2024 Cambridge IAS-Level Chemistry (9701) 模擬試題連答案詳解

The decomposition equation for a Group 2 metal nitrate is:
\(2\text{M(NO}_3)_2(s) \rightarrow 2\text{MO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)\)

From the equation, \(2\text{ mol}\) of \(\text{M(NO}_3)_2\) yields \(5\text{ mol}\) of gas (\(4\text{ mol}\) of \(\text{NO}_2\) and \(1\text{ mol}\) of \(\text{O}_2\)), giving a ratio of \(1 : 2.5\).

The total moles of gas collected at r.t.p. is:
\(n(\text{gas}) = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0250\text{ mol}\)

The moles of \(\text{M(NO}_3)_2\) decomposed is:
\(n(\text{M(NO}_3)_2) = \frac{0.0250\text{ mol}}{2.5} = 0.0100\text{ mol}\)

The molar mass of \(\text{M(NO}_3)_2\) is:
\(M_r = \frac{1.64\text{ g}}{0.0100\text{ mol}} = 164\text{ g mol}^{-1}\)

Using the formula mass:
\(A_r(\text{M}) + 2 \times [14.0 + (3 \times 16.0)] = 164\)
\(A_r(\text{M}) + 124 = 164\)
\(A_r(\text{M}) = 40.0\)

This relative atomic mass corresponds to Calcium (\(\text{Ca}\)).

[1 mark] Correctly identifies B. Award 1 mark for calculating correct moles of gas (0.0250 mol), moles of nitrate (0.0100 mol), and Mr of nitrate (164), leading to Ca.

The equation for the reaction of sodium carbonate with hydrochloric acid is:
\(\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)

Note that sodium chloride does not react with hydrochloric acid.

Moles of \(\text{CO}_2\) gas evolved:
\(n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0100\text{ mol}\)

Since the mole ratio of \(\text{Na}_2\text{CO}_3\) to \(\text{CO}_2\) is \(1 : 1\):
\(n(\text{Na}_2\text{CO}_3) = 0.0100\text{ mol}\)

Mass of \(\text{Na}_2\text{CO}_3\) in the mixture:
\(m(\text{Na}_2\text{CO}_3) = 0.0100\text{ mol} \times 106.0\text{ g mol}^{-1} = 1.06\text{ g}\)

Mass of sodium chloride in the mixture:
\(m(\text{NaCl}) = 2.00\text{ g} - 1.06\text{ g} = 0.94\text{ g}\)

Percentage by mass of \(\text{NaCl}\):
\(\%\text{NaCl} = \frac{0.94\text{ g}}{2.00\text{ g}} \times 100 = 47\%\)

[1 mark] Award 1 mark for option A. Distractors: B represents the percentage of sodium carbonate (53%), and C and D represent errors in stoichiometry or molar mass calculation.

The atomic number of Cobalt (\(\text{Co}\)) is 27. Its ground state electronic configuration is \([\text{Ar}] 3\text{d}^7 4\text{s}^2\).
For the \(\text{Co}^{2+}\) ion, the two \(4\text{s}\) electrons are lost, giving the configuration \([\text{Ar}] 3\text{d}^7\).
Using Hund's rule to fill the five \(3\text{d}\) orbitals: 2 orbitals contain paired electrons and 3 orbitals contain single unpaired electrons (total of 3 unpaired electrons).

Let's check the options:
- \(\text{Fe}^{2+}\) (atomic number 26) has configuration \([\text{Ar}] 3\text{d}^6\), which has 4 unpaired electrons.
- \(\text{Cr}^{3+}\) (atomic number 24) has configuration \([\text{Ar}] 3\text{d}^3\), which has 3 unpaired electrons.
- \(\text{Ni}^{2+}\) (atomic number 28) has configuration \([\text{Ar}] 3\text{d}^8\), which has 2 unpaired electrons.
- \(\text{Cu}^{2+}\) (atomic number 29) has configuration \([\text{Ar}] 3\text{d}^9\), which has 1 unpaired electron.

Thus, \(\text{Cr}^{3+}\) has the same number of unpaired d-electrons (3) as \(\text{Co}^{2+}\).

[1 mark] Award 1 mark for option B.

Transition metal complexes are colored because the d-orbitals split into different energy levels in the presence of ligands. When light is incident, d-electrons absorb specific frequencies of visible light to be promoted from lower-energy to higher-energy d-orbitals (d-d transitions). The complementary color to the absorbed wavelength is transmitted. The energy gap \(\Delta E\) depends on the nature of the ligands and the geometry of the complex. Changing the ligand from water to chloride alters the splitting of the d-orbitals, thus changing the absorbed frequency of light and the observed color.

[1 mark] Award 1 mark for option A. Option B is incorrect because color arises from absorption, not emission. Option C is incorrect because blue light is transmitted, not absorbed, to show blue. Option D is incorrect because d-d transitions occur, not 3d to 4s.

The equation for the standard enthalpy change of formation of liquid propan-1-ol is:
\(3\text{C(graphite)} + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\)

According to Hess's Law using enthalpies of combustion:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

\(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 4 \times \Delta H_c^\ominus(\text{H}_2)] - [\Delta H_c^\ominus(\text{propan-1-ol})]\)

\(\Delta H_f^\ominus = [3 \times (-394) + 4 \times (-286)] - [-2021]\)
\(\Delta H_f^\ominus = [-1182 - 1144] + 2021\)
\(\Delta H_f^\ominus = -2326 + 2021 = -305\text{ kJ mol}^{-1}\)

[1 mark] Award 1 mark for option A. Correct application of Hess's Law with correct stoichiometric coefficients.

Total volume of mixture = \(50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\).
Mass of mixture, \(m = 100.0\text{ g}\).
Temperature change, \(\Delta T = 28.2\text{ }^\circ\text{C} - 21.5\text{ }^\circ\text{C} = 6.7\text{ K}\).
Heat released, \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ}\).

Moles of acid used: \(n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
Moles of alkali used: \(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
Moles of water formed: \(n(\text{H}_2\text{O}) = 0.0500\text{ mol}\).

Enthalpy of neutralisation:
\(\Delta H = -\frac{q}{n(\text{H}_2\text{O})} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\)

[1 mark] Award 1 mark for option A. Award 0 marks for distractors such as using wrong total mass (e.g., 50.0 g instead of 100.0 g) or incorrect signs.

Let's analyze the properties of the options:
- \(\text{P}_4\text{O}_{10}\) is a Period 3 oxide that reacts with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), which is a moderately strong acid (giving a pH of approximately 2). This matches \(\text{X}\).
- \(\text{SiCl}_4\) is a Period 3 chloride that reacts vigorously with water (hydrolyzes) to form silicon dioxide and hydrogen chloride gas (which forms misty white fumes in air). The dissolved \(\text{HCl}\) forms a strongly acidic solution (pH 1). This matches \(\text{Y}\).

Other distractors are incorrect because:
- \(\text{Al}_2\text{O}_3\) and \(\text{SiO}_2\) are insoluble in water and do not lower the pH of water.
- \(\text{NaCl}\) dissolves in water to form a neutral solution (pH 7) without reacting or forming misty fumes.

[1 mark] Correctly identifies option A. All observations perfectly align with the typical reactions of Period 3 oxides and chlorides with water.

