2023 OCR A-Level Chemistry A - H432 模擬試題連答案詳解

The coordination number is 6 because six monodentate ligands (four water molecules and two chloride ions) are bonded to the central transition metal ion.

Since water is a neutral ligand (charge of 0) and chlorine forms a chloro ligand with a charge of \(-1\), the overall charge of \(+1\) on the complex requires the metal ion to have an oxidation state of \(+3\) (since \(3 + 4(0) + 2(-1) = +1\)).

Cis-trans isomerism is possible, but neither the cis nor the trans isomer is chiral (both contain planes of symmetry), so optical isomerism is not observed. Because the chloride ions are coordinated inside the inner sphere as ligands, they do not easily dissociate in aqueous solution to react with silver nitrate.

Award 1 mark for the correct option B.
Reject other options:
- A is incorrect because the coordination number is 6.
- C is incorrect because the complex can display cis-trans isomerism but not optical isomerism.
- D is incorrect because the chloride ligands are inside the coordination sphere and do not readily precipitate.

1. Calculate the initial amount, in moles, of propanoic acid (\(\text{HA}\)):
\(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\)

2. Calculate the amount, in moles, of \(\text{NaOH}\) added:
\(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 0.00300\text{ mol}\)

3. Propanoic acid reacts with sodium hydroxide in a 1:1 ratio:
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)

4. Calculate the remaining moles of propanoic acid (\(\text{HA}\)) after neutralization:
\(n(\text{HA})_{\text{remaining}} = 0.00500 - 0.00300 = 0.00200\text{ mol}\)

5. Calculate the moles of conjugate base (\(\text{A}^-\)) formed:
\(n(\text{A}^-) = 0.00300\text{ mol}\)

6. Use the Henderson-Hasselbalch equation (or the \(K_a\) expression) to find pH:
\(\text{pH} = -\log_{10}(K_a) + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)
\(\text{pH} = -\log_{10}(1.30 \times 10^{-5}) + \log_{10}\left(\frac{0.00300}{0.00200}\right)\)
\(\text{pH} = 4.886 + 0.176 = 5.06\)

Award 1 mark for the correct option D.
Reject:
- A (4.61) and B (4.71) which result from incorrect buffer ratio calculations (e.g. inverted ratio of acid/base).
- C (4.89) which is the value of pKa without considering the addition of base.

Step 1: hot ethanolic \(\text{NaOH}\) (elimination) converts 1-bromobutane to but-1-ene.

Step 2: reaction of but-1-ene with \(\text{HBr(g)}\) (electrophilic addition) occurs via the more stable secondary carbocation intermediate, yielding 2-bromobutane as the major product.

Step 3: excess ethanolic ammonia under pressure (nucleophilic substitution) converts 2-bromobutane to 2-aminobutane.

Option A produces 1-aminobutane. Option B produces butanoic acid. Option D is incorrect as alkenes do not react with aqueous ammonia.

Award 1 mark for the correct option C.
Reject options A, B, and D because they do not lead to the formation of the target secondary amine (2-aminobutane) as the major product.

To calculate the enthalpy change of the reaction:
\(\Delta H = \Sigma(\text{bond enthalpies of reactants}) - \Sigma(\text{bond enthalpies of products})\)

Alternatively, consider only the bonds that change:
Bonds broken:
- 1 \(\text{C=C}\) bond = \(+612\text{ kJ mol}^{-1}\)
- 1 \(\text{H–H}\) bond = \(+436\text{ kJ mol}^{-1}\)
Total energy input = \(612 + 436 = +1048\text{ kJ mol}^{-1}\)

Bonds formed:
- 1 \(\text{C–C}\) bond = \(+347\text{ kJ mol}^{-1}\)
- 2 \(\text{C–H}\) bonds = \(2 \times 413 = +826\text{ kJ mol}^{-1}\)
Total energy released = \(347 + 826 = +1173\text{ kJ mol}^{-1}\)

\(\Delta H = 1048 - 1173 = -125\text{ kJ mol}^{-1}\)

Award 1 mark for the correct option A.
Reject option B (positive sign of the correct calculation), and options C and D which use incorrect stoichiometric ratios of bond enthalpies.

1. The IR spectrum shows a strong absorption at \(1740\text{ cm}^{-1}\) which corresponds to a carbonyl group (\(\text{C=O}\)). There is no broad peak above \(3000\text{ cm}^{-1}\), indicating the absence of an \(\text{O–H}\) group. This rules out butanoic acid (option D), so the compound must be an ester.

2. In the \(^1\text{H}\) NMR spectrum:
- The singlet at \(\delta = 2.0\text{ ppm}\) (3H) corresponds to a methyl group next to a carbonyl group (\(\text{CH}_3\text{-CO-}\)).
- The quartet at \(\delta = 4.1\text{ ppm}\) (2H) and triplet at \(\delta = 1.2\text{ ppm}\) (3H) form a characteristic spin system for an ethyl group adjacent to an oxygen atom in an ester (\(-\text{O-CH}_2\text{CH}_3\)).

Combining these segments yields ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)).

Award 1 mark for the correct option A.
Reject:
- B (methyl propanoate would show a quartet around 2.3 ppm and a singlet around 3.7 ppm).
- C (propyl methanoate would show a singlet at around 8.0 ppm for the ester proton).
- D (butanoic acid is ruled out by the lack of an OH peak in the IR spectrum).

Using the ideal gas equation: \(PV = nRT\)

Convert units to SI units:
- \(P = 1.01 \times 10^5\text{ Pa}\)
- \(V = 350\text{ cm}^3 = 350 \times 10^{-6}\text{ m}^3 = 3.50 \times 10^{-4}\text{ m}^3\)
- \(T = 293\text{ K}\)

Calculate the amount of gas in moles, \(n\):
\(n = \frac{PV}{RT} = \frac{(1.01 \times 10^5\text{ Pa}) \times (3.50 \times 10^{-4}\text{ m}^3)}{(8.314\text{ J mol}^{-1}\text{ K}^{-1}) \times (293\text{ K})} = \frac{35.35}{2435.998} = 0.01451\text{ mol}\)

According to the stoichiometric equation, the molar ratio of \(\text{CaCO}_3 : \text{CO}_2\) is 1:1.
Therefore, the moles of \(\text{CaCO}_3\) decomposed is \(0.01451\text{ mol}\).

Calculate the mass of \(\text{CaCO}_3\):
\(m = n \times M_r = 0.01451\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.452\text{ g} \approx 1.45\text{ g}\)

Award 1 mark for the correct option D.
Reject:
- A (which results from using an incorrect molar volume approximation of 24.4 dm3 mol-1).
- B and C which are incorrect by factors of 10 due to incorrect unit conversions of volume.

The nitration of methylbenzene is an electrophilic aromatic substitution reaction (ruling out option A).

Concentrated sulfuric acid acts as a catalyst by protonating concentrated nitric acid to form the active electrophile, the nitronium ion (\(\text{NO}_2^+\)), which makes option B correct.

The methyl group is an electron-donating group (by inductive effect), which activates the benzene ring (ruling out option D) and directs electrophilic attack to the 2- and 4-positions (ortho/para-directing), making the major mono-nitrated products 2-nitromethylbenzene and 4-nitromethylbenzene (ruling out option C).

Award 1 mark for the correct option B.
Reject other options:
- A is incorrect as the reaction mechanism is electrophilic substitution.
- C is incorrect because the methyl group is 2,4-directing.
- D is incorrect because the methyl group is electron-donating and activates the ring.

The value of the equilibrium constant \(K_c\) is affected only by a change in temperature.

Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right to produce more heat. This increases both the equilibrium yield of \(\text{N}_2\text{O}_4\text{(g)}\) and the value of \(K_c\).

Increasing the pressure (option A) shifts the equilibrium to the right and increases the yield of \(\text{N}_2\text{O}_4\), but does not change the value of \(K_c\). Adding a catalyst (option C) has no effect on either. Increasing the concentration of reactants (option D) increases the yield of products but does not alter \(K_c\).

Award 1 mark for the correct option B.
Reject options A, C, and D because they do not change the value of the equilibrium constant \(K_c\).

1. Calculate initial moles: n(CH3CH2COOH) = 0.0500 dm3 x 0.200 mol dm-3 = 0.0100 mol. n(NaOH) = 0.0500 dm3 x 0.100 mol dm-3 = 0.0050 mol. 2. Propanoic acid reacts with NaOH in a 1:1 ratio: CH3CH2COOH + NaOH -> CH3CH2COONa + H2O. 3. Moles remaining after reaction: n(CH3CH2COOH) = 0.0100 - 0.0050 = 0.0050 mol. n(CH3CH2COO-) formed = 0.0050 mol. 4. Since the moles of weak acid and its conjugate base are equal, the ratio [CH3CH2COO-]/[CH3CH2COOH] = 1. Therefore, [H+] = Ka = 1.35 x 10^-5 mol dm-3. 5. pH = -log10(1.35 x 10^-5) = 4.87.

1 mark for selecting the correct option B. Award 1 mark for correct pH calculation or identifying that [H+] = Ka when acid is half-neutralised.

When excess concentrated HCl is added to aqueous cobalt(II) ions, a ligand substitution reaction occurs to form a blue tetrahedral complex: [Co(H2O)6]2+ + 4Cl- <-> [CoCl4]2- + 6H2O. The complex ion has the formula [CoCl4]2-. It has 4 chloride ligands, meaning its coordination number is 4. The large size of the chloride ligands causes steric hindrance, which prevents 6-coordination and leads to a tetrahedral geometry rather than square planar.

1 mark for selecting the correct option A. Award 1 mark for identifying the correct complex formula, coordination number, and tetrahedral geometry.

1. Calculate cell potential: E_theta_cell = E_theta_red - E_theta_ox = +0.77 - (-0.74) = +1.51 V. 2. Determine direction: The half-cell with the more positive E_theta (+0.77 V) undergoes reduction: Fe3+(aq) + e- -> Fe2+(aq). The half-cell with the more negative E_theta (-0.74 V) undergoes oxidation: Cr(s) -> Cr3+(aq) + 3e-. 3. Balance electrons by multiplying the iron half-equation by 3: 3Fe3+(aq) + 3e- -> 3Fe2+(aq). 4. Combine half-equations: 3Fe3+(aq) + Cr(s) -> 3Fe2+(aq) + Cr3+(aq).

1 mark for selecting the correct option A. Award 1 mark for the correct calculation of standard cell potential and the correctly balanced overall spontaneous equation.

According to VSEPR theory, XeF4 has 6 electron pairs around the central xenon atom (4 bonding pairs and 2 lone pairs). To minimise repulsion, the 2 lone pairs are positioned opposite each other, resulting in a square planar shape. SF4 is seesaw (4 bonding pairs, 1 lone pair). NH4+ and BF4- are tetrahedral (4 bonding pairs, 0 lone pairs).

1 mark for selecting the correct option C. Award 1 mark for identifying the square planar shape based on valence shell electron pair repulsion theory.

1. Empirical formula: %H = 100 - 85.7 = 14.3%. Ratio of C to H = (85.7 / 12.0) : (14.3 / 1.0) = 7.14 : 14.3 = 1 : 2. Empirical formula is CH2 (formula mass = 14.0). 2. Molar mass: moles of gas = 24.0 / 24,000 = 0.00100 mol. Mr = mass / moles = 0.042 / 0.00100 = 42.0 g mol-1. 3. Molecular formula: 42.0 / 14.0 = 3. Therefore, the molecular formula is (CH2)x3 = C3H6.

1 mark for selecting the correct option C. Award 1 mark for correct calculation of empirical formula, molar mass, and molecular formula.

The first ionisation energy of sulfur is lower than that of phosphorus because sulfur's outer electron configuration is 3s2 3p4, which contains one paired set of electrons in a 3p orbital. The repulsion between these paired electrons makes it easier to remove one of them compared to removing an unpaired electron from phosphorus's half-filled 3p3 subshell. Shielding is similar as they are in the same period, so A is incorrect. The outer electron is in 3p, not 3d, so C is incorrect. Phosphorus has a half-filled 3p subshell, not completely filled, so D is incorrect.

1 mark for selecting the correct option B. Award 1 mark for explaining the effect of spin-pairing repulsion on first ionisation energy across Period 3.

Using Hess's Law: DrH_theta = Sum(DfH_theta(products)) - Sum(DfH_theta(reactants)). Products: Sum(DfH_theta(products)) = [2 x (+135.0)] + [6 x (-242.0)] = 270.0 - 1452.0 = -1182.0 kJ mol-1. Reactants: Sum(DfH_theta(reactants)) = [2 x (-46.0)] + [2 x (-75.0)] = -92.0 - 150.0 = -242.0 kJ mol-1 (note DfH_theta(O2) = 0). Calculation: DrH_theta = -1182.0 - (-242.0) = -940.0 kJ mol-1.

1 mark for selecting the correct option A. Award 1 mark for correct calculation of reactants sum, products sum, and overall enthalpy change of reaction.

1. Calculate moles of \(\text{MnO}_4^-\): \(\text{n}(\text{MnO}_4^-) = 0.0200 \times \frac{45.00}{1000} = 9.00 \times 10^{-4}\text{ mol}\). 2. Calculate moles of \(\text{FeC}_2\text{O}_4 \cdot x\text{H}_2\text{O}\): Using the 5:3 ratio: \(\text{n}(\text{salt}) = 9.00 \times 10^{-4} \times \frac{5}{3} = 1.50 \times 10^{-3}\text{ mol}\). 3. Calculate molar mass (\(M_r\)) of the hydrated salt: \(M_r = \frac{0.2697\text{ g}}{1.50 \times 10^{-3}\text{ mol}} = 179.8\text{ g mol}^{-1}\). 4. Calculate \(x\): \(M_r(\text{FeC}_2\text{O}_4) = 55.8 + 24.0 + 64.0 = 143.8\text{ g mol}^{-1}\). Mass of water = \(179.8 - 143.8 = 36.0\text{ g mol}^{-1}\). \(x = \frac{36.0}{18.0} = 2\).

M1: Calculates \(\text{n}(\text{MnO}_4^-) = 9.00 \times 10^{-4}\text{ mol}\) and \(\text{n}(\text{salt}) = 1.50 \times 10^{-3}\text{ mol}\) (1 mark). M2: Calculates molar mass of the hydrated salt as \(179.8\text{ g mol}^{-1}\) (or sets up correct algebraic equation) (1 mark). M3: Deduces \(x = 2\) (1 mark).

1. Initial moles: \(\text{n}(\text{HA}) = 0.0400 \times 0.150 = 6.00 \times 10^{-3}\text{ mol}\); \(\text{n}(\text{OH}^-) = 0.0200 \times 0.180 = 3.60 \times 10^{-3}\text{ mol}\). 2. Post-reaction moles: \(\text{n}(\text{A}^-) = 3.60 \times 10^{-3}\text{ mol}\); \(\text{n}(\text{HA})_{\text{remaining}} = 6.00 \times 10^{-3} - 3.60 \times 10^{-3} = 2.40 \times 10^{-3}\text{ mol}\). 3. Calculate \([\text{H}^+]\): \([\text{H}^+] = K_a \times \frac{\text{n}(\text{HA})}{\text{n}(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{2.40 \times 10^{-3}}{3.60 \times 10^{-3}} = 9.00 \times 10^{-6}\text{ mol dm}^{-3}\). 4. Calculate pH: \(\text{pH} = -\log_{10}(9.00 \times 10^{-6}) = 5.0457 \approx 5.05\).

