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

Let the hydrocarbon be \(C_xH_y\). The equation for combustion is: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). Since all measurements are at r.t.p., \(H_2O\) is liquid with negligible volume. The decrease in volume when passed through aqueous NaOH is due to the absorption of \(CO_2\). Thus, volume of \(CO_2 = 85 - 45 = 40\text{ cm}^3\). Since 10 cm³ of hydrocarbon was used, \(10x = 40\), so \(x = 4\). The remaining gas (45 cm³) is the unreacted excess oxygen. The volume of oxygen reacted is \(100 - 45 = 55\text{ cm}^3\). From the stoichiometry, \(10(x + \frac{y}{4}) = 55 \Rightarrow x + \frac{y}{4} = 5.5\). Substituting \(x = 4\) gives \(4 + \frac{y}{4} = 5.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\). The molecular formula is \(C_4H_6\).

1 mark for the correct option B.

Using Hess's Law, the standard enthalpy change of formation of propan-1-ol, \(3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\), is calculated as: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). Therefore, \(\Delta H_f^\ominus = 3 \times \Delta H_c^\ominus[\text{C(s)}] + 4 \times \Delta H_c^\ominus[\text{H}_2\text{(g)}] - \Delta H_c^\ominus[\text{C}_3\text{H}_7\text{OH(l)}] = 3(-394) + 4(-286) - (-2021) = -1182 - 1144 + 2021 = -305\text{ kJ mol}^{-1}\).

1 mark for the correct option A.

In the dimer \(\text{Al}_2\text{Cl}_6\), each aluminium atom is bonded to four chlorine atoms in a tetrahedral arrangement. Two of the chlorine atoms act as bridging atoms, forming one normal covalent bond to one aluminium atom and sharing a lone pair to form a coordinate bond to the other aluminium atom. The other four terminal chlorine atoms only form single covalent bonds. Thus, each aluminium atom has four bonds in total: three normal covalent bonds and one coordinate bond, completing an octet of 8 electrons. The coordination number of aluminium is 4.

1 mark for the correct option A.

Using the ideal gas equation, \(pV = nRT\), the volume is proportional to \(\frac{T}{p}\) when the number of moles \(n\) is constant. Therefore, \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\). Substituting the given values: \(\frac{p V}{T} = \frac{0.6p \times V_2}{1.5T}\). Solving for \(V_2\) gives: \(V_2 = V \times \frac{1.5}{0.6} = 2.5V\).

1 mark for the correct option D.

The equilibrium constant \(K_c\) is only affected by temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature will shift the equilibrium position to the right to release heat (Le Chatelier's principle). This increases the concentration of products and decreases the concentration of reactants at equilibrium, thereby increasing the value of \(K_c\).

1 mark for the correct option B.

In nitrogen monoxide (\(\text{NO}\)), nitrogen has an oxidation state of +2. In nitrogen gas (\(\text{N}_2\)), the oxidation state is 0. Since the oxidation state of nitrogen decreases, nitrogen is reduced. Carbon monoxide (\(\text{CO}\)) is oxidized as the carbon oxidation state increases from +2 to +4. Therefore, \(\text{NO}\) acts as an oxidizing agent, not a reducing agent. A catalyst only speeds up the rate of both forward and reverse reactions and does not shift the position of equilibrium.

1 mark for the correct option B.

An oxide that reacts with both acids and bases is amphoteric. Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is the only amphoteric oxide in Period 3. Magnesium oxide is basic and only reacts with acids. Silicon dioxide is weakly acidic and only reacts with strong bases (like hot concentrated \(\text{NaOH}\)). Phosphorus oxides are acidic and react with water to form acidic solutions.

1 mark for the correct option B.

The rate of hydrolysis of halogenoalkanes depends on two factors: the strength of the carbon-halogen bond and the mechanism (\(S_N1\) vs \(S_N2\)). Since the C-Cl bond is stronger than the C-Br bond, bromoalkanes react faster than chloroalkanes. Furthermore, tertiary halogenoalkanes (like 2-bromo-2-methylpropane) react extremely rapidly via the \(S_N1\) mechanism because they form a highly stable tertiary carbocation intermediate. Primary bromoalkanes (like 1-bromobutane) and secondary ones (like 2-bromobutane) react much more slowly via \(S_N2\) or mixed mechanisms. Therefore, 2-bromo-2-methylpropane produces the precipitate of silver bromide the fastest.

1 mark for the correct option D.

According to the balanced equation, 2 moles of \(Ca(NO_3)_2\) produce 5 moles of gas (4 moles of \(NO_2\) and 1 mole of \(O_2\)).
Therefore, 0.010 mol of \(Ca(NO_3)_2\) will produce:
\(0.010 \times \frac{5}{2} = 0.025\text{ mol}\) of gas.

At r.t.p., the volume of gas is:
\(0.025\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.60\text{ dm}^3 = 600\text{ cm}^3\).

[1 mark]: Award for C.
Method: Determine the mole ratio of reactant to total gas (2 : 5), calculate the moles of gas produced (0.025 mol), and multiply by 24,000 to convert to cm3.

We can construct a Hess cycle where both sides are combusted:
\(N_2(g) + 2H_2(g) \rightarrow N_2H_4(l)\)

Combustion of reactants:
\(2 \times \Delta H_c^\ominus [H_2(g)] = 2 \times (-286) = -572\text{ kJ mol}^{-1}\)
(Note: \(N_2\) does not burn under standard combustion conditions, so its contribution is 0).

Combustion of products:
\(\Delta H_c^\ominus [N_2H_4(l)] = -622\text{ kJ mol}^{-1}\)

According to Hess's law:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus (\text{reactants}) - \sum \Delta H_c^\ominus (\text{products})\)
\(\Delta H_f^\ominus = -572 - (-622) = +50\text{ kJ mol}^{-1}\).

[1 mark]: Award for C.
Method: Apply Hess's Law correctly using \(2 \times \Delta H_c[H_2] - \Delta H_c[N_2H_4]\).