The description identifies the oxide of \(\text{Z}\) as amphoteric, since it reacts with both acids (hydrochloric acid) and bases (sodium hydroxide). Among the Period 3 oxides, \(\text{Al}_2\text{O}_3\) (aluminium oxide) is amphoteric. It has a high melting point due to its giant ionic structure with some covalent character and is insoluble in water. Thus, element \(\text{Z}\) is Aluminium (\(\text{Al}\)).

- Magnesium oxide (\(\text{MgO}\)) is basic, reacting only with acids.
- Silicon dioxide (\(\text{SiO}_2\)) is acidic, reacting with concentrated alkalis but not acids.
- Phosphorus oxides are acidic molecular solids with low melting points.

[1 mark] Award 1 mark for option B, based on the amphoteric nature and high melting point of aluminium oxide.

First, calculate the moles of hydrogen gas produced: moles of \( \text{H}_2 = 112\text{ cm}^3 / 24000\text{ cm}^3\text{ mol}^{-1} = 0.004667\text{ mol} \). Since the reaction stoichiometry is \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \), the moles of \( \text{Mg} \) reacted = \( 0.004667\text{ mol} \). Calculate the mass of pure magnesium: mass of \( \text{Mg} = 0.004667\text{ mol} \times 24.3\text{ g mol}^{-1} = 0.1134\text{ g} \). Finally, determine the percentage purity: \( (0.1134\text{ g} / 0.120\text{ g}) \times 100\% = 94.5\% \).

1 mark for the correct calculation of percentage purity.

The total volume of gases (\( \text{CO}_2 \) and \( \text{H}_2\text{O} \)) collected is \( 600\text{ cm}^3 \). Moles of gaseous products = \( 600 / 24000 = 0.025\text{ mol} \). According to the equation, 2 moles of \( \text{MHCO}_3 \) decompose to produce 2 moles of gas (1 mole of \( \text{CO}_2 \) and 1 mole of \( \text{H}_2\text{O} \)). Therefore, the mole ratio of \( \text{MHCO}_3 \) to total gas is 1:1. Moles of \( \text{MHCO}_3 \) reacted = \( 0.025\text{ mol} \). Molar mass of \( \text{MHCO}_3 = 2.10\text{ g} / 0.025\text{ mol} = 84.0\text{ g mol}^{-1} \). Since \( M_r(\text{MHCO}_3) = A_r(\text{M}) + 1.0 + 12.0 + (3 \times 16.0) = A_r(\text{M}) + 61.0 = 84.0 \), we find \( A_r(\text{M}) = 23.0 \).

1 mark for the correct determination of the relative atomic mass of metal M.

The electronic configuration of chromium is \( [\text{Ar}] 3d^5 4s^1 \). The \( \text{Cr}^{3+} \) ion is formed by losing one 4s electron and two 3d electrons, resulting in the configuration \( [\text{Ar}] 3d^3 \). According to Hund's rule, these three d-electrons will occupy separate, degenerate d-orbitals with parallel spins, meaning there are exactly three unpaired electrons. Titanium(III) has a \( 3d^1 \) configuration (1 unpaired electron). Iron(III) has a \( 3d^5 \) configuration (5 unpaired electrons). Cobalt(III) has a \( 3d^6 \) configuration (4 unpaired electrons in a high-spin state).

1 mark for identifying the correct transition metal ion with three unpaired electrons.

An aqueous solution of cobalt(II) contains the octahedral hexaaquacobalt(II) ion, \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \), which is pink. Upon addition of excess concentrated hydrochloric acid, chloride ligands displace the water ligands in a ligand exchange reaction to form the tetrahedral tetrachlorocobaltate(II) ion, \( [\text{CoCl}_4]^{2-} \), which is blue. Thus, the color change is pink to blue.

1 mark for selecting the correct color change and formula of the new complex.

Using Hess's Law: \( \Delta H^\theta_f = 3 \times \Delta H^\theta_c[\text{C}(\text{s})] + 3 \times \Delta H^\theta_c[\text{H}_2(\text{g})] - \Delta H^\theta_c[\text{CH}_3\text{CH}_2\text{CO}_2\text{H}(\text{l})] \). Substituting the values: \( \Delta H^\theta_f = 3(-394) + 3(-286) - (-1527) = -1182 - 858 + 1527 = -2040 + 1527 = -513\text{ kJ mol}^{-1} \).

1 mark for the correct application of Hess's Law and calculation of standard enthalpy of formation.

Total mass of solution \( m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g} \). Temperature change \( \Delta T = 28.1 - 21.4 = 6.7\text{ K} \). Heat released \( q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ} \). Moles of water formed = moles of acid reacted = \( 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol} \). Enthalpy change of neutralisation = \( -q / n = -2.80\text{ kJ} / 0.0500\text{ mol} = -56.0\text{ kJ mol}^{-1} \).

1 mark for the correct calculation of the standard enthalpy change of neutralisation.

Oxide \( X \) must be \( \text{P}_4\text{O}_{10} \) (or \( \text{SO}_3 \)) because it reacts vigorously with water to form phosphoric acid, giving a strongly acidic solution (pH around 2). Oxide \( Y \) is \( \text{Al}_2\text{O}_3 \) because it is amphoteric, meaning it reacts with both strong acids and strong bases, and it is insoluble in water. Oxide \( Z \) is \( \text{Na}_2\text{O} \) because it reacts readily with water to produce sodium hydroxide, which is a strongly alkaline solution (pH ~ 13-14). \( \text{MgO} \) is only sparingly soluble in water, forming a weakly alkaline solution (pH ~ 9).

1 mark for identifying all three Period 3 oxides correctly based on their reactions with water, acids, and bases.

\( \text{SiCl}_4 \) is a covalent chloride that undergoes rapid, violent hydrolysis when added to water to produce hydrogen chloride gas (seen as white fumes) and silicon dioxide (which forms a white precipitate or suspension): \( \text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g}) \). \( \text{NaCl} \) dissolves to give a completely neutral clear solution (pH 7). \( \text{MgCl}_2 \) dissolves to form a weakly acidic solution (pH ~ 6.5) due to slight hydration-shell hydrolysis. \( \text{AlCl}_3 \) undergoes significant hydrolysis to produce a strongly acidic solution (pH ~ 3).

1 mark for selecting the statement that correctly describes the physical and chemical behavior of a Period 3 chloride in water.

1. Determine the volume of \(\text{CO}_2\) produced: The gas absorbed by \(\text{NaOH}\)(aq) is \(\text{CO}_2\). Therefore, the volume of \(\text{CO}_2 = 60\text{ cm}^3 - 20\text{ cm}^3 = 40\text{ cm}^3\). 2. Determine the volume of unreacted and reacted \(\text{O}_2\): The gas remaining after treatment with \(\text{NaOH}\)(aq) is excess \(\text{O}_2\), which is \(20\text{ cm}^3\). Volume of \(\text{O}_2\) reacted = \(80\text{ cm}^3 - 20\text{ cm}^3 = 60\text{ cm}^3\). 3. Calculate \(x\) and \(y\): From Avogadro's hypothesis, the mole ratio equals the volume ratio: - \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(40\text{ cm}^3\) of \(\text{CO}_2\), so \(x = 40 / 10 = 4\). - \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) reacts with \(60\text{ cm}^3\) of \(\text{O}_2\). The combustion equation is: \(\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O}\). Therefore, \(x + y/4 = 60 / 10 = 6\). Substituting \(x = 4\): \(4 + y/4 = 6\) which gives \(y/4 = 2\) and \(y = 8\). Thus, the formula is \(\text{C}_4\text{H}_8\).