M1: Calculates initial moles of propanoic acid (\(6.00 \times 10^{-3}\)) and NaOH (\(3.60 \times 10^{-3}\)) (1 mark). M2: Calculates remaining propanoic acid (\(2.40 \times 10^{-3}\text{ mol}\)) and formed propanoate (\(3.60 \times 10^{-3}\text{ mol}\)) (1 mark). M3: Calculates \([\text{H}^+] = 9.00 \times 10^{-6}\text{ mol dm}^{-3}\) (1 mark). M4: Deduces \(\text{pH} = 5.05\) (2 decimal places required) (1 mark).

Bonds broken: \(1 \times (\text{C}=\text{C}) + 4 \times (\text{C}-\text{H}) + 1 \times (\text{F}-\text{F}) = 612 + 1652 + 158 = 2422\text{ kJ mol}^{-1}\). Bonds formed: \(1 \times (\text{C}-\text{C}) + 4 \times (\text{C}-\text{H}) + 2 \times (\text{C}-\text{F}) = 347 + 1652 + 934 = 2933\text{ kJ mol}^{-1}\). \(\Delta H = \text{Bonds broken} - \text{Bonds formed} = 2422 - 2933 = -511\text{ kJ mol}^{-1}\).

M1: Calculates total energy of bonds broken as \(2422\text{ kJ mol}^{-1}\) (or \(770\text{ kJ mol}^{-1}\) if \(\text{C}-\text{H}\) bonds are excluded) (1 mark). M2: Calculates total energy of bonds formed as \(2933\text{ kJ mol}^{-1}\) (or \(1281\text{ kJ mol}^{-1}\) if \(\text{C}-\text{H}\) bonds are excluded) (1 mark). M3: Calculates overall enthalpy change \(\Delta H = -511\text{ kJ mol}^{-1}\) (sign required) (1 mark).

Born-Haber cycle equation: \(\Delta H_f^\ominus = \Delta H_{at}^\ominus(\text{Mg}) + \text{IE}_1 + \text{IE}_2 + \text{Bond Enthalpy}(\text{F}_2) + 2 \times \text{EA}(\text{F}) + \Delta H_{latt}^\ominus\). Substitute values: \(-1124 = 148 + 738 + 1451 + 158 + 2(-328) + \Delta H_{latt}^\ominus\). \(-1124 = 2495 - 656 + \Delta H_{latt}^\ominus = 1839 + \Delta H_{latt}^\ominus\). \(\Delta H_{latt}^\ominus = -1124 - 1839 = -2963\text{ kJ mol}^{-1}\).

M1: Multiplies electron affinity of fluorine by two: \(2 \times (-328) = -656\text{ kJ mol}^{-1}\) (1 mark). M2: Utilises the full bond enthalpy of \(\text{F}_2\) (\(+158\text{ kJ mol}^{-1}\)) for forming 2 moles of gaseous F atoms (1 mark). M3: Sets up the correct equation linking all enthalpy terms (1 mark). M4: Calculates final lattice enthalpy value as \(-2963\text{ kJ mol}^{-1}\) (accept \(+2963\text{ kJ mol}^{-1}\) only if clearly defined as lattice dissociation) (1 mark).

1. Standard cell potential: \(E_{\text{cell}}^\ominus = E^\ominus(\text{cathode}) - E^\ominus(\text{anode}) = 1.33 - 0.77 = +0.56\text{ V}\). 2. For the overall equation, multiply the iron half-equation by 6 to balance electrons: \(6\text{Fe}^{2+}(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 6\text{e}^-\). Combining gives: \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) + 6\text{Fe}^{3+}(\text{aq})\).

M1: Calculates \(E_{\text{cell}}^\ominus = +0.56\text{ V}\) (1 mark). M2: Correctly balances the stoichiometry (1 \(\text{Cr}_2\text{O}_7^{2-}\) : 14 \(\text{H}^+\) : 6 \(\text{Fe}^{2+}\) to produce 2 \(\text{Cr}^{3+}\) : 7 \(\text{H}_2\text{O}\) : 6 \(\text{Fe}^{3+}\)) (1 mark). M3: Complete overall equation is balanced and has correct ionic charges (1 mark).

- At pH 1 (highly acidic), both carboxyl groups remain protonated (\(\text{-COOH}\)) and the amino group is protonated to form a cation: \(\text{HOOC}-\text{CH}_2-\text{CH}(\text{NH}_3^+)-\text{COOH}\). - At pH 13 (highly alkaline), both carboxyl groups are deprotonated to form carboxylate anions (\(\text{-COO}^-\)) and the amino group remains neutral (\(\text{-NH}_2\)): \(\text{OOC}^--\text{CH}_2-\text{CH}(\text{NH}_2)-\text{COO}^-\). - There is 1 chiral carbon in aspartic acid (the central CH carbon bonded to four different groups).

M1: Correct structure of aspartic acid at pH 1 with protonated amine (1 mark). M2: Correct structure of aspartic acid at pH 13 with both carboxyl groups deprotonated (1 mark). M3: Identifies 1 chiral carbon atom (1 mark).

1. Molar masses: \(M_r(\text{C}_7\text{H}_8) = 92.0\text{ g mol}^{-1}\), \(M_r(\text{C}_9\text{H}_{10}\text{O}_2) = 150.0\text{ g mol}^{-1}\). 2. Initial moles of methylbenzene = \(\frac{18.4}{92.0} = 0.200\text{ mol}\). 3. Overall yield = \(75.0\% \times 60.0\% = 45.0\%\). 4. Actual moles of ethyl benzoate = \(0.200 \times 0.450 = 0.0900\text{ mol}\). 5. Actual mass of ethyl benzoate = \(0.0900 \times 150.0 = 13.5\text{ g}\).

M1: Calculates molar masses of reactant (\(92.0\)) and product (\(150.0\)) (1 mark). M2: Calculates initial moles of methylbenzene as \(0.200\text{ mol}\) (1 mark). M3: Calculates overall fractional yield as \(0.450\) (or step-wise moles: \(0.150\text{ mol}\) after Step 1) (1 mark). M4: Calculates final mass of ethyl benzoate as \(13.5\text{ g}\) (1 mark).

1. \(\text{ClO}_3^-\) central Cl has 7 valence electrons, plus 1 from the negative charge = 8. It forms 3 bonding regions with oxygens and has 1 lone pair remaining. 2. 4 electron-pair regions arrange tetrahedrally to minimise repulsion, resulting in a pyramidal (or trigonal pyramidal) shape with a bond angle of approximately \(107^\circ\) (accept range \(104^\circ - 108^\circ\)). 3. Repulsion: electron pairs repel to be as far apart as possible; lone pairs repel more than bonding pairs, forcing the bonds closer together.

M1: Identifies shape as pyramidal / trigonal pyramidal AND bond angle in range \(104^\circ - 108^\circ\) (1 mark). M2: Identifies that there are 3 bonding regions and 1 lone pair on the central chlorine atom (1 mark). M3: Explains that electron pairs repel to minimise repulsion, with lone pairs repelling more than bonding pairs (1 mark).

1. **Calculate the moles of chloride ions in the volumetric flask:**
- Molar mass of \(\text{AgCl} = 107.9 + 35.5 = 143.4\text{ g mol}^{-1}\).
- Moles of \(\text{AgCl}\) in \(25.00\text{ cm}^3\) portion = \(\frac{1.076}{143.4} = 0.00750\text{ mol}\).
- Moles of \(\text{Cl}^-\)\ in the original \(100.0\text{ cm}^3\) solution = \(0.00750 \times 4 = 0.0300\text{ mol}\).

2. **Determine \(n\) in \(\text{CrCl}_n\):**
- Mass of anhydrous salt = \(1.585\text{ g}\).
- Since the moles of \(\text{CrCl}_n\) is \(\frac{0.0300}{n}\):
\(\text{Mass} = \text{moles} \times M_r \implies 1.585 = \frac{0.0300}{n} \times (52.0 + 35.5n)\)
\(1.585 = \frac{1.56}{n} + 1.065\)
\(0.52 = \frac{1.56}{n} \implies n = 3\).

3. **Determine \(x\) (water of crystallisation):**
- Mass of water lost = \(2.665\text{ g} - 1.585\text{ g} = 1.080\text{ g}\).
- Moles of \(\text{H}_2\text{O}\) lost = \(\frac{1.080}{18.0} = 0.0600\text{ mol}\).
- Moles of anhydrous salt (\(\text{CrCl}_3\)) = \(\frac{0.0300}{3} = 0.0100\text{ mol}\).
- Ratio of \(\text{H}_2\text{O} : \text{CrCl}_3 = \frac{0.0600}{0.0100} = 6\).
- Therefore, \(x = 6\).

4. **Formula:** \(\text{CrCl}_3 \cdot 6\text{H}_2\text{O}\).

M1: Calculates moles of chloride ions in 100 cm3 solution (0.0300 mol). [1 mark]
M2: Sets up algebraic equation using anhydrous mass and moles of chloride to calculate n = 3. [1 mark]
M3: Calculates moles of water lost (0.0600 mol) and moles of CrCl3 (0.0100 mol). [1 mark]
M4: Correctly identifies x = 6 and states formula as CrCl3.6H2O. [1 mark]

1. **Calculate initial moles:**
- \(n(\text{HCOOH}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
- \(n(\text{HCOO}^-) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.00750\text{ mol}\)

2. **Calculate added moles of hydroxide ions:**
- \(n(\text{OH}^-) = 0.00200\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00020\text{ mol}\)

3. **Calculate moles at equilibrium after addition:**
- \(n(\text{HCOOH}) = 0.0125 - 0.00020 = 0.0123\text{ mol}\)
- \(n(\text{HCOO}^-) = 0.00750 + 0.00020 = 0.00770\text{ mol}\)

4. **Calculate pH:**
- \([\text{H}^+] = K_a \times \frac{n(\text{HCOOH})}{n(\text{HCOO}^-)} = 1.80 \times 10^{-4} \times \frac{0.0123}{0.00770} = 2.8753 \times 10^{-4}\text{ mol dm}^{-3}\)
- \(\text{pH} = -\log_{10}(2.8753 \times 10^{-4}) = 3.5413 \approx 3.54\)

M1: Calculates initial moles of HCOOH (0.0125 mol) and HCOO- (0.00750 mol) and added OH- (0.00020 mol). [1 mark]
M2: Calculates new moles after reaction: HCOOH = 0.0123 mol and HCOO- = 0.00770 mol. [1 mark]
M3: Calculates pH correctly as 3.54 (must be 2 decimal places). [1 mark]
Note: Allow ecf from incorrect moles.

1. **Reagent for Step 2:** Concentrated aqueous ammonia (\(\text{NH}_3(\text{aq})\)).
2. **Reason for excess:** If a stoichiometric amount of ammonia is used, the formed primary amine (2-aminopropanoic acid) can act as a nucleophile and react with more 2-chloropropanoic acid, leading to secondary and tertiary amines or quaternary ammonium salt impurities. A large excess ensures that the primary amine is the major product. (Alternatively, excess ammonia is needed to react with/neutralise the acidic \(\text{HCl}\) byproduct to prevent protonation of the nucleophile).
3. **Mechanism type:** Nucleophilic substitution.

M1: Identifies reagent as concentrated ammonia / NH3. [1 mark]
M2: Explains that excess prevents further substitution reactions to form secondary/tertiary amines / maximises yield of primary amine. [1 mark]
M3: States mechanism type is nucleophilic substitution. [1 mark]

(a) The half-system with the more positive \(E^\theta\) value undergoes reduction, while the other undergoes oxidation. Thus, \(\text{Fe}^{3+}\) is reduced and \(\text{I}^-\)
is oxidized:
\(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

(b) \(E^\theta_{\text{cell}} = E^\theta(\text{reduction}) - E^\theta(\text{oxidation}) = +0.77 - (+0.54) = +0.23\text{ V}\).

(c) Platinum (\(\text{Pt}\)) is used because both half-cells contain aqueous species without solid metals.

M1: Correct overall equation: 2Fe3+(aq) + 2I-(aq) -> 2Fe2+(aq) + I2(aq) (state symbols not required but must be balanced). [1 mark]
M2: Correct cell potential: +0.23 V (unit required, sign optional but preferred). [1 mark]
M3: Correct electrode material: Platinum / Pt. [1 mark]

By constructing a Born-Haber cycle for \(\text{MgF}_2(\text{s})\):

\(\Delta H_f^\theta[\text{MgF}_2(\text{s})] = \Delta H_{\text{at}}^\theta[\text{Mg}] + \text{IE}_1[\text{Mg}] + \text{IE}_2[\text{Mg}] + \Delta H_{\text{bond}}^\theta[\text{F}_2] + 2 \times \text{EA}_1[\text{F}] + \Delta H_{\text{LE}}^\theta[\text{MgF}_2(\text{s})]\)

Substitute the values into the equation:
\(-1124 = 148 + 738 + 1451 + 158 + 2(-328) + \Delta H_{\text{LE}}^\theta\)

Calculate the intermediate sum of gaseous ion generation:
\(\text{Sum} = 148 + 738 + 1451 + 158 - 656 = +1839\text{ kJ mol}^{-1}\)

Now, solve for \(\Delta H_{\text{LE}}^\theta\):
\(-1124 = 1839 + \Delta H_{\text{LE}}^\theta\)
\(\Delta H_{\text{LE}}^\theta = -1124 - 1839 = -2963\text{ kJ mol}^{-1}\)

M1: Identifies that 1 mol of F-F bond enthalpy is required (to form 2 moles of F atoms) i.e. +158 kJ mol-1. [1 mark]
M2: Accounts for 2 moles of electron affinity of fluorine, i.e., 2 x (-328) = -656 kJ mol-1. [1 mark]
M3: Shows a complete and correct mathematical expression/ Born-Haber cycle working. [1 mark]
M4: Evaluates final answer correctly to -2963 kJ mol-1 (must have negative sign and correct value). [1 mark]

1. **Determine equilibrium moles using ICE method:**
- Initial moles: \(\text{SO}_2 = 1.50\), \(\text{O}_2 = 1.00\), \(\text{SO}_3 = 0.00\)
- Change in moles: \(\text{SO}_3 = +0.80\), \(\text{SO}_2 = -0.80\), \(\text{O}_2 = -0.40\)
- Equilibrium moles: \(\text{SO}_2 = 0.70\), \(\text{O}_2 = 0.60\), \(\text{SO}_3 = 0.80\)

2. **Convert moles to concentration (Volume = \(2.00\text{ dm}^3\)):**
- \([\text{SO}_2] = \frac{0.70}{2.00} = 0.35\text{ mol dm}^{-3}\)
- \([\text{O}_2] = \frac{0.60}{2.00} = 0.30\text{ mol dm}^{-3}\)
- \([\text{SO}_3] = \frac{0.80}{2.00} = 0.40\text{ mol dm}^{-3}\)

3. **Calculate \(K_c\):**
- \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.40)^2}{(0.35)^2 \times 0.30} = \frac{0.16}{0.1225 \times 0.30} = \frac{0.16}{0.03675} = 4.3537 \approx 4.35\)

4. **Determine units:**
- \(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\)

M1: Calculates equilibrium moles of SO2 (0.70 mol) and O2 (0.60 mol). [1 mark]
M2: Divides by volume (2.00 dm3) and substitutes correct concentrations into Kc expression. [1 mark]
M3: Correctly evaluates numerical value as 4.35 (to 3 sf) and units as dm3 mol-1. [1 mark]

(a) The equation for the generation of the nitronium ion electrophile is:
\(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)
(or \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\))

(b) Mechanism steps:
1. A curly arrow from the pi-system of the benzene ring of methyl benzoate to the nitrogen atom of the \(\text{NO}_2^+\)
electrophile.
2. Formation of the intermediate: A positively charged species where the ring has a broken pi-system (horseshoe shape spanning 5 carbons, open and positive charge pointing towards carbon-3) and carbon-3 is bonded to both a hydrogen atom and a nitro (\(-\text{NO}_2\)) group.
3. A curly arrow from the \(\text{C}-\text{H}\) bond of carbon-3 back into the ring to reform the delocalised pi-system and release an \(\text{H}^+\)
ion.

