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Write down the balanced chemical equations for the complete combustion of both hydrocarbons:
1. Ethane:
\(\text{C}_2\text{H}_6\text{(g)} + 3.5\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\)
For \(10\text{ cm}^3\) of \(\text{C}_2\text{H}_6\):
- Oxygen consumed = \(10 \times 3.5 = 35\text{ cm}^3\)
- Carbon dioxide produced = \(10 \times 2 = 20\text{ cm}^3\)

2. Propane:
\(\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\)
For \(10\text{ cm}^3\) of \(\text{C}_3\text{H}_8\):
- Oxygen consumed = \(10 \times 5 = 50\text{ cm}^3\)
- Carbon dioxide produced = \(10 \times 3 = 30\text{ cm}^3\)

Total volume of oxygen consumed = \(35 + 50 = 85\text{ cm}^3\).
Remaining volume of oxygen = \(100 - 85 = 15\text{ cm}^3\).
Total volume of carbon dioxide produced = \(20 + 30 = 50\text{ cm}^3\).
At room temperature and pressure, water is a liquid, so its volume is negligible.
Therefore, the total volume of dry gas remaining = \(15\text{ cm}^3\text{ (oxygen)} + 50\text{ cm}^3\text{ (carbon dioxide)} = 65\text{ cm}^3\).

1 mark: Correctly calculate the volume of oxygen remaining (15 cm³) and carbon dioxide produced (50 cm³), and sum them to obtain 65 cm³.

In sulfuric acid, \(\text{H}_2\text{SO}_4\), the oxidation number of sulfur is \(+6\).
Potassium iodide is a strong reducing agent and reduces sulfuric acid to several sulfur-containing species:
- \(\text{SO}_2\): The oxidation number of sulfur is \(+4\) (decrease of 2).
- \(\text{S}\): The oxidation number of sulfur is \(0\) (decrease of 6).
- \(\text{H}_2\text{S}\): The oxidation number of sulfur is \(-2\) (decrease of 8).

Comparing these values, the greatest decrease in oxidation number for sulfur occurs during the formation of \(\text{H}_2\text{S}\) (from \(+6\) to \(-2\), which is a decrease of 8).

1 mark: Identify the oxidation states of sulfur in each of the listed sulfur-containing products, and choose the one with the greatest reduction (H₂S, decrease of 8).

Using Hess's Law and standard enthalpies of combustion:
\(\Delta H_r^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\)

\(\Delta H_r^\theta = [\Delta H_c^\theta(\text{C}_4\text{H}_6\text{(g)}) + 2 \times \Delta H_c^\theta(\text{H}_2\text{(g)})] - [\Delta H_c^\theta(\text{C}_4\text{H}_{10}\text{(g)})]\)

\(\Delta H_r^\theta = [-2540 + 2 \times (-286)] - [-2878]\)
\(\Delta H_r^\theta = [-2540 - 572] + 2878\)
\(\Delta H_r^\theta = -3112 + 2878 = -234\text{ kJ mol}^{-1}\)

1 mark: Correctly set up the Hess's Law cycle using standard enthalpies of combustion, remembering to multiply the value for H₂ by 2, and accurately calculate -234 kJ mol⁻¹.

Analyze the structural formula of 4-methylhex-2-ene to identify stereogenic centers:
1. The double bond between C2 and C3 (\(\text{-CH}=\text{CH-}\)) is bonded to \(\text{-H}\) and \(\text{-CH}_3\) on C2, and \(\text{-H}\) and \(\text{-CH(CH}_3)\text{CH}_2\text{CH}_3\) on C3. Since both carbons in the double bond have two different groups attached, it exhibits cis-trans (E/Z) isomerism (2 configurations).
2. Carbon 4 (\(\text{-CH(CH}_3)-\)) is a chiral center because it is bonded to four different groups: \(\text{-H}\), \(\text{-CH}_3\), \(\text{-CH}_2\text{CH}_3\), and \(\text{-CH}=\text{CHCH}_3\). Thus, it exhibits optical isomerism (2 configurations: R and S).

Total number of stereoisomers = \(2 \text{ (cis/trans)} \times 2 \text{ (R/S)} = 4\).

1 mark: Correctly identify the presence of exactly one double bond with cis-trans isomerism and one chiral carbon, and determine the total number of stereoisomers is 4.

According to VSEPR theory:
1. \(\text{NF}_3\): Nitrogen has 5 valence electrons, forms 3 bonding pairs and has 1 lone pair. This gives a trigonal pyramidal shape with a bond angle of approximately \(102^\circ\) (less than the tetrahedral angle of \(109.5^\circ\) due to lone pair-bonding pair repulsion).
2. \(\text{CF}_4\): Carbon has 4 valence electrons, forms 4 bonding pairs and has 0 lone pairs. This gives a regular tetrahedral shape with a bond angle of \(109.5^\circ\).
3. \(\text{BF}_3\): Boron has 3 valence electrons, forms 3 bonding pairs and has 0 lone pairs. This gives a trigonal planar shape with a bond angle of \(120^\circ\).

Thus, the correct order of increasing bond angle is \(\text{NF}_3 < \text{CF}_4 < \text{BF}_3\).

1 mark: Correctly identify the shape and bond angle of each species and list them in order of increasing bond angle.

Alkaline hydrolysis of an ester using aqueous sodium hydroxide (NaOH) yields a sodium carboxylate salt and an alcohol:
\(\text{R-COOR'} + \text{NaOH} \rightarrow \text{R-COO}^-\text{Na}^+ + \text{R'-OH}\)

From the formula \(\text{CH}_3\text{CH}_2\text{COOCH(CH}_3)_2\):
- The acid part (acyl group) is \(\text{CH}_3\text{CH}_2\text{CO-}\), which yields the sodium salt: sodium propanoate, \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\).
- The alcohol part (alkoxy group) is \(-\text{CH(CH}_3)_2\), which yields the branched alcohol: propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\).

Therefore, the products are \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\) and \(\text{CH}_3\text{CH(OH)CH}_3\).