[1 mark] C is correct. Award 1 mark for calculating the correct values of \(x = 4\) and \(y = 8\) using gas volume relationships.

1. Calculate the amount in moles of \(\text{Na}_2\text{CO}_3\): \(\text{moles of Na}_2\text{CO}_3 = 0.0500\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 1.25 \times 10^{-3}\text{ mol}\). 2. Determine the required moles of \(\text{HCl}\): From the stoichiometry of the equation, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\). \(\text{moles of HCl} = 2 \times 1.25 \times 10^{-3}\text{ mol} = 2.50 \times 10^{-3}\text{ mol}\). 3. Calculate the concentration of \(\text{HCl}\): \(\text{concentration} = \text{moles} / \text{volume in dm}^3 = 2.50 \times 10^{-3}\text{ mol} / 0.02150\text{ dm}^3 \approx 0.116\text{ mol dm}^{-3}\).

[1 mark] C is correct. Award 1 mark for the correct calculation of concentration of hydrochloric acid to 3 significant figures.

A is incorrect because the highest common and stable oxidation state of copper is +2. B is incorrect because titanium has a higher melting point than calcium due to the involvement of both \(3d\) and \(4s\) electrons in metallic bonding. C is correct because in a free ion, all five \(3d\) orbitals have the same energy (degenerate). In an octahedral ligand field, the repulsion from electrostatic charges of the ligands causes the \(3d\) orbitals to split into two groups of different energy levels. D is incorrect because the +1 oxidation state is only common and stable for copper; other early transition metals do not form stable +1 ions.

[1 mark] C is correct. Award 1 mark for identifying the correct behavior of \(3d\) orbitals in a transition metal complex.

The complex \([\text{Co(C}_2\text{O}_4)_2\text{Cl}_2]^{3-}\) is an octahedral complex of the type \([\text{M(AA)}_2\text{X}_2]\). It exhibits cis-trans isomerism: 1. In the trans isomer, the two \(\text{Cl}^-\) ligands are arranged at an angle of \(180^\circ\) to each other. This isomer has a plane of symmetry, making it optically inactive (achiral). Thus, there is only 1 trans isomer. 2. In the cis isomer, the two \(\text{Cl}^-\) ligands are arranged at an angle of \(90^\circ\) to each other. This structure lacks a plane of symmetry, making it chiral. It therefore exists as a pair of non-superimposable mirror images (enantiomers), which are the d- and l- forms. Thus, there are 2 cis isomers. Total number of stereoisomers = 1 (trans) + 2 (cis enantiomers) = 3.

[1 mark] B is correct. Award 1 mark for identifying the existence of one trans isomer and two cis enantiomers.

The chemical equation for the standard enthalpy change of formation of propanoic acid is: \(3\text{C(graphite)} + 3\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{CO}_2\text{H(l)}\). Using Hess's law with standard enthalpy changes of combustion: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). \(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 3 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{propanoic acid})\). \(\Delta H_f^\ominus = [3(-394) + 3(-286)] - (-1527)\). \(\Delta H_f^\ominus = [-1182 - 858] + 1527 = -2040 + 1527 = -513\text{ kJ mol}^{-1}\).

[1 mark] B is correct. Award 1 mark for applying Hess's Law correctly with correct stoichiometric coefficients and signs to find \(-513\text{ kJ mol}^{-1}\).

The enthalpy change of solution (\(\Delta H_{sol}^\ominus\)) is related to the lattice energy (\(\Delta H_{latt}^\ominus\)) and the hydration enthalpies (\(\Delta H_{hyd}^\ominus\)) by the equation: \(\Delta H_{sol}^\ominus = -\Delta H_{latt}^\ominus + \Delta H_{hyd}^\ominus(\text{Mg}^{2+}) + 2 \times \Delta H_{hyd}^\ominus(\text{Cl}^-)\). Substitute the given values into the equation: \(\Delta H_{sol}^\ominus = -(-2526) + (-1920) + 2 \times (-364) = 2526 - 1920 - 728 = -122\text{ kJ mol}^{-1}\).

[1 mark] B is correct. Award 1 mark for the correct calculation, including doubling the hydration enthalpy of the chloride ion.

Let's review the properties of Period 3 oxides: Aluminium oxide (\(\text{Al}_2\text{O}_3\)) and silicon dioxide (\(\text{SiO}_2\)) are white solids but are insoluble in water. Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) is a white solid at room temperature and reacts vigorously with water to form phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)), which is a strongly acidic solution. Sulfur trioxide (\(\text{SO}_3\)) reacts vigorously with water to form a strongly acidic solution, but it is a liquid at room temperature, not a solid. Therefore, element \(X\) is phosphorus.

[1 mark] C is correct. Award 1 mark for identifying phosphorus as the element whose oxide is a white solid and reacts with water to form a strongly acidic solution.

Let's analyze the behavior of Period 3 chlorides in water: Sodium chloride (\(\text{NaCl}\), chloride of \(\text{Na}\)) dissolves in water to form a neutral solution with a pH of approximately 7. This matches the properties of chloride \(Y\). Silicon tetrachloride (\(\text{SiCl}_4\), chloride of \(\text{Si}\)) reacts rapidly with water to form silicon dioxide (\(\text{SiO}_2\)) and hydrogen chloride (\(\text{HCl}\)) gas, which is seen as white fumes. The resulting solution is strongly acidic with a pH of approximately 1-2. This matches the properties of chloride \(Z\). Therefore, \(Y\) is sodium (\(\text{Na}\)) and \(Z\) is silicon (\(\text{Si}\)).

[1 mark] A is correct. Award 1 mark for identifying the correct chlorides based on pH and reaction with water.

The balanced equation for the complete combustion of butane is:
\(\text{C}_4\text{H}_{10}(\text{g}) + 6.5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\)

From the stoichiometry of the reaction:
1. \(10\text{ cm}^3\) of butane reacts with \(10 \times 6.5 = 65\text{ cm}^3\) of oxygen.
2. The volume of oxygen remaining in excess is \(80 - 65 = 15\text{ cm}^3\).
3. The volume of carbon dioxide gas produced is \(10 \times 4 = 40\text{ cm}^3\).
4. Water is a liquid at room temperature, so its volume is negligible.

Before passing through the aqueous sodium hydroxide, the total gas volume is \(15\text{ cm}^3\) (excess oxygen) + \(40\text{ cm}^3\) (carbon dioxide product) = \(55\text{ cm}^3\).

When the gaseous mixture is passed through aqueous sodium hydroxide, the acidic carbon dioxide gas is completely absorbed:
\(\text{CO}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

Therefore, only the unreacted excess oxygen remains, which has a volume of \(15\text{ cm}^3\).

Award 1 mark for the correct option A.
- Method: Determine the volume of reacting oxygen (65 cm3) and the volume of CO2 produced (40 cm3).
- Identify that water condenses at room temperature and CO2 is completely absorbed by the NaOH solution.
- Compute final remaining gas volume = 15 cm3.