Nitrogen monoxide forms in internal combustion engines due to the very high temperatures created by spark plug ignition, which provides enough energy to break the strong triple bond in \(N_2\). In a catalytic converter, \(NO\) is reduced by carbon monoxide to form harmless nitrogen gas and carbon dioxide: \(2CO + 2NO \rightarrow 2CO_2 + N_2\).

[1 mark]: Award for A.
Identify high temperature as the cause of nitrogen oxide formation and the redox equation representing its removal.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. Because the C–Br bond has a lower bond enthalpy than the C–Cl bond, it breaks more easily. Therefore, 1-bromobutane reacts faster than 1-chlorobutane, despite the C–Cl bond being more polar.

[1 mark]: Award for B.
Correctly link bond enthalpy (not bond polarity) to the relative rate of nucleophilic substitution of halogenoalkanes.

The mechanism is nucleophilic addition. The cyanide ion, \(CN^-\), acts as a nucleophile and attacks the electron-deficient carbonyl carbon. The intermediate has a negative charge on the oxygen atom (not carbon). The hybridization of the carbonyl carbon changes from \(sp^2\) (planar reactant) to \(sp^3\) (tetrahedral intermediate and product).

[1 mark]: Award for C.
Recall details of the nucleophilic addition mechanism including nucleophile identity, intermediate structure, and hybridization changes.

The equilibrium constant, \(K_c\), is only affected by changes in temperature. Because the forward reaction is exothermic (\(\Delta H < 0\)), a decrease in temperature shifts the equilibrium position to the right to favor the exothermic process. This increases the concentration of the products and decreases the concentration of reactants, thereby increasing the value of \(K_c\).

[1 mark]: Award for C.
State that temperature is the only factor affecting Kc, and deduce how decreasing temperature affects Kc for an exothermic reaction.

\(Al_2O_3\) is amphoteric with a high-lattice-energy giant structure, making it insoluble in water and unreactive towards it. \(P_4O_{10}\) is an acidic molecular oxide that reacts violently with water to form phosphoric(V) acid, \(H_3PO_4\), which yields a strongly acidic solution.

[1 mark]: Award for B.
Recognize the insolubility of aluminium oxide and the reaction of phosphorus(V) oxide with water to produce an acidic solution.

Going down Group 2, the solubility of hydroxides increases (magnesium hydroxide is sparingly soluble, barium hydroxide is soluble), while the solubility of sulfates decreases (magnesium sulfate is highly soluble, barium sulfate is completely insoluble). Thermal stability of nitrates increases down the group because the metal cation radius increases, reducing polarising power.

[1 mark]: Award for B.
Recall the solubility trends of Group 2 hydroxides and sulfates correctly.

Step 1: Calculate mass of Carbon in \(CO_2\): \(1.76 \times (12.0 / 44.0) = 0.48\) g. Step 2: Calculate mass of Hydrogen in \(H_2O\): \(0.72 \times (2.0 / 18.0) = 0.08\) g. Step 3: Calculate mass of Oxygen by subtraction: \(1.20 - (0.48 + 0.08) = 0.64\) g. Step 4: Convert masses to moles: C: \(0.48 / 12.0 = 0.04\) mol, H: \(0.08 / 1.0 = 0.08\) mol, O: \(0.64 / 16.0 = 0.04\) mol. Step 5: Find the simplest whole-number ratio: C:H:O = 1:2:1. Therefore, the empirical formula is \(CH_2O\).

1 mark for the correct option B.

By Hess's Law, the formation reaction is \(3C(s) + 4H_2(g) \rightarrow C_3H_8(g)\). Therefore, \(\Delta H_f^{\ominus}[C_3H_8(g)] = 3 \times \Delta H_c^{\ominus}[C(s)] + 4 \times \Delta H_c^{\ominus}[H_2(g)] - \Delta H_c^{\ominus}[C_3H_8(g)] = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1}\).

1 mark for the correct option A.

In the aluminium chloride dimer, \(Al_2Cl_6\), each aluminium atom is bonded to three chlorine atoms by simple covalent bonds, and to a fourth chlorine atom (originally belonging to the other \(AlCl_3\) unit) via a coordinate (dative covalent) bond where the chlorine atom donates a lone pair of electrons.

1 mark for the correct option A.

Using the ideal gas equation: \(pV = nRT\). Convert units to SI: \(p = 120 \times 10^3\text{ Pa}\), \(V = 400 \times 10^{-6}\text{ m}^3\), \(T = 27 + 273 = 300\text{ K}\). Calculate moles: \(n = pV / RT = (120 \times 10^3 \times 400 \times 10^{-6}) / (8.31 \times 300) \approx 0.01925\text{ mol}\). Calculate mass: \(m = n \times M = 0.01925 \times 32.0 \approx 0.616\text{ g}\).

1 mark for the correct option A.

According to Le Chatelier's principle, decreasing the temperature of an exothermic system shifts the equilibrium position in the exothermic (forward) direction to release heat. Thus, the yield of \(Z\) increases. Since temperature is the only factor that alters the equilibrium constant, and the equilibrium has shifted to favor the products, the value of \(K_c\) increases.

1 mark for the correct option A.

The rate of hydrolysis of halogenoalkanes is determined by the strength (bond enthalpy) of the carbon-halogen bond. The C-I bond has the lowest bond enthalpy among the three, making it the easiest to break. Therefore, 1-iodobutane reacts the fastest.

1 mark for the correct option A.

Reaction with 2,4-dinitrophenylhydrazine indicates Y is a carbonyl compound (aldehyde or ketone). The negative test with Tollens' reagent shows Y is not an aldehyde, and must be a ketone. With four carbon atoms, the only possible ketone is butanone.

1 mark for the correct option A.

When the temperature is increased, the Maxwell-Boltzmann distribution curve flattens and shifts to the right. This means the peak (most probable energy) moves to a higher energy. The total area under the curve represents the total number of molecules, which remains constant.

1 mark for the correct option A.

