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Let the percentage abundance of \(^{32}\text{S}\) be \(x\%\). Since the abundance of \(^{33}\text{S}\) is \(2.0\%\), the abundance of \(^{34}\text{S}\) is \((100 - 2.0 - x)\% = (98 - x)\%\). Using the relative atomic mass formula: \(A_r = \frac{32x + 33(2.0) + 34(98 - x)}{100} = 32.20\). Multiplying both sides by 100: \(32x + 66 + 3332 - 34x = 3220\). Simplifying gives: \(-2x + 3398 = 3220 \implies -2x = -178 \implies x = 89\%\). Thus, the abundance of \(^{34}\text{S}\) is \(98 - 89 = 9.0\%\).

1 mark for the correct answer of 9.0% (B).

The equation for the formation of ethane is: \(2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)\). Using standard enthalpies of combustion, \(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products}) = [2 \times (-394) + 3 \times (-286)] - [-1560] = [-788 - 858] + 1560 = -1646 + 1560 = -86\text{ kJ mol}^{-1}\).

1 mark for correctly calculating the enthalpy change as -86 kJ mol^-1 (A).

Reactivity of Group 2 elements with water increases down the group, so barium reacts more vigorously than calcium (C is correct). Magnesium hydroxide is less soluble in water than barium hydroxide, as solubility of Group 2 hydroxides increases down the group (A is incorrect). First ionisation energy decreases down the group due to increased shielding and atomic radius (B is incorrect). Carbonate thermal stability increases down the group, so barium carbonate is more stable than calcium carbonate (D is incorrect).

1 mark for selecting option C.

First, find the total moles of gas: \(n(\text{gas}) = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0250\text{ mol}\). According to the chemical equation, 2 moles of \(\text{X(NO}_3)_2\) produce 5 moles of gas. Therefore, the moles of \(\text{X(NO}_3)_2\) reacted is \(0.0250 \times \frac{2}{5} = 0.0100\text{ mol}\). Now find the molar mass of the compound: \(M_r = \frac{1.64\text{ g}}{0.0100\text{ mol}} = 164\text{ g mol}^{-1}\). The formula mass of \(2 \times \text{NO}_3\) is \(2 \times [14.0 + (3 \times 16.0)] = 124.0\text{ g mol}^{-1}\). The atomic mass of metal \(\text{X}\) is \(164.0 - 124.0 = 40.0\text{ g mol}^{-1}\), which corresponds to calcium.

1 mark for identifying Calcium (B).

When temperature is increased, the activation energy remains unchanged (only a catalyst can change it). The peak of the Boltzmann distribution shifts to a higher energy (to the right) and becomes flatter/lower to keep the area (total number of molecules) constant. This shifting of the curve results in a much greater proportion of molecules having energy greater than or equal to the activation energy.

1 mark for selecting option D.

For a compound to exhibit E/Z isomerism, each carbon atom of the double bond must be bonded to two different groups. In 1,1-dichlorobut-2-ene, \(\text{CHCl}_2-\text{CH}=\text{CH}-\text{CH}_3\), carbon-2 is bonded to \(\text{-H}\) and \(\text{-CHCl}_2\) (two different groups), and carbon-3 is bonded to \(\text{-H}\) and \(\text{-CH}_3\) (two different groups). Therefore, it can show stereoisomerism. All other options have at least one double-bonded carbon attached to two identical groups (e.g., two methyl groups or two bromine atoms).

1 mark for selecting option A.

The evolution of ammonia gas (turning red litmus blue) on heating with \(\text{NaOH}(aq)\) confirms the presence of ammonium ions, \(\text{NH}_4^+\). The formation of a cream-coloured precipitate with \(\text{AgNO}_3\) that dissolves only in concentrated ammonia confirms the presence of bromide ions, \(\text{Br}^-\). Thus, the salt is ammonium bromide, \(\text{NH}_4\text{Br}\).

1 mark for selecting option B.

An alcohol with the formula \(\text{C}_5\text{H}_{12}\text{O}\) being oxidized to a carboxylic acid with formula \(\text{C}_5\text{H}_{10}\text{O}_2\) must be a primary alcohol. Among the given options: 2-methylbutan-2-ol is a tertiary alcohol (not oxidized), pentan-3-ol and 3-methylbutan-2-ol are secondary alcohols (oxidized to ketones with formula \(\text{C}_5\text{H}_{10}\text{O}\)), and 3-methylbutan-1-ol is a primary alcohol, which oxidizes to 3-methylbutanoic acid (\(\text{C}_5\text{H}_{10}\text{O}_2\)).

1 mark for selecting option D.

