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A sulfide ion, \(\text{S}^{2-}\), has \(16 + 2 = 18\) electrons. Its electronic configuration is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\). A scandium atom has \(21\) electrons, so a \(\text{Sc}^{3+}\) ion has \(21 - 3 = 18\) electrons, giving it the identical electronic configuration of \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\). \(\text{P}^{3+}\) has 12 electrons, \(\text{Se}^{2-}\) has 36 electrons, and \(\text{Ar}^{+}\) has 17 electrons.

1 mark for identifying \(\text{Sc}^{3+}\) as the species with 18 electrons, matching the electron configuration of the sulfide ion.

Halogens act as oxidizing agents, and their oxidizing ability decreases down Group 17 (Group 7): \(\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2\). A halogen higher in the group can displace a halide ion lower in the group. Bromine (\(\text{Br}_2\)) is a stronger oxidizing agent than iodine, so it can oxidize iodide ions (\(\text{I}^-\)) to iodine (\(\text{I}_2\)), resulting in a displacement reaction: \(\text{Br}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Br}^-\text{(aq)} + \text{I}_2\text{(aq)}\). The other pairs will not react because the free halogen is a weaker oxidizing agent than the halogen corresponding to the halide ion.

1 mark for identifying that bromine can oxidize iodide to iodine (Option D) because bromine is a stronger oxidizing agent than iodine.

Butan-1-ol and 2-methylpropan-2-ol both experience hydrogen bonding, which is much stronger than the permanent dipole-dipole forces in ethoxyethane and the instantaneous dipole-induced dipole (London) forces in butane. Between the two isomeric alcohols, butan-1-ol has a straight carbon chain, which allows for greater surface contact between molecules and therefore stronger instantaneous dipole-induced dipole forces than the highly branched 2-methylpropan-2-ol. Thus, butan-1-ol has the highest boiling point.

1 mark for selecting Butan-1-ol as the compound with the highest boiling point due to a combination of hydrogen bonding and a straight-chain structure that maximizes London forces.

Step 1: Calculate the amount in moles of octane:
\(n(\text{C}_8\text{H}_{18}) = \frac{\text{mass}}{M_r} = \frac{5.70}{114} = 0.050\text{ mol}\).

Step 2: Use the stoichiometric ratio from the balanced equation to find the moles of carbon dioxide produced:
\(n(\text{CO}_2) = 8 \times n(\text{C}_8\text{H}_{18}) = 8 \times 0.050 = 0.40\text{ mol}\).

Step 3: Calculate the volume of \(\text{CO}_2\) at r.t.p.:
\(V = n \times 24.0 = 0.40 \times 24.0 = 9.60\text{ dm}^3\).

1 mark for the correct volume of 9.60 dm3 (Option B).

In the thiosulfate ion, \(\text{S}_2\text{O}_3^{2-}\), let \(x\) be the oxidation state of sulfur:
\(2x + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).

In the tetrathionate ion, \(\text{S}_4\text{O}_6^{2-}\), let \(y\) be the oxidation state of sulfur:
\(4y + 6(-2) = -2 \Rightarrow 4y - 12 = -2 \Rightarrow 4y = +10 \Rightarrow y = +2.5\).

Therefore, the oxidation state of sulfur changes from \(+2\) to \(+2.5\).

1 mark for calculating the initial (+2) and final (+2.5) oxidation states of sulfur correctly (Option A).

The infrared spectrum displays two characteristic features:
1. A very broad absorption band in the range \(2500\text{--}3300\text{ cm}^{-1}\), which is characteristic of the \(\text{O--H}\) bond stretch of a carboxylic acid.
2. A strong absorption band at \(1715\text{ cm}^{-1}\), characteristic of a carbonyl \(\text{C=O}\) stretch.

These features confirm that the compound is a carboxylic acid. Among the choices, butanoic acid is the only carboxylic acid (with the formula \(\text{C}_4\text{H}_8\text{O}_2\)). Ethyl ethanoate and methyl propanoate are esters and do not have an \(\text{O--H}\) stretch. 4-hydroxybutanal has an aldehyde group and an alcohol group, but alcohol \(\text{O--H}\) peaks are typically much narrower and found at higher wavenumbers (\(3200\text{--}3600\text{ cm}^{-1}\)).

1 mark for identifying Butanoic acid (Option C) based on the combined presence of a carboxylic acid O-H band and a C=O band.

The equation for the formation of propane is:
\(3\text{C(graphite)} + 4\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_8(\text{g})\)

According to Hess's Law, when using enthalpy changes of combustion:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)

\(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C}) + 4 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{C}_3\text{H}_8)\)

\(\Delta_f H^\theta = [3 \times (-394) + 4 \times (-286)] - (-2220)\)

\(\Delta_f H^\theta = [-1182 - 1144] + 2220\)

\(\Delta_f H^\theta = -2326 + 2220 = -106\text{ kJ mol}^{-1}\).

1 mark for calculating the correct standard enthalpy change of formation of propane as -106 kJ mol-1.

Let us analyze the steps:
- Option A represents an initiation step where UV radiation breaks a C-Cl bond homolytically to form a chlorine radical.
- Option B represents a propagation step, where a highly reactive chlorine radical attacks ozone, yielding a chlorine monoxide radical and oxygen gas. (The other propagation step is \(\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2\)).
- Option C represents a termination step where two chlorine radicals combine.
- Option D represents the dimerization of chlorine monoxide radicals, which is not one of the standard propagation steps in the main ozone-depleting catalytic cycle.

1 mark for identifying the correct propagation step in the ozone depletion mechanism (Option B).

The electronic configuration of an iron atom is \( [\text{Ar}] 3d^6 4s^2 \). For the \( \text{Fe}^{3+} \) ion, three electrons are removed (two from the \( 4s \) subshell and one from the \( 3d \) subshell), giving the configuration \( [\text{Ar}] 3d^5 \). According to Hund's rule, all 5 electrons in the \( 3d \) subshell will occupy separate orbitals with parallel spins, meaning there are 5 unpaired electrons. Now check the options: (A) \( \text{Mn}^{2+} \) has the configuration \( [\text{Ar}] 3d^5 \), which also contains 5 unpaired electrons. (B) \( \text{Co}^{2+} \) is \( [\text{Ar}] 3d^7 \) (3 unpaired). (C) \( \text{Cr}^{3+} \) is \( [\text{Ar}] 3d^3 \) (3 unpaired). (D) \( \text{Ni}^{2+} \) is \( [\text{Ar}] 3d^8 \) (2 unpaired). Therefore, option A is correct.