1. Write the thermal decomposition equation: \(\text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g})\). Note that magnesium oxide does not decompose on heating. 2. Calculate the moles of \(\text{CO}_2\) released: \(n(\text{CO}_2) = \frac{1.76\text{ g}}{44.0\text{ g mol}^{-1}} = 0.0400\text{ mol}\). 3. Relate moles of \(\text{CO}_2\) to \(\text{MgCO}_3\): Since the mole ratio is 1:1, there are \(0.0400\text{ mol}\) of \(\text{MgCO}_3\) in the original mixture. 4. Calculate the mass of \(\text{MgCO}_3\): \(m(\text{MgCO}_3) = 0.0400\text{ mol} \times 84.3\text{ g mol}^{-1} = 3.372\text{ g}\). 5. Calculate the mass of \(\text{MgO}\) in the original mixture: \(m(\text{MgO}) = 5.22\text{ g} - 3.372\text{ g} = 1.848\text{ g}\). 6. Calculate the percentage by mass of \(\text{MgO}\): \(\% \text{MgO} = \frac{1.848}{5.22} \times 100\% \approx 35.4\%\).

1 mark for the correct calculation of the percentage by mass of MgO (35.4%). Reject 64.6% (which is the percentage of magnesium carbonate).

The equation for the standard enthalpy change of formation of propan-1-ol is: \(3\text{C}(\text{graphite}) + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_3\text{H}_8\text{O}(\text{l})\). Using Hess's Law and standard enthalpies of combustion: \(\Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). Therefore, \(\Delta H^\theta_f = [3 \times \Delta H^\theta_c(\text{C}) + 4 \times \Delta H^\theta_c(\text{H}_2)] - \Delta H^\theta_c(\text{C}_3\text{H}_8\text{O})\). Substituting the given values: \(\Delta H^\theta_f = [3(-394) + 4(-286)] - (-2021) = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\).

1 mark for the correct calculation of standard enthalpy of formation (-305 kJ mol^-1). Reject positive value (+305 kJ mol^-1) and arithmetic errors like -4347 kJ mol^-1.

Convert all variables to SI units: \(T = 127 + 273 = 400\text{ K}\), \(V = 110\text{ cm}^3 = 110 \times 10^{-6}\text{ m}^3 = 1.10 \times 10^{-4}\text{ m}^3\), and \(P = 1.01 \times 10^5\text{ Pa}\). Use the ideal gas equation, \(pV = nRT\), to find the number of moles: \(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 1.10 \times 10^{-4}}{8.31 \times 400} = \frac{11.11}{3324} \approx 0.003342\text{ mol}\). Finally, calculate the relative molecular mass: \(M_r = \frac{\text{mass}}{n} = \frac{0.281\text{ g}}{0.003342\text{ mol}} \approx 84.1\).

1 mark for the correct molecular mass of 84.1. Reject 26.7 (where temperature in Celsius was used instead of Kelvin).

The rate of hydrolysis of halogenoalkanes is determined by the strength of the C-X bond: C-I is weaker than C-Cl, so iodoalkanes react faster. The mechanism (\(\text{S}_\text{N}1\) vs \(\text{S}_\text{N}2\)) depends on the structure: tertiary halogenoalkanes react primarily via the \(\text{S}_\text{N}1\) pathway because of the stability of the tertiary carbocation intermediate due to the electron-donating inductive effect of three alkyl groups. Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest and primarily via an \(\text{S}_\text{N}1\) mechanism.

1 mark for identifying 2-iodo-2-methylpropane as the correct option. Reject options containing primary halogenoalkanes (1-iodobutane, 1-chlorobutane) which undergo S_N2, or chloroalkanes which react more slowly.

Calculate the equilibrium moles: Initial moles are \(2.00\text{ mol}\) for \(\text{SO}_2\) and \(1.00\text{ mol}\) for \(\text{O}_2\). Since \(70.0\%\) of \(\text{SO}_2\) reacts, \(2.00 \times 0.70 = 1.40\text{ mol}\) of \(\text{SO}_2\) is converted. At equilibrium, \(n(\text{SO}_2) = 2.00 - 1.40 = 0.60\text{ mol}\), \(n(\text{O}_2) = 1.00 - 0.70 = 0.30\text{ mol}\), and \(n(\text{SO}_3) = 1.40\text{ mol}\). Expressing concentrations in terms of volume \(V\): \([\text{SO}_2] = \frac{0.60}{V}\), \([\text{O}_2] = \frac{0.30}{V}\), and \([\text{SO}_3] = \frac{1.40}{V}\). Substituting these into the equilibrium expression: \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(1.40/V)^2}{(0.60/V)^2 \times (0.30/V)} = \frac{1.96}{0.108} V \approx 18.1 V\).

1 mark for the correct algebraic expression showing 18.1V. Reject expressions containing 1/V or incorrect coefficient calculations.

Reaction with 2,4-DNPH indicates the presence of a carbonyl group, ruling out the alcohol but-3-en-2-ol. The lack of reaction with Tollens' reagent indicates a ketone rather than an aldehyde, which rules out butanal and methylpropanal. The positive reaction with alkaline aqueous iodine (triiodomethane test) indicates a methyl ketone, containing the \(\text{CH}_3\text{CO}-\) group. Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) has this structure, so it is compound \(\mathbf{X}\).

1 mark for identifying butanone. Reject aldehydes (butanal, methylpropanal) and alcohols (but-3-en-2-ol).

The solubility of Group 2 hydroxides increases down the group, so statement C is correct. The solubility of Group 2 sulfates decreases down the group, meaning statement A is incorrect. The thermal stability of Group 2 nitrates increases down the group as the larger cation size decreases polarization of the nitrate anion, making statement B incorrect. Barium reacts more vigorously with cold water than magnesium because its outer electrons are further from the nucleus and more easily lost, making statement D incorrect.

1 mark for selecting the correct statement regarding hydroxide solubility (C). Reject other choices based on incorrect trends.

In the atmospheric catalytic oxidation of sulfur dioxide, nitrogen monoxide is oxidized to nitrogen dioxide in the first step: \(2\text{NO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}_2(\text{g})\). In the second step, nitrogen dioxide oxidizes sulfur dioxide and is reduced back to nitrogen monoxide: \(\text{NO}_2(\text{g}) + \text{SO}_2(\text{g}) \rightarrow \text{NO}(\text{g}) + \text{SO}_3(\text{g})\). Thus, nitrogen monoxide is regenerated in this step.