First, calculate the moles of each reactant. Moles of \(\text{Mg} = \frac{0.1215\text{ g}}{24.3\text{ g mol}^{-1}} = 0.00500\text{ mol}\). Moles of \(\text{HCl} = 0.0200\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.00500\text{ mol}\). According to the equation, 1 mole of \(\text{Mg}\) requires 2 moles of \(\text{HCl}\). Therefore, \(\text{HCl}\) is the limiting reactant. The moles of \(\text{H}_2\) gas produced will be half the moles of the reacting \(\text{HCl}\): \(\text{moles of H}_2 = \frac{0.00500\text{ mol}}{2} = 0.00250\text{ mol}\). Finally, calculate the volume of hydrogen gas at RTP: \(\text{Volume} = 0.00250\text{ mol} \times 24,000\text{ cm}^3\text{ mol}^{-1} = 60.0\text{ cm}^3\).

1 mark for the correct answer B. Correctly identifying HCl as the limiting reactant and calculating the correct volume of hydrogen gas.

The equation for the standard enthalpy of formation of methanol is: \(\text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\). Using Hess's law and standard enthalpies of combustion: \(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\). Therefore, \(\Delta H_f^\theta = [\Delta H_c^\theta(\text{C}) + 2 \times \Delta H_c^\theta(\text{H}_2)] - \Delta H_c^\theta(\text{CH}_3\text{OH})\). Substituting the values: \(\Delta H_f^\theta = [-394 + 2(-286)] - [-726] = [-394 - 572] + 726 = -966 + 726 = -240\text{ kJ mol}^{-1}\).

1 mark for the correct answer B. Correctly applying the Hess's Law cycle using standard enthalpies of combustion.

To identify both ions without interference, \(\text{Ba(NO}_3)_2\text{(aq)}\) is added first to precipitate only the sulfate ions as white \(\text{BaSO}_4\text{(s)}\). Chloride ions do not precipitate with barium. The mixture is filtered to remove \(\text{BaSO}_4\). Adding \(\text{AgNO}_3\text{(aq)}\) to the filtrate then precipitates the remaining chloride ions as white \(\text{AgCl(s)}\). Adding barium chloride first would introduce chloride ions, creating a false positive. Adding hydrochloric acid would also introduce chloride ions. Adding silver nitrate first would precipitate silver chloride and some silver sulfate, leading to confusion.

1 mark for the correct answer C. Identifies the correct reagents and correct logical sequence.

When temperature increases, the average kinetic energy of the molecules increases, which shifts the peak of the curve to the right (higher energy). Because the total number of molecules remains constant, the area under the curve must be the same, which means the peak must flatten out and become lower.

1 mark for the correct answer D. Understanding the physical features of the Boltzmann distribution at a higher temperature.

The rate of hydrolysis of haloalkanes is determined by bond strength (bond enthalpy) rather than bond polarity. Since the C–I bond has the lowest bond enthalpy (is the weakest), it is broken most easily. Therefore, 1-iodobutane reacts the fastest.

1 mark for the correct answer A. Correctly links rate of hydrolysis to carbon-halogen bond strength/enthalpy.

All three species have four electron pairs around the central nitrogen atom, based on a tetrahedral arrangement. Under Electron Pair Repulsion Theory, lone pair-lone pair repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion. \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs, giving a regular tetrahedral angle of \(109.5^\circ\). \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, reducing the angle to around \(107^\circ\). \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs, reducing the angle further to around \(104.5^\circ\). Thus, the order of increasing bond angle is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

1 mark for the correct answer A. Demonstrates correct application of electron pair repulsion rules.

The solubility of Group 2 hydroxides increases down the group, whereas the solubility of Group 2 sulfates decreases down the group. Reactivity of Group 2 metals with water increases down the group. The first ionisation energy of Group 2 metals decreases down the group due to increased shielding and atomic radius.

1 mark for the correct answer B. Recalls the correct trends in physical properties of Group 2 elements and their compounds.

2-methylbutane-1,3-diol has two alcohol groups: a primary alcohol at C1 (\(-\text{CH}_2\text{OH}\)) and a secondary alcohol at C3 (\(-\text{CH(OH)}-\)). Heating under reflux with excess acidified potassium dichromate(VI) fully oxidises the primary alcohol group to a carboxylic acid group (\(-\text{COOH}\)) and the secondary alcohol group to a ketone group (\(-\text{CO}-\)). Therefore, the product contains both a carboxylic acid and a ketone group.

1 mark for the correct answer A. Identifies that primary alcohols oxidise to carboxylic acids under reflux and secondary alcohols oxidise to ketones.

