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The addition of concentrated hydrochloric acid results in a ligand exchange reaction where water ligands are replaced by chloride ligands:

\([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\)

The larger chloride ions cause steric hindrance, meaning only four can fit around the cobalt(II) ion, resulting in a tetrahedral geometry for the blue \([CoCl_4]^{2-}\) complex.

1 mark: Correct option chosen (B).
0 marks: Any other choice or blank.

To find the empirical formula, calculate the molar ratio of the elements:

\(n(\text{Cu}) = \frac{40.0}{63.5} = 0.630\text{ mol}\)

\(n(\text{S}) = \frac{20.0}{32.1} = 0.623\text{ mol}\)

\(n(\text{O}) = \frac{40.0}{16.0} = 2.50\text{ mol}\)

Divide each value by the smallest value (0.623):

\(\text{Cu} = \frac{0.630}{0.623} \approx 1\)

\(\text{S} = \frac{0.623}{0.623} = 1\)

\(\text{O} = \frac{2.50}{0.623} \approx 4\)

The empirical formula is \(\text{CuSO}_4\).

1 mark: Correct option chosen (B).
0 marks: Any other choice or blank.

First, methylbenzene is synthesized via a Friedel-Crafts alkylation of benzene with chloromethane in the presence of an anhydrous aluminium chloride catalyst.

Then, methylbenzene is oxidized to benzoic acid using alkaline potassium manganate(VII), heating under reflux, followed by acidification with dilute hydrochloric acid.

1 mark: Correct option chosen (A).
0 marks: Any other choice or blank.

Since the reaction is zero order with respect to \(B\), changing the concentration of \(B\) has no effect on the rate.

Since the reaction is second order with respect to \(A\), halving the concentration of \(A\) scales the rate by a factor of \(\left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25\).

1 mark: Correct option chosen (A).
0 marks: Any other choice or blank.

In the Contact Process, \(\text{V}_2\text{O}_5\) first oxidizes \(\text{SO}_2\) to \(\text{SO}_3\) and is reduced to \(\text{VO}_2\) (where vanadium is in the +4 oxidation state):

\(\text{V}_2\text{O}_5 + \text{SO}_2 \rightarrow 2\text{VO}_2 + \text{SO}_3\)

In the second step, vanadium(IV) in \(\text{VO}_2\) is oxidized back to the +5 state in \(\text{V}_2\text{O}_5\) by reacting with oxygen:

\(2\text{VO}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{V}_2\text{O}_5\)

1 mark: Correct option chosen (B).
0 marks: Any other choice or blank.

1. Calculate moles of \(\text{CaCO}_3\):
\(n(\text{CaCO}_3) = \frac{10.0}{100.1} = 0.0999\text{ mol}\)

2. Moles of \(\text{CO}_2\) produced = \(0.0999\text{ mol}\)

3. Use the ideal gas equation \(PV = nRT\) where:
\(P = 100\text{ kPa} = 100,000\text{ Pa}\)
\(T = 300\text{ K}\)

\(V = \frac{nRT}{P} = \frac{0.0999 \times 8.31 \times 300}{100,000} = 2.49 \times 10^{-3}\text{ m}^3\)

1 mark: Correct option chosen (A).
0 marks: Any other choice or blank.

Acyl chlorides (such as propanoyl chloride) are highly reactive towards nucleophilic acyl substitution due to the strongly electron-withdrawing chlorine atom and the fact that chloride is a very good leaving group. Propanoyl chloride reacts rapidly with alcohols like ethanol at room temperature without requiring a catalyst.

1 mark: Correct option chosen (B).
0 marks: Any other choice or blank.

From the logarithmic form of the Arrhenius equation:

\(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\)

Therefore, the gradient of the line is \(-\frac{E_a}{R}\).

\(\text{gradient} = -1.20 \times 10^4\text{ K}\)

\(-\frac{E_a}{R} = -1.20 \times 10^4\)

\(E_a = 1.20 \times 10^4 \times 8.31 = 99,720\text{ J mol}^{-1}\)

To convert to \(\text{kJ mol}^{-1}\):

\(E_a = \frac{99,720}{1000} = +99.7\text{ kJ mol}^{-1}\)

1 mark: Correct option chosen (B).
0 marks: Any other choice or blank.

First, calculate the amount of \(\text{CO}_2\) gas collected in moles: \(n = \frac{72.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.00300\text{ mol}\). Since the reaction stoichiometry between \(\text{MCO}_3\) and \(\text{CO}_2\) is 1:1, the amount of \(\text{MCO}_3\) reacted is also \(0.00300\text{ mol}\). Next, calculate the molar mass of \(\text{MCO}_3\): \(M_r = \frac{0.300\text{ g}}{0.00300\text{ mol}} = 100.0\text{ g\ mol}^{-1}\). Subtract the relative formula mass of the carbonate group (\(\text{CO}_3^{2-}\) is \(12.0 + 3 \times 16.0 = 60.0\text{ g\ mol}^{-1}\)) to find the relative atomic mass of M: \(A_r(\text{M}) = 100.0 - 60.0 = 40.0\text{ g\ mol}^{-1}\). This corresponds to calcium.

1 mark: Correctly identifies Calcium (option B) based on calculation of the relative atomic mass of the metal.

Vanadium has an atomic number of 23, so a neutral vanadium atom has 23 electrons with the ground state electronic configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2\) or \([\text{Ar}] 3d^3 4s^2\). To form the \(\text{V}^{3+}\) ion, three electrons must be removed. Outer 4s electrons are removed first, followed by one 3d electron, which yields the configuration \([\text{Ar}] 3d^2\).

1 mark: Correctly identifies the electronic configuration of the \(\text{V}^{3+}\) ion as \([\text{Ar}] 3d^2\) (option A).

To convert propan-1-ol to the chloroalkane 1-chloropropane, phosphorus(V) chloride (\(\text{PCl}_5\)) is a standard reagent (Reagent X). To convert 1-chloropropane to butanenitrile, a nucleophilic substitution reaction is carried out using potassium cyanide (\(\text{KCN}\)) dissolved in ethanol under reflux (Reagent Y). To convert the nitrile to a carboxylic acid with the same number of carbons, acid hydrolysis is performed by heating with dilute hydrochloric acid under reflux (Reagent Z).

1 mark: Correctly identifies all three reagents in the pathway (option A).

Comparing Experiments 1 and 2: when \([\text{A}]\) is doubled from 0.10 to 0.20 \(\text{mol dm}^{-3}\) while keeping \([\text{B}]\) constant at 0.10 \(\text{mol dm}^{-3}\), the initial rate increases by a factor of 4 (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). This indicates that the reaction is second order with respect to A: \(\text{Rate} \propto [\text{A}]^2\). Comparing Experiments 1 and 3: when \([\text{B}]\) is doubled from 0.10 to 0.20 \(\text{mol dm}^{-3}\) while keeping \([\text{A}]\) constant at 0.10 \(\text{mol dm}^{-3}\), the initial rate remains unchanged at \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). This indicates that the reaction is zero order with respect to B: \(\text{Rate} \propto [\text{B}]^0\). Therefore, the overall rate equation is \(\text{Rate} = k[\text{A}]^2\).

1 mark: Correctly determines the order for both reactants to identify the rate equation (option C).

First, calculate the molecular masses of the reactants and products. The desired product is 1-chlorobutane, \(\text{C}_4\text{H}_9\text{Cl}\): \(M_r = 4(12.0) + 9(1.0) + 35.5 = 92.5\text{ g\ mol}^{-1}\). The reactants are butane (\(\text{C}_4\text{H}_{10}\): \(M_r = 4(12.0) + 10(1.0) = 58.0\text{ g\ mol}^{-1}\)) and chlorine (\(\text{Cl}_2\): \(M_r = 2(35.5) = 71.0\text{ g\ mol}^{-1}\)). The total mass of reactants is \(58.0 + 71.0 = 129.0\text{ g\ mol}^{-1}\). The percentage atom economy is \(\frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\% = \frac{92.5}{129.0} \times 100\% = 71.7\%\).

1 mark: Correctly calculates the percentage atom economy of the reaction (option B).

When excess concentrated hydrochloric acid is added to \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), a ligand substitution reaction occurs: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\). Due to the larger size of the chloride ligands and their mutual electrostatic repulsion, only four chloride ligands coordinate to the copper ion. Thus, the coordination number decreases from 6 to 4, and the geometry changes from octahedral to tetrahedral.

1 mark: Correctly identifies that the coordination number decreases from 6 to 4 and the geometry becomes tetrahedral (option B).

2-methylbutan-2-ol is a tertiary alcohol with the structure \(\text{CH}_3\text{C(OH)(CH}_3)\text{CH}_2\text{CH}_3\). Synthesizing a tertiary alcohol using a Grignard reagent requires reacting a ketone with a Grignard reagent. Reacting the ketone butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) with the Grignard reagent methylmagnesium bromide (\(\text{CH}_3\text{MgBr}\)) followed by hydrolysis introduces a methyl group to the carbonyl carbon, producing the tertiary alcohol 2-methylbutan-2-ol. Reacting aldehydes (A or C) would yield secondary alcohols. Reacting propanone with propylmagnesium bromide (D) would yield 2-methylpentan-2-ol.

1 mark: Correctly identifies butanone and methylmagnesium bromide (option B) as the correct reactants.

The direct reaction between \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) has a high activation energy due to electrostatic repulsion between two negatively charged anions. Because iron is a transition metal with variable oxidation states, \(\text{Fe}^{2+}\) ions can act as a homogeneous catalyst. The \(\text{Fe}^{2+}\) ions react first with \(\text{S}_2\text{O}_8^{2-}\) to form \(\text{Fe}^{3+}\) and sulphate ions. Then, the newly formed \(\text{Fe}^{3+}\) ions react with \(\text{I}^-\) to yield iodine and regenerate the \(\text{Fe}^{2+}\) catalyst. These steps involve reaction between oppositely charged species, which occurs with a much lower activation energy.

1 mark: Correctly identifies that variable oxidation states allow a two-step mechanism that avoids anion-anion repulsion (option B).

1. Find the moles of hydrogen gas produced: \( n(\text{H}_2) = \frac{384\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0160\text{ mol} \). 2. Write the chemical equation for a divalent metal: \( \text{M(s)} + 2\text{HCl(aq)} \rightarrow \text{MCl}_2\text{(aq)} + \text{H}_2\text{(g)} \). Thus, \( 1\text{ mol} \) of \( \text{M} \) produces \( 1\text{ mol} \) of \( \text{H}_2 \), which means \( n(\text{M}) = 0.0160\text{ mol} \). 3. Calculate the mass of pure metal: \( m(\text{M}) = 0.0160\text{ mol} \times 65.4\text{ g mol}^{-1} = 1.0464\text{ g} \). 4. Calculate percentage purity: \( \text{Purity} = \frac{1.0464\text{ g}}{1.20\text{ g}} \times 100\% = 87.2\% \).

1 Mark: Correct calculation of moles of hydrogen, conversion to mass of pure metal, and calculation of percentage purity (87.2%).

Let us look at the outer electronic configurations in their ground state: - \( \text{Fe}^{3+} \) has the configuration \( [\text{Ar}] 3\text{d}^5 \). According to Hund's rule, all 5 electrons occupy the five d-orbitals singly, resulting in 5 unpaired d-electrons. - \( \text{Co}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^7 \), giving 3 unpaired d-electrons (two paired orbitals, three singly occupied). - \( \text{Ni}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^8 \), giving 2 unpaired d-electrons (three paired orbitals, two singly occupied). - \( \text{Cu}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^9 \), giving 1 unpaired d-electron. Thus, \( \text{Fe}^{3+} \) has the greatest number of unpaired d-electrons.

1 Mark: Correct identification of the ground state electronic configurations and selecting the species with the highest number of unpaired d-electrons.

Let the original rate be: \( \text{Rate}_1 = k[\text{A}][\text{B}]^2 \). Under the new conditions: \( [\text{A}]_{\text{new}} = 2[\text{A}] \) and \( [\text{B}]_{\text{new}} = 0.5[\text{B}] \). Substituting these into the rate equation: \( \text{Rate}_2 = k(2[\text{A}])(0.5[\text{B}])^2 = k \cdot 2[\text{A}] \cdot 0.25[\text{B}]^2 = 0.5 k[\text{A}][\text{B}]^2 = 0.5 \text{Rate}_1 \). Therefore, the initial rate of reaction is halved.

1 Mark: Deducing that the change in concentration of A multiplies the rate by 2, and B multiplies it by 0.25, giving an overall factor of 0.5 (halved).

The overall yield is found by multiplying the fractional yields of the individual consecutive steps together: \( \text{Overall Yield} = 0.80 \times 0.60 \times 0.75 = 0.36 \). Converting this back into a percentage gives: \( 0.36 \times 100\% = 36\% \).

1 Mark: Correctly multiplying the fractional yields to obtain the overall cumulative yield of 36%.

The reaction that takes place is: \( [\text{Co}(\text{H}_2\text{O})_6]^{2+}\text{(aq)} + 4\text{Cl}^{-}\text{(aq)} \rightleftharpoons [\text{CoCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)} \). The pink reactant has a coordination number of 6 and an octahedral shape. The blue product has a coordination number of 4 and a tetrahedral shape. The oxidation state of cobalt remains +2 in both species.

1 Mark: Identifying that the coordination number changes from 6 to 4 and the shape changes from octahedral to tetrahedral.

1. Find mass of carbon: \( m(\text{C}) = \frac{12.0}{44.0} \times 4.40\text{ g} = 1.20\text{ g} \). 2. Find mass of hydrogen: \( m(\text{H}) = \frac{2.0}{18.0} \times 1.80\text{ g} = 0.20\text{ g} \). 3. Find mass of oxygen by subtraction: \( m(\text{O}) = 3.00 - 1.20 - 0.20 = 1.60\text{ g} \). 4. Calculate moles: \( n(\text{C}) = \frac{1.20}{12.0} = 0.10\text{ mol} \); \( n(\text{H}) = \frac{0.20}{1.0} = 0.20\text{ mol} \); \( n(\text{O}) = \frac{1.60}{16.0} = 0.10\text{ mol} \). 5. Finding the simplest whole-number ratio: \( \text{C} : \text{H} : \text{O} = 1 : 2 : 1 \). Thus, the empirical formula is \( \text{CH}_2\text{O} \).

1 Mark: Correctly determining the mass/moles of C, H, and O to find the correct empirical formula ratio (CH2O).

Benzaldehyde is \( \text{C}_6\text{H}_5\text{CHO} \), an aldehyde. 1-Phenylethanol is \( \text{C}_6\text{H}_5\text{CH(OH)CH}_3 \), a secondary alcohol with an additional methyl group. The conversion of an aldehyde to a secondary alcohol with one extra carbon atom is achieved by nucleophilic addition of a Grignard reagent like methylmagnesium iodide (\( \text{CH}_3\text{MgI} \)) followed by hydrolysis with a dilute acid.

1 Mark: Identifying that the Grignard reagent followed by acid treatment is required for this conversion.

Using the Arrhenius equation: \( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \). Substitute \( E_a = 50000\text{ J mol}^{-1} \), \( T_1 = 293\text{ K} \), and \( T_2 = 303\text{ K} \): \( \ln\left(\frac{k_2}{k_1}\right) = \frac{50000}{8.31} \times \left(\frac{1}{293} - \frac{1}{303}\right) \approx 6016.85 \times 0.00011264 \approx 0.6777 \). Taking the exponential: \( \frac{k_2}{k_1} = e^{0.6777} \approx 1.97 \approx 2.0 \).

1 Mark: Correctly applying the Arrhenius equation with values converted into matching units to calculate the factor of 2.0.

First, calculate the moles of hydrogen gas produced: \( n(\text{H}_2) = \frac{80.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 3.33 \times 10^{-3}\text{ mol} \). Since the metal \( \text{X} \) forms a divalent cation \( \text{X}^{2+} \), the stoichiometry of the reaction is 1:1, meaning \( n(\text{X}) = n(\text{H}_2) = 3.33 \times 10^{-3}\text{ mol} \). Now calculate the molar mass of \( \text{X} \): \( M_r = \frac{0.218\text{ g}}{3.33 \times 10^{-3}\text{ mol}} = 65.4\text{ g mol}^{-1} \). This molar mass corresponds to zinc (\( A_r = 65.4 \)). Therefore, the correct option is D.

1 mark for the correct calculation of moles and molar mass to identify Zinc, and selecting option D.

Calculate the moles of the starting material: \( n(\text{CoCl}_2\cdot6\text{H}_2\text{O}) = \frac{4.12\text{ g}}{237.9\text{ g mol}^{-1}} = 0.01732\text{ mol} \). Since there is 1 mole of cobalt atom per formula unit in both the reactant and the product complex, the theoretical yield of \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \) is also \( 0.01732\text{ mol} \). Calculate the theoretical mass of the complex: \( \text{mass} = 0.01732\text{ mol} \times 267.5\text{ g mol}^{-1} = 4.633\text{ g} \). Finally, calculate the percentage yield: \( \text{Yield} = \frac{3.50\text{ g}}{4.633\text{ g}} \times 100\% = 75.55\% \approx 75.6\% \). This matches option B.

1 mark for calculating the theoretical mass and the correct percentage yield, selecting option B.

Determine the ground-state electron configuration for each gaseous metal ion: \( \text{Fe}^{3+} \) has the configuration \( [\text{Ar}]3d^5 \), which contains 5 unpaired d-electrons (due to Hund's rule of maximum multiplicity). \( \text{Cr}^{3+} \) has the configuration \( [\text{Ar}]3d^3 \) (3 unpaired d-electrons). \( \text{Co}^{2+} \) has the configuration \( [\text{Ar}]3d^7 \) (3 unpaired d-electrons, as two pairs are formed). \( \text{Cu}^{2+} \) has the configuration \( [\text{Ar}]3d^9 \) (1 unpaired d-electron). Therefore, \( \text{Fe}^{3+} \) contains the highest number of unpaired d-electrons. This corresponds to option A.

1 mark for evaluating the d-electron configuration of each ion and choosing the species with 5 unpaired electrons, option A.

When excess aqueous ammonia is added to \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \), ligand exchange occurs where four water molecules are replaced by four ammonia molecules. Due to the Jahn-Teller effect, the replacement is limited to the four equatorial ligands, keeping two axial water ligands intact. The reaction is: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \). The species responsible for the deep blue colour is therefore \( [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \), matching option B.

1 mark for identifying the correct ligand-exchange product formula with four ammonia and two water ligands, selecting option B.

To find the units of the rate constant \( k \), rearrange the rate equation: \( k = \frac{\text{rate}}{[\text{NO}]^2[\text{H}_2]} \). Substituting the units: \( \text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \). If the concentration of both reactants is doubled, the new rate becomes: \( \text{rate}' = k(2[\text{NO}])^2(2[\text{H}_2]) = 8 \times k[\text{NO}]^2[\text{H}_2] = 8 \times \text{rate} \). Thus, the rate increases by a factor of 8. This corresponds to option A.

1 mark for evaluating the overall units of k and the mathematical scaling factor of the rate when concentrations are doubled, selecting option A.

Comparing the Arrhenius equation with the equation for a straight line (\( y = mx + c \)), we find that the gradient \( m = -\frac{E_a}{R} \). Therefore: \( -1.20 \times 10^4\text{ K} = -\frac{E_a}{8.31\text{ J K}^{-1}\text{ mol}^{-1}} \). Solving for \( E_a \) gives: \( E_a = 1.20 \times 10^4 \times 8.31 = 99720\text{ J mol}^{-1} \). Convert to \( \text{kJ mol}^{-1} \) by dividing by 1000: \( E_a = +99.72\text{ kJ mol}^{-1} \approx +99.7\text{ kJ mol}^{-1} \). Activation energy must be positive. This corresponds to option B.

1 mark for the correct calculation of activation energy including unit conversion to kJ/mol, selecting option B.

To synthesize 2-hydroxypropanoic acid from ethanal: Step 1: Add \( \text{KCN} \) in the presence of dilute acid (to generate \( \text{HCN} \) nucleophiles in situ) which adds to the carbonyl carbon of ethanal to form 2-hydroxypropanenitrile. Step 2: Hydrolyze the nitrile functional group to a carboxylic acid functional group by heating under reflux with a dilute mineral acid like hydrochloric acid (\( \text{HCl}(aq) \)). This gives the correct product and matches option A. Option B uses an ethanolic medium which would be used for substitution on halogenoalkanes. Option C uses a reducing agent that converts nitriles to amines. Option D uses an oxidizing agent that would oxidize the secondary alcohol group of the hydroxynitrile.

1 mark for identifying the correct sequential conditions for nucleophilic addition and acid hydrolysis, and selecting option A.

Compound \( \text{Y} \) has a degree of unsaturation of 1 and reacts with 2,4-DNPH, indicating it contains a carbonyl group (aldehyde or ketone). Since it does not react with Fehling's solution, it is a ketone, not an aldehyde. The only ketone with the molecular formula \( \text{C}_4\text{H}_8\text{O} \) is butanone. Reducing butanone with the reducing agent \( \text{NaBH}_4 \) yields a secondary alcohol, butan-2-ol. Therefore, compound \( \text{Z} \) is butan-2-ol. This corresponds to option B.

1 mark for deducing that Y is butanone and its reduction product Z is butan-2-ol, selecting option B.