1 mark for the correct option (A). 0 marks for any other option or if left blank.

Step 1: Calculate the amount of \( \text{CO}_2 \) gas in moles: \( n(\text{CO}_2) = \frac{0.240 \text{ dm}^3}{24.0 \text{ dm}^3\text{ mol}^{-1}} = 0.0100 \text{ mol} \). Step 2: From the 1:1 stoichiometry of the equation, the amount of \( \text{MCO}_3 \) is also \( 0.0100 \text{ mol} \). Step 3: Calculate the molar mass of \( \text{MCO}_3 \): \( M = \frac{\text{mass}}{\text{moles}} = \frac{1.97 \text{ g}}{0.0100 \text{ mol}} = 197 \text{ g mol}^{-1} \). Step 4: Determine the relative atomic mass of \( \text{M} \): \( A_r(\text{M}) = M_r(\text{MCO}_3) - [A_r(\text{C}) + 3 \times A_r(\text{O})] = 197 - [12.0 + 48.0] = 137 \text{ g mol}^{-1} \). The element with this relative atomic mass is barium (Ba).

1 mark for the correct option (D). 0 marks for any other option or if left blank.

Concentrated sulfuric acid acts as both an acid and an oxidizing agent. With potassium fluoride and chloride, no redox reaction occurs because \( \text{F}^- \) and \( \text{Cl}^- \) are weak reducing agents. With potassium bromide, \( \text{Br}^- \) reduces \( \text{H}_2\text{SO}_4 \) to \( \text{SO}_2 \) (a colorless gas) and orange bromine vapour is seen. With potassium iodide, \( \text{I}^- \) is a very strong reducing agent and reduces \( \text{H}_2\text{SO}_4 \) to solid yellow sulfur (\( \text{S} \)) and gaseous hydrogen sulfide (\( \text{H}_2\text{S} \)), while also forming acidic hydrogen iodide (\( \text{HI} \)) gas that turns damp blue litmus paper red.

1 mark for the correct option (D). 0 marks for any other option or if left blank.

The reactant chlorine is in the element form \( \text{Cl}_2 \), where its oxidation state is 0. In the chlorate(V) ion, \( \text{ClO}_3^- \), oxygen has an oxidation state of -2. Let \( x \) be the oxidation state of chlorine: \( x + 3(-2) = -1 \) which gives \( x - 6 = -1 \) so \( x = +5 \). Therefore, the change in oxidation state of chlorine is from 0 to +5.

1 mark for the correct option (C). 0 marks for any other option or if left blank.

The broad peak in the region \( 2500 - 3300 \text{ cm}^{-1} \) is characteristic of the O-H stretch in a carboxylic acid. The strong, sharp peak at \( 1715 \text{ cm}^{-1} \) corresponds to the C=O stretch of a carbonyl group. Together, these peaks confirm the presence of a carboxylic acid functional group. Since the molecular formula has 3 carbon atoms (\( \text{C}_3\text{H}_6\text{O}_2 \)), the compound must be propanoic acid.

1 mark for the correct option (B). 0 marks for any other option or if left blank.

The equation for the formation of propene is: \( 3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_6(\text{g}) \). Using a Hess cycle based on enthalpies of combustion: \( \Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) \). Therefore: \( \Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C}) + 3 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{C}_3\text{H}_6) \), \( \Delta_f H^\theta = [3(-394) + 3(-286)] - (-2058) \), \( \Delta_f H^\theta = [-1182 - 858] + 2058 = -2040 + 2058 = +18 \text{ kJ mol}^{-1} \).

1 mark for the correct option (C). 0 marks for any other option or if left blank.

To compare boiling points, examine the intermolecular forces of each compound. Butane and 2-methylpropane only have weak instantaneous dipole-induced dipole forces. Propanone has permanent dipole-permanent dipole forces in addition to instantaneous dipole-induced dipole forces due to its polar carbonyl bond (\( \text{C}=\text{O} \)). Propan-1-ol has a highly polar hydroxyl group (\( \text{-OH} \)) which can form hydrogen bonds, the strongest type of intermolecular force. Therefore, propan-1-ol requires the most thermal energy to overcome these intermolecular forces, giving it the highest boiling point.

1 mark for the correct option (C). 0 marks for any other option or if left blank.

Let us analyze the statements: (A) \( \text{ClO}^\bullet \) is produced in Step 1 and consumed in Step 2, so it is an intermediate, not a catalyst. (B) Summing the two steps yields the overall equation: \( \text{O}_3 + \text{O} \rightarrow 2\text{O}_2 \) (the catalyst \( \text{Cl}^\bullet \) and intermediate \( \text{ClO}^\bullet \) cancel out). (C) \( \text{Cl}^\bullet \) is consumed in Step 1 and regenerated in Step 2, so it behaves as a catalyst, not an intermediate. (D) Both steps start with a radical and produce a radical, meaning they are both propagation steps. Thus, D is the correct option.

1 mark for the correct option (D). 0 marks for any other option or if left blank.

First, calculate the amount in moles of Mg: moles = mass / Ar = 0.120 / 24.3 = 0.004938 mol. From the stoichiometry of the equation, 1 mole of Mg produces 1 mole of H2, so moles of H2 = 0.004938 mol. The volume of H2 gas at RTP = moles * molar gas volume = 0.004938 * 24000 cm3 mol-1 = 118.5 cm3, which rounds to 119 cm3.

Award 1 mark for correct selection of option A. Reject all other options.

The cream-coloured precipitate with silver nitrate indicates the presence of bromide ions (AgBr is cream). When chlorine water is added, chlorine (being more reactive than bromine) oxidises bromide to bromine: Cl2 + 2Br- -> 2Cl- + Br2. The molecular bromine dissolved in the organic cyclohexane layer gives an orange colour. Since cyclohexane is less dense than water, it forms the upper layer, which becomes orange.

Award 1 mark for correct selection of option B. Reject all other options.

Using Hess's law, the standard enthalpy change of formation is calculated from enthalpy of combustion data using: Delta_f H = sum of Delta_c H(reactants) - sum of Delta_c H(products). For the reaction 3C(s) + 4H2(g) -> C3H8(g): Delta_f H = [3 * (-394) + 4 * (-286)] - [-2220] = [-1182 - 1144] + 2220 = -2326 + 2220 = -106 kJ mol-1.

Award 1 mark for correct selection of option A. Reject all other options.

