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

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

2. Calculate the moles of sulfuric acid reacted:
\(\text{Moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Moles of H}_2\text{SO}_4 = 0.0500\text{ mol/dm}^3 \times \frac{18.8}{1000}\text{ dm}^3 = 0.000940\text{ mol}\)

3. Use the stoichiometric ratio (1 mole of acid reacts with 2 moles of base) to find the moles of sodium hydroxide:
\(\text{Moles of NaOH} = 2 \times 0.000940\text{ mol} = 0.00188\text{ mol}\)

4. Calculate the concentration of the sodium hydroxide solution:
\(\text{Concentration of NaOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00188\text{ mol}}{0.0250\text{ dm}^3} = 0.0752\text{ mol/dm}^3\)

[1 mark] B is correct.
Award 1 mark for the correct calculation of concentration based on 1:2 acid-to-base stoichiometry.

1. The reaction of the cation with aqueous ammonia produces a green precipitate that is insoluble in excess. This is the characteristic test for iron(II) ions, \(\text{Fe}^{2+}\).
2. The reaction of the anion with dilute nitric acid followed by aqueous barium nitrate produces a white precipitate of barium sulfate. This is the characteristic test for sulfate ions, \(\text{SO}_4^{2-}\).
3. Therefore, the ionic compound is iron(II) sulfate.

[1 mark] C is correct.
Award 1 mark for identifying the cation as iron(II) and the anion as sulfate.

Terylene is a polyester formed by condensation polymerisation of a dicarboxylic acid and a diol. The reaction between carboxyl groups (\(-\text{COOH}\)) and alcohol groups (\(-\text{OH}\)) forms ester linkages (\(-\text{COO}-\)) with the elimination of a molecule of water (\(\text{H}_2\text{O}\)) for each linkage formed.

[1 mark] C is correct.
Award 1 mark for identifying that Terylene contains ester linkages and that water is eliminated during its condensation polymerisation.

1. Calculate the number of moles of \(\text{H}_2\text{SO}_4\) used: \(\text{moles} = \text{concentration} \times \text{volume in dm}^3 = 0.0500\text{ mol/dm}^3 \times (15.0 / 1000)\text{ dm}^3 = 0.00075\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation: 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{KOH}\). Therefore, \(\text{moles of KOH} = 2 \times 0.00075\text{ mol} = 0.00150\text{ mol}\). 3. Calculate the concentration of \(\text{KOH}\): \(\text{concentration} = \text{moles} / \text{volume in dm}^3 = 0.00150\text{ mol} / (20.0 / 1000)\text{ dm}^3 = 0.0750\text{ mol/dm}^3\).

Correct option is B. 1 mark for the correct mathematical calculation of the concentration of KOH based on the stoichiometric ratio.

In Test 1, the formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). Chromium(III) ions, \(\text{Cr}^{3+}\), also form a green precipitate, but it is soluble in excess sodium hydroxide to give a green solution. In Test 2, the addition of dilute nitric acid followed by aqueous barium nitrate forms a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these results, salt X is iron(II) sulfate.

Correct option is D. 1 mark for identifying both the iron(II) cation and the sulfate anion correctly based on the analytical test observations.

A polyester is formed by a condensation polymerization reaction between a diol and a dicarboxylic acid, eliminating water molecules. To find the monomers, we break the ester linkages (\(-O-CO-\)) and add water components: - The diol part \(-O-CH_2-O-\) becomes \(HO-CH_2-OH\). - The dicarboxylic acid part \(-CO-CH_2-CO-\) becomes \(HOOC-CH_2-COOH\). Thus, the correct monomers are \(HO-CH_2-OH\) and \(HOOC-CH_2-COOH\).

Correct option is A. 1 mark for identifying the correct diol and dicarboxylic acid structures that form the given repeating unit.

1. Calculate the number of moles of \(\text{NaOH}\) used: \(\text{moles of NaOH} = \frac{20.0}{1000} \times 0.150 = 0.00300\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation (\(2\text{ mol of NaOH} : 1\text{ mol of H}_2\text{SO}_4\)) to find the moles of \(\text{H}_2\text{SO}_4\): \(\text{moles of H}_2\text{SO}_4 = \frac{0.00300}{2} = 0.00150\text{ mol}\). 3. Calculate the concentration of the sulfuric acid: \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00150}{25.0 / 1000} = 0.0600\text{ mol/dm}^3\).

Award 1 mark for the correct option A. Deduct 1 mark for incorrect stoichiometric ratio (which leads to B). Deduct 1 mark for multiplying moles of NaOH by 2 instead of dividing by 2 (which leads to D).

The polymer shown is poly(chloroethene), which has a repeating carbon-carbon single bond backbone with chlorine and hydrogen atoms attached. This is an addition polymer formed from the monomer chloroethene (\(\text{CH}_2=\text{CHCl}\)). Addition polymerization does not involve condensation, the polymer is non-biodegradable, and the empirical formula of an addition polymer is identical to that of its monomer.

Award 1 mark for the correct option B. Option A is incorrect because it is an addition polymer. Option C is incorrect because addition polymers are generally non-biodegradable. Option D is incorrect because the empirical formula is unchanged during addition polymerization.

A more reactive metal displaces a less reactive metal from its oxide. Metal \(X\) reacts with the oxides of \(W\), \(Y\), and \(Z\), meaning \(X\) is the most reactive. Metal \(Z\) reacts with the oxides of \(W\) and \(Y\) but not \(X\), meaning \(Z\) is more reactive than \(W\) and \(Y\). Metal \(W\) reacts with the oxide of \(Y\) but not \(X\) or \(Z\), meaning \(W\) is more reactive than \(Y\). Metal \(Y\) does not react with any of the oxides, making it the least reactive. Therefore, the order of reactivity from most to least reactive is \(X \rightarrow Z \rightarrow W \rightarrow Y\).

Award 1 mark for the correct option A. Option B is the reverse order (least to most reactive). Options C and D have incorrect orderings of intermediate reactivities.

To ensure the accuracy of a titration: 1. The pipette must be rinsed with distilled water to clean it, and then with the solution it is measuring (aqueous sodium hydroxide). This ensures that any remaining water droplets do not dilute the sodium hydroxide aliquot, which would change the number of moles transferred. Therefore, statement A is correct. 2. The burette must be rinsed with distilled water and then with the solution it will contain (dilute nitric acid) to prevent dilution. Rinsing with distilled water only would leave droplets of water that dilute the acid, leading to an artificially high titre. Thus, statement B is incorrect. 3. The conical flask should only be rinsed with distilled water. Rinsing it with sodium hydroxide (statement C) or nitric acid (statement D) would introduce unknown extra amounts of reactants, making the titration calculation inaccurate.

[1 mark] for A. A is correct because rinsing the pipette with the solution it will measure prevents dilution of the measured volume. B is incorrect because the burette must be rinsed with the acid to prevent dilution. C and D are incorrect because the conical flask must only be rinsed with distilled water to avoid introducing extra moles of reactants.

1. Test for anions: The addition of dilute nitric acid followed by aqueous barium nitrate is the standard test for sulfate ions, \(SO_4^{2-}\). The white precipitate formed is barium sulfate, \(BaSO_4\). This rules out chloride ions, which would give a white precipitate with silver nitrate rather than barium nitrate. 2. Test for cations: The reaction with aqueous ammonia to give a light blue precipitate that dissolves in excess to form a deep blue solution is characteristic of copper(II) ions, \(Cu^{2+}\). Iron(II) ions would give a green precipitate that is insoluble in excess ammonia. Combining these results, the ions present in the mixture are \(Cu^{2+}\) and \(SO_4^{2-}\).

[1 mark] for B. B is correct because barium nitrate forms a white precipitate with sulfate ions and excess ammonia forms a deep blue solution with copper(II) ions. A is incorrect because chloride ions do not form a precipitate with barium nitrate. C and D are incorrect because iron(II) forms a green precipitate with ammonia, which does not dissolve to form a deep blue solution.

1. Type of reaction: The reaction involves two different monomers, each containing two functional groups (a diol and a dicarboxylic acid). When these react, a molecule of water is eliminated for each linkage formed. This is a condensation polymerization reaction. Addition polymerization occurs with monomers containing carbon-carbon double bonds and does not eliminate small molecules. 2. Type of linkage: The reaction of an alcohol group (\(-OH\)) with a carboxylic acid group (\(-COOH\)) forms an ester linkage (\(-COO-\)). An amide linkage (\(-CONH-\)) is formed between amine and carboxylic acid groups.

[1 mark] for D. D is correct because reacting a diol with a dicarboxylic acid eliminates water (condensation) and forms ester groups (ester linkage). A and B are incorrect because addition polymerization requires unsaturated monomer units. C is incorrect because an amide linkage requires an amine reactant.

If the pipette is rinsed with distilled water and not the sodium hydroxide solution, the remaining water droplets dilute the sodium hydroxide. As a result, fewer moles of sodium hydroxide are transferred to the conical flask. Consequently, a smaller volume of hydrochloric acid is required for neutralisation. Since the calculation assumes the full volume of undiluted sodium hydroxide was used, the calculated concentration of sodium hydroxide will be lower than the true concentration (an underestimate). Rinsing the conical flask with alkali (Option A) or rinsing the burette with water (Option B) would cause an overestimate of the concentration. Reading the meniscus consistently at the top (Option D) does not affect the measured titre volume.

1 mark for the correct option C.

The polymer contains ester links (\(-\text{O}-\text{CO}-\)) in the main chain, which indicates that it is a polyester formed by condensation polymerisation. During condensation polymerisation, a diol and a dicarboxylic acid react with the elimination of water molecules. The repeating unit contains a two-carbon block (from a diol) and a three-carbon block with two carbonyl groups (from a dicarboxylic acid). Therefore, the monomers are the diol \(\text{HOCH}_2\text{CH}_2\text{OH}\) (ethane-1,2-diol) and the dicarboxylic acid \(\text{HOOCCH}_2\text{COOH}\) (propanedioic acid).

1 mark for the correct option A.

Iron does not react with cold water but reacts with steam to form iron oxide and hydrogen gas. Iron oxide can be reduced by heating with carbon because carbon is more reactive than iron. Iron is above copper in the reactivity series, so it displaces copper from copper(II) sulfate, forming a pink-brown deposit of copper. Sodium reacts violently with cold water, magnesium is more reactive than carbon so its oxide cannot be reduced by carbon, and copper does not react with steam or dilute acid.

1 mark for the correct option B.

1. Find the number of moles of sulfuric acid reacted: \(n(\text{H}_2\text{SO}_4) = 0.100\text{ mol/dm}^3 \times \frac{12.50}{1000}\text{ dm}^3 = 0.00125\text{ mol}\). 2. Use the stoichiometric ratio from the balanced chemical equation to find moles of sodium hydroxide: From the equation, 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{NaOH}\). Thus, \(n(\text{NaOH}) = 2 \times 0.00125\text{ mol} = 0.00250\text{ mol}\). 3. Calculate the concentration of the sodium hydroxide solution: \(c(\text{NaOH}) = \frac{n}{V} = \frac{0.00250\text{ mol}}{0.0250\text{ dm}^3} = 0.100\text{ mol/dm}^3\).

Award 1 mark for the correct option B. Reject option A (which results from using an incorrect 1:1 mole ratio) and options C or D (which result from incorrect stoichiometric division/multiplication or volume calculations).

1. Analyze the cation test: Reacting the aqueous solution with sodium hydroxide produces a green precipitate that is insoluble in excess. This confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). Note that chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to give a green solution.
2. Analyze the anion test: Reacting the acidified solution with aqueous barium nitrate produces a white precipitate of barium sulfate, \(\text{BaSO}_4\). This confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
3. Combine both ions: The identity of the solid is iron(II) sulfate.

Award 1 mark for the correct option C. Reject option A (chromium(III) sulfate, because chromium(III) hydroxide is soluble in excess sodium hydroxide), B (iron(II) chloride, because chloride ions would not precipitate with barium nitrate), and D (iron(III) sulfate, which produces a red-brown precipitate with sodium hydroxide).