The electronic configurations of the transition metal ions are:
- \( \text{Cr}^{3+} \): \( [\text{Ar}] 3d^3 \), which has 3 unpaired d-electrons.
- \( \text{Fe}^{3+} \): \( [\text{Ar}] 3d^5 \), which has 5 unpaired d-electrons.
- \( \text{Co}^{2+} \): \( [\text{Ar}] 3d^7 \), which has 3 unpaired d-electrons.
- \( \text{Cu}^{2+} \): \( [\text{Ar}] 3d^9 \), which has 1 unpaired d-electron.
Therefore, \( \text{Fe}^{3+} \) has the maximum number of unpaired d-electrons.

1 mark: Correctly identifies B as the option containing the ion with the maximum number of unpaired d-electrons.

1. Calculate the number of moles of each reactant:
- Moles of \( \text{Zn} = \frac{3.27 \text{ g}}{65.4 \text{ g mol}^{-1}} = 0.050 \text{ mol} \)
- Moles of \( \text{HCl} = 1.00 \text{ mol dm}^{-3} \times 0.100 \text{ dm}^3 = 0.100 \text{ mol} \)

2. Determine the limiting reactant:
According to the equation, \( 1 \text{ mol} \) of \( \text{Zn} \) reacts with \( 2 \text{ mol} \) of \( \text{HCl} \). Since \( 0.050 \text{ mol} \) of \( \text{Zn} \) reacts exactly with \( 0.100 \text{ mol} \) of \( \text{HCl} \), both reactants are in stoichiometric quantities.

3. Calculate the volume of \( \text{H}_2 \) produced:
- Moles of \( \text{H}_2 \) produced = \( 0.050 \text{ mol} \)
- Volume of \( \text{H}_2 = 0.050 \text{ mol} \times 24.0 \text{ dm}^3 \text{ mol}^{-1} = 1.20 \text{ dm}^3 = 1200 \text{ cm}^3 \).

1 mark: Correctly calculates the volume of hydrogen gas as \( 1200 \text{ cm}^3 \) (Option B).

1. Find the starting moles of benzene:
- Moles of benzene = \( \frac{7.8 \text{ g}}{78.0 \text{ g mol}^{-1}} = 0.10 \text{ mol} \)

2. Determine the theoretical yield of phenylethene:
- Theoretical moles of phenylethene = \( 0.10 \text{ mol} \)
- Theoretical mass of phenylethene = \( 0.10 \text{ mol} \times 104.0 \text{ g mol}^{-1} = 10.4 \text{ g} \)

3. Calculate the percentage yield:
- Percentage yield = \( \frac{4.16 \text{ g}}{10.4 \text{ g}} \times 100\% = 40.0\% \).

1 mark: Correctly calculates the overall percentage yield as \( 40.0\% \) (Option B).

Let the initial rate be \( R_1 = k [\text{X}]^2 [\text{Y}] \).
If the concentration of \( \text{X} \) is doubled to \( 2[\text{X}] \) and the concentration of \( \text{Y} \) is halved to \( 0.5[\text{Y}] \), the new rate \( R_2 \) is:
\( R_2 = k (2[\text{X}])^2 (0.5[\text{Y}]) = k (4[\text{X}]^2) (0.5[\text{Y}]) = 2 k [\text{X}]^2 [\text{Y}] = 2 R_1 \).
Therefore, the rate of reaction is doubled.

1 mark: Correctly identifies that the rate of reaction doubles (Option C).

Aqueous copper(II) ions exist as the pale blue hexaaquacopper(II) complex, \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \).
When excess ammonia is added, ligand substitution occurs to form the deep blue tetraamminediandaquacopper(II) complex, \( [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \).
Thus, the color change is from a pale blue solution to a deep blue solution.

1 mark: Correctly identifies the color change as from a pale blue solution to a deep blue solution (Option B).

1. Find the moles of \( \text{AgCl} \) precipitate:
- Moles of \( \text{AgCl} = \frac{1.435 \text{ g}}{143.4 \text{ g mol}^{-1}} = 0.0100 \text{ mol} \)

2. Relate the moles to \( \text{NaCl} \):
Since \( \text{NaCl(aq)} + \text{AgNO}_3\text{(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)} \), the mole ratio is 1:1.
- Moles of \( \text{NaCl} = 0.0100 \text{ mol} \n
3. Calculate the mass and percentage of \) \text{NaCl} \):
- Mass of \( \text{NaCl} = 0.0100 \text{ mol} \times 58.5 \text{ g mol}^{-1} = 0.585 \text{ g} \)
- Percentage of \( \text{NaCl} = \frac{0.585 \text{ g}}{1.50 \text{ g}} \times 100\% = 39.0\% \).

1 mark: Correctly calculates the percentage by mass of sodium chloride as \( 39.0\% \) (Option B).

The correct sequence of steps is:
1. Alkylation: React benzene with chloromethane in the presence of anhydrous aluminium chloride catalyst to produce methylbenzene.
2. Oxidation: Reflux methylbenzene with hot alkaline potassium manganate(VII), followed by acidification with dilute hydrochloric acid, to produce benzoic acid.
3. Acylation: React benzoic acid with phosphorus(V) chloride to produce benzoyl chloride.
4. Amidation: React benzoyl chloride with concentrated aqueous ammonia to yield benzamide.

1 mark: Correctly identifies the reagent sequence in Option A as the correct pathway.

The gradient of the Arrhenius plot is \( m = -\frac{E_a}{R} \).
Therefore:
\( -\frac{E_a}{R} = -1.20 \times 10^4 \text{ K} \)
\( E_a = 1.20 \times 10^4 \times 8.31 \text{ J mol}^{-1} = 99720 \text{ J mol}^{-1} \)
Convert to \( \text{kJ mol}^{-1} \):
\( E_a = 99.7 \text{ kJ mol}^{-1} \).

1 mark: Correctly determines the activation energy as \( +99.7 \text{ kJ mol}^{-1} \) (Option B).

First, calculate the moles and mass of carbon:
\( n(\text{C}) = n(\text{CO}_2) = \frac{4.40 \text{ g}}{44.0 \text{ g mol}^{-1}} = 0.10 \text{ mol} \)
\( m(\text{C}) = 0.10 \text{ mol} \times 12.0 \text{ g mol}^{-1} = 1.20 \text{ g} \)

Next, calculate the moles and mass of hydrogen:
\( n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{1.80 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.20 \text{ mol} \)
\( m(\text{H}) = 0.20 \text{ mol} \times 1.0 \text{ g mol}^{-1} = 0.20 \text{ g} \)

Calculate the mass of oxygen by subtraction:
\( m(\text{O}) = 2.20 \text{ g} - 1.20 \text{ g} - 0.20 \text{ g} = 0.80 \text{ g} \)

Calculate the moles of oxygen:
\( n(\text{O}) = \frac{0.80 \text{ g}}{16.0 \text{ g mol}^{-1}} = 0.05 \text{ mol} \)

Determine the simplest whole number molar ratio of the elements:
\( \text{C} : \text{H} : \text{O} = 0.10 : 0.20 : 0.05 = 2 : 4 : 1 \)

Thus, the empirical formula of the organic compound is \( \text{C}_2\text{H}_4\text{O} \).

1 mark for correct option B. Reject any other option.

Calculate the amount of nitric acid reacted:
\( n(\text{HNO}_3) = c \times V = 1.50 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.0750 \text{ mol} \)

From the stoichiometry of the chemical equation, 2 moles of \( \text{HNO}_3 \) react to form 1 mole of \( \text{Cu(NO}_3)_2 \). Therefore, the theoretical moles of copper(II) nitrate (and consequently hydrated crystals) is:
\( n(\text{Cu(NO}_3)_2\cdot3\text{H}_2\text{O})_{\text{theoretical}} = \frac{0.0750 \text{ mol}}{2} = 0.0375 \text{ mol} \)

Apply the percentage yield to find the actual amount of hydrated crystals produced:
\( n(\text{Cu(NO}_3)_2\cdot3\text{H}_2\text{O})_{\text{actual}} = 0.0375 \text{ mol} \times 0.800 = 0.0300 \text{ mol} \)

Calculate the mass of the hydrated crystals obtained:
\( \text{mass} = 0.0300 \text{ mol} \times 241.6 \text{ g mol}^{-1} = 7.248 \text{ g} \approx 7.25 \text{ g} \).

1 mark for correct option A. Reject any other option.

The rate-determining step is the slowest elementary step, which is Step 1. The reactants in Step 1 are two molecules of \( \text{NO}_2 \). Therefore, the rate equation depends only on Step 1 reactants, giving \( \text{rate} = k[\text{NO}_2]^2 \). Because \( \text{CO} \) only participates in the fast step (Step 2), it is zero-order and doubling its concentration will not affect the rate of reaction.

1 mark for correct option A. Reject any other option.

From the Arrhenius equation, \( \ln k = -\frac{E_a}{RT} + \ln A \).
Thus, a plot of \( \ln k \) against \( \frac{1}{T} \) yields a straight line with a gradient of \( -\frac{E_a}{R} \).

\( \text{Gradient} = -\frac{E_a}{R} \implies -1.20 \times 10^4 \text{ K} = -\frac{E_a}{8.31 \text{ J K}^{-1} \text{ mol}^{-1}} \)
\( E_a = 1.20 \times 10^4 \times 8.31 = 99720 \text{ J mol}^{-1} \)

Convert Joules to kilojoules:
\( E_a = \frac{99720}{1000} \text{ kJ mol}^{-1} = 99.7 \text{ kJ mol}^{-1} \).

1 mark for correct option B. Reject any other option.

In aqueous solution, cobalt(II) exists as the octahedral hexaaquacobalt(II) complex, \( [\text{Co(H}_2\text{O)}_6]^{2+} \), which is pink (coordination number of 6). When excess concentrated hydrochloric acid is added, ligand substitution occurs to yield the tetrahedral tetrachlorocobaltate(II) complex, \( [\text{CoCl}_4]^{2-} \), which is blue (coordination number of 4).

1 mark for correct option A. Reject any other option.

Determine the ground-state d-electron configuration for each transition metal ion:
- \( \text{Fe}^{3+} \) has configuration \( [\text{Ar}] 3d^5 \), which has 5 unpaired electrons.
- \( \text{Co}^{2+} \) has configuration \( [\text{Ar}] 3d^7 \), which has 3 unpaired electrons.
- \( \text{Ni}^{2+} \) has configuration \( [\text{Ar}] 3d^8 \), which has 2 unpaired electrons.
- \( \text{Cu}^{2+} \) has configuration \( [\text{Ar}] 3d^9 \), which has 1 unpaired electron.

Thus, \( \text{Fe}^{3+} \) has the greatest number of unpaired d-electrons.

1 mark for correct option A. Reject any other option.

In Step 1, nitration of methylbenzene occurs. The methyl group (\( -\text{CH}_3 \)) is an electron-donating, activating group, which directs incoming electrophiles specifically to the 2- (ortho) and 4- (para) positions. Therefore, nitration will produce 2-nitromethylbenzene and 4-nitromethylbenzene as major products, rather than the 3-isomer. Subsequent oxidation of these intermediates would yield 2-nitrobenzoic acid and 4-nitrobenzoic acid. To make 3-nitrobenzoic acid, methylbenzene should be oxidized to benzoic acid first (as the carboxylic acid group is 3-directing) and then nitrated.

1 mark for correct option B. Reject any other option.

In the first step, propanone, \( \text{CH}_3\text{COCH}_3 \), undergoes nucleophilic addition with cyanide (from \( \text{NaCN} \) in acidic medium) to form the hydroxynitrile compound, 2-hydroxy-2-methylpropanenitrile (Compound X), which has the formula \( (\text{CH}_3)_2\text{C(OH)CN} \).

In the second step, the nitrile group (\( -\text{CN} \)) is hydrolysed to a carboxylic acid group (\( -\text{COOH} \)) by heating with dilute sulfuric acid. This yields the hydroxy acid, 2-hydroxy-2-methylpropanoic acid (Compound Y), which has the formula \( (\text{CH}_3)_2\text{C(OH)COOH} \).

1 mark for correct option A. Reject any other option.

First, calculate the number of moles of \(\text{NaHCO}_3\) reacted:
\[
\text{Moles of NaHCO}_3 = \frac{4.20 \text{ g}}{84.0 \text{ g mol}^{-1}} = 0.050 \text{ mol}
\]

Next, use the stoichiometric ratio from the balanced equation. Two moles of \(\text{NaHCO}_3\) yield one mole of \(\text{CO}_2\):
\[
\text{Moles of CO}_2 = \frac{0.050 \text{ mol}}{2} = 0.025 \text{ mol}
\]

Finally, calculate the volume of \(\text{CO}_2\) gas at r.t.p.:
\[
\text{Volume of CO}_2 = 0.025 \text{ mol} \times 24.0 \text{ dm}^3\text{ mol}^{-1} = 0.60 \text{ dm}^3
\]

• Correct calculation of moles of sodium hydrogencarbonate (0.050 mol) and stoichiometric division by 2 to yield 0.025 mol of carbon dioxide (1 mark).
• Correct calculation of gas volume as 0.60 dm³ (1 mark).

The atomic number of nickel (Ni) is 28. Its ground-state electronic configuration is:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^8 4\text{s}^2\)

When transition metals form positive ions, they lose electrons from the outer \(4\text{s}\) subshell first. Therefore, forming a \(\text{Ni}^{2+}\) ion requires removing the two \(4\text{s}\) electrons:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^8\)

• Identifies the correct ground state electron configuration of the Ni²⁺ ion by removing the 4s electrons first (1 mark).

The species that appear in the rate equation must be involved in the rate-determining step (or in steps prior to it). Since the rate equation is first order with respect to \(\text{P}\) and first order with respect to \(\text{Q}\), the rate-determining step must involve one molecule of \(\text{P}\) and one molecule of \(\text{Q}\). Therefore, \(\text{P} + \text{Q} \rightarrow \text{X}\) is a possible rate-determining step.

• Correctly relates the orders of reaction in the rate equation to the coefficients of the reactants in the rate-determining step (1 mark).

To convert 1-bromopropane (3 carbons) into butylamine (4 carbons), the length of the carbon chain must be increased by one carbon atom.

1. Nucleophilic substitution of 1-bromopropane using potassium cyanide, \(\text{KCN}\), in aqueous ethanol produces butanenitrile, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\).
2. Reduction of the nitrile using a strong reducing agent such as lithium tetrahydridoaluminate, \(\text{LiAlH}_4\), in dry ether yields pure butylamine (a primary amine) as the sole amine product, avoiding the mixture of primary, secondary, and tertiary amines formed during direct substitution with ammonia.

• Identifies the need to step up the carbon chain using a nitrile and reduces it with LiAlH₄ in dry ether to form the primary amine selectively (1 mark).

When excess aqueous ammonia is added to a solution of copper(II) ions, a ligand substitution reaction occurs. Four of the water ligands in the octahedral hexaaquacopper(II) complex, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), are replaced by ammonia ligands to form the deep blue tetraamminedihydracopper(II) complex, \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

• Identifies the correct formula and charge of the tetraamminedihydracopper(II) complex ion (1 mark).

1. Calculate the percentage of oxygen by mass:
\[
\%\text{ O} = 100.0\% - 70.0\% = 30.0\%
\]

2. Determine the ratio of moles of atoms:
\[
\text{Moles of Fe} = \frac{70.0}{55.8} = 1.254 \text{ mol}
\]
\[
\text{Moles of O} = \frac{30.0}{16.0} = 1.875 \text{ mol}
\]

3. Divide by the smallest value to obtain a whole number ratio:
\[
\text{Fe} = \frac{1.254}{1.254} = 1
\]
\[
\text{O} = \frac{1.875}{1.254} = 1.495 \approx 1.5
\]

Multiplying both numbers by 2 gives a whole-number ratio of \(2:3\). Hence, the empirical formula is \(\text{Fe}_2\text{O}_3\).

• Correct division of mass percentages by relative atomic masses to get molar ratio 1.254 : 1.875 (1 mark).
• Simplification of ratio to give Fe₂O₃ (1 mark).

1. The positive reaction with 2,4-dinitrophenylhydrazine indicates that compound \(\text{W}\) contains a carbonyl group (either an aldehyde or a ketone).
2. The negative test with Tollens' reagent indicates that the compound is not an aldehyde; hence, it must be a ketone.
3. The only ketone with the molecular formula \(\text{C}_4\text{H}_8\text{O}\) is butanone.

• Identifies the carbonyl group from the 2,4-DNPH test and ketone from the Tollens' test, leading to the name butanone (1 mark).

Comparing the Arrhenius equation to the equation of a straight line, \(y = mx + c\):

• \(y = \ln k\)
• \(x = \frac{1}{T}\)
• \(m = -\frac{E_a}{R}\)
• \(c = \ln A\)

Therefore, the gradient (slope) of the line is equal to \(-\frac{E_a}{R}\).

• Identifies the gradient as -E_a/R by comparison with the standard straight-line equation (1 mark).

First, calculate the molar mass of anhydrous \( \text{MgSO}_4 \) which is \( 24.3 + 32.1 + (16.0 \times 4) = 120.4\text{ g mol}^{-1} \). Next, calculate the moles of anhydrous residue: \( 2.40\text{ g} / 120.4\text{ g mol}^{-1} = 0.01993\text{ mol} \). Then, calculate the mass of water lost: \( 4.92\text{ g} - 2.40\text{ g} = 2.52\text{ g} \). Calculate the moles of water lost: \( 2.52\text{ g} / 18.0\text{ g mol}^{-1} = 0.140\text{ mol} \). Find the molar ratio: \( 0.140 / 0.01993 \approx 7.02 \), which rounds to 7.

1 mark for correct calculation and identification of C as the correct option. Reject other options.

A chromium atom in its ground state has the electronic configuration \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1 \). When forming the \( \text{Cr}^{3+} \) ion, three electrons are removed, starting with the outer \( 4\text{s} \) electron and then two \( 3\text{d} \) electrons, which gives \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3 \).

1 mark for correct electronic configuration identification of A. Reject configurations with electrons remaining in 4s or incorrect number of d electrons.

Comparing Experiments 1 and 2, doubling \( [\text{A}] \) while keeping \( [\text{B}] \) constant quadruples the rate, showing the reaction is second order with respect to A. Comparing Experiments 1 and 3, doubling \( [\text{B}] \) while keeping \( [\text{A}] \) constant does not change the rate, showing the reaction is zero order with respect to B. Therefore, the rate equation is \( \text{Rate} = k[\text{A}]^2 \).

1 mark for the correct determination of the reaction orders and selecting option B.

The nitro group in nitrobenzene is meta-directing (3-directing). Therefore, bromination with \( \text{Br}_2 \) and a halogen carrier like \( \text{FeBr}_3 \) must be carried out first to yield 3-bromonitrobenzene. This is then followed by reduction of the nitro group to an amine group using tin (Sn) and concentrated hydrochloric acid to form 3-bromophenylamine. If reduction were done first, the resulting phenylamine group would direct the bromine to the 2, 4, or 6 positions.

1 mark for identifying that bromination must precede reduction to achieve meta substitution, selecting option B.

Concentrated hydrochloric acid supplies a high concentration of chloride ions which undergo ligand substitution with the hexaaquacopper(II) complex: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O} \). The tetrachlorocuprate(II) ion, \( [\text{CuCl}_4]^{2-} \), is yellow, which mixes with the remaining blue hexaaqua complex to yield a yellow-green solution.

1 mark for identifying the correct tetrahedral complex ion, option B.

Using the ideal gas equation: \( pV = nRT \), where \( p = 1.01 \times 10^5\text{ Pa} \), \( V = 1.25 \times 10^{-4}\text{ m}^3 \), \( T = 373\text{ K} \). Therefore, \( n = \frac{1.01 \times 10^5 \times 1.25 \times 10^{-4}}{8.31 \times 373} = 0.004073\text{ mol} \). The molar mass \( M = \frac{\text{mass}}{n} = \frac{0.355\text{ g}}{0.004073\text{ mol}} \approx 87.2\text{ g mol}^{-1} \).

1 mark for correct unit conversions, calculation of moles and molar mass, choosing C.

Heating a nitrile under reflux with a dilute acid causes complete acid hydrolysis of the nitrile group (\( -\text{CN} \)) to form a carboxylic acid group (\( -\text{COOH} \)). Propanenitrile contains three carbon atoms, so it hydrolyses to form propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)).

1 mark for identifying the product of nitrile acid hydrolysis as a carboxylic acid with the same carbon chain length, option B.

From the logarithmic form of the Arrhenius equation: \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \). Thus, the gradient of the plot is \( -\frac{E_a}{R} \). Given the gradient is \( -1.20 \times 10^4\text{ K} \), we have: \( -\frac{E_a}{R} = -1.20 \times 10^4 \), so \( E_a = 1.20 \times 10^4 \times 8.31 = 99720\text{ J mol}^{-1} \approx 99.7\text{ kJ mol}^{-1} \).

1 mark for the correct application of the Arrhenius gradient relationship and unit conversion to kJ/mol, choosing C.

1. Calculate the mass of water lost: \(4.93\text{ g} - 2.41\text{ g} = 2.52\text{ g}\).
2. Calculate the moles of water: \(2.52\text{ g} / 18.0\text{ g mol}^{-1} = 0.14\text{ mol}\).
3. Calculate the moles of anhydrous \(\text{MgSO}_4\): \(2.41\text{ g} / 120.4\text{ g mol}^{-1} = 0.020\text{ mol}\).
4. Determine the molar ratio: \(x = 0.14 / 0.020 = 7\).

1 mark for the correct answer C (7).
- Incorrect options represent arithmetic errors or inversion of ratio calculations.