1 mark: Correctly identify that alkaline hydrolysis produces the salt (sodium propanoate) and a secondary alcohol (propan-2-ol).

Let's examine consecutive Period 3 elements and their properties:
- Sodium (Na), Magnesium (Mg), Aluminum (Al), Silicon (Si).
- Element P must form an ionic oxide and have high electrical conductivity. Both Na and Mg fit this, but Q must be consecutive to P and have the highest conductivity in the period.
- Aluminum (Al) has the highest electrical conductivity in Period 3 due to having 3 valence electrons per atom that are free to delocalise in its metallic structure. Therefore, Q must be Aluminum (Al).
- Since P, Q, and R are consecutive, and Q is Al (atomic number 13), P must be Magnesium (Mg, atomic number 12) and R must be Silicon (Si, atomic number 14).
- This perfectly fits: Mg (P) has a giant ionic oxide (MgO) and high electrical conductivity; Al (Q) has the highest electrical conductivity; Si (R) is a metalloid semiconductor with a giant covalent structure.

1 mark: Identify P = Mg, Q = Al, R = Si based on electrical conductivity and bonding/structure trends in Period 3 elements.

Write down the half-equations for the redox processes:
1. Reduction of dichromate(VI):
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Each dichromate(VI) ion gains 6 electrons.

2. Oxidation of tin(II):
\(\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^-\)
Each tin(II) ion loses 2 electrons.

To balance the electron transfer, multiply the tin oxidation half-equation by 3:
\(3\text{Sn}^{2+} \rightarrow 3\text{Sn}^{4+} + 6\text{e}^-\)

Combining the equations gives:
\(\text{Cr}_2\text{O}_7^{2-} + 3\text{Sn}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{Sn}^{4+} + 7\text{H}_2\text{O}\)

The ratio of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Sn}^{2+}\) is therefore \(1 : 3\).

1 mark: Balance the reduction and oxidation half-equations to show that 1 mole of dichromate(VI) reacts with 3 moles of tin(II) ions.

1. Calculate the moles of \(\text{CO}_2\) produced: \(\text{Moles} = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol}\). 2. From the stoichiometry of the reaction, \(\text{MCO}_3 + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2\text{O} + \text{CO}_2\), the moles of pure \(\text{MCO}_3\) is also \(0.010\text{ mol}\). 3. Calculate the mass of pure \(\text{MCO}_3\): \(\text{Mass} = 0.010\text{ mol} \times 100.0\text{ g mol}^{-1} = 1.00\text{ g}\). 4. Calculate the percentage purity: \(\text{Purity} = \frac{1.00\text{ g}}{1.25\text{ g}} \times 100\% = 80.0\%\).

1 mark for the correct calculation of percentage purity leading to 80.0% (Option C).

When solid potassium bromide, \(\text{KBr}\), is reacted with concentrated sulfuric acid, hydrogen bromide gas is initially formed, which fumes in moist air. Because bromide ions are strong enough reducing agents, they reduce the concentrated sulfuric acid to sulfur dioxide, \(\text{SO}_2\) (a gas that turns acidified potassium dichromate(VI) paper green), while being oxidized to bromine, \(\text{Br}_2\) (a dark brown liquid).

1 mark for identifying the halide as bromide (Option C) based on its characteristic reduction of concentrated sulfuric acid to sulfur dioxide and oxidation to bromine.

The combustion of propan-1-ol is represented by: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(\text{l}) + 4.5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\). Using Hess's law: \(\Delta H^\theta_\text{c} = [3 \times \Delta H^\theta_\text{f}(\text{CO}_2) + 4 \times \Delta H^\theta_\text{f}(\text{H}_2\text{O})] - \Delta H^\theta_\text{f}(\text{propan-1-ol}) = [3(-394) + 4(-286)] - [-303] = [-1182 - 1144] + 303 = -2023\text{ kJ mol}^{-1}\).

1 mark for the correct calculation of standard enthalpy of combustion to give -2023 kJ/mol (Option B).

The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the exothermic (forward) direction, thereby increasing the concentration of products and decreasing the concentration of reactants. This increases both the yield of \(\text{SO}_3\) and the value of the equilibrium constant, \(K_c\). Changes in pressure or the use of a catalyst do not alter the value of \(K_c\).

1 mark for identifying that decreasing the temperature increases both the yield and the value of Kc for an exothermic reaction (Option A).

There is a very large jump between the 4th and 5th ionisation energies (from \(4360\) to \(16100\text{ kJ mol}^{-1}\)), which indicates the 5th electron is removed from an inner quantum shell. Thus, \(\text{Y}\) has 4 valence electrons and belongs to Group 14. In Period 3, this is silicon. Its highest oxidation state is +4, forming the oxide \(\text{YO}_2\).

1 mark for identifying that Y belongs to Group 14 and thus its oxide in the highest oxidation state has the formula YO2 (Option C).

There are exactly four structural isomers with the formula \(\text{C}_3\text{H}_6\text{Cl}_2\): 1,1-dichloropropane (\(\text{CHCl}_2\text{CH}_2\text{CH}_3\)), 1,2-dichloropropane (\(\text{CH}_2\text{ClCHClCH}_3\)), 1,3-dichloropropane (\(\text{CH}_2\text{ClCH}_2\text{CH}_2\text{Cl}\)), and 2,2-dichloropropane (\(\text{CH}_3\text{CCl}_2\text{CH}_3\)).

1 mark for identifying that there are exactly 4 structural isomers (Option B).

Rearranging the ideal gas equation \(pV = nRT = \frac{m}{M} RT\) gives the molar mass \(M = \frac{\rho RT}{p}\), where density \(\rho = 2.40\text{ g dm}^{-3} = 2.40\text{ kg m}^{-3}\). Substituting the values: \(M = \frac{2.40 \times 8.31 \times 300}{1.50 \times 10^5} = 0.0399\text{ kg mol}^{-1} = 39.9\text{ g mol}^{-1}\). This molar mass corresponds to argon (atomic mass 39.9).

1 mark for calculating the molar mass of the gas (approx. 39.9 g/mol) and identifying it as argon (Option C).