M1: Writes correct equation for the generation of the NO2+ electrophile. [1 mark]
M2: Draws curly arrow from the benzene ring to the NO2+ electrophile at the meta-position. [1 mark]
M3: Draws the correct structure of the positively charged intermediate with curly arrow from the C-H bond of carbon-3 back into the ring. [1 mark]

(a) Calculate \(q\):
\(\Delta T = 50.5 - 18.5 = 32.0\text{ }^\circ\text{C}\) (or \(32.0\text{ K}\))
\(q = mc\Delta T = 200.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 32.0\text{ K} = 26752\text{ J} = 26.752\text{ kJ}\) (accept \(26.8\text{ kJ}\))

(b) Calculate \(\Delta_c H\):
\(n(\text{pentan-1-ol}) = \frac{0.880\text{ g}}{88.0\text{ g mol}^{-1}} = 0.0100\text{ mol}\)
\(\Delta_c H = -\frac{q}{n} = -\frac{26.752\text{ kJ}}{0.0100\text{ mol}} = -2675.2\text{ kJ mol}^{-1}\) (accept \(-2680\text{ kJ mol}^{-1}\) or \(-2675\text{ kJ mol}^{-1}\))

(c) One reason other than heat loss:
- Incomplete combustion of pentan-1-ol (indicated by soot formation).
- Evaporation of the alcohol from the wick after extinguishing.

M1: Calculates q as 26.8 kJ or 26.75 kJ. [1 mark]
M2: Calculates moles of pentan-1-ol as 0.0100 mol. [1 mark]
M3: Calculates Delta_c H as -2680 or -2675 kJ mol-1 (must have negative sign and match working, allow ecf from M1 and M2). [1 mark]
M4: Suggests incomplete combustion OR evaporation of alcohol/water (Do not accept 'heat loss' or 'non-standard conditions'). [1 mark]

1. Calculate initial moles: Moles of propanoic acid = \(0.0500 \times 0.250 = 0.0125\text{ mol}\). Moles of NaOH = \(0.0300 \times 0.150 = 0.00450\text{ mol}\). 2. Determine moles at equilibrium: Propanoate ions (A⁻) formed = \(0.00450\text{ mol}\). Propanoic acid remaining (HA) = \(0.0125 - 0.00450 = 0.00800\text{ mol}\). 3. Calculate \([H^+]\): \([H^+] = K_a \times \frac{[HA]}{[A^-]} = 1.35 \times 10^{-5} \times \frac{0.00800}{0.00450} = 2.40 \times 10^{-5}\text{ mol dm}^{-3}\). 4. Calculate pH: \(\text{pH} = -\log_{10}(2.40 \times 10^{-5}) = 4.62\).

M1: Moles of propanoic acid remaining = 0.00800 mol AND moles of propanoate ions = 0.00450 mol. M2: Correct substitution into the \(K_a\) equation to find \([H^+] = 2.40 \times 10^{-5}\text{ mol dm}^{-3}\). M3: Correct calculation of pH to 2 decimal places = 4.62.

1. Calculate moles of Cl⁻: \(n(\text{Cl}) = n(\text{AgCl}) = \frac{2.151}{143.4} = 0.0150\text{ mol}\). 2. Calculate moles of \(\text{NH}_3\): \(n(\text{NH}_3) = n(\text{HCl}) = 0.0500 \times 0.500 = 0.0250\text{ mol}\). 3. Calculate mass of cobalt: Mass of Cl = \(0.0150 \times 35.5 = 0.5325\text{ g}\). Mass of \(\text{NH}_3\) = \(0.0250 \times 17.0 = 0.4250\text{ g}\). Mass of Co = \(1.252 - 0.5325 - 0.4250 = 0.2945\text{ g}\). 4. Calculate moles of Co: \(n(\text{Co}) = \frac{0.2945}{58.9} = 0.00500\text{ mol}\). 5. Determine mole ratio: Co : \(\text{NH}_3\) : Cl = 0.00500 : 0.0250 : 0.0150 = 1 : 5 : 3. The empirical formula is \(\text{Co}(\text{NH}_3)_5\text{Cl}_3\).

M1: Calculate moles of chloride ions = 0.0150 mol. M2: Calculate moles of ammonia = 0.0250 mol. M3: Calculate mass of cobalt = 0.2945 g and hence moles of cobalt = 0.00500 mol. M4: State the correct empirical formula of the complex as Co(NH3)5Cl3 (or [Co(NH3)5Cl]Cl2).

Step 1: Convert propan-1-ol to 1-bromopropane using sodium bromide and concentrated sulfuric acid, heating under reflux. Step 2: Convert 1-bromopropane to butanenitrile by reacting with potassium cyanide in ethanol under reflux. Step 3: Convert butanenitrile to butanoic acid by acid hydrolysis, heating under reflux with dilute hydrochloric acid.

M1: Step 1: NaBr + H2SO4 (or HBr / PBr3) to form 1-bromopropane. M2: Step 2: KCN (or NaCN) in ethanol (reflux) to form butanenitrile. M3: Step 3: Dilute aqueous acid (e.g., HCl / H2SO4) under reflux to form butanoic acid.

Step 1: Reduction of \(\text{Fe}^{3+}\) by \(\text{I}^-\). Equation: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\). \(E^\ominus_{\text{cell}} = 0.77 - 0.54 = +0.23\text{ V}\). Step 2: Oxidation of \(\text{Fe}^{2+}\) by \(\text{S}_2\text{O}_8^{2-}\). Equation: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\). \(E^\ominus_{\text{cell}} = 2.01 - 0.77 = +1.24\text{ V}\). Both steps are thermodynamically feasible since both cell potentials are positive. Direct reaction between \(\text{I}^-\) and \(\text{S}_2\text{O}_8^{2-}\) has high activation energy due to repulsion between negatively charged ions. Catalysis bypasses this repulsion.

M1: Calculate \(E^\ominus_{\text{cell}}\) for Step 1 = +0.23 V AND Step 2 = +1.24 V. M2: State that both steps are thermodynamically feasible because both \(E^\ominus_{\text{cell}}\) values are positive. M3: Explain that the uncatalysed reaction involves collision of two negatively charged ions which repel each other, whereas catalysis avoids this high activation energy barrier.

Using a Born-Haber cycle: \(\Delta H^\ominus_f = \Delta H^\ominus_{at}(\text{Mg}) + 1\text{st IE}(\text{Mg}) + 2\text{nd IE}(\text{Mg}) + 2 \times \Delta H^\ominus_{at}(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) + \Delta H^\ominus_{\text{lattice}}\). Substitute values: \(-642 = 148 + 738 + 1451 + 2(121) + 2(-349) + \Delta H^\ominus_{\text{lattice}}\). \(-642 = 2337 + 242 - 698 + \Delta H^\ominus_{\text{lattice}}\). \(-642 = 1881 + \Delta H^\ominus_{\text{lattice}}\). \(\Delta H^\ominus_{\text{lattice}} = -642 - 1881 = -2523\text{ kJ mol}^{-1}\).

M1: Account for 2 moles of chlorine atoms: \(2 \times 121 = 242\text{ kJ}\). M2: Account for 2 moles of chloride electron affinity: \(2 \times (-349) = -698\text{ kJ}\). M3: Set up correct algebraic energy cycle: \(\Delta H^\ominus_{\text{lattice}} = -642 - (148 + 738 + 1451 + 242 - 698)\). M4: Correct final evaluation of \(-2523\text{ kJ mol}^{-1}\) (must include negative sign).

The singlet at \(\delta = 11.5\text{ ppm}\) represents the carboxylic acid proton (\(-\text{COOH}\)). With 4 carbons in total, the compound is butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\). Splitting analysis: 1. The peak at 1.0 ppm (\(\text{CH}_3\)) is a triplet because it is adjacent to a \(\text{CH}_2\) group containing 2 protons (\(n+1 = 3\) peaks). 2. The peak at 1.7 ppm (\(\text{CH}_2\)) is a sextet because it is adjacent to 5 protons (3 from \(\text{CH}_3\) and 2 from \(\text{CH}_2\)), giving \(n+1 = 6\) peaks.

M1: Identify Compound X as butanoic acid (either by name or drawn structural formula). M2: State that the triplet at 1.0 ppm arises because the adjacent carbon has 2 protons (a \(\text{CH}_2\) group). M3: State that the sextet at 1.7 ppm arises because the adjacent carbons have a total of 5 protons (adjacent to both \(\text{CH}_3\) and \(\text{CH}_2\) groups).

Equation: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\) (or \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\)). Mechanism: 1. A curly arrow starts from the delocalised \(\pi\) ring of methylbenzene and points to the nitrogen atom of \(\text{NO}_2^+\). 2. This forms a positively charged intermediate with a broken ring (horseshoe shape open towards carbon-4) containing a positive charge inside, with both H and \(\text{NO}_2\) attached to carbon-4. 3. A curly arrow starts from the C-H bond at carbon-4 and points back into the ring to restore aromaticity, yielding 4-nitromethylbenzene and releasing \(\text{H}^+\).

M1: Correct equation for the generation of \(\text{NO}_2^+\). M2: Correct curly arrow from the benzene ring of methylbenzene to the nitrogen of \(\text{NO}_2^+\), AND correct drawing of the positively charged intermediate. M3: Correct curly arrow from the C-H bond back into the ring, forming 4-nitromethylbenzene.

1. Set up an ICE table: Initial moles: \(\text{SO}_2 = 1.50\), \(\text{O}_2 = 1.00\), \(\text{SO}_3 = 0.00\). At equilibrium: \(\text{SO}_3 = 0.80\text{ mol}\). Change: \(\text{SO}_3 = +0.80\), \(\text{SO}_2 = -0.80\), \(\text{O}_2 = -0.40\). Equilibrium moles: \(\text{SO}_2 = 1.50 - 0.80 = 0.70\text{ mol}\); \(\text{O}_2 = 1.00 - 0.40 = 0.60\text{ mol}\). Total moles = \(0.70 + 0.60 + 0.80 = 2.10\text{ mol}\). 2. Calculate partial pressures: \(p(\text{SO}_2) = \frac{0.70}{2.10} \times 3.00 = 1.00\text{ atm}\); \(p(\text{O}_2) = \frac{0.60}{2.10} \times 3.00 = 0.8571\text{ atm}\); \(p(\text{SO}_3) = \frac{0.80}{2.10} \times 3.00 = 1.1429\text{ atm}\). 3. Calculate \(K_p\): \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)} = \frac{(1.1429)^2}{(1.00)^2 \times 0.8571} = 1.52\text{ atm}^{-1}\).

M1: Correct calculation of equilibrium moles (0.70 mol for \(\text{SO}_2\) and 0.60 mol for \(\text{O}_2\); total 2.10 mol). M2: Correct calculation of partial pressures (\(p(\text{SO}_2) = 1.00\text{ atm}\), \(p(\text{O}_2) = 0.857\text{ atm}\), \(p(\text{SO}_3) = 1.14\text{ atm}\)). M3: Correct expression of \(K_p\) and substitution of values. M4: Correct final calculated value of 1.52 (accept 1.5) AND units of \(\text{atm}^{-1}\).

By applying Hess's Law to the Born-Haber cycle for magnesium sulfide, the standard enthalpy change of formation can be equated to the sum of the enthalpy changes of all the steps in the cycle:

\(\Delta_fH^{\ominus}[\text{MgS(s)}] = \Delta_{at}H^{\ominus}[\text{Mg(s)}] + I_1[\text{Mg(g)}] + I_2[\text{Mg(g)}] + \Delta_{at}H^{\ominus}[\text{S(s)}] + EA_1[\text{S(g)}] + EA_2[\text{S}^{-}(\text{g})] + \Delta_{LE}H^{\ominus}\)

Substituting the values from the table into this equation:

\(-346 = (+148) + (+738) + (+1451) + (+279) + (-200) + (+640) + \Delta_{LE}H^{\ominus}\)

Simplify the sum of the terms representing the formation of gaseous ions:

\(-346 = 3056 + \Delta_{LE}H^{\ominus}\)

Rearrange to solve for the lattice enthalpy (\(\Delta_{LE}H^{\ominus}\)):

\(\Delta_{LE}H^{\ominus} = -346 - 3056 = -3402 \text{ kJ mol}^{-1}\)

M1: Summing the intermediate enthalpy changes to form gaseous ions (atomisation, ionisation energies, and electron affinities):
148 + 738 + 1451 + 279 - 200 + 640 = (+)3056 (kJ mol^-1)

M2: Showing correct application of Born-Haber cycle / Hess's Law equation:
Lattice Enthalpy = -346 - 3056

M3: Correct final value with negative sign: -3402 (kJ mol^-1)
(Allow 2 marks for +3402, which represents the lattice dissociation enthalpy)

The reaction of an amino acid with an alcohol in the presence of a concentrated acid catalyst results in the esterification of the carboxyl group to form an ester group (\(\text{-COOCH}_3\)). Because the reaction mixture is highly acidic due to the concentrated sulfuric acid catalyst, the basic amine group (\(\text{-NH}_2\)) is protonated to form an ammonium cation (\(\text{-NH}_3^+\)). Thus, the major organic species in solution is \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOCH}_3\).

Award 1 mark for the correct option (B).
- Reject option A (neutral ester is not the major species in strongly acidic conditions).
- Reject option C (zwitterion is not formed in strongly acidic esterification conditions).
- Reject option D (carboxylate anion is only formed in alkaline conditions).

The ester group (\(\text{-COOCH}_3\)) is attached to the benzene ring via the carbonyl carbon. Carbonyl groups are electron-withdrawing and direct incoming electrophiles to the 3-position (meta-directing). Therefore, the major mono-nitration product is methyl 3-nitrobenzoate.

Award 1 mark for the correct option (B).
- Reject option A (2-position is ortho, which is not directed by the ester group).
- Reject option C (4-position is para, which is not directed by the ester group).
- Reject option D (mononitration occurs under standard conditions; dinitration is not the major product).

The triplet-quartet splitting pattern (with peak areas of 3 and 2 respectively) is characteristic of an ethyl group, \(\text{-CH}_2\text{CH}_3\). The chemical shift of the quartet (\(\delta = 4.1\text{ ppm}\)) indicates that the \(\text{-CH}_2-\) group is directly attached to an electronegative ester oxygen (\(\text{-OCH}_2\text{CH}_3\)). The singlet at \(\delta = 2.0\text{ ppm}\) with relative peak area 3 represents a methyl group next to a carbonyl (\(\text{CH}_3\text{CO-}\)). Combining these structural pieces gives ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

Award 1 mark for the correct option (B).
- Reject option A (methyl propanoate would have the quartet at a lower chemical shift of around 2.3 ppm as it is next to a carbonyl group, and the singlet at 3.7 ppm).
- Reject option C (propyl methanoate would have four proton environments).
- Reject option D (butanoic acid would have four proton environments, including a highly deshielded singlet for the carboxylic proton around 11-12 ppm).