1. At room temperature and pressure, the water produced in the combustion is liquid and has negligible volume.
2. The remaining gas after reaction is \(110\text{ cm}^3\). This consists of unreacted \(\text{O}_2\) and the product \(\text{CO}_2\).
3. Aqueous sodium hydroxide absorbs acidic gases like \(\text{CO}_2\). The decrease in volume is the volume of \(\text{CO}_2\):
\(V(\text{CO}_2) = 110\text{ cm}^3 - 30\text{ cm}^3 = 80\text{ cm}^3\).
4. The remaining gas of \(30\text{ cm}^3\) is unreacted \(\text{O}_2\). Thus, the volume of \(\text{O}_2\) that reacted is:
\(V(\text{O}_2\text{ reacted}) = 150\text{ cm}^3 - 30\text{ cm}^3 = 120\text{ cm}^3\).
5. Use Avogadro's hypothesis (gas mole ratio = volume ratio):
Ratio of \(\text{C}_x\text{H}_y\) to \(\text{CO}_2\) is \(20 : 80 = 1 : 4\), which means \(x = 4\).
Ratio of \(\text{C}_x\text{H}_y\) to reacted \(\text{O}_2\) is \(20 : 120 = 1 : 6\).
6. The balanced equation is:
\(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\)
Since \(x = 4\), then \(4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\).
7. The molecular formula is \(\text{C}_4\text{H}_8\).

[1 mark] B is the correct answer. Award 1 mark for correct selection. No partial marks.

To determine the molecular shape, count the bonding pairs and lone pairs around the central atom using valence shell electron pair repulsion (VSEPR) theory:
- \(\text{BF}_3\): Boron has 3 valence electrons, all bonded to fluorine atoms. 3 bonding pairs, 0 lone pairs. Shape is trigonal planar (bond angle \(120^\circ\)).
- \(\text{NH}_4^+\): Nitrogen effectively has 4 valence electrons in this cation, all forming single covalent bonds. 4 bonding pairs, 0 lone pairs. Shape is tetrahedral (bond angle \(109.5^\circ\)).
- \(\text{H}_3\text{O}^+\): Oxygen has 6 valence electrons. Losing 1 electron to become \(\text{O}^+\) leaves 5 valence electrons. 3 form bonding pairs with hydrogen, leaving 1 lone pair. 3 bonding pairs + 1 lone pair gives a trigonal pyramidal shape (bond angle \(\approx 107^\circ\)).
- \(\text{CO}_3^{2-}\): The central carbon atom has 3 electron regions (one double bond, two single bonds) and no lone pairs. Shape is trigonal planar (bond angle \(120^\circ\)).

[1 mark] C is the correct answer. Award 1 mark for correct selection. No partial marks.

1. Write the equation representing the formation of propane:
\(3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\)
2. Construct a Hess cycle using enthalpy of combustion values. The cycle routes from reactants to combustion products, and then from combustion products to propane:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
3. Substitute the given values:
\(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C}) + 4 \times \Delta_c H^\theta(\text{H}_2)] - [\Delta_c H^\theta(\text{C}_3\text{H}_8)]\)
\(\Delta_f H^\theta = [3 \times (-394) + 4 \times (-286)] - (-2220)\)
\(\Delta_f H^\theta = [-1182 - 1144] + 2220\)
\(\Delta_f H^\theta = -2326 + 2220 = -106\text{ kJ mol}^{-1}\).

[1 mark] A is the correct answer. Award 1 mark for correct selection. No partial marks.

1. Determine oxidation numbers of the elements that undergo changes:
- Manganese in \(\text{KMnO}_4\): \(\text{K} = +1\), \(\text{O} = -2\). So, \(+1 + \text{Mn} + 4(-2) = 0 \implies \text{Mn} = +7\).
- Manganese in \(\text{MnSO}_4\): Since the sulfate ion is \(\text{SO}_4^{2-}\), manganese exists as \(\text{Mn}^{2+}\), with oxidation number \(+2\).
- Carbon in \(\text{H}_2\text{C}_2\text{O}_4\): \(\text{H} = +1\), \(\text{O} = -2\). So, \(2(+1) + 2(\text{C}) + 4(-2) = 0 \implies 2\text{C} = +6 \implies \text{C} = +3\).
- Carbon in \(\text{CO}_2\): \(\text{O} = -2\). So, \(\text{C} + 2(-2) = 0 \implies \text{C} = +4\).
2. Identify oxidation vs reduction:
- Manganese is reduced as its oxidation number decreases from \(+7\) to \(+2\).
- Carbon is oxidized as its oxidation number increases from \(+3\) to \(+4\).
3. Match with the options: Option B correctly describes the reduction of manganese and its corresponding change.

[1 mark] B is the correct answer. Award 1 mark for correct selection. No partial marks.