For an alkene to exhibit E/Z isomerism, both carbon atoms of the double bond must be attached to two different groups. In 3-methylpent-2-ene, C2 is bonded to -H and -CH3 (different) and C3 is bonded to -CH3 and -CH2CH3 (different). 2-methylbut-2-ene and 2,3-dimethylbut-2-ene have at least one carbon bonded to two identical methyl groups, and 2-methylpent-1-ene has C1 bonded to two identical hydrogen atoms.

Award 1 mark for correct selection of option A. Reject all other options.

Mass of water lost = \(4.41\text{ g} - 3.33\text{ g} = 1.08\text{ g}\). Moles of \(\text{H}_2\text{O}\) lost = \(1.08\text{ g} / 18.0\text{ g mol}^{-1} = 0.060\text{ mol}\). Moles of anhydrous \(\text{CaCl}_2\) = \(3.33\text{ g} / 111.0\text{ g mol}^{-1} = 0.030\text{ mol}\). Ratio of moles of water to calcium chloride = \(0.060 / 0.030 = 2\). Therefore, \(x = 2\).

M1: Mass of water lost (1.08 g) and its moles (0.060 mol) calculated [1 mark]; M2: Moles of anhydrous calcium chloride (0.030 mol) calculated [1 mark]; M3: Ratio of water to salt calculated to give the final integer value of x = 2 [1 mark].

The reaction of chlorine with cold, dilute sodium hydroxide is a disproportionation reaction. The ionic equation is \(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\). Chlorine starts with an oxidation state of 0 in \(\text{Cl}_2\). It is reduced to -1 in the chloride ion (\(\text{Cl}^-\)) and oxidized to +1 in the chlorate(I) ion (\(\text{ClO}^-\)).

M1: Correct ionic equation with balanced charges [1 mark]; M2: Correctly identifies chlorine being reduced from 0 to -1 [1 mark]; M3: Correctly identifies chlorine being oxidized from 0 to +1 [1 mark].

Boiling point increases as the carbon chain length increases from pentane to heptane. This is because longer-chain alkanes have a larger molecular size and more electrons, resulting in a larger surface contact area. Consequently, the strength of the instantaneous dipole-induced dipole forces (London forces) between molecules increases, requiring more energy to overcome.

M1: Trend stated correctly: boiling point increases from pentane to heptane [1 mark]; M2: Explanation of more electrons / larger surface area of contact [1 mark]; M3: Stronger instantaneous dipole-induced dipole forces (London forces) needing more thermal energy to break [1 mark].

Heat transferred to the water: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 22.5\text{ K} = 9405\text{ J} = 9.405\text{ kJ}\). Moles of methanol burned: \(n = 0.80\text{ g} / 32.0\text{ g mol}^{-1} = 0.025\text{ mol}\). Enthalpy change: \(\Delta_c H = -q / n = -9.405\text{ kJ} / 0.025\text{ mol} = -376.2\text{ kJ mol}^{-1}\). Expressed to 3 significant figures, this is \(-376\text{ kJ mol}^{-1}\).

M1: Correct calculation of heat transferred, \(q = 9405\text{ J}\) or \(9.405\text{ kJ}\) [1 mark]; M2: Correct calculation of moles of methanol, \(n = 0.025\text{ mol}\) [1 mark]; M3: Correct calculation of molar enthalpy change with negative sign, \(-376\text{ kJ mol}^{-1}\) (allow -380 or -376.2) [1 mark].

The infrared spectrum shows a broad, strong peak at \(3000\text{ cm}^{-1}\), which is characteristic of an \(\text{O-H}\) stretch in carboxylic acids (range \(2500\text{-}3300\text{ cm}^{-1}\)). The sharp peak at \(1715\text{ cm}^{-1}\) corresponds to the \(\text{C=O}\) double bond stretch (range \(1680\text{-}1750\text{ cm}^{-1}\)). With the molecular formula \(\text{C}_3\text{H}_6\text{O}_2\), this compound is propanoic acid, with structural formula \(\text{CH}_3\text{CH}_2\text{COOH}\).

M1: Identifies the compound as a carboxylic acid / provides structural formula \(\text{CH}_3\text{CH}_2\text{COOH}\) [1 mark]; M2: Assigns the peak at \(1715\text{ cm}^{-1}\) to the \(\text{C=O}\) bond stretch [1 mark]; M3: Assigns the broad peak at \(3000\text{ cm}^{-1}\) to the carboxylic acid \(\text{O-H}\) bond stretch [1 mark].

The destruction of ozone is catalyzed by chlorine radicals via two propagation steps. In the first propagation step, a chlorine radical reacts with ozone to form a chlorine monoxide radical and oxygen: \(\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2\). In the second propagation step, the chlorine monoxide radical reacts with a monoatomic oxygen radical, regenerating the chlorine catalyst: \(\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2\). The overall reaction is obtained by summing the two steps: \(\text{O}_3 + \text{O} \rightarrow 2\text{O}_2\).

M1: Correct equation for propagation step 1 with radical dot [1 mark]; M2: Correct equation for propagation step 2 with radical dot [1 mark]; M3: Correct overall equation [1 mark].

Bromide ions are oxidized to bromine gas: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\). Sulfuric acid acts as the oxidizing agent and is reduced to sulfur dioxide: \(\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\) (or \(\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\)). Combining these two half-equations gives: \(2\text{Br}^- + \text{H}_2\text{SO}_4 + 2\text{H}^+ \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\). The oxidizing agent is concentrated sulfuric acid (or sulfate ions) because the oxidation state of sulfur decreases from +6 to +4.

M1: Identifies the oxidizing agent as \(\text{H}_2\text{SO}_4\) / sulfuric acid / \(\text{SO}_4^{2-}\) with explanation based on oxidation state decrease [1 mark]; M2: Correctly writes balanced half-equations or showing correct species [1 mark]; M3: Writes the fully correct and balanced ionic equation [1 mark].

According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium in the direction of the endothermic reaction to absorb the added thermal energy. Since the forward reaction is endothermic (\(\Delta H = +57\text{ kJ mol}^{-1}\)), the equilibrium shifts to the right, favoring the production of \(\text{NO}_2\). Since \(\text{NO}_2\) is dark brown and \(\text{N}_2\text{O}_4\) is colorless, the color intensity of the gas mixture will increase (it becomes a darker brown).