- \(\text{Fe}^{3+}\) has the electron configuration \([\text{Ar}] 3\text{d}^5\). It has 5 unpaired d-electrons.
- \(\text{Co}^{2+}\) has the electron configuration \([\text{Ar}] 3\text{d}^7\). It has 3 unpaired d-electrons.
- \(\text{Ni}^{2+}\) has the electron configuration \([\text{Ar}] 3\text{d}^8\). It has 2 unpaired d-electrons.
- \(\text{Cu}^{2+}\) has the electron configuration \([\text{Ar}] 3\text{d}^9\). It has 1 unpaired d-electron.

1 mark for the correct option A.

In mechanism A:
- Step 1: \(\text{X} + \text{Y} \rightleftharpoons \text{XY}\) (fast equilibrium), so \([\text{XY}] = K_c [\text{X}][\text{Y}]\).
- Step 2: \(\text{XY} + \text{Y} \rightarrow \text{Products}\) (slow rate-determining step), so \(\text{Rate} = k_2 [\text{XY}][\text{Y}]\).
- Substituting the equilibrium expression for \([\text{XY}]\) into the rate expression gives: \(\text{Rate} = k_2 K_c [\text{X}][\text{Y}][\text{Y}] = k[\text{X}][\text{Y}]^2\). This perfectly matches the experimental rate equation.

1 mark for correct selection A.

1. Stage 1: Oxidize propan-1-ol completely to propanoic acid using acidified potassium dichromate(VI) under reflux. Distillation would yield propanal instead of the acid.
2. Stage 2: Heat propanoic acid under reflux with ethanol in the presence of concentrated sulfuric acid as a catalyst to form ethyl propanoate.

1 mark for correct answer B.

Use the ideal gas equation: \(PV = nRT\)
- \(n(\text{O}_2) = \frac{3.20}{32.0} = 0.100\text{ mol}\)
- \(T = 50.0 + 273.15 = 323.15\text{ K}\)
- \(P = 100 \times 10^3\text{ Pa}\)
- \(V = \frac{nRT}{P} = \frac{0.100 \times 8.31 \times 323.15}{100 \times 10^3} = 0.002685\text{ m}^3 = 2.685\text{ dm}^3 \approx 2.68\text{ dm}^3\).

1 mark for correct selection B.

The addition of excess concentrated \(\text{HCl}\) to aqueous copper(II) ions results in a ligand exchange reaction where water ligands are replaced by chloride ligands:

\([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)

Due to the larger size and mutual electrostatic repulsion of the chloride ligands, the complex adopts a tetrahedral geometry rather than octahedral or square planar.

1 mark for correct option A.

- The methyl group in methylbenzene is ortho- and para-directing. If you nitrate first, you get 2-nitrotoluene and 4-nitrotoluene.
- The carboxylic acid group (\(-\text{COOH}\)) is meta-directing. Therefore, you must oxidize the methyl group to a carboxylic acid first using alkaline \(\text{KMnO}_4\) (followed by acid workup), then perform electrophilic nitration using concentrated \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\).

1 mark for correct selection B.

1. Comparing Exp 1 and Exp 2: \([\text{B}]\) is kept constant, \([\text{A}]\) is doubled (\(0.10 \rightarrow 0.20\)). The rate increases by a factor of 4 (\(2.0 \times 10^{-4} \rightarrow 8.0 \times 10^{-4}\)), which is \(2^2\). Hence, the order with respect to \(\text{A}\) is 2.
2. Comparing Exp 2 and Exp 3: \([\text{A}]\) is kept constant, \([\text{B}]\) is doubled (\(0.10 \rightarrow 0.20\)). The rate increases by a factor of 2 (\(8.0 \times 10^{-4} \rightarrow 1.6 \times 10^{-3}\)), which is \(2^1\). Hence, the order with respect to \(\text{B}\) is 1.

1 mark for the correct answer B.

The hexaaquacobalt(II) ion, \([Co(H_2O)_6]^{2+}\), is a pink octahedral complex with a coordination number of 6. Addition of concentrated hydrochloric acid introduces a high concentration of chloride ligands, which replace the water ligands to form the tetrahedral tetrachlorocobaltate(II) ion, \([CoCl_4]^{2-}\). This complex is blue and has a coordination number of 4. Therefore, the coordination number changes from 6 to 4, and the colour changes from pink to blue.

1 mark for the correct option. A is correct.

First, find the number of moles of gas using the ideal gas equation, \(PV = nRT\): \(n = \frac{PV}{RT}\). Converting the units: \(P = 100 \times 10^3\text{ Pa}\), \(V = 620 \times 10^{-6}\text{ m}^3\), and \(T = 298\text{ K}\). This gives \(n = \frac{100 \times 10^3 \times 620 \times 10^{-6}}{8.31 \times 298} \approx 0.02504\text{ mol}\). Next, calculate the molar mass: \(M = \frac{\text{mass}}{n} = \frac{1.15\text{ g}}{0.02504\text{ mol}} \approx 45.9\text{ g mol}^{-1}\). Comparing this to the molar masses of the options: \(NO = 30.0\text{ g mol}^{-1}\), \(N_2O = 44.0\text{ g mol}^{-1}\), \(NO_2 = 46.0\text{ g mol}^{-1}\), and \(N_2O_4 = 92.0\text{ g mol}^{-1}\). The calculated molar mass corresponds to \(NO_2\).

1 mark for the correct option. B is correct.

Nitrobenzene is reduced to phenylamine using tin (Sn) and concentrated hydrochloric acid, followed by sodium hydroxide to liberate the free amine. Phenylamine is then converted to benzenediazonium chloride by reacting with nitrous acid (generated in situ from sodium nitrate(III) and hydrochloric acid) at a low temperature of 5 degrees Celsius to prevent decomposition of the diazonium salt.

1 mark for the correct option. A is correct.

The overall order of the reaction is the sum of the powers of the concentration terms, which is \(1 + 1 = 2\) (second order overall). Because iodine does not appear in the rate equation, the reaction is zero order with respect to iodine, meaning it is not involved in the rate-determining step.

1 mark for the correct option. A is correct.

Let us determine the electronic configurations: \(Fe^{3+}\) is \([Ar] 3d^5\) (5 unpaired d-electrons); \(Cr^{3+}\) is \([Ar] 3d^3\) (3 unpaired d-electrons); \(Ni^{2+}\) is \([Ar] 3d^8\) (2 unpaired d-electrons); \(Cu^{2+}\) is \([Ar] 3d^9\) (1 unpaired d-electron). Therefore, \(Fe^{3+}\) has the highest number of unpaired d-electrons.

1 mark for the correct option. A is correct.

Atom economy is defined as \(\frac{\text{Total molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\%\). Here, desired product is iron: \(2 \times \text{Fe} = 2 \times 55.8 = 111.6\text{ g mol}^{-1}\). The reactants are \(\text{Fe}_2\text{O}_3\) (\(159.6\text{ g mol}^{-1}\)) and \(3\text{CO}\) (\(3 \times 28.0 = 84.0\text{ g mol}^{-1}\)). The total mass of reactants is \(159.6 + 84.0 = 243.6\text{ g mol}^{-1}\). The atom economy is \(\frac{111.6}{243.6} \times 100\% = 45.81\%\).

1 mark for the correct option. B is correct.

The primary amine product (butylamine) has a nucleophilic lone pair on the nitrogen atom and can react further with 1-bromobutane to form secondary and tertiary amines. Using a large excess of ammonia ensures that 1-bromobutane is much more likely to react with ammonia molecules than with the product, suppressing further substitution.

1 mark for the correct option. A is correct.

According to the logarithmic form of the Arrhenius equation: \(\ln(k) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln(A)\). The gradient of the plot is \(-\frac{E_a}{R}\). Therefore, \(-\frac{E_a}{R} = -6.5 \times 10^3\text{ K}\), which gives \(E_a = 6.5 \times 10^3 \times 8.31 = 54015\text{ J mol}^{-1} \approx +54.0\text{ kJ mol}^{-1}\).

1 mark for the correct option. A is correct.

1. Calculate the total moles of gas produced:
\( n(\text{gas}) = \frac{1.50\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0625\text{ mol} \)

2. Use the stoichiometric ratio from the decomposition equation:
\( 2\text{M(NO}_3)_2\text{(s)} \rightarrow 2\text{MO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)} \)
The ratio of \( \text{M(NO}_3)_2 \) to total gas (\( 4\text{NO}_2 + \text{O}_2 \)) is \( 2:5 \).
\( n(\text{M(NO}_3)_2) = 0.0625\text{ mol} \times \frac{2}{5} = 0.0250\text{ mol} \)

3. Calculate the molar mass (\( M_r \)) of the nitrate and the relative atomic mass (\( A_r \)) of \( \text{M} \):
\( M_r(\text{M(NO}_3)_2) = \frac{4.10\text{ g}}{0.0250\text{ mol}} = 164.0\text{ g mol}^{-1} \)
\( A_r(\text{M}) = 164.0 - [2 \times (14.01 + 3 \times 16.00)] = 164.0 - 124.02 = 39.98 \approx 40.0 \)
The Group 2 metal with \( A_r \approx 40.0 \) is Calcium (\( \text{Ca} \)).

• M1: Correct calculation of moles of gas (0.0625 mol) (1 mark)
• M2: Correct calculation of moles of nitrate (0.0250 mol) (1 mark)
• M3: Correct calculation of Ar (40.0) and identifying Calcium / Ca (1 mark)

1. Write the balanced equilibrium equation for ligand substitution:
\( \text{[Cu(H}_2\text{O)}_6\text{]}^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons \text{[CuCl}_4\text{]}^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)} \)

2. Identify the color change:
Hexaaquacopper(II) is pale blue; tetrachlorocuprate(II) is yellow-green (or yellow/green).

3. Explain the geometry change:
The geometry changes from octahedral (6-coordinate) to tetrahedral (4-coordinate) because the chloride ligands are larger than water molecules and experience steric hindrance / repulsion, preventing six chloride ions from coordinating around the central copper(II) ion.

• M1: Correct balanced ionic equation, including charges and state of hydration (1 mark)
• M2: Correct color change (from pale blue to yellow-green/green/yellow) (1 mark)
• M3: Explanation of change from octahedral to tetrahedral due to steric hindrance / larger size of Cl- ligands (1 mark)

Use the logarithmic form of the Arrhenius equation:
\( \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \)

1. Calculate the logarithmic term:
\( \ln\left(\frac{4.98 \times 10^{-4}}{3.46 \times 10^{-5}}\right) = \ln(14.393) = 2.6667 \)

2. Calculate the temperature term:
\( \left(\frac{1}{318} - \frac{1}{298}\right) = 0.0031447 - 0.0033557 = -2.110 \times 10^{-4}\text{ K}^{-1} \)

3. Rearrange and solve for \( E_a \):
\( 2.6667 = -\frac{E_a}{8.31} \times (-2.110 \times 10^{-4}) \)
\( E_a = \frac{2.6667 \times 8.31}{2.110 \times 10^{-4}} = 105024\text{ J mol}^{-1} \approx 105\text{ kJ mol}^{-1} \)

• M1: Correct setup of the Arrhenius relation with substituted values (1 mark)
• M2: Correct calculation of intermediate terms (ln ratio and temperature terms) (1 mark)
• M3: Final value of Ea between 104.9 and 105.1 kJ mol^-1 with appropriate units (1 mark)

• **Step 1:** React methylbenzene with chlorine (or bromine) in the presence of ultraviolet (UV) light to undergo free radical substitution.
Product: (chloromethyl)benzene or (bromomethyl)benzene.

• **Step 2:** React the (chloromethyl)benzene with potassium cyanide (or sodium cyanide) dissolved in ethanol under reflux.
Product: phenylethanenitrile.

• **Step 3:** Heat the phenylethanenitrile with dilute hydrochloric acid (or dilute sulfuric acid) under reflux to hydrolyze the nitrile group.
Product: phenylethanoic acid.

• M1: Step 1 reagents (Cl2 or Br2), conditions (UV light), and correct halogenated intermediate (1 mark)
• M2: Step 2 reagents (KCN/NaCN), solvent (ethanol), reflux, and nitrile intermediate (1 mark)
• M3: Step 3 reagents (dilute acid, e.g., HCl/H2SO4), conditions (heat under reflux) to yield carboxylic acid (1 mark)

1. Calculate the cell potential:
\( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.33\text{ V} - (+0.77\text{ V}) = +0.56\text{ V} \)

2. Construct the overall balanced ionic equation:
Multiply the iron half-equation by 6 to balance electrons and reverse it since it undergoes oxidation:
\( 6\text{Fe}^{2+}\text{(aq)} + \text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} \rightarrow 6\text{Fe}^{3+}\text{(aq)} + 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)} \)

3. Determine the anode:
Oxidation happens at the anode, so the iron half-cell (or the platinum electrode in contact with the \( \text{Fe}^{3+}/\text{Fe}^{2+} \) mixture) acts as the anode.

• M1: Correct standard cell potential calculation (+0.56 V) with sign and units (1 mark)
• M2: Correctly balanced overall ionic equation with correct species and stoichiometry (1 mark)
• M3: Identification of the Fe3+/Fe2+ half-cell (or the Platinum electrode within it) as the anode (1 mark)

1. Calculate the initial moles of each reactant:
\( n(\text{HCOOH}) = 0.0500\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol} \)
\( n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.0050\text{ mol} \)

2. Determine moles remaining after neutralization:
\( \text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O} \)
\( n(\text{HCOOH})_{\text{remaining}} = 0.0100 - 0.0050 = 0.0050\text{ mol} \)
\( n(\text{HCOO}^-)_{\text{produced}} = 0.0050\text{ mol} \)

3. Apply the buffer equation:
Since \( n(\text{HCOOH}) = n(\text{HCOO}^-) \), the concentration ratio is 1.
\( \text{pH} = \text{p}K_a = -\log_{10}(1.80 \times 10^{-4}) = 3.74 \)

• M1: Calculate initial moles of acid (0.0100 mol) and base (0.0050 mol) (1 mark)
• M2: Determine that remaining acid and conjugate base concentrations are equal (0.0050 mol each) (1 mark)
• M3: Correctly calculate pH = pKa = 3.74 (accept 3.7) (1 mark)

Establish the Born-Haber cycle equation:
\( \Delta H_f^\ominus[\text{MgCl}_2\text{(s)}] = \Delta H_{\text{at}}^\ominus[\text{Mg(s)}] + I.E._1[\text{Mg(g)}] + I.E._2[\text{Mg(g)}] + 2 \times \Delta H_{\text{at}}^\ominus[\text{Cl}] + 2 \times E.A._1[\text{Cl}] + \Delta H_{\text{latt}}^\ominus \)

Substitute the known values:
\( -642 = +148 + 738 + 1451 + 2(+122) + 2(-349) + \Delta H_{\text{latt}}^\ominus \)

Simplify the terms:
\( -642 = 148 + 738 + 1451 + 244 - 698 + \Delta H_{\text{latt}}^\ominus \)
\( -642 = 1883 + \Delta H_{\text{latt}}^\ominus \)
\( \Delta H_{\text{latt}}^\ominus = -642 - 1883 = -2525\text{ kJ mol}^{-1} \)

(Note: If defined as lattice dissociation, \( +2525\text{ kJ mol}^{-1} \) is also acceptable if specified.)

• M1: Correct algebraic expression for the cycle, showing stoichiometry (especially the factor of 2 for Cl atomisation and electron affinity) (1 mark)
• M2: Correct substitution of values into equation (1 mark)
• M3: Final calculated value of -2525 kJ mol^-1 with correct units (accept +2525 if clear) (1 mark)

1. **Generation of the electrophile:**
\( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \) (or equivalent balanced equation)

2. **Mechanism:**
• Curly arrow from the delocalised \( \pi \)-system of the benzene ring to the nitrogen atom of the \( \text{NO}_2^+ \) electrophile.
• Formation of the intermediate: A hexagon with a partially open circle/horseshoe containing a positive charge (facing the carbon that is bonded to both hydrogen and the nitro group).
• Curly arrow from the \( \text{C-H} \) bond back into the ring to restore the aromatic system, releasing a proton (\( \text{H}^+ \)).

• M1: Correctly balanced equation for the generation of the NO2+ electrophile (1 mark)
• M2: Correct first mechanism step: curly arrow from benzene ring to NO2+ (1 mark)
• M3: Correct intermediate representation (horseshoe with positive charge) and curly arrow from C-H bond restoring aromaticity (1 mark)

First, calculate the mass of water of crystallization lost: \(5.56\text{ g} - 3.04\text{ g} = 2.52\text{ g}\). Next, calculate the number of moles of water lost: \(n(\text{H}_2\text{O}) = \frac{2.52\text{ g}}{18.02\text{ g mol}^{-1}} = 0.140\text{ mol}\). Then, calculate the number of moles of anhydrous iron(II) sulfate: \(n(\text{FeSO}_4) = \frac{3.04\text{ g}}{151.91\text{ g mol}^{-1}} = 0.0200\text{ mol}\). Finally, determine the ratio of moles of water to moles of anhydrous salt: \(x = \frac{0.140}{0.0200} = 7\).

M1: Calculation of mass of water lost (2.52 g) and moles of water (0.140 mol). [1 mark] M2: Calculation of moles of anhydrous iron(II) sulfate (0.0200 mol). [1 mark] M3: Derivation of the final integer value of x = 7. [1 mark]

The complex salt consists of the complex cation \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) and a chloride counterion. Let the oxidation state of Co be \(y\). Since ammonia is neutral and chloride has a -1 charge: \(y + 4(0) + 2(-1) = +1\), which gives \(y = +3\). The coordination number is 6 because there are 6 monodentate ligands (4 ammonia and 2 chloride) bonded to the central cobalt ion. This octahedral complex with formula \([\text{MA}_4\text{B}_2]^+\) exhibits geometric (cis-trans) isomerism.

M1: Cobalt oxidation state of +3 (or III). [1 mark] M2: Coordination number of 6. [1 mark] M3: Geometric / cis-trans isomerism. [1 mark]

Rearrange the rate equation to solve for the rate constant: \(k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]}\). Substitute the given values into the equation: \(k = \frac{1.2 \times 10^{-4}}{(0.025)^2 \times 0.015} = \frac{1.2 \times 10^{-4}}{9.375 \times 10^{-6}} = 12.8\). To find the units: \(\text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

M1: Correct rearrangement of rate equation and substitution of values. [1 mark] M2: Evaluation of the rate constant as 12.8 (or 13 to 2 s.f.). [1 mark] M3: Correct units of dm^6 mol^-2 s^-1. [1 mark]

Step 1: Reduce nitrobenzene to phenylamine using tin (Sn) and concentrated hydrochloric acid (HCl), heated under reflux. Sodium hydroxide (NaOH) is then added to liberate the amine. Step 2: Convert phenylamine to benzenediazonium chloride by reacting with sodium nitrite (NaNO2) and hydrochloric acid (HCl) at a low temperature maintained between 0 and 10 degrees Celsius.

M1: Step 1 reagents and conditions: Sn / conc. HCl, heated under reflux, followed by addition of NaOH. [1 mark] M2: Step 2 reagents: NaNO2 and HCl (or HNO2). [1 mark] M3: Step 2 condition: temperature of 0-10 degrees Celsius (or below 10 degrees Celsius). [1 mark]

Convert temperature and volume to SI units: \(T = 97 + 273 = 370\text{ K}\) and \(V = 38.5 \times 10^{-6}\text{ m}^3\). Use the ideal gas equation to find moles of gas: \(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 38.5 \times 10^{-6}}{8.31 \times 370} = 1.2647 \times 10^{-3}\text{ mol}\). Calculate the molar mass: \(M = \frac{\text{mass}}{n} = \frac{0.116\text{ g}}{1.2647 \times 10^{-3}\text{ mol}} = 91.7\text{ g mol}^{-1}\) (to 3 s.f.).

M1: Conversions of T to 370 K and V to 3.85 x 10^-5 m^3. [1 mark] M2: Calculate moles of gas as 1.26 x 10^-3 mol. [1 mark] M3: Correct calculation of molar mass as 91.7 g mol^-1 (allow range 91.5 - 92.0). [1 mark]

The ionic equation is: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\). The coordination number of the copper ion changes from 6 in the octahedral hexaaqua complex to 4 in the tetrahedral tetrachlorocuprate(II) complex. The color of the solution changes from pale blue to yellow (a green solution is often observed due to the mixture of the blue and yellow complexes).

M1: Correct ionic equation with correct formulae and balancing. [1 mark] M2: Coordination number changes from 6 to 4. [1 mark] M3: Color change from pale blue to yellow (accept green). [1 mark]

Step 1: Bromoethane is reacted with magnesium metal (Mg) in dry ether to form the Grignard reagent, ethylmagnesium bromide (\(\text{CH}_3\text{CH}_2\text{MgBr}\)). Step 2: Ethylmagnesium bromide is reacted with methanal (\(\text{HCHO}\)), followed by hydrolysis with a dilute aqueous acid (such as \(\text{HCl}\)) to produce propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)).

M1: Step 1 reagent Mg and condition dry ether to form CH3CH2MgBr. [1 mark] M2: Step 2 reagents: methanal (HCHO) followed by dilute acid (H+ / H2O). [1 mark] M3: Correct structures: intermediate ethylmagnesium bromide (CH3CH2MgBr) and product propan-1-ol (CH3CH2CH2OH). [1 mark]

First, calculate the moles of manganate(VII) ions used: \(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{22.50}{1000}\text{ dm}^3 = 4.50 \times 10^{-4}\text{ mol}\). According to the equation, 1 mole of \(\text{MnO}_4^-\) reacts with 5 moles of \(\text{Fe}^{2+}\). Therefore, moles of \(\text{Fe}^{2+}\) in the sample = \(5 \times 4.50 \times 10^{-4}\text{ mol} = 2.25 \times 10^{-3}\text{ mol}\). Finally, calculate the concentration of \(\text{Fe}^{2+}\): \([\text{Fe}^{2+}] = \frac{2.25 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0900\text{ mol dm}^{-3}\).