Alkaline hydrolysis of an ester \(\text{RCOOR'}\) yields \(\text{RCOONa}\) and \(\text{R'OH}\). Since the product is sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)), the acid portion of the ester has 3 carbons (\(\text{CH}_3\text{CH}_2\text{CO}-\)). Since the ester has 5 carbons in total, the alcohol portion must have 2 carbons (\(-\text{CH}_2\text{CH}_3\)). Thus, the starting ester is ethyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\).

1 mark for identifying the correct ester structure (ethyl propanoate, Option B) based on the molecular formula and the hydrolysed product.

The reaction of the hydrocarbon \(\text{C}_x\text{H}_y\) with oxygen is:
\(\text{C}_x\text{H}_y(\text{g}) + (x + \frac{y}{4})\text{O}_2(\text{g}) \to x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})\)

1. The volume of the residual gas after cooling is \(50\text{ cm}^3\). This gas consists of unreacted \(\text{O}_2\) and the produced \(\text{CO}_2\).
2. Shaking with excess aqueous sodium hydroxide absorbs \(\text{CO}_2\). The remaining gas is unreacted \(\text{O}_2\), which is \(20\text{ cm}^3\).
3. Therefore, the volume of \(\text{CO}_2\) produced is:
\(50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3\).
4. The volume of \(\text{O}_2\) reacted is:
\(70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3\).

Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) reacted:
- For Carbon: \(x = \frac{30\text{ cm}^3}{10\text{ cm}^3} = 3\).
- For Oxygen: \(x + \frac{y}{4} = \frac{50\text{ cm}^3}{10\text{ cm}^3} = 5\).
Substituting \(x = 3\):
\(3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8\).

Thus, the molecular formula of the hydrocarbon is \(\text{C}_3\text{H}_8\).

[1 mark] Deduce the volume of carbon dioxide produced (30 cm³) and oxygen reacted (50 cm³). Apply Gay-Lussac's law of combining volumes to find x = 3 and y = 8, leading to the correct answer D.

The equation for the standard enthalpy change of formation of butane is:
\(4\text{C}(\text{s}) + 5\text{H}_2(\text{g}) \to \text{C}_4\text{H}_{10}(\text{g})\)

Using Hess's law and standard enthalpy changes of combustion:
\(\Delta H_f^\theta = \sum \Delta H_c^\theta[\text{reactants}] - \sum \Delta H_c^\theta[\text{products}]\)
\(\Delta H_f^\theta = 4 \times \Delta H_c^\theta[\text{C}(\text{s})] + 5 \times \Delta H_c^\theta[\text{H}_2(\text{g})] - \Delta H_c^\theta[\text{C}_4\text{H}_{10}(\text{g})]\)

Substitute the given values:
\(\Delta H_f^\theta = 4(-393.5) + 5(-285.8) - (-2877.5)\)
\(\Delta H_f^\theta = -1574.0 - 1429.0 + 2877.5\)
\(\Delta H_f^\theta = -3003.0 + 2877.5 = -125.5\text{ kJ mol}^{-1}\)

[1 mark] Construct a correct Hess's Law cycle or expression using combustion values: 4(Combustion of C) + 5(Combustion of H2) - (Combustion of butane). Calculate the correct value of -125.5 kJ mol⁻¹ to select answer A.

When concentrated sulfuric acid is added to solid sodium iodide, \(\text{NaI}\), several reactions occur because iodide ions are powerful reducing agents:
1. Acid-base reaction: \(\text{NaI} + \text{H}_2\text{SO}_4 \to \text{NaHSO}_4 + \text{HI}\)
2. Reduction of \(\text{H}_2\text{SO}_4\) by \(\text{HI}\):
- Iodide is oxidised to iodine, \(\text{I}_2\), which is observed as a purple vapour.
- Sulfuric acid is reduced past \(\text{SO}_2\) and \(\text{S}\) all the way to hydrogen sulfide, \(\text{H}_2\text{S}\).
- \(\text{H}_2\text{S}\) is a highly toxic gas with a characteristic rotten-egg smell that forms black lead(II) sulfide on moist lead acetate paper.

Sodium bromide is only strong enough to reduce sulfuric acid to sulfur dioxide, \(\text{SO}_2\) (no \(\text{H}_2\text{S}\) is formed), and produces brown bromine vapour. Sodium chloride and sodium fluoride are not strong enough reducing agents to reduce sulfuric acid.

[1 mark] Correctly identify that only iodide (I⁻) has a strong enough reducing power to reduce sulfuric acid to hydrogen sulfide (rotten-egg smell, turns lead acetate paper black) and produce iodine (purple vapour). This points uniquely to sodium iodide, D.

Let's determine the equilibrium moles using an ICE table:
- Initial: \(n(\text{N}_2) = 1.00\text{ mol}\), \(n(\text{H}_2) = 3.00\text{ mol}\), \(n(\text{NH}_3) = 0\text{ mol}\).
- Change: \(n(\text{NH}_3) = +0.40\text{ mol}\). From stoichiometry, \(n(\text{N}_2)\) decreases by \(0.20\text{ mol}\) and \(n(\text{H}_2)\) decreases by \(0.60\text{ mol}\).
- Equilibrium moles:
- \(n(\text{N}_2) = 1.00 - 0.20 = 0.80\text{ mol}\)
- \(n(\text{H}_2) = 3.00 - 0.60 = 2.40\text{ mol}\)
- \(n(\text{NH}_3) = 0.40\text{ mol}\)

Since the volume \(V = 2.0\text{ dm}^3\), we convert moles to concentrations:
- \([\text{N}_2] = \frac{0.80\text{ mol}}{2.0\text{ dm}^3} = 0.40\text{ mol dm}^{-3}\)
- \([\text{H}_2] = \frac{2.40\text{ mol}}{2.0\text{ dm}^3} = 1.20\text{ mol dm}^{-3}\)
- \([\text{NH}_3] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}\)

Substitute concentrations into the expression for \(K_c\):
\(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.20)^2}{(0.40)(1.20)^3} = \frac{0.040}{0.40 \times 1.728} = \frac{0.040}{0.6912} \approx 0.058\text{ dm}^6\text{ mol}^{-2}\)

[1 mark] Determine the equilibrium concentrations of reactants and products by dividing the equilibrium moles by the volume (2.0 dm³). Calculate the value of Kc correctly to obtain 0.058 dm⁶ mol⁻² (B).