1. Calculate the number of moles of each reactant:
- Moles of \(\text{Mg} = \frac{12.0}{24.3} = 0.494\text{ mol}\)
- Moles of \(\text{Cl}_2 = \frac{14.2}{2 \times 35.5} = \frac{14.2}{71.0} = 0.200\text{ mol}\)

2. Determine the limiting reactant from the 1:1 stoichiometry of the equation:
- Since \(0.200\text{ mol}\) of \(\text{Cl}_2\) requires \(0.200\text{ mol}\) of \(\text{Mg}\), and we have \(0.494\text{ mol}\) of \(\text{Mg}\), magnesium is in excess.

3. Calculate the mass of product based on the limiting reactant (\(\text{Cl}_2\)):
- Moles of \(\text{MgCl}_2\) formed = \(0.200\text{ mol}\)
- Molar mass of \(\text{MgCl}_2 = 24.3 + 2(35.5) = 95.3\text{ g mol}^{-1}\)
- Mass of \(\text{MgCl}_2 = 0.200 \times 95.3 = 19.1\text{ g}\).

Award 1 mark for the correct option A.
- Method: Calculate the moles of both reactants to identify that Mg is in excess.
- Accuracy: Calculate the mass of magnesium chloride based on the limiting reactant to obtain 19.1 g.

The equation for the standard enthalpy change of formation of propanoic acid is:
\(3\text{C}(\text{graphite}) + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{COOH}(\text{l})\)

Using Hess's Law and combustion data:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)
\(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 3 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{C}_2\text{H}_5\text{COOH})\)
\(\Delta H_f^\ominus = [3(-394) + 3(-286)] - (-1527)\)
\(\Delta H_f^\ominus = [-1182 - 858] + 1527\)
\(\Delta H_f^\ominus = -2040 + 1527 = -513\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A.
- Method: State the correct enthalpy cycle or mathematical formula derived from Hess's law.
- Multiply enthalpies of combustion of carbon and hydrogen by 3.
- Correctly subtract the enthalpy of combustion of propanoic acid to get -513 kJ mol-1.

- The chloride of magnesium, \(\text{MgCl}_2\), undergoes slight hydrolysis in water due to the high charge density of the hydrated \(\text{Mg}^{2+}\) ion, yielding a weakly acidic solution of \(\text{pH} \approx 6.5\).
- The chloride of silicon, \(\text{SiCl}_4\), reacts vigorously and completely with water to form silicon dioxide and hydrogen chloride gas (white fumes), resulting in a strongly acidic solution of \(\text{pH} \approx 2\).
- \(\text{NaCl}\) (the chloride of sodium) dissolves to form a neutral solution of \(\text{pH} = 7\), so \(X\) cannot be sodium.

Award 1 mark for the correct option B.
- Method: Deduce that slight hydrolysis to pH 6.5 corresponds to magnesium chloride (element X is Mg).
- Deduce that vigorous reaction yielding white fumes of HCl and a strongly acidic solution of pH 2 corresponds to silicon tetrachloride (element Y is Si).

1. Identify the oxidation state of cobalt in \([\text{Co}(\text{NH}_3)_6]^{3+}\). Since ammonia (\(\text{NH}_3\)) is a neutral ligand, the charge of the complex is equal to the charge of the metal ion: cobalt has an oxidation state of +3, forming a \(\text{Co}^{3+}\) ion.
2. The atomic number of cobalt (\text{Co}) is 27. The electronic configuration of a neutral Co atom is \([\text{Ar}] 3\text{d}^7 4\text{s}^2\).
3. When forming transition metal cations, electrons are lost from the outermost \(4\text{s}\) subshell before any electrons are removed from the \(3\text{d}\) subshell.
4. Removing three electrons: two from \(4\text{s}\) and one from \(3\text{d}\) gives the electronic configuration of \(\text{Co}^{3+}\) as \([\text{Ar}] 3\text{d}^6\).

Award 1 mark for the correct option A.
- Method: Determine the +3 oxidation state of cobalt in the complex.
- Use the periodic table to find the electronic configuration of the neutral cobalt atom ([Ar] 3d7 4s2).
- Remove electrons from the 4s subshell first, then the 3d subshell to obtain [Ar] 3d6.

Adding concentrated hydrochloric acid (containing high concentration of \(\text{Cl}^-\)) to aqueous copper(II) ions results in ligand substitution:
\([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)
- The reactant complex \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) has 6-fold coordination and is octahedral, with a pale blue color.
- The product complex \([\text{CuCl}_4]^{2-}\) has 4-fold coordination (due to the larger size and repulsion of the chloride ligands) and is tetrahedral, with a yellow-green (or yellow) color.
- Thus, the geometry changes from octahedral to tetrahedral, and the color changes from pale blue to yellow-green.

Award 1 mark for correct option A.
- Method: State the formulas of the initial hexaaquacopper(II) complex and the final tetrachlorocuprate(II) complex.
- Deduce that the geometry changes from octahedral to tetrahedral and the color changes from pale blue to yellow-green.

1. Calculate the heat energy released, \(q\):
\(q = m c \Delta T\)
\(q = 100\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.6\text{ K} = 4012.8\text{ J} = 4.0128\text{ kJ}\)

2. Calculate the moles of solute \(\text{CaCl}_2\):
\(n = \frac{5.55\text{ g}}{111.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\)

3. Calculate the molar enthalpy change of solution:
\(\Delta H_{sol}^\ominus = -\frac{q}{n} = -\frac{4.0128\text{ kJ}}{0.0500\text{ mol}} = -80.3\text{ kJ mol}^{-1}\)
(The negative sign indicates an exothermic process, as evidenced by the temperature rise of the water).

Award 1 mark for the correct option A.
- Method: Compute heat energy transferred (4.01 kJ) using q = mcΔT.
- Compute moles of solute (0.0500 mol).
- Correctly divide energy by moles and assign the negative sign to denote an exothermic reaction (-80.3 kJ mol-1).

1. Use the density and the molar volume of a gas at room temperature and pressure to find the molar mass (\(M_r\)) of the hydrocarbon:
\(\text{Molar mass} = \text{density} \times \text{molar volume}\)
\(M_r = 2.41\text{ g dm}^{-3} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 57.84\text{ g mol}^{-1} \approx 58\text{ g mol}^{-1}\)

2. Calculate the molar masses of the given options to find the match:
- \(\text{C}_3\text{H}_6\): \(3 \times 12.0 + 6 \times 1.0 = 42.0\text{ g mol}^{-1}\)
- \(\text{C}_3\text{H}_8\): \(3 \times 12.0 + 8 \times 1.0 = 44.0\text{ g mol}^{-1}\)
- \(\text{C}_4\text{H}_{10}\): \(4 \times 12.0 + 10 \times 1.0 = 58.0\text{ g mol}^{-1}\)
- \(\text{C}_5\text{H}_{12}\): \(5 \times 12.0 + 12 \times 1.0 = 72.0\text{ g mol}^{-1}\)

Therefore, the molecular formula of \(Z\) is \(\text{C}_4\text{H}_{10}\).

Award 1 mark for the correct option C.
- Method: Calculate the relative molecular mass of the gas using Mr = density * molar volume.
- Identify that 58 g mol-1 corresponds to butane (C4H10).