(i) Heat absorbed by water: q = m * c * \Delta T = 120.0 g * 4.18 J g^-1 K^-1 * 36.5 K = 18308.4 J = 18.3084 kJ.
Moles of pentan-1-ol burned: n = mass / M_r = 1.65 / 88.0 = 0.01875 mol.
Enthalpy of combustion: \Delta H_c = -q / n = -18.3084 / 0.01875 = -976.448 kJ mol^-1. To 3 sig figs: -976 kJ mol^-1.
(ii) Incomplete combustion of pentan-1-ol (forming soot/CO instead of CO2); evaporation of water or alcohol from the wick; non-standard conditions.
(iii) Equation: C5H11OH(l) + 7.5 O2(g) -> 5 CO2(g) + 6 H2O(l).
\Delta_c H^\theta = \sum \Delta_f H^\theta(products) - \sum \Delta_f H^\theta(reactants)
\Delta_c H^\theta = [5 * (-394) + 6 * (-286)] - [-354] = [-1970 - 1716] + 354 = -3332 kJ mol^-1.

Part (i): [4 marks]
- m * c * \Delta T calculation: 120.0 * 4.18 * 36.5 = 18308.4 J (or 18.31 kJ) (1)
- moles of pentan-1-ol: 1.65 / 88.0 = 0.01875 mol (1)
- division of q by n with negative sign: -18.31 / 0.01875 (1)
- final answer to 3 sig figs: -976 (kJ mol^-1) (1)

Part (ii): [2 marks]
- Incomplete combustion (1)
- Evaporation of alcohol from the wick (1)

Part (iii): [2 marks]
- Correct expression using 5 * CO2 and 6 * H2O: 5(-394) + 6(-286) = -3686 (1)
- Correct final answer: -3332 (kJ mol^-1) (1)

(i) Mass of water lost = 4.38 - 2.22 = 2.16 g.
Moles of CaCl2 = 2.22 / 111.1 = 0.0200 mol.
Moles of H2O = 2.16 / 18.0 = 0.120 mol.
Ratio H2O : CaCl2 = 0.120 / 0.0200 = 6.
Therefore, x = 6.
(ii) Moles of CaCl2 = 5.55 / 111.1 = 0.0500 mol.
Moles of Cl^- ions = 2 * 0.0500 = 0.100 mol.
Concentration of Cl^- ions = 0.100 / (250.0 / 1000) = 0.400 mol dm^-3.
(iii) The reaction between silver ions and chloride ions produces a precipitate of silver chloride: Ag^+(aq) + Cl^-(aq) -> AgCl(s).

Part (i): [3 marks]
- Mass of H2O = 2.16 g AND moles of H2O = 0.120 mol (1)
- Moles of CaCl2 = 0.0200 mol (1)
- Correct ratio leading to x = 6 (1)

Part (ii): [3 marks]
- Moles of CaCl2 = 0.0500 mol (1)
- Moles of Cl^- = 0.100 mol (1)
- Concentration of Cl^- = 0.400 mol dm^-3 (1)

Part (iii): [2 marks]
- Correct species in equation: Ag^+ + Cl^- -> AgCl (1)
- Correct state symbols: (aq) + (aq) -> (s) (1)

(i) Compound A is 3-methylpent-2-ene. Its two stereoisomers are (E)-3-methylpent-2-ene and (Z)-3-methylpent-2-ene.
(ii) It exhibits E/Z stereoisomerism because: 1) There is restricted rotation around the carbon-carbon double bond (C=C) due to the presence of a \pi bond. 2) Each carbon atom in the double bond is attached to two different groups (C2 is attached to -H and -CH3; C3 is attached to -CH3 and -CH2CH3).
(iii) When 3-methylpent-2-ene reacts with HBr, the major product is 3-bromo-3-methylpentane. This is because the addition of H^+ to C2 forms a highly stable tertiary carbocation intermediate on C3. In contrast, adding H^+ to C3 would form a less stable secondary carbocation intermediate on C2. Tertiary carbocations are more stable than secondary carbocations due to the greater electron-donating inductive effect of three alkyl groups.

Part (i): [4 marks]
- Correct skeletal structure for (E)-isomer (1)
- Correct skeletal structure for (Z)-isomer (1)
- Correct name: (E)-3-methylpent-2-ene (1)
- Correct name: (Z)-3-methylpent-2-ene (1)

Part (ii): [2 marks]
- Restricted rotation about C=C bond (due to pi bond) (1)
- Each carbon of the C=C is attached to two different groups (1)

Part (iii): [2 marks]
- Correct skeletal structure of 3-bromo-3-methylpentane (1)
- Explanation in terms of tertiary carbocation intermediate being more stable than secondary (1)

(i) K_c = [HI]^2 / ([H2][I2]).
(ii) Initial moles: H2 = 1.00, I2 = 1.00, HI = 0.00.
Equilibrium moles: HI = 1.56 mol.
Moles of H2 reacted = 1.56 / 2 = 0.78 mol.
Moles of I2 reacted = 0.78 mol.
Equilibrium moles: H2 = 1.00 - 0.78 = 0.22 mol; I2 = 1.00 - 0.78 = 0.22 mol.
Equilibrium concentrations: [H2] = 0.22 / 5.00 = 0.044 mol dm^-3; [I2] = 0.22 / 5.00 = 0.044 mol dm^-3; [HI] = 1.56 / 5.00 = 0.312 mol dm^-3.
K_c = (0.312)^2 / (0.044 * 0.044) = 50.28 = 50.3 (to 3 s.f.).
(iii) The forward reaction is exothermic (\Delta H is negative). Increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the heat. Because the equilibrium shifts to the left, the concentration of HI decreases and the concentrations of H2 and I2 increase, resulting in a lower value of K_c.