The oxidation product of alcohol \(Y\) does not react with Tollens' reagent, indicating that the oxidation product is a ketone, and therefore \(Y\) must be a secondary alcohol (ruling out pentan-1-ol).
Dehydration of pentan-3-ol (secondary) yields only pent-2-ene, which has 2 stereoisomers (E-pent-2-ene and Z-pent-2-ene).
Dehydration of 3-methylbutan-2-ol (secondary) yields 3-methylbut-1-ene and 2-methylbut-2-ene, neither of which exhibits stereoisomerism, giving 2 isomers in total.
Dehydration of pentan-2-ol (secondary) can remove a hydrogen from C1 to give pent-1-ene, or from C3 to give pent-2-ene (which exists as E and Z stereoisomers). This produces a mixture of 3 isomeric alkenes: pent-1-ene, (E)-pent-2-ene, and (Z)-pent-2-ene.

Award 1 mark for the correct option (B).
- Reject option A (pentan-1-ol is a primary alcohol which oxidizes to an aldehyde and reacts with Tollens' reagent).
- Reject option C (pentan-3-ol dehydration only yields 2 isomers).
- Reject option D (3-methylbutan-2-ol dehydration only yields 2 isomers).

The repeating unit is split into two components: the dicarboxylic acid residue, \(\text{-CO-(CH}_2)_4\text{-CO-}\), and the diamine residue, \(\text{-NH-CH}_2\text{CH}_2\text{-NH-}\).
The dicarboxylic acid contains \(2 + 4 = 6\) carbon atoms in total, which corresponds to hexanedioic acid, \(\text{HOOC(CH}_2)_4\text{COOH}\).
The diamine contains 2 carbon atoms, corresponding to ethane-1,2-diamine, \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\).

Award 1 mark for the correct option (A).
- Reject option B (butanedioic acid has only 4 carbons).
- Reject option C (1,2-aminoethanol would form a polyesteramide rather than a polyamide).
- Reject option D (pentanedioic acid has 5 carbons and propane-1,3-diamine has 3 carbons).

The reaction of a carbonyl compound with \(\text{HCN}\) is a nucleophilic addition to form a hydroxynitrile. For the product to have a chiral center, the central carbon must be bonded to four different groups.
- Propanone (symmetrical) forms \(\text{CH}_3\text{C(OH)(CN)CH}_3\) (achiral).
- Pentan-3-one (symmetrical) forms \(\text{CH}_3\text{CH}_2\text{C(OH)(CN)CH}_2\text{CH}_3\) (achiral).
- Methanal forms \(\text{H}_2\text{C(OH)(CN)}\) (achiral).
- Butanone (unsymmetrical) forms \(\text{CH}_3\text{C(OH)(CN)CH}_2\text{CH}_3\), which has a chiral carbon bonded to \(\text{-OH}\), \(\text{-CN}\), \(\text{-CH}_3\), and \(\text{-CH}_2\text{CH}_3\).

Award 1 mark for the correct option (C).
- Reject options A, B, and D because they form symmetrical products without chiral centers.

Let \(x\) be the number of carbon atoms in acid \(P\), and \(y\) be the number of carbon atoms in alcohol \(Q\). Since the total number of carbon atoms in the ester is 6, we have \(x + y = 6\).
Oxidation of primary alcohol \(Q\) yields carboxylic acid \(P\) with the same number of carbon atoms (same carbon skeleton), which means \(x = y\).
Solving \(x + y = 6\) with \(x = y\) yields \(x = 3\) and \(y = 3\).
Therefore, \(P\) is propanoic acid and \(Q\) is propan-1-ol. The ester formed from these is propyl propanoate.

Award 1 mark for the correct option (B).
- Reject option A (ethyl butanoate: alcohol has 2 carbons, acid has 4; oxidation of ethanol yields ethanoic acid, not butanoic acid).
- Reject option C (methyl pentanoate: alcohol has 1 carbon, acid has 5; oxidation of methanol yields methanoic acid, not pentanoic acid).
- Reject option D (butyl ethanoate: alcohol has 4 carbons, acid has 2; oxidation of butan-1-ol yields butanoic acid, not ethanoic acid).

To determine the number of structural isomers, we find the possible arrangements of the carbon chain and chlorine atom for 4 carbon atoms:
1. Based on the straight-chain butane skeleton:
- 1-chlorobutane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\))
- 2-chlorobutane (\(\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3\))
2. Based on the branched methylpropane skeleton:
- 1-chloro-2-methylpropane (\(\text{(CH}_3)_2\text{CHCH}_2\text{Cl}\))
- 2-chloro-2-methylpropane (\(\text{(CH}_3)_3\text{CCl}\))

There are exactly 4 structural isomers. (Note: stereoisomers/optical isomers are excluded when counting structural isomers).

Award 1 mark for the correct option (B).
- Reject option A (missed one of the structural isomers).
- Reject option C (this would include optical isomerism, but the question specifies structural isomers).
- Reject option D (incorrect count).

Let's analyse the options:

* **A (Ethyl 2-aminopropanoate):** Undergoes acid hydrolysis to form the chiral amino acid 2-aminopropanoic acid (alanine) and the alcohol ethanol. The starting ester is also chiral due to the central carbon of the alanyl group. This matches all criteria.
* **B (Methyl 2-amino-2-methylpropanoate):** Undergoes acid hydrolysis to form 2-amino-2-methylpropanoic acid and methanol. However, 2-amino-2-methylpropanoic acid is not chiral because the central carbon is bonded to two identical methyl groups.
* **C (Propyl 2-aminoacetate):** Undergoes acid hydrolysis to form glycine (2-aminoacetic acid) and propan-1-ol. Glycine is not chiral.
* **D (3-Aminopentanoic acid):** This is a carboxylic acid, not an ester, and does not undergo hydrolysis to yield an alcohol.

[1 mark] - Correct option A chosen.
[0 marks] - Any other option chosen.

* **A is incorrect:** The methyl group in methylbenzene is activating and directing to the 2- (ortho) and 4- (para) positions. Nitration of methylbenzene gives 2-nitromethylbenzene and 4-nitromethylbenzene as major products, not 3-nitromethylbenzene.
* **B is correct:** Acylation of methylbenzene using \(\text{CH}_3\text{COCl}\) and an anhydrous \(\text{AlCl}_3\) catalyst undergoes electrophilic substitution. The methyl group directs the incoming electrophile predominantly to the 4-position (due to steric hindrance at the 2-position), yielding 1-(4-methylphenyl)ethanone.
* **C is incorrect:** The reaction of methylbenzene with \(\text{Cl}_2\) in the presence of UV light is a free-radical substitution that occurs on the alkyl side-chain, producing (chloromethyl)benzene, not electrophilic ring substitution.
* **D is incorrect:** Hydrogenation of methylbenzene using \(\text{H}_2\) and a \(\text{Ni}\) catalyst fully reduces the aromatic ring to produce methylcyclohexane, not 1-methylcyclohexene.

[1 mark] - Correct option B chosen.
[0 marks] - Any other option chosen.

Let's analyse the splitting and chemical shifts for each candidate isomer:

* **A (Ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\)):**
- The \(\text{CH}_3\) of the ethyl group is adjacent to a \(\text{CH}_2\) group, splitting it into a triplet (3H) at \(\delta \approx 1.3\text{ ppm}\).
- The \(\text{CH}_3\) attached directly to the carbonyl group (\(\text{CH}_3\text{CO}-\)) has no adjacent protons, giving a singlet (3H) at \(\delta \approx 2.0\text{ ppm}\).
- The \(\text{CH}_2\) group is bonded to oxygen (\(-\text{OCH}_2-\)) which shifts it downfield to \(\delta \approx 4.1\text{ ppm}\), and is split by the adjacent \(\text{CH}_3\) into a quartet (2H). This matches the spectrum perfectly.
* **B (Methyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_3\)):** The singlet (3H) would be the ester methyl group \(-\text{OCH}_3\) at around \(\delta \approx 3.7\text{ ppm}\) and the quartet (2H) would be at \(\delta \approx 2.3\text{ ppm}\) (adjacent to carbonyl).
* **C (Propyl methanoate, \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)):** This compound has 4 proton environments and contains a formyl proton (\(\text{HCO}-\)) at \(\delta \approx 8.0\text{ ppm}\).
* **D (Butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)):** This has 4 proton environments, including a highly downfield carboxylic acid peak at \(\delta \approx 11.0\text{ ppm}\).

[1 mark] - Correct option A chosen.
[0 marks] - Any other option chosen.

The reduction of a carbonyl by aqueous \(\text{NaBH}_4\) proceeds via a nucleophilic addition mechanism:

* **Step 1:** The nucleophilic hydride ion (\(\text{H}^-\)) from \(\text{NaBH}_4\) attacks the partially positive carbonyl carbon (\(\text{C}^{\delta+}\)). Therefore, a curly arrow starts from a lone pair of electrons on the hydride ion and points directly to the carbonyl carbon. Simultaneously, the carbon-oxygen \(\pi\)-bond undergoes heterolytic fission, with a curly arrow starting from the \(\text{C}=\text{O}\) double bond and pointing to the oxygen atom.
* **Step 2:** The negative intermediate (an alkoxide ion) is protonated by a water molecule (not acting as a nucleophile but as a proton donor) to yield propan-2-ol.

Evaluating the options:
* **A** is incorrect because the reaction is an addition, not a substitution.
* **B** is correct as it describes the first step of the nucleophilic attack.
* **C** is incorrect because the intermediate is a negatively charged alkoxide ion, not a carbocation.
* **D** is incorrect because water is a proton donor in the second step, not the nucleophile in the first.

[1 mark] - Correct option B chosen.
[0 marks] - Any other option chosen.

Phenol is less nucleophilic than aliphatic alcohols because the lone pair of electrons on the oxygen atom is partially delocalised into the benzene ring \(\pi\)-system.

* **Statement 1 is incorrect:** Phenol is not reactive enough to undergo esterification with carboxylic acids (such as ethanoic acid) directly, even with an acid catalyst.
* **Statement 2 is correct:** Phenol reacts readily with highly reactive acyl chlorides (like ethanoyl chloride) at room temperature to form phenyl ethanoate and \(\text{HCl}\).
* **Statement 3 is correct:** Phenol reacts with acid anhydrides (like ethanoic anhydride) when warmed, particularly in the presence of a base or acid catalyst, to yield phenyl ethanoate and ethanoic acid.

Therefore, only 2 and 3 are correct.

[1 mark] - Correct option C chosen.
[0 marks] - Any other option chosen.

Dehydration of alcohols with concentrated acid catalyst involves elimination of water. Let's analyse each alcohol:

* **Hexan-2-ol:** Dehydration can eliminate hydrogen from C1 or C3, producing:
- hex-1-ene (no E/Z stereoisomerism)
- *E*-hex-2-ene and *Z*-hex-2-ene
This gives 3 isomeric alkenes.

* **Hexan-3-ol:** Dehydration can eliminate hydrogen from C2 or C4:
- Elimination from C2 yields hex-2-ene, which exists as *E*-hex-2-ene and *Z*-hex-2-ene.
- Elimination from C4 yields hex-3-ene, which exists as *E*-hex-3-ene and *Z*-hex-3-ene.
This yields exactly 4 distinct isomeric alkenes. This matches the question.

* **3-Methylpentan-3-ol:** Dehydration yields 3-methylpent-2-ene (which has *E* and *Z* stereoisomers) as the only product, giving 2 isomeric alkenes.

* **2-Methylpentan-3-ol:** Dehydration can eliminate from C2 or C4:
- Elimination from C2 yields 2-methylpent-2-ene (no E/Z isomerism as one double-bonded carbon has two methyl groups).
- Elimination from C4 yields 4-methylpent-2-ene (which has *E* and *Z* stereoisomers).
This gives 3 isomeric alkenes.

[1 mark] - Correct option B chosen.
[0 marks] - Any other option chosen.

Base strength depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:

1. **Phenylamine** is the weakest base because the lone pair of electrons on the nitrogen atom is partially delocalised into the aromatic ring's \(\pi\)-system, making it much less available to accept a proton.
2. **Ammonia** is a stronger base than phenylamine because it has no delocalisation, but is weaker than aliphatic amines.
3. **Ethylamine** (primary aliphatic amine) is a stronger base than ammonia because the ethyl group is electron-donating via the inductive effect, which increases the electron density on the nitrogen atom and makes the lone pair more available.
4. **Diethylamine** (secondary aliphatic amine) has two electron-donating ethyl groups, making the lone pair even more available than in ethylamine.

Thus, the correct order is: Phenylamine < Ammonia < Ethylamine < Diethylamine.

[1 mark] - Correct option A chosen.
[0 marks] - Any other option chosen.

1. Moles of propanoic acid = \(50.0/1000 \times 0.250 = 0.0125\text{ mol}\). 2. Moles of NaOH = \(30.0/1000 \times 0.150 = 0.00450\text{ mol}\). 3. Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) remaining = \(0.0125 - 0.00450 = 0.00800\text{ mol}\). Moles of propanoate ions formed = \(0.00450\text{ mol}\). 4. \([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = (1.35 \times 10^{-5}) \times \frac{0.00800}{0.00450} = 2.40 \times 10^{-5}\text{ mol dm}^{-3}\). 5. \(\text{pH} = -\log_{10}(2.40 \times 10^{-5}) = 4.62\).

Mark 1: Correct moles of propanoic acid (0.0125 mol) AND NaOH (0.00450 mol). Mark 2: Correct calculation of remaining propanoic acid (0.00800 mol) and formed propanoate (0.00450 mol). Mark 3: Correct \([\text{H}^+]\) calculation of \(2.40 \times 10^{-5}\text{ mol dm}^{-3}\). Mark 4: pH value of 4.62 (to 2 decimal places, allow ecf from calculated \([\text{H}^+]\)).

1. Moles of \(\text{MnO}_4^-\text{ in titration} = \frac{22.40}{1000} \times 0.0150 = 3.36 \times 10^{-4}\text{ mol}\). 2. Moles of \(\text{Fe}^{2+}\text{ in } 25.0\text{ cm}^3 = 5 \times 3.36 \times 10^{-4} = 1.68 \times 10^{-3}\text{ mol}\). 3. Moles of \(\text{Fe}^{2+}\text{ in } 250.0\text{ cm}^3 = 1.68 \times 10^{-3} \times 10 = 1.68 \times 10^{-2}\text{ mol}\). 4. Mass of iron = \(1.68 \times 10^{-2} \times 55.8 = 0.93744\text{ g}\). 5. % by mass = \(\frac{0.93744}{1.84} \times 100 = 50.9\%\).

Mark 1: Moles of \(\text{MnO}_4^-\text{ calculated as } 3.36 \times 10^{-4}\text{ mol}\). Mark 2: Moles of \(\text{Fe}^{2+}\text{ in } 25.0\text{ cm}^3\) titration scaled by 5 to \(1.68 \times 10^{-3}\text{ mol}\). Mark 3: Scaled to \(250.0\text{ cm}^3\) to find total Fe moles = \(1.68 \times 10^{-2}\text{ mol}\). Mark 4: Correct final percentage of 50.9% (allow 51.0% if Fe=56 used; allow ecf throughout).