The stoichiometry of the reaction is \(2\text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}\). 30 cm\(^3\) of NO requires 15 cm\(^3\) of O\(_2\) to react completely. Since 25 cm\(^3\) of O\(_2\) is present, NO is the limiting reactant and 10 cm\(^3\) of O\(_2\) remains unreacted (\(25 - 15 = 10\text{ cm}^3\)). The reaction produces 30 cm\(^3\) of NO\(_2\) gas. Therefore, the total volume of gas remaining is 10 cm\(^3\) of O\(_2\) + 30 cm\(^3\) of NO\(_2\) = 40 cm\(^3\).

1 mark for identifying NO as the limiting reactant, calculating the leftover volume of oxygen (10 cm\(^3\)) and the volume of NO\(_2\) produced (30 cm\(^3\)), and summing them to get 40 cm\(^3\).

Transition metals and their complexes show high catalytic activity primarily due to their ability to adopt variable oxidation states, which allows them to form stable intermediates. Option A is incorrect because scandium and zinc are not transition elements as they do not form stable ions with partially filled d-subshells. Option B is incorrect because copper has the ground-state electronic configuration [Ar] 3d\(^{10}\) 4s\(^1\). Option D is incorrect because chromium atoms ([Ar] 3d\(^5\) 4s\(^1\)) have six unpaired electrons in their ground state.

1 mark for identifying that the catalytic activity of transition metals is due to their ability to change oxidation states, and recognizing that the other choices contain incorrect facts about d-block configurations or definitions of transition metals.

The equation for the formation of methanol is: \(\text{C}(\text{s}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l})\). Using Hess\'s law: \(\Delta H^\ominus_\text{f} = \sum \Delta H^\ominus_\text{c}(\text{reactants}) - \sum \Delta H^\ominus_\text{c}(\text{products}) = [\Delta H^\ominus_\text{c}(\text{C}) + 2 \times \Delta H^\ominus_\text{c}(\text{H}_2)] - [\Delta H^\ominus_\text{c}(\text{CH}_3\text{OH})] = [-394 + 2(-286)] - [-726] = -966 + 726 = -240\text{ kJ mol}^{-1}\).

1 mark for setting up the Hess\'s cycle correctly, scaling the enthalpy of combustion of hydrogen by 2, and accurately calculating the value of -240 kJ mol\(^{-1}\).

Silicon tetrachloride, \(\text{SiCl}_4\), is a simple molecular covalent liquid at room temperature. It undergoes rapid and complete hydrolysis with water to form silicon dioxide (an insoluble white precipitate) and hydrochloric acid (strongly acidic, pH 1): \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\). \(\text{PCl}_5\) is a solid, \(\text{AlCl}_3\) is a solid, and \(\text{MgCl}_2\) dissolves to give a weakly acidic solution without forming a precipitate.

1 mark for identifying SiCl\(_4\) based on its liquid state and characteristic hydrolysis reaction forming a white precipitate of SiO\(_2\) and a strongly acidic solution of HCl.

The equation for the precipitation is \(Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\). Moles of Ba\(^{2+}\) = 0.0500 dm\(^3\) \(\times\) 0.200 mol dm\(^{-3}\) = 0.0100 mol. Moles of SO\(_4^{2-}\) = 0.0750 dm\(^3\) \(\times\) 0.150 mol dm\(^{-3}\) = 0.01125 mol. Since the reaction ratio is 1:1, Ba\(^{2+}\) is the limiting reactant. Therefore, 0.0100 mol of BaSO\(_4\) precipitate is formed. Mass of BaSO\(_4\) = 0.0100 mol \(\times\) 233.4 g mol\(^{-1}\) = 2.33 g.

1 mark for calculating the amount of both reactants, identifying barium chloride as the limiting reagent, and calculating the mass of barium sulfate to 3 significant figures.

When ligands bind to a transition metal ion, the d-orbitals split into two groups of different energy levels. When visible light shines on the complex, an electron can be promoted from a lower energy d-orbital to a higher energy d-orbital, absorbing a specific wavelength of light. The complementary color of the absorbed light is transmitted or reflected, which is the color we observe.

1 mark for identifying the correct explanation involving d-orbital splitting and d-d transition via absorption of specific wavelengths of visible light.

The standard enthalpy change of combustion is defined for one mole of a substance burning completely in oxygen under standard conditions, with all reactants and products in their standard states. Under standard conditions (298 K, 1 atm), methanol is a liquid, oxygen is a gas, carbon dioxide is a gas, and water is a liquid. Therefore, the correct equation is: \(\text{CH}_3\text{OH(l)} + 1.5\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\).

1 mark for selecting the equation where exactly 1 mole of liquid methanol reacts with gaseous oxygen to form gaseous carbon dioxide and liquid water.

Aluminium oxide, \(\text{Al}_2\text{O}_3\), is amphoteric. It reacts with acids like HCl to form aluminium chloride and water, and with strong bases like NaOH to form sodium aluminate. \(\text{MgO}\) is basic, while \(\text{SiO}_2\) and \(\text{P}_4\text{O}_{10}\) are acidic oxides.

1 mark for identifying aluminium oxide as the amphoteric Period 3 oxide that reacts with both acids and bases.

(a) (i) Dissolve the \(1.753\text{ g}\) of solid in a beaker using a smaller volume of distilled/deionised water (e.g. \(100\text{ cm}^3\)) and stir with a glass rod. Transfer the solution quantitatively to a \(250.0\text{ cm}^3\) volumetric flask, rinsing the beaker and glass rod and adding the washings to the flask. Make up to the graduation mark with distilled water (until the bottom of the meniscus is on the line) and then invert the flask several times to ensure thorough mixing.

(ii) \(2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)

(iii) \(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{24.00}{1000}\text{ dm}^3 = 4.80 \times 10^{-4}\text{ mol}\)

(iv) Using the mole ratio of \(2 : 5\) from the equation:
\(n(\text{C}_2\text{O}_4^{2-}) = 4.80 \times 10^{-4}\text{ mol} \times 2.5 = 1.20 \times 10^{-3}\text{ mol}\)

(v) Moles of \(\text{C}_2\text{O}_4^{2-}\) in \(250.0\text{ cm}^3\) = \(1.20 \times 10^{-3}\text{ mol} \times 10 = 1.20 \times 10^{-2}\text{ mol}\)
Since \(1\text{ mol}\) of \(\text{MC}_2\text{O}_4 \cdot n\text{H}_2\text{O}\) contains \(1\text{ mol}\) of \(\text{C}_2\text{O}_4^{2-}\):
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.753\text{ g}}{1.20 \times 10^{-2}\text{ mol}} = 146.1\text{ g mol}^{-1}\) (or \(146.08\text{ g mol}^{-1}\)).

(b) (i) Number of moles of hydrated salt decomposed:
\(n(\text{salt}) = \frac{3.506\text{ g}}{146.1\text{ g mol}^{-1}} = 0.0240\text{ mol}\)
According to the stoichiometric equation, \(1\text{ mol}\) of hydrated salt decomposes to yield \(1\text{ mol}\) of \(\text{MO}\).
Therefore, \(n(\text{MO}) = 0.0240\text{ mol}\).
\(M_r(\text{MO}) = \frac{1.346\text{ g}}{0.0240\text{ mol}} = 56.1\text{ g mol}^{-1}\)
\(A_r(\text{M}) = 56.1 - 16.0 = 40.1\text{ g mol}^{-1}\)
This corresponds to calcium, so \(\text{M} = \text{Calcium (Ca)}\).
To find \(n\):
\(M_r(\text{CaC}_2\text{O}_4 \cdot n\text{H}_2\text{O}) = 146.1\)
\(40.1 + 24.0 + 64.0 + n(18.0) = 146.1\)
\(128.1 + 18.0n = 146.1\)
\(18.0n = 18.0 \implies n = 1\).