M1: Calculate moles of MnO4- = 4.50 x 10^-4 mol. [1 mark] M2: Calculate moles of Fe2+ = 2.25 x 10^-3 mol. [1 mark] M3: Calculate concentration of Fe2+ = 0.0900 mol dm^-3 (allow 0.09). [1 mark]

First, calculate the moles of magnesium nitrate: \( n(\text{Mg(NO}_3)_2) = \frac{2.22}{148.3} = 0.01497\text{ mol} \). Second, use the stoichiometry of the reaction to find the total moles of gas produced. 2 moles of anhydrous magnesium nitrate produce 5 moles of gas (4 moles of \( \text{NO}_2 \) and 1 mole of \( \text{O}_2 \)). Therefore, total moles of gas \( n_{\text{gas}} = 0.01497 \times 2.5 = 0.03742\text{ mol} \). Third, use the ideal gas equation \( pV = nRT \) to calculate the volume: \( V = \frac{nRT}{p} = \frac{0.03742 \times 8.31 \times 373}{101000} = 0.001148\text{ m}^3 \). Convert to \( \text{dm}^3 \) by multiplying by 1000: \( V = 1.15\text{ dm}^3 \).

Mark 1: Calculate moles of magnesium nitrate (\( 0.01497\text{ mol} \)) OR correctly identify the 2.5 gas ratio.
Mark 2: Calculate total moles of gas produced (\( 0.03742\text{ mol} \)).
Mark 3: Calculate the volume in \( \text{dm}^3 \) showing correct substitution into \( pV=nRT \) (\( 1.15\text{ dm}^3 \), accept 1.14 to 1.16).

Comparing Experiments 1 and 2: [B] is constant, [A] doubles, and the rate increases by a factor of 4 (\( 4.80 \times 10^{-3} / 1.20 \times 10^{-3} = 4 \)), so the reaction is second order with respect to A. Comparing Experiments 1 and 3: [A] is constant, [B] doubles, and the rate increases by a factor of 2 (\( 2.40 \times 10^{-3} / 1.20 \times 10^{-3} = 2 \)), so the reaction is first order with respect to B. Therefore, the rate equation is: \( \text{Rate} = k[\text{A}]^2[\text{B}] \). To calculate the rate constant \( k \), use Experiment 1 values: \( 1.20 \times 10^{-3} = k (0.100)^2 (0.100) \), which gives \( k = 1.20 \). The units of \( k \) are calculated as: \( \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

Mark 1: Correct rate equation: \( \text{Rate} = k[\text{A}]^2[\text{B}] \) (with or without working).
Mark 2: Correct calculation of the numerical value of \( k = 1.20 \).
Mark 3: Correct units for \( k \): \( \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

First, calculate the moles of \( \text{MnO}_4^- \): \( n(\text{MnO}_4^-) = \frac{16.80}{1000} \times 0.0150 = 2.52 \times 10^{-4}\text{ mol} \). The stoichiometry of the reaction is \( 5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \), meaning 1 mole of manganate reacts with 5 moles of iron(II). Thus, moles of \( \text{Fe}^{2+} = 5 \times 2.52 \times 10^{-4} = 1.26 \times 10^{-3}\text{ mol} \). Next, find the mass of iron: \( m(\text{Fe}) = 1.26 \times 10^{-3} \times 55.8 = 0.07031\text{ g} \). Finally, calculate the percentage by mass: \( \frac{0.07031}{0.250} \times 100 = 28.12\% \).

Mark 1: Calculate moles of manganate(VII) (\( 2.52 \times 10^{-4}\text{ mol} \)).
Mark 2: Use the 1:5 ratio to calculate moles of iron(II) (\( 1.26 \times 10^{-3}\text{ mol} \)).
Mark 3: Calculate the mass of iron and percentage by mass (\( 28.1\% \), accept 28.1% to 28.2%).

First, calculate the actual moles of ethyl benzoate needed: \( n = \frac{10.0}{150.0} = 0.06667\text{ mol} \). Second, calculate the overall yield of the two-step synthesis: \( 0.850 \times 0.700 = 0.595 \) (or \( 59.5\% \)). Third, calculate the theoretical moles of benzoic acid required: \( n_{\text{theoretical}} = \frac{0.06667}{0.595} = 0.1121\text{ mol} \). Finally, calculate the mass of benzoic acid: \( m = 0.1121 \times 122.0 = 13.68\text{ g} \), which rounds to 13.7 g (3 sig figs).

Mark 1: Calculate the moles of ethyl benzoate (\( 0.0667\text{ mol} \)).
Mark 2: Account for the multi-step yield (either by calculating intermediate moles of benzoyl chloride, \( 0.0952\text{ mol} \), or calculating overall yield of \( 59.5\% \)).
Mark 3: Calculate the final mass of benzoic acid (\( 13.7\text{ g} \), accept 13.68 g to 13.7 g).

The reaction with the more positive electrode potential will proceed as a reduction (forward direction): \( \text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+} \). The reaction with the less positive potential will proceed as an oxidation (reverse direction): \( 2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^- \). The cell potential is: \( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 - 0.54 = +0.23\text{ V} \). To write the balanced ionic equation, multiply the iron half-equation by 2 to balance electrons and combine: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \).

Mark 1: Correct calculation of \( E^\ominus_{\text{cell}} = +0.23\text{ V} \) (state sign and value).
Mark 2: Correctly balanced reactants and products in the ionic equation.
Mark 3: Correct state symbols and stoichiometry for the overall equation: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \).

First, find the mass of water of crystallization lost: \( m(\text{H}_2\text{O}) = 4.76 - 2.60 = 2.16\text{ g} \). Next, calculate the moles of anhydrous cobalt(II) chloride: \( M_{\text{r}}(\text{CoCl}_2) = 58.9 + (2 \times 35.5) = 129.9\text{ g mol}^{-1} \). \( n(\text{CoCl}_2) = \frac{2.60}{129.9} = 0.0200\text{ mol} \). Calculate the moles of water: \( n(\text{H}_2\text{O}) = \frac{2.16}{18.0} = 0.120\text{ mol} \). Determine the mole ratio: \( x = \frac{n(\text{H}_2\text{O})}{n(\text{CoCl}_2)} = \frac{0.120}{0.0200} = 6 \). Therefore, \( x = 6 \).

Mark 1: Calculate the mass of water lost (\( 2.16\text{ g} \)) and the molar mass of anhydrous cobalt(II) chloride (\( 129.9\text{ g mol}^{-1} \)).
Mark 2: Calculate the moles of \( \text{CoCl}_2 \) (\( 0.0200\text{ mol} \)) and moles of \( \text{H}_2\text{O} \) (\( 0.120\text{ mol} \)).
Mark 3: Find the integer ratio of \( \text{H}_2\text{O} : \text{CoCl}_2 \) to deduce \( x = 6 \).

Step 1: Convert ethanol into a halogenoalkane (such as bromoethane or chloroethane). A suitable reagent is phosphorus tribromide (\( \text{PBr}_3 \)) or phosphorus pentachloride (\( \text{PCl}_5 \)). The intermediate is bromoethane (or chloroethane). Step 2: Convert the halogenoalkane into ethylamine by nucleophilic substitution. This requires reaction with excess ammonia (\( \text{NH}_3 \)) in ethanol as solvent, heated in a sealed tube/under pressure.

Mark 1: Reagent for Step 1: \( \text{PBr}_3 \) or \( \text{PCl}_5 \) or concentrated hydrobromic acid.
Mark 2: Name (or formula) of intermediate: bromoethane (or chloroethane).
Mark 3: Reagents and conditions for Step 2: excess ammonia (excess needed to prevent further substitution), dissolved in ethanol, heated in a sealed tube (or under pressure).

Using the Arrhenius equation: \( \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_{\text{a}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \). Substitute the given values: \( \ln\left(\frac{6.00 \times 10^{-3}}{1.50 \times 10^{-3}}\right) = -\frac{E_{\text{a}}}{8.31} \left(\frac{1}{320} - \frac{1}{300}\right) \). This simplifies to: \( \ln(4) = -\frac{E_{\text{a}}}{8.31} \left(-2.0833 \times 10^{-4}\right) \). \( 1.3863 = E_{\text{a}} \times 2.507 \times 10^{-5} \). \( E_{\text{a}} = \frac{1.3863}{2.507 \times 10^{-5}} = 55296\text{ J mol}^{-1} \). Convert to \( \text{kJ mol}^{-1} \): \( E_{\text{a}} = 55.3\text{ kJ mol}^{-1} \).

Mark 1: Correct substitution of values into the Arrhenius equation.
Mark 2: Calculate \( E_{\text{a}} \) as \( 55300\text{ J mol}^{-1} \).
Mark 3: Correct conversion to \( \text{kJ mol}^{-1} \) and appropriate rounding to 3 significant figures (\( 55.3\text{ kJ mol}^{-1} \), accept range 55.2 to 55.4).

Step 1: Calculate the moles of manganate(VII) ions used:
\(n(\text{MnO}_4^-) = 0.02250\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 4.50 \times 10^{-4}\text{ mol}\)

Step 2: Use the stoichiometry of the reaction (\(1\text{ mol of } \text{MnO}_4^-\text{ reacts with } 5\text{ mol of } \text{Fe}^{2+}\)) to find moles of iron:
\(n(\text{Fe}^{2+}) = 5 \times 4.50 \times 10^{-4}\text{ mol} = 2.25 \times 10^{-3}\text{ mol}\)

Step 3: Calculate the mass of iron and the percentage by mass in the ore:
\(\text{Mass of Fe} = 2.25 \times 10^{-3}\text{ mol} \times 55.8\text{ g mol}^{-1} = 0.12555\text{ g}\)
\(\text{Percentage by mass} = \frac{0.12555\text{ g}}{1.25\text{ g}} \times 100\% = 10.044\% \approx 10.0\%\)

1 mark: Correct calculation of moles of manganate(VII) (\(4.50 \times 10^{-4}\text{ mol}\)).
1 mark: Correct calculation of moles of iron(II) using 1:5 ratio (\(2.25 \times 10^{-3}\text{ mol}\)).
1 mark: Correct calculation of percentage by mass to 3 significant figures (\(10.0\%\)). Accept \(10\%\) or \(10.04\%\).

Step 1: Calculate the moles of \(\text{M}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\):
\(n(\text{carbonate}) = \frac{1.431\text{ g}}{286.2\text{ g mol}^{-1}} = 0.00500\text{ mol}\)

Step 2: Since the stoichiometry between the carbonate and \(\text{CO}_2\) is 1:1, moles of \(\text{CO}_2\) = \(0.00500\text{ mol}\).

Step 3: Calculate the volume of \(\text{CO}_2\) using \(PV = nRT\):
\(V = \frac{nRT}{P} = \frac{0.00500 \times 8.31 \times 298}{101000} = 1.226 \times 10^{-4}\text{ m}^3\)
Convert volume to \(\text{cm}^3\):
\(V = 1.226 \times 10^{-4} \times 10^6 = 122.6\text{ cm}^3 \approx 123\text{ cm}^3\)

1 mark: Correct moles of hydrated metal carbonate (\(0.00500\text{ mol}\)).
1 mark: Rearranging ideal gas equation correctly with pressure in Pa (\(101000\text{ Pa}\)).
1 mark: Calculation of final volume in \(\text{cm}^3\) to 3 significant figures (\(123\text{ cm}^3\) or \(122.6\text{ cm}^3\)).

Step 1: Calculate the initial moles of toluene:
\(n(\text{toluene}) = \frac{4.60\text{ g}}{92.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\)

Step 2: Calculate the overall yield fraction:
\(\text{Overall Yield} = 0.700 \times 0.600 = 0.420\) (or \(42.0\%\))

Step 3: Calculate the expected moles and mass of ethyl benzoate:
\(n(\text{ethyl benzoate}) = 0.0500\text{ mol} \times 0.420 = 0.0210\text{ mol}\)
\(\text{Mass of ethyl benzoate} = 0.0210\text{ mol} \times 150.0\text{ g mol}^{-1} = 3.15\text{ g}\)

1 mark: Correct calculation of starting moles of toluene (\(0.0500\text{ mol}\)).
1 mark: Correct determination of overall yield (\(42.0\%\)) or step-by-step moles (\(0.0350\text{ mol}\) intermediate benzoic acid).
1 mark: Correct final mass of ethyl benzoate (\(3.15\text{ g}\)).

Step 1: Rearrange the rate equation to solve for \(k\):
\(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}\)

Step 2: Substitute the experimental values into the equation:
\(k = \frac{2.56 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.0400\text{ mol dm}^{-3})(0.0800\text{ mol dm}^{-3})} = 0.0800\)

Step 3: Determine the units for a second-order reaction:
\(\text{Units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

1 mark: Correct rearrangement of the rate equation and substitution.
1 mark: Correct numerical value of \(k\) as \(0.0800\) or \(8.00 \times 10^{-2}\).
1 mark: Correct units of \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (independent of numerical calculation).

Step 1: Calculate moles of EDTA\(^{4-}\) used:
\(n(\text{EDTA}^{4-}) = 0.01840\text{ dm}^3 \times 0.0150\text{ mol dm}^{-3} = 2.76 \times 10^{-4}\text{ mol}\)

Step 2: Since the stoichiometry is 1:1, moles of \(\text{Cu}^{2+}\) in \(25.0\text{ cm}^3\) is \(2.76 \times 10^{-4}\text{ mol}\). Calculate the concentration in \(\text{mol dm}^{-3}\):
\(c(\text{Cu}^{2+}) = \frac{2.76 \times 10^{-4}\text{ mol}}{0.0250\text{ dm}^3} = 0.01104\text{ mol dm}^{-3}\)

Step 3: Convert the concentration to \(\text{g dm}^{-3}\):
\(\text{Concentration} = 0.01104\text{ mol dm}^{-3} \times 63.5\text{ g mol}^{-1} = 0.70104\text{ g dm}^{-3} \approx 0.701\text{ g dm}^{-3}\)

1 mark: Correct calculation of moles of EDTA\(^{4-}\) or \(\text{Cu}^{2+}\) (\(2.76 \times 10^{-4}\text{ mol}\)).
1 mark: Correct molar concentration of \(\text{Cu}^{2+}\) (\(0.01104\text{ mol dm}^{-3}\)).
1 mark: Correct mass concentration in \(\text{g dm}^{-3}\) to 3 significant figures (\(0.701\text{ g dm}^{-3}\)).

Step 1: Calculate moles of \(\text{HCl}\) used in titration:
\(n(\text{HCl}) = 0.02240\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.24 \times 10^{-3}\text{ mol}\)

Step 2: Calculate moles of \(\text{Na}_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) aliquot using the 1:2 ratio:
\(n(\text{Na}_2\text{CO}_3\text{ in aliquot}) = \frac{2.24 \times 10^{-3}}{2} = 1.12 \times 10^{-3}\text{ mol}\)
Multiply by 10 to find moles in \(250.0\text{ cm}^3\):
\(n(\text{Na}_2\text{CO}_3\text{ total}) = 1.12 \times 10^{-2}\text{ mol}\)

Step 3: Calculate mass of pure \(\text{Na}_2\text{CO}_3\) and percentage purity:
\(\text{Mass of pure } \text{Na}_2\text{CO}_3 = 1.12 \times 10^{-2}\text{ mol} \times 106.0\text{ g mol}^{-1} = 1.1872\text{ g}\)
\(\text{Percentage purity} = \frac{1.1872\text{ g}}{1.45\text{ g}} \times 100\% = 81.875\% \approx 81.9\%\)

1 mark: Correct moles of \(\text{HCl}\) (\(2.24 \times 10^{-3}\text{ mol}\)) and calculation of moles of \(\text{Na}_2\text{CO}_3\) in aliquot (\(1.12 \times 10^{-3}\text{ mol}\)).
1 mark: Scaled up moles of \(\text{Na}_2\text{CO}_3\) to \(250.0\text{ cm}^3\) (\(1.12 \times 10^{-2}\text{ mol}\)) and mass of pure compound (\(1.1872\text{ g}\)).
1 mark: Correct calculation of percentage purity to 3 significant figures (\(81.9\%\)). Accept \(81.9\%\) or \(81.88\%\).

Step 1: Calculate the moles of benzoic acid required:
\(n(\text{benzoic acid}) = \frac{5.00\text{ g}}{122.0\text{ g mol}^{-1}} = 0.04098\text{ mol}\)

Step 2: Account for the percentage yield of \(85.0\%\) to find the theoretical moles of benzamide required:
\(n(\text{benzamide}) = \frac{0.04098\text{ mol}}{0.850} = 0.04821\text{ mol}\)

Step 3: Calculate the mass of benzamide required:
\(\text{Mass of benzamide} = 0.04821\text{ mol} \times 121.0\text{ g mol}^{-1} = 5.833\text{ g} \approx 5.83\text{ g}\)

1 mark: Correct moles of benzoic acid (\(0.04098\text{ mol}\)).
1 mark: Correct scaling of moles by dividing by efficiency (0.850) (\(0.04821\text{ mol}\)).
1 mark: Correct final mass of benzamide to 3 significant figures (\(5.83\text{ g}\)).

Step 1: Calculate the left-hand side of the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \ln\left(\frac{1.20 \times 10^{-2}}{1.50 \times 10^{-3}}\right) = \ln(8.00) = 2.0794\)

Step 2: Calculate the temperature difference term:
\(\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \left(\frac{1}{318} - \frac{1}{298}\right) = 0.0031447 - 0.0033557 = -2.110 \times 10^{-4}\text{ K}^{-1}\)

Step 3: Solve for \(E_a\):
\(2.0794 = -\frac{E_a}{8.31} \times (-2.110 \times 10^{-4})\)
\(E_a = \frac{2.0794 \times 8.31}{2.110 \times 10^{-4}} = 81888\text{ J mol}^{-1} = 81.9\text{ kJ mol}^{-1}\)

1 mark: Correct evaluation of the logarithmic ratio (\(2.079\)) and reciprocal temperature term (\(-2.11 \times 10^{-4}\text{ K}^{-1}\)).
1 mark: Rearranging expression correctly to calculate \(E_a\) in \(\text{J mol}^{-1}\) (\(81888\text{ J mol}^{-1}\)).
1 mark: Conversion to \(\text{kJ mol}^{-1}\) with correct value to 3 significant figures (\(81.9\text{ kJ mol}^{-1}\)).

First, calculate the moles of \(\text{AgCl}\) precipitated:
\(\text{moles of AgCl} = \frac{0.804}{107.9 + 35.5} = \frac{0.804}{143.4} = 5.61 \times 10^{-3}\text{ mol}\)

Since the formula is \([Co(NH_3)_x]Cl_3\), each mole of complex contains 3 moles of chloride ions. Therefore:
\(\text{moles of complex} = \frac{5.61 \times 10^{-3}}{3} = 1.87 \times 10^{-3}\text{ mol}\)

Now, calculate the molar mass of the complex:
\(\text{Molar Mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.500}{1.87 \times 10^{-3}} = 267.4\text{ g mol}^{-1}\)

Subtract the molar masses of cobalt and chlorine:
\(267.4 - 58.9 - (3 \times 35.5) = 102.0\text{ g mol}^{-1}\)

This mass corresponds to the ammonia ligands:
\(x \times 17.0 = 102.0 \Rightarrow x = 6\)

1. Moles of \(\text{AgCl}\) calculated correctly as \(5.61 \times 10^{-3}\text{ mol}\) (1 mark)
2. Moles of complex and molar mass of complex calculated correctly as \(267.4\text{ g mol}^{-1}\) (1 mark)
3. Final integer value of \(x = 6\) obtained with working shown (1 mark).

Mass of water lost = \(2.36\text{ g} - 1.64\text{ g} = 0.72\text{ g}\)

Determine the number of moles of water:
\(\text{moles of H}_2\text{O} = \frac{0.72}{18.0} = 0.040\text{ mol}\)

Determine the number of moles of anhydrous calcium nitrate:
\(\text{moles of Ca(NO}_3\text{)}_2 = \frac{1.64}{164.1} = 0.010\text{ mol}\)

Find the mole ratio:
\(y = \frac{0.040}{0.010} = 4\)

1. Mass of water lost calculated as \(0.72\text{ g}\) and moles of water calculated as \(0.040\text{ mol}\) (1 mark)
2. Moles of anhydrous calcium nitrate calculated as \(0.010\text{ mol}\) (1 mark)
3. Simplest ratio of water to anhydrous salt calculated to give \(y = 4\) (1 mark).

Calculate the moles of benzoic acid used:
\(\text{moles of benzoic acid} = \frac{6.10}{122.0} = 0.0500\text{ mol}\)

Since the stoichiometry is 1:1, the theoretical yield of ethyl benzoate is \(0.0500\text{ mol}\).

Calculate the theoretical mass of ethyl benzoate:
\(\text{theoretical mass} = 0.0500\text{ mol} \times 150.0\text{ g mol}^{-1} = 7.50\text{ g}\)

Calculate the percentage yield:
\(\text{percentage yield} = \frac{5.25}{7.50} \times 100\% = 70.0\%\)

1. Moles of benzoic acid calculated as \(0.0500\text{ mol}\) (1 mark)
2. Theoretical mass of ethyl benzoate calculated as \(7.50\text{ g}\) (1 mark)
3. Correct calculation of percentage yield as \(70.0\%\) (accept 70%) (1 mark).

