(26.0\,^\circ\text{C}\) [1]
(ALLOW \(25.8\text{--}26.2\,^\circ\text{C}\); ECF from candidate's plotted line)

(a) Plot time on the horizontal axis ($0$ to $140\text{ s}$) and volume on the vertical axis ($0$ to $60\text{ cm}^3$). Plot all 8 points. Draw a single smooth curve passing through the valid points and starting from the origin \((0,0)\).

(b) The point at \(t = 80\text{ s}\) has a recorded volume of \(35\text{ cm}^3\), which falls significantly below the curve trend. Circling this point and reading the value on the curve of best fit at \(t = 80\text{ s}\) gives \(46\text{ cm}^3\).

(c)(i) Find $50\text{ s}$ on the horizontal axis, draw a vertical line up to the curve, and read across to the vertical axis: \(\text{Volume} = 37\text{ cm}^3\).
(c)(ii) The reaction is complete when the curve first becomes completely horizontal / flat: \(t = 120\text{ s}\).

(a) [1.5 marks]
- M1: 7 or 8 points plotted correctly to within half a small square [1]
- M2: Smooth curve of best fit drawn ignoring the anomalous point, starting at \((0,0)\) [0.5]
(REJECT thick/multiple lines or straight point-to-point lines)

(b) [1.0 mark]
- M1: Point at \((80, 35)\) circled / identified [0.5]
- M2: Expected volume read from candidate's curve at \(t = 80\text{ s}\) \(= 46\text{ cm}^3\) (ALLOW \(45\text{--}47\text{ cm}^3\)) [0.5]

(c) [1.0 mark]
- (i) Volume at \(50\text{ s}\) read correctly from graph (\(37\text{ cm}^3\), range \(36\text{--}38\text{ cm}^3\)) AND tie line visible on grid [0.5]
- (ii) Time at which reaction finished read correctly (\(120\text{ s}\), range \(115\text{--}125\text{ s}\)) [0.5]

1. Addition of aqueous \(\text{NaOH}\) producing a green precipitate insoluble in excess indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)).
2. Addition of dilute nitric acid followed by aqueous silver nitrate producing a yellow precipitate confirms the presence of iodide ions (\(\text{I}^{-}\)).

• Cation: iron(II) / \(\text{Fe}^{2+}\) [1]
• Anion: iodide / \(\text{I}^{-}\) [0.5]
(Total: 1.5 marks)
ALLOW: \(\text{Fe(II)}\) for iron(II)
REJECT: iron / \(\text{Fe}^{3+}\) / iodine / \(\text{I}_2\)

(a) The pungent alkaline gas that turns damp red litmus paper blue is ammonia (\(\text{NH}_3\)).
(b) The formation of ammonia when a substance is heated with aqueous sodium hydroxide and aluminium foil is the diagnostic test for the nitrate ion (\(\text{NO}_3^-\)).

(a) ammonia / \(\text{NH}_3\) [0.5]
(b) nitrate / \(\text{NO}_3^-\)
ALLOW: \(\text{NO}_3\) [1]
REJECT: ammonium for (b)

(a) Copper(II) ions (\(\text{Cu}^{2+}\)) react with aqueous ammonia to give a light blue precipitate of copper(II) hydroxide, which dissolves in excess ammonia to form a characteristic deep/dark blue solution.
(b) The flame test for copper(II) ions produces a distinctive blue-green colour.

(a) copper(II) / \(\text{Cu}^{2+}\) [1]
(b) blue-green [0.5]
ALLOW: green-blue / green for flame colour
REJECT: copper / \(\text{Cu}^{+}\) for (a)

(a) An orange-red flame indicates the presence of calcium cations (\(\text{Ca}^{2+}\)).
(b) Effervescence with acid producing a gas that turns limewater milky confirms carbon dioxide gas (\(\text{CO}_2\)), which indicates the presence of carbonate anions (\(\text{CO}_3^{2-}\)) or hydrogencarbonate.

(a) calcium / \(\text{Ca}^{2+}\) [0.5]
(b) Gas: carbon dioxide / \(\text{CO}_2\) [0.5] AND Anion: carbonate / \(\text{CO}_3^{2-}\) / hydrogencarbonate / \(\text{HCO}_3^-\)
ALLOW: \(\text{CO}_3\) [0.5]

1. Add aqueous ammonia dropwise and then in excess to separate samples of each solution.
2. With aqueous zinc sulfate, a white precipitate forms that dissolves in excess aqueous ammonia to give a colourless solution.
3. With aqueous aluminium sulfate, a white precipitate forms that is insoluble in excess aqueous ammonia.