1. Equilibrium moles: \(\text{PCl}_5 = 0.350\text{ mol}\); \(\text{PCl}_3 = 1.50 - 0.350 = 1.15\text{ mol}\); \(\text{Cl}_2 = 1.50 - 0.350 = 1.15\text{ mol}\). 2. Equilibrium concentrations: \([\text{PCl}_5] = \frac{0.350}{2.50} = 0.140\text{ mol dm}^{-3}\); \([\text{PCl}_3] = \frac{1.15}{2.50} = 0.460\text{ mol dm}^{-3}\); \([\text{Cl}_2] = \frac{1.15}{2.50} = 0.460\text{ mol dm}^{-3}\). 3. \(K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} = \frac{0.140}{0.460 \times 0.460} = 0.6616\text{ dm}^3\text{ mol}^{-1} \approx 0.662\text{ dm}^3\text{ mol}^{-1}\).

Mark 1: Calculating equilibrium moles of \(\text{PCl}_3\) and \(\text{Cl}_2\) as 1.15 mol. Mark 2: Converting moles to concentrations by dividing by 2.50 dm3. Mark 3: Correct value of \(K_c = 0.662\) (or 0.66) and correct units \(\text{dm}^3\text{ mol}^{-1}\).

1. Sulfur has 6 valence electrons, and forms 4 covalent bonds with fluorine, leaving 1 lone pair (total 5 electron pairs). 2. These 5 pairs adopt a trigonal bipyramidal arrangement to minimize repulsion. Since there are 4 bonding pairs and 1 lone pair, the molecular shape is seesaw. 3. Lone pairs repel more strongly than bonding pairs, which distorts the shape, reducing the bond angles from the ideal \(120^\circ\) and \(90^\circ\) to approximately \(117^\circ\) and \(87^\circ\) (accept less than \(120^\circ\) and less than \(90^\circ\)).

Mark 1: Predicts shape as 'seesaw' AND bond angles as less than \(120^\circ\) and less than \(90^\circ\). Mark 2: Identifies 4 bonding pairs and 1 lone pair around the central sulfur atom. Mark 3: States that electron pairs repel to be as far apart as possible, and lone pairs repel more than bonding pairs.

1. Feasibility: Yes, because the electrode potential of the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell (\(+0.77\text{ V}\)) is more positive than that of the \(\text{I}_2/\text{I}^-\) half-cell (\(+0.54\text{ V}\)). 2. Overall equation: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\). 3. Cell potential: \(E^{\theta}_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\).

Mark 1: Concludes reaction is feasible because the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) system has a more positive electrode potential than the \(\text{I}_2/\text{I}^-\) system. Mark 2: Writes correct balanced equation: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\). Mark 3: Correct value of \(E^{\theta}_{\text{cell}} = +0.23\text{ V}\) (must include the '+' sign).

1. IR: broad peak at \(3350\text{ cm}^{-1}\) indicates \(\text{O-H}\) alcohol group. Peak at \(1715\text{ cm}^{-1}\) indicates \(\text{C=O}\) carbonyl group. 2. 1H NMR: Singlet at \(2.2\text{ ppm}\) (3H) indicates a \(\text{CH}_3\text{CO-}\) ketone fragment. Doublet at \(1.4\text{ ppm}\) (3H) and quartet at \(4.2\text{ ppm}\) (1H) show a \(\text{CH}_3\text{-CH(O)-}\) spin-system. Peak at \(3.4\text{ ppm}\) that disappears with \(\text{D}_2\text{O}\) confirms the \(\text{O-H}\) proton. 3. 13C NMR: 4 peaks confirm 4 non-equivalent carbon environments. Combining these gives 3-hydroxybutan-2-one, \(\text{CH}_3\text{COCH(OH)CH}_3\).

Mark 1: Identifies \(\text{O-H}\) group from absorption at \(3350\text{ cm}^{-1}\) AND \(\text{C=O}\) group from absorption at \(1715\text{ cm}^{-1}\). Mark 2: Assigns proton environments: singlet at \(2.2\text{ ppm}\) as methyl next to carbonyl, doublet/quartet system as \(\text{CH}_3\text{-CH-}\), and \(\text{D}_2\text{O}\) exchangeable peak as alcohol. Mark 3: Correct structure of 3-hydroxybutan-2-one (structural formula \(\text{CH}_3\text{COCH(OH)CH}_3\) or drawn correctly).

1. Step 1: Phenylethanone is a ketone; it can be reduced to the secondary alcohol 1-phenylethan-1-ol using \(\text{NaBH}_4\) in aqueous/ethanolic conditions. 2. Step 2: 1-phenylethan-1-ol is esterified using ethanoic acid in the presence of concentrated sulfuric acid catalyst under heat (or ethanoyl chloride/ethanoic anhydride). 3. Step 3: Acidified potassium dichromate(VI) will oxidize the secondary alcohol (1-phenylethan-1-ol) but not the ketone (phenylethanone). The alcohol turns the orange solution green, while the ketone shows no reaction (remains orange).

Mark 1: Step 1 reagents: \(\text{NaBH}_4\) in aqueous or ethanol solvent (or \(\text{LiAlH}_4\) in dry ether). Mark 2: Step 2 reagents: \(\text{CH}_3\text{COOH}\) AND concentrated \(\text{H}_2\text{SO}_4\) catalyst with heat (allow ethanoyl chloride or ethanoic anhydride). Mark 3: Distinguishing test: Acidified potassium dichromate(VI) (\(\text{H}^+/\text{Cr}_2\text{O}_7^{2-}\)) and heat; 1-phenylethan-1-ol turns green, phenylethanone remains orange (accept 2,4-DNP test with orange ppt for phenylethanone and no reaction for 1-phenylethan-1-ol).

1. Alanine contains a chiral carbon. The 3D structures must show a tetrahedral arrangement around the central carbon atom using wedge and dash representations, forming non-superimposable mirror images. 2. When reacted with excess acid (HCl), the basic amino group (\(\text{-NH}_2\)) is protonated to form a cation: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\). 3. When reacted with excess base (NaOH), the carboxylic acid group (\(\text{-COOH}\)) is deprotonated to form an anion: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

Mark 1: Draws one correct 3D tetrahedral representation of alanine using wedges and dashes. Mark 2: Draws the correct mirror image of the first structure. Mark 3: Correct structure of protonated alanine: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\) (or \(\text{CH}_3\text{CH}(\text{NH}_3^+\text{Cl}^-)\text{COOH}\)). Mark 4: Correct structure of deprotonated alanine: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\). (or \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COONa}\)).

1. Calculate the initial amount in moles of propanoic acid and sodium hydroxide:
\(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
\(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.00500\text{ mol}\)

2. Determine the amounts of acid and salt at equilibrium:
Since propanoic acid is in excess, all \(\text{NaOH}\) reacts to form propanoate salt.
\(n(\text{CH}_3\text{CH}_2\text{COO}^-) = 0.00500\text{ mol}\)
Remaining \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0125 - 0.00500 = 0.00750\text{ mol}\)

3. Calculate the hydrogen ion concentration, \([\text{H}^+]\):
\([\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{CH}_2\text{COOH}]}{[\text{CH}_3\text{CH}_2\text{COO}^-]} = 1.35 \times 10^{-5} \times \frac{0.00750}{0.00500} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\)

4. Calculate the pH:
\(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6935 \approx 4.69\)

• M1: Correct calculation of initial moles: \(n(\text{acid}) = 0.0125\text{ mol}\) AND \(n(\text{NaOH}) = 0.00500\text{ mol}\) (1 mark)
• M2: Correct buffer ratio calculation (remaining \(n(\text{acid}) = 0.00750\text{ mol}\) and formed \(n(\text{salt}) = 0.00500\text{ mol}\)) OR setting up correct \([\text{H}^+]\) expression (1 mark)
• M3: Correct evaluation of \(\text{pH} = 4.69\) (allow 4.70 if rounding error occurred earlier, but 4.69 is the correct 2 d.p. answer) (1 mark)

1. Moles of manganate(VII) used in titration:
\(n(\text{MnO}_4^-) = 0.02500\text{ dm}^3 \times 0.0150\text{ mol dm}^{-3} = 3.75 \times 10^{-4}\text{ mol}\)

2. Moles of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) sample (1:5 ratio):
\(5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
\(n(\text{Fe}^{2+}) = 5 \times 3.75 \times 10^{-4}\text{ mol} = 1.875 \times 10^{-3}\text{ mol}\)

3. Moles of \(\text{Fe}^{2+}\) in the total \(100.0\text{ cm}^3\) original solution:
\(n(\text{Fe}^{2+})_{\text{total}} = 1.875 \times 10^{-3} \times 4 = 7.50 \times 10^{-3}\text{ mol}\)

4. Molar mass of the hydrated salt:
\(M_r = \frac{2.940\text{ g}}{7.50 \times 10^{-3}\text{ mol}} = 392.0\text{ g mol}^{-1}\)

5. Calculate \(x\):
\(M_r((\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2) = 284.0\text{ g mol}^{-1}\)
Mass of water of crystallisation = \(392.0 - 284.0 = 108.0\text{ g mol}^{-1}\)
\(x = \frac{108.0}{18.0} = 6\)

• M1: Calculate \(n(\text{MnO}_4^-) = 3.75 \times 10^{-4}\text{ mol}\) (1 mark)
• M2: Calculate \(n(\text{Fe}^{2+})\text{ in } 100.0\text{ cm}^3 = 7.50 \times 10^{-3}\text{ mol}\) (1 mark)
• M3: Calculate \(M_r\) of hydrated salt = \(392\text{ g mol}^{-1}\) (1 mark)
• M4: Deduces \(x = 6\) based on anhydrous mass of \(284\text{ g mol}^{-1}\) (1 mark)

Using the Born-Haber cycle:
\(\Delta_f H^{\ominus}(\text{CaCl}_2\text{(s)}) = \Delta_{at} H^{\ominus}(\text{Ca}) + IE_1(\text{Ca}) + IE_2(\text{Ca}) + 2 \times \Delta_{at} H^{\ominus}(\text{Cl}) + 2 \times EA_1(\text{Cl}) + \Delta_{LE} H^{\ominus}\)

Substitute the values into the equation:
\(-796 = 178 + 590 + 1145 + 2(121) + 2(-349) + \Delta_{LE} H^{\ominus}\)
\(-796 = 178 + 590 + 1145 + 242 - 698 + \Delta_{LE} H^{\ominus}\)
\(-796 = 1457 + \Delta_{LE} H^{\ominus}\)
\(\Delta_{LE} H^{\ominus} = -796 - 1457 = -2253\text{ kJ mol}^{-1}\)

• M1: Correctly accounts for two moles of Cl atoms in both atomisation and electron affinity steps, i.e., \(2 \times 121 = 242\) and \(2 \times (-349) = -698\) (1 mark)
• M2: Constructing a correct thermodynamic cycle expression, e.g., \(\Delta_{LE} H^{\ominus} = \Delta_f H - (\Delta_{at} H(\text{Ca}) + IE_1 + IE_2 + 2\Delta_{at} H(\text{Cl}) + 2EA_1)\) (1 mark)
• M3: Final evaluation of \(-2253\text{ kJ mol}^{-1}\) with correct sign (1 mark)

1. Calculate the cell potential:
\(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{reduction}} - E^{\ominus}_{\text{oxidation}} = +0.80 - (-0.74) = +1.54\text{ V}\)

2. Write the spontaneous overall equation:
\(3\text{Ag}^+\text{(aq)} + \text{Cr(s)} \rightarrow 3\text{Ag(s)} + \text{Cr}^{3+}\text{(aq)}\)

3. Calculate the moles of Ag deposited:
Since the stoichiometry ratio of \(\text{Cr} : \text{Ag}\) is \(1:3\):
\(n(\text{Ag}) = 3 \times 0.0500\text{ mol} = 0.150\text{ mol}\)

4. Calculate the mass of Ag deposited:
\(\text{Mass} = 0.150\text{ mol} \times 107.9\text{ g mol}^{-1} = 16.185\text{ g}\)
To 3 significant figures, this is \(16.2\text{ g}\).

• M1: Correct standard cell potential calculation: \(E^{\ominus}_{\text{cell}} = +1.54\text{ V}\) (1 mark)
• M2: Deducing the \(3:1\) molar ratio of silver to chromium, obtaining \(0.150\text{ mol}\) of silver (1 mark)
• M3: Correct calculation of mass of silver to 3 significant figures: \(16.2\text{ g}\) (1 mark)

1. Calculate equilibrium moles:
- Moles of \(\text{D}\) at equilibrium = \(0.120\text{ mol dm}^{-3} \times 5.00\text{ dm}^3 = 0.600\text{ mol}\).
- Using stoichiometry:
- Moles of \(\text{C}\) formed = \(2 \times 0.600 = 1.20\text{ mol}\).
- Moles of \(\text{B}\) reacted = \(2 \times 0.600 = 1.20\text{ mol}\), so remaining \(\text{B} = 2.00 - 1.20 = 0.800\text{ mol}\).
- Moles of \(\text{A}\) reacted = \(0.600\text{ mol}\), so remaining \(\text{A} = 1.50 - 0.600 = 0.900\text{ mol}\).

2. Calculate equilibrium concentrations:
- \([\text{A}] = 0.900 / 5.00 = 0.180\text{ mol dm}^{-3}\)
- \([\text{B}] = 0.800 / 5.00 = 0.160\text{ mol dm}^{-3}\)
- \([\text{C}] = 1.20 / 5.00 = 0.240\text{ mol dm}^{-3}\)
- \([\text{D}] = 0.120\text{ mol dm}^{-3}\)

3. Calculate \(K_c\):
\(K_c = \frac{[\text{C}]^2 [\text{D}]}{[\text{A}] [\text{B}]^2} = \frac{(0.240)^2 \times 0.120}{0.180 \times (0.160)^2} = \frac{0.006912}{0.004608} = 1.50\)

4. Determine units:
\(\text{Units} = \frac{(\text{mol dm}^{-3})^3}{(\text{mol dm}^{-3})^3} = \text{no units}\).

• M1: Correct equilibrium moles for all species: \(n(\text{A}) = 0.900\), \(n(\text{B}) = 0.800\), \(n(\text{C}) = 1.20\) (1 mark)
• M2: Correct equilibrium concentrations for all species: \([\text{A}] = 0.180\), \([\text{B}] = 0.160\), \([\text{C}] = 0.240\) (1 mark)
• M3: Correct evaluation of \(K_c = 1.50\) (1 mark)
• M4: States that there are no units (1 mark)

1. The functional group connecting the two amino acids in a dipeptide is an amide group (or peptide linkage/peptide bond).
2. Ala-Ser has two chiral carbon atoms: the \(\alpha\)-carbon of the alanine residue (bonded to \(-\text{H}\), \(-\text{CH}_3\), \(-\text{NH}_2\), and the carbonyl) and the \(\alpha\)-carbon of the serine residue (bonded to \(-\text{H}\), \(-\text{CH}_2\text{OH}\), \(-\text{NH}-\), and \(-\text{COOH}\)).
3. The maximum number of stereoisomers is \(2^n\), where \(n\) is the number of chiral centres: \(2^2 = 4\).