1. Under condensation polymerisation, a dicarboxylic acid containing two carboxylic acid groups (\(-\text{COOH}\)) reacts with a diol containing two alcohol groups (\(-\text{OH}\)).
2. The reaction of an alcohol with a carboxylic acid produces an ester linkage (\(-\text{COO}-\)), and thus the polymer formed is a polyester.
3. The formation of each ester linkage involves the elimination of a small water molecule (\(\text{H}_2\text{O}\)) as a byproduct.

Award 1 mark for the correct option D. Reject options A and B because reacting a carboxylic acid with an alcohol forms an ester linkage, not an amide linkage. Reject option C because the byproduct of the condensation reaction between carboxylic acids and alcohols is water, not hydrogen chloride.

Step 1: Calculate the number of moles of \(\text{NaOH}\) used: \(\text{moles of NaOH} = \text{concentration} \times \text{volume in dm}^3 = 0.100\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.00200\text{ mol}\). Step 2: Use the stoichiometric ratio from the balanced equation. \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Thus, \(\text{moles of H}_2\text{SO}_4 = \frac{0.00200}{2} = 0.00100\text{ mol}\). Step 3: Calculate the concentration of \(\text{H}_2\text{SO}_4\): \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00100\text{ mol}}{0.0250\text{ dm}^3} = 0.0400\text{ mol/dm}^3\). This corresponds to option A.

Award 1 mark for the correct option A. Reject B (if the 1:2 mole ratio was neglected). Reject C (if the titration ratio was inverted). Reject D (if the mole ratio was incorrectly doubled).

Step 1: Identify the linking group in the backbone of the polymer. The link shown is \(-\text{O}-\text{C}(=\text{O})-\), which represents an ester linkage. Step 2: Classify the polymer. Polymers containing ester linkages are polyesters. Step 3: Identify the type of polymerization. Polyesters are formed via condensation polymerization with the elimination of water. Therefore, option C is the correct statement.

Award 1 mark for the correct option C. Reject A because addition polymers only have carbon-carbon single bonds in their backbone. Reject B and D because amide linkages are not present (this is not a polyamide).

Step 1: Identify the cation. The formation of a light blue precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of copper(II) ions, \(\text{Cu}^{2+}\). Step 2: Identify the anion. The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Step 3: Combine the ions to find the formula of salt \(X\), which is copper(II) sulfate, \(\text{CuSO}_4\). Therefore, option B is correct.

Award 1 mark for the correct option B. Reject A because chloride ions do not form a precipitate with barium nitrate. Reject C and D because iron(II) and iron(III) ions yield green and red-brown precipitates with sodium hydroxide respectively.

First, calculate the number of moles of sulfuric acid used: \(\text{moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3 = 0.100\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.00200\text{ mol}\). From the stoichiometry of the balanced equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Therefore, the number of moles of \(\text{NaOH}\) present is: \(0.00200\text{ mol} \times 2 = 0.00400\text{ mol}\). Finally, calculate the concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00400\text{ mol}}{0.0250\text{ dm}^3} = 0.160\text{ mol/dm}^3\). This corresponds to option C.

1 mark for the correct option C.

In Test 1, the formation of a light blue precipitate with aqueous sodium hydroxide indicates the presence of copper(II) ions, \(\text{Cu}^{2+}\). In Test 2, the formation of a white precipitate upon addition of dilute nitric acid and aqueous barium nitrate indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these two findings, the compound \(X\) is copper(II) sulfate.

1 mark for the correct option B.

The repeating unit of the polymer is \(-\text{CH}_2 - \text{CH}(\text{CH}_3)-\). This corresponds to the monomer propene, \(\text{CH}_2=\text{CH}-\text{CH}_3\), because the double bond breaks during polymerization to form single bonds in the backbone. Since the monomer units join together without the loss of any small molecules, this is an addition polymerization reaction.

1 mark for the correct option A.

1. Calculate the number of moles of sulfuric acid used: \(\text{moles of } \text{H}_2\text{SO}_4 = 0.100\text{ mol/dm}^3 \times \frac{18.5}{1000}\text{ dm}^3 = 0.00185\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation to find the moles of sodium hydroxide: according to the equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Therefore, \(\text{moles of } \text{NaOH} = 2 \times 0.00185\text{ mol} = 0.00370\text{ mol}\). 3. Calculate the concentration of the sodium hydroxide solution: \(\text{concentration of } \text{NaOH} = \frac{0.00370\text{ mol}}{0.0250\text{ dm}^3} = 0.148\text{ mol/dm}^3\).

[1 mark] B - Correct calculation of the sodium hydroxide concentration. Award 1 mark for the correct option.

The polymer contains the ester linkage, \(-\text{O}-\text{CO}-\), repeating throughout the chain, which classifies it as a polyester. Polyesters are synthetic condensation polymers formed by the reaction between a diol (providing the \(-\text{O}-\text{Y}-\text{O}-\) part) and a dicarboxylic acid (providing the \(-\text{CO}-\text{Z}-\text{CO}-\) part). During this condensation polymerisation reaction, a water molecule, \(\text{H}_2\text{O}\), is eliminated for each ester linkage formed.

[1 mark] D - Correct identification of the polyester linkage and the elimination of water. Award 1 mark for the correct option.

1. Metal \(\text{X}\) reduces the oxide of \(\text{Y}\), which means \(\text{X}\) is more reactive than \(\text{Y}\) (\(\text{X} > \text{Y}\)). 2. Metal \(\text{Y}\) cannot reduce the oxide of \(\text{Z}\), which means \(\text{Z}\) is more reactive than \(\text{Y}\) (\(\text{Z} > \text{Y}\)). 3. Metal \(\text{Z}\) reduces the oxide of \(\text{X}\), which means \(\text{Z}\) is more reactive than \(\text{X}\) (\(\text{Z} > \text{X}\)). Combining these relationships gives: \(\text{Z} > \text{X} > \text{Y}\).

[1 mark] C - Correct deduction of the reactivity order of the metals. Award 1 mark for the correct option.

The polymer contains amide links, \(-\text{NH}-\text{CO}-\), which means it is a polyamide. Polyamides are formed by condensation polymerisation from two different monomers: a diamine (which provides the \(-\text{NH}-\) groups) and a dicarboxylic acid (which provides the \(-\text{CO}-\) groups). Thus, option C is correct.

Award 1 mark for the correct option C.

First, the cation test: Iron(II) ions, \(\text{Fe}^{2+}\), form a green precipitate with aqueous sodium hydroxide that is insoluble in excess. Chromium(III) ions also form a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. Therefore, the cation is \(\text{Fe}^{2+}\). Second, the anion test: The reaction of the solution with barium nitrate in the presence of acid to form a white precipitate of barium sulfate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these, solid X is iron(II) sulfate.

Award 1 mark for the correct option D.

1. Calculate the number of moles of \(\text{HCl}\) used: \(\text{moles of HCl} = 0.100\text{ mol/dm}^3 \times (20.0 / 1000)\text{ dm}^3 = 0.00200\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation to find the moles of \(\text{M(OH)}_2\): \(\text{moles of M(OH)}_2 = 0.00200 / 2 = 0.00100\text{ mol}\). 3. Calculate the concentration of \(\text{M(OH)}_2\): \(\text{concentration} = 0.00100\text{ mol} / (25.0 / 1000)\text{ dm}^3 = 0.0400\text{ mol/dm}^3\).

Award 1 mark for the correct option B.

Step 1: Write down the balanced chemical equation:
\( 2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \)

Step 2: Calculate the number of moles of \( \text{KOH} \):
\( \text{moles of KOH} = \frac{20.0}{1000} \times 0.150 = 0.00300\text{ mol} \)

Step 3: Determine the moles of \( \text{H}_2\text{SO}_4 \) required based on the stoichiometry (2:1 ratio):
\( \text{moles of H}_2\text{SO}_4 = \frac{0.00300}{2} = 0.00150\text{ mol} \)

Step 4: Calculate the volume of \( \text{H}_2\text{SO}_4 \) needed:
\( \text{Volume} = \frac{\text{moles}}{\text{concentration}} = \frac{0.00150}{0.100} = 0.0150\text{ dm}^3 = 15.0\text{ cm}^3 \).

1 mark for the correct option A.

- A white precipitate with aqueous sodium hydroxide that dissolves in excess indicates either \( \text{Al}^{3+} \) or \( \text{Zn}^{2+} \) ions.
- A white precipitate with aqueous ammonia that is insoluble in excess confirms the presence of \( \text{Al}^{3+} \) ions (as \( \text{Zn}^{2+} \) forms a precipitate that dissolves in excess ammonia to give a colourless solution).
- A white precipitate with acidified silver nitrate confirms the presence of chloride, \( \text{Cl}^- \), ions.

Therefore, salt \( \text{X} \) is aluminium chloride.

1 mark for the correct option A.

During addition polymerisation of propene (\( \text{CH}_2=\text{CHCH}_3 \)), the \( \text{C}=\text{C} \) double bond opens up to form a single bond, linking the monomer units together:

\( \dots-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-\dots \)

Each repeating unit must consist of two carbon atoms in the main polymer backbone, with one of the carbon atoms bearing a methyl group (\( -\text{CH}_3 \)). Option A correctly shows this repeating structure where every second carbon atom in the main chain has a methyl branch.

1 mark for the correct option A.

The given polymer contains the amide linkage, \(-CO-NH-\), which makes it a polyamide. The monomers used to synthesize a polyamide are a dicarboxylic acid and a diamine. Hexane-1,6-diamine, \(H_2N-(CH_2)_6-NH_2\), is a diamine that has 6 carbon atoms, matching the \(-(CH_2)_6-\) section of the chain. Hence, option A is correct.

1 mark: Correct option A selected.

Step 1: Calculate moles of \(\text{NaOH}\): \(n(\text{NaOH}) = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}\). Step 2: Use the stoichiometric ratio. 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Thus, \(n(\text{H}_2\text{SO}_4) = 0.00250\text{ mol} / 2 = 0.00125\text{ mol}\). Step 3: Calculate the concentration: \(\text{Concentration} = 0.00125\text{ mol} / 0.0200\text{ dm}^3 = 0.0625\text{ mol/dm}^3\). Therefore, B is correct.

1 mark: Correct option B selected.

The white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). The light blue precipitate with sodium hydroxide indicates copper(II) ions, \(\text{Cu}^{2+}\). Therefore, substance X is copper(II) sulfate, \(\text{CuSO}_4\).

1 mark: Correct option B selected.

First, the green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also gives a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution). Second, the formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the unknown solid is iron(II) sulfate, \(\text{FeSO}_4\).

Award 1 mark for identifying the correct option C. Reject all other options.

In addition polymerisation, the double bond of an unsaturated monomer breaks to form single bonds that link the repeating units. To find the monomer, we locate the two carbon atoms that form the backbone of the repeating unit and convert the single carbon-carbon bond back into a double bond. This gives \(\text{CH}_2=\text{CH}-\text{CN}\).

Award 1 mark for the correct option B.

A more reactive metal displaces a less reactive metal from its compound. From observation 3, Y is the most reactive because it displaces all other three metal ions (W, X, Z). From observation 1, W is more reactive than X and Z, but less reactive than Y. From observation 4, Z is more reactive than X, but less reactive than W. From observation 2, X is the least reactive because it cannot displace any other metal. Therefore, the decreasing order of reactivity is Y -> W -> Z -> X.

Award 1 mark for the correct option A.

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

Step 2: Calculate the number of moles of sulfuric acid used:
\(\text{moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume (dm}^3\)\)
\(\text{moles of H}_2\text{SO}_4 = 0.100\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00250\text{ mol}\)

Step 3: Determine the number of moles of sodium hydroxide required using the stoichiometric ratio (1 mole of acid reacts with 2 moles of alkali):
\(\text{moles of NaOH} = 2 \times 0.00250\text{ mol} = 0.00500\text{ mol}\)

Step 4: Calculate the concentration of the sodium hydroxide solution:
\(\text{concentration of NaOH} = \frac{\text{moles}}{\text{volume (dm}^3\)} = \frac{0.00500\text{ mol}}{20.0 / 1000\text{ dm}^3} = 0.250\text{ mol/dm}^3\)

Therefore, option C is correct.