• Add aqueous ammonia dropwise and then in excess [0.5]
• Zinc sulfate: white precipitate dissolves / forms a colourless solution in excess [0.5]
• Aluminium sulfate: white precipitate does not dissolve / is insoluble in excess [0.5]

(a) The standard qualitative test for sulfur dioxide gas (\(\text{SO}_2\)) is the decolourisation of acidified aqueous potassium manganate(VII) from purple to colourless.
(b) Sulfur dioxide acts as a reducing agent (it reduces \(\text{MnO}_4^-\text{(aq)}\) to \(\text{Mn}^{2+}\text{(aq)}\)).

(a) sulfur dioxide / \(\text{SO}_2\) [1]
(b) reducing agent / reductant [0.5]
ALLOW: it causes reduction / it is oxidised
REJECT: oxidising agent

(a) The formation of a red-brown precipitate with aqueous sodium hydroxide insoluble in excess confirms the presence of iron(III) ions (\(\text{Fe}^{3+}\)).
(b) The formation of a white precipitate upon adding dilute nitric acid followed by aqueous silver nitrate confirms the presence of chloride ions (\(\text{Cl}^{-}\)).

(a) iron(III) / \(\text{Fe}^{3+}\) [0.75]
(b) chloride / \(\text{Cl}^{-}\) [0.75]
ALLOW: \(\text{Fe(III)}\) for iron(III)
REJECT: iron / \(\text{Fe}^{2+}\) / chlorine / \(\text{Cl}_2\)

(a) Chlorine gas (\(\text{Cl}_2\)) bleaches damp litmus paper white (turning it briefly red first due to acidic hypochlorous/hydrochloric acid formation).
(b) Hydrogen gas (\(\text{H}_2\)) produces a characteristic 'pop' sound with a lighted splint.

(a) chlorine / \(\text{Cl}_2\) [0.75]
(b) hydrogen / \(\text{H}_2\) [0.75]
REJECT: chloride / \(\text{Cl}^-\); hydride / \(\text{H}^+\)

A suitable plan includes:

1. Apparatus:
- Balance (to measure masses)
- Measuring cylinder (or burette/pipette, to measure acid volume)
- Conical flask and cotton wool plug (to allow gas to escape while preventing liquid spray loss)
- Stopwatch / timer (optional, to monitor completion)

2. Procedure:
- Use a balance to weigh out a known mass of crushed eggshells from Feed X (e.g. $5.0\text{ g}$).
- Use a measuring cylinder to measure a known volume of excess dilute hydrochloric acid (e.g. $50\text{ cm}^3$) and place it in a conical flask.
- Place the cotton wool plug in the neck and measure the total initial mass of the flask, acid, cotton wool, and eggshell sample on the balance.
- Add the eggshell sample to the acid, immediately replace the cotton wool plug, and allow the mixture to react.
- Swirl the flask and wait until effervescence stops / until the mass reading becomes constant on the balance.
- Record the final mass of the flask and its contents.
- Calculate the mass of carbon dioxide lost: $\text{mass loss} = \text{initial mass} - \text{final mass}$.
- Repeat the identical procedure using the crushed eggshells from Feed Y.

3. Fair Testing / Controlled Variables:
- Same initial mass of eggshell sample.
- Same volume and concentration of dilute hydrochloric acid.
- Same degree of crushing / particle size of eggshell.

4. Conclusion / Comparison:
- Compare the mass of $\text{CO}_2$ lost for both samples.
- The sample that produces the greater loss in mass evolved more $\text{CO}_2$, and therefore contains a higher percentage of calcium carbonate.

Award [1 mark] for each of the following marking points (up to a maximum of 6 marks):

- M1 (Weighing sample): Weighs a specified / known mass of crushed eggshell using a balance.
- M2 (Acid measurement): Measures a specified / known volume of dilute hydrochloric acid using a measuring cylinder / pipette / burette.
- M3 (Initial mass measurement): Measures the initial total mass of the flask, acid, cotton wool, and eggshell sample before reaction (or flask + acid + cotton wool and adds known mass of eggshell).
- M4 (Reaction completion & final mass): Mixes reactants, allows reaction to finish (until bubbling ceases / mass is constant), and records the final mass.
- M5 (Fair testing / control): Repeats the experiment using the other eggshell sample keeping at least one relevant variable constant (e.g. mass of eggshell / volume of acid / concentration of acid / temperature).
- M6 (Calculation & conclusion): Calculates mass loss ($=\text{initial mass} - \text{final mass}$) and states that the sample with the greater loss in mass contains the higher percentage / mass of calcium carbonate.