Part (i): [1 mark]
- Correct expression for K_c (1)

Part (ii): [4 marks]
- Moles of H2 and I2 reacted = 0.78 mol (1)
- Equilibrium moles of H2 and I2 = 0.22 mol (1)
- Calculating concentrations by dividing by 5.00 dm^3 (or realizing volumes cancel) (1)
- Final calculated K_c value = 50.3 (accept 50 or 50.28) (1)

Part (iii): [3 marks]
- Equilibrium shifts to the left / in reverse direction (1)
- Explanation: Forward reaction is exothermic / reverse reaction is endothermic so system opposes temperature increase by shifting in endothermic direction (1)
- K_c value decreases (1)

(i) To compare rates, add equal volumes of 1-chlorobutane, 1-bromobutane, and 1-iodobutane to separate test tubes. Add ethanol (which acts as a mutual solvent) followed by aqueous silver nitrate (source of water for hydrolysis and silver ions to form precipitate). Place the tubes in a water bath at a constant temperature (e.g., 50 \(^\circ\)C). Measure the time taken for a precipitate to appear in each test tube.
(ii) Rate of hydrolysis: 1-iodobutane (fastest) > 1-bromobutane > 1-chlorobutane (slowest).
Reason: The rate depends on the strength of the C-X bond (bond enthalpy). The C-I bond is the longest and weakest (has the lowest bond enthalpy), so it breaks most easily, leading to the fastest rate. The C-Cl bond is the shortest and strongest (highest bond enthalpy), so it is hardest to break, leading to the slowest rate.
(iii) The silver ions react with the released halide ions to form a silver halide precipitate: Ag^+(aq) + Br^-(aq) -> AgBr(s).

Part (i): [4 marks]
- Use of silver nitrate (aq) AND ethanol (1)
- Warm / use a water bath (1)
- Measure the time taken for the precipitate to appear (1)
- Identify precipitate colours (white for Cl, cream for Br, yellow for I) (1)

Part (ii): [3 marks]
- Correct order of rates: 1-iodobutane > 1-bromobutane > 1-chlorobutane (1)
- Rate depends on C-Halogen bond enthalpy / strength (NOT bond polarity) (1)
- C-I bond is weakest / has lowest bond enthalpy, so breaks easiest (1)

Part (iii): [1 mark]
- Ag^+(aq) + Br^-(aq) -> AgBr(s) (including correct state symbols) (1)

(i) Magnesium reacts with oxygen to form ionic magnesium oxide: 2Mg(s) + O2(g) -> 2MgO(s). The bonding in magnesium oxide is ionic (electrostatic attraction between Mg^2+ and O^2- ions).
(ii) Reactivity of Group 2 elements with water increases down the group. This is because down the group, atomic radius increases and shielding increases. These factors outweigh the increase in nuclear charge, resulting in a weaker attraction between the nucleus and the two outer s-electrons. Therefore, the first and second ionisation energies decrease, making it easier for the atoms to lose electrons and react.
(iii) The solubility of Group 2 hydroxides increases down the group (e.g., Mg(OH)2 is sparingly soluble, while Ba(OH)2 is soluble). As solubility increases, more OH^- ions dissolve in water, increasing the concentration of hydroxide ions, which results in a higher pH (more alkaline solution) down the group.
(iv) Heating calcium carbonate strongly causes thermal decomposition: CaCO3(s) -> CaO(s) + CO2(g).

Part (i): [2 marks]
- Correct balanced equation: 2Mg(s) + O2(g) -> 2MgO(s) with state symbols (1)
- Bonding: Ionic (1)

Part (ii): [3 marks]
- Reactivity increases down the group (1)
- Atomic radius and shielding increase down the group (1)
- Nuclear attraction decreases / ionisation energies decrease, so outer electrons are lost more easily (1)

Part (iii): [2 marks]
- Solubility increases down the group (1)
- pH increases / solutions become more alkaline (1)

Part (iv): [1.33 marks]
- Type of reaction: Thermal decomposition (0.33)
- Equation: CaCO3(s) -> CaO(s) + CO2(g) (1)

(a) \(K_c = \frac{[\text{CH}_3\text{OH(g)}]}{[\text{CO(g)}][\text{H}_2\text{(g)}]^2}\)
Units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\)