[1 mark] for the correct option C.
- Award 1 mark for selecting C.
- Award 0 marks for selecting A, B, or D.
- Distractor B (0.125) is obtained if the 1:2 stoichiometric ratio is incorrectly taken as 1:1.

Analysis of Test 1:
The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. This confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) ions, \(\text{Cr}^{3+}\), also form a green precipitate with aqueous sodium hydroxide, but this precipitate dissolves in excess sodium hydroxide to give a green solution).

Analysis of Test 2:
Adding dilute nitric acid followed by aqueous barium nitrate yields a white precipitate of barium sulfate, \(\text{BaSO}_4\). This confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).

Combining these results, solid \(X\) is iron(II) sulfate.

[1 mark] for the correct option D.
- Award 1 mark for identifying both the iron(II) cation and the sulfate anion correctly.
- Reject A and B because chromium(III) hydroxide is soluble in excess sodium hydroxide.
- Reject C because chloride ions do not form a white precipitate with barium nitrate.

The polymer shown is a polyester, containing ester linkages (\(\text{-O-CO-}\)) in the main chain. Polyesters are formed via condensation polymerisation.

To find the monomers, we can conceptually split the ester link (\(\text{-O-CO-}\)) by adding a water molecule (\(\text{-H}\) to the oxygen atom and \(\text{-OH}\) to the carbonyl carbon atom):
1. The diol segment \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\) becomes \(\text{HO-CH}_2\text{-CH}_2\text{-OH}\).
2. The dicarboxylic acid segment \(\text{-CO-CH}_2\text{-CH}_2\text{-CO-}\) becomes \(\text{HOOC-CH}_2\text{-CH}_2\text{-COOH}\).

Therefore, the correct pair of monomers is given in option A.

[1 mark] for the correct option A.
- Award 1 mark for identifying the correct diol and dicarboxylic acid monomers.
- Reject B because a single monomer like \(\text{HO-CH}_2\text{-CH}_2\text{-COOH}\) would produce a polymer with a different repeating unit structure: \(\text{[-O-CH}_2\text{-CH}_2\text{-CO-]_n}\).

Metal \(Z\) reacts violently with water, while metal \(X\) reacts slowly, indicating that \(Z\) is more reactive than \(X\) (\(Z > X\)). Both \(Z\) and \(X\) are highly reactive metals located above carbon in the reactivity series, meaning their oxides cannot be reduced by carbon. On the other hand, the oxides of \(W\) and \(Y\) are reduced by carbon, making them less reactive. Since metal \(W\) displaces \(Y\) from its solution, \(W\) is more reactive than \(Y\) (\(W > Y\)). Combining these observations gives the reactivity order: \(Z \rightarrow X \rightarrow W \rightarrow Y\).

1 mark for identifying the correct reactivity order from most reactive to least reactive as Z > X > W > Y.

In Test 1, the reaction of a solid with dilute nitric acid to produce carbon dioxide (which turns limewater cloudy) confirms the presence of carbonate ions (\(\text{CO}_3^{2-}\)). In Test 2, a yellow precipitate formed upon addition of acidified silver nitrate to an aqueous halide solution confirms the presence of iodide ions (\(\text{I}^-\)). Therefore, the two anions present in the mixture are carbonate and iodide.

1 mark for selecting the option containing both carbonate and iodide.

First, calculate the number of moles of sodium hydroxide used:
\(\text{moles of NaOH} = \text{volume in dm}^3 \times \text{concentration}\)
\(\text{moles of NaOH} = 0.0250\text{ dm}^3 \times 0.100\text{ mol/dm}^3 = 0.00250\text{ mol}\)

According to the balanced equation, \(2\text{ moles of NaOH}\) react with \(1\text{ mole of H}_2\text{SO}_4\). Therefore, calculate the moles of sulfuric acid needed:
\(\text{moles of H}_2\text{SO}_4 = \frac{0.00250\text{ mol}}{2} = 0.00125\text{ mol}\)

Finally, calculate the concentration of the sulfuric acid using the titration volume of \(20.0\text{ cm}^3\) (\(0.0200\text{ dm}^3\)):
\(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00125\text{ mol}}{0.0200\text{ dm}^3} = 0.0625\text{ mol/dm}^3\)

1 mark for the correct calculation showing the concentration is 0.0625 mol/dm3.

The polymer contains the ester linkage, \(-CO-O-\), which is formed when a dicarboxylic acid reacts with a diol. This is a condensation polymerisation reaction, which eliminates a small molecule of water for each linkage formed. Therefore, it is a polyester. Polyamides and proteins contain amide linkages, not ester linkages.

1 mark for the correct option B.

To ensure accurate concentrations of solutions: The pipette must be rinsed with sodium hydroxide after washing with water to avoid dilution. The burette must be rinsed with hydrochloric acid after washing with water to avoid dilution. The conical flask should only be rinsed with distilled water because rinsing it with the acid or alkali would introduce extra unmeasured moles of reactant, causing an error.

1 mark for the correct option B.

The reaction of solution X with aqueous sodium hydroxide yields a green precipitate insoluble in excess, indicating the presence of iron(II) ions, \(Fe^{2+}\). Chromium(III) also forms a green precipitate but dissolves in excess sodium hydroxide to give a green solution. The reaction with dilute nitric acid and aqueous barium nitrate yields a white precipitate, indicating the presence of sulfate ions, \(SO_4^{2-}\). Thus, the solution is iron(II) sulfate.

1 mark for the correct option B.

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (While chromium(III) also forms a green precipitate, it dissolves in excess sodium hydroxide to form a green solution). A cream precipitate formed upon addition of dilute nitric acid and aqueous silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\). Therefore, salt X is iron(II) bromide.

1 mark for the correct choice B. Award 0 marks for any other option.

During condensation polymerisation to form a polyester, each diol molecule (Monomer X) loses a hydrogen atom from its hydroxyl groups (\(\text{-OH}\)) to form \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\). Each dicarboxylic acid molecule (Monomer Y) loses a hydroxyl group from its carboxyl groups (\(\text{-COOH}\)) to form \(\text{-CO-CH}_2\text{-CH}_2\text{-CO-}\). Combining these two fragments via ester links gives the repeating unit: \(\text{-(O-CH}_2\text{-CH}_2\text{-O-CO-CH}_2\text{-CH}_2\text{-CO)-}\).

1 mark for the correct choice A. Award 0 marks for any other option.

First, calculate the moles of \(\text{H}_2\text{SO}_4\) reacted: \(\text{moles} = \text{concentration} \times \text{volume} = 0.0500\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 1.25 \times 10^{-3}\text{ mol}\). According to the chemical equation, 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{NaOH}\). Therefore, the moles of \(\text{NaOH}\) required = \(2 \times 1.25 \times 10^{-3}\text{ mol} = 2.50 \times 10^{-3}\text{ mol}\). Finally, calculate the concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{2.50 \times 10^{-3}\text{ mol}}{0.0200\text{ dm}^3} = 0.125\text{ mol/dm}^3\).

1 mark for the correct choice C. Award 0 marks for any other option.

To identify the monomer of an addition polymer, locate the smallest repeating unit in the main carbon backbone. The repeating unit here is \( -\text{CH}_2-\text{CH}(\text{COOCH}_3)- \). Replacing the single bond between the two backbone carbon atoms with a carbon-to-carbon double bond gives the structure of the monomer: \( \text{CH}_2=\text{CH}-\text{COOCH}_3 \). Therefore, option B is correct.

1 mark for the correct option B.

The green precipitate formed with aqueous sodium hydroxide that remains insoluble in excess is characteristic of iron(II) ions, \( \text{Fe}^{2+} \). Note that chromium(III) ions also form a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. The reaction with dilute nitric acid followed by aqueous barium nitrate produces a white precipitate of insoluble barium sulfate, which confirms the presence of sulfate ions, \( \text{SO}_4^{2-} \). Therefore, the salt \( \mathbf{X} \) is iron(II) sulfate.

1 mark for the correct option D.

Analyzing each experiment:
- From Experiment 1: Metal \( \text{P} \) displaces metal \( \text{Q} \) from its solution, so \( \text{P} \) is more reactive than \( \text{Q} \) (\( \text{P} > \text{Q} \)).
- From Experiment 2: Metal \( \text{R} \) cannot displace metal \( \text{Q} \), so \( \text{Q} \) is more reactive than \( \text{R} \) (\( \text{Q} > \text{R} \)).
- From Experiment 3: Metal \( \text{S} \) reacts with cold water, meaning it is highly reactive. Metal \( \text{P} \) only reacts with dilute acid, meaning \( \text{S} \) is more reactive than \( \text{P} \) (\( \text{S} > \text{P} \)).

Combining these relationships gives the reactivity order: \( \text{S} > \text{P} > \text{Q} > \text{R} \).

1 mark for the correct option A.

To ensure maximum accuracy in an acid-base titration:
1. The pipette must be rinsed with distilled water to clean it, followed by the solution it will measure (sodium hydroxide) so that any remaining water droplets do not dilute the transferred alkali.
2. The burette must be rinsed with distilled water, followed by the solution it will deliver (hydrochloric acid) so that any remaining water droplets do not dilute the acid.
3. The conical flask should be rinsed with distilled water only. Rinsing it with sodium hydroxide or hydrochloric acid would introduce an unknown number of additional moles of reactant, altering the volume of acid needed to reach the exact endpoint.

Award 1 mark for the correct option A.
- Reject B because using only distilled water in the pipette and burette will dilute the reagents, introducing error.
- Reject C because rinsing the conical flask with sodium hydroxide introduces extra moles of alkali.
- Reject D because rinsing the pipette with distilled water only dilutes the alkali.

1. Cation Test: The reaction with aqueous sodium hydroxide produces a green precipitate which is insoluble in excess. This is characteristic of the iron(II) ion, \(\text{Fe}^{2+}\). Note that chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to yield a green solution.

2. Anion Test: The reaction with dilute nitric acid and aqueous barium nitrate produces a white precipitate, which is barium sulfate. This confirms the presence of the sulfate ion, \(\text{SO}_4^{2-}\).

Combining these observations, salt \(X\) is iron(II) sulfate.

Award 1 mark for selecting B.
- Option A is incorrect because chromium(III) forms a precipitate that is soluble in excess sodium hydroxide.
- Option C is incorrect because iron(III) ions form a red-brown precipitate with sodium hydroxide.
- Option D is incorrect because chloride ions do not form a precipitate with barium nitrate.

The linkage \(- \text{CO} - \text{NH} -\) is an amide (or peptide) linkage, meaning the macromolecule is a polyamide (such as Nylon).
- Polyamides are condensation polymers that contain the same amide linkages linking amino acids together in proteins.
- Option A is incorrect because it is a condensation polymer, not an addition polymer.
- Option C is incorrect because the reaction between a diol and a dicarboxylic acid yields a polyester containing ester linkages (\(- \text{CO} - \text{O} -\)).
- Option D is incorrect because polyamide monomers contain nitrogen-bearing and oxygen-bearing functional groups (diamines and dicarboxylic acids), so they are not hydrocarbons.

Award 1 mark for choosing B.
- Reject A because amide link formation involves condensation polymerization with the elimination of small molecules (like water).
- Reject C because diols and dicarboxylic acids react to form ester linkages.
- Reject D because hydrocarbons consist of hydrogen and carbon atoms only, whereas these monomers contain nitrogen and oxygen.