(ii) The gaseous products are \(\text{CO}\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\). When cooled to room temperature, steam condenses, leaving only \(\text{CO}\) and \(\text{CO}_2\) as gases.
For every mole of salt decomposed, \(1\text{ mol}\) of \(\text{CO}\) and \(1\text{ mol}\) of \(\text{CO}_2\) are produced (total \(2\text{ mol}\) of gas).
\(n(\text{gas}) = 2 \times 0.0240\text{ mol} = 0.0480\text{ mol}\)
\(V(\text{gas}) = 0.0480\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.15\text{ dm}^3\) (or \(1.152\text{ dm}^3\)).

(a) (i)
- M1: Dissolve the solid in a beaker using distilled/deionised water and stir with a glass rod. [1]
- M2: Transfer the solution quantitatively to a 250.0 cm3 volumetric flask, rinsing the beaker and glass rod and adding the washings. [1]
- M3: Make up to the graduation mark with distilled water (until the bottom of the meniscus is on the line) and invert several times to mix. [1]

(ii)
- M1: Correct reactants and products: MnO4^- + C2O4^2- + H^+ -> Mn^2+ + CO2 + H2O [1]
- M2: Correctly balanced: 2MnO4^- + 5C2O4^2- + 16H^+ -> 2Mn^2+ + 10CO2 + 8H2O [1]

(iii)
- M1: n(MnO4^-) = 4.80 x 10^-4 mol [1]

(iv)
- M1: n(C2O4^2-) = (5/2) x 4.80 x 10^-4 = 1.20 x 10^-3 mol (Allow ecf from (iii) and (ii)) [1]

(v)
- M1: Moles in 250.0 cm3 = 1.20 x 10^-2 mol [1]
- M2: Mr = 1.753 / 0.0120 = 146.1 g mol^-1 (Accept 146) [1]

(b) (i)
- M1: n(salt) = 3.506 / 146.1 = 0.0240 mol (Allow ecf from (a)(v)) [1]
- M2: Mr(MO) = 1.346 / 0.0240 = 56.1 g mol^-1 [1]
- M3: Ar(M) = 56.1 - 16.0 = 40.1 g mol^-1, so M is Calcium (Ca) [1]
- M4: 128.1 + 18n = 146.1, so n = 1 [1]

(ii)
- M1: Total moles of gas at r.t.p. = 2 x 0.0240 = 0.0480 mol (steam condenses) [1]
- M2: Volume of gas = 0.0480 x 24.0 = 1.15 dm3 (or 1.152 dm3) [1]
*(Note: If student includes steam as a gas, total moles = 3 x 0.0240 = 0.0720 mol, volume = 1.73 dm3 - max 1 mark out of 2 if explicitly stated)*

(a) (i) When sodium is heated and reacted with oxygen, it burns with a bright yellow (or yellow-orange) flame to form a white solid. The main product is sodium oxide (or sodium peroxide).
Equation: \(4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}\)
(Accept: \(2\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow \text{Na}_2\text{O}_2\text{(s)}\))

(ii) Phosphorus reacts with excess oxygen to form phosphorus(V) oxide, which exists as \(\text{P}_4\text{O}_{10}\).
Equation: \(\text{P}_4\text{(s)} + 5\text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}\)
Structure: Simple molecular
Bonding: Covalent

(b) (i) Sulfur trioxide reacts vigorously with water to form sulfuric acid:
\(\text{SO}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)}\)
pH of the resulting solution: \(0 - 2\) (highly acidic).

(ii) Magnesium oxide, \(\text{MgO}\), is sparingly soluble in water and reacts slightly to form a weakly alkaline solution (pH \(8 - 10\)) of magnesium hydroxide, \(\text{Mg(OH)}_2\).
Aluminium oxide, \(\text{Al}_2\text{O}_3\), is completely insoluble in water and the resulting mixture is neutral (pH 7).
Both oxides have giant ionic structures. However, \(\text{Al}_2\text{O}_3\) is insoluble because it has a much higher lattice energy (due to the highly charged \(\text{Al}^{3+}\) and \(\text{O}^{2-}\) ions) which cannot be overcome by hydration, or because of significant covalent character due to polarization of the oxide ion by the small, highly charged \(\text{Al}^{3+}\) ion.

(c) (i) Silicon dioxide, \(\text{SiO}_2\), has a giant covalent macromolecular structure with strong, directional covalent \(\text{Si}-\text{O}\) bonds extending throughout. Water molecules do not have enough energy to break these strong covalent bonds, so it is insoluble and unreactive with water.
However, \(\text{SiO}_2\) is an acidic oxide and reacts with hot concentrated sodium hydroxide (a strong base) to form sodium silicate and water.
Equation: \(\text{SiO}_2\text{(s)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SiO}_3\text{(aq)} + \text{H}_2\text{O(l)}\)

(ii) Aluminium oxide reacts with hydrochloric acid to form aluminium chloride and water:
\(\text{Al}_2\text{O}_3\text{(s)} + 6\text{HCl(aq)} \rightarrow 2\text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{O(l)}\)

(a) (i)
- M1: Yellow (or yellow-orange) flame [1]
- M2: White solid (or sodium oxide / sodium peroxide) [1]
- M3: Balanced equation: 4Na + O2 -> 2Na2O or 2Na + O2 -> Na2O2 [1]

(ii)
- M1: Balanced equation: P4 + 5O2 -> P4O10 (Accept: 2P2 + 5O2 -> 2P2O5 or P2O5) [1]
- M2: Simple molecular [1]
- M3: Covalent [1]

(b) (i)
- M1: Equation: SO3 + H2O -> H2SO4 [1]
- M2: pH range: 0 to 2 [1]

(ii)
- M1: MgO is sparingly soluble/reacts slightly to form an alkaline solution (pH 8-10) [1]
- M2: Al2O3 is insoluble (pH 7) [1]
- M3: Explanation: Al2O3 has very high lattice energy (due to highly charged Al3+ and O2- ions) or has high covalent character/polarization which prevents water from breaking the structure. [1]

(c) (i)
- M1: Giant covalent / macromolecular structure with strong covalent bonds that cannot be broken by water. [1]
- M2: SiO2 is an acidic oxide, so it reacts with hot, concentrated alkali. [1]
- M3: Equation: SiO2 + 2NaOH -> Na2SiO3 + H2O [1]

(ii)
- M1: Al2O3 + 6HCl -> 2AlCl3 + 3H2O (Accept ionic equation: Al2O3 + 6H+ -> 2Al3+ + 3H2O) [1]

(a) The standard enthalpy change of combustion, \(\Delta H^\ominus_c\), is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (\(298\text{ K}\) and \(100\text{ kPa}\)), with all reactants and products in their standard states.