Look at the differences between successive ionisation energies of element \(Y\):
- 1st to 2nd: \(1903 - 1012 = 891\)
- 2nd to 3rd: \(2912 - 1903 = 1009\)
- 3rd to 4th: \(4957 - 2912 = 2045\)
- 4th to 5th: \(6274 - 4957 = 1317\)
- 5th to 6th: \(21269 - 6274 = 14995\) (A huge increase!)

The huge jump between the 5th and 6th ionisation energies indicates that the 6th electron is being removed from a lower, inner shell closer to the nucleus. Therefore, \(Y\) has 5 outer-shell (valence) electrons and belongs to Group 15 of the Periodic Table.

The Period 3 element in Group 15 is phosphorus (\(\text{P}\)). Phosphorus forms a stable chloride, phosphorus trichloride (\(\text{PCl}_3\)), representing the formula \(Y\text{Cl}_3\).

[1 mark] Identify that the massive jump in successive ionisation energies is between the 5th and 6th ionisation, indicating five valence electrons. Determine that the Group 15 element in Period 3 forms a stable chloride of the form YCl₃ (C).

An ester has the general structural formula \(\text{R}-\text{COO}-\text{R}'\), where \(\text{R}'\) must be an alkyl group (at least 1 carbon, e.g., \(-\text{CH}_3\)).

Let's systematically construct the esters with 4 carbons total:
1. Acid with 1 Carbon (methanoic acid derivative, \(\text{H}-\text{COO}-\text{R}'\)):
The alcohol part must contain 3 carbons (\(-\text{C}_3\text{H}_7\)). There are two possible alkyl groups:
- Propyl group: \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\) (propyl methanoate)
- Isopropyl group: \(\text{HCOOCH}(\text{CH}_3)_2\) (isopropyl methanoate)

2. Acid with 2 Carbons (ethanoic acid derivative, \(\text{CH}_3-\text{COO}-\text{R}'\)):
The alcohol part must contain 2 carbons (\(-\text{CH}_2\text{CH}_3\)).
- Ethyl group: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (ethyl ethanoate)

3. Acid with 3 Carbons (propanoic acid derivative, \(\text{CH}_3\text{CH}_2-\text{COO}-\text{R}'\)):
The alcohol part must contain 1 carbon (\(-\text{CH}_3\)).
- Methyl group: \(\text{CH}_3\text{CH}_2\text{COOCH}_3\) (methyl propanoate)

There are no other combinations since the alcohol portion must have at least 1 carbon. Thus, there are exactly 4 structural isomers that are esters with this molecular formula.

[1 mark] List all structural isomers of esters with the molecular formula C₄H₈O₂ systematically. Show that there are four unique esters, concluding that C is the correct answer.

We use the ideal gas equation: \(pV = nRT\)

First, express all units in SI units:
- Temperature, \(T = 97.0 + 273.15 = 370.15\text{ K}\) (using \(370\text{ K}\) is also acceptable for calculations)
- Pressure, \(p = 1.01 \times 10^5\text{ Pa}\)
- Volume, \(V = 41.5\text{ cm}^3 = 41.5 \times 10^{-6}\text{ m}^3\)
- Mass, \(m = 0.130\text{ g}\)

We know that the number of moles \(n = \frac{m}{M_{\text{r}}}\). Substituting this into the ideal gas equation:
\(pV = \frac{m}{M_{\text{r}}} RT \implies M_{\text{r}} = \frac{mRT}{pV}\)

Now substitute the values:
\(M_{\text{r}} = \frac{0.130 \times 8.31 \times 370.15}{1.01 \times 10^5 \times 41.5 \times 10^{-6}}\)
\(M_{\text{r}} = \frac{399.87}{4.1915} \approx 95.4\)

[1 mark] Convert all variables into correct SI units (T to Kelvin, V to m³). Use the rearranged ideal gas equation to solve for the molar mass, yielding approximately 95.4 (C).

The ester given is isopropyl propanoate (or 1-methylethyl propanoate). Under alkaline hydrolysis conditions (heating with aqueous \(\text{NaOH}\) under reflux):
- The ester bond \(\text{C}-\text{O}\) single bond is cleaved.
- The carboxylic acid portion, \(\text{CH}_3\text{CH}_2\text{COO}-\), is converted into its salt form, which is sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)).
- The alcohol portion, \(-\text{CH}(\text{CH}_3)_2\), is hydrolysed to propan-2-ol (\((\text{CH}_3)_2\text{CHOH}\)).

[1 mark] Identify the cleavage of the ester bond in isopropyl propanoate. Recognize that alkaline hydrolysis forms the carboxylate salt (sodium propanoate) and the secondary alcohol (propan-2-ol), pointing to option D.

First, determine the mass of oxygen in the oxide by subtracting the mass of iron from the total mass of the sample:
\(\text{Mass of O} = 1.20\text{ g} - 0.84\text{ g} = 0.36\text{ g}\).

Next, calculate the number of moles of iron and oxygen atoms:
\(\text{Moles of Fe} = \frac{0.84\text{ g}}{55.8\text{ g mol}^{-1}} = 0.0151\text{ mol}\)
\(\text{Moles of O} = \frac{0.36\text{ g}}{16.0\text{ g mol}^{-1}} = 0.0225\text{ mol}\).

Determine the simplest whole-number ratio of \(\text{Fe}\) to \(\text{O}\) by dividing both by the smaller value:
\(\text{Ratio of Fe} = \frac{0.0151}{0.0151} = 1\)
\(\text{Ratio of O} = \frac{0.0225}{0.0151} = 1.49 \approx 1.5\).