Step 1: Calculate the amount in moles of sulfuric acid used. \(\text{Moles of H}_2\text{SO}_4 = 0.100\text{ mol/dm}^3 \times (18.5 / 1000)\text{ dm}^3 = 0.00185\text{ mol}\). Step 2: Use the balanced chemical equation, \(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\), to find the reaction ratio. The ratio of \(\text{NaOH}\) to \(\text{H}_2\text{SO}_4\) is \(2:1\). Step 3: Calculate the moles of \(\text{NaOH}\) neutralized. \(\text{Moles of NaOH} = 2 \times 0.00185\text{ mol} = 0.00370\text{ mol}\). Step 4: Calculate the concentration of the \(\text{NaOH}\) solution. \(\text{Concentration} = 0.00370\text{ mol} / (25.0 / 1000)\text{ dm}^3 = 0.148\text{ mol/dm}^3\). Therefore, option C is correct.

[1 mark] for the correct answer C. Award 0 marks for distractors arising from a 1:1 mole ratio (B) or other calculation errors.

Monomer X contains carboxylic acid groups (-\(\text{COOH}\)) and Monomer Y contains alcohol groups (-\(\text{OH}\)). During condensation polymerization, these functional groups react to form ester linkages (-\(\text{COO}-\)) with the elimination of water (\(\text{H}_2\text{O}\)) molecules. This produces a polyester. Therefore, option B is correct.

[1 mark] for the correct answer B. Award 0 marks for choosing amide linkages or hydrogen chloride as the eliminated molecule.

The reaction of the solution with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess, which confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also produces a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution. The reaction with barium nitrate in the presence of dilute nitric acid produces a white precipitate of barium sulfate, confirming the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, the salt is iron(II) sulfate.

[1 mark] for the correct answer B. Reject A because chromium(III) precipitate is soluble in excess NaOH. Reject C because iron(III) gives a red-brown precipitate. Reject D because chloride ions would not form a white precipitate with barium nitrate.

The addition of aqueous sodium hydroxide to a solution containing \(\text{Fe}^{2+}\) ions produces a green precipitate of iron(II) hydroxide, which is insoluble in excess. While chromium(III) ions also form a green precipitate, their precipitate is soluble in excess sodium hydroxide to form a green solution. The addition of dilute nitric acid followed by aqueous barium nitrate is the qualitative test for sulfate ions (\(\text{SO}_4^{2-}\)), producing a white precipitate of insoluble barium sulfate. Therefore, salt \(X\) is iron(II) sulfate.

Award 1 mark for the correct option B. Reject A because chromium(III) hydroxide is soluble in excess aqueous sodium hydroxide. Reject C because chloride ions do not form a precipitate with acidified barium nitrate. Reject D because iron(III) ions produce a red-brown precipitate with aqueous sodium hydroxide.

For the species \(^{34}_{16}\text{S}^{2-}\): The atomic number (proton number) is 16 and the nucleon number (mass number) is 34. The number of neutrons is calculated as \(34 - 16 = 18\). The species has a \(2-\u0000\) charge, meaning it has gained two electrons, so the number of electrons is \(16 + 2 = 18\). Sulfur has 16 protons, giving it an electronic configuration of 2,8,6, which places it in Group VI of the Periodic Table. Therefore, the correct row is 18 neutrons, 18 electrons, Group VI.

Award 1 mark for selecting B. Reject A because the number of electrons in a \(2-\u0000\) ion of sulfur must be 18, not 16. Reject C because the number of neutrons is 18 (not 16) and sulfur belongs to Group VI (not IV). Reject D because all three parameters are incorrect.

The polymer backbone consists entirely of carbon-carbon single bonds, indicating that it is an addition polymer formed from the monomer phenylethene, \(\text{C}_6\text{H}_5\text{CH}=\text{CH}_2\). The molecular formula of this monomer is \(\text{C}_8\text{H}_8\). Because the monomer is an unsaturated hydrocarbon (an alkene derivative), it undergoes an addition reaction with aqueous bromine, decolorizing it. Thus, statement B is correct. Statement A is incorrect because there are no ester or amide links in the backbone. Statement C is incorrect because addition polymers are generally non-biodegradable and inert to acid hydrolysis. Statement D is incorrect because addition polymerization involves no loss of atoms, meaning the monomer and polymer share the exact same empirical formula.

Award 1 mark for selecting B. Reject A because addition polymerization is used to make this polymer. Reject C because addition polymers with carbon-only backbones are non-biodegradable and resistant to chemical hydrolysis. Reject D because the empirical formula of an addition polymer is always identical to that of its monomer.

First, write the balanced chemical equation for the reaction: \(\text{H}_{2}\text{SO}_{4} + 2\text{NaOH} \rightarrow \text{Na}_{2}\text{SO}_{4} + 2\text{H}_{2}\text{O}\). Calculate the moles of \(\text{NaOH}\) reacted: \(\text{moles} = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^{3} \times 0.0200\text{ dm}^{3} = 0.00200\text{ mol}\). From the stoichiometry of the equation, the mole ratio of \(\text{H}_{2}\text{SO}_{4}\) to \(\text{NaOH}\) is 1:2. Thus, the moles of \(\text{H}_{2}\text{SO}_{4}\) present = \(0.00200 / 2 = 0.00100\text{ mol}\). Finally, calculate the concentration of \(\text{H}_{2}\text{SO}_{4}\): \(\text{concentration} = \text{moles} / \text{volume} = 0.00100\text{ mol} / 0.0250\text{ dm}^{3} = 0.0400\text{ mol/dm}^{3}\).

1 mark for the correct choice A. Reject other options which arise from incorrect stoichiometry or calculation errors.

The reaction with aqueous sodium hydroxide produces a green precipitate. Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates with sodium hydroxide. However, chromium(III) hydroxide is soluble in excess NaOH to form a green solution, whereas iron(II) hydroxide remains insoluble in excess NaOH. This identifies the cation as \(\text{Fe}^{2+}\). The test with barium nitrate in the presence of dilute nitric acid yields a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_{4}^{2-}\). Therefore, the compound is iron(II) sulfate.

1 mark for the correct choice B. Reject A because chromium(III) hydroxide dissolves in excess NaOH. Reject C because iron(III) forms a red-brown precipitate. Reject D because chloride ions would not give a precipitate with barium nitrate.

To find the monomer of an addition polymer, identify the repeating unit in the polymer chain. Here, the repeating unit is \(-\text{CH}_{2}-\text{CH(CH}_{3})-\). By replacing the single bond in the backbone of the repeating unit with a double bond, we get the monomer: \(\text{CH}_{2}=\text{CH-CH}_{3}\). This compound is propene.

1 mark for the correct choice B. Reject A, C, and D as they would yield different repeating unit structures.

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 sodium hydroxide used: \(\text{moles of NaOH} = \frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol}\). 3. Determine the moles of sulfuric acid reacting, using the stoichiometric 2:1 ratio: \(\text{moles of H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol}\). 4. Calculate the concentration of the sulfuric acid: \(\text{concentration} = \frac{0.00125\text{ mol}}{0.0125\text{ dm}^3} = 0.100\text{ mol/dm}^3\).

1 mark for the correct option B. Award 1 mark for calculating the concentration as 0.100 mol/dm³ based on the correct stoichiometric mole ratio (2:1 of NaOH to H₂SO₄).

1. The reaction with aqueous sodium hydroxide produces a light blue precipitate that is insoluble in excess, which is the characteristic test identifying copper(II) cations, \(\text{Cu}^{2+}\). 2. The reaction with dilute nitric acid followed by aqueous barium nitrate produces a white precipitate of barium sulfate, identifying the sulfate anion, \(\text{SO}_4^{2-}\). Therefore, solid X is copper(II) sulfate.

1 mark for the correct option B. Correctly identifying copper(II) from the sodium hydroxide test and sulfate from the barium nitrate test.

The given polymer structure contains ester linkages, represented by \(\text{--O--C(=O)--}\), making it a polyester. Polyesters are a group of condensation polymers that are formed by the reaction of a diol and a dicarboxylic acid, releasing a water molecule for each ester linkage made. Therefore, option C is the correct statement.

1 mark for the correct option C. Correctly identifying that the polymer contains ester linkages (polyester) and is synthesized via condensation with the elimination of water molecules.

To find the concentration of \(\text{KOH}\):

1. Calculate the number of moles of \(\text{H}_2\text{SO}_4\) used:
\(\text{moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3\)
\(\text{moles of H}_2\text{SO}_4 = 0.100\text{ mol/dm}^3 \times \frac{12.5}{1000}\text{ dm}^3 = 0.00125\text{ mol}\)

2. Use the stoichiometric ratio from the balanced equation to find the moles of \(\text{KOH}\):
From the equation, \(1\text{ mol of H}_2\text{SO}_4\) reacts with \(2\text{ mol of KOH}\).
\(\text{moles of KOH} = 2 \times 0.00125\text{ mol} = 0.00250\text{ mol}\)

3. Calculate the concentration of the \(\text{KOH}\) solution:
\(\text{concentration of KOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00250\text{ mol}}{0.0250\text{ dm}^3} = 0.100\text{ mol/dm}^3\)

Therefore, the correct choice is B.

1 mark for the correct option B.
- Award 1 mark for correct selection of B.
- Incorrect calculations that lead to A (failing to multiply moles by 2), C (multiplying by 2 incorrectly or using wrong volume ratio), or D are incorrect.

Let's analyze each statement:

- **A is incorrect**: Nylon is a polyamide containing amide linkages (\(\text{-CONH-}\)), not a polyester.
- **B is incorrect**: Terylene is a polyester containing ester linkages (\(\text{-COO-}\)), not a polyamide.
- **C is correct**: Condensation polymerization involves the reaction of monomers to form a polymer chain with the elimination of a small molecule, most commonly water (\(\text{H}_2\text{O}\)) or hydrogen chloride (\(\text{HCl}\)).
- **D is incorrect**: Monomers for addition polymerization contain a carbon-carbon double bond (\(\text{C=C}\)) and do not require functional groups like carboxylic acid or alcohol at both ends. The latter are used for condensation polymerization.

Thus, option C is correct.

1 mark for the correct option C.
- Award 1 mark for identifying that condensation polymerization produces a small molecule like water.
- Reject other options because they mismatch the polymers (Nylon/Terylene) with their linkage types, or confuse addition polymer monomers with condensation monomers.

We can identify the salt by examining the results of both tests:

- **Test 1**: Reaction with aqueous sodium hydroxide (\(\text{NaOH(aq)}\)):
- A green precipitate is formed that is insoluble in excess.
- Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates with \(\text{NaOH(aq)}\).
- However, the precipitate of \(\text{Cr(OH)}_3\) is **soluble** in excess \(\text{NaOH(aq)}\) to form a green solution, whereas the precipitate of \(\text{Fe(OH)}_2\) is **insoluble** in excess.
- This confirms that the cation present in salt \(X\) is iron(II), \(\text{Fe}^{2+}\).

- **Test 2**: Reaction with dilute nitric acid followed by aqueous barium nitrate:
- A white precipitate is formed, which is barium sulfate (\(\text{BaSO}_4\)).
- This is the diagnostic test for the sulfate ion, \(\text{SO}_4^{2-}\).

Combining these results, salt \(X\) is iron(II) sulfate (\(\text{FeSO}_4\)).

Therefore, the correct option is B.

1 mark for the correct option B.
- Award 1 mark for identifying Iron(II) sulfate based on the insoluble green precipitate with \(\text{NaOH}\) (confirming \(\text{Fe}^{2+}\)) and the white precipitate with \(\text{Ba(NO}_3)_2\) (confirming \(\text{SO}_4^{2-}\)).
- Reject A because chromium(III) hydroxide is soluble in excess sodium hydroxide.
- Reject C because iron(III) forms a red-brown precipitate.
- Reject D because chloride ions would give a white precipitate with silver nitrate, not barium nitrate.

The polymer shown is a polyester, containing repeating ester linkages (\(-\text{O}-\text{CO}-\)). Polyesters are formed by the condensation reaction between a diol (a molecule with two alcohol \(-\text{OH}\) groups) and a dicarboxylic acid (a molecule with two carboxylic acid \(-\text{COOH}\) groups).

Splitting the ester linkages reveals the two distinct monomers:
1. The diol component: \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\)
2. The dicarboxylic acid component: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\)

1 mark: Correct identification of both the diol and dicarboxylic acid monomers corresponding to Option A.