1 mark for selecting the correct step showing regeneration of NO (B). Reject choices representing standard formation or consumption of NO.

1. Understand that aqueous sodium hydroxide absorbs carbon dioxide. The decrease in volume from \(50\text{ cm}^3\) to \(20\text{ cm}^3\) means that \(30\text{ cm}^3\) of \(CO_2\) was produced. The remaining \(20\text{ cm}^3\) is unreacted excess oxygen. 2. Calculate the volume of oxygen reacted: \(70\text{ cm}^3\text{ (initial)} - 20\text{ cm}^3\text{ (excess)} = 50\text{ cm}^3\) of oxygen reacted. 3. Use the general equation for the combustion of a hydrocarbon: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). Since volumes of gases at constant temperature and pressure are proportional to their moles: \(10\text{ cm}^3\) of \(C_xH_y\) reacts to produce \(30\text{ cm}^3\) of \(CO_2\). Therefore, \(x = 3\). 4. Use the oxygen stoichiometry: \(10(x + \frac{y}{4}) = 50\), which simplifies to \(x + \frac{y}{4} = 5\). Since \(x = 3\), \(3 + \frac{y}{4} = 5\), which gives \(y = 8\). The molecular formula is \(C_3H_8\).

[1 mark] awarded for selecting option D. Awarded for correctly identifying the volumes of carbon dioxide produced (30 cm3) and oxygen reacted (50 cm3), then using stoichiometry to determine the molecular formula.

The target equation is the formation of liquid ethanol: \(2C(graphite) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\). According to Hess's Law: \(\Delta H^\ominus_f = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})\). Therefore: \(\Delta H^\ominus_f = [2 \times (-393.5) + 3 \times (-285.8)] - [-1367.3] = [-787.0 - 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

[1 mark] awarded for selecting option A. Awarded for setting up the Hess's Law cycle correctly multiplying the enthalpy values of graphite and hydrogen by their stoichiometric coefficients, subtracting the combustion of ethanol, and arriving at -277.1 kJ mol-1.

Element \(X\) is aluminium. Aluminium oxide, \(Al_2O_3\), is amphoteric, meaning it is insoluble in water but reacts with both acids (dilute \(HCl\)) and bases (aqueous \(NaOH\)). Aluminium chloride, \(AlCl_3\), undergoes vigorous hydrolysis with water, liberating hydrogen chloride gas (which is seen as white fumes) and forming an acidic solution of hexaaquaaluminium ions and chloride ions.

[1 mark] awarded for selecting option B. Correctly identifying aluminium based on the amphoteric nature of its oxide and the hydrolysing behavior of its chloride.

The rate of hydrolysis of halogenoalkanes is determined by the C-X bond strength (C-Br bonds are weaker than C-Cl bonds and therefore break more rapidly). This narrows the choices to the two bromoalkanes: 1-bromobutane and 2-bromo-2-methylpropane. Under neutral/polar solvent conditions, tertiary halogenoalkanes like 2-bromo-2-methylpropane undergo nucleophilic substitution via the \(S_N1\) mechanism, which is much faster than the \(S_N2\) mechanism undergone by primary halogenoalkanes due to the stable tertiary carbocation intermediate formed in the rate-determining step. Thus, 2-bromo-2-methylpropane reacts most rapidly to yield a cream precipitate of silver bromide.

[1 mark] awarded for selecting option C. Deducting that bromoalkanes react faster than chloroalkanes and tertiary halogenoalkanes undergo faster hydrolysis via S_N1 than primary halogenoalkanes.

1. The reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH) to form an orange precipitate indicates the presence of a carbonyl group (aldehyde or ketone). 2. The negative reaction with Tollens' reagent rules out aldehydes, meaning \(Y\) must be a ketone. 3. The positive triiodomethane (iodoform) test with alkaline aqueous iodine (yellow precipitate of \(CHI_3\)) indicates the presence of a methyl ketone group (\(CH_3CO-\)). 4. The only 4-carbon ketone is butanone (\(CH_3COCH_2CH_3\)), which is a methyl ketone and satisfies all requirements.

[1 mark] awarded for selecting option B. Explaining that 2,4-DNPH identifies a carbonyl group, Tollens' rules out aldehydes, and alkaline aqueous iodine confirms a methyl ketone structure.

Using the ideal gas equation: \(pV = nRT = \frac{m}{M_r}RT\), which rearranges to \(M_r = \frac{mRT}{pV}\). Convert all parameters to SI units: mass \(m = 0.282\text{ g}\), temperature \(T = 97 + 273 = 370\text{ K}\), pressure \(p = 1.01 \times 10^5\text{ Pa}\), volume \(V = 85.0 \times 10^{-6}\text{ m}^3\). Calculate: \(M_r = \frac{0.282 \times 8.31 \times 370}{1.01 \times 10^5 \times 85.0 \times 10^{-6}} = \frac{866.99}{8.585} = 101.0\text{ g mol}^{-1}\).

[1 mark] awarded for selecting option C. Correct conversion of units (especially temperature to Kelvin and volume to cubic meters) and precise calculation using the ideal gas equation.

Set up an ICE (Initial, Change, Equilibrium) table:
- Initial moles: \(SO_2 = 2.00\), \(O_2 = 2.00\), \(SO_3 = 0.00\).
- Change in moles: since \(1.60\text{ mol}\) of \(SO_3\) is formed, \(\Delta SO_3 = +1.60\), \(\Delta SO_2 = -1.60\), and \(\Delta O_2 = -0.80\).
- Equilibrium moles: \(SO_3 = 1.60\), \(SO_2 = 2.00 - 1.60 = 0.40\), \(O_2 = 2.00 - 0.80 = 1.20\).
- Equilibrium concentrations in \(\text{mol dm}^{-3}\): \([SO_3] = \frac{1.60}{V}\), \([SO_2] = \frac{0.40}{V}\), \([O_2] = \frac{1.20}{V}\).
- Expression for \(K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(1.60/V)^2}{(0.40/V)^2 \times (1.20/V)} = \frac{2.56 / V^2}{(0.16 / V^2) \times (1.20 / V)} = \frac{2.56}{0.192 / V} = 13.3 V\text{ dm}^3\text{ mol}^{-1}\).