M1: Predicts shift of equilibrium position to the right / forward direction [1 mark]; M2: Explains that because the forward reaction is endothermic, increasing temperature favors this direction [1 mark]; M3: Deduces that the brown color intensity will increase / become darker due to more \(\text{NO}_2\) [1 mark].

When chlorine reacts with cold, dilute aqueous sodium hydroxide, it undergoes disproportionation. One chlorine atom is reduced and another is oxidized: \( \text{Cl}_2(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cl}^-(aq) + \text{ClO}^-(aq) + \text{H}_2\text{O}(l) \). The oxidation state of chlorine in \( \text{Cl}_2 \) is 0. In \( \text{Cl}^- \), the oxidation state is -1. In \( \text{ClO}^- \), the oxidation state is +1.

1 mark: Correctly balanced ionic equation with state symbols. 1 mark: Identification of starting oxidation state of chlorine as 0 and ending states as -1 and +1. 0.7 mark: Explicit mention that this is a disproportionation reaction.

Bonds broken: 3(\( \text{C-H} \)) + 1(\( \text{C-O} \)) + 1(\( \text{O-H} \)) + 1.5(\( \text{O=O} \)) = 3(413) + 358 + 463 + 1.5(498) = 1239 + 358 + 463 + 747 = +2807 \( \text{kJ mol}^{-1} \). Bonds formed: 2(\( \text{C=O} \)) + 4(\( \text{O-H} \)) = 2(805) + 4(463) = 1610 + 1852 = +3462 \( \text{kJ mol}^{-1} \). \( \Delta_c H^\theta \) = Bonds broken - Bonds formed = 2807 - 3462 = -655 \( \text{kJ mol}^{-1} \).

1 mark: Correct calculation of total bonds broken (+2807). 1 mark: Correct calculation of total bonds formed (+3462). 0.7 mark: Correct final calculation with negative sign (-655 kJ mol^-1).

Step 1: Calculate moles of reactant. Moles of \( \text{CuO} = 15.90 / 79.5 = 0.200\text{ mol} \). Step 2: Determine theoretical yield. Based on 2:2 stoichiometry, 0.200 mol of \( \text{CuO} \) should produce 0.200 mol of \( \text{Cu} \). Theoretical mass of \( \text{Cu} = 0.200 \times 63.5 = 12.70\text{ g} \). Step 3: Calculate percentage yield. Percentage yield = (Actual mass / Theoretical mass) \\times 100 = (10.80 / 12.70) \\times 100 = 85.039% \\approx 85.0%.

1 mark: Calculate correct moles of CuO (0.200 mol). 1 mark: Calculate correct theoretical mass of Cu (12.70 g). 0.7 mark: Correct final yield of 85.0% (accept 85% or 85.0%).

Chlorine is more reactive than iodine and displaces iodide ions from solution. The ionic equation is \( \text{Cl}_2(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{I}_2(aq) \). The iodide ions (\( \text{I}^- \)) lose electrons (oxidation state changes from -1 to 0) and thus act as the reducing agent.

1 mark: Correct balanced ionic equation. 1 mark: Correct state symbols. 0.7 mark: Correct identification of iodide ion/I^- as the reducing agent (do not accept I2 or KI).

In the first step, a chlorine radical reacts with an ozone molecule: \( \text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2 \). In the second step, the chlorine monoxide radical reacts with a monoatomic oxygen atom: \( \text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2 \). The net reaction is \( \text{O}_3 + \text{O} \rightarrow 2\text{O}_2 \). The chlorine radical is regenerated at the end, meaning it is not consumed and acts as a catalyst.

1 mark: Correct first propagation step. 1 mark: Correct second propagation step with radical dots. 0.7 mark: Correct explanation of catalytic role (regenerated at the end, not consumed overall).

An absorption peak at \( 1715\text{ cm}^{-1} \) indicates the presence of a carbonyl group (\( \text{C=O} \)). The absence of any peak above \( 3200\text{ cm}^{-1} \) confirms the absence of an alcohol group (\( \text{O-H} \)). Since the formula is \( \text{C}_3\text{H}_6\text{O} \) and it contains a carbonyl group but no hydroxyl group, the molecule is propanone, a ketone. The skeletal structure consists of three carbons in a chain with a double bond to an oxygen atom on the middle carbon.

1 mark: Correctly identifies carbonyl / ketone group from the peak at 1715 cm^-1. 1 mark: Correctly rules out alcohol due to lack of O-H peak above 3200 cm^-1. 0.7 mark: Correct skeletal structure of propanone.

As you descend Group 2, the thermal stability of the carbonates increases. Going down the group, the ionic radius of the Group 2 cation increases while the 2+ charge remains the same, meaning charge density decreases. As a result, the larger cations have a lower polarizing power and are less able to polarize/distort the carbonate ion (\( \text{CO}_3^{2-} \)). Less polarization means the carbon-oxygen bonds within the carbonate ion are weakened less, requiring more thermal energy to decompose the compound.

1 mark: Stating thermal stability increases down the group. 1 mark: Explaining that cation ionic radius increases / charge density decreases down the group. 0.7 mark: Linking this to less polarization/distortion of the carbonate ion.

Both molecules have similar relative molecular masses, so they have similar strengths of instantaneous dipole-induced dipole (London) forces. However, propan-1-ol contains a highly polar \( \text{O-H} \) bond which allows it to form intermolecular hydrogen bonds. Hydrogen bonds are significantly stronger than London forces and require a much larger amount of thermal energy to overcome, resulting in a much higher boiling point for propan-1-ol compared to butane, which only has weak London forces.

1 mark: Identifying that butane only has instantaneous dipole-induced dipole (London) forces while propan-1-ol also has hydrogen bonding. 1 mark: Stating that hydrogen bonding is much stronger than London forces. 0.7 mark: Explaining that more energy is needed to separate propan-1-ol molecules, leading to a higher boiling point.

(i) Aqueous chlorine oxidises iodide ions to iodine. The initial solution of potassium iodide is colourless, and the resulting aqueous iodine is brown or orange-brown.

(ii) The ionic equation is:
\(\text{Cl}_2(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\)

(iii) Iodine is much more soluble in cyclohexane (an organic solvent) than in water. When shaken with cyclohexane, iodine dissolves in the upper organic layer, turning it purple/violet.