In Test 1, the formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess is characteristic of the iron(II) ion, \(\text{Fe}^{2+}\). (Note: Chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution).

In Test 2, adding dilute nitric acid followed by aqueous barium nitrate is the standard test for sulfate ions, \(\text{SO}_4^{2-}\). The formation of a white precipitate of barium sulfate confirms their presence.

Combining these deductions, salt \(X\) is iron(II) sulfate.

1 mark: Identify iron(II) from the insoluble green precipitate and sulfate from the white precipitate formed with acidified barium nitrate, leading uniquely to Option C.

We can determine the relative reactivity by analyzing each statement:
- Metal \(X\) reacts with the oxide of metal \(Y\): \(X\) is more reactive than \(Y\) (\(X > Y\)).
- Metal \(Y\) reacts with \(Z\) ions: \(Y\) is more reactive than \(Z\) (\(Y > Z\)).
- Metal \(W\) does not react with the oxide of metal \(X\): \(X\) is more reactive than \(W\) (\(X > W\)).
- Metal \(W\) reacts with \(Y\) ions: \(W\) is more reactive than \(Y\) (\(W > Y\)).

Combining these results:
- From \(X > W\) and \(W > Y\), we establish \(X > W > Y\).
- Since \(Y > Z\), the overall order of decreasing reactivity is \(X \rightarrow W \rightarrow Y \rightarrow Z\).

1 mark: Correctly process the four experimental statements to deduce the reactivity sequence as Option A.

1. Calculate the number of moles of \(\text{H}_2\text{SO}_4\) used: \(\text{moles} = \text{concentration} \times \text{volume in dm}^3\) \(\text{moles} = 0.100\text{ mol/dm}^3 \times \frac{15.0}{1000}\text{ dm}^3 = 0.00150\text{ mol}\) 2. Determine the moles of \(\text{KOH}\) that reacted using the stoichiometric ratio from the balanced chemical equation: From the equation, \(2\) moles of \(\text{KOH}\) react with \(1\) mole of \(\text{H}_2\text{SO}_4\). \(\text{moles of KOH} = 2 \times 0.00150\text{ mol} = 0.00300\text{ mol}\) 3. Calculate the concentration of the \(\text{KOH}\) solution: \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3}\) \(\text{concentration} = \frac{0.00300\text{ mol}}{0.0250\text{ dm}^3} = 0.120\text{ mol/dm}^3\). Therefore, the correct choice is B.

1 mark for the correct option B. Correctly identifies the 1:2 molar ratio between acid and alkali and performs calculations to arrive at 0.120 mol/dm³.

1. Test 1: The formation of a white precipitate upon addition of dilute nitric acid and silver nitrate confirms the presence of chloride ions (\(\text{Cl}^-\)). This eliminates B (aluminium sulfate) and D (zinc sulfate). 2. Test 2: Addition of dropwise aqueous ammonia to a solution containing zinc ions (\(\text{Zn}^{2+}\)) produces a white precipitate of zinc hydroxide, which dissolves in excess ammonia to form a colorless complex solution. In contrast, aluminium ions (\(\text{Al}^{3+}\)) form a white precipitate that is insoluble in excess aqueous ammonia. Therefore, salt \(X\) is zinc chloride.

1 mark for the correct option C. Correctly deduces the anion as chloride and the cation as zinc from the given analytical chemical tests.

1. Examine the backbone of the polymer: it consists entirely of a chain of carbon-carbon single bonds with no heteroatoms (such as oxygen or nitrogen) in the main chain. This indicates it is an addition polymer. 2. Identify the repeating unit of the polymer: the repeating pattern is \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\). 3. Determine the monomer: a repeating unit of \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\) comes from the monomer propene (\(\text{CH}_2=\text{CH}-\text{CH}_3\)), which has three carbon atoms and a double bond. 4. Therefore, it is formed by addition polymerization of propene (Row B).

1 mark for the correct option B. Correctly identifies addition polymerization and propene as the monomer from the given structural representation of the polymer.

The polymer contains ester linkages (\(-O-CO-\)), which indicates that it is a polyester. Polyesters are formed by condensation polymerisation of a diol and a dicarboxylic acid, during which water is eliminated as a small molecule product.

1 mark: Correctly identifies condensation polymerisation and water as the other product (Option B).

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(Fe^{2+}\). (Chromium(III) also gives a green precipitate, but it is soluble in excess NaOH). A cream precipitate with dilute nitric acid followed by silver nitrate indicates the presence of bromide ions, \(Br^-\). Therefore, salt X is iron(II) bromide.

1 mark: Correctly deduces iron(II) bromide based on cation and anion test observations (Option C).

(a) Step 1: Rinse the burette thoroughly with distilled or deionised water.
Step 2: Rinse the burette with a small volume of the hydrochloric acid solution to be used.
Step 3: Fill the burette with the acid above the zero line, then open the tap to fill the jet space below the tap completely (ensuring no air bubbles are trapped) and adjust the meniscus to a starting reading.

(b) A conical flask can be swirled vigorously to mix the reactants during the titration without the risk of liquid splashing or spilling out.

(c) (i) Yellow (since sodium hydroxide is alkaline).
(ii) Orange / peach (the transition colour at neutrality).

(d) (i) \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\)
(ii)
- Moles of \(\text{NaOH}\) reacted: \(n = C \times V = 0.120\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00300\text{ mol}\)
- Since the mole ratio is \(1:1\), moles of \(\text{HCl}\) reacted = \(0.00300\text{ mol}\).
- Concentration of \(\text{HCl}\) in \(\text{mol/dm}^3\): \(C = \frac{n}{V} = \frac{0.00300\text{ mol}}{0.0200\text{ dm}^3} = 0.150\text{ mol/dm}^3\).
- Concentration of \(\text{HCl}\) in \(\text{g/dm}^3\): \(0.150\text{ mol/dm}^3 \times 36.5\text{ g/mol} = 5.475\text{ g/dm}^3\) (accept \(5.48\text{ g/dm}^3\)).

(e) Methyl orange is a weak acid, so it will react with and neutralise some of the sodium hydroxide in the flask. Consequently, a smaller volume of hydrochloric acid is needed from the burette to reach the end-point. Since less acid volume is recorded, the calculated concentration of the hydrochloric acid will be lower than its actual true value.

(a) [3 marks]
- 1 mark for rinsing with distilled/deionised water.
- 1 mark for rinsing with the hydrochloric acid solution.
- 1 mark for filling the jet space / removing air bubbles below the tap.

(b) [1 mark]
- 1 mark for: allows swirling without splashing / spilling.

(c) [2 marks]
- (i) 1 mark: Yellow.
- (ii) 1 mark: Orange / peach (reject red, pink, or yellow).

(d) [5 marks total]
- (i) [1 mark]: \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\). Deduct 1 mark if state symbols are missing or incorrect.
- (ii) [4 marks]:
- 1 mark for calculating moles of \(\text{NaOH}\) as \(0.00300\text{ mol}\).
- 1 mark for identifying 1:1 ratio, meaning moles of \(\text{HCl}\) = \(0.00300\text{ mol}\).
- 1 mark for calculating concentration in \(\text{mol/dm}^3\) as \(0.150\text{ mol/dm}^3\).
- 1 mark for multiplying by \(36.5\) to get \(5.475\text{ g/dm}^3\) (or \(5.48\text{ g/dm}^3\)).

(e) [1.3 marks]
- 1 mark for: Indicator acts as an acid and reacts with some of the NaOH.
- 0.3 marks for: Calculated concentration is lower than the true value.

(a) (i) Monomers must contain a carbon-carbon double bond (\(\text{C}=\text{C}\)).
(ii) Addition polymerisation produces only one product, which is the polymer. Condensation polymerisation produces both the polymer and a small molecule (such as water, \(\text{H}_2\text{O}\), or hydrogen chloride, \(\text{HCl}\)).

(b) (i) The monomer is acrylonitrile (propenenitrile). Its fully displayed formula is:
H H
| |
C = C
| |
H C ≡ N
(Where Carbon 1 is double-bonded to Carbon 2. Carbon 1 is single-bonded to two hydrogen atoms. Carbon 2 is single-bonded to one hydrogen atom and one nitrile carbon atom, which has a triple bond to nitrogen).
(ii) The polymer contains only single carbon-carbon bonds (\(\text{C--C}\)) in its main backbone chain; the double bonds of the monomer open up during polymerisation.

(c) (i) The repeat unit is drawn by removing the \(\text{--OH}\) from the carboxylic acid groups and the \(\text{--H}\) from the alcohol groups to form ester linkages:
\(\text{--O--}\bigcirc\text{--O--CO--}\square\text{--CO--}\)
(The ester linkage \(\text{--O--C(=O)--}\) must show the double bond between the carbon and oxygen atom).
(ii) Landfill: They are non-biodegradable, so they persist indefinitely, occupying valuable space and potentially trapping wildlife.
Incineration: Burning plastic waste releases toxic gases (e.g., hydrogen cyanide, carbon monoxide) and carbon dioxide, a greenhouse gas that contributes to climate change.
(iii) Photodegradable plastics undergo chemical breakdown (cleavage of polymer chains) when exposed to ultraviolet (UV) radiation from sunlight, preventing them from remaining intact in the environment for decades and reducing visible plastic pollution.

(a) [3 marks]
- (i) 1 mark: Carbon-to-carbon double bond / \(\text{C}=\text{C}\).
- (ii) 2 marks: 1 mark for addition polymerisation forming only the polymer; 1 mark for condensation polymerisation producing a small molecule / water / hydrogen chloride as well.

(b) [3 marks]
- (i) 2 marks: 1 mark for drawing \(\text{C}=\text{C}\) double bond with correct hydrogens; 1 mark for drawing the nitrile group attached to carbon with a \(\text{C}\equiv\text{N}\) triple bond (accept \(\text{-CN}\) if all other bonds are fully displayed).
- (ii) 1 mark: Main chain contains only single carbon-carbon bonds / no carbon-carbon double bonds remain.

(c) [6.3 marks total]
- (i) [3 marks]:
- 1 mark for correct ester linkage shown showing \(\text{C}=\text{O}\) double bond.
- 1 mark for correct alternating block diagram (circle and square connected via ester group).
- 1 mark for continuation / open bonds shown at both ends of the repeat unit.
- (ii) [2 marks]:
- 1 mark for landfill issue: non-biodegradable / persists / fills landfill space / ingestion hazard.
- 1 mark for incineration issue: releases toxic fumes / releases greenhouse gases / releases \(\text{CO}_2\).
- (iii) [1.3 marks]:
- 1 mark for: polymers break up / degrade when exposed to sunlight / UV radiation.
- 0.3 marks for: reduces volume of long-term litter / accumulation in environment.

(a) (i) Iron(II) ion (\(\text{Fe}^{2+}\)) causes a green precipitate with sodium hydroxide that is insoluble in excess.
(ii) Copper(II) ion (\(\text{Cu}^{2+}\)) forms a light blue precipitate with ammonia that dissolves in excess to form a deep blue solution.
(iii) \(\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq)} \rightarrow \text{Cu(OH)}_2(\text{s})\)

(b) (i) Sulfate ion (\(\text{SO}_4^{2-}\)).
(ii) Dilute nitric acid reacts with and removes other anions (such as carbonate, \(\text{CO}_3^{2-}\), or sulfite, \(\text{SO}_3^{2-}\)) that might be present. This prevents them from forming an alternative white precipitate (like barium carbonate) with barium ions, which would give a false-positive result for sulfate.

(c) (i) Brown gas: Nitrogen dioxide (\(\text{NO}_2\)).
Colorless gas: Oxygen (\(\text{O}_2\)).
(ii) Test: Pass the brown gas through aqueous potassium iodide (KI) solution.
Observation: The colorless solution turns brown / a grey precipitate is formed (due to the oxidation of iodide to iodine by nitrogen dioxide).
(iii) Nitrate ion (\(\text{NO}_3^-\)) undergoes thermal decomposition in metal nitrates to produce metal oxide, nitrogen dioxide, and oxygen.