[1 mark] awarded for selecting option B. Calculated the equilibrium moles of each species correctly, established the concentration terms containing V, and simplified the algebra to get 13.3 V.

The reaction occurring is: \(2NO(g) + 2CO(g) \rightarrow N_2(g) + 2CO_2(g)\). Looking at oxidation states: \(N\) in \(NO\) is +2, and in \(N_2\) it is 0. Thus, nitrogen is reduced and its oxidation state decreases from +2 to 0, which makes option B correct. Carbon monoxide acts as a reducing agent (as it is oxidised from +2 to +4 in \(CO_2\)), ruling out option A. The catalysts used in catalytic converters are transition metals (solid phase), so the reaction is an example of heterogeneous catalysis, ruling out option C. Nitrogen monoxide is reduced to nitrogen gas, not oxidised, ruling out option D.

[1 mark] awarded for selecting option B. Demonstrated clear understanding of the redox changes in a catalytic converter reaction.

(a) (i) \(n(\text{MnO}_4^-) = 0.0200 \times \frac{25.00}{1000} = 5.00 \times 10^{-4}\text{ mol}\)
(ii) From the equation, 1 mole of \(\text{MnO}_4^-\) reacts with 5 moles of \(\text{Fe}^{2+}\).
\(n(\text{Fe}^{2+})\text{ in } 25.0\text{ cm}^3 = 5 \times 5.00 \times 10^{-4} = 2.50 \times 10^{-3}\text{ mol}\)
(iii) \(n(\text{Fe}^{2+})\text{ in } 250.0\text{ cm}^3 = 2.50 \times 10^{-3} \times 10 = 2.50 \times 10^{-2}\text{ mol}\)
(iv) Mass of pure hydrated iron(II) sulfate in the sample: \(7.31 \times 0.950 = 6.9445\text{ g}\)
Molar mass of pure salt: \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{6.9445}{2.50 \times 10^{-2}} = 277.78\text{ g mol}^{-1}\) (accept \(278\text{ g mol}^{-1}\))
(v) \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (16.0 \times 4) = 151.9\)
Mass of water of crystallization: \(277.78 - 151.9 = 125.88\)
\(x = \frac{125.88}{18.0} = 6.99 \approx 7\)

(b) (i) To balance the equation:
Chromium is reduced: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Iodide is oxidized: \(2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-\)
Multiplying the oxidation half-equation by 3 and adding gives:
\(\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}\)
(ii) Oxidizing agent: \(\text{Cr}_2\text{O}_7^{2-}\) (or dichromate(VI) ion)
Explanation: The oxidation state of chromium decreases from \(+6\) in \(\text{Cr}_2\text{O}_7^{2-}\) to \(+3\) in \(\text{Cr}^{3+}\) (decrease in oxidation number indicates reduction, which means it behaves as an oxidizing agent).

(a) (i) [1] Award 1 mark for correct calculation of moles of \(\text{MnO}_4^-\): \(5.00 \times 10^{-4}\text{ mol}\).
(ii) [1] Award 1 mark for multiplying moles of \(\text{MnO}_4^-\right.\) by 5: \(2.50 \times 10^{-3}\text{ mol}\).
(iii) [1] Award 1 mark for multiplying by 10: \(2.50 \times 10^{-2}\text{ mol}\).
(iv) [3] Award 1 mark for calculating the mass of pure salt (\(6.9445\text{ g}\)). Award 1 mark for dividing pure mass by total moles. Award 1 mark for obtaining the correct molar mass (\(277.8\text{ g mol}^{-1}\) or \(278\text{ g mol}^{-1}\)).
(v) [2] Award 1 mark for subtracting 151.9 from their calculated molar mass. Award 1 mark for dividing by 18.0 to obtain \(x = 7\) (must be an integer).

(b) (i) [2] Award 1 mark for correct formulas and species. Award 1 mark for correct stoichiometric coefficients: \(1, 6, 14 \rightarrow 2, 3, 7\).
(ii) [2] Award 1 mark for identifying \(\text{Cr}_2\text{O}_7^{2-}\) (or dichromate(VI)). Award 1 mark for explaining that chromium's oxidation number decreases from \(+6\) to \(+3\).

(a) (i) Standard enthalpy change of formation is the enthalpy change when 1 mole of a compound [1] is formed from its constituent elements in their standard states under standard conditions [1].
(ii) \(3\text{C}(\text{s}) + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{CH(OH)CH}_3(\text{l})\) (State symbols required [1], correct balanced equation [1]).
(iii) According to Hess's Law:
\(\Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\)
\(\Delta H^\theta_f = [3 \times \Delta H^\theta_c(\text{C}) + 4 \times \Delta H^\theta_c(\text{H}_2)] - \Delta H^\theta_c(\text{propan-2-ol})\)
\(\Delta H^\theta_f = [3 \times (-393.5) + 4 \times (-285.8)] - (-2006.0)\)
\(\Delta H^\theta_f = [-1180.5 - 1143.2] + 2006.0\)
\(\Delta H^\theta_f = -2323.7 + 2006.0 = -317.7\text{ kJ mol}^{-1}\)

(b) (i) Moles of propan-2-ol = \(\frac{1.20}{60.0} = 0.0200\text{ mol}\)
Heat released, \(q = \text{moles} \times \Delta H^\theta_c = 0.0200 \times 2006.0\text{ kJ mol}^{-1} = 40.12\text{ kJ} = 40120\text{ J}\)
Using \(q = m c \Delta T\):
\(40120 = 150 \times 4.18 \times \Delta T\)
\(40120 = 627 \times \Delta T\)
\ \Delta T = \frac{40120}{627} = 64.0\text{ K}\) (or \(64.0^\circ\text{C}\))
(ii) Any two of:
- Heat is lost to the surrounding air or to the glass calorimeter.
- Incomplete combustion of the propan-2-ol (forming carbon or carbon monoxide).
- Evaporation of the fuel (propan-2-ol) from the wick before/after burning.
- Non-standard conditions.