Multiplying by 2 to obtain whole numbers gives a ratio of \(2:3\). Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

Award 1 mark for the correct answer C. Award 0 marks for incorrect options. Method: calculates moles of Fe and O correctly, deduces the integer ratio, and selects the corresponding formula.

When concentrated sulfuric acid is added to solid sodium bromide (\(\text{NaBr}\)), \(\text{HBr}\) gas is formed (an acidic gas). Because bromide ions are strong enough reducing agents, they reduce some of the concentrated sulfuric acid to sulfur dioxide (\(\text{SO}_2\)), a colourless gas which reduces orange acidified dichromate(VI) ions to green chromium(III) ions. The bromide ions are oxidised to bromine (\(\text{Br}_2\)), which is seen as a brown vapour. No yellow solid (sulfur) or hydrogen sulfide (\(\text{H}_2\text{S}\), rotten-egg smell) is produced because bromide is not a strong enough reducing agent to reduce sulfuric acid beyond sulfur dioxide, unlike iodide.

Award 1 mark for the correct answer C. Award 0 marks for other options. Deduces that the bromide ion reduces sulfuric acid only to sulfur dioxide without forming deeper reduction products.

The equation for the standard enthalpy change of formation of ethanol is:
\(2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\).

According to Hess's Law:
\(\Delta H^\ominus_f = 2 \times \Delta H^\ominus_c[\text{C}] + 3 \times \Delta H^\ominus_c[\text{H}_2] - \Delta H^\ominus_c[\text{C}_2\text{H}_5\text{OH}]\)
\(\Delta H^\ominus_f = 2(-393.5) + 3(-285.8) - (-1367.3)\)
\(\Delta H^\ominus_f = -787.0 - 857.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer A. Award 0 marks for incorrect calculations.

First, construct an ICE table for the moles of each species at equilibrium:
Reaction: \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\)
Initial moles: \(\text{N}_2 = 1.00\), \(\text{H}_2 = 3.00\), \(\text{NH}_3 = 0\)
Change in moles: \(\text{NH}_3 = +0.40\), \(\text{N}_2 = -0.20\), \(\text{H}_2 = -0.60\)
Equilibrium moles: \(\text{N}_2 = 0.80\), \(\text{H}_2 = 2.40\), \(\text{NH}_3 = 0.40\)

Now, divide equilibrium moles by the volume (\(2.00\text{ dm}^3\)) to find concentrations:
\([\text{N}_2]_{eq} = 0.40\text{ mol dm}^{-3}\)
\([\text{H}_2]_{eq} = 1.20\text{ mol dm}^{-3}\)
\([\text{NH}_3]_{eq} = 0.20\text{ mol dm}^{-3}\)

Calculate \(K_c\):
\(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.20)^2}{(0.40)(1.20)^3} = \frac{0.04}{0.40 \times 1.728} = 0.0579\text{ dm}^6\text{ mol}^{-2}\).

Award 1 mark for the correct answer B. Award 0 marks for using moles instead of concentrations or incorrect stoichiometry.

There is a very large jump in ionisation energy between the 6th and 7th ionisation energies (from \(8500\) to \(27100\text{ kJ mol}^{-1}\)). This indicates that the 7th electron is removed from an inner quantum shell, meaning there are 6 valence electrons in the outermost shell of element \(Z\). Element \(Z\) is in Group 16 (VI) of Period 3, which is sulfur. The stable hydride of Group 16 elements has the general formula \(\text{H}_2\text{Z}\) (e.g., \(\text{H}_2\text{S}\)).

Award 1 mark for the correct answer B. Award 0 marks for incorrect options based on wrong group identification.

An ester has the general formula \(\text{R}_1\text{COOR}_2\). The total number of carbon atoms must be 4, so \(\text{R}_1\) and \(\text{R}_2\) must contain a combined total of 3 carbon atoms (excluding the carbonyl carbon).

We can systematically list the possibilities:
1. If \(\text{R}_1 = \text{H}\) (methanoate ester): \(\text{R}_2\) must be a propyl group. This can be propyl methanoate, \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\), or isopropyl methanoate, \(\text{HCOOCH(CH}_3)_2\). (2 isomers)
2. If \(\text{R}_1 = \text{CH}_3\) (ethanoate ester): \(\text{R}_2\) must be an ethyl group. This is ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\). (1 isomer)
3. If \(\text{R}_1 = \text{CH}_3\text{CH}_2\) (propanoate ester): \(\text{R}_2\) must be a methyl group. This is methyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_3\). (1 isomer)

Summing these up gives \(2 + 1 + 1 = 4\) structural isomers.

Award 1 mark for the correct answer C. Award 0 marks for other values.

An intensity ratio of \(3:1\) for the \(M\) and \(M+2\) peaks is characteristic of a compound containing a single chlorine atom, due to the natural abundance of \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\). This eliminates bromomethane and bromoethane (which would show a \(1:1\) ratio). The major fragment peak at \(m/z = 29\) corresponds to the ethyl cation, \([\text{CH}_3\text{CH}_2]^+\). Chloromethane would yield a methyl cation fragment at \(m/z = 15\). Therefore, the compound is chloroethane.

Award 1 mark for the correct answer B. Award 0 marks for incorrect compound assignments.

Alkaline hydrolysis followed by acidification of an ester yields a carboxylic acid and an alcohol. Since the alcohol can be oxidised to a ketone, it must be a secondary alcohol.
- Option A (propyl ethanoate) yields propan-1-ol, a primary alcohol (oxidises to an aldehyde/carboxylic acid).
- Option B (ethyl propanoate) yields ethanol, a primary alcohol (oxidises to an aldehyde/carboxylic acid).
- Option C (methyl butanoate) yields methanol, a primary-type alcohol (oxidises to methanal/methanoic acid).
- Option D (isopropyl ethanoate) yields propan-2-ol, a secondary alcohol. Propan-2-ol is oxidised by acidified potassium dichromate(VI) to propanone (a ketone). Thus, the original ester is isopropyl ethanoate.

Award 1 mark for the correct answer D. Award 0 marks for options yielding primary alcohols.