(a) [4 marks]
- (i) 1 mark: Iron(II) / \(\text{Fe}^{2+}\) (reject Iron / \(\text{Fe}^{3+}\)).
- (ii) 1 mark: Copper(II) / \(\text{Cu}^{2+}\) (reject Copper / \(\text{Cu}^+\)).
- (iii) 2 marks: 1 mark for correct formulae and balancing (\(\text{Cu}^{2+} + 2\text{OH}^- \rightarrow \text{Cu(OH)}_2\)); 1 mark for correct state symbols (\(\text{aq}\) for reactants, \(\text{s}\) for product).

(b) [3 marks]
- (i) 1 mark: Sulfate / \(\text{SO}_4^{2-}\).
- (ii) 2 marks:
- 1 mark for: reacts with / removes carbonate (or sulfite) ions / prevents other barium precipitates.
- 1 mark for: prevents false-positive results.

(c) [5.3 marks total]
- (i) 2 marks: 1 mark for nitrogen dioxide / \(\text{NO}_2\); 1 mark for oxygen / \(\text{O}_2\).
- (ii) 2 marks:
- 1 mark for test: bubble through / react with aqueous potassium iodide (or starch-iodide paper).
- 1 mark for observation: turns brown (or paper turns blue-black).
- (Accept alternative test: pass over heated copper turning; observation: copper turns black).
- (iii) 1.3 marks: Nitrate / \(\text{NO}_3^-\).

### Step-by-step Solution

**(a)**
(i) Methyl orange in an alkaline solution (\(\text{NaOH}\)) is **yellow**.
(ii) At the neutralisation end-point, methyl orange turns **orange** (or peach).

**(b)**
(i) The balanced equation is:
\(\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH}\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O}\text{(l)}\)
- Reactants and products correctly written: 1 mark
- Balancing: 1 mark
- State symbols: 1 mark

(ii) \(\text{Moles of NaOH} = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Moles of NaOH} = 0.150\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00375\text{ mol}\)

(iii) From the balanced equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\).
\(\text{Moles of H}_2\text{SO}_4 = \frac{0.00375}{2} = 0.001875\text{ mol}\)

(iv) \(\text{Concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3}\)
\(\text{Volume of acid} = 18.75\text{ cm}^3 = 0.01875\text{ dm}^3\)
\(\text{Concentration} = \frac{0.001875\text{ mol}}{0.01875\text{ dm}^3} = 0.100\text{ mol/dm}^3\)

**(c)**
To convert concentration from \(\text{mol/dm}^3\) to \(\text{g/dm}^3\):
\(\text{Concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r\)
\(\text{Concentration} = 0.100\text{ mol/dm}^3 \times 98.0\text{ g/mol} = 9.80\text{ g/dm}^3\)

**(d)**
A volumetric pipette has a much higher degree of accuracy and lower percentage uncertainty (or error) compared to a measuring cylinder, ensuring a precise volume of \(25.0\text{ cm}^3\) is delivered.

**(a)**
(i) Yellow [1]
(ii) Orange / peach [1] (Reject: red, pink)

**(b)**
(i) \(\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH}\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O}\text{(l)}\)
- Correct formulae: [1]
- Correct balancing: [1]
- Correct state symbols: [1]

(ii) \(0.00375\text{ mol}\) (or \(3.75 \times 10^{-3}\)) [1]

(iii) \(0.001875\text{ mol}\) (or \(1.875 \times 10^{-3}\)) [1] (Accept: ecf from (b)(ii) divided by 2)

(iv) \(0.100\text{ mol/dm}^3\) [2]
- \(\text{Volume in dm}^3 = 0.01875\) [1]
- \(\text{Concentration} = 0.100\) [1] (Accept: ecf from (b)(iii))

**(c)**
\(9.80\text{ g/dm}^3\) [2]
- Multiplication of concentration by 98: [1]
- Correct evaluation to \(9.80\) (or \(9.8\)): [1] (Accept: ecf from (b)(iv))

**(d)**
- It is more accurate / has smaller percentage uncertainty / delivers a precise, fixed volume. [1.3] (Reject: 'easy to use' or 'cleaner')

### Step-by-step Solution

**(a)**
(i) Monomer \(A\) is propene, \(\text{C}_3\text{H}_6\). Its fully displayed structure shows a carbon-carbon double bond, and each carbon has the appropriate number of hydrogen atoms:
```
H H H
| | |
H - C = C - C - H
|
H
```
(ii) In the addition polymerisation of propene, the double bond opens up. Showing two repeat units:
```
H H H H
| | | |
- - C - C - C - C - -
| | | |
H CH3 H CH3
```
- Single bonds between all carbon atoms in the chain: 1 mark
- Correct side-groups (methyl and hydrogen) in correct repeating positions: 1 mark

(iii) The polymer is called **poly(propene)** or **polypropylene**.

**(b)**
(i) Nylon is a polyamide, made from the reaction of a **diamine** and a **dicarboxylic acid**. The two functional groups are the **amine group** (or amino group) and the **carboxylic acid group** (or carboxyl group).
(ii) The linkage is an **amide** linkage (or peptide linkage).
(iii) The small molecule eliminated is **water** (\(\text{H}_2\text{O}\)).

**(c)**
(i) The ester linkage consists of a carbonyl carbon double-bonded to an oxygen, which is single-bonded to another oxygen: \(-\text{C}(=\text{O})-\text{O}-\).
Showing all bonds:
```
O
||
- C - O -
```
(ii) Synthetic polymers are **non-biodegradable** because microbes/decomposers cannot break them down. Consequently, they accumulate in landfills, taking up valuable space for hundreds of years. Recycling is preferred because it conserves crude oil reserves (the non-renewable resource from which monomers are derived) and reduces landfill waste and greenhouse gas emissions from incineration.

**(a)**
(i) Correct fully displayed structure of propene (showing \(\text{C}=\text{C}\), three carbons, and six \(\text{H}\) atoms with all single bonds shown). [1]
(ii) Two repeat units of poly(propene) correctly drawn with open single bonds on each end. [2]
- Correct continuous carbon backbone with open-ended bonds: [1]
- Correctly positioned \(-\text{CH}_3\) groups on alternate carbon atoms: [1]
(iii) Poly(propene) / polypropylene [1]

**(b)**
(i) Amine (or amino) [1] AND carboxylic acid (or carboxyl) [1]
(ii) Amide linkage / peptide linkage [1]
(iii) Water / \(\text{H}_2\text{O}\) [1]

**(c)**
(i) Structure of ester linkage: \(-\text{C}(=\text{O})-\text{O}-\) showing all bonds (the carbon must show a double bond to one oxygen and a single bond to another). [2]
(ii)
- Landfill issue: Non-biodegradable / persists in the environment for a long time / takes up landfill space / harms wildlife. [1.3]
- Reason for recycling: Conserves fossil fuels / crude oil OR reduces waste / decreases litter. [1.0]

### Step-by-step Solution

**(a)**
(i) When aqueous sodium hydroxide is added to a solution containing \(\text{Fe}^{2+}\) ions, a **green precipitate** is formed. This precipitate is **insoluble in excess** sodium hydroxide.
(ii) When aqueous ammonia is added to a solution containing \(\text{Fe}^{2+}\) ions, a **green precipitate** is formed. This precipitate is **insoluble in excess** ammonia.
(iii) The cation is the **iron(II) ion** (\(\text{Fe}^{2+}\)). Note: Specifying the oxidation state \((II)\) is essential as iron can also exist as \(\text{Fe}^{3+}\).

**(b)**
(i) The gas produced by the reaction of a carbonate with acid is carbon dioxide, \(\text{CO}_2\). This gas turns limewater **cloudy** (or milky/chalky) due to the formation of a calcium carbonate precipitate. The anion present in \(Y\) is the **carbonate ion** (\(\text{CO}_3^{2-}\)).
(ii) The ionic equation is:
\(\text{CO}_3^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O}\text{(l)}\)
*(Note: \(\text{CO}_3^{2-}\text{(s)}\) is also accepted as the test was performed on solid \(Y\))*.
- Correct reactants and products: 1 mark
- Correct balancing: 1 mark
- Correct state symbols: 1 mark

**(c)**
To distinguish between \(\text{CO}_2\) and \(\text{SO}_2\):
- **Reagent**: Acidified aqueous potassium manganate(VII) (\(\text{KMnO}_4\)) or acidified potassium dichromate(VI).
- **With sulfur dioxide (\(\text{SO}_2\))**: The purple solution turns **colourless** (it is decolourised) because \(\text{SO}_2\) is a reducing agent.
- **With carbon dioxide (\(\text{CO}_2\))**: There is **no change** (the solution remains purple).

**(a)**
(i) Green precipitate [1]; insoluble in excess [1]
(ii) Green precipitate [1]; insoluble in excess [1]
(iii) Iron(II) / \(\text{Fe}^{2+}\) [1] (Reject: 'iron' or 'iron(III)')

**(b)**
(i) Limewater turns cloudy / milky / chalky [1]; Carbonate / \(\text{CO}_3^{2-}\) [1]
(ii) \(\text{CO}_3^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O}\text{(l)}\)
(Accept solid state symbol for carbonate: \(\text{CO}_3^{2-}\text{(s)}\))
- Correct reactant and product formulae: [1]
- Correct balancing: [1]
- Correct state symbols: [1]

**(c)**
- Test reagent: Acidified aqueous potassium manganate(VII) [1] (Accept: Acidified potassium dichromate(VI))
- Observation with \(\text{SO}_2\): Solution turns from purple to colourless (decolourised) [1] (Accept: turns from orange to green with dichromate)
- Observation with \(\text{CO}_2\): No change / remains purple [0.3] (Accept: remains orange with dichromate)

(a) A standard solution is a solution of known concentration.

(b) Practical procedure:
1. Measure exactly \(25.0\text{ cm}^3\) of the sodium hydroxide solution using a volumetric pipette (fitted with a pipette filler) and transfer it into a clean conical flask.
2. Add a few drops of methyl orange indicator to the conical flask; the solution will turn yellow.
3. Fill a clean burette with the hydrochloric acid solution (after rinsing it with some acid), ensuring there are no air bubbles in the jet, and record the initial burette reading.
4. Slowly add the hydrochloric acid from the burette to the conical flask while swirling continuously.
5. As the end-point is approached, add the acid drop-by-drop until the indicator changes colour from yellow to orange (or peach/pink) permanently. Record the final burette reading.

(c) Calculation:
1. The balanced chemical equation is:
\(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\)
2. Calculate the moles of \(\text{HCl}\) used:
\(n(\text{HCl}) = \text{concentration} \times \text{volume} = 0.120\text{ mol/dm}^3 \times \frac{18.50}{1000}\text{ dm}^3 = 0.00222\text{ mol}\)
3. Since the reaction stoichiometry is 1:1, the moles of \(\text{NaOH}\) in \(25.0\text{ cm}^3\) is also \(0.00222\text{ mol}\).
4. Calculate the concentration of the sodium hydroxide solution:
\(\text{Concentration of NaOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00222\text{ mol}}{0.0250\text{ dm}^3} = 0.0888\text{ mol/dm}^3\).

(d) A volumetric pipette is designed to measure a single, highly accurate, and fixed volume (e.g. \(25.0\text{ cm}^3\)), which is ideal for the aliquot of sodium hydroxide. A burette is designed to deliver a variable volume of solution accurately, allowing the user to control the flow drop-by-drop and stop exactly at the neutralisation endpoint.

Part (a) [1 mark]:
- 1 mark: Solution of known concentration.

Part (b) [5 marks]:
- 1 mark: Use of a volumetric pipette to transfer the sodium hydroxide to a conical flask.
- 1 mark: Addition of methyl orange showing a yellow colour in NaOH.
- 1 mark: Running acid from a burette and swirling the flask.
- 1 mark: Dropwise addition near the endpoint.
- 1 mark: Yellow to orange (accept peach/pink, reject red) colour change.