**(i)** Colourless to brown / orange-brown / yellow [1 mark]
*Do not accept 'red' or 'clear'*

**(ii)** \(\text{Cl}_2(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\) [1 mark]
*Allow correct state symbols even if species are incorrect, but the mark is for the full correct equation with state symbols.*

**(iii)** Purple / pink / violet [1 mark]
*Do not accept blue*

(i) The balanced chemical equation for the complete combustion of butane is:
\(\text{C}_4\text{H}_{10}(\text{g}) + 6.5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\) or \(2\text{C}_4\text{H}_{10}(\text{g}) + 13\text{O}_2(\text{g}) \rightarrow 8\text{CO}_2(\text{g}) + 10\text{H}_2\text{O}(\text{l})\).

(ii) From the mole ratio in the balanced equation, \(1\text{ mol}\) of \(\text{C}_4\text{H}_{10}\) reacts with \(6.5\text{ mol}\) of \(\text{O}_2\).
Since volume is proportional to moles at constant temperature and pressure:
\(\text{Volume of } \text{O}_2 = 150\text{ cm}^3 \times 6.5 = 975\text{ cm}^3\).

(iii) From the mole ratio, \(1\text{ mol}\) of \(\text{C}_4\text{H}_{10}\) produces \(4\text{ mol}\) of \(\text{CO}_2\).
\(\text{Volume of } \text{CO}_2 = 150\text{ cm}^3 \times 4 = 600\text{ cm}^3\).
Converting to \(\text{dm}^3\):
\(600\text{ cm}^3 = 0.600\text{ dm}^3\).
\(\text{Moles of } \text{CO}_2 = \frac{0.600\text{ dm}^3}{24\text{ dm}^3\text{ mol}^{-1}} = 0.025\text{ mol}\).

**(i)** \(\text{C}_4\text{H}_{10}(\text{g}) + 6.5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\) [1 mark]
*Allow multiples. Ignore state symbols unless incorrect.*

**(ii)** \(975\text{ cm}^3\) [1 mark]
*Allow error carried forward (ECF) from (i), e.g. \(150 \times \text{ratio from (i)}\)*

**(iii)** \(0.025\text{ mol}\) (or \(2.50 \times 10^{-2}\)) [1 mark]
*Allow ECF from (i) or (ii)*

1. **Experimental Procedure**:
- Weigh a spirit burner containing propan-1-ol.
- Measure a known volume (e.g., \(100\text{ cm}^3\)) of water into a copper calorimeter/beaker and measure its initial temperature.
- Support the calorimeter above the burner, shield it to minimize drafts, and light the burner.
- Burn the alcohol until the temperature rises by a suitable amount (e.g., about \(20\text{ }^\circ\text{C}\)).
- Extinguish the burner and measure the maximum temperature reached by the water.
- Reweigh the spirit burner to determine the mass of propan-1-ol burned.

2. **Calculation**:
- Use \(q = mc\Delta T\) to calculate the heat energy transferred to the water (where \(m\) is the mass of water, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T\) is the temperature rise).
- Calculate the moles of propan-1-ol burned: \(n = \frac{\text{mass of alcohol burned}}{M_r}\) (where \(M_r \text{ of propan-1-ol} = 60.0\text{ g mol}^{-1}\)).
- Calculate the enthalpy change of combustion: \(\Delta_c H = -\frac{q}{n}\) (converting \(q\) to \(\text{kJ}\)).

3. **Reasons for Discrepancy (less exothermic) and Improvements**:
- **Heat loss to the surroundings**: Use draft shields, or a bomb calorimeter.
- **Incomplete combustion of the alcohol** (evidenced by soot): Ensure adequate supply of oxygen or use a pure oxygen stream.
- **Evaporation of alcohol from the wick**: Put the cap back on the spirit burner immediately after extinguishing.
- **Heat capacity of the container ignored**: Include the heat capacity of the calorimeter in the calculation.

**Level 3 (5–6 marks)**: Thorough description of the experimental procedure including all key measurements. Shows a clear, step-by-step mathematical method to determine the enthalpy change. Identifies at least two sources of heat loss/incomplete combustion and suggests realistic improvements.

**Level 2 (3–4 marks)**: Reasonable description of the experimental setup and measurements. Outlines the calculation using \(q = mc\Delta T\) and moles. Identifies at least one source of error and one improvement.

**Level 1 (1–2 marks)**: Identifies some measurements needed (e.g. mass of alcohol, temperature change) or mentions the formula \(q = mc\Delta T\). Gives a vague reason for the difference in values.

1. **Infrared Analysis**:
- The broad peak at \(3350\text{ cm}^{-1}\) corresponds to an \(\text{O-H}\) alcohol stretch.
- The absence of a peak around \(1700\text{ cm}^{-1}\) indicates that there is no \(\text{C=O}\) carbonyl group present. Thus, X is an alcohol.

2. **Mass Spectrometry Analysis**:
- The molecular ion peak \(M^+\) at \(m/z = 74\) gives the molar mass of the compound as \(74\text{ g mol}^{-1}\).
- An aliphatic alcohol with \(1\) oxygen atom has formula \(\text{C}_n\text{H}_{2n+2}\text{O}\).
- \(12n + (2n+2) + 16 = 74 \implies 14n + 18 = 74 \implies 14n = 56 \implies n = 4\).
- Therefore, X is an isomer of butanol (formula \(\text{C}_4\text{H}_{10}\text{O}\)).
- The fragment peak at \(m/z = 43\) corresponds to the alkyl carbocation \([\text{C}_3\text{H}_7]^+\) (or \([\text{CH}_3\text{CH}_2\text{CH}_2]^+\) or \([\text{CH(CH}_3)_2]^+\)), which is formed by the loss of a \(\text{CH}_2\text{OH}\) group (mass \(31\)) from the molecular ion: \(74 - 31 = 43\).

3. **Identification and Skeletal Structure**:
- Isomers like butan-1-ol or butan-2-ol are consistent with this. For butan-1-ol, the skeletal structure is a 4-carbon chain with an -OH group at the end.