Upon heating, only the sodium hydrogencarbonate undergoes thermal decomposition according to the equation:
\(2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g})\)

The mass loss is due to the escape of gaseous \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

Let the mass of \(\text{NaHCO}_3\) in the original mixture be \(x\) g.
From the stoichiometry:
\(2 \text{ mol}\) of \(\text{NaHCO}_3\) (\(2 \times 84.0 = 168.0\text{ g}\)) produces \(1 \text{ mol}\) of \(\text{CO}_2\) (\(44.0\text{ g}\)) and \(1 \text{ mol}\) of \(\text{H}_2\text{O}\) (\(18.0\text{ g}\)), representing a total mass loss of \(44.0 + 18.0 = 62.0\text{ g}\).

Therefore, the mass loss from \(x\) g of \(\text{NaHCO}_3\) is:
\(\text{Mass loss} = x \times \frac{62.0}{168.0}\)

The actual mass loss is:
\(5.00\text{ g} - 4.07\text{ g} = 0.93\text{ g}\)

Setting these equal:
\(x \times \frac{62.0}{168.0} = 0.93\)
\(x = 0.93 \times \frac{168.0}{62.0} = 2.52\text{ g}\)

The percentage by mass of \(\text{NaHCO}_3\) is:
\(\% \text{ mass} = \frac{2.52\text{ g}}{5.00\text{ g}} \times 100\% = 50.4\%\)

1 mark for the correct option C.
- Method: Identify the decomposition equation and relate the mass loss of 0.93 g to the stoichiometry of \(\text{NaHCO}_3\).
- Accuracy: Correctly calculate the mass of \(\text{NaHCO}_3\) as 2.52 g, yielding 50.4%.

The halide ions act as reducing agents towards concentrated sulfuric acid, with their reducing ability increasing down Group 17 (\(\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-\)).

- \(\text{NaCl}\) and \(\text{NaF}\) are not strong enough reducing agents to reduce sulfuric acid; they only undergo acid-base reactions to form \(\text{HCl}\) and \(\text{HF}\).
- \(\text{NaBr}\) reduces sulfuric acid to sulfur dioxide, \(\text{SO}_2\) (a colorless gas).
- \(\text{NaI}\) is a very strong reducing agent and reduces sulfur in sulfuric acid (oxidation state +6) to several species, including sulfur dioxide (\(\text{SO}_2\)), elemental sulfur (\(\text{S}\), which is a yellow solid), and hydrogen sulfide (\(\text{H}_2\text{S}\), a colorless gas with a rotten-egg odor).

1 mark for the correct option C.
- Reject A and D: chloride and fluoride do not reduce concentrated sulfuric acid.
- Reject B: bromide only reduces sulfuric acid to \(\text{SO}_2\).

Using Hess's Law and a cycle based on enthalpy changes of combustion:

\(\Delta H^\ominus_r = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})\)

\(\Delta H^\ominus_r = [\Delta H^\ominus_c(\text{C}_2\text{H}_4(\text{g})) + \Delta H^\ominus_c(\text{H}_2(\text{g}))] - \Delta H^\ominus_c(\text{C}_2\text{H}_6(\text{g}))\)

\(\Delta H^\ominus_r = [(-1411) + (-286)] - (-1560)\)
\(\Delta H^\ominus_r = -1697 + 1560 = -137\text{ kJ mol}^{-1}\)

1 mark for the correct option A.
- Method: Apply Hess's Law correctly using \(\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})\).
- Accuracy: Arrive at \(-137\text{ kJ mol}^{-1}\).

Construct an ICE (Initial, Change, Equilibrium) table for moles:

- **Initial moles**:
\(\text{SO}_2 = 2.00\text{ mol}\), \(\text{O}_2 = 1.00\text{ mol}\), \(\text{SO}_3 = 0\text{ mol}\)
- **Change in moles**:
\(\text{SO}_3\) increased by \(+1.20\text{ mol}\).
By stoichiometry, \(\text{SO}_2\) decreased by \(1.20\text{ mol}\).
\(\text{O}_2\) decreased by \(1.20 / 2 = 0.60\text{ mol}\).
- **Equilibrium moles**:
\(\text{SO}_2 = 2.00 - 1.20 = 0.80\text{ mol}\)
\(\text{O}_2 = 1.00 - 0.60 = 0.40\text{ mol}\)
\(\text{SO}_3 = 1.20\text{ mol}\)

- **Equilibrium concentrations** (divide by \(V = 2.00\text{ dm}^3\)):
\([\text{SO}_2] = 0.80 / 2.00 = 0.40\text{ mol dm}^{-3}\)
\([\text{O}_2] = 0.40 / 2.00 = 0.20\text{ mol dm}^{-3}\)
\([\text{SO}_3] = 1.20 / 2.00 = 0.60\text{ mol dm}^{-3}\)

- **Equilibrium constant \(K_c\)**:
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.60)^2}{(0.40)^2 \times 0.20} = \frac{0.36}{0.16 \times 0.20} = \frac{0.36}{0.032} = 11.25 \approx 11.3\text{ dm}^3\text{ mol}^{-1}\)

1 mark for the correct option B.
- Method: Correctly find the equilibrium moles of each species, convert them to concentrations using \(V = 2.00\text{ dm}^3\), and apply the correct \(K_c\) expression.
- Reject A: This is the value obtained if the volume division is omitted.

Looking at the successive ionisation energies, we identify where the largest jump occurs. This indicates the removal of an electron from an inner shell closer to the nucleus.
- 1st to 6th ionisation energies increase steadily from 1000 to 8500 \(\text{kJ mol}^{-1}\).
- The 7th ionisation energy (27100 \(\text{kJ mol}^{-1}\)) is more than three times the 6th ionisation energy (8500 \(\text{kJ mol}^{-1}\)).
- This large jump indicates that the 7th electron is removed from an inner core shell, meaning there are 6 electrons in the outermost shell.
- Therefore, the element is in Group 16.

1 mark for the correct option D.
- Method: Spot the major jump between the 6th and 7th ionisation energy values.
- Deduction: Determine 6 valence electrons, hence Group 16.