Part (c) [3.3 marks]:
- 1 mark: Correct calculation of moles of HCl (\(0.00222\text{ mol}\)).
- 1 mark: Stating 1:1 mole ratio of reactants (moles of NaOH = \(0.00222\text{ mol}\)).
- 1.3 marks: Correct final concentration of \(0.0888\text{ mol/dm}^3\) (allow 1.3 marks if correctly rounded to 3 significant figures; deduct 0.3 marks for incorrect significant figures or math errors but allow error-carried-forward from step 2).

Part (d) [3 marks]:
- 1.5 marks: Pipette measures a fixed/single specific volume extremely accurately.
- 1.5 marks: Burette measures variable/flexible volumes and allows controlled/dropwise addition.

(a) (i) Tetrafluoroethene.
(ii) The repeating unit has a single bond between carbon atoms with four fluorine atoms attached, and open bonds on each side: \(-(\text{CF}_2-\text{CF}_2)-\).
(iii) Addition polymerisation.

(b) (i) Nylon structure representation showing amide linkages (peptide bonds) between block chains:
\(-\text{HN}-\square-\text{NH}-\text{CO}-\square-\text{CO}-\text{HN}-\square-\text{NH}-\text{CO}-\square-\text{CO}-\)
Alternatively, showing two repeating units: \(-[\text{NH}-\text{block}_1-\text{NH}-\text{CO}-\text{block}_2-\text{CO}]_2-\).
(ii) Water (\(\text{H}_2\text{O}\)).

(c) (i) Amide link (or peptide link / peptide bond).
(ii) Amino acids.
(iii) Hydrolysis (acid hydrolysis or enzymatic hydrolysis).

Part (a) [4 marks]:
- (i) 1 mark: Tetrafluoroethene.
- (ii) 2 marks: Correct structure of the repeating unit with a single C-C bond, 4 F atoms attached, and continuation bonds on both sides (1 mark for C-C single bond and F atoms, 1 mark for extension bonds).
- (iii) 1 mark: Addition (polymerisation).

Part (b) [4.3 marks]:
- (i) 3.3 marks:
- 1.3 marks for drawing correct amide linkages (\(-\text{NH}-\text{CO}-\) or \(-\text{CO}-\text{NH}-\)).
- 1 mark for alternating monomer blocks (represented by different or same boxes/rectangles).
- 1 mark for showing continuation bonds at the ends of two repeating units.
- (ii) 1 mark: Water (or \(\text{H}_2\text{O}\)).

Part (c) [4 marks]:
- (i) 1 mark: Peptide link / peptide bond / amide linkage.
- (ii) 1 mark: Amino acids.
- (iii) 2 marks: Hydrolysis (accept heating with concentrated hydrochloric acid).

(a) (i) Add aqueous sodium hydroxide to a solution of Compound X. A green precipitate is formed (which is insoluble in excess sodium hydroxide).
(ii) Add aqueous ammonia to a solution of Compound X. A green precipitate is formed (which is insoluble in excess ammonia).

(b) (i) A white precipitate is formed.
(ii) The ionic equation is:
\(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\)
(iii) Nitric acid is added to react with and remove any carbonate (or sulfite) impurities present. This prevents a false positive white precipitate of barium carbonate (or barium sulfite) from forming.

(c) (i) Add the liquid to anhydrous copper(II) sulfate; it turns from white to blue. (Alternatively, add to anhydrous cobalt(II) chloride; it turns from blue to pink).
(ii) Water of crystallisation.

Part (a) [4 marks]:
- (i) 2 marks: 1 mark for adding aqueous sodium hydroxide, 1 mark for green precipitate (insoluble in excess).
- (ii) 2 marks: 1 mark for adding aqueous ammonia, 1 mark for green precipitate (insoluble in excess).

Part (b) [5.3 marks]:
- (i) 1 mark: White precipitate.
- (ii) 2.3 marks: 1 mark for correct formulas of reactants and products, 1 mark for correct state symbols \(\text{(aq)}\) and \(\text{(s)}\), 0.3 marks for correct ionic charges and balancing.
- (iii) 2 marks: 1 mark for stating that it reacts with/removes carbonate ions, 1 mark for explaining that this prevents a false positive/white precipitate of barium carbonate.

Part (c) [3 marks]:
- (i) 2 marks: 1 mark for using anhydrous copper(II) sulfate (or anhydrous cobalt(II) chloride), 1 mark for correct colour change from white to blue (or blue to pink).
- (ii) 1 mark: Water of crystallisation.

(a) The sulfuric acid is in the conical flask, so the initial solution is acidic. Phenolphthalein is colorless in acidic solution and turns pink at the neutralization endpoint when a tiny excess of alkali is present. Therefore, the color change is from colorless to pink.

(b) Volumetric pipettes are calibrated to deliver exactly one specific volume (e.g. \(25.0\text{ cm}^3\)) with high precision. Burettes are used to deliver variable volumes of liquids precisely, allowing the user to add dropwise until the end-point is observed.

(c) Sulfuric acid react with sodium hydroxide to form sodium sulfate and water:
\(\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}\)

(d)(i) \(\text{Moles of NaOH} = \text{concentration} \times \text{volume in dm}^3 = 0.100\text{ mol/dm}^3 \times \frac{18.80}{1000}\text{ dm}^3 = 1.88 \times 10^{-3}\text{ mol}\)

(d)(ii) According to the stoichiometry of the equation, \(1\text{ mole}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ moles}\) of \(\text{NaOH}\).
\(\text{Moles of H}_2\text{SO}_4 = \frac{1.88 \times 10^{-3}\text{ mol}}{2} = 9.40 \times 10^{-4}\text{ mol}\)

(d)(iii) \(\text{Concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{9.40 \times 10^{-4}\text{ mol}}{0.0250\text{ dm}^3} = 0.0376\text{ mol/dm}^3\) (3 significant figures).

(d)(iv) \(\text{Concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r = 0.0376\text{ mol/dm}^3 \times 98.0\text{ g/mol} = 3.6848 \approx 3.68\text{ g/dm}^3\).

(a) [1 mark] Colorless to pink (reject: clear to pink, pink to colorless).

(b) [2 marks]
- 1 mark for stating that a pipette measures a single/fixed/constant volume accurately.
- 1 mark for stating that a burette measures a variable volume / allows dropwise/slow addition.

(c) [2 marks]
- 1 mark for correct chemical formulas of reactants and products with correct balancing: \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)
- 1 mark for all state symbols correct: \(\text{(aq)}\) for acid, alkali, and salt, and \(\text{(l)}\) for water.

(d)(i) [1 mark] \(1.88 \times 10^{-3}\text{ mol}\) or \(0.00188\text{ mol}\).

(d)(ii) [1 mark] \(9.40 \times 10^{-4}\text{ mol}\) or \(0.00094\text{ mol}\) (allow error carried forward from d(i) divided by 2).

(d)(iii) [2.3 marks]
- 1 mark for dividing moles from d(ii) by \(0.0250\text{ dm}^3\).
- 1 mark for calculation accuracy: \(0.0376\text{ mol/dm}^3\).
- 0.3 marks for correct rounding to 3 significant figures.

(d)(iv) [1 mark] \(3.68\text{ g/dm}^3\) (allow error carried forward from d(iii) multiplied by 98.0, rounded appropriately).

(a)(i) Chloroethene is an alkene derivative. Its structure features a carbon-carbon double bond, \(\text{C}=\text{C}\). Three positions are occupied by hydrogen atoms, and one by a chlorine atom:
H H
\ /
C=C
/ \
H Cl

(a)(ii) In the repeat unit of an addition polymer, the double bond is converted to a single covalent bond with open-ended bonds continuing on either side:
[ H H ]
[ | | ]
--[--C - C--]--
[ | | ]
[ H Cl ]

(b)(i) In addition polymerization, monomers containing carbon-carbon double bonds link up to form a polymer as the sole product. In condensation polymerization, monomers containing functional groups react to form a polymer along with the elimination of a small molecule, typically water (\(\text{H}_2\text{O}\)) or hydrogen chloride (\(\text{HCl}\)).

(b)(ii) The ester linkage consists of a carbonyl group bonded to an oxygen atom: \(-\text{C}(=\text{O})-\text{O}-\).

(b)(iii) When the dicarboxylic acid (loss of \(-\text{OH}\)) reacts with the diol (loss of \(-\text{H}\)), water is eliminated and an ester link is formed. The repeat unit must contain one residue of each monomer connected by the ester link:
\(\text{[-O-CH}_2\text{-CH}_2\text{-O-CO-CH}_2\text{-CH}_2\text{-CO-]}\) with open-ended bonds at both ends.

(a)(i) [2 marks]
- 1 mark for correct \(\text{C}=\text{C}\) double bond.
- 1 mark for showing three \(\text{C-H}\) bonds and one \(\text{C-Cl}\) bond.

(a)(ii) [2 marks]
- 1 mark for correct single bond between carbons, containing brackets and open bonds extending outside the brackets: \(\text{[-CH}_2\text{-CHCl-]}_n\) or equivalent.
- 1 mark for correct substituents matching the monomer.

(b)(i) [2 marks]
- 1 mark for stating addition polymerization forms only one product (the polymer).
- 1 mark for stating condensation polymerization forms the polymer and a small molecule / water / hydrogen chloride.

(b)(ii) [2 marks]
- 1 mark for showing \(\text{C}=\text{O}\) double bond.
- 1 mark for showing single bond from that carbon to an oxygen, with correct open bonds on the carbonyl carbon and the single-bonded oxygen: \(-\text{C}(=\text{O})-\text{O}-\).

(b)(iii) [4.3 marks]
- 1 mark for correct diol residue block \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\).
- 1 mark for correct dicarboxylic acid residue block \(\text{-CO-CH}_2\text{-CH}_2\text{-CO-}\).
- 1 mark for correct ester linkage joining the monomer blocks.
- 1.3 marks for correct open bonds at both ends of the polymer chain, enclosed within brackets.

(a) To prepare an aqueous solution of compound \(\text{Z}\), the student should place a sample of the solid crystals in a suitable container (such as a beaker or test tube), add distilled/deionized water, and stir or shake the mixture until the solid has dissolved completely. If there are any insoluble impurities, the mixture should be filtered.

(b)(i) The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess is the characteristic test for iron(II) ions, \(\text{Fe}^{2+}\).

(b)(ii) The reaction involves the combination of iron(II) ions and hydroxide ions to form insoluble iron(II) hydroxide:
\(\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^{-}\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}\)

(c)(i) Barium nitrate reacts with sulfate ions to form barium sulfate, which is highly insoluble and appears as a white precipitate.

(c)(ii) The addition of dilute nitric acid destroys any carbonate ions (\(\text{CO}_3^{2-}\)) or sulfite ions (\(\text{SO}_3^{2-}\)) present. If not removed, these ions would also react with barium ions to form white precipitates (barium carbonate or barium sulfite), giving a false-positive result for sulfate.

(d) To test for ammonium ions: add aqueous sodium hydroxide to the compound, warm the mixture gently, and test any evolved gas with damp red litmus paper. Ammonium ions react with hydroxide ions on heating to release ammonia gas (\(\text{NH}_3\)), which is alkaline and will turn the damp red litmus paper blue.

(a) [1.3 marks]
- 1 mark for adding distilled/deionized water to the solid crystals and stirring/agitating.
- 0.3 marks for specifying use of distilled/deionized water (as opposed to tap water) and/or filtration if needed.

(b)(i) [1 mark] Iron(II) ion / \(\text{Fe}^{2+}\) (reject: iron, iron(III), \(\text{Fe}^{3+}\)).

(b)(ii) [2 marks]
- 1 mark for correct formulas of reactants and products: \(\text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe(OH)}_2\).
- 1 mark for correct state symbols: \(\text{(aq)}\) for reactants, \(\text{(s)}\) for the precipitate.

(c)(i) [1 mark] White precipitate (allow: white solid).

(c)(ii) [2 marks]
- 1 mark for stating it removes/reacts with carbonate/sulfite ions.
- 1 mark for explaining that this prevents a false-positive result / prevents the formation of other insoluble barium salts (like barium carbonate).