M1: Identifies the broad peak at \(3350\text{ cm}^{-1}\) as an \(\text{O-H}\) alcohol stretch, and the absence of a peak at \(1700\text{ cm}^{-1}\) as no carbonyl (\(\text{C=O}\)) group. [1 mark]
M2: Uses the molecular ion peak \(m/z = 74\) to deduce the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). [1 mark]
M3: Draws a correct skeletal structure of any butanol isomer (e.g., butan-1-ol, butan-2-ol, methylpropan-1-ol, or methylpropan-2-ol). [1 mark]
M4: Identifies the fragment at \(m/z = 43\) as \([\text{C}_3\text{H}_7]^+\) (positive charge must be included). [1 mark]
M5: Explains the formation of the \(m/z = 43\) fragment by the loss of a \(\text{CH}_2\text{OH}\) radical (mass 31) from \(M^+\), or shows a reasonable fragmentation equation. [1 mark]

The mechanism for ozone depletion catalyzed by chlorine radicals is:
1. **Propagation Step 1**:
\(\text{Cl}^\bullet\text{(g)} + \text{O}_3\text{(g)} \rightarrow \text{ClO}^\bullet\text{(g)} + \text{O}_2\text{(g)}\)

2. **Propagation Step 2**:
\(\text{ClO}^\bullet\text{(g)} + \text{O(g)} \rightarrow \text{Cl}^\bullet\text{(g)} + \text{O}_2\text{(g)}\)

3. **Overall Equation**:
By adding these two steps, we get:
\(\text{O}_3\text{(g)} + \text{O(g)} \rightarrow 2\text{O}_2\text{(g)}\)

4. **Role as a Catalyst**:
- A catalyst increases the rate of reaction and is chemically unchanged at the end of the reaction.
- \(\text{Cl}^\bullet\) is consumed in the first step and regenerated in the second step, so it is not consumed overall.

M1: Correct equation for Propagation Step 1: \(\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2\). [1 mark]
M2: Correct equation for Propagation Step 2: \(\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2\). [1 mark]
M3: Correct overall equation: \(\text{O}_3 + \text{O} \rightarrow 2\text{O}_2\). [1 mark]
M4: Correctly explains that \(\text{Cl}^\bullet\) acts as a catalyst because it is used up in Step 1 and regenerated in Step 2, so its concentration is unchanged overall. [1 mark]
(Accept without radical dots if consistent, but dots are preferred).

1. **Observation**:
- The colorless solution turns brown (or yellow/brown) due to the formation of aqueous iodine (\(\text{I}_2\text{(aq)}\)).

2. **Ionic Equation**:
\(\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\)

3. **Oxidizing Agent and Oxidation Numbers**:
- The oxidizing agent is chlorine, \(\text{Cl}_2\).
- The oxidation number of chlorine decreases from \(0\) (in \(\text{Cl}_2\)) to \(-1\) (in \(\text{Cl}^-\)), meaning it has been reduced.
- It has oxidized iodide ions (\(\text{I}^-\), oxidation number \(-1\)) to iodine (\(\text{I}_2\), oxidation number \(0\)).

4. **Addition of Cyclohexane**:
- Iodine is more soluble in organic solvents than in water.
- Shaking with cyclohexane will result in a two-phase mixture.
- The top cyclohexane layer will turn purple (or violet/pink).

M1: Solution turns brown / orange-brown (reject yellow alone, accept brown). [1 mark]
M2: Correct ionic equation: \(\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\) (states not required but must be balanced). [1 mark]
M3: Identifies chlorine (or \(\text{Cl}_2\)) as the oxidizing agent. [1 mark]
M4: Explanation using oxidation numbers: \(\text{Cl}\) goes from \(0\) to \(-1\) (reduction / gaining electrons) OR \(\text{I}\) goes from \(-1\) to \(0\) (oxidation / losing electrons) so \(\text{Cl}_2\) causes the oxidation. [1 mark]
M5: Cyclohexane layer (upper layer) turns purple / pink / violet. [1 mark]

Let's perform the calculation step-by-step:

1. **Calculate the initial moles of HCl added to the limestone sample**:
\(n(\text{HCl})_{\text{initial}} = c \times V = 0.500\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0250\text{ mol}\)

2. **Calculate moles of NaOH used to neutralize the excess acid in the 25.0 cm³ aliquot**:
\(n(\text{NaOH}) = c \times V = 0.100\text{ mol dm}^{-3} \times 0.01840\text{ dm}^3 = 1.84 \times 10^{-3}\text{ mol}\)

3. **Calculate excess moles of HCl in the 25.0 cm³ aliquot**:
Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the ratio is 1:1.
\(n(\text{HCl})_{\text{excess in 25 mL}} = 1.84 \times 10^{-3}\text{ mol}\)

4. **Calculate excess moles of HCl in the total 100.0 cm³ solution**:
\(n(\text{HCl})_{\text{excess in 100 mL}} = 1.84 \times 10^{-3}\text{ mol} \times \frac{100.0}{25.0} = 7.36 \times 10^{-3}\text{ mol}\)

5. **Calculate the moles of HCl that reacted with CaCO₃**:
\(n(\text{HCl})_{\text{reacted}} = n(\text{HCl})_{\text{initial}} - n(\text{HCl})_{\text{excess in 100 mL}}\)
\(n(\text{HCl})_{\text{reacted}} = 0.0250\text{ mol} - 0.00736\text{ mol} = 0.01764\text{ mol}\)

6. **Calculate moles of CaCO₃ that reacted**:
The equation for the reaction is:
\(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\)
So, 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\).
\(n(\text{CaCO}_3) = \frac{1}{2} \times n(\text{HCl})_{\text{reacted}} = \frac{0.01764}{2} = 8.82 \times 10^{-3}\text{ mol}\)

7. **Calculate the mass of CaCO₃ in the sample**:
\(\text{mass of CaCO}_3 = 8.82 \times 10^{-3}\text{ mol} \times 100.1\text{ g mol}^{-1} = 0.88288\text{ g}\)

8. **Calculate the percentage purity of CaCO₃**:
\(\text{Percentage purity} = \frac{\text{mass of CaCO}_3}{\text{mass of sample}} \times 100 = \frac{0.88288}{1.25} \times 100 = 70.63\% \approx 70.6\%\)

M1: Calculates initial moles of HCl: \(0.500 \times 0.0500 = 0.0250\text{ mol}\). [1 mark]
M2: Calculates moles of NaOH in aliquot: \(0.100 \times 0.01840 = 1.84 \times 10^{-3}\text{ mol}\). [1 mark]
M3: Scales up to find excess moles of HCl in 100 cm³: \(1.84 \times 10^{-3} \times 4 = 7.36 \times 10^{-3}\text{ mol}\). [1 mark]
M4: Calculates moles of HCl reacted: \(0.0250 - 7.36 \times 10^{-3} = 0.01764\text{ mol}\). [1 mark]
M5: Deduces moles of \(\text{CaCO}_3\) is half of moles of reacted HCl: \(8.82 \times 10^{-3}\text{ mol}\). [1 mark]
M6: Calculates mass of \(\text{CaCO}_3\) and percentage purity to 3 significant figures: \(70.6\%\) (accept \(70.6\%\) or \(70.7\%\) depending on rounding of intermediate steps). [1 mark]

- **Trend**: Thermal stability of Group 2 carbonates increases down the group (higher temperature is required for decomposition).
- **Cation Size and Charge**: Down the group, the ionic radius of the Group 2 cations increases (more electron shells), while the charge remains constant at \(2+\).
- **Charge Density**: This results in a decrease in charge density of the cation down the group.
- **Polarization**: A cation with lower charge density has a weaker polarizing effect on the carbonate ion (it distorts the carbonate electron cloud less).
- **Decomposition**: Less distortion means the carbon-oxygen double/single bonds in the carbonate ion are weakened less, requiring more thermal energy to break them and release carbon dioxide.