(a) (i) [2] Award 1 mark for "formation of 1 mole of compound". Award 1 mark for "from elements in their standard states under standard conditions".
(ii) [2] Award 1 mark for correct formulas and balancing. Award 1 mark for correct state symbols: \(\text{C}(\text{s})\), \(\text{H}_2(\text{g})\), \(\text{O}_2(\text{g})\), and \(\text{CH}_3\text{CH(OH)CH}_3(\text{l})\).
(iii) [3] Award 1 mark for setting up the correct cycle or mathematical pathway. Award 1 mark for substituting values correctly: \(3(-393.5) + 4(-285.8) - (-2006.0)\). Award 1 mark for the correct calculation value of \(-317.7\text{ kJ mol}^{-1}\).

(b) (i) [3] Award 1 mark for calculating \(0.0200\text{ mol}\) of propan-2-ol. Award 1 mark for calculating the heat energy released, \(40120\text{ J}\) (or \(40.12\text{ kJ}\)). Award 1 mark for correct calculation of temperature rise, \(64.0\text{ K}\) (or \(64.0^\circ\text{C}\)).
(ii) [2] Award 1 mark each for any two reasonable sources of experimental error (e.g., heat loss to surroundings, incomplete combustion, fuel evaporation).

(a) (i) \(K_p = \frac{p^2(\text{SO}_3)}{p^2(\text{SO}_2) \cdot p(\text{O}_2)}\)
Units: \(\frac{\text{kPa}^2}{\text{kPa}^2 \cdot \text{kPa}} = \text{kPa}^{-1}\)
(ii) Position of equilibrium: shifts to the left / reactants side [1], because the forward reaction is exothermic (and an increase in temperature favors the endothermic direction) [1].
Value of \(K_p\): decreases [1], because the concentration of products at equilibrium decreases while the concentration of reactants increases.
(iii) A high pressure of \(200\text{ atm}\) is not used because:
- Generating and maintaining high pressures is very expensive (high capital/operating costs) and requires reinforced equipment [1].
- The conversion/yield of \(\text{SO}_3\) is already very high (approx. \(98\%\)) at low pressures of \(1-2\text{ atm}\), so higher pressure is unnecessary [1].

(b) Initial concentrations:
\([\text{SO}_2]_{\text{initial}} = \frac{0.400\text{ mol}}{2.00\text{ dm}^3} = 0.200\text{ mol dm}^{-3}\)
\([\text{O}_2]_{\text{initial}} = \frac{0.300\text{ mol}}{2.00\text{ dm}^3} = 0.150\text{ mol dm}^{-3}\)

At equilibrium:
\([\text{SO}_3]_{\text{eq}} = 0.150\text{ mol dm}^{-3}\)
Since \(2\text{ moles of } \text{SO}_2\) react to form \(2\text{ moles of } \text{SO}_3\):
\([\text{SO}_2]_{\text{eq}} = 0.200 - 0.150 = 0.050\text{ mol dm}^{-3}\)

Since \(1\text{ mole of } \text{O}_2\) reacts to form \(2\text{ moles of } \text{SO}_3\):
\([\text{O}_2]_{\text{eq}} = 0.150 - \frac{0.150}{2} = 0.150 - 0.075 = 0.075\text{ mol dm}^{-3}\)

Expression for \(K_c\):
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\)

Substituting equilibrium concentrations:
\(K_c = \frac{(0.150)^2}{(0.050)^2 \times 0.075} = \frac{0.0225}{0.0025 \times 0.075} = \frac{0.0225}{0.0001875} = 120\)

Units:
\(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \cdot (\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\)

(a) (i) [2] Award 1 mark for the correct expression for \(K_p\). Award 1 mark for correct units of \(text{kPa}^{-1}\).
(ii) [3] Award 1 mark for stating equilibrium shifts left. Award 1 mark for stating that the forward reaction is exothermic. Award 1 mark for stating that \(K_p\) decreases.
(iii) [2] Award 1 mark for mentioning high costs/economic factors or safety risks of high pressure. Award 1 mark for stating that high conversion is already achieved at near-atmospheric pressure.

(b) [5] Award 1 mark for calculating both initial concentrations (\([\text{SO}_2] = 0.200\text{ mol dm}^{-3}\) and \([\text{O}_2] = 0.150\text{ mol dm}^{-3}\)). Award 1 mark for correct calculation of equilibrium concentration of \(\text{SO}_2\) (\(0.050\text{ mol dm}^{-3}\)). Award 1 mark for correct calculation of equilibrium concentration of \(\text{O}_2\) (\(0.075\text{ mol dm}^{-3}\)). Award 1 mark for correct calculated value of \(120\). Award 1 mark for the correct units (\(\text{dm}^3\text{ mol}^{-1}\)).

(a) (i) Reaction of sodium oxide with water:
\(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq})\) [1]
Approximate pH: \(13 - 14\) [1]
Reaction of phosphorus(V) oxide with water:
\(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\) [1]
Approximate pH: \(1 - 2\) [1]

(ii) Amphoteric means the oxide can react with both acids and bases to form salts [1].
Acidic behavior (reacts with bases):
\(\text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaAl(OH)}_4(\text{aq})\)
(or \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\))
OR
Basic behavior (reacts with acids):
\(\text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)
(or \(\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}\)) [1]
Explanation: Aluminium oxide has intermediate ionic-covalent bonding character due to high charge density of \(\text{Al}^{3+}\) polarizing oxide ions [1].

(b) (i) Trend: Thermal stability increases down the group [1].
Explanation: Cationic radius increases down the group while ionic charge remains \(+2\), leading to a decrease in cationic charge density [1]. As a result, the larger cations down the group have less polarizing effect on the carbonate ion (the \(\text{CO}_3^{2-}\) electron cloud is distorted less), which makes the carbon-oxygen bond stronger and harder to break [1].
(ii) \(2\text{Ca(NO}_3)_2(\text{s}) \rightarrow 2\text{CaO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\) [2] (1 mark for correct products/reactants, 1 mark for balancing).