We must draw all possible structural isomers of butyl bromide, which has a skeleton of 4 carbons and 1 bromine atom:
1. **Based on a butane (straight-chain) skeleton**:
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\) (1-bromobutane)
- \(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\) (2-bromobutane)

2. **Based on a methylpropane (branched-chain) skeleton**:
- \((\text{CH}_3)_2\text{CHCH}_2\text{Br}\) (1-bromo-2-methylpropane)
- \((\text{CH}_3)_3\text{CBr}\) (2-bromo-2-methylpropane)

There are exactly 4 structural isomers. Note that while 2-bromobutane exhibits optical isomerism (stereoisomerism), the question specifically asks for *structural* isomers.

1 mark for the correct option B.
- Accuracy: Correctly identify the 4 structural isomers.
- Reject C (5): 5 includes the two enantiomers of 2-bromobutane, but these are stereoisomers, not structural isomers.

Let the probability of finding a \(^{79}\text{Br}\) atom be \(p = 0.5\) and a \(^{81}\text{Br}\) atom be \(q = 0.5\).

For a molecule with two bromine atoms, the distribution of isotopes can be calculated using the expansion \((p + q)^2 = p^2 + 2pq + q^2\):

- **\(M\) peak** (both atoms are \(^{79}\text{Br}\)):
\(p^2 = (0.5)^2 = 0.25\)
- **\(M+2\) peak** (one \(^{79}\text{Br}\) and one \(^{81}\text{Br}\)):
\(2pq = 2(0.5)(0.5) = 0.50\)
- **\(M+4\) peak** (both atoms are \(^{81}\text{Br}\)):
\(q^2 = (0.5)^2 = 0.25\)

Simplifying the ratio \(0.25 : 0.50 : 0.25\) gives \(1 : 2 : 1\).

1 mark for the correct option C.
- Method: Apply the binomial probability expansion to the 1:1 ratio of bromine isotopes.
- Reject D: This is the ratio for two chlorine atoms (where \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\) are in a 3:1 ratio).

1. The ester has the formula \(\text{C}_5\text{H}_{10}\text{O}_2\).
2. Hydrolysis of the ester produces carboxylic acid \(P\) and alcohol \(Q\).
3. Alcohol \(Q\) is oxidized to a ketone by acidified potassium dichromate(VI), which means \(Q\) must be a secondary alcohol.
4. The smallest possible secondary alcohol is propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\) (3 carbon atoms).
5. If \(Q\) is propan-2-ol, the carboxylic acid \(P\) must have \(5 - 3 = 2\) carbon atoms, which is ethanoic acid, \(\text{CH}_3\text{COOH}\).
6. The ester formed from ethanoic acid and propan-2-ol is propan-2-yl ethanoate, which has the structural formula \(\text{CH}_3\text{COOCH(CH}_3)_2\).

1 mark for the correct option B.
- Method: Deduce that only a secondary alcohol can oxidize to a ketone, identifying \(Q\) as propan-2-ol.
- Accuracy: Link this to the correct ester formula of \(\text{CH}_3\text{COOCH(CH}_3)_2\).

(a)(i)
- Sodium has a giant metallic structure with metallic bonding (strong electrostatic attraction between \(\text{Na}^+\)
cations and delocalised electrons).
- Silicon has a giant covalent macromolecular structure with strong covalent bonds between Si atoms.
- Sulfur has a simple molecular structure with weak intermolecular (London/dispersion) forces between \(\text{S}_8\) molecules.
- Consequently, silicon has the highest melting point because breaking many strong covalent bonds requires a large amount of energy, whereas sulfur has a relatively low melting point because only weak intermolecular forces need to be overcome.

(a)(ii)
- Sulfur exists as \(\text{S}_8\) molecules while phosphorus exists as \(\text{P}_4\) molecules.
- \(\text{S}_8\) has a larger molecular size and more electrons than \(\text{P}_4\), which leads to stronger London dispersion (instantaneous dipole-induced dipole) forces between sulfur molecules, requiring more energy to overcome.

(b)(i)
- Equation: \(\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}\)
- pH: 13 (accept any value in the range 12–14).

(b)(ii)
- Equation: \(\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\) (or \(\text{P}_2\text{O}_5\text{(s)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{H}_3\text{PO}_4\text{(aq)}\))
- pH: 2 (accept any value in the range 0–3).

(b)(iii)
- Ionic Equation: \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2[\text{Al(OH)}_4]^-\text{(aq)}\) or \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} \rightarrow 2\text{AlO}_2^-\text{(aq)} + \text{H}_2\text{O(l)}\)

(c)(i)
- Observation: Steamy / misty white fumes (of \(\text{HCl}\)) OR the mixture gets hot / reacts vigorously.
- Equation: \(\text{PCl}_5\text{(s)} + 4\text{H}_2\text{O(l)} \rightarrow \text{H}_3\text{PO}_4\text{(aq)} + 5\text{HCl(aq)}\) (accept \(\text{PCl}_5\text{(s)} + \text{H}_2\text{O(l)} \rightarrow \text{POCl}_3\text{(l)} + 2\text{HCl(aq)}\))

(c)(ii)
- \(\text{NaCl}\) has a giant ionic structure with strong electrostatic attractions between oppositely charged ions, whereas \(\text{SiCl}_4\) has a simple molecular structure with weak intermolecular (London/dispersion) forces.