Use the Arrhenius equation in the form:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

Substitute the values into the equation:
\(\ln\left(\frac{1.60 \times 10^{-3}}{3.20 \times 10^{-4}}\right) = \frac{E_a}{8.31} \left(\frac{1}{298} - \frac{1}{318}\right)\)
\(\ln(5.0) = \frac{E_a}{8.31} \left(3.356 \times 10^{-3} - 3.145 \times 10^{-3}\right)\)
\(1.6094 = \frac{E_a}{8.31} \left(2.110 \times 10^{-4}\right)\)

Solve for \(E_a\):
\(E_a = \frac{1.6094 \times 8.31}{2.110 \times 10^{-4}} = 63371\text{ J mol}^{-1} = 63.4\text{ kJ mol}^{-1}\)

1. Correct substitution of rate constants and temperatures into the Arrhenius equation (1 mark)
2. Correct evaluation of \(\ln(k_2/k_1)\) and temperature reciprocal difference (1 mark)
3. Final value of \(E_a\) in the range \(63.2\text{ - }63.5\text{ kJ mol}^{-1}\) with appropriate units (1 mark).

The half-reaction with the more positive potential is reduced: \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2e^- \rightarrow 2\text{SO}_4^{2-}(\text{aq})\)

The half-reaction with the less positive potential is oxidized: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + e^-\)

Multiply the oxidation half-reaction by 2 to balance electrons and combine:
\(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\)

Calculate \(E^\ominus_{\text{cell}}\):
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 2.01\text{ V} - 0.77\text{ V} = +1.24\text{ V}\)

1. Identifies that the iron half-reaction must be reversed and multiplied by 2 (1 mark)
2. Writes the correct overall balanced equation with reactants on the left and products on the right (1 mark)
3. Calculates \(E^\ominus_{\text{cell}} = +1.24\text{ V}\) (including standard units and sign) (1 mark).

A reaction is thermodynamically feasible when \(\Delta G \le 0\).
At the minimum temperature for feasibility, \(\Delta G = 0\), so \(\Delta H - T\Delta S = 0\) which gives \(T = \frac{\Delta H}{\Delta S}\).

Convert \(\Delta H\) to \(\text{J mol}^{-1}\):
\(\Delta H = 178 \times 10^3\text{ J mol}^{-1}\)

Calculate \(T\):
\(T = \frac{178000}{160} = 1112.5\text{ K}\) (or \(1113\text{ K}\) to 4 significant figures).

Assumption: We assume that both \(\Delta H^\ominus\) and \(\Delta S^\ominus\) do not change with temperature.

1. Uses condition \(\Delta G = 0\) or \(\Delta G \le 0\) to state relation \(T = \Delta H / \Delta S\) (1 mark)
2. Correct unit conversion for enthalpy or entropy and calculates temperature as \(1112.5\text{ K}\) or \(1113\text{ K}\) (1 mark)
3. States the assumption that enthalpy and entropy values remain constant at different temperatures (1 mark).

For a weak acid, \(K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}\).
Therefore, \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\).

Substitute the values:
\([\text{H}^+] = \sqrt{1.60 \times 10^{-4} \times 0.150} = \sqrt{2.40 \times 10^{-5}} = 4.90 \times 10^{-3}\text{ mol dm}^{-3}\)

Calculate pH:
\(\text{pH} = -\log_{10}(4.90 \times 10^{-3}) = 2.31\)

1. Correctly states or uses the expression \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\) (1 mark)
2. Correctly calculates \([\text{H}^+] = 4.90 \times 10^{-3}\text{ mol dm}^{-3}\) (1 mark)
3. Correctly calculates pH to 2 decimal places as \(2.31\) (1 mark).

Use the ideal gas equation: \(pV = nRT\)

Convert units to SI units:
\(p = 101 \times 10^3\text{ Pa}\)
\(V = 58.2 \times 10^{-6}\text{ m}^3\)
\(T = 373\text{ K}\)

Calculate moles, \(n\):
\(n = \frac{pV}{RT} = \frac{101 \times 10^3 \times 58.2 \times 10^{-6}}{8.31 \times 373} = \frac{5.8782}{3099.63} = 1.896 \times 10^{-3}\text{ mol}\)

Calculate molar mass, \(M\):
\(M = \frac{\text{mass}}{n} = \frac{0.120\text{ g}}{1.896 \times 10^{-3}\text{ mol}} = 63.3\text{ g mol}^{-1}\)

1. Correctly converts pressure to \(\text{Pa}\) and volume to \(\text{m}^3\) (1 mark)
2. Correctly calculates the amount of substance \(n = 1.90 \times 10^{-3}\text{ mol}\) (or \(1.896 \times 10^{-3}\)) (1 mark)
3. Correctly calculates molar mass in range \(63.0\) to \(63.5\text{ g mol}^{-1}\) (1 mark).

Mass of water lost = \(0.500 \text{ g} - 0.273 \text{ g} = 0.227 \text{ g}\). Moles of anhydrous \(\text{FeSO}_4 = 0.273 / 151.9 = 1.80 \times 10^{-3} \text{ mol}\). Moles of \(\text{H}_2\text{O} = 0.227 / 18.0 = 1.26 \times 10^{-2} \text{ mol}\). Ratio \(x = (1.26 \times 10^{-2}) / (1.80 \times 10^{-3}) = 7.00\). Thus, the value of \(x\) is 7.

1 Mark: Calculate moles of anhydrous \(\text{FeSO}_4\) (\(1.80 \times 10^{-3} \text{ mol}\)). 1 Mark: Calculate moles of \(\text{H}_2\text{O}\) (\(1.26 \times 10^{-2} \text{ mol}\)). 1 Mark: Find the correct whole-number ratio of \(x = 7\).

Adding excess concentrated HCl introduces chloride ligands which replace water ligands. The reaction is: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\). The tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\), is tetrahedral due to steric hindrance of the larger chloride ligands. The color changes from the pale blue of the hexaaqua complex to yellow-green (or green/yellow) of the tetrachloro complex.

1 Mark: Formula of complex: \([\text{CuCl}_4]^{2-}\). 1 Mark: Shape: Tetrahedral. 1 Mark: Color change: Pale blue to yellow-green / green / yellow.

Step 1: React bromomethane with potassium cyanide (\(\text{KCN}\)) in aqueous ethanol under reflux. This undergoes nucleophilic substitution to form ethanenitrile (\(\text{CH}_3\text{CN}\)). Step 2: Reduce the ethanenitrile using lithium tetrahydridoaluminate (\(\text{LiAlH}_4\)) in dry ether (or hydrogen gas with a nickel catalyst) to produce ethylamine.

1 Mark: Step 1 reagent and conditions: Potassium cyanide (\(\text{KCN}\) / \(\text{NaCN}\)) in aqueous ethanol and heated under reflux. 1 Mark: Identify the intermediate as ethanenitrile (\(\text{CH}_3\text{CN}\)) (explicitly or implicitly). 1 Mark: Step 2 reagent and conditions: Lithium tetrahydridoaluminate (\(\text{LiAlH}_4\)) in dry ether (or \(\text{H}_2\) and nickel catalyst).

Integrated rate equation for a first-order reaction is \(\ln([A]_0 / [A]_t) = k t\). Here, \([A]_0 = 0.800\) and \([A]_t = 0.100\). \(\ln(0.800 / 0.100) = \ln(8) = 2.0794\). Therefore, \(t = 2.0794 / (3.45 \times 10^{-3}) = 602.7 \text{ s}\). Alternatively, a reduction from 0.800 to 0.100 is exactly 3 half-lives. The half-life is \(t_{1/2} = \ln(2) / k = 0.693 / (3.45 \times 10^{-3}) = 200.9 \text{ s}\). Total time = \(3 \times 200.9 = 602.7 \text{ s}\). Rounding to 3 significant figures gives 603 s.

1 Mark: Use of first-order integrated rate expression or recognition that the change represents exactly 3 half-lives. 1 Mark: Calculation of half-life as 201 s or setting up the expression \(t = \ln(8)/k\). 1 Mark: Correct calculation of time as 603 s (accept 600 to 604 s).

The electronic configurations are: \(\text{Fe}^{3+}\) is \([\text{Ar}] 3\text{d}^5\) and \(\text{Fe}^{2+}\) is \([\text{Ar}] 3\text{d}^6\). \(\text{Mn}^{2+}\) is \([\text{Ar}] 3\text{d}^5\) and \(\text{Mn}^{3+}\) is \([\text{Ar}] 3\text{d}^4\). A \(3\text{d}^5\) configuration represents a half-filled d-subshell. A half-filled d-subshell has extra stability because of a symmetric distribution of charge and minimized electron-electron repulsion. Hence, iron easily oxidizes from \(\text{Fe}^{2+}\) to the more stable \(\text{Fe}^{3+}\), while manganese is stable in the \(\text{Mn}^{2+}\) state and resistant to oxidation to \(\text{Mn}^{3+}\).

1 Mark: State that both \(\text{Fe}^{3+}\) and \(\text{Mn}^{2+}\) have the \(3\text{d}^5\) electronic configuration. 1 Mark: Explain that \(3\text{d}^5\) is a half-filled d-subshell. 1 Mark: Explain that a half-filled subshell offers extra stability, making these states more stable than \(\text{Fe}^{2+}\) (\(3\text{d}^6\)) and \(\text{Mn}^{3+}\) (\(3\text{d}^4\)) respectively.

First, find moles of nitrogen gas: \(n(\text{N}_2) = 60.0 \text{ dm}^3 / 24.0 \text{ dm}^3 \text{ mol}^{-1} = 2.50 \text{ mol}\). According to the stoichiometric equation, 2 moles of \(\text{NaN}_3\) produce 3 moles of \(\text{N}_2\). Therefore, \(n(\text{NaN}_3) = 2/3 \times n(\text{N}_2) = 2/3 \times 2.50 \text{ mol} = 1.667 \text{ mol}\). Finally, calculate the mass of sodium azide: \(m(\text{NaN}_3) = 1.667 \text{ mol} \times 65.0 \text{ g mol}^{-1} = 108.33 \text{ g}\). To three significant figures, this is 108 g.

1 Mark: Moles of \(\text{N}_2 = 2.50 \text{ mol}\). 1 Mark: Moles of \(\text{NaN}_3 = 1.67 \text{ mol}\) (or 1.667 mol). 1 Mark: Correct final mass of \(108 \text{ g}\) (accept 108.3 g).

In the first step, the catalyst \(\text{Mn}^{2+}\) is oxidized to \(\text{Mn}^{3+}\) by reacting with manganate(VII) ions: \(\text{MnO}_4^- + 4\text{Mn}^{2+} + 8\text{H}^+ \to 5\text{Mn}^{3+} + 4\text{H}_2\text{O}\). In the second step, the intermediate \(\text{Mn}^{3+}\) oxidizes the ethanedioate ions, regenerating the catalyst \(\text{Mn}^{2+}\): \(2\text{Mn}^{3+} + \text{C}_2\text{O}_4^{2-} \to 2\text{Mn}^{2+} + 2\text{CO}_2\).

1 Mark: Identify \(\text{Mn}^{3+}\) as the intermediate ion species. 1 Mark: Write a balanced equation for the formation of \(\text{Mn}^{3+}\) from \(\text{MnO}_4^-\). 1 Mark: Write a balanced equation for the reduction of \(\text{Mn}^{3+}\) back to \(\text{Mn}^{2+}\) by \(\text{C}_2\text{O}_4^{2-}\).

Step 1: Methylbenzene is oxidized to benzoic acid. Reagent: alkaline potassium manganate(VII) (\(\text{KMnO}_4\)), heated under reflux. This forms benzoate ions, which are then acidified with dilute acid (such as hydrochloric acid) to precipitate benzoic acid. Intermediate: Benzoic acid. Step 2: Benzoic acid is esterified to methyl benzoate. Reagents: methanol (\(\text{CH}_3\text{OH}\)) and a concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) catalyst, heated under reflux.

1 Mark: Step 1 reagents and conditions: alkaline potassium manganate(VII) / \(\text{KMnO}_4\), heated under reflux, followed by addition of dilute acid. 1 Mark: Name intermediate: benzoic acid. 1 Mark: Step 2 reagents and conditions: methanol and concentrated sulfuric acid, heated under reflux.

1. Find moles of Al: \(n(\text{Al}) = \frac{0.108}{27.0} = 0.00400\text{ mol}\). 2. Use the stoichiometric ratio from the equation \(2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\) to find the moles of hydrogen gas: \(n(\text{H}_2) = 1.5 \times 0.00400 = 0.00600\text{ mol}\). 3. Apply the ideal gas equation: \(V = \frac{nRT}{P} = \frac{0.00600 \times 8.31 \times 293}{101000} = 1.4464 \times 10^{-4}\text{ m}^3\). 4. Convert this volume from \(\text{m}^3\) to \(\text{cm}^3\): \(1.4464 \times 10^{-4} \times 10^6 = 145\text{ cm}^3\) (to 3 significant figures).

M1: Calculates the moles of hydrogen gas as 0.00600 mol. (1) M2: Correctly rearranges the ideal gas equation to make V the subject, converting units correctly (P = 101000 Pa, T = 293 K). (1) M3: Evaluates the volume as 145 cm3 (or 144.6 cm3). (1)

1. Calculate the mass of water lost during heating: \(4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\). 2. Find the moles of water lost: \(n(\text{H}_2\text{O}) = \frac{1.80}{18.0} = 0.100\text{ mol}\). 3. Find the moles of anhydrous salt remaining: \(n(\text{CuSO}_4) = \frac{3.19}{159.6} = 0.0200\text{ mol}\). 4. Determine the ratio of water moles to anhydrous salt moles: \(x = \frac{0.100}{0.0200} = 5\).

M1: Calculates mass of water (1.80 g) and moles of water (0.100 mol). (1) M2: Calculates moles of anhydrous copper(II) sulfate (0.0200 mol). (1) M3: Divides to obtain the correct integer value x = 5. (1)

1. Calculate the left-hand side: \(\ln(k_2 / k_1) = \ln(9.60 \times 10^{-4} / 1.20 \times 10^{-4}) = \ln(8) = 2.0794\). 2. Calculate the temperature difference term: \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \left(\frac{1}{320} - \frac{1}{300}\right) = -2.0833 \times 10^{-4}\text{ K}^{-1}\). 3. Substitute values to solve for \(E_a\): \(2.0794 = -\frac{E_a}{8.31} \times (-2.0833 \times 10^{-4})\). This yields: \(2.0794 = E_a \times 2.5070 \times 10^{-5}\). 4. Calculate \(E_a\): \(E_a = 82944\text{ J mol}^{-1} = 82.9\text{ kJ mol}^{-1}\) (to 3 significant figures).

M1: Evaluates ln(k2/k1) = 2.08 and calculates the temperature term (1/T2 - 1/T1) = -2.08 x 10^-4 K^-1. (1) M2: Correctly rearranges to solve for Ea in J mol^-1, obtaining a value of approximately 82900 J mol^-1. (1) M3: Converts to kJ mol^-1 and rounds correctly to 3 significant figures to give 82.9 kJ mol^-1 (allow 83.0 if rounding occurred earlier). (1)

1. Compare Experiment 1 and Experiment 2: \([B]\) is constant while \([A]\) is doubled. The initial rate increases by a factor of 4 (from \(2.0 \times 10^{-3}\) to \(8.0 \times 10^{-3}\)). Since \(2^2 = 4\), the reaction is second order with respect to A. 2. Compare Experiment 2 and Experiment 3: \([A]\) is constant while \([B]\) is doubled. The initial rate doubles (from \(8.0 \times 10^{-3}\) to \(1.6 \times 10^{-2}\)). Since \(2^1 = 2\), the reaction is first order with respect to B. 3. Combining these gives the overall rate equation: \(\text{Rate} = k[A]^2[B]\).

M1: States that the reaction is second order with respect to A and explains this using Experiments 1 and 2. (1) M2: States that the reaction is first order with respect to B and explains this using Experiments 2 and 3. (1) M3: Writes the correct rate equation: Rate = k[A]^2[B]. (1)

(a) The balanced equation for the ligand exchange is: \([\text{Cu(H}_2\text{O)}_6]^{2+} + 3\text{en} \rightarrow [\text{Cu(en)}_3]^{2+} + 6\text{H}_2\text{O}\). (b) There are more moles of particles on the product side (7 moles) than on the reactant side (4 moles). This increase in the number of species in solution leads to greater disorder, which results in a significant positive change in system entropy (\(\Delta S_{\text{system}} > 0\)), making the reaction highly feasible.

M1: Writes the correct balanced equation (state symbols not required). (1) M2: Explains that there is an increase in the number of particles in solution (from 4 on the left to 7 on the right). (1) M3: Explains that this leads to an increase in system entropy (disorder) or refers to the chelate effect. (1)

1. Calculate moles of manganate(VII) used: \(n(\text{MnO}_4^-) = 0.0150 \times \frac{18.50}{1000} = 2.775 \times 10^{-4}\text{ mol}\). 2. Use the stoichiometry (1:5 ratio) to find moles of \(\text{Fe}^{2+}\) in the 25.0 \(\text{cm}^3\) sample: \(n(\text{Fe}^{2+}) = 5 \times 2.775 \times 10^{-4} = 1.3875 \times 10^{-3}\text{ mol}\). 3. Scale up to the total 250.0 \(\text{cm}^3\) volume: \(n(\text{Fe}^{2+})_{\text{total}} = 1.3875 \times 10^{-3} \times 10 = 1.3875 \times 10^{-2}\text{ mol}\). 4. Calculate the mass of iron: \(m(\text{Fe}) = 1.3875 \times 10^{-2} \times 55.8 = 0.7742\text{ g}\). 5. Calculate percentage by mass of iron in the original 1.85 g fertilizer sample: \(\% = \frac{0.7742}{1.85} \times 100 = 41.85\% \approx 41.9\%\).

M1: Calculates the moles of Fe2+ in the 25.0 cm3 aliquot (1.3875 x 10^-3 mol). (1) M2: Determines the total mass of Fe in the 250.0 cm3 solution (0.7742 g). (1) M3: Calculates the correct mass percentage to 3 significant figures: 41.9% (accept 41.8%). (1)

(a) Reagent: (Excess) ammonia dissolved in ethanol (ethanolic ammonia). Conditions: Heated under pressure / in a sealed tube. (b) Propylamine is a primary amine which contains a nitrogen atom with a lone pair of electrons. It acts as a nucleophile and can react further with remaining 1-chloropropane, leading to a mixture of secondary/tertiary amines and quaternary ammonium salts, reducing the yield of the primary amine.

M1: Reagent: (Excess) ethanolic ammonia / ammonia in ethanol (do not accept aqueous ammonia alone unless under pressure). (1) M2: Conditions: Heated under pressure / in a sealed container/tube. (1) M3: Explanation: The product (propylamine) is a nucleophile and can react further with 1-chloropropane to yield a mixture of amine products (secondary/tertiary/quaternary). (1)

Step 1: React bromoethane with potassium cyanide (KCN) dissolved in a mixture of ethanol and water, heating under reflux. This nucleophilic substitution forms the intermediate propanenitrile, \(\text{CH}_3\text{CH}_2\text{CN}\). Step 2: Heat the propanenitrile under reflux with a dilute strong acid (such as \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)) to hydrolyze the nitrile group into a carboxylic acid group, forming propanoic acid.

M1: Step 1: Reagents and conditions (KCN in aqueous ethanol AND heat under reflux). (1) M2: Name of intermediate: propanenitrile (accept formula CH3CH2CN). (1) M3: Step 2: Reagents and conditions (dilute acid / HCl / H2SO4 AND heat under reflux). (1)

1. Calculate the amount of the chromium complex in moles:
\(n(\text{complex}) = \frac{1.332\text{ g}}{266.5\text{ g mol}^{-1}} = 0.00500\text{ mol}\)

2. Calculate the amount of \(\text{AgCl}\) precipitated in moles:
\(n(\text{AgCl}) = \frac{0.717\text{ g}}{143.4\text{ g mol}^{-1}} = 0.00500\text{ mol}\)

3. Determine the ratio of free chloride ions to the complex:
\(\text{Ratio} = \frac{n(\text{AgCl})}{n(\text{complex})} = \frac{0.00500}{0.00500} = 1\)
This means that only 1 chloride ion is ionic (outside the coordination sphere) per formula unit. The remaining 2 chloride ions and 4 water molecules must be ligands within the coordination sphere to maintain octahedral geometry (coordination number of 6). Thus, the complex is \([\text{Cr(H}_2\text{O)}_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}\), and the complex cation is \([\text{Cr(H}_2\text{O)}_4\text{Cl}_2]^+\).