(b) (i) \(q = mc\Delta T\)
\(m = 150.0\text{ g}\)
\(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
\(\Delta T = 62.3 - 21.5 = 40.8\text{ K}\)
\(q = 150.0 \times 4.18 \times 40.8 = 25581.6\text{ J} = 25.58\text{ kJ}\) (or \(25.6\text{ kJ}\))

(ii) \(M_r(\text{C}_3\text{H}_7\text{OH}) = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\)
\(n = \frac{\text{mass}}{M_r} = \frac{0.930\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0155\text{ mol}\)

(iii) \(\Delta H_c = -\frac{q}{n} = -\frac{25.5816\text{ kJ}}{0.0155\text{ mol}} = -1650.4 \approx -1650\text{ kJ mol}^{-1}\) (to 3 sig figs)

(iv) 1. Heat loss to the surroundings (e.g., from the water, copper calorimeter, or flame).
2. Incomplete combustion of propan-1-ol (visible soot on the bottom of the calorimeter).
(Accept: Evaporation of propan-1-ol from the wick after extinguishing the burner; heating of the calorimeter itself is not accounted for.)

(c) (i) \(3\text{C(s, graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\)

(ii) According to Hess's Law, the enthalpy change of the reaction can be calculated from the standard enthalpy changes of combustion:
\(\Delta H^\ominus_f + \Delta H^\ominus_c(\text{propan-1-ol}) = 3 \times \Delta H^\ominus_c(\text{C}) + 4 \times \Delta H^\ominus_c(\text{H}_2)\)
\(\Delta H^\ominus_f + (-2021.0) = 3(-393.5) + 4(-285.8)\)
\(\Delta H^\ominus_f - 2021.0 = -1180.5 - 1143.2 = -2323.7\)
\(\Delta H^\ominus_f = -2323.7 + 2021.0 = -302.7\text{ kJ mol}^{-1}\).

(iii) For the isomerization:
\(\Delta H = \Delta H^\ominus_f(\text{propan-2-ol}) - \Delta H^\ominus_f(\text{propan-1-ol}) = -318.1 - (-302.7) = -15.4\text{ kJ mol}^{-1}\).
Since \(\Delta H < 0\) (exothermic), propan-2-ol is thermodynamically more stable because it lies at a lower energy/enthalpy level (or has a more negative standard enthalpy change of formation).

(a)
- M1: Enthalpy change when one mole of a substance is burned completely in excess oxygen. [1]
- M2: Under standard conditions of 100 kPa and 298 K (with all species in standard states). [1]

(b) (i)
- M1: q = 150.0 x 4.18 x 40.8 = 25.6 kJ (or 25.58 kJ) [1]

(ii)
- M1: n = 0.930 / 60.0 = 0.0155 mol [1]

(iii)
- M1: division: 25.58 / 0.0155 [1]
- M2: correct value and sign to 3 s.f.: -1650 kJ mol^-1 (Allow -1650 to -1654 depending on rounding) [1]

(iv)
- M1: Heat loss to the surroundings / copper calorimeter [1]
- M2: Incomplete combustion of propan-1-ol / evaporation of propan-1-ol [1]

(c) (i)
- M1: Correct species: 3C + 4H2 + 1/2O2 -> C3H7OH [1]
- M2: Correct state symbols: C(s) + H2(g) + O2(g) -> C3H7OH(l) [1]

(ii)
- M1: Sum of combustion of elements = 3(-393.5) + 4(-285.8) = -2323.7 kJ mol^-1 [1]
- M2: Cycle setup: ΔHf = Sum(combustion of reactants) - Sum(combustion of products) [1]
- M3: Correct calculation: -2323.7 - (-2021.0) = -302.7 kJ mol^-1 [1]

(iii)
- M1: ΔH = -318.1 - (-302.7) = -15.4 kJ mol^-1 [1]
- M2: Propan-2-ol is more stable because it is more exothermic/has a more negative enthalpy of formation / lower enthalpy. [1]

(a) A transition element is defined as a d-block element that forms at least one stable ion with a partially filled (or incomplete) d-subshell.

(b) (i) Cobalt has an atomic number of 27.
Co: \([\text{Ar}] 3\text{d}^7 4\text{s}^2\)
For \(\text{Co}^{2+}\), two electrons are lost from the \(4\text{s}\) subshell first:
\(\text{Co}^{2+}\): \([\text{Ar}] 3\text{d}^7\)

(ii) The ligand exchange reaction is:
\([\text{Co(H}_2\text{O)}_6]^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{CoCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)}\)

(iii) Formula: \([\text{CoCl}_4]^{2-}\)
Geometry: Tetrahedral
Explanation: Chloride ligands (\(\text{Cl}^-\)) are larger than water molecules (\(\text{H}_2\text{O}\)). There is greater steric hindrance (and electrostatic repulsion between the negatively charged chloride ions) when attempting to fit six chloride ligands around the central cobalt ion. Thus, only four chloride ligands can fit, changing the coordination number from 6 to 4.

(c) (i) Dropwise addition: A pale green precipitate of iron(II) hydroxide is formed.
In excess sodium hydroxide: The green precipitate remains insoluble.
Ionic equation: \(\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}\)

(ii) The green precipitate turns brown (or red-brown) on standing in air.
This is because iron(II) hydroxide, \(\text{Fe(OH)}_2\), is oxidized by oxygen in the air to iron(III) hydroxide, \(\text{Fe(OH)}_3\).
During this process, the oxidation state of iron changes from +2 (in \(\text{Fe(OH)}_2\)) to +3 (in \(\text{Fe(OH)}_3\)).

(a)
- M1: d-block element that forms at least one stable ion [1]
- M2: with a partially filled / incomplete d-subshell [1]

(b) (i)
- M1: Co: [Ar] 3d7 4s2 [1]
- M2: Co2+: [Ar] 3d7 [1]

(ii)
- M1: Correct species: [Co(H2O)6]^2+ + Cl^- -> [CoCl4]^2- + H2O [1]
- M2: Correctly balanced: [Co(H2O)6]^2+ + 4Cl^- -> [CoCl4]^2- + 6H2O [1]

(iii)
- M1: Formula: [CoCl4]^2- [1]
- M2: Geometry: Tetrahedral [1]
- M3: Explanation: Cl^- ligands are larger than H2O molecules / steric hindrance / greater electrostatic repulsion between negatively charged chloride ligands prevents accommodation of six ligands. [1]

(c) (i)
- M1: Observation: Green precipitate [1]
- M2: Observation: Insoluble in excess [1]
- M3: Equation: Fe2+(aq) + 2OH^-(aq) -> Fe(OH)2(s) [1]

(ii)
- M1: Color change: turns brown / red-brown [1]
- M2: Reason: oxidation of Fe(OH)2 by oxygen / air [1]
- M3: Change in species: Fe2+ (or +2 state) oxidized to Fe3+ (or +3 state) / formation of Fe(OH)3 [1]

Let us assume the mean titre obtained is exactly 25.00 cm³.

(a) Mean volume of FA 2 used:
\(\text{Mean titre} = \frac{25.00 + 25.00}{2} = 25.00\text{ cm}^3\).

(b) Moles of \(\text{HCl}\) in mean volume:
\(n(\text{HCl}) = 0.100 \times \frac{25.00}{1000} = 0.00250\text{ mol}\).

(c) Reaction equation: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\).
Thus, mole ratio is 1:2.
\(n(\text{Na}_2\text{CO}_3) \text{ in 25.0 cm}^3 = \frac{0.00250}{2} = 0.00125\text{ mol}\).