• M1: Identifies the linkage functional group as amide (accept peptide) (1 mark)
• M2: Identifies that there are 2 chiral carbon atoms (1 mark)
• M3: Calculates that there are 4 stereoisomers (allow ecf from M2 as \(2^{M2}\)) (1 mark)

1. Convert units to SI units for the ideal gas equation:
- \(p = 101\text{ kPa} = 1.01 \times 10^5\text{ Pa}\)
- \(V = 245\text{ cm}^3 = 2.45 \times 10^{-4}\text{ m}^3\)
- \(T = 25.0^{\circ}\text{C} = 298.15\text{ K}\) (accept \(298\text{ K}\))

2. Use the ideal gas equation to find moles of \(\text{CO}_2\):
\(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 2.45 \times 10^{-4}}{8.314 \times 298} = 9.988 \times 10^{-3}\text{ mol}\)

3. Since \(\text{MCO}_3 : \text{CO}_2 = 1:1\):
\(n(\text{MCO}_3) = 9.988 \times 10^{-3}\text{ mol}\)

4. Calculate molar mass of \(\text{MCO}_3\):
\(M_r = \frac{0.843\text{ g}}{9.988 \times 10^{-3}\text{ mol}} = 84.4\text{ g mol}^{-1}\) (to 3 s.f.)

5. Identify metal \(\text{M}\):
\(M_r(\text{M}) = 84.4 - [12.0 + (3 \times 16.0)] = 84.4 - 60.0 = 24.4\text{ g mol}^{-1}\)
The closest Group 2 metal is Magnesium (\(24.3\text{ g mol}^{-1}\)).

• M1: Correct conversion of pressure, volume, and temperature to SI units (1 mark)
• M2: Correct calculation of moles of \(\text{CO}_2\) as \(9.99 \times 10^{-3}\text{ mol}\) (1 mark)
• M3: Correct calculation of molar mass of \(\text{MCO}_3\) as \(84.4\text{ g mol}^{-1}\) (1 mark)
• M4: Correct identification of \(\text{M}\) as Magnesium (allow ecf from calculated molar mass) (1 mark)

1. Calculate \(\Delta G^{\ominus}\) at \(298\text{ K}\):
Convert \(\Delta S^{\ominus}\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\):
\(\Delta S^{\ominus} = +0.161\text{ kJ K}^{-1}\text{ mol}^{-1}\)
\(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\)
\(\Delta G^{\ominus} = +178 - (298 \times 0.161) = 178 - 47.978 = +130\text{ kJ mol}^{-1}\) (to 3 s.f.)

2. Minimum temperature for thermodynamic feasibility (where \(\Delta G^{\ominus} \le 0\)):
\(T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\)
\(T = \frac{178 \times 10^3\text{ J mol}^{-1}}{161\text{ J K}^{-1}\text{ mol}^{-1}} = 1105.59\text{ K} \approx 1110\text{ K}\) (to 3 s.f.)

• M1: Correct conversion of units and calculation of \(\Delta G^{\ominus} = +130\text{ kJ mol}^{-1}\) (or \(+130.0\)) (1 mark)
• M2: Setting up the inequality/expression for feasibility: \(T = \frac{\Delta H}{\Delta S}\) (1 mark)
• M3: Correct calculation of temperature as \(1106\text{ K}\) or \(1110\text{ K}\) (1 mark)

1. Calculate the initial amount, in moles, of propanoic acid (\(\text{HA}\)) and hydroxide ions (\(\text{OH}^-\)):
\(n(\text{HA})_{\text{initial}} = 0.150\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0300\text{ mol}\)
\(n(\text{OH}^-)_{\text{added}} = 0.100\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 0.0120\text{ mol}\)

2. Determine the remaining moles of \(\text{HA}\) and the moles of the conjugate base (\(\text{A}^-\)) formed after neutralization:
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)
\(n(\text{HA})_{\text{remaining}} = 0.0300 - 0.0120 = 0.0180\text{ mol}\)
\(n(\text{A}^-)_{\text{formed}} = 0.0120\text{ mol}\)

3. Use the buffer equation or \(K_a\) expression to find \([\text{H}^+]\):
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{0.0180}{0.0120} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\)

4. Calculate pH:
\(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6936... \approx 4.69\)

• M1: Calculates \(n(\text{HA})_{\text{initial}} = 0.0300\text{ mol}\) AND \(n(\text{OH}^-)_{\text{added}} = 0.0120\text{ mol}\) [1]
• M2: Calculates remaining \(n(\text{HA}) = 0.0180\text{ mol}\) AND \(n(\text{A}^-) = 0.0120\text{ mol}\) [1]
• M3: Calculates pH = 4.69 (allow 4.70 if rounded slightly early, but reject if from incorrect chemical calculation) [1]

1. Write the equation or use the reacting ratio for the titration:
\(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)
Ratio of \(\text{MnO}_4^- : \text{Fe}^{2+} = 1 : 5\)

2. Calculate moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.02250\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 4.50 \times 10^{-4}\text{ mol}\)

3. Calculate moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\):
\(n(\text{Fe}^{2+}) = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\text{ mol}\)

4. Scale up to the total \(250.0\text{ cm}^3\) solution:
\(n(\text{Fe}^{2+})_{\text{total}} = 2.25 \times 10^{-3} \times \frac{250.0}{25.0} = 2.25 \times 10^{-2}\text{ mol}\)

5. Calculate mass of Fe in the sample:
\(\text{Mass of Fe} = 2.25 \times 10^{-2}\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.2555\text{ g}\)

6. Calculate percentage by mass:
\(\%\text{ Fe} = \frac{1.2555}{1.420} \times 100 = 88.415...\% \approx 88.4\%\)

• M1: Calculates moles of \(\text{MnO}_4^-\) as \(4.50 \times 10^{-4}\text{ mol}\) [1]
• M2: Calculates moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\) as \(2.25 \times 10^{-3}\text{ mol}\) [1]
• M3: Scales up to \(250.0\text{ cm}^3\) to get \(2.25 \times 10^{-2}\text{ mol}\) [1]
• M4: Calculates final mass of \(1.2555\text{ g}\) and percentage as 88.4% (allow 3 s.f., range 88.3 - 88.5) [1]

1. Write the balanced equation for the complete combustion of propan-1-ol:
\(\text{C}_3\text{H}_7\text{OH}(\text{l}) + 4.5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)

2. Setup the Hess's Law cycle calculation:
\(\Delta_c H^\theta = \sum \Delta_f H^\theta(\text{products}) - \sum \Delta_f H^\theta(\text{reactants})\)
\(\Delta_c H^\theta = [3 \times \Delta_f H^\theta(\text{CO}_2) + 4 \times \Delta_f H^\theta(\text{H}_2\text{O})] - [\Delta_f H^\theta(\text{C}_3\text{H}_7\text{OH})]\)
Note: \(\Delta_f H^\theta(\text{O}_2) = 0\)

3. Substitute values:
\(\Delta_c H^\theta = [3 \times (-394) + 4 \times (-286)] - [-303]\)
\Delta_c H^\theta = [-1182 - 1144] + 303
\Delta_c H^\theta = -2326 + 303 = -2023\text{ kJ mol}^{-1}\)

• M1: Correct balanced combustion equation (allow fractional coefficients like 4.5 or 9/2) [1]
• M2: Correct application of Hess's Law formula: \(3(-394) + 4(-286) - (-303)\) [1]
• M3: Final answer of \(-2023\text{ kJ mol}^{-1}\) (must include negative sign; units not required but must not be incorrect) [1]

1. Convert all parameters to SI units:
\(p = 102\text{ kPa} = 1.02 \times 10^5\text{ Pa}\)
\(V = 78.0\text{ cm}^3 = 7.80 \times 10^{-5}\text{ m}^3\)
\(T = 97 + 273.15 = 370.15\text{ K}\) (or \(370\text{ K}\))

2. Use the ideal gas equation \(pV = nRT\) to find \(n\):
\(n = \frac{pV}{RT} = \frac{1.02 \times 10^5 \times 7.80 \times 10^{-5}}{8.314 \times 370.15} = \frac{7.956}{3077.43} = 2.585... \times 10^{-3}\text{ mol}\)

3. Calculate molar mass \(M = \frac{m}{n}\):
\(M = \frac{0.205}{2.585 \times 10^{-3}} = 79.29...\text{ g mol}^{-1}\)
Rounding to 3 significant figures gives \(79.3\text{ g mol}^{-1}\) (or \(79.3\)).

• M1: Correct conversion of volume to \(\text{m}^3\) and temperature to K (either 370 or 370.15 K) [1]
• M2: Correct calculation of moles: \(2.58 \times 10^{-3}\text{ to } 2.59 \times 10^{-3}\text{ mol}\) [1]
• M3: Correct final molar mass to 3 s.f.: \(79.3\text{ g mol}^{-1}\) (Accept 79.2 to 79.4 to allow for minor rounding variations) [1]

(i) Generation of electrophile:
\(\text{CH}_3\text{Br} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_3\text{Br}^-\)
(Accept \(\text{AlCl}_3\text{Br}^-\))

(ii) In the carbocation intermediate:
There is a positive charge inside a broken ring (representing 4 pi electrons, open towards the \(\text{sp}^3\) carbon).
The \(\text{C-H}\) bond breaks, with the pair of electrons moving from the \(\text{C-H}\) bond back into the ring to restore the aromatic system.
The \(\text{H}^+\) ion released reacts with \(\text{AlCl}_3\text{Br}^-\), forming \(\text{HBr}\) and regenerating the \(\text{AlCl}_3\) catalyst.

• M1: Correct equation for electrophile generation: \(\text{CH}_3\text{Br} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_3\text{Br}^-\) [1]
• M2: Curly arrow from the \(\text{C-H}\) bond back into the partially broken benzene ring in the intermediate [1]
• M3: Statement or equation showing regeneration of the catalyst: \(\text{AlCl}_3\text{Br}^- + \text{H}^+ \rightarrow \text{AlCl}_3 + \text{HBr}\) [1]

(i) Standard cell potential:
\(E^\theta_{\text{cell}} = E^\theta(\text{reduction}) - E^\theta(\text{oxidation}) = 2.01\text{ V} - 0.54\text{ V} = +1.47\text{ V}\)

(ii) Catalysis explanation:
- First, \(\text{Fe}^{3+}\) can oxidize \(\text{I}^-\):
\(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) > E^\theta(\text{I}_2/\text{I}^-)\) (i.e. \(+0.77\text{ V} > +0.54\text{ V}\)), so the reaction \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) is feasible.
- Second, \(\text{S}_2\text{O}_8^{2-}\) can oxidize the generated \(\text{Fe}^{2+}\) back to \(\text{Fe}^{3+}\):
\(E^\theta(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) > E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+})\) (i.e. \(+2.01\text{ V} > +0.77\text{ V}\)), so the reaction \(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\) is feasible.
- Both steps involve oppositely charged ions, bypasses the high activation energy barrier of the direct reaction between negatively charged \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\).

• M1: Calculates \(E^\theta_{\text{cell}} = +1.47\text{ V}\) (including unit and sign) [1]
• M2: Explains that \(\text{Fe}^{3+}\) can oxidize \(\text{I}^-\)\ because \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) > E^\theta(\text{I}_2/\text{I}^-)\) (or comparing \(+0.77\text{ V}\) and \(+0.54\text{ V}\)) [1]
• M3: Explains that \(\text{S}_2\text{O}_8^{2-}\) can oxidize \(\text{Fe}^{2+}\) back to \(\text{Fe}^{3+}\) because \(E^\theta(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) > E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+})\) (or comparing \(+2.01\text{ V}\) and \(+0.77\text{ V}\)) [1]

1. Analyze the triplet and quartet:
- The triplet (\(3\text{H}\)) at \(\delta = 1.25\text{ ppm}\) and quartet (\(2\text{H}\)) at \(\delta = 4.12\text{ ppm}\) show a \(-\text{CH}_2-\text{CH}_3\) ethyl group because they split each other (n+1 rule: neighboring carbon of \(\text{CH}_3\) has 2 protons so it splits to triplet, neighboring carbon of \(\text{CH}_2\) has 3 protons so it splits to quartet).
- The chemical shift of the quartet (\(\delta = 4.12\text{ ppm}\)) is in the range for a \(\text{H}-\text{C}-\text{O}-\) group, indicating the ethyl group is attached directly to the single-bonded oxygen of an ester.

2. Analyze the singlet:
- The singlet (\(3\text{H}\)) at \(\delta = 2.03\text{ ppm}\) indicates a isolated \(\text{CH}_3\) group with no neighboring protons.
- The shift of \(2.03\text{ ppm}\) is characteristic of \(\text{H}-\text{C}-\text{C}=\text{O}\), indicating a methyl group adjacent to a carbonyl group.

3. Put parts together:
- Combine \(\text{CH}_3-\text{C}=\text{O}\) with \(-\text{O}-\text{CH}_2-\text{CH}_3\).
- This forms ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

• M1: Identifies the presence of an ethyl group, \(-\text{CH}_2\text{CH}_3\), from triplet-quartet splitting pattern [1]
• M2: Associates the quartet at \(\delta = 4.12\text{ ppm}\) with \(-\text{O}-\text{CH}_2\text{CH}_3\) (attached to oxygen) [1]
• M3: Associates the singlet at \(\delta = 2.03\text{ ppm}\) with \(\text{CH}_3\text{C}=\text{O}\) (methyl next to carbonyl) [1]
• M4: Deduces the correct structure as ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (either name or displayed/structural formula) [1]

1. Determine equilibrium amounts using ICE table:
- Initial: \(n(\text{H}_2) = 1.50\text{ mol}\), \(n(\text{I}_2) = 1.50\text{ mol}\), \(n(\text{HI}) = 0\text{ mol}\)
- Change: \(n(\text{HI})\) increased by \(+2.20\text{ mol}\), so \(n(\text{H}_2)\) and \(n(\text{I}_2)\) both decreased by \(\frac{2.20}{2} = 1.10\text{ mol}\).
- Equilibrium:
\(n(\text{H}_2) = 1.50 - 1.10 = 0.40\text{ mol}\)
\(n(\text{I}_2) = 1.50 - 1.10 = 0.40\text{ mol}\)
\(n(\text{HI}) = 2.20\text{ mol}\)

2. Calculate equilibrium concentrations (divided by volume \(V = 2.00\text{ dm}^3\)):
- \([\text{H}_2] = \frac{0.40}{2.00} = 0.20\text{ mol dm}^{-3}\)
- \([\text{I}_2] = \frac{0.40}{2.00} = 0.20\text{ mol dm}^{-3}\)
- \([\text{HI}] = \frac{2.20}{2.00} = 1.10\text{ mol dm}^{-3}\)

3. Calculate \(K_c\):
- \(K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\)
- \(K_c = \frac{(1.10)^2}{(0.20)(0.20)} = \frac{1.21}{0.04} = 30.25 \approx 30.3\)

• M1: Deduuces that \(1.10\text{ mol}\) of \(\text{H}_2\) and \(\text{I}_2\) react [1]
• M2: Calculates equilibrium moles of \(\text{H}_2\) and \(\text{I}_2\) as \(0.40\text{ mol}\) [1]
• M3: Converts moles to concentrations by dividing by 2.00 (or correctly cancels volumes in the expression) [1]
• M4: Obtains correct value for \(K_c\) of \(30.3\) (or \(30.25\); ignore units since they cancel) [1]

1. **Calculate the initial amount, in moles, of lactic acid (HA) and sodium hydroxide (NaOH):**
\(n(\text{HA}) = \frac{35.0}{1000} \times 0.250 = 8.75 \times 10^{-3}\text{ mol}\)
\(n(\text{NaOH}) = \frac{15.0}{1000} \times 0.180 = 2.70 \times 10^{-3}\text{ mol}\)

2. **Determine the equilibrium amounts of HA and the conjugate base (\(\text{A}^-\)):**
\(n(\text{HA})_{\text{excess}} = 8.75 \times 10^{-3} - 2.70 \times 10^{-3} = 6.05 \times 10^{-3}\text{ mol}\)
\(n(\text{A}^-) = 2.70 \times 10^{-3}\text{ mol}\)

3. **Calculate the concentration of hydrogen ions, \([\text{H}^+]\):**
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.38 \times 10^{-4} \times \frac{6.05 \times 10^{-3}}{2.70 \times 10^{-3}} = 3.0922 \times 10^{-4}\text{ mol dm}^{-3}\)

4. **Calculate pH:**
\(\text{pH} = -\log_{10}(3.0922 \times 10^{-4}) = 3.51\)

**Mark 1:** Calculation of excess moles of lactic acid (\(6.05 \times 10^{-3}\text{ mol}\)) AND moles of conjugate base formed (\(2.70 \times 10^{-3}\text{ mol}\)).
**Mark 2:** Calculation of \([\text{H}^+]\) as \(3.09 \times 10^{-4}\text{ mol dm}^{-3}\) (or correct substitution of values into the Henderson-Hasselbalch equation).
**Mark 3:** Correct final pH value to 2 decimal places (\(3.51\)).