- **M1**: Calculates the moles of the chromium complex and the moles of \(\text{AgCl}\) as \(0.00500\text{ mol}\) (1)
- **M2**: Deduces that the ratio is 1:1, meaning 1 chloride ion is outside the sphere, and 2 chloride ions must be bonded as ligands inside the coordination sphere (1)
- **M3**: Gives the correct formula of the complex cation: \([\text{Cr(H}_2\text{O)}_4\text{Cl}_2]^+\) (or \([\text{Cr(Cl)}_2\text{(H}_2\text{O)}_4]^+\)) (1)

*Note: Accept charges written as \(+1\) or \(+\). Do not accept formulas with water of crystallization inside the bracket.*

(i) The hexaaquairon(III) ion, \([\text{Fe(H}_2\text{O)}_6]^{3+}\), appears yellow or pale brown in aqueous solution due to slight hydrolysis (though it is pale violet when pure).
(ii) Addition of aqueous sodium hydroxide deprotonates the complex to form a red-brown precipitate of iron(III) hydroxide, \(\text{Fe(H}_2\text{O)}_3\text{(OH)}_3(\text{s})\).
(iii) The balanced ionic equation is:
\([\text{Fe(H}_2\text{O)}_6]^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(H}_2\text{O)}_3\text{(OH)}_3(\text{s}) + 3\text{H}_2\text{O(l)}\)

- **M1**: Yellow / brown / pale violet (1)
- **M2**: Red-brown (precipitate) (1)
- **M3**: \([\text{Fe(H}_2\text{O)}_6]^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(H}_2\text{O)}_3\text{(OH)}_3(\text{s}) + 3\text{H}_2\text{O(l)}\) with correct state symbols (1)
*Allow: \(\text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_3(\text{s})\) for M3.*

1. Convert values to SI units:
\(p = 101 \times 10^3\text{ Pa}\)
\(V = 36.8 \times 10^{-6}\text{ m}^3\)
\(T = 298\text{ K}\)

2. Calculate moles of \(\text{CO}_2\) using the ideal gas equation:
\(n(\text{CO}_2) = \frac{pV}{RT} = \frac{101000 \times 36.8 \times 10^{-6}}{8.31 \times 298} = 0.00150\text{ mol}\)

3. Since \(\text{MCO}_3 + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\), the mole ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is 1:1.
\(n(\text{MCO}_3) = 0.00150\text{ mol}\)

4. Calculate the molar mass of \(\text{MCO}_3\):
\(M(\text{MCO}_3) = \frac{0.150\text{ g}}{0.00150\text{ mol}} = 100\text{ g mol}^{-1}\)

5. Determine the molar mass of metal \(\text{M}\):
\(M(\text{M}) = 100 - (12.0 + 3 \times 16.0) = 40.0\text{ g mol}^{-1}\)
This corresponds to Calcium (\(\text{Ca}\)).

- **M1**: Uses \(pV = nRT\) with correct conversions to calculate \(n(\text{CO}_2) = 0.00150\text{ mol}\) (1)
- **M2**: Calculates the molar mass of \(\text{MCO}_3\) as \(100\text{ g mol}^{-1}\) or the molar mass of \(\text{M}\) as \(40.0\text{ g mol}^{-1}\) (1)
- **M3**: Identifies the metal as Calcium (or \(\text{Ca}\)) based on a calculated molar mass close to \(40.1\text{ g mol}^{-1}\) (1)

1. Find mass of Carbon in \(2.64\text{ g}\) of \(\text{CO}_2\):
\(m(\text{C}) = 2.64 \times \frac{12.0}{44.0} = 0.72\text{ g}\)

2. Find mass of Hydrogen in \(0.72\text{ g}\) of \(\text{H}_2\text{O}\):
\(m(\text{H}) = 0.72 \times \frac{2.0}{18.0} = 0.08\text{ g}\)

3. Find mass of Oxygen by subtraction:
\(m(\text{O}) = 1.44 - 0.72 - 0.08 = 0.64\text{ g}\)

4. Convert masses to moles:
\(n(\text{C}) = \frac{0.72}{12.0} = 0.060\text{ mol}\)
\(n(\text{H}) = \frac{0.08}{1.0} = 0.080\text{ mol}\)
\(n(\text{O}) = \frac{0.64}{16.0} = 0.040\text{ mol}\)

5. Find the simplest whole-number mole ratio:
\(\text{C} : \text{H} : \text{O} = \frac{0.060}{0.040} : \frac{0.080}{0.040} : \frac{0.040}{0.040} = 1.5 : 2 : 1 = 3 : 4 : 2\)
The empirical formula is \(\text{C}_3\text{H}_4\text{O}_2\).

- **M1**: Calculates the masses of carbon (\(0.72\text{ g}\)) and hydrogen (\(0.08\text{ g}\)) (1)
- **M2**: Calculates the mass of oxygen (\(0.64\text{ g}\)) and converts all masses to moles (\(0.060\text{ mol C}\), \(0.080\text{ mol H}\), \(0.040\text{ mol O}\)) (1)
- **M3**: Obtains the correct simplest whole-number ratio and states the empirical formula: \(\text{C}_3\text{H}_4\text{O}_2\) (1)

Because the methyl group on methylbenzene is 2,4-directing and the carboxylic acid group is 3-directing, we must oxidize the methyl group first to make benzoic acid before performing nitration.

- **Step 1**: React methylbenzene with alkaline potassium manganate(VII) (\(\text{KMnO}_4\)), heat under reflux, followed by acidification with a dilute acid (such as hydrochloric or sulfuric acid). This forms the intermediate **benzoic acid** (\(\text{C}_6\text{H}_5\text{COOH}\)).
- **Step 2**: React benzoic acid with a mixture of concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) and heat (e.g., above \(50\text{ }^\circ\text{C}\)) to obtain 3-nitrobenzoic acid.

- **M1**: Step 1 reagents and conditions: alkaline \(\text{KMnO}_4\), heat under reflux, followed by addition of dilute acid (1)
- **M2**: Identifies intermediate as benzoic acid / \(\text{C}_6\text{H}_5\text{COOH}\) (1)
- **M3**: Step 2 reagents and conditions: concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\), heat (1)

To synthesize 2-phenylbutan-2-ol (a tertiary alcohol) from phenylethanone (\(\text{C}_6\text{H}_5\text{COCH}_3\)), we must add an ethyl group (\(-\text{CH}_2\text{CH}_3\)).
1. The Grignard reagent required is ethylmagnesium bromide (or chloride/iodide), \(\text{CH}_3\text{CH}_2\text{MgBr}\).
2. The reaction must be carried out in dry ether (ethoxyethane) solvent to prevent hydrolysis of the nucleophilic Grignard reagent.
3. The second stage requires dilute aqueous acid (e.g. \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)) to protonate the intermediate alkoxide salt to form the final alcohol product.

- **M1**: Grignard reagent: ethylmagnesium bromide / ethylmagnesium chloride / ethylmagnesium iodide / \(\text{CH}_3\text{CH}_2\text{MgBr}\) / \(\text{CH}_3\text{CH}_2\text{MgCl}\) / \(\text{CH}_3\text{CH}_2\text{MgI}\) (1)
- **M2**: Solvent: dry ether / ethoxyethane / anhydrous diethyl ether (1)
- **M3**: Second stage: dilute (aqueous) hydrochloric acid / dilute sulfuric acid / \(\text{H}_3\text{O}^+\) / \(\text{H}^+(\text{aq})\) (1)

Use the two-point form of the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)

1. Calculate the left-hand side:
\(\ln\left(\frac{1.25 \times 10^{-3}}{2.50 \times 10^{-4}}\right) = \ln(5.00) = 1.6094\)

2. Calculate the temperature reciprocal term:
\(\left(\frac{1}{320} - \frac{1}{300}\right) = 0.003125 - 0.0033333 = -2.0833 \times 10^{-4}\text{ K}^{-1}\)

3. Rearrange and solve for \(E_a\):
\(1.6094 = -\frac{E_a}{8.31} \times (-2.0833 \times 10^{-4})\)
\(E_a = \frac{1.6094 \times 8.31}{2.0833 \times 10^{-4}} = 64198\text{ J mol}^{-1}\)

4. Convert to \(\text{kJ mol}^{-1}\):
\(E_a = 64.2\text{ kJ mol}^{-1}\)

- **M1**: Correct substitution of values into the Arrhenius equation (1)
- **M2**: Correct calculation of intermediate terms, e.g., \(\ln(5) = 1.61\) and \(\Delta(\frac{1}{T}) = -2.08 \times 10^{-4}\) (1)
- **M3**: Calculates \(E_a = 64.2\text{ kJ mol}^{-1}\) (accept range \(64.0\) to \(64.3\)) (1)

1. Determine order with respect to \([\text{S}_2\text{O}_8^{2-}]\):
Comparing Exp 1 and Exp 2, \([\text{I}^-]\) is constant, \([\text{S}_2\text{O}_8^{2-}]\) doubles, and the rate doubles (\(1.5 \times 10^{-5}\) to \(3.0 \times 10^{-5}\)). Thus, the order with respect to \([\text{S}_2\text{O}_8^{2-}]\) is 1.

2. Determine order with respect to \([\text{I}^-]\):
Comparing Exp 2 and Exp 3, \([\text{S}_2\text{O}_8^{2-}]\) is constant, \([\text{I}^-]\) doubles, and the rate doubles (\(3.0 \times 10^{-5}\) to \(6.0 \times 10^{-5}\)). Thus, the order with respect to \([\text{I}^-]\) is 1.

3. Rate Equation:
\(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

4. Calculate the rate constant \(k\) using Exp 1 data:
\(1.5 \times 10^{-5} = k \times 0.010 \times 0.010\)
\(k = \frac{1.5 \times 10^{-5}}{1.0 \times 10^{-4}} = 0.15\)

5. Determine the units:
\(\text{Units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

- **M1**: Deduces both orders are 1 and writes the correct rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (1)
- **M2**: Calculates the value of \(k = 0.15\) (1)
- **M3**: States correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(\text{dm}^3 / \text{mol} / \text{s}\)) (1)

First, calculate the number of moles of \(\text{CO}_2\) produced using the ideal gas equation:
\(pV = nRT \implies n = \frac{pV}{RT}\)
Convert units: \(p = 1.01 \times 10^5\text{ Pa}\), \(V = 367 \times 10^{-6}\text{ m}^3\), \(T = 298\text{ K}\).
\(n = \frac{1.01 \times 10^5 \times 3.67 \times 10^{-4}}{8.31 \times 298} = 0.0150\text{ mol}\).

Since the stoichiometry of \(M\text{CO}_3\) to \(\text{CO}_2\) is 1:1, there are \(0.0150\text{ mol}\) of \(M\text{CO}_3\).

Molar mass of \(M\text{CO}_3 = \frac{\text{mass}}{\text{moles}} = \frac{1.50\text{ g}}{0.0150\text{ mol}} = 100.0\text{ g mol}^{-1}\).

Molar mass of \(M = M_r(M\text{CO}_3) - M_r(\text{CO}_3^{2-}) = 100.0 - (12.0 + 3 \times 16.0) = 40.0\text{ g mol}^{-1}\).

Looking at the Periodic Table, the group 2 metal with a molar mass close to \(40.0\text{ g mol}^{-1}\) is Calcium (Ca).

M1: Calculates the number of moles of \(\text{CO}_2\) using the ideal gas equation: \(0.0150\text{ mol}\) (allow 1 sig fig change).
M2: Calculates the molar mass of \(M\text{CO}_3\) as \(100.0\text{ g mol}^{-1}\).
M3: Deduces the molar mass of \(M\) as \(40.0\text{ g mol}^{-1}\) and identifies \(M\) as Calcium (Ca).

First, calculate the moles of the chromium complex and \(\text{AgCl}\) precipitate:
\(\text{Moles of complex} = \frac{2.665}{266.5} = 0.0100\text{ mol}\).
\(\text{Moles of AgCl} = \frac{2.866}{143.4} = 0.0200\text{ mol}\).

The ratio of free chloride ions (outside the coordination sphere) to the complex is \(\frac{0.0200}{0.0100} = 2\).
This indicates that there are 2 chloride ions outside the coordination sphere acting as counter-ions. Therefore, 1 chloride ion must reside inside the coordination sphere as a ligand.

Since the coordination number of chromium is 6, the remaining 5 coordination sites are occupied by water molecules. One water molecule resides outside the sphere as water of crystallization.
Formula: \([\text{Cr(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\).
Systematic name: pentaaquachloridochromium(III) chloride hydrate.

M1: Calculates moles of complex (\(0.0100\text{ mol}\)) and moles of \(\text{AgCl}\) precipitate (\(0.0200\text{ mol}\)).
M2: Deduces the mole ratio of free \(\text{Cl}^-\text{ to complex}\) is 2:1, showing 2 chlorides are outside the coordination sphere.
M3: Gives the correct coordination formula: \([\text{Cr(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\) and name: pentaaquachloridochromium(III) chloride (monohydrate).

Stage 1 of the reduction involves heating nitrobenzene under reflux with tin (Sn) and concentrated hydrochloric acid (HCl). The phenylammonium ion formed is then treated with sodium hydroxide to liberate phenylamine.
The equation for the reduction of nitrobenzene is:
\(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\).
Phenylamine is a liquid that is codistilled with water via steam distillation, followed by solvent extraction and final distillation.

M1: Reagents and conditions: Tin / \(\text{Sn}\) and concentrated \(\text{HCl}\), heated under reflux / heated.
M2: Equation: \(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\) (balanced correctly with 6[H] and 2H2O).
M3: Separation method: Steam distillation (accept: solvent extraction / ether extraction followed by distillation).

We use the two-temperature form of the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)

Substitute the given values:
\(k_1 = 3.46 \times 10^{-5}\text{ s}^{-1}\), \(T_1 = 298\text{ K}\)
\(k_2 = 4.87 \times 10^{-3}\text{ s}^{-1}\), \(T_2 = 338\text{ K}\)

\(\ln\left(\frac{4.87 \times 10^{-3}}{3.46 \times 10^{-5}}\right) = -\frac{E_a}{8.31} \left(\frac{1}{338} - \frac{1}{298}\right)\)

\(\ln(140.75) = -\frac{E_a}{8.31} (0.0029586 - 0.0033557)\)

\(4.947 = -\frac{E_a}{8.31} (-3.971 \times 10^{-4})\)

\(4.947 = \frac{E_a \times 3.971 \times 10^{-4}}{8.31}\)

\(E_a = \frac{4.947 \times 8.31}{3.971 \times 10^{-4}} = 1.035 \times 10^5\text{ J mol}^{-1}\).

Convert to \(\text{kJ mol}^{-1}\):
\(E_a = 103.5\text{ kJ mol}^{-1}\) or \(104\text{ kJ mol}^{-1}\) (to 3 sig figs).

M1: Correct substitution into the Arrhenius equation: \(\ln(4.87 \times 10^{-3} / 3.46 \times 10^{-5}) = -E_a / 8.31 \times (1/338 - 1/298)\) or equivalent.
M2: Evaluates the terms: \(4.947\) and \(-3.971 \times 10^{-4}\) (or equivalent difference in inverse temperatures).
M3: Evaluates \(E_a\) correctly to 3 significant figures with correct units: \(104\text{ kJ mol}^{-1}\) (allow \(103\text{ to }104\)).

The atomic number of iron is 26. The electronic configuration of neutral iron is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2\).
To form \(\text{Fe}^{3+}\), three electrons are removed (two from \(4\text{s}\) and one from \(3\text{d}\)). This yields the electronic configuration:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5\).

\(\text{Fe}^{3+}\) is more stable than \(\text{Fe}^{2+}\) because the \(3\text{d}\) subshell of \(\text{Fe}^{3+}\) is half-filled (containing five unpaired electrons, one in each orbital). A half-filled subshell has a symmetrical distribution of electron density and minimized electron-electron repulsion, making it particularly stable.

M1: Correct electronic configuration for \(\text{Fe}^{3+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5\) (or \([\text{Ar}] 3\text{d}^5\)).
M2: Identifies \(\text{Fe}^{3+}\) as the more stable ion.
M3: Explains that the half-filled d-subshell (\(3\text{d}^5\)) has a symmetrical electron distribution or reduced electron-electron repulsion compared to the \(3\text{d}^6\) configuration of \(\text{Fe}^{2+}\).

Step 1: Write down the stoichiometric relationships:
\(\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}\)
\(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\)
Thus, \(1\text{ mol of ClO}^- \equiv 2\text{ mol of S}_2\text{O}_3^{2-}\).

Step 2: Calculate moles of thiosulfate used in titration:
\(n(\text{S}_2\text{O}_3^{2-}) = 0.100 \times \frac{18.40}{1000} = 1.84 \times 10^{-3}\text{ mol}\).

Step 3: Calculate moles of \(\text{ClO}^-\) in the \(25.0\text{ cm}^3\) aliquot:
\(n(\text{ClO}^-) = \frac{1.84 \times 10^{-3}}{2} = 9.20 \times 10^{-4}\text{ mol}\).

Step 4: Scale up to the total volume of \(250.0\text{ cm}^3\):
\(n(\text{ClO}^-\text{ in 250 cm}^3) = 9.20 \times 10^{-4} \times \frac{250.0}{25.0} = 9.20 \times 10^{-3}\text{ mol}\).

Step 5: Calculate concentration in original \(10.0\text{ cm}^3\) bleach:
\(c(\text{NaClO}) = \frac{9.20 \times 10^{-3}\text{ mol}}{0.0100\text{ dm}^3} = 0.920\text{ mol dm}^{-3}\).

M1: Calculates the moles of thiosulfate: \(1.84 \times 10^{-3}\text{ mol}\) and relates to moles of \(\text{ClO}^-\): \(9.20 \times 10^{-4}\text{ mol}\).
M2: Scales up to the total diluted volume (\(250.0\text{ cm}^3\)) to find moles: \(9.20 \times 10^{-3}\text{ mol}\).
M3: Calculates the correct concentration of original bleach: \(0.920\text{ mol dm}^{-3}\) (with 3 sig figs).

Stage 2 is an acid-catalyzed hydrolysis of a nitrile. The nitrile group (\(-\text{CN}\)) is hydrolyzed to a carboxylic acid group (\(-\text{COOH}\)) by heating under reflux with dilute hydrochloric acid (or dilute sulfuric acid).
Stage 1 is the addition of a nucleophile (\(\text{CN}^-\)) to a carbonyl group (\(\text{C=O}\)), which is classified as nucleophilic addition.
The product of Stage 1 is 2-hydroxy-2-methylpropanenitrile, which has the structural formula \((\text{CH}_3)_2\text{C(OH)CN}\).

M1: Reagents and conditions for hydrolysis: Dilute hydrochloric acid / dilute sulfuric acid / \(\text{HCl(aq)}\) AND heat / reflux.
M2: Name of mechanism: Nucleophilic addition.
M3: Correct structural formula: \((\text{CH}_3)_2\text{C(OH)CN}\) or drawn correctly.

Step 1: Determine orders of reaction:
- Comparing Exp 1 and 2: [A] doubles, [B] and [C] are constant. The rate doubles (\(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\)). Therefore, order with respect to A is 1.
- Comparing Exp 1 and 3: [B] doubles, [A] and [C] are constant. The rate quadruples (\(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Therefore, order with respect to B is 2.
- Comparing Exp 1 and 4: [C] doubles, [A] and [B] are constant. The rate remains unchanged (\(2.0 \times 10^{-4}\)). Therefore, order with respect to C is 0.

Rate Equation: \(\text{Rate} = k[A][B]^2\).

Step 2: Calculate the rate constant, \(k\), using Exp 1:
\(2.0 \times 10^{-4} = k (0.10)(0.10)^2\)
\(2.0 \times 10^{-4} = k (0.0010)\)
\(k = 0.20\).

Step 3: Determine the units of \(k\):
\(\text{Units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

M1: Deduces the correct rate equation: \(\text{Rate} = k[A][B]^2\) (or clearly shows orders: A=1, B=2, C=0).
M2: Calculates \(k = 0.20\) (allow error carried forward from incorrect rate equation).
M3: States units: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) (or \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)) corresponding to the calculated rate equation.

1. Addition of concentrated hydrochloric acid to aqueous copper(II) sulfate:
- The pale blue hexaaquacopper(II) ion, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), undergoes ligand substitution with chloride ions.
- Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)\)
- Observation: Pale blue solution turns yellow-green.
- Shape and Coordination: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) is octahedral with a coordination number of 6. \([\text{CuCl}_4]^{2-}\) is tetrahedral with a coordination number of 4. This change in shape and coordination number is due to the larger size of the chloride ligands, which cause greater steric repulsion around the central copper ion.

2. Addition of 1,2-diaminoethane (en) to the tetrachlorocuprate(II) solution:
- The bidentate 1,2-diaminoethane ligand displaces the monodentate chloride ligands.
- Equation: \([\text{CuCl}_4]^{2-}(aq) + 3\text{en}(aq) \rightleftharpoons [\text{Cu}(\text{en})_3]^{2+}(aq) + 4\text{Cl}^-(aq)\)
- Observation: Yellow-green solution turns deep blue-violet.
- Shape and Coordination: \([\text{Cu}(\text{en})_3]^{2+}\) is octahedral with a coordination number of 6.

3. Thermodynamic driving force (the Chelate Effect):
- During the second reaction, 4 reacting particles (1 complex ion and 3 bidentate ligand molecules) produce 5 product particles (1 complex ion and 4 chloride ions).
- This results in an increase in the number of particles in solution, leading to an increase in the disorder of the system.
- Therefore, the entropy change of the system is positive (\(\Delta S_{\text{sys}} > 0\)).
- Since the enthalpy change (\(\Delta H\)) for ligand substitution reactions involving similar coordinate bonds is usually very small (close to zero), the Gibbs free energy change (\(\Delta G = \Delta H - T\Delta S\)) becomes negative.
- This thermodynamic feasibility makes the chelated complex extremely stable compared to the monodentate complex.

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Explains both reactions with correct equations, colors, shapes, and coordination numbers.
- Explains the change in coordination number from 6 to 4 in terms of ligand size.
- Explains the chelate effect clearly, detailing the increase in the number of particles, positive entropy of system (\(\Delta S > 0\)), and why \(\Delta G < 0\).
- Well-structured, logical flow, and correct chemical terminology throughout.