(b) Calculate concentrations at equilibrium:
\([\text{CO}] = \frac{0.400\text{ mol}}{2.00\text{ dm}^3} = 0.200\text{ mol dm}^{-3}\)
\([\text{H}_2] = \frac{0.300\text{ mol}}{2.00\text{ dm}^3} = 0.150\text{ mol dm}^{-3}\)
\([\text{CH}_3\text{OH}] = \frac{0.150\text{ mol}}{2.00\text{ dm}^3} = 0.0750\text{ mol dm}^{-3}\)

Substitute values into \(K_c\):
\(K_c = \frac{0.0750}{0.200 \times (0.150)^2} = \frac{0.0750}{0.0045} = 16.67 \approx 16.7\text{ dm}^6\text{ mol}^{-2}\)

(c)(i) An increase in temperature shifts the equilibrium in the endothermic direction to absorb heat. Since the forward reaction is exothermic (\(\Delta H = -91\text{ kJ mol}^{-1}\)), the position of equilibrium shifts to the left. The concentration of reactants increases and the concentration of products decreases, so the value of \(K_c\) decreases.

(c)(ii) An increase in pressure shifts the equilibrium to the side with fewer gas molecules to decrease pressure. There are 3 moles of gas on the left and 1 mole on the right, so the position of equilibrium shifts to the right. The value of \(K_c\) remains unchanged because only temperature can change the value of \(K_c\).

(a) 2 marks:
- Correct Kc expression (1)
- Correct units of dm^6 mol^-2 (1)

(b) 4 marks:
- All equilibrium concentrations calculated correctly (1)
- Correct substitution into Kc expression (1)
- Correct value of 16.7 (1)
- Answer quoted to 3 significant figures (1)

(c)(i) 3 marks:
- Equilibrium shifts left / in endothermic direction to absorb heat (1)
- Decreased concentration of products / increased concentration of reactants (1)
- Kc decreases (1)

(c)(ii) 2.66 marks:
- Equilibrium shifts right to the side with fewer gas moles (3 to 1) (1)
- Kc remains unchanged (1)
- Explains that Kc is only affected by temperature changes (0.66)

(a) \(\text{NH}_3\) shape: Trigonal pyramidal. Bond angle: \(107^\circ\) (accept range \(106^\circ - 108^\circ\)).
Explanation: The central nitrogen atom has 4 electron pairs in its outer shell (3 bonding pairs, 1 lone pair). Electron pairs repel each other to get as far apart as possible. Lone pairs repel more than bonding pairs, which reduces the tetrahedral bond angle of \(109.5^\circ\) by approximately \(2.5^\circ\) to \(107^\circ\).

(b)(i) Requirements: A hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) which has at least one lone pair of electrons. In \(\text{NH}_3\), N is highly electronegative, creating a polar \(\text{N}^\delta-\text{-H}^\delta+\) bond. In \(\text{HF}\), F is highly electronegative, creating a polar \(\text{H}^\delta+\text{-F}^\delta-\) bond. The hydrogen bond forms between the lone pair on the nitrogen atom of \(\text{NH}_3\) and the electron-deficient hydrogen atom (\(\delta^+\)) of \(\text{HF}\).

(b)(ii) \(\text{HF}\) has a significantly higher boiling point than \(\text{F}_2\). \(\text{HF}\) has polar molecules and can form hydrogen bonds as well as London forces. \(\text{F}_2\) is non-polar and only has weak London forces between its molecules. Hydrogen bonds are much stronger than London forces and require more energy to overcome.

(b)(iii) Each water molecule can form up to 4 hydrogen bonds (averaging 2 hydrogen bonds per molecule) because it has 2 lone pairs on the oxygen and 2 polar hydrogen atoms. Each \(\text{HF}\) molecule can only form on average 1 hydrogen bond because it is limited by having only 1 hydrogen atom (despite having 3 lone pairs on fluorine). Thus, water forms a more extensive hydrogen-bonded network, requiring more energy to break.

(a) 4 marks:
- Trigonal pyramidal shape (1)
- Bond angle of 107 (1)
- Explanation: 4 electron pairs repel to get as far apart as possible (1)
- Lone pairs repel more than bonding pairs (1)

(b)(i) 3 marks:
- Mentions need for H bonded to highly electronegative element (N/O/F) with lone pair (1)
- Identifies N as having lone pair and H in HF as being electron-deficient / delta+ (1)
- Explains hydrogen bond formed between N lone pair and H of HF (1)

(b)(ii) 2.66 marks:
- Identifies HF has hydrogen bonding, F2 has only London forces (1)
- States hydrogen bonds are stronger than London forces (1)
- More energy required to break IMF in HF (0.66)

(b)(iii) 2 marks:
- H2O forms on average 2 hydrogen bonds per molecule, whereas HF forms only 1 (1)
- More energy is needed to break the greater number of hydrogen bonds in H2O (1)

(a)(i) Stereoisomers have the same structural formula but a different arrangement of atoms in space.