(a) (i) [4] Award 1 mark for the correct balanced equation for sodium oxide with water. Award 1 mark for correct pH (13 or 14). Award 1 mark for the correct balanced equation for phosphorus(V) oxide with water. Award 1 mark for correct pH (1 or 2).
(ii) [3] Award 1 mark for defining amphoteric (reacts with acids and bases). Award 1 mark for a correct balanced equation with either acid or base. Award 1 mark for mentioning the polarization/bonding character explanation.

(b) (i) [3] Award 1 mark for identifying the trend (stability increases down the group). Award 1 mark for explaining the increase in cationic radius / decrease in charge density of Group 2 cations. Award 1 mark for explaining that the carbonate ion is polarized less by the larger cation down the group.
(ii) [2] Award 1 mark for correct reactants and products (\(\text{CaO}\), \(\text{NO}_2\), \(\text{O}_2\)). Award 1 mark for correct balancing.

(a) (i) The reaction with 2,4-DNPH indicates the presence of a carbonyl group, \(\text{C}=\text{O}\) (it is an aldehyde or ketone) [1].
The lack of reaction with Tollens' reagent indicates that **X** is not an aldehyde (so it must be a ketone) [1].
(ii) Structure: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) [1]
IUPAC name: butanone (or butan-2-one) [1]
(iii) Equation: \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)} \text{CH}_2\text{CH}_3\) [1]
Organic product: butan-2-ol [1]

(b) (i)
- The broad absorption at \(3350\text{ cm}^{-1}\) indicates the presence of an alcohol group (\(\text{O}-\text{H}\) bond) [1].
- The sharp absorption at \(1645\text{ cm}^{-1}\) indicates the presence of an alkene group (\(\text{C}=\text{C}\) bond) [1].
- (The lack of absorption at \(1680 - 1750\text{ cm}^{-1}\) confirms no carbonyl \(\text{C}=\text{O}\) group is present.)
(ii) Possible structures include:
\(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}\) (but-3-en-1-ol),
\(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2\text{OH}\) (but-2-en-1-ol), or
\(\text{CH}_2=\text{C(CH}_3)\text{CH}_2\text{OH}\) (2-methylprop-2-en-1-ol).
Award 1 mark for any valid unsaturated alcohol formula matching \(\text{C}_4\text{H}_8\text{O}\).

(c) (i) **Z** must be 2-chlorobutane, \(\text{CH}_3\text{CH(Cl)CH}_2\text{CH}_3\) [1].
(ii) Step 1: React **Z** with aqueous sodium hydroxide (\(\text{NaOH}(\text{aq})\)), heat under reflux (nucleophilic substitution to form butan-2-ol) [1].
Step 2: Oxidize butan-2-ol using acidified potassium dichromate(VI) (\(\text{H}^+ / \text{Cr}_2\text{O}_7^{2-}\)), heat under reflux (to yield butanone) [1].

(a) (i) [2] Award 1 mark for identifying the carbonyl group (\(\text{C}=\text{O}\)) from 2,4-DNPH. Award 1 mark for concluding that it is a ketone / not an aldehyde due to the negative Tollens' test.
(ii) [2] Award 1 mark for drawing the correct structure of butanone. Award 1 mark for the correct name "butanone" or "butan-2-one".
(iii) [2] Award 1 mark for the balanced equation using \(2[\text{H}]\) (or \([\text{H}]\) if showing reduction process schematically). Award 1 mark for identifying the product as "butan-2-ol".

(b) (i) [2] Award 1 mark for identifying \(\text{O}-\text{H}\) / alcohol at \(3350\text{ cm}^{-1}\). Award 1 mark for identifying \(\text{C}=\text{C}\) / alkene at \(1645\text{ cm}^{-1}\).
(ii) [1] Award 1 mark for drawing any valid acyclic structure that contains both a carbon-carbon double bond and a hydroxyl group.

(c) (i) [1] Award 1 mark for drawing the structural formula of 2-chlorobutane.
(ii) [2] Award 1 mark for Step 1 reagents and conditions: \(\text{NaOH}(\text{aq})\) / heat under reflux. Award 1 mark for Step 2 reagents and conditions: acidified potassium dichromate(VI) (\(\text{H}^+/\text{Cr}_2\text{O}_7^{2-}\)) / heat under reflux (or distillation).

Detailed solution steps: (a) \(\text{H}_2\text{C}_3\text{H}_2\text{O}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{C}_3\text{H}_2\text{O}_4\text{(aq)} + 2\text{H}_2\text{O(l)}\). (b) Moles of \(\text{NaOH}\) = concentration \(\times\) volume = \(0.100\text{ mol dm}^{-3} \times (25.00 / 1000)\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\). (c) Since malonic acid is diprotic, 1 mole of \(\text{H}_2\text{A}\) reacts with 2 moles of \(\text{NaOH}\). Moles of \(\text{H}_2\text{C}_3\text{H}_2\text{O}_4\) in \(25.0\text{ cm}^3 = 2.50 \times 10^{-3}\text{ mol} / 2 = 1.25 \times 10^{-3}\text{ mol}\). (d) Concentration of FA 1 = \((1.25 \times 10^{-3}\text{ mol} / 25.0\text{ cm}^3) \times 1000\text{ cm}^3\text{ dm}^{-3} = 0.0500\text{ mol dm}^{-3}\). (e) Mass of pure malonic acid in \(1.00\text{ dm}^3 = \text{moles} \times \text{molar mass} = 0.0500\text{ mol} \times 104.0\text{ g mol}^{-1} = 5.20\text{ g}\). Percentage purity = \((\text{mass of pure acid} / \text{mass of impure sample}) \times 100\% = (5.20\text{ g} / 6.50\text{ g}) \times 100\% = 80.0\%\). (f) A titration titre requires two burette readings (initial and final). Total uncertainty in the titre = \(2 \times (\pm 0.05\text{ cm}^3) = \pm 0.10\text{ cm}^3\). Maximum percentage uncertainty = \((0.10 / 25.00) \times 100\% = 0.40\%\).