**Level 2 (3-4 marks):**
- Explains both reactions with mostly correct equations and observations.
- Identifies the shapes of the complex ions.
- Mentions that entropy increases due to more particles on the product side but may lack detail on \(\Delta G\).

**Level 1 (1-2 marks):**
- Describes at least one reaction with some correct observations or complex shapes.
- Shows limited understanding of ligand substitution or entropy changes.
- Fragmented explanation with poor structure.

1. Experimental Method to Determine the Rate Equation:
- Use the initial rates method, specifically an 'iodine clock' reaction.
- Mix known volumes of peroxodisulfate and iodide ions with a fixed, small volume of sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)) and starch indicator.
- Measure the time (\(t\)) taken for the solution to turn blue-black (which happens when all thiosulfate is consumed and free iodine reacts with starch).
- Initial rate is proportional to \(1/t\).
- Repeat the experiment, systematically varying the concentration of \(\text{S}_2\text{O}_8^{2-}\) while keeping the concentration of \(\text{I}^-\), total volume, and temperature constant.
- Repeat again, varying the concentration of \(\text{I}^-\) while keeping the concentration of \(\text{S}_2\text{O}_8^{2-}\) constant.
- Plot graphs of initial rate against concentration or use log-log plots to determine the reaction orders with respect to each reactant, giving the rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

2. Role of the Catalyst:
- The uncatalyzed reaction involves the collision between two negatively charged ions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)). High electrostatic repulsion results in a very high activation energy.
- The \(\text{Fe}^{2+}\) catalyst provides an alternative pathway involving transitions between different oxidation states of iron (\(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\)).
- Step 1: Iron(II) ions reduce peroxodisulfate to sulfate ions, being oxidized to iron(III):
\(\text{S}_2\text{O}_8^{2-}(aq) + 2\text{Fe}^{2+}(aq) \rightarrow 2\text{SO}_4^{2-}(aq) + 2\text{Fe}^{3+}(aq)\)
- Step 2: Iron(III) ions then oxidize iodide to iodine, regenerating the iron(II) catalyst:
\(2\text{I}^-(aq) + 2\text{Fe}^{3+}(aq) \rightarrow \text{I}_2(aq) + 2\text{Fe}^{2+}(aq)\)
- Both steps involve collisions between oppositely charged ions, which has a significantly lower activation energy than the collision between two negative ions.

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Describes a valid experimental setup (initial rates/iodine clock) including key reagents (thiosulfate, starch) and control of variables (temperature, total volume).
- Explains how rate is calculated (\(1/t\)) and how orders are deduced from the data.
- Explains the mechanism of homogeneous catalysis using \(\text{Fe}^{2+}\), including two balanced chemical equations and explaining why the activation energy is lowered (avoidance of negative-negative ion repulsion).

**Level 2 (3-4 marks):**
- Mentions an experimental method to measure rates by varying concentrations.
- Explains the catalytic pathway with correct equations but may fail to fully link the lower activation energy to electrostatic attraction/repulsion.

**Level 1 (1-2 marks):**
- Briefly describes changing concentration and measuring time.
- Identifies that \(\text{Fe}^{2+}\) acts as a catalyst but fails to provide balanced equations or clear explanations for why the rate increases.

1. Method 1: Thermal Decomposition (Gravimetric Analysis):
- Weigh an empty, dry crucible, then weigh the crucible containing a sample of the hydrated salt.
- Heat the crucible gently at first (to avoid spitting), then strongly using a Bunsen burner to drive off the water of crystallization.
- Allow to cool in a desiccator and reweigh.
- Repeat the cycle of heating, cooling, and weighing until a constant mass is reached, ensuring all water has been evaporated.
- Processing Data:
- Mass of anhydrous salt = final mass - empty crucible mass.
- Mass of water lost = initial mass of hydrated salt - mass of anhydrous salt.
- Calculate moles of water lost: \(n(\text{H}_2\text{O}) = \frac{\text{mass of water}}{18.0}\).
- Calculate moles of anhydrous salt: \(n(\text{anhydrous}) = \frac{\text{mass of anhydrous}}{284.1}\).
- The ratio \(x = \frac{n(\text{H}_2\text{O})}{n(\text{anhydrous})}\).

2. Method 2: Titration with Potassium Manganate(VII):
- Accurately weigh a known mass of the hydrated salt and dissolve it in dilute sulfuric acid to prevent oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\).
- Transfer to a volumetric flask and make up to \(250.0\text{ cm}^3\) with distilled water, mixing thoroughly.
- Pipette a \(25.0\text{ cm}^3\) aliquot of this solution into a conical flask and add extra dilute sulfuric acid.
- Titrate against a standard solution of potassium manganate(VII), \(\text{KMnO}_4\), of known concentration until a permanent pale pink end point is observed.
- Processing Data:
- Use the titration volume of \(\text{MnO}_4^-\): \(n(\text{MnO}_4^-) = C \times V\).
- The reaction ratio is \(1 \text{ mol MnO}_4^- \equiv 5 \text{ mol Fe}^{2+}\):
\(\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)\)
- Find moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\), then scale up by 10 to find total moles of \(\text{Fe}^{2+}\) in the original weighed sample.
- Since \(1 \text{ mol}\) of anhydrous salt contains \(1 \text{ mol}\) of \(\text{Fe}^{2+}\), this gives \(n(\text{anhydrous})\).
- Mass of anhydrous salt = \(n(\text{anhydrous}) \times 284.1\).
- Mass of water in sample = initial mass of hydrated salt - mass of anhydrous salt.
- Calculate \(n(\text{H}_2\text{O}) = \frac{\text{mass of water}}{18.0}\) and find the ratio \(x = \frac{n(\text{H}_2\text{O})}{n(\text{anhydrous})}\).

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Describes both methods with precise experimental steps (constant mass heating for thermal decomposition; acidification/volumetric flask/titrant for titration).
- Shows complete processing pathways for both methods, including formulas, molar masses, stoichiometric ratios (especially the 1:5 ratio of manganate to iron), and how to calculate \(x\).
- Highly structured and logically rigorous.

**Level 2 (3-4 marks):**
- Describes both methods with minor omissions in experimental steps (e.g., forgets to heat to constant mass or forgets acidification in titration).
- Outlines the calculation pathway for at least one of the methods correctly.

**Level 1 (1-2 marks):**
- Outlines one method with basic experimental details.
- Shows minimal calculation processing or fails to correctly identify the 1:5 stoichiometry in the titration.

To synthesize 2-aminobenzoic acid from methylbenzene, a three-step organic pathway is required:

1. Step 1: Nitration of methylbenzene
- **Reagents:** Concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\) catalyst).
- **Conditions:** Keep temperature below \(50^\circ\text{C}\) to prevent dinitration.
- **Intermediate Organic Product:** 2-nitromethylbenzene (separated from the 4-nitromethylbenzene isomer by distillation/chromatography).
- **Type of Reaction:** Electrophilic aromatic substitution.

2. Step 2: Oxidation of the methyl group
- **Reagents:** Alkaline potassium manganate(VII) (\(\text{KMnO}_4\) in \(\text{NaOH}(aq)\)), followed by acidification with dilute hydrochloric acid (\(\text{HCl}\)) or sulfuric acid (\(\text{H}_2\text{SO}_4\)).
- **Conditions:** Heat under reflux.
- **Intermediate Organic Product:** 2-nitrobenzoic acid.
- **Type of Reaction:** Oxidation.

3. Step 3: Reduction of the nitro group
- **Reagents:** Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), followed by addition of aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate the free amine from its salt.
- **Conditions:** Heat under reflux.
- **Final Product:** 2-aminobenzoic acid.
- **Type of Reaction:** Reduction.

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Describes a fully viable 3-step synthesis from methylbenzene to 2-aminobenzoic acid.
- Correctly identifies all reagents and conditions for each step (including acidification in the oxidation step and addition of NaOH in the reduction step).
- Correctly names or describes the structure of all intermediates.
- Correctly identifies all reaction types (electrophilic substitution, oxidation, reduction).

**Level 2 (3-4 marks):**
- Proposes a mostly correct 3-step route, but may have minor omissions in reagents or conditions (e.g., omitting acidification or the final alkali addition).
- Most intermediate structures and reaction types are correct.

**Level 1 (1-2 marks):**
- Suggests a route that may contain incorrect order of steps (e.g., oxidizing first, which makes the carboxylic acid meta-directing for nitration).
- Identifies some correct reagents but with major gaps in understanding of directing groups or reaction conditions.

1. Addition of dropwise, then excess sodium hydroxide to \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}\):
- **Observations:** Dropwise addition forms a green/grey-green precipitate. Addition in excess dissolves the precipitate to form a dark green solution.
- **Equations:**
- Dropwise: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3(s) + 3\text{H}_2\text{O}(l)\)
- Excess: \(\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3(s) + 3\text{OH}^-(aq) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(aq) + 3\text{H}_2\text{O}(l)\)
- **Reaction Type:** Acid-base (deprotonation) followed by ligand exchange showing amphoteric behavior.

2. Addition of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) and warming:
- **Observations:** The dark green solution turns into a bright yellow solution.
- **Equation:** \(2[\text{Cr}(\text{OH})_6]^{3-}(aq) + 3\text{H}_2\text{O}_2(aq) \rightarrow 2\text{CrO}_4^{2-}(aq) + 2\text{OH}^-(aq) + 8\text{H}_2\text{O}(l)\)
- **Species and Color:** The yellow species is the chromate(VI) ion, \(\text{CrO}_4^{2-}\). Chromium is oxidized from +3 to +6.
- **Reaction Type:** Redox (oxidation).

3. Acidification of the chromate(VI) solution:
- **Observations:** The yellow solution turns orange.
- **Equation:** \(2\text{CrO}_4^{2-}(aq) + 2\text{H}^+(aq) \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}(aq) + \text{H}_2\text{O}(l)\)
- **Species and Color:** The orange species is the dichromate(VI) ion, \(\text{Cr}_2\text{O}_7^{2-}\).
- **Reaction Type:** Acid-base / non-redox condensation equilibrium (chromium remains in the +6 oxidation state).

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Covers all three stages with correct observations (green ppt -> green solution -> yellow solution -> orange solution).
- Provides correct chemical species formulas and balanced equations for all reactions.
- Classifies all reaction types correctly (deprotonation/amphoterism, oxidation/redox, acid-base equilibrium).
- High clarity of logical sequencing.

**Level 2 (3-4 marks):**
- Covers most stages with correct color changes and formulas.
- Equations may have minor balancing errors but the core reacting species are correct.
- Most reaction types are identified.

**Level 1 (1-2 marks):**
- Describes a few color changes or identifies some chromium species (e.g., chromate is yellow, dichromate is orange).
- Equations are missing or incorrect. Reaction types are vague.

1. Identification of **X**:
- Reaction with 2,4-DNPH indicates the presence of a carbonyl (C=O) group, meaning **X** is either an aldehyde or a ketone.
- The negative test with Tollens' reagent rules out an aldehyde, confirming **X** is a ketone.
- For molecular formula \(\text{C}_4\text{H}_8\text{O}\), the only possible ketone is butanone: \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

2. Identification of **Y**:
- Reduction of butanone with \(\text{LiAlH}_4\) yields a secondary alcohol, butan-2-ol: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
- Equation: \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)

3. Origin of Optical Activity in **Y**:
- Carbon-2 in butan-2-ol is a chiral (asymmetric) carbon because it is bonded to four different groups: a methyl group (\(-\text{CH}_3\)), an ethyl group (\(-\text{CH}_2\text{CH}_3\)), a hydroxyl group (\(-\text{OH}\)), and a hydrogen atom (\(-\text{H}\)).
- Because of this, it exists as two non-superimposable mirror images (enantiomers).
- Individual enantiomers rotate the plane of plane-polarized light in opposite directions (one clockwise, one counterclockwise).

4. Structural Drawings:
- **X**: \(\text{CH}_3\text{-CO-CH}_2\text{-CH}_3\)
- Enantiomers of **Y** (draw in tetrahedral 3D representation around the central chiral carbon C2):
- Enantiomer 1: Central C bonded to \(-\text{OH}\) (vertical), \(-\text{H}\) (dashed wedge), \(-\text{CH}_3\) (solid wedge), and \(-\text{CH}_2\text{CH}_3\) (horizontal).
- Enantiomer 2: The exact mirror image of Enantiomer 1.

This is a Level of Response question evaluated on the following criteria:

**Level 3 (5-6 marks):**
- Deduces that **X** is butanone and **Y** is butan-2-ol, fully justifying both structures using the chemical tests (2,4-DNPH, Tollens').
- Draws clear, correct 3D tetrahedral structures of both enantiomers of **Y** showing they are non-superimposable mirror images.
- Explains optical activity by identifying the chiral carbon and its four different attached groups, explaining how they interact with plane-polarized light.

**Level 2 (3-4 marks):**
- Correctly identifies **X** as butanone and **Y** as butan-2-ol.
- Identifies the chiral center but may fail to draw clear 3D tetrahedral representations of the enantiomers.

**Level 1 (1-2 marks):**
- Identifies the functional groups from the tests (ketone and alcohol).
- May identify butanone or butan-2-ol but lacks explanation of chiral chemistry or optical activity.

1. Calculate the amount of manganate(VII) ions used: \(n(\text{MnO}_4^-) = 0.0200 \times 0.02240 = 4.48 \times 10^{-4}\text{ mol}\). 2. Use the stoichiometric ratio of \(5\text{Fe}^{2+} : 1\text{MnO}_4^-\). The amount of iron(II) in the \(25.0\text{ cm}^3\) aliquot is: \(n(\text{Fe}^{2+}) = 5 \times 4.48 \times 10^{-4} = 2.24 \times 10^{-3}\text{ mol}\). 3. Scale up to the total volume of \(250.0\text{ cm}^3\): \(n(\text{Fe}^{2+})_{\text{total}} = 2.24 \times 10^{-3} \times 10 = 2.24 \times 10^{-2}\text{ mol}\). 4. Calculate the mass of iron: \(m(\text{Fe}) = 2.24 \times 10^{-2} \times 55.8 = 1.24992\text{ g}\). 5. Calculate the percentage purity: \(\text{Percentage purity} = \frac{1.24992}{1.45} \times 100 = 86.201...\\%\). To three significant figures, this is \(86.2\\%\).

1 mark for calculating the moles of manganate(VII) ions. 1 mark for multiplying by 5 to find moles of iron(II) in the aliquot, and scaling up by 10 for the total moles. 1 mark for calculating the mass of iron. 1 mark for calculating the final percentage purity to three significant figures with correct rounding.

1. Calculate the theoretical moles of cyclohexanol used: \(n(\text{cyclohexanol}) = \frac{12.0}{100.2} = 0.11976\text{ mol}\). Since the reaction has 1:1 stoichiometry, the theoretical yield of cyclohexene is also \(0.11976\text{ mol}\). 2. Calculate the theoretical mass of cyclohexene: \(m(\text{theoretical}) = 0.11976 \times 82.1 = 9.8323\text{ g}\). 3. Calculate the percentage yield: \(\text{Percentage yield} = \frac{5.82}{9.8323} \times 100 = 59.191...\\%\). To three significant figures, this is \(59.2\\%\).

1 mark for calculating the theoretical yield of cyclohexene in moles or grams. 1 mark for setting up the percentage yield calculation. 1 mark for the final answer of 59.2%.

1. Calculate the temperature change: \(\Delta T = 25.8 - 19.4 = 6.4\text{ }^\circ\text{C}\). 2. Since two temperature readings are made (initial and maximum), the total uncertainty in \(\Delta T\) is: \(\text{Total uncertainty} = 2 \times 0.05 = 0.10\text{ }^\circ\text{C}\). 3. Calculate the percentage uncertainty: \(\text{Percentage uncertainty} = \frac{0.10}{6.4} \times 100 = 1.5625\\%\). To two significant figures, this is \(1.6\\%\).

1 mark for calculating the temperature change as 6.4 degrees Celsius. 1 mark for doubling the individual uncertainty to get a total uncertainty of 0.10 degrees Celsius. 1 mark for calculating the final percentage uncertainty to two significant figures (1.6%).

1. Calculate the initial rate of reaction: \(\text{Rate} = \frac{\Delta [\text{I}_2]}{\Delta t} = \frac{8.00 \times 10^{-4}\text{ mol dm}^{-3}}{42.5\text{ s}} = 1.8824 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). To three significant figures, this is \(1.88 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). 2. Starch is used purely as an indicator to signal the completion of the thiosulfate reaction. Since it is not a reactant in the rate-determining step, doubling its concentration will have no effect on the reaction rate, and thus no effect on the time taken for the colour to appear.

1 mark for correct calculation of the rate value (1.88 x 10^-5). 1 mark for correct units of rate (mol dm^-3 s^-1). 1 mark for stating that there is no effect on the time taken. 1 mark for explaining that starch is an indicator and does not participate in the reaction steps affecting the rate.

1. Calculate the mass of water lost: \(m(\text{H}_2\text{O}) = 2.350 - 1.284 = 1.066\text{ g}\). 2. Calculate the moles of anhydrous \(\text{CoCl}_2\): \(n(\text{CoCl}_2) = \frac{1.284}{129.9} = 0.009885\text{ mol}\). 3. Calculate the moles of water lost: \(n(\text{H}_2\text{O}) = \frac{1.066}{18.0} = 0.05922\text{ mol}\). 4. Determine the ratio \(x\): \(x = \frac{n(\text{H}_2\text{O})}{n(\text{CoCl}_2)} = \frac{0.05922}{0.009885} = 5.991 \approx 6\).

1 mark for calculating the mass of water lost (1.066 g) and finding the moles of anhydrous cobalt(II) chloride. 1 mark for finding the moles of water. 1 mark for calculating the mole ratio and rounding to the nearest whole number (6).

1. When hydrolysis occurs, halide ions are released which react with silver ions to form an insoluble silver halide precipitate. For 1-bromobutane, this is a cream precipitate of silver bromide. 2. The ionic equation is \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\). 3. 1-iodobutane reacts the fastest because the C-I bond is the weakest (has the lowest bond enthalpy) and is broken most easily.

1 mark for identifying the formation of a precipitate (accept cream precipitate or precipitate formation). 1 mark for correct chemical symbols in the ionic equation. 1 mark for correct state symbols in the ionic equation. 1 mark for identifying 1-iodobutane as the fastest reactant due to the weaker C-I bond.

1. Convert all units to SI units: \(p = 101 \times 10^3\text{ Pa}\), \(V = 38.4 \times 10^{-6}\text{ m}^3\), \(T = 98 + 273 = 371\text{ K}\). 2. Rearrange the ideal gas equation \(pV = nRT\) to solve for moles: \(n = \frac{pV}{RT} = \frac{101 \times 10^3 \times 38.4 \times 10^{-6}}{8.31 \times 371} = 0.001258\text{ mol}\). 3. Calculate molar mass \(M\): \(M = \frac{m}{n} = \frac{0.116}{0.001258} = 92.21\text{ g mol}^{-1}\). To three significant figures, this is \(92.2\text{ g mol}^{-1}\).

1 mark for correct conversions of pressure, volume, and temperature to SI units. 1 mark for calculating the moles of gas (0.00126 mol). 1 mark for calculating the molar mass to three significant figures with appropriate unit (92.2 g mol^-1).

1. Washing with sodium hydrogencarbonate neutralises and removes acidic impurities (the carboxylic acid reactant and acid catalyst) by converting them into water-soluble salts. 2. The ionic equation for the neutralisation reaction is: \(\text{H}^+(\text{aq}) + \text{HCO}_3^-(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). 3. A suitable anhydrous salt to dry the organic layer is anhydrous magnesium sulfate (\(\text{MgSO}_4\)) or anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)).

1 mark for stating that the wash neutralises the acidic impurities. 2 marks for the ionic equation (1 mark for species, 1 mark for state symbols). 1 mark for naming a suitable drying agent (e.g., anhydrous magnesium sulfate, anhydrous sodium sulfate, or anhydrous calcium chloride).

1. Calculate mass of water lost:
\( 4.886 \text{ g} - 4.166 \text{ g} = 0.720 \text{ g} \).

2. Calculate moles of water lost:
\( n(\text{H}_2\text{O}) = \frac{0.720 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.0400 \text{ mol} \).

3. Calculate moles of anhydrous \( \text{BaCl}_2 \) (using \( A_r \) values: \( \text{Ba} = 137.3 \), \( \text{Cl} = 35.5 \), so \( M_r(\text{BaCl}_2) = 208.3 \text{ g mol}^{-1} \)):
\( n(\text{BaCl}_2) = \frac{4.166 \text{ g}}{208.3 \text{ g mol}^{-1}} = 0.0200 \text{ mol} \).

4. Determine the ratio of \( \text{H}_2\text{O} \) to \( \text{BaCl}_2 \):
\( \text{Ratio} = \frac{0.0400}{0.0200} = 2.00 \), so \( x = 2 \).

5. Calculate percentage uncertainty in the mass of water lost:
Two weighings are required to find the mass of water (initial and final mass of crucible + contents).
\( \text{Absolute uncertainty} = 2 \times 0.01 \text{ g} = 0.02 \text{ g} \).
\( \text{Percentage uncertainty} = \frac{0.02 \text{ g}}{0.720 \text{ g}} \times 100\% = 2.78\% \).

- Moles of both water (0.0400 mol) and barium chloride (0.0200 mol) calculated correctly (1 mark)
- Correct deduction of x = 2 (0.5 mark)
- Absolute uncertainty of water lost recognized as \( \pm 0.02 \text{ g} \) (1 mark)
- Percentage uncertainty of 2.78% (or 2.8%) correctly calculated (1 mark)

1. Find the concentration of \( \text{Cu}^{2+} \) using the equation of the line:
\( 0.449 = 12.4 \times [\text{Cu}^{2+}] + 0.015 \)
\( [\text{Cu}^{2+}] = \frac{0.449 - 0.015}{12.4} = 0.0350 \text{ mol dm}^{-3} \).