M1: State that thermal stability increases down the group. [1 mark]
M2: State that down the group, the ionic radius of the Group 2 cation increases while the ionic charge remains the same (\(2+\)). [1 mark]
M3: Explain that this leads to a decrease in charge density of the cation down the group. [1 mark]
M4: Link lower charge density to less polarization/distortion of the carbonate ion (or its electron cloud), weakening the \(\text{C-O}\) bond less. [1 mark]

1. **\(K_c\) Expression**:
\(K_c = \frac{[\text{ICl}]^2}{[\text{I}_2][\text{Cl}_2]}\)

2. **Effect of Increasing Temperature on the Position of Equilibrium**:
- The forward reaction is exothermic (\(\Delta H = -26\text{ kJ mol}^{-1}\)).
- According to Le Chatelier's principle, an increase in temperature will shift the position of equilibrium to favor the endothermic direction to absorb the added heat.
- Therefore, the equilibrium shifts to the left (towards the reactants).

3. **Effect of Increasing Temperature on \(K_c\)**:
- Since the equilibrium shifts to the left, the concentration of the product, \([\text{ICl}]\), decreases, and the concentrations of the reactants, \([\text{I}_2]\) and \([\text{Cl}_2]\), increase.
- Therefore, the value of \(K_c\) decreases.

M1: Correct \(K_c\) expression: \(K_c = \frac{[\text{ICl}]^2}{[\text{I}_2][\text{Cl}_2]}\). (Square brackets required, ignore state symbols). [1 mark]
M2: States that position of equilibrium shifts to the left (towards reactants). [1 mark]
M3: Explains that the forward reaction is exothermic, so the system shifts in the endothermic direction to oppose the temperature increase. [1 mark]
M4: States that the value of \(K_c\) decreases. [1 mark]
M5: Explains that since equilibrium shifts to the left, the product concentration decreases relative to reactant concentrations, reducing \(K_c\). [1 mark]

1. **Maxwell-Boltzmann Curve features**:
- y-axis labeled as 'Number of molecules' (or 'Fraction of molecules' / 'Number of particles').
- x-axis labeled as 'Energy' (or 'Kinetic energy').
- The curve must start at the origin \((0,0)\), rise to a single peak, and decay asymptotically towards the x-axis (never touching the x-axis at high energy).

2. **Labeling Activation Energies**:
- A vertical line labeled \(E_a\) represents the activation energy of the uncatalyzed reaction.
- A vertical line labeled \(E_c\) (or \(E_a\text{(cat)}\)) to the left of \(E_a\) represents the lower activation energy of the catalyzed reaction.

3. **Explanation of Catalyst Action**:
- A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_c < E_a\)).
- The shaded area under the curve to the right of the activation energy represents the fraction of molecules with sufficient energy to react.
- Because \(E_c\) is lower than \(E_a\), a significantly larger fraction of molecules now possess energy greater than or equal to the activation energy (represented by the larger shaded area under the curve to the right of \(E_c\)).
- This results in a higher frequency of successful collisions, thereby increasing the rate of reaction.

M1: Draw a correct Maxwell-Boltzmann distribution curve (starts at origin, skewed to the left, never touches x-axis at high energy) with both axes correctly labeled ('Number of molecules' on y-axis and 'Energy' on x-axis). [1 mark]
M2: Correctly label \(E_a\) (uncatalyzed activation energy) and show a lower activation energy, \(E_c\) (catalyzed activation energy), to the left of \(E_a\). [1 mark]
M3: State that a catalyst provides an alternative pathway with a lower activation energy. [1 mark]
M4: Explain that at a lower activation energy, a larger fraction/proportion of molecules have energy \(\ge\) the activation energy. [1 mark]
M5: State that this leads to a higher frequency of successful collisions (or more successful collisions per unit time). [1 mark]

Step 1: Calculate the heat energy absorbed by the water.
\(q = m c \Delta T\)
\(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 20.9\text{ K} = 13104.3\text{ J} = 13.1\text{ kJ}\)

Step 2: Calculate the number of moles of propan-1-ol burned.
\(n = \frac{\text{mass}}{\text{Molar Mass}} = \frac{0.56\text{ g}}{60.0\text{ g mol}^{-1}} = 0.00933\text{ mol}\)

Step 3: Calculate the enthalpy change of combustion (with negative sign for exothermic reaction).
\(\Delta_c H = -\frac{q}{n} = -\frac{13.1043\text{ kJ}}{0.00933\text{ mol}} = -1404\text{ kJ mol}^{-1}\)

Step 4: Round to two significant figures (due to mass being 0.56 g, 2 sig figs).
\(\Delta_c H = -1400\text{ kJ mol}^{-1}\)

- Mark 1: Correct calculation of heat absorbed by water: \(13.1\text{ kJ}\) (or \(13104\text{ J}\)).
- Mark 2: Correct calculation of moles of propan-1-ol: \(0.00933\text{ mol}\).
- Mark 3: Division of heat by moles to find \(\Delta_c H\).
- Mark 4: Correct final answer to 2 significant figures: \(-1400\text{ kJ mol}^{-1}\) (must include negative sign for this mark).

1. The reaction is: \(Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)\).
2. Chlorine (\(Cl_2\)) is the oxidizing agent.
- In terms of electrons, chlorine gains electrons to become chloride ions: \(Cl_2 + 2e^- \rightarrow 2Cl^-\).
- In terms of oxidation states, chlorine is reduced as its oxidation state decreases from \(0\) (in \(Cl_2\)) to \(-1\) (in \(Cl^-\)).