Marking Scheme: (a) [2 marks] Correct balanced molecular equation with appropriate formulas and balancing. (b) [2 marks] Moles of NaOH calculated correctly as \(2.50 \times 10^{-3}\text{ mol}\) with working shown. (c) [2 marks] Correct use of 1:2 stoichiometry to find moles of malonic acid as \(1.25 \times 10^{-3}\text{ mol}\). (d) [2 marks] Correct concentration of malonic acid as \(0.0500\text{ mol dm}^{-3}\) expressed to 3 significant figures. (e) [2 marks] Mass of pure malonic acid calculated as \(5.20\text{ g}\) and percentage purity correctly evaluated as \(80.0\%\). (f) [3.33 marks] Correct recognition of two readings needed for a titre (giving total uncertainty of \(\pm 0.10\text{ cm}^3\)) and calculation of percentage uncertainty as \(0.40\%\). (Award 1.5 marks if single-reading uncertainty of 0.05 is used to get 0.20%).

Detailed solution steps: (a) From t = 5 to 10 min, the temperature decreases linearly by \(1.0\text{ }^\circ\text{C}\) per minute. Extrapolating back from 5 min (\(42.5\text{ }^\circ\text{C}\)) to 4 min gives an extrapolated maximum temperature of \(42.5 + 1.0 = 43.5\text{ }^\circ\text{C}\). The initial temperature is \(21.0\text{ }^\circ\text{C}\). Therefore, \(\Delta T = 43.5 - 21.0 = 22.5\text{ }^\circ\text{C}\). (b) Heat energy \(q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 22.5\text{ K} = 4702.5\text{ J} = 4.70\text{ kJ}\) (to 3 sig figs). (c) Moles of \(\text{CuSO}_4\) = volume \(\times\) concentration = \((50.0 / 1000) \times 0.800 = 0.0400\text{ mol}\). Enthalpy change \(\Delta H = -q / n = -4.7025\text{ kJ} / 0.0400\text{ mol} = -117.56\text{ kJ mol}^{-1}\), which rounds to \(-118\text{ kJ mol}^{-1}\) (to 3 sig figs). (d) Polystyrene is an excellent thermal insulator compared to glass, which minimizes heat loss to the surroundings during the experiment. Without correcting for heat loss, the uncorrected maximum recorded temperature (at t = 5 min) would be lower than the true maximum possible temperature, thereby underestimating \(\Delta T\) and making the calculated \(\Delta H\) less exothermic (less negative) than its actual value.

Marking Scheme: (a) [3.33 marks] Correct extrapolation of the linear cooling curve back to 4 min to get maximum temperature of \(43.5\text{ }^\circ\text{C}\) and correct temperature rise \(\Delta T = 22.5\text{ }^\circ\text{C}\). (b) [2 marks] Correct calculation of heat energy \(q = 4.70\text{ kJ}\) (or \(4.7025\text{ kJ}\)) using mass of 50.0 g and correct formula. (c) [3 marks] Correct calculation of moles of CuSO4 (\(0.0400\text{ mol}\)) and correct division to find \(\Delta H = -118\text{ kJ mol}^{-1}\) (must include negative sign and be to 3 significant figures). (d) [5 marks] Explains that polystyrene is a poor thermal conductor/good insulator that reduces heat loss; explains that heat loss lowers the recorded maximum temperature rise, leading to a calculated \(\Delta H\) that is less exothermic than the true value.

Detailed analysis of observations: (a) FA 1: The white precipitate formed with both \(\text{NaOH}\) and \(\text{NH}_3\), which is insoluble in excess of either reagent, indicates the presence of magnesium ions, \(\text{Mg}^{2+}\). The white precipitate with \(\text{AgNO}_3\) that dissolves in dilute \(\text{NH}_3\) indicates chloride ions, \(\text{Cl}^-\). Hence, FA 1 is magnesium chloride, \(\text{MgCl}_2\). FA 2: The green precipitate formed with both \(\text{NaOH}\) and \(\text{NH}_3\), turning brown on standing, indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The white precipitate with \(\text{Ba(NO}_3)_2\) which is insoluble in dilute \(\text{HNO}_3\) indicates sulfate ions, \(\text{SO}_4^{2-}\). Hence, FA 2 is iron(II) sulfate, \(\text{FeSO}_4\). FA 3: No precipitate with either hydroxide or ammonia, but the evolution of an alkaline gas on warming with \(\text{NaOH}\), indicates ammonium ions, \(\text{NH}_4^+\). The cream precipitate with \(\text{AgNO}_3\), insoluble in dilute ammonia but soluble in concentrated ammonia, indicates bromide ions, \(\text{Br}^-\). Hence, FA 3 is ammonium bromide, \(\text{NH}_4\text{Br}\). (b) Ionic equation for Test 1 with FA 1: \(\text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Mg(OH)}_2\text{(s)}\). (c) Ionic equation for Test 3 with FA 2: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). (d) The evolved gas is ammonia, \(\text{NH}_3\). The ionic equation for its formation is: \(\text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{NH}_3\text{(g)} + \text{H}_2\text{O(l)}\).

Marking Scheme: (a) [6 marks] 1 mark for each correctly identified ion with its correct chemical formula: FA 1 cation = \(\text{Mg}^{2+}\), anion = \(\text{Cl}^-\); FA 2 cation = \(\text{Fe}^{2+}\), anion = \(\text{SO}_4^{2-}\); FA 3 cation = \(\text{NH}_4^+\), anion = \(\text{Br}^-\). (b) [2 marks] Correct ionic equation with state symbols: \(\text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Mg(OH)}_2\text{(s)}\). (c) [2 marks] Correct ionic equation with state symbols: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). (d) [3.33 marks] Correct identification of ammonia gas, \(\text{NH}_3\) (1.33 marks), and correct ionic equation for its formation: \(\text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{NH}_3\text{(g)} + \text{H}_2\text{O(l)}\). (2 marks).