2. Calculate the amount in moles of copper in the volumetric flask:
\( n(\text{Cu}^{2+}) = 0.0350 \text{ mol dm}^{-3} \times 0.2500 \text{ dm}^3 = 8.75 \times 10^{-3} \text{ mol} \).

3. Calculate the mass of copper in the sample:
\( m(\text{Cu}) = 8.75 \times 10^{-3} \text{ mol} \times 63.5 \text{ g mol}^{-1} = 0.5556 \text{ g} \).

4. Calculate the percentage by mass of copper in the brass sample:
\( \% \text{ mass} = \frac{0.5556 \text{ g}}{1.200 \text{ g}} \times 100\% = 46.3\% \) (to 3 significant figures).

- Correct calculation of copper(II) concentration (1 mark)
- Correct calculation of moles of copper in the 250 cm\(^3\) flask (1 mark)
- Mass of copper calculated (0.5 mark)
- Percentage by mass of copper to 3 s.f. (1 mark)

1. Calculate the moles of cyclohexanol reactant:
\( n(\text{cyclohexanol}) = \frac{10.0 \text{ g}}{100.2 \text{ g mol}^{-1}} = 0.0998 \text{ mol} \).

2. Since the stoichiometry of the dehydration is 1:1, the theoretical moles of cyclohexene is \( 0.0998 \text{ mol} \).

3. Calculate the theoretical mass of cyclohexene:
\( \text{Theoretical mass} = 0.0998 \text{ mol} \times 82.1 \text{ g mol}^{-1} = 8.194 \text{ g} \).

4. Calculate the actual mass of cyclohexene obtained:
\( \text{Actual mass} = 4.50 \text{ cm}^3 \times 0.811 \text{ g cm}^{-3} = 3.6495 \text{ g} \).

5. Calculate the percentage yield:
\( \% \text{ Yield} = \frac{3.6495 \text{ g}}{8.194 \text{ g}} \times 100\% = 44.5\% \) (to 3 significant figures).

- Calculation of moles of reactant (1 mark)
- Calculation of the actual mass of cyclohexene using density (1 mark)
- Calculation of theoretical mass of cyclohexene (0.5 mark)
- Correct percentage yield to 3 s.f. (1 mark)

1. The Arrhenius equation in logarithmic form where rate is proportional to \( 1/t \) is given by:
\( \ln(1/t) = -\frac{E_a}{R}\frac{1}{T} + \text{constant} \).

2. Therefore, the gradient of the line is:
\( \text{Gradient} = -\frac{E_a}{R} \).

3. Rearranging to find \( E_a \):
\( -6150 = -\frac{E_a}{8.31} \implies E_a = 6150 \times 8.31 = 51106.5 \text{ J mol}^{-1} \).

4. Convert from \( \text{J mol}^{-1} \) to \( \text{kJ mol}^{-1} \):
\( E_a = \frac{51106.5}{1000} = 51.1 \text{ kJ mol}^{-1} \) (to 3 significant figures).

- Identification of the relationship between gradient and activation energy (1.5 marks)
- Calculation of the value in J mol\(^{-1}\) (1 mark)
- Conversion to kJ mol\(^{-1}\) and rounding to 3 significant figures (1 mark)

1. Calculate the average rate of reaction between 30 and 90 seconds:
\( \text{Average Rate} = \frac{\Delta V}{\Delta t} = \frac{42.0 \text{ cm}^3 - 18.0 \text{ cm}^3}{90 \text{ s} - 30 \text{ s}} = \frac{24.0 \text{ cm}^3}{60 \text{ s}} = 0.400 \text{ cm}^3 \text{ s}^{-1} \).

2. The total potential volume of gas corresponding to 100% of the reactant is \( 60.0 \text{ cm}^3 \).
At \( 90 \text{ s} \), \( 42.0 \text{ cm}^3 \) of gas has already been produced.

3. The volume of oxygen yet to be evolved (which is directly proportional to the amount of remaining \( \text{H}_2\text{O}_2 \)) is:
\( 60.0 \text{ cm}^3 - 42.0 \text{ cm}^3 = 18.0 \text{ cm}^3 \).

4. Calculate the percentage of hydrogen peroxide remaining:
\( \% \text{ remaining} = \frac{18.0 \text{ cm}^3}{60.0 \text{ cm}^3} \times 100\% = 30.0\% \).

- Calculation of average rate with correct units (1.5 marks)
- Calculation of the remaining volume of gas to be evolved (1 mark)
- Calculation of the percentage of reactant remaining (1 mark)

1. Write the balanced ionic equation for the titration:
\( \text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \).

2. Calculate the moles of \( \text{MnO}_4^- \) in the titre:
\( n(\text{MnO}_4^-) = 0.0100 \text{ mol dm}^{-3} \times \frac{22.40}{1000} \text{ dm}^3 = 2.24 \times 10^{-4} \text{ mol} \).

3. Calculate the moles of \( \text{Fe}^{2+} \) in the \( 25.0 \text{ cm}^3 \) aliquot:
\( n(\text{Fe}^{2+}) = 5 \times 2.24 \times 10^{-4} \text{ mol} = 1.12 \times 10^{-3} \text{ mol} \).

4. Calculate the moles of \( \text{Fe}^{2+} \) in the total \( 250.0 \text{ cm}^3 \) solution:
\( n(\text{total}) = 1.12 \times 10^{-3} \text{ mol} \times \frac{250.0}{25.0} = 1.12 \times 10^{-2} \text{ mol} \).

5. Calculate the total mass of iron in the five tablets:
\( m(\text{total}) = 1.12 \times 10^{-2} \text{ mol} \times 55.8 \text{ g mol}^{-1} = 0.62496 \text{ g} \).

6. Calculate the mass of iron in a single tablet in milligrams:
\( m(\text{tablet}) = \frac{0.62496 \text{ g}}{5} \times 1000 \text{ mg g}^{-1} = 125 \text{ mg} \) (to 3 significant figures).

- Moles of manganate(VII) calculated correctly (1 mark)
- Moles of iron scaled to the total 250 cm\(^3\) flask (1 mark)
- Total mass of iron in 5 tablets calculated (0.5 mark)
- Final mass of iron in one tablet calculated in mg and rounded to 3 s.f. (1 mark)

1. Calculate mass of solution:
\( m = 50.0 \text{ cm}^3 + 50.0 \text{ cm}^3 = 100.0 \text{ cm}^3 \implies 100.0 \text{ g} \).

2. Calculate temperature change:
\( \Delta T = 26.8 - 20.2 = 6.6 \ ^\circ\text{C} \) (or \( 6.6 \text{ K} \)).

3. Calculate heat energy released:
\( q = m c \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 6.6 \text{ K} = 2758.8 \text{ J} = 2.7588 \text{ kJ} \).

4. Determine limiting reactant and moles of water formed:
- \( n(\text{HCl}) = 0.0500 \text{ dm}^3 \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol} \).
- \( n(\text{NaOH}) = 0.0500 \text{ dm}^3 \times 1.05 \text{ mol dm}^{-3} = 0.0525 \text{ mol} \).
Since \( \text{HCl} \) is the limiting reactant, \( 0.0500 \text{ mol} \) of water is formed.

5. Calculate \( \Delta H_{\text{neut}} \) (reaction is exothermic, so sign is negative):
\( \Delta H_{\text{neut}} = -\frac{2.7588 \text{ kJ}}{0.0500 \text{ mol}} = -55.2 \text{ kJ mol}^{-1} \) (to 3 significant figures).

- Heat energy released calculated correctly (1 mark)
- Identification of HCl as limiting reactant and correct moles of water (1 mark)
- Correct calculation of enthalpy of neutralisation with negative sign (1 mark)
- Value rounded to 3 significant figures with units (0.5 mark)

1. At the equivalence point, the moles of \( \text{NaOH} \) added equal the initial moles of the monoprotic acid \( \text{HA} \):
\( n(\text{NaOH}) = 0.100 \text{ mol dm}^{-3} \times \frac{20.0}{1000} \text{ dm}^3 = 0.00200 \text{ mol} \).

2. Calculate the concentration of \( \text{HA} \) in the original \( 25.0 \text{ cm}^3 \) sample:
\( [\text{HA}] = \frac{0.00200 \text{ mol}}{0.0250 \text{ dm}^3} = 0.0800 \text{ mol dm}^{-3} \).

3. At the half-equivalence point, exactly half of the weak acid has been neutralised, so \( [\text{HA}] = [\text{A}^-] \). Substituting this into the acid dissociation constant expression:
\( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \implies K_a = [\text{H}^+] \).
Taking negative logarithms of both sides gives:
\( \text{pH} = \text{p}K_a \).

- Moles of NaOH at equivalence calculated correctly (1 mark)
- Initial concentration of HA calculated as 0.0800 mol dm\(^{-3}\) (1.5 marks)
- Correct relationship stated as pH = pK\(_a\) (1 mark)

1. Calculate mass of anhydrous \(\text{CoCl}_2\):
\(20.19\text{ g} - 18.24\text{ g} = 1.95\text{ g}\)

2. Calculate mass of water lost:
\(21.81\text{ g} - 20.19\text{ g} = 1.62\text{ g}\)

3. Calculate moles of \(\text{CoCl}_2\):
\(M_r(\text{CoCl}_2) = 58.9 + (2 \times 35.5) = 129.9\text{ g mol}^{-1}\)
\(n(\text{CoCl}_2) = \frac{1.95}{129.9} = 0.01501\text{ mol}\)

4. Calculate moles of \(\text{H}_2\text{O}\):
\(n(\text{H}_2\text{O}) = \frac{1.62}{18.0} = 0.0900\text{ mol}\)

5. Calculate ratio:
\(x = \frac{0.0900}{0.01501} = 5.996 \approx 6\)

- Mass of anhydrous salt and water lost correctly calculated: 1 mark
- Moles of anhydrous salt and water correctly calculated: 1 mark
- Value of x calculated as 6 (accept 5.99 to 6.00) with clear showing of working: 1.5 marks

1. Calculate moles of \(\text{Fe}^{2+}\) reacted:
\(n(\text{Fe}^{2+}) = 0.02500\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 5.00 \times 10^{-4}\text{ mol}\)

2. Determine moles of \(\text{MnO}_4^-\) using the stoichiometry (1:5):
\(n(\text{MnO}_4^-) = \frac{5.00 \times 10^{-4}}{5} = 1.00 \times 10^{-4}\text{ mol}\)

3. Calculate concentration of \(\text{KMnO}_4\):
\(\text{Volume} = 19.65\text{ cm}^3 = 0.01965\text{ dm}^3\)
\(\text{Concentration} = \frac{1.00 \times 10^{-4}\text{ mol}}{0.01965\text{ dm}^3} = 5.089 \times 10^{-3}\text{ mol dm}^{-3} \approx 5.09 \times 10^{-3}\text{ mol dm}^{-3}\)

- Moles of \(\text{Fe}^{2+}\) correctly calculated: 1 mark
- Moles of \(\text{MnO}_4^-\) correctly calculated using 1:5 ratio: 1 mark
- Final concentration correctly calculated to 3 significant figures: 1.5 marks

1. Calculate moles of cyclohexanol reactant:
\(n(\text{cyclohexanol}) = \frac{12.0}{100.2} = 0.1198\text{ mol}\)

2. Calculate the theoretical yield of cyclohexene:
Since the stoichiometry is 1:1, theoretical moles of cyclohexene = \(0.1198\text{ mol}\).
Theoretical mass = \(0.1198\text{ mol} \times 82.1\text{ g mol}^{-1} = 9.836\text{ g}\)

3. Calculate the percentage yield:
\(\text{Percentage yield} = \frac{6.12\text{ g}}{9.836\text{ g}} \times 100\% = 62.22\% \approx 62.2\%\)

- Moles of cyclohexanol: 1 mark
- Theoretical mass of cyclohexene: 1 mark
- Percentage yield to 3 significant figures: 1.5 marks (Accept 62.1% to 62.3% depending on rounding at intermediate stages)

1. Determine order with respect to \(\text{S}_2\text{O}_8^{2-}\):
Comparing Exp 1 and Exp 2, when \([\text{S}_2\text{O}_8^{2-}]\) doubles (while \([\text{I}^-]\) is constant), the rate doubles. Therefore, the reaction is first order with respect to \(\text{S}_2\text{O}_8^{2-}\).

2. Determine order with respect to \(\text{I}^-\):
Comparing Exp 1 and Exp 3, when \([\text{I}^-]\) doubles (while \([\text{S}_2\text{O}_8^{2-}]\) is constant), the rate doubles. Therefore, the reaction is first order with respect to \(\text{I}^-\).

3. Write the rate equation:
\(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

4. Calculate the rate constant \(k\) using data from Experiment 1:
\(2.80 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times 0.0400\text{ mol dm}^{-3} \times 0.0500\text{ mol dm}^{-3}\)
\(k = \frac{2.80 \times 10^{-5}}{0.00200} = 0.0140\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(1.40 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\))

- Deduces first order for both reactants with explanation: 1 mark
- Correct rate equation written: 1 mark
- Calculation of rate constant k as 0.0140 (or 1.40 x 10^-2) with correct units (dm3 mol-1 s-1): 1.5 marks

1. Calculate mass of solution:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\) (since density = \(1.00\text{ g cm}^{-3}\))

2. Calculate temperature rise:
\(\Delta T = 26.2 - 19.5 = 6.7\text{ }^\circ\text{C}\)

3. Calculate heat energy released (q):
\(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 6.7\text{ }^\circ\text{C} = 2800.6\text{ J} = 2.8006\text{ kJ}\)

4. Calculate moles of water produced:
\(n(\text{H}_2\text{O}) = n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)

5. Calculate \(\Delta H_{\text{neut}}\):
\(\Delta H_{\text{neut}} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.012\text{ kJ mol}^{-1} \approx -56.0\text{ kJ mol}^{-1}\)

- Correct calculation of heat energy q (2.80 kJ): 1 mark
- Correct determination of moles of water formed (0.0500 mol): 1 mark
- Correct final value of -56.0 kJ mol-1 with negative sign and to 3 sig figs: 1.5 marks

1. Calculate moles of thiosulfate used in titration:
\(n(\text{S}_2\text{O}_3^{2-}) = 0.02240\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.24 \times 10^{-3}\text{ mol}\)

2. Relate moles of thiosulfate to moles of copper ions in \(25.0\text{ cm}^3\):
From the equations, \(1\text{ mol of } \text{I}_2 \equiv 2\text{ mol of } \text{S}_2\text{O}_3^{2-}\) and \(2\text{ mol of } \text{Cu}^{2+} \equiv 1\text{ mol of } \text{I}_2\).
Therefore, \(n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-}) = 2.24 \times 10^{-3}\text{ mol}\) in the \(25.0\text{ cm}^3\) aliquot.

3. Calculate moles of copper in the original \(250\text{ cm}^3\) solution:
\(n(\text{Cu}^{2+})_{\text{total}} = 2.24 \times 10^{-3}\text{ mol} \times 10 = 2.24 \times 10^{-2}\text{ mol}\)

4. Calculate mass and percentage by mass of copper:
\(\text{Mass of Cu} = 2.24 \times 10^{-2}\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.4224\text{ g}\)
\(\text{Percentage of Cu} = \frac{1.4224\text{ g}}{2.50\text{ g}} \times 100\% = 56.896\% \approx 56.9\%\)

- Moles of thiosulfate and hence moles of Cu2+ in aliquot correctly calculated: 1 mark
- Moles of total Cu2+ in 250 cm3 calculated: 1 mark
- Final percentage of Cu in brass sample calculated to 3 significant figures (56.9%): 1.5 marks

1. Calculate theoretical yield of aspirin:
\(n(\text{salicylic acid}) = \frac{5.00}{138.0} = 0.03623\text{ mol}\)
Theoretical mass of aspirin = \(0.03623\text{ mol} \times 180.0\text{ g mol}^{-1} = 6.521\text{ g}\)

2. Calculate percentage yield:
\(\text{Percentage yield} = \frac{4.15}{6.521} \times 100\% = 63.64\% \approx 63.6\%\)

3. Reason for lower yield:
Some product remains dissolved in the cold solvent during recrystallisation, or some product is lost during transfer/filtration.

- Theoretical mass of aspirin calculated correctly: 1 mark
- Percentage yield calculated correctly (63.6%): 1 mark
- A valid reason given for loss of product (e.g. some solid remains dissolved in solvent, loss on transferring between containers, loss during recrystallisation / filtration): 1.5 marks

1. A straight line graph of concentration of a reactant against time has a constant gradient. This means the rate of reaction is constant and does not change as the concentration of iodine decreases.

2. Since the rate is independent of the concentration of iodine, the reaction is zero order with respect to iodine.

3. The rate of reaction is given by the negative of the gradient of the concentration-time graph (since concentration decreases over time, the gradient is negative, but rate must be a positive value): \(\text{Rate} = -\text{gradient}\).

- Correctly identifies zero order: 1 mark
- Explains that the straight line indicates a constant rate independent of iodine concentration: 1.5 marks
- States that the rate of reaction is equal to the negative of the gradient: 1 mark

Mass of hydrated salt = \(24.88 - 20.00 = 4.88\text{ g}\). Mass of anhydrous residue = \(24.16 - 20.00 = 4.16\text{ g}\). Mass of water lost = \(4.88 - 4.16 = 0.72\text{ g}\). Moles of anhydrous \(\text{BaCl}_2\) (\(M_r = 208.2\text{ g mol}^{-1}\)) = \(4.16 / 208.2 = 0.0200\text{ mol}\). Moles of \(\text{H}_2\text{O}\) (\(M_r = 18.0\text{ g mol}^{-1}\)) = \(0.72 / 18.0 = 0.0400\text{ mol}\). Ratio \(x = 0.0400 / 0.0200 = 2\). If heating is incomplete, some water remains in the residue, making the calculated mass of water lost too small, leading to a lower value of \(x\).

1 mark: Correct calculation of moles of anhydrous salt and water. 1 mark: Correct calculation of x = 2. 1 mark: Correctly identifying incomplete heating / failure to heat to constant mass. 0.5 marks: Correctly linking this error to a lower calculated mass of water lost.

Theoretical moles of cyclohexanol = \(15.0 / 100.0 = 0.150\text{ mol}\). Theoretical mass of cyclohexene = \(0.150 \times 82.0 = 12.3\text{ g}\). Percentage yield = \((7.38 / 12.3) \times 100 = 60.0\%\). Saturated sodium chloride solution increases the ionic strength of the aqueous phase, reducing the solubility of cyclohexene in the aqueous layer (salting out) and helping to break emulsions for better layer separation. Since cyclohexene has a density of \(0.81\text{ g cm}^{-3}\), which is lower than the density of the aqueous saline solution (\(> 1.0\text{ g cm}^{-3}\)), cyclohexene will be found in the upper layer.

1.5 marks: Correct yield calculation (1 mark for working showing theoretical mass, 0.5 marks for 60.0%). 1 mark: Explanation of using saturated saline (salting out / decreases organic solubility in water / breaks emulsions). 1 mark: Correctly identifying the upper layer with justification based on density.

Moles of \(\text{MnO}_4^-\text{ used} = 10.20 \times 10^{-3} \times 0.0100 = 1.02 \times 10^{-4}\text{ mol}\). Reacting ratio of \(\text{MnO}_4^-\text{ to Fe}^{2+}\text{ is 1:5}\). Moles of \(\text{Fe}^{2+}\text{ in 25.0 cm}^3 = 5.10 \times 10^{-4}\text{ mol}\). Moles of \(\text{Fe}^{2+}\text{ in 250 cm}^3 = 5.10 \times 10^{-3}\text{ mol}\). Mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O} = 5.10 \times 10^{-3} \times 277.9 = 1.417\text{ g}\). Percentage purity = \((1.417 / 1.50) \times 100 = 94.5\%\). Dilute sulfuric acid is used because hydrochloric acid contains chloride ions (\(\text{Cl}^-\)), which are easily oxidized to chlorine gas (\(\text{Cl}_2\)) by the strong oxidizing agent manganate(VII). This oxidation would consume extra manganate(VII) solution, leading to an artificially high titre value.

1 mark: Correct calculation of moles of iron(II) in the entire flask. 1 mark: Correct calculation of percentage purity (94.5%). 1 mark: Identifying that chloride ions are oxidized to chlorine. 0.5 marks: Stating that this would lead to an overestimation of the titre / purity.

Relative rate in Experiment 1 = \(1/45 = 0.0222\text{ s}^{-1}\). Relative rate in Experiment 2 = \(1/90 = 0.0111\text{ s}^{-1}\). When the concentration of peroxodisulfate ions is halved (from \(0.040\) to \(0.020\text{ mol dm}^{-3}\)), the rate also halves. Therefore, the reaction is first-order with respect to peroxodisulfate ions. Sodium thiosulfate reacts rapidly with any iodine formed, converting it back to iodide. This delays the appearance of the blue-black starch-iodine complex until all sodium thiosulfate has been consumed, providing a clear and constant endpoint for measuring the initial rate.

1 mark: Shows working using rate proportional to 1/t. 1 mark: Correctly deduces first-order kinetics. 1 mark: Explains that sodium thiosulfate temporarily prevents the starch-iodine color by reacting with iodine. 0.5 marks: Explains that this allows the measurement of the time taken to produce a fixed, small amount of product.