(d) Concentration of FA 1:
\([\text{Na}_2\text{CO}_3] = 0.00125 \times \frac{1000}{25.0} = 0.0500\text{ mol dm}^{-3}\).

(e) Moles of hydrated salt in 250 cm³:
\(n(\text{total}) = 0.0500 \times 0.250 = 0.0125\text{ mol}\).
\(M_r\text{ of } \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} = \frac{\text{mass}}{\text{moles}} = \frac{3.58}{0.0125} = 286.4\text{ g mol}^{-1}\).
\(M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\).
\(x \times M_r(\text{H}_2\text{O}) = 286.4 - 106.0 = 180.4\text{ g mol}^{-1}\).
\(x = \frac{180.4}{18.0} = 10.02 \approx 10\).

(f) Percentage error in pipette volume:
\(\text{Percentage error} = \frac{0.06}{25.0} \times 100\% = 0.24\%\).

1. **Burette readings table format** (2 marks): Headings with correct units and all readings recorded to 2 decimal places with consistent precision (either .00 or .05).
2. **Accuracy of titration** (3 marks): Award 3 marks if the student's titre is within \(\pm 0.10\text{ cm}^3\) of the reference value (25.00 cm³); 2 marks if within \(\pm 0.20\text{ cm}^3\); 1 mark if within \(\pm 0.30\text{ cm}^3\).
3. **Mean titre calculation** (1 mark): Correctly identifies concordant titres and calculates the average, showing the working.
4. **Moles of HCl and Na2CO3** (2 marks): Correctly calculates moles of \(\text{HCl}\) (1 mark) and correctly applies 1:2 mole ratio to find moles of \(\text{Na}_2\text{CO}_3\) (1 mark).
5. **Molar mass and value of x** (2 marks): Correctly scales up concentration to 250 cm³ and calculates \(M_r\) (1 mark), and solves for \(x\) to the nearest integer (1 mark).
6. **Pipette error and significant figures** (3.33 marks): Correct calculation of percentage error (1.33 marks) and all final numerical answers in (b), (c), (d), and (e) are quoted to 3 or 4 significant figures (2 marks).

**FA 3 Identification**:
- Test (a) with NaOH(aq) produces a blue precipitate that turns pink/beige on standing, and is insoluble in excess. This is characteristic of cobalt(II), \(\text{Co}^{2+}\).
- Test (b) with NH₃(aq) produces a blue precipitate that dissolves in excess to form a yellow-brown/brown solution, confirming \(\text{Co}^{2+}\).
- Test (c) with HNO₃ and AgNO₃ produces a white precipitate of \(\text{AgCl}\) which is soluble in dilute ammonia, confirming chloride, \(\text{Cl}^-\).

**FA 4 Identification**:
- Test (d) with NaOH(aq) produces a pale blue precipitate which is insoluble in excess, characteristic of copper(II), \(\text{Cu}^{2+}\).
- Test (e) with NH₃(aq) produces a pale blue precipitate that dissolves in excess to give a deep blue solution, confirming \(\text{Cu}^{2+}\).
- Test (f) with HCl and BaCl₂ produces a white precipitate of \(\text{BaSO}_4\) which is insoluble in dilute acid, confirming sulfate, \(\text{SO}_4^{2-}\).

**Ionic Equation**:
\(\text{Cu}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cu(OH)}_2(s)\)

1. **FA 3 Observations** (3 marks):
- 1 mark for blue precipitate with NaOH turning pink/beige, insoluble in excess.
- 1 mark for blue precipitate with NH₃ dissolving in excess to give a brown/yellow-brown solution.
- 1 mark for white precipitate with AgNO₃ dissolving in dilute ammonia.
2. **FA 4 Observations** (3 marks):
- 1 mark for pale blue precipitate with NaOH, insoluble in excess.
- 1 mark for pale blue precipitate with NH₃ dissolving in excess to give a deep blue solution.
- 1 mark for white precipitate with BaCl₂ that is insoluble in acid.
3. **Ion Identification** (4 marks):
- 1 mark for each correctly identified ion: \(\text{Co}^{2+}\), \(\text{Cl}^-\), \(\text{Cu}^{2+}\), \(\text{SO}_4^{2-}\).
4. **Ionic Equation** (2 marks):
- 1 mark for correct formulas: \(\text{Cu}^{2+} + 2\text{OH}^- \rightarrow \text{Cu(OH)}_2\).
- 1 mark for correct state symbols: \((aq)\) for reactants, \((s)\) for product.
5. **Safety Precaution** (1.33 marks):
- State a reasonable safety precaution, e.g., wear chemical-resistant gloves because transition metal solutions (cobalt and copper salts) are toxic/irritating.

(a) & (b) The student plots the temperature values. The initial temperatures should be constant (e.g., 21.0 °C). The maximum temperature reached is followed by a slow cooling trend. Drawing best-fit lines and extrapolating to \(t = 3.0\) min gives \(\Delta T = 18.5\text{ °C}\) (from 21.0 °C to 39.5 °C).

(c) Heat energy change, \(q = m \cdot c \cdot \Delta T\)
\(m = 25.0\text{ g}\) (since density is 1.00 g cm⁻³ and volume is 25.0 cm³)
\(q = 25.0 \times 4.18 \times 18.5 = 1933.25\text{ J}\).

(d) Moles of \(\text{Cu}^{2+}\):
\(n(\text{Cu}^{2+}) = 0.800 \times \frac{25.0}{1000} = 0.0200\text{ mol}\).

(e) Enthalpy change, \(\Delta H = -\frac{q}{n(\text{Cu}^{2+}) \times 1000} = -\frac{1933.25}{0.0200 \times 1000} = -96.7\text{ kJ mol}^{-1}\).

(f) Modification: Add a lid to the plastic cup (or wrap the plastic cup in cotton wool / bubble wrap to increase insulation).
Uncertainty of thermometer = \(\pm 0.2\text{ °C}\) per reading.
For a difference of temperature (two readings), total uncertainty = \(2 \times 0.2 = \pm 0.4\text{ °C}\).
\(\text{Percentage uncertainty} = \frac{0.4}{18.5} \times 100\% = 2.16\%\).

1. **Data Table** (2 marks): Headings with correct units, all temperature readings recorded consistently to 1 decimal place.
2. **Graph Plotting & Extrapolation** (3 marks): Correct axes and scaling (1 mark); accurate plotting of points (1 mark); drawing of two straight lines of best fit and clear extrapolation to find \(\Delta T\) at \(t = 3.0\) minutes (1 mark).
3. **Heat Energy Calculation (q)** (1.33 marks): Correctly applies \(q = m c \Delta T\) with \(m = 25.0\text{ g}\) and retrieves correct value with units.
4. **Moles of Reactant** (1 mark): Calculates moles of \(\text{Cu}^{2+}\) correctly as 0.0200 mol.
5. **Enthalpy Change Calculation** (2 marks): Correctly divides \(q / n\) and converts to kJ mol⁻¹ (1 mark), and includes the negative sign since the reaction is exothermic (1 mark).
6. **Evaluation & Uncertainty** (3 marks):
- 1 mark for proposing a valid insulation improvement (e.g., plastic lid, outer insulating layer).
- 2 marks for calculating percentage uncertainty: 1 mark for realizing total uncertainty is \(\pm 0.4\text{ °C}\) (due to two readings) and 1 mark for arriving at 2.16% (or correct value based on their own temperature rise).