(d) [5 marks]
- 1 mark for adding aqueous sodium hydroxide.
- 1 mark for warming/heating the mixture.
- 1 mark for testing the evolved gas with damp red litmus paper.
- 1 mark for observing that the damp red litmus paper turns blue.
- 1 mark for identifying the gas as ammonia / \(\text{NH}_3\).

(a)
- When aqueous ammonia is added dropwise, a light blue (or blue) precipitate of copper(II) hydroxide forms.
- When excess aqueous ammonia is added, the light blue precipitate dissolves/is soluble.
- It forms a deep blue (or dark blue) solution.

(b)
- To the test solution, add aqueous sodium hydroxide (\(\text{NaOH}\)) and aluminum foil (or Devarda's alloy).
- Warm/heat the mixture gently.
- Test the gas produced with damp red litmus paper; it will turn blue (due to the production of ammonia gas).

(c)
(i) The solid residue left is copper(II) oxide (\(\text{CuO}\)), which is a black solid.
(ii) The other gas produced is oxygen (\(\text{O}_2\)). The test is to insert a glowing splint into the test tube. The positive result is that the glowing splint relights (rekindles).

(d)
The balanced equation for the decomposition is:
\[ 2\text{Cu(NO}_3\text{)}_2\text{(s)} \rightarrow 2\text{CuO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)} \]

(e)
1. Calculate moles of \(\text{Cu(NO}_3\text{)}_2\):
\[ \text{Moles} = \frac{\text{mass}}{M_r} = \frac{4.70 \text{ g}}{188 \text{ g/mol}} = 0.025 \text{ mol} \]
2. From the balanced equation, the mole ratio of \(\text{Cu(NO}_3\text{)}_2 : \text{NO}_2\) is \(2 : 4\) (or \(1 : 2\)).
\[ \text{Moles of } \text{NO}_2 = 0.025 \times 2 = 0.050 \text{ mol} \]
3. Calculate the volume of \(\text{NO}_2\) gas at r.t.p.:
\[ \text{Volume} = 0.050 \text{ mol} \times 24.0 \text{ dm}^3/\text{mol} = 1.2 \text{ dm}^3 \text{ (or } 1200 \text{ cm}^3\text{)} \]

(a) [3 marks total]
- 1 mark: light blue precipitate / blue precipitate formed dropwise.
- 1 mark: precipitate dissolves / is soluble in excess.
- 1 mark: to form a deep blue / dark blue solution.

(b) [3 marks total]
- 1 mark: Add aqueous sodium hydroxide AND aluminum foil / Devarda's alloy.
- 1 mark: Heat / warm the mixture.
- 1 mark: Gas evolved turns damp red litmus paper blue / ammonia gas is produced.

(c) [3 marks total]
- (i) 1 mark: black.
- (ii) 2 marks: use a glowing splint (1 mark); splint relights / rekindles (1 mark). (Reject: 'burning splint' or 'pop test').

(d) [2 marks total]
- 1 mark: Correct formulas of products (\(\text{CuO}\), \(\text{NO}_2\), \(\text{O}_2\)).
- 1 mark: Correct balancing coefficients (2, 2, 4, 1).

(e) [1.3 marks total]
- 0.5 marks: Correctly calculating the moles of \(\text{Cu(NO}_3\text{)}_2\) as 0.025 mol and using the 1:2 ratio to get 0.050 mol of \(\text{NO}_2\).
- 0.8 marks: Correct final volume of \(1.2\text{ dm}^3\) (or \(1200\text{ cm}^3\)) with correct units.

(a) Calculate the differences: Titre 1: 24.35 - 0.50 = 23.85 cm3; Titre 2: 47.60 - 24.35 = 23.25 cm3; Titre 3: 23.95 - 0.20 = 23.75 cm3; Titre 4: 47.85 - 24.05 = 23.80 cm3.
(b) Concordant titres are within 0.10 cm3 of each other: Titres 1, 3, and 4. Average = (23.85 + 23.75 + 23.80) / 3 = 23.80 cm3. (Titre 2 is excluded as it is an outlier).
(c) The titration starts with lime juice (acid) in the flask, so the initial color of phenolphthalein is colorless. At the end-point, the addition of excess alkali turns it pale pink / pink.
(d) The white tile provides a uniform background to see the faint color change clearly.
(e) Moles of NaOH = (23.80 / 1000) * 0.100 = 0.00238 mol. From the mole ratio 1:3, moles of citric acid in 25.0 cm3 = 0.00238 / 3 = 0.0007933 mol. Concentration in mol/dm3 = 0.0007933 * (1000 / 25.0) = 0.03173 mol/dm3. Concentration in g/dm3 = 0.03173 * 192 = 6.09 g/dm3 (accept 6.1 g/dm3).

(a) All 4 values calculated correctly [2 marks]. (3 correct [1 mark], fewer than 3 [0 marks]).
(b) Identifies Titres 1, 3, and 4 as concordant [1 mark]. Calculates average as 23.80 cm3 [1 mark].
(c) Colorless [1 mark] to pink / pale pink [1 mark]. (Reject: red / purple).
(d) To see the color change more clearly / easily [1 mark].
(e) Moles of NaOH calculated correctly (2.38 x 10^-3 mol) [1 mark]. Moles of citric acid calculated by dividing moles of NaOH by 3 (7.93 x 10^-4 mol) [1 mark]. Final concentration in g/dm3 calculated correctly (6.09 g/dm3) with units [1 mark].

(a) (i) Soluble iron(II) salts react with sodium hydroxide to form a dirty-green precipitate of iron(II) hydroxide, which does not dissolve in excess NaOH.
(a) (ii) Similarly, with aqueous ammonia, a green precipitate of iron(II) hydroxide forms, which is insoluble in excess ammonia.
(a) (iii) Adding barium nitrate in the presence of dilute nitric acid yields a white precipitate of barium sulfate, confirming the presence of sulfate ions (SO4^2-).
(b) (i) The transition from blue to pink in anhydrous cobalt(II) chloride paper confirms the presence of water.
(b) (ii) Acidified potassium manganate(VII) is a strong oxidizing agent; it is reduced and decolorized (purple to colorless) by reducing gases such as sulfur dioxide (SO2).
(b) (iii) Based on the cation (Iron(II)) and the anion (sulfate), Compound X is iron(II) sulfate (FeSO4).

(a)(i) green [1 mark], insoluble [1 mark].
(a)(ii) green [1 mark], insoluble [1 mark].
(a)(iii) white [1 mark], sulfate / SO4^(2-) [1 mark].
(b)(i) Presence of water [1 mark].
(b)(ii) Sulfur dioxide / SO2 [1 mark].
(b)(iii) Iron(II) [1 mark] sulfate [1 mark] (must specify the Roman numeral (II) for the full cation mark; accept FeSO4).

To establish a reactivity series using calorimetry, the four metals must be reacted under identical conditions with a solution containing the ions of the least reactive metal (or each metal reacted against all solutions systematically). The most efficient approach is to react the four metals (Mg, Zn, Fe, Cu) with copper(II) sulfate solution, since copper is the least reactive and will be displaced by the other three metals, producing varying temperature rises.

Experimental Method:
1. Measure a fixed volume (e.g., 25 cm3) of 0.5 mol/dm3 copper(II) sulfate solution using a measuring cylinder and pour it into an insulated polystyrene cup.
2. Support the cup in a beaker and place a thermometer in the solution to record the initial temperature.
3. Weigh out a fixed mass (e.g., 1.0 g - ensuring it is in excess) of magnesium powder.
4. Add the magnesium powder to the cup, stir the mixture continuously, and record the maximum temperature reached.
5. Repeat steps 1-4 using the same mass and particle size of zinc, iron, and copper powders respectively.

Variables to control:
- Volume of copper(II) sulfate solution (25 cm3).
- Concentration of copper(II) sulfate solution (0.5 mol/dm3).
- Mass of metal powder (1.0 g).
- Surface area / particle size of the metal powders (all should be fine powders).
- Insulation level of the cup.

Processing and Analysis:
- Calculate the temperature change (\(\Delta T = T_{\text{max}} - T_{\text{initial}}\)) for each metal.
- Rank the metals by the magnitude of temperature change: the metal causing the greatest temperature rise (Magnesium) is the most reactive, followed by Zinc, then Iron.
- Copper will show a temperature change of 0, indicating it is the least reactive of the group.

Method [Max 5 marks]:
- Measure a specified, fixed volume of copper(II) sulfate solution into an insulated / polystyrene cup [1 mark].
- Measure and record the initial temperature of the solution using a thermometer [1 mark].
- Add a specified, fixed mass (or excess) of a metal powder [1 mark].
- Stir the mixture [1 mark] and record the maximum temperature achieved [1 mark].
- Repeat the experiment using the other three metals (Zn, Fe, Cu) with fresh copper(II) sulfate solution [1 mark].

Control Variables [Max 2 marks]:
- Any two from: volume of sulfate solution, concentration of sulfate solution, mass of metal, particle size/surface area of metal, starting temperature [2 marks].

Data Processing & Conclusion [Max 3 marks]:
- Calculate the temperature change for each metal (\(T_{\text{max}} - T_{\text{initial}}\)) [1 mark].
- State that the larger the temperature rise, the more reactive the metal [1 mark].
- Correctly predict the expected order of reactivity based on temperature rise: Mg > Zn > Fe > Cu (no reaction/no temperature change for Cu) [1 mark].

To find the mixture with the highest percentage of sodium chloride:
1. Weigh a known mass of mixture A using a balance and record this starting mass (\(m_1\)).
2. Transfer the mixture to a beaker, add distilled water, and stir thoroughly with a glass rod to dissolve all the sodium chloride.
3. Filter the mixture through a filter paper and funnel into a flask.
4. Wash the residue (insoluble sand) remaining on the filter paper with a small amount of distilled water to ensure all salt solution is washed through.
5. Place the wet sand residue in a warm oven to dry completely, then weigh the dry sand and record this mass (\(m_2\)).
6. Repeat the exact same steps for mixtures B and C, using the same starting mass of each mixture (or recording their respective starting masses).
7. For each mixture, calculate the mass of dissolved sodium chloride by subtracting the mass of the dry sand from the initial mass of the mixture (\(\text{mass of NaCl} = m_1 - m_2\)).
8. Calculate the percentage of sodium chloride by mass in each mixture: \(\% \text{NaCl} = \frac{m_1 - m_2}{m_1} \times 100\).
9. Compare the percentages of mixtures A, B, and C. The mixture with the largest percentage contains the highest percentage of sodium chloride by mass.
10. Safety precaution: Wear safety goggles to protect eyes from solid particles and splashes during stirring and washing.

Award 1 mark for each of the following points, up to a maximum of 10 marks:
1. Weighing: Weigh a known starting mass of mixture A using a balance [1]
2. Dissolving: Add distilled water to the mixture in a beaker [1]
3. Dissolving technique: Stir/agitate the mixture to ensure the soluble salt dissolves completely [1]
4. Separation: Filter the mixture to separate insoluble sand as residue and salt solution as filtrate [1]
5. Washing: Wash/rinse the residue on the filter paper with distilled water to recover all dissolved salt [1]
6. Drying: Dry the residue (sand) in an oven/warm place OR evaporate the filtrate to dryness using an evaporating basin and Bunsen burner [1]
7. Final weighing: Weigh the dry sand residue OR weigh the dry evaporated salt [1]
8. Control/Repetition: Repeat the entire procedure for mixtures B and C [1]
9. Mass Calculation: Subtract the dry residue mass from the starting mixture mass to find the mass of salt, OR weigh the dry salt directly [1]
10. Processing & Comparison: Calculate \(\% \text{NaCl} = \frac{\text{mass of NaCl}}{\text{starting mass}} \times 100\) and identify the mixture with the highest percentage [1]

Accept: Alternative methods such as evaporating the filtrate to dryness and weighing the dry sodium chloride directly.
Reject: Any attempt to separate the dry mixture by heating directly (sand and salt do not separate or evaporate by dry heating).