*Allow ECF at each step, including calculation of concentrations instead of using mole ratios direct: \([\text{HA}] = 0.121\text{ mol dm}^{-3}\) and \([\text{A}^-] = 0.0540\text{ mol dm}^{-3}\).*

1. Calculate initial moles:
\(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
\(n(\text{A}^-) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.00750\text{ mol}\)

2. Calculate moles of \(\text{H}^+\) added:
\(n(\text{H}^+) = 0.00200\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.000200\text{ mol}\)

3. Update moles after reaction (\(\text{A}^- + \text{H}^+ \rightarrow \text{HA}\)):
\(n(\text{HA})_{\text{new}} = 0.0125 + 0.000200 = 0.0127\text{ mol}\)
\(n(\text{A}^-)_{\text{new}} = 0.00750 - 0.000200 = 0.00730\text{ mol}\)

4. Calculate new \([\text{H}^+]\):
\([\text{H}^+] = K_a \times \frac{n(\text{HA})_{\text{new}}}{n(\text{A}^-)_{\text{new}}} = 1.35 \times 10^{-5} \times \frac{0.0127}{0.00730} = 2.3486 \times 10^{-5}\text{ mol dm}^{-3}\)

5. Calculate pH:
\(\text{pH} = -\log_{10}(2.3486 \times 10^{-5}) = 4.629 \approx 4.63\)

M1: Moles of initial propanoic acid (0.0125 mol), propanoate ions (0.00750 mol), and added H+ (0.000200 mol) [1]
M2: New moles of propanoic acid (0.0127 mol) and propanoate ions (0.00730 mol) [1]
M3: Calculation of [H+] = 2.35 x 10^-5 mol dm^-3 [1]
M4: Correct pH to 2 decimal places = 4.63 [0.5]

1. Calculate moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.0200 \times 0.00500 = 1.00 \times 10^{-4}\text{ mol}\)

2. Determine moles of \(\text{Fe}^{2+}\) in the titration using the stoichiometry \(5\text{Fe}^{2+} : 1\text{MnO}_4^-\):
\(n(\text{Fe}^{2+})\text{ in } 25.00\text{ cm}^3 = 5 \times 1.00 \times 10^{-4} = 5.00 \times 10^{-4}\text{ mol}\)

3. Scale up to find moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\):
\(n(\text{Fe}^{2+})\text{ in } 250.0\text{ cm}^3 = 5.00 \times 10^{-3}\text{ mol}\)

4. Calculate molar mass of the hydrated salt:
\(M = \frac{1.96\text{ g}}{5.00 \times 10^{-3}\text{ mol}} = 392\text{ g mol}^{-1}\)

5. Calculate mass of anhydrous salt and determine water content:
\(M((\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2) = 2(18.0) + 55.8 + 2(96.1) = 284.0\text{ g mol}^{-1}\)
\(M(\text{water of crystallisation}) = 392 - 284 = 108\text{ g mol}^{-1}\)
\(z = \frac{108}{18.0} = 6\)

M1: Correct calculation of moles of Fe2+ in 250 cm3 (5.00 x 10^-3 mol) [1]
M2: Calculation of the molar mass of the hydrated salt (392 g mol^-1) [1]
M3: Calculation of molar mass of anhydrous salt (284 g mol^-1) OR mass of water per mole (108 g) [1]
M4: Deducing z = 6 [0.5]

1. Moles of starting methylbenzene:
\(n(\text{methylbenzene}) = \frac{9.20}{92.0} = 0.100\text{ mol}\)

2. Theoretical maximum moles of ethyl benzoate that can be formed from \(0.100\text{ mol}\) of methylbenzene is \(0.100\text{ mol}\).

3. Actual moles of ethyl benzoate obtained in Step 2:
\(n(\text{ethyl benzoate}) = \frac{9.00}{150.0} = 0.0600\text{ mol}\)

4. Overall percentage yield:
\(\text{Overall Yield} = \frac{\text{Actual Moles}}{\text{Theoretical Moles}} \times 100 = \frac{0.0600}{0.100} \times 100 = 60.0\%\)

Alternative method via masses:
Mass of benzoic acid obtained = \(0.100 \times 0.750 \times 122.0 = 9.15\text{ g}\).
Theoretical yield of ethyl benzoate from \(9.15\text{ g}\) of benzoic acid = \(\frac{9.15}{122.0} \times 150.0 = 11.25\text{ g}\).
But original theoretical mass of ethyl benzoate from \(0.100\text{ mol}\) of methylbenzene is \(0.100 \times 150.0 = 15.0\text{ g}\).
Overall Yield = \ \frac{9.00}{15.0} \times 100 = 60.0\%\).

M1: Calculates initial moles of methylbenzene = 0.100 mol [1]
M2: Calculates actual moles of ethyl benzoate = 0.0600 mol (or theoretical mass of 15.0 g) [1]
M3: Correct overall percentage yield expression [1]
M4: Correct overall yield = 60.0% (accept 60%) [0.5]

1. Use Hess's Law cycle relationship for dissolving an ionic lattice:
\(\Delta_{sol}H^{\ominus} = -\Delta_{LE}H^{\ominus} + \Delta_{hyd}H^{\ominus}(\text{Ca}^{2+}) + 2\Delta_{hyd}H^{\ominus}(\text{Cl}^-)\)

2. Substitute the values into the equation:
\(-82.0 = -(-2258) + (-1579) + 2\Delta_{hyd}H^{\ominus}(\text{Cl}^-)\)
\(-82.0 = 2258 - 1579 + 2\Delta_{hyd}H^{\ominus}(\text{Cl}^-)\)
\(-82.0 = 679 + 2\Delta_{hyd}H^{\ominus}(\text{Cl}^-)\)

3. Rearrange to solve for the hydration enthalpy of chloride ions:
\(2\Delta_{hyd}H^{\ominus}(\text{Cl}^-) = -82.0 - 679 = -761\text{ kJ mol}^{-1}\)
\(\Delta_{hyd}H^{\ominus}(\text{Cl}^-) = \frac{-761}{2} = -380.5\text{ kJ mol}^{-1}\)

M1: Correct algebraic expression of the energy cycle [1]
M2: Correct substitution of values including signs [1]
M3: Correct calculation of total hydration of chloride ions as -761 kJ mol^-1 [1]
M4: Correct division by 2 to give -380.5 kJ mol^-1 (must have negative sign) [0.5]

1. Determine which species is reduced: The silver half-cell has the more positive potential (\(+0.80\text{ V} > +0.77\text{ V}\)), so \(\text{Ag}^+(aq)\) is reduced:
\(\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)

2. The iron half-cell undergoes oxidation (reversed equation):
\(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\)

3. Combine to form the overall feasible equation:
\(\text{Ag}^+(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Ag}(s) + \text{Fe}^{3+}(aq)\)

4. Calculate \(E^{\ominus}_{\text{cell}}\):
\(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{red}} - E^{\ominus}_{\text{ox}} = +0.80 - (+0.77) = +0.03\text{ V}\)

5. Identify the oxidizing agent:
The species being reduced is \(\text{Ag}^+(aq)\), so it is the oxidizing agent.

M1: Correct overall equation with correct species and balancing [1]
M2: State symbols included correctly in the overall equation [0.5]
M3: Correct calculation of E_cell = +0.03 V (value and sign required) [1]
M4: Correctly identifies Ag+(aq) (or silver(I) ions / silver ions) as the oxidizing agent [1]

1. Calculate moles of methyl benzoate used:
\(n(\text{methyl benzoate}) = \frac{4.08\text{ g}}{136.0\text{ g mol}^{-1}} = 0.0300\text{ mol}\)

2. Calculate theoretical maximum yield of methyl 3-nitrobenzoate in moles:
Since the stoichiometry is 1:1, theoretical moles = \(0.0300\text{ mol}\)

3. Calculate theoretical maximum mass of product:
\(\text{Theoretical Mass} = 0.0300\text{ mol} \times 181.0\text{ g mol}^{-1} = 5.43\text{ g}\)

4. Calculate the percentage yield:
\(\text{Percentage Yield} = \frac{4.34\text{ g}}{5.43\text{ g}} \times 100 = 79.926\% \approx 79.9\%\)

M1: Calculates moles of methyl benzoate as 0.0300 mol [1]
M2: Calculates theoretical maximum mass of product as 5.43 g [1]
M3: Correct substitution of values into percentage yield formula [1]
M4: Correct final percentage yield value of 79.9% (must be to 3 sig figs) [0.5]

1. Determine equilibrium moles using ICE method:
\(\text{CO} = 1.00 - 0.400 = 0.600\text{ mol}\)
\(\text{H}_2 = 2.00 - 2(0.400) = 1.20\text{ mol}\)
\(\text{CH}_3\text{OH} = 0.400\text{ mol}\)

Total equilibrium moles = \(0.600 + 1.20 + 0.400 = 2.20\text{ mol}\)

2. Calculate partial pressures:
\(p(\text{CO}) = \frac{0.600}{2.20} \times 12.0 = 3.2727\text{ kPa}\)
\(p(\text{H}_2) = \frac{1.20}{2.20} \times 12.0 = 6.5454\text{ kPa}\)
\(p(\text{CH}_3\text{OH}) = \frac{0.400}{2.20} \times 12.0 = 2.1818\text{ kPa}\)

3. Calculate \(K_p\):
\(K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times p(\text{H}_2)^2} = \frac{2.1818}{3.2727 \times (6.5454)^2} = 0.01556\text{ kPa}^{-2}\)

With 3 significant figures, \(K_p = 0.0156\text{ kPa}^{-2}\) (or \(1.56 \times 10^{-2}\text{ kPa}^{-2}\)).

M1: Correct calculation of equilibrium moles of CO (0.60 mol), H2 (1.20 mol), and total moles (2.20 mol) [1]
M2: Calculation of the partial pressures of all three components [1]
M3: Correct Kp expression and substitution to give 0.0156 (or 0.0155 - 0.0157) [1]
M4: Correct units of kPa^-2 [0.5]

1. Determine empirical formula from combustion data:
\(n(\text{C}) = n(\text{CO}_2) = \frac{0.352}{44.0} = 0.00800\text{ mol}\) (mass of \(\text{C} = 0.0960\text{ g}\))
\(n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{0.144}{18.0} = 0.0160\text{ mol}\) (mass of \(\text{H} = 0.0160\text{ g}\))
\(\text{Mass of O} = 0.176 - 0.0960 - 0.0160 = 0.0640\text{ g}\)
\(n(\text{O}) = \frac{0.0640}{16.0} = 0.00400\text{ mol}\)

Ratio of \(\text{C} : \text{H} : \text{O} = 0.00800 : 0.0160 : 0.00400 = 2 : 4 : 1\).
Empirical formula = \(\text{C}_2\text{H}_4\text{O}\).

2. Determine molecular formula:
Empirical formula mass = 44.0.
Since \(M_r = 88\), molecular formula = \(\text{C}_4\text{H}_8\text{O}_2\).

3. Deduce structure from IR and NMR:
- IR absorption at \(1740\text{ cm}^{-1}\) indicates a carbonyl (\(\text{C}=\text{O}\)) group. No broad peak above \(3000\text{ cm}^{-1}\) indicates the absence of an \(\text{O}-\text{H}\) bond, so it is an ester, not a carboxylic acid.
- NMR quartet at \(\delta = 4.1\text{ ppm}\) (2H) and triplet at \(\delta = 1.2\text{ ppm}\) (3H) indicates a \(\text{-OCH}_2\text{CH}_3\) ethyl ester group.
- NMR singlet at \(\delta = 2.0\text{ ppm}\) (3H) indicates an isolated methyl group next to the carbonyl, \(\text{CH}_3\text{CO-}\).

Thus, the structural formula is \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (ethyl ethanoate).

M1: Correct calculation to determine empirical formula C2H4O from combustion data [1]
M2: Deduces molecular formula as C4H8O2 using molecular ion peak at m/z = 88 [1]
M3: Identifies ester group (C=O at 1740 cm^-1, no O-H) and ethyl group from triplet/quartet NMR splitting [1]
M4: Correct structural formula: CH3COOCH2CH3 [0.5]

1. Calculate the amount in moles of the chromium compound: \(n(\text{compound}) = \frac{2.665}{266.5} = 0.0100\text{ mol}\).
2. Calculate the amount in moles of silver chloride precipitated: \(n(\text{AgCl}) = \frac{2.868}{143.4} = 0.0200\text{ mol}\).
3. Determine the ratio of precipitated chloride ions to chromium complex: \(\frac{0.0200}{0.0100} = 2\). This means there are 2 ionic chloride ions outside the coordination sphere, leaving 1 chloride ion within the coordination sphere.
4. Since the coordination number is 6, there must be 5 water ligands in the coordination sphere. Thus, the formula of the complex ion is \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]^{2+}\).

M1: Calculates moles of compound as 0.0100 mol and moles of AgCl as 0.0200 mol (1)
M2: Deduces that the ratio is 2 chloride ions outside the coordination sphere per Cr atom (1)
M3: Writes correct formula for the complex ion: [Cr(H2O)5Cl]2+ (1). (Accept [Cr(Cl)(H2O)5]2+)

1. Calculate initial moles:
\(n(\text{HCOOH}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
\(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.300\text{ mol dm}^{-3} = 0.0075\text{ mol}\)
2. Calculate remaining moles of acid and moles of conjugate base formed after reaction:
\(\text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O}\)
\(n(\text{HCOOH})_{\text{remaining}} = 0.0125 - 0.0075 = 0.0050\text{ mol}\)
\(n(\text{HCOO}^-)_{\text{formed}} = 0.0075\text{ mol}\)
3. Use the acid dissociation constant expression:
\([\text{H}^+] = K_a \times \frac{[\text{HCOOH}]}{[\text{HCOO}^-]} = 1.78 \times 10^{-4} \times \frac{0.0050}{0.0075} = 1.187 \times 10^{-4}\text{ mol dm}^{-3}\)
4. Calculate pH:
\(\text{pH} = -\log_{10}(1.187 \times 10^{-4}) = 3.9256 \approx 3.93\).