(a)(ii) Identify the groups attached to each carbon of the double bond. C2 has \(\text{-CH}_3\) and \(\text{-H}\). C3 has \(\text{-CH}_3\) and \(\text{-H}\). Priority is determined by the atomic number of the atom directly bonded: Carbon (atomic number 6) has higher priority than Hydrogen (atomic number 1), so \(\text{-CH}_3\) is the high-priority group on both carbons. In \(Z\)-but-2-ene, the two high-priority groups are on the same side of the C=C double bond. In \(E\)-but-2-ene, they are on opposite sides.

(b)(i) Electrophilic addition.

(b)(ii) Mechanism steps:
1. Hydrogen bromide has a permanent dipole: \(\text{H}^{\delta+}\text{-Br}^{\delta-}\).
2. A curly arrow goes from the C=C pi-bond of \(Z\)-but-2-ene to the \(\text{H}^{\delta+}\) of \(\text{HBr}\).
3. The \(\text{H-Br}\) bond breaks, shown by a curly arrow from the bond to the \(\text{Br}\) atom, forming a bromide ion (\(\text{Br}^-\)) and a secondary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\)).
4. A curly arrow goes from a lone pair on the \(\text{Br}^-\) ion to the positive carbon atom of the carbocation, forming 2-bromobutane.

(b)(iii) But-1-ene is unsymmetrical, so \(\text{H}^+\) can add to either C1 or C2. Adding \(\text{H}^+\) to C1 forms a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\)), whereas adding \(\text{H}^+\) to C2 forms a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{C}^+\text{H}_2\)). Secondary carbocations are more stable than primary carbocations due to the inductive electron-donating effect of two alkyl groups rather than one. The major product is therefore 2-bromobutane, formed via the more stable secondary carbocation intermediate.

(a)(i) 1 mark:
- Same structural formula but different spatial arrangement of atoms (1)

(a)(ii) 3 marks:
- High priority groups determined by highest atomic number (C > H) (1)
- Identifies -CH3 as high priority and -H as low priority (1)
- E-isomer has high priority on opposite sides, Z-isomer has high priority on same side (1)

(b)(i) 1 mark:
- Electrophilic addition (1)

(b)(ii) 3.66 marks:
- Correct dipole on H-Br and curly arrow from C=C to H (1)
- Curly arrow from H-Br bond to Br (1)
- Correct secondary carbocation structure (1)
- Curly arrow from Br- lone pair to C+ of carbocation (0.66)

(b)(iii) 3 marks:
- States 2-bromobutane is the major product (1)
- Explains formation via secondary carbocation, which is more stable than primary carbocation (1)
- Explains stability via the electron-donating inductive effect of alkyl groups (1)

(a) Salt **A**: Chloride (\(\text{Cl}^-\)) because it forms a white precipitate of \(\text{AgCl}\) that dissolves in dilute ammonia.
Salt **B**: Bromide (\(\text{Br}^-\)) because it forms a cream precipitate of \(\text{AgBr}\) that is insoluble in dilute ammonia but dissolves in concentrated ammonia.
Salt **C**: Carbonate (\(\text{CO}_3^{2-}\)) because it reacts with acid to release a gas (\(\text{CO}_2\)) that turns limewater cloudy, meaning no free anion was available to form a silver halide precipitate.

(b)(i) \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\)

(b)(ii) \(\text{CO}_3^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

(c) Carry out a flame test. Dip a clean nichrome wire (dipped in concentrated \(\text{HCl}\) first) into a sample of the solid or solution, then place it into a blue Bunsen burner flame. Observation: A bright yellow (or golden-yellow) flame.

(d) Nitric acid (\(\text{HNO}_3\)) must be used because it does not introduce any competing ions. If hydrochloric acid (\(\text{HCl}\)) were used, it would introduce chloride ions (\(\text{Cl}^-\)), forming a white precipitate of \(\text{AgCl}\) and giving a false-positive result. If sulfuric acid (\(\text{H}_2\text{SO}_4\)) were used, it could form a precipitate of silver sulfate (\(\text{Ag}_2\text{SO}_4\)) if concentrated, interfering with the results.

(a) 4 marks:
- Identifies Salt A as chloride with reason (dissolves in dilute NH3) (1)
- Identifies Salt B as bromide with reason (dissolves only in conc NH3) (1)
- Identifies Salt C as carbonate with reason (bubbling with acid, turns limewater cloudy) (2)

(b)(i) 2 marks:
- Ag+(aq) + Cl-(aq) -> AgCl(s) (1 mark for species, 1 mark for all correct state symbols)

(b)(ii) 2 marks:
- CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l) (1 mark for species, 1 mark for all correct state symbols)

(c) 1.66 marks:
- Describes flame test using a clean nichrome/platinum wire (1)
- Observes a yellow/golden-yellow flame (0.66)