- Mark 1: Correct balanced ionic equation (species and balancing).
- Mark 2: Correct state symbols: \(aq\) for reactants and products (allow \(aq\) or \(l\) for \(Br_2\)).
- Mark 3: Identifies chlorine (\(Cl_2\)) as the oxidizing agent because it gains electrons / is reduced.
- Mark 4: Explains oxidation state change from \(0\) to \(-1\).

1. Equation: \(Cl_2 + H_2O \rightleftharpoons HClO + HCl\) (allow single arrow, accept ionic equivalents).
2. Disproportionation is a reaction in which the same element is simultaneously oxidized and reduced.
- The oxidation state of chlorine in \(Cl_2\) is \(0\).
- In \(HClO\), the oxidation state of chlorine is \(+1\) (oxidized).
- In \(HCl\), the oxidation state of chlorine is \(-1\) (reduced).
3. Risk: Chlorine gas is toxic / can react with organic material in water to form toxic chlorinated hydrocarbons (trihalomethanes) which are carcinogenic.
Benefit: Effectively sanitizes water by killing harmful bacteria, preventing major outbreaks of waterborne diseases (like cholera).

- Mark 1: Correct equation for chlorine reacting with water.
- Mark 2: Defines disproportionation showing chlorine is both oxidized and reduced.
- Mark 3: Identifies correct oxidation states: 0 in \(Cl_2\), +1 in \(HClO\), -1 in \(HCl\).
- Mark 4: Valid risk of chlorination (e.g. formation of toxic organochlorine compounds / toxicity of gas).
- Mark 5: Valid benefit of chlorination (kills disease-causing bacteria / sterilizes water).

1. Functional group identification:
- The broad absorption band at \(2500 - 3300\text{ cm}^{-1}\) indicates an \(O-H\) stretch of a carboxylic acid.
- The strong peak at \(1715\text{ cm}^{-1}\) indicates a \(C=O\) stretch.
- Therefore, the functional group is a carboxylic acid.

2. Name of compound:
- A 3-carbon carboxylic acid with molecular formula \(C_3H_6O_2\) is propanoic acid (\(CH_3CH_2COOH\)).

3. Fragment at \(m/z = 45\):
- Carboxylic acids often fragment to yield the \([COOH]^+\) ion.
- Mass calculation: \(C = 12.0\), \(O \times 2 = 32.0\), \(H = 1.0\); Total mass = \(12.0 + 32.0 + 1.0 = 45.0\).
- The positive charge is required for detection in mass spectrometry: \([COOH]^+\) (or \(CO_2H^+\)).

- Mark 1: Identifies the \(O-H\) broad absorption (2500-3300 cm-1) and/or \(C=O\) peak (1715 cm-1) as characteristic of a carboxylic acid.
- Mark 2: Names the functional group as a carboxylic acid.
- Mark 3: Deduces the IUPAC name as propanoic acid.
- Mark 4: Identifies the fragment at \(m/z = 45\) as \([COOH]^+\) or \([CO_2H]^+\).
- Mark 5: Includes the positive charge on the fragment ion formula (reject formulas without a positive charge).

Step 1: Calculate the mass of water lost.
\(\text{Mass of } H_2O = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\)

Step 2: Calculate the number of moles of anhydrous \(CuSO_4\).
\(n(CuSO_4) = \frac{3.19\text{ g}}{159.6\text{ g mol}^{-1}} = 0.0200\text{ mol}\)

Step 3: Calculate the number of moles of water lost.
\(n(H_2O) = \frac{1.80\text{ g}}{18.0\text{ g mol}^{-1}} = 0.100\text{ mol}\)

Step 4: Find the simplest molar ratio.
\(x = \frac{n(H_2O)}{n(CuSO_4)} = \frac{0.100}{0.0200} = 5.00\)

Therefore, the value of \(x = 5\).

- Mark 1: Calculates the mass of water lost: \(1.80\text{ g}\).
- Mark 2: Calculates the amount of anhydrous \(CuSO_4\): \(0.0200\text{ mol}\).
- Mark 3: Calculates the amount of \(H_2O\): \(0.100\text{ mol}\).
- Mark 4: Expresses the ratio as a whole number: \(x = 5\).

1. Initiation: Under UV radiation, a C-Cl bond breaks homolytically:
\(CF_2Cl_2 \xrightarrow{UV} \cdot CF_2Cl + Cl\cdot\)

2. Propagation steps:
- Step 1: \(Cl\cdot + O_3 \rightarrow ClO\cdot + O_2\)
- Step 2: \(ClO\cdot + O \rightarrow Cl\cdot + O_2\)

3. Role as catalyst:
- The chlorine radical (\(Cl\cdot\)) is consumed in propagation step 1 but is regenerated in propagation step 2.
- Consequently, it is not used up in the overall chemical reaction and can go on to destroy thousands more ozone molecules.

- Mark 1: Correct initiation equation showing radical dots on both \(CF_2Cl\) and \(Cl\) (allow dots placed anywhere appropriate on the radical).
- Mark 2: Propagation step 1 correct.
- Mark 3: Propagation step 2 correct.
- Mark 4: Mentions that \(Cl\cdot\) is used up/reacted in step 1 and regenerated/reformed in step 2.
- Mark 5: States that because it is regenerated, it is not consumed overall (acting as a catalyst).

1. Intermolecular forces:
- Both pentane and 2,2-dimethylpropane are non-polar hydrocarbons that rely entirely on instantaneous dipole-induced dipole (London / dispersion) forces between molecules.

2. Shape and contact surface area:
- Pentane is a straight-chain (linear) molecule. This shape allows molecules to pack closely together, maximizing the contact surface area between adjacent molecules.
- 2,2-dimethylpropane is a highly branched, spherical molecule. Its compact shape limits the amount of surface contact possible between adjacent molecules.

3. Energy requirement:
- Because pentane has a larger contact surface area, its London forces are stronger and more numerous.
- Consequently, more thermal energy is needed to overcome these intermolecular forces and separate the molecules during boiling, resulting in a higher boiling point.

- Mark 1: Identifies the intermolecular forces in both compounds as instantaneous dipole-induced dipole (London / dispersion) forces.
- Mark 2: Describes pentane as straight-chain/linear with a larger contact surface area.
- Mark 3: Describes 2,2-dimethylpropane as branched/spherical with a smaller contact surface area.
- Mark 4: Relates larger contact surface area to stronger/more intermolecular forces in pentane (or weaker in 2,2-dimethylpropane).
- Mark 5: Explains that more thermal energy is needed to overcome the stronger intermolecular forces in pentane, leading to its higher boiling point.