M1: Calculates initial moles of HCOOH (0.0125 mol) and NaOH (0.0075 mol) (1)
M2: Calculates remaining moles of HCOOH (0.0050 mol) and moles of HCOO- formed (0.0075 mol) (1)
M3: Substitutes values correctly into Ka expression to find [H+] = 1.19 x 10^-4 mol dm^-3 (1)
M4: Calculates pH as 3.93 (must be 2 decimal places) (1)

1. Calculate \(\Delta H^\ominus\):
\(\Delta H^\ominus = [(-635.1) + (-393.5)] - [-1207.0] = -1028.6 + 1207.0 = +178.4\text{ kJ mol}^{-1}\).
2. Calculate \(\Delta S^\ominus\):
\(\Delta S^\ominus = [39.7 + 213.6] - 92.9 = +160.4\text{ J K}^{-1}\text{mol}^{-1} = +0.1604\text{ kJ K}^{-1}\text{mol}^{-1}\).
3. Find the minimum temperature for feasibility (where \(\Delta G \le 0\)):
\(T = \frac{\Delta H^\ominus}{\Delta S^\ominus} = \frac{178.4}{0.1604} = 1112.2\text{ K}\).
4. Round to 3 significant figures: \(1110\text{ K}\).

M1: Correct calculation of enthalpy change: +178.4 kJ mol^-1 (1)
M2: Correct calculation of entropy change: +160.4 J K^-1 mol^-1 (1)
M3: Uses T = delta H / delta S with correct conversion of units (dividing entropy by 1000) (1)
M4: Calculates temperature as 1110 K (or 1112 K) (1)

1. Calculate the standard cell potential:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77\text{ V} - 0.34\text{ V} = +0.43\text{ V}\).
2. Write the overall equation:
Since the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell has a more positive potential, it will undergo reduction: \(\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}\).
The \(\text{Cu}^{2+}/\text{Cu}\) half-cell has a less positive potential, so it will undergo oxidation: \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\).
Combining the two half-equations gives: \(\text{Cu}(\text{s}) + 2\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq})\).
3. Identify the reducing agent:
\(\text{Cu}\) (or copper metal) is oxidized and therefore acts as the reducing agent.

M1: Standard cell potential = +0.43 V (1)
M2: Correct balanced overall equation: Cu(s) + 2Fe3+(aq) -> Cu2+(aq) + 2Fe2+(aq) (state symbols not required) (1)
M3: Identifies Cu (or copper) as the reducing agent (1)

1. Calculate \(R_f\) of Spot B:
\(R_f = \frac{\text{distance traveled by substance}}{\text{distance traveled by solvent front}} = \frac{5.8}{8.4} = 0.6905 \approx 0.69\).
2. Explanation of relative movement:
Silica gel on the TLC plate acts as the stationary phase and is highly polar because of surface silanol (Si-OH) groups.
Spot B has a higher \(R_f\) value than Spot A, meaning it has traveled further up the plate.
This indicates that Spot B is less polar than Spot A, so it has weaker adsorption to the polar stationary phase and greater solubility in the less polar mobile phase.

M1: Calculates Rf of Spot B as 0.69 (must be to 2 decimal places) (1)
M2: Explains that silica stationary phase is polar (1)
M3: Explains that Spot B is less polar than Spot A, so it adsorbs less strongly to the stationary phase / dissolves more in the mobile phase (1)

1. Calculate moles of each reactant:
\(n(\text{salicylic acid}) = \frac{4.14}{138.0} = 0.0300\text{ mol}\)
\(n(\text{ethanoic anhydride}) = \frac{6.12}{102.0} = 0.0600\text{ mol}\)
2. Since the reaction stoichiometry is 1:1, salicylic acid is the limiting reactant because there are fewer moles of it.
3. Calculate theoretical yield of aspirin:
\(n(\text{aspirin})_{\text{theoretical}} = 0.0300\text{ mol}\)
Theoretical mass = \(0.0300\text{ mol} \times 180.0\text{ g mol}^{-1} = 5.40\text{ g}\).
4. Calculate percentage yield:
Percentage yield = \frac{3.84}{5.40} \times 100\% = 71.11\% \approx 71.1\%.

M1: Calculates moles of salicylic acid as 0.0300 mol AND ethanoic anhydride as 0.0600 mol (1)
M2: Identifies salicylic acid as limiting reactant with a valid reason (e.g. 1:1 ratio and fewer moles of salicylic acid) (1)
M3: Determines theoretical mass of aspirin as 5.40 g (1)
M4: Calculates percentage yield of aspirin as 71.1% (accept 71.1% - 71.2%) (1)

1. Find empirical formula:
\(n(\text{C}) = n(\text{CO}_2) = \frac{3.52}{44.0} = 0.080\text{ mol}\) (mass of C = \(0.080 \times 12.0 = 0.96\text{ g}\))
\(n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{1.44}{18.0} = 0.16\text{ mol}\) (mass of H = \(0.16 \times 1.0 = 0.16\text{ g}\))
\(\text{Mass of O} = 1.76 - (0.96 + 0.16) = 0.64\text{ g}\)
\(n(\text{O}) = \frac{0.64}{16.0} = 0.040\text{ mol}\)
Ratio C : H : O = \(0.080 : 0.16 : 0.040 = 2 : 4 : 1\). Empirical formula = \(\text{C}_2\text{H}_4\text{O}\).
2. Find molecular formula:
Empirical formula mass = 44.0. Since \(m/z = 88\), the molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).
3. Deduce structure from NMR:
- Triplet (3H) and quartet (2H) show an ethyl group adjacent to an electronegative atom: \(\text{-O-CH}_2\text{-CH}_3\).
- Singlet (3H) shows a methyl group adjacent to a carbonyl group: \(\text{CH}_3\text{-CO-}\).
- Therefore, X is ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

M1: Correct calculation of empirical formula as C2H4O (1)
M2: Identifies molecular formula as C4H8O2 using m/z = 88 (1)
M3: Uses NMR splitting to identify the ethyl group (-CH2CH3) and singlet to identify CH3CO- group (1)
M4: Deduces correct structural formula CH3COOCH2CH3 (or ethyl ethanoate) (1)

1. Set up equilibrium table using concentrations (in \(\text{mol dm}^{-3}\)):
Flask volume = \(2.00\text{ dm}^3\).
Initial concentrations:
\([\text{SO}_2]_{\text{initial}} = \frac{1.00}{2.00} = 0.500\text{ mol dm}^{-3}\)
\([\text{O}_2]_{\text{initial}} = \frac{1.00}{2.00} = 0.500\text{ mol dm}^{-3}\)
\([\text{SO}_3]_{\text{initial}} = 0.00\text{ mol dm}^{-3}\)
At equilibrium:
\([\text{SO}_3]_{\text{equilibrium}} = 0.350\text{ mol dm}^{-3}\)
Change in \([\text{SO}_3] = +0.350\text{ mol dm}^{-3}\)
Change in \([\text{SO}_2] = -0.350\text{ mol dm}^{-3}\)
Change in \([\text{O}_2] = -0.175\text{ mol dm}^{-3}\)
Equilibrium concentrations:
\([\text{SO}_2]_{\text{equilibrium}} = 0.500 - 0.350 = 0.150\text{ mol dm}^{-3}\)
\([\text{O}_2]_{\text{equilibrium}} = 0.500 - 0.175 = 0.325\text{ mol dm}^{-3}\)
2. Calculate \(K_c\):
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{0.350^2}{0.150^2 \times 0.325} = 16.75\text{ dm}^3\text{ mol}^{-1}\).
Rounding to 3 sig figs: \(16.8\text{ dm}^3\text{ mol}^{-1}\).

M1: Determines equilibrium concentrations of [SO2] = 0.150 mol dm^-3 and [O2] = 0.325 mol dm^-3 (1)
M2: Correct Kc expression and calculation to yield 16.8 (1)
M3: Correct units: dm3 mol-1 (1)

First, calculate the initial moles of propanoic acid and sodium hydroxide: Moles of \(\text{C}_2\text{H}_5\text{COOH} = \frac{40.0}{1000} \times 0.150 = 6.00 \times 10^{-3}\text{ mol}\); Moles of \(\text{NaOH} = \frac{60.0}{1000} \times 0.0800 = 4.80 \times 10^{-3}\text{ mol}\). The sodium hydroxide reacts completely with the propanoic acid to form sodium propanoate: \(\text{C}_2\text{H}_5\text{COOH} + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{COONa} + \text{H}_2\text{O}\). Moles of \(\text{C}_2\text{H}_5\text{COO}^-\text{ formed} = 4.80 \times 10^{-3}\text{ mol}\); Moles of remaining \(\text{C}_2\text{H}_5\text{COOH} = 6.00 \times 10^{-3} - 4.80 \times 10^{-3} = 1.20 \times 10^{-3}\text{ mol}\). Using the acid dissociation constant expression: \(K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_5\text{COO}^-]}{[\text{C}_2\text{H}_5\text{COOH}]}\). Rearranging to solve for \([\text{H}^+]\): \([\text{H}^+] = K_a \times \frac{[\text{C}_2\text{H}_5\text{COOH}]}{[\text{C}_2\text{H}_5\text{COO}^-]}\). Since the volumes cancel out, we can use the mole ratio directly: \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{1.20 \times 10^{-3}}{4.80 \times 10^{-3}} = 3.375 \times 10^{-6}\text{ mol dm}^{-3}\). Calculate pH: \(\text{pH} = -\log_{10}(3.375 \times 10^{-6}) = 5.4717 \approx 5.47\).

1. Moles of acid AND moles of \(\text{NaOH}\) correctly calculated: \(\text{mol(HA)} = 6.00 \times 10^{-3}\) and \(\text{mol(NaOH)} = 4.80 \times 10^{-3}\) (1 mark). 2. Moles of \(\text{A}^- = 4.80 \times 10^{-3}\) and remaining moles of \(\text{HA} = 1.20 \times 10^{-3}\) correctly determined (1 mark). 3. Correct substitution into the \([\text{H}^+]\) or \(\text{pH}\) equation (1 mark). 4. Final pH of 5.47 (1 mark) - must be given to 2 decimal places. Do not accept 5.5 or 5.472.

First, calculate the entropy change of the reaction, \(\Delta S^\theta\): \(\Delta S^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})\) so \(\Delta S^\theta = (39.7 + 213.6) - 92.9 = 253.3 - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\). For the reaction to be feasible, the Gibbs free energy change must be less than or equal to zero (\(\Delta G \le 0\)): \(\Delta G = \Delta H - T\Delta S \le 0\). At the minimum temperature for feasibility, \(\Delta G = 0\), so: \(T = \frac{\Delta H}{\Delta S}\). Convert \(\Delta H\) to \(\text{J mol}^{-1}\) (or \(\Delta S\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\)): \(\Delta H = 178 \times 10^3 = 178000\text{ J mol}^{-1}\). Thus, \(T = \frac{178000}{160.4} = 1109.7\text{ K}\). To 3 significant figures, this is \(1110\text{ K}\).

1. Correct calculation of \(\Delta S^\theta = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\) (1 mark). 2. Correct rearrangement of Gibbs free energy equation and conversion of units (e.g. dividing 178000 by 160.4 or 178 by 0.1604) (1 mark). 3. Final temperature of \(1110\text{ K}\) (1 mark) - must be 3 significant figures. Do not accept 1109.7 or 1100.

First, find the moles of manganate(VII) ions used in the titration: \(n(\text{MnO}_4^-) = \text{concentration} \times \text{volume} = 0.0200 \times \frac{24.80}{1000} = 4.96 \times 10^{-4}\text{ mol}\). Using the stoichiometry of the equation, find the moles of \(\text{Fe}^{2+}\) in the \(25.00\text{ cm}^3\) sample: \(n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-) = 5 \times 4.96 \times 10^{-4} = 2.48 \times 10^{-3}\text{ mol}\). Scale up to find the moles of \(\text{Fe}^{2+}\) in the original \(100.0\text{ cm}^3\) solution: \(n(\text{Fe}^{2+})_{100} = 2.48 \times 10^{-3} \times \frac{100.0}{25.00} = 9.92 \times 10^{-3}\text{ mol}\). Since \(1\text{ mol}\) of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) contains \(1\text{ mol}\) of \(\text{Fe}^{2+}\), the moles of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) is \(9.92 \times 10^{-3}\text{ mol}\). Calculate the mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\): \(\text{mass} = 9.92 \times 10^{-3} \times 278.0 = 2.75776\text{ g}\). Calculate percentage purity: \(\text{percentage purity} = \frac{2.75776}{3.20} \times 100 = 86.18\% \approx 86.2\%\).

1. Correct calculation of moles of \(\text{MnO}_4^- = 4.96 \times 10^{-4}\text{ mol}\) (1 mark). 2. Correctly multiplies by 5 and scales up by 4 to get moles of \(\text{Fe}^{2+}\) in \(100.0\text{ cm}^3 = 9.92 \times 10^{-3}\text{ mol}\) (1 mark). 3. Correct calculation of the mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O} = 2.76\text{ g}\) (or 2.758 g) (1 mark). 4. Correct calculation of percentage purity as \(86.2\%\) (or \(86.18\%\)) (1 mark). Do not accept rounding to 86% without a shown calculation.

First, find the moles of \(\text{AgCl}\) precipitated: \(n(\text{AgCl}) = \frac{0.430}{143.4} = 2.9986 \times 10^{-3}\text{ mol} \approx 3.00 \times 10^{-3}\text{ mol}\). Since there are 3 chloride ions per formula unit of \(\text{MCl}_3 \cdot 6\text{H}_2\text{O}\), the moles of the complex in the \(25.00\text{ cm}^3\) portion is: \(n(\text{complex})_{25} = \frac{1}{3} \times n(\text{AgCl}) = 1.00 \times 10^{-3}\text{ mol}\). Scale up to find the total moles in the original \(250.0\text{ cm}^3\) solution: \(n(\text{complex})_{250} = 1.00 \times 10^{-3} \times 10 = 1.00 \times 10^{-2}\text{ mol}\). Calculate the molar mass of \(\text{MCl}_3 \cdot 6\text{H}_2\text{O}\): \(M = \frac{2.665\text{ g}}{1.00 \times 10^{-2}\text{ mol}} = 266.5\text{ g mol}^{-1}\). Subtract the molar masses of the chlorine atoms and water molecules to find the atomic mass of \(\text{M}\): \(M_{\text{M}} = 266.5 - (3 \times 35.5) - (6 \times 18.0) = 266.5 - 106.5 - 108.0 = 52.0\text{ g mol}^{-1}\). The transition metal with an atomic mass of \(52.0\text{ g mol}^{-1}\) is Chromium (\(\text{Cr}\)).

1. Correct calculation of moles of \(\text{AgCl} = 3.00 \times 10^{-3}\text{ mol}\) and scaling to find moles of complex in \(250.0\text{ cm}^3 = 1.00 \times 10^{-2}\text{ mol}\) (1 mark). 2. Correct determination of the molar mass of \(\text{MCl}_3 \cdot 6\text{H}_2\text{O} = 266.5\text{ g mol}^{-1}\) (1 mark). 3. Correct calculation of the atomic mass of \(\text{M} = 52.0\text{ g mol}^{-1}\) and identifying \(\text{M}\) as Chromium / \(\text{Cr}\) (1 mark).