(d) 2 marks:
- HCl contains Cl- ions which would react with Ag+ to form a false positive precipitate (1)
- H2SO4 contains SO4^2- which might form a precipitate of silver sulfate (1)

(a) Moles of \(\text{NaOH} = \text{volume} \times \text{concentration} = \frac{28.50}{1000} \times 0.100 = 2.85 \times 10^{-3}\text{ mol}\.\n\n(b) Since the reaction stoichiometry is \)1:1\) between \(\text{HCl}\) and \(\text{NaOH}\), the moles of \(\text{HCl}\) in \(25.0\text{ cm}^3\) of Solution **S** \(= 2.85 \times 10^{-3}\text{ mol}\).
Thus, the total moles of unreacted \(\text{HCl}\) in the \(250.0\text{ cm}^3\) volumetric flask \(= 2.85 \times 10^{-3} \times \frac{250.0}{25.0} = 2.85 \times 10^{-2}\text{ mol}\) (or \(0.0285\text{ mol}\)).

(c) Initial moles of \(\text{HCl} = \frac{50.0}{1000} \times 1.00 = 0.0500\text{ mol}\.\n\n(d) Moles of \)\text{HCl}\) reacted with \(\text{CaCO}_3 = \text{initial moles} - \text{unreacted moles} = 0.0500 - 0.0285 = 0.0215\text{ mol}\.\n\n(e) From the equation, \)1\text{ mol}\) of \(\text{CaCO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).
Moles of \(\text{CaCO}_3 = \frac{0.0215}{2} = 0.01075\text{ mol}\.\nMass of \)\text{CaCO}_3 = 0.01075\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.076\text{ g}\.\nPercentage by mass \(= \frac{1.076}{1.25} \times 100 = 86.08\% \approx 86.1\%\).

(a) 1.66 marks:
- Correct calculation of NaOH moles: 2.85 x 10^-3 mol (1.66)

(b) 2 marks:
- Identifies 1:1 ratio for titration to find HCl in 25 cm^3 (1)
- Scales up by 10 to find moles in 250 cm^3: 2.85 x 10^-2 mol (1)

(c) 1 mark:
- Moles of initial HCl = 0.0500 mol (1)

(d) 2 marks:
- Subtracts unreacted HCl from initial HCl to get 0.0215 mol (2)

(e) 5 marks:
- Moles of CaCO3 = moles of reacted HCl / 2 = 0.01075 mol (1)
- Mass of CaCO3 = 1.076 g (1)
- Mass percentage calculation (1)
- Correct final percentage of 86.1% (1)
- All final values rounded to 3 significant figures (1)

(a) Temperature rise: \(\Delta T = 55.5 - 20.5 = 35.0^\circ\text{C}\) (or \(35.0\text{ K}\))
Using \(q = m c \Delta T\):
\(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 35.0\text{ K} = 21945\text{ J} = 21.945\text{ kJ}\) (or \(21.9\text{ kJ}\)).

(b) Mass of propan-1-ol burned \(= 124.80 - 123.95 = 0.85\text{ g}\).
Moles of propan-1-ol burned \(= \frac{0.85}{60.0\text{ g mol}^{-1}} = 0.01417\text{ mol}\).

(c) \(\Delta_c H = -\frac{q}{n} = -\frac{21.945\text{ kJ}}{0.01417\text{ mol}} = -1549\text{ kJ mol}^{-1} \approx -1550\text{ kJ mol}^{-1}\) (to 3 significant figures). The sign is negative because the temperature of the water increased, meaning the reaction is exothermic.

(d)(i) 1. Heat loss to the surroundings / draughts.
2. Incomplete combustion of propan-1-ol, producing soot/carbon monoxide (less energy released than complete combustion).
3. Evaporation of the fuel from the wick of the burner after extinguishing but before weighing.
4. Heat capacity of the copper calorimeter not taken into account.

(d)(ii) A bomb calorimeter uses high-pressure pure oxygen, which ensures complete combustion. It is also highly insulated, reducing heat loss to the surroundings, meaning almost all the heat is captured by the water.

(a) 2 marks:
- Temperature difference calculated as 35.0 (1)
- Correct value of q = 21.9 kJ or 21.945 kJ (1)

(b) 2 marks:
- Mass of propan-1-ol burned calculated as 0.85 g (1)
- Moles of propan-1-ol = 0.01417 mol (1)

(c) 3 marks:
- Divides heat energy by moles (1)
- Value of 1550 or 1549 (1)
- Correct negative sign indicating exothermic reaction (1)

(d)(i) 2.66 marks:
- Suggests heat loss (1)
- Suggests incomplete combustion (1)
- Suggests fuel evaporation / heat capacity of calorimeter (0.66)

(d)(ii) 2 marks:
- Bomb calorimeter provides pure oxygen for complete combustion (1)
- Sealed / insulated to minimize heat loss to surroundings (1)