(a)(i) [3 marks]
- M1: Sodium has giant metallic structure / metallic bonding AND Silicon has giant covalent macromolecular structure / covalent bonding. [1]
- M2: Sulfur has simple molecular structure / weak intermolecular forces (or London dispersion / van der Waals forces). [1]
- M3: Silicon melting point is very high because strong covalent bonds must be broken, whereas sulfur is lower as only weak intermolecular forces are broken (or similar comparative explanation of energy needed). [1]

(a)(ii) [2 marks]
- M1: Identifies sulfur as \(\text{S}_8\) and phosphorus as \(\text{P}_4\). [1]
- M2: Explains that \(\text{S}_8\) has more electrons / is a larger molecule, leading to stronger London dispersion (or van der Waals) forces. [1]

(b)(i) [2 marks]
- M1: Correct balanced equation: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\) [1]
- M2: pH value between 12 and 14. [1]

(b)(ii) [2 marks]
- M1: Correct balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) (or \(\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4\)) [1]
- M2: pH value between 0 and 3. [1]

(b)(iii) [2 marks]
- M1: Correct formulas of reactants (\(\text{Al}_2\text{O}_3 + \text{OH}^- + \text{H}_2\text{O}\)) and products (\([\text{Al(OH)}_4]^-\)) OR (\(\text{Al}_2\text{O}_3 + \text{OH}^-\)) and products (\(\text{AlO}_2^- + \text{H}_2\text{O}\)). [1]
- M2: Correctly balanced equation. [1]

(c)(i) [2 marks]
- M1: Observation: Steamy / misty white fumes (or solid dissolves / vigorous reaction). [1] (Reject: white smoke)
- M2: Equation: \(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\) (or partial hydrolysis equation). [1]

(c)(ii) [1 mark]
- M1: \(\text{NaCl}\) has a giant ionic lattice with strong ionic bonds / electrostatic forces, while \(\text{SiCl}_4\) has a simple molecular structure with weak intermolecular forces. [1]

(a) In \(\text{SCl}_2\), sulfur (Group 16) is the central atom with 6 valence electrons, and chlorine (Group 17) has 7 valence electrons. S forms a single covalent bond with each Cl atom.
- S has 2 bonding pairs and 2 lone pairs (total 8 electrons in outer shell).
- Each Cl has 1 bonding pair and 3 lone pairs (total 8 electrons in outer shell).

(b) Shape: Bent / non-linear / V-shaped.
- Bond angle: \(103^\circ\) (accept any value from \(101^\circ\) to \(105^\circ\)).
- Explanation: There are 2 bonding pairs and 2 lone pairs around the central sulfur atom. These four pairs arrange tetrahedrally to minimize repulsion. Since lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which is greater than bond pair-bond pair repulsion, the bond angle is compressed below the standard tetrahedral angle of \(109.5^\circ\).

(c) (i) The balanced chemical equation is:
\(2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_2 + \text{S} + 4\text{HCl}\)

(ii) Sulfur is in Period 3 and has vacant, low-lying 3d orbitals in its valence shell. These 3d orbitals can accept a lone pair of electrons from the oxygen atom of water to initiate nucleophilic attack (hydrolysis). Carbon is in Period 2 and has no d-orbitals in its valence shell, preventing it from expanding its coordination shell.

(a)
- M1: Correct shared pairs of electrons (one pair of S-Cl electrons for each bond) [1]
- M2: Correct lone pairs shown (two lone pairs on S and three lone pairs on each Cl) [1]

(b)
- M1: Bent / non-linear / V-shaped [1]
- M2: Any angle in the range \(101^\circ\) to \(105^\circ\) [1]
- M3: Explanation stating that there are 2 lone pairs and 2 bonding pairs around S, and lone pairs repel more than bonding pairs [1]

(c) (i)
- M1: Correctly balanced equation: \(2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_2 + \text{S} + 4\text{HCl}\) [1]

(c) (ii)
- M1: Sulfur has vacant, low-lying 3d orbitals to accept a lone pair of electrons from water (or can expand its octet), whereas carbon does not have d-orbitals / cannot expand its octet. [1]

(a) Down Group 17, the atomic radius of the halogen increases, resulting in a longer H–X bond. This increases the bond length and decreases the orbital overlap, making the H–X bond weaker. Consequently, the H–X bond enthalpy decreases down the group, so less thermal energy is required to break the bond, meaning thermal stability decreases from \(\text{HF}\) to \(\text{HI}\).

(b) (i) For \(\text{HCl}\), no visible change is observed because it is thermally stable and does not decompose.
(ii) For \(\text{HI}\), purple/violet vapour (fumes) of iodine are produced as the compound decomposes.

(c) (i) The expression for the equilibrium constant is:
\[K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}\]
Since there are equal numbers of moles of gaseous reactants and products, the units cancel out, so there are no units.

(ii) Let us determine the equilibrium moles of each species:
- Initial: \(n(\text{HI}) = 0.800\text{ mol}\), \(n(\text{H}_2) = 0\text{ mol}\), \(n(\text{I}_2) = 0\text{ mol}\)
- Change: \(n(\text{HI}) = -2x\), \(n(\text{H}_2) = +x\), \(n(\text{I}_2) = +x\)
- At equilibrium, \(n(\text{I}_2) = x = 0.088\text{ mol}\)
- Thus, \(n(\text{H}_2) = 0.088\text{ mol}\)
- Equilibrium \(n(\text{HI}) = 0.800 - 2(0.088) = 0.624\text{ mol}\)

Now, calculate equilibrium concentrations in the \(2.00\text{ dm}^3\) flask:
- \([\text{HI}] = \frac{0.624\text{ mol}}{2.00\text{ dm}^3} = 0.312\text{ mol dm}^{-3}\)
- \([\text{H}_2] = \frac{0.088\text{ mol}}{2.00\text{ dm}^3} = 0.044\text{ mol dm}^{-3}\)
- \([\text{I}_2] = \frac{0.088\text{ mol}}{2.00\text{ dm}^3} = 0.044\text{ mol dm}^{-3}\)

Substitute these concentrations into the \(K_c\) expression:
\[K_c = \frac{(0.044)(0.044)}{(0.312)^2} = \frac{0.001936}{0.097344} \approx 0.0199\]

(a)
- M1: Thermal stability decreases down the group (from \(\text{HF}\) to \(\text{HI}\)). [1]
- M2: Halogen atomic radius increases, so the H–X bond length increases / bond becomes weaker. [1]
- M3: Thus, the H–X bond energy/enthalpy decreases (requires less energy to break). [1]

(b)
- M1 (for HCl): No change / no reaction. [1]
- M2 (for HI): Purple/violet fumes / vapour (or dark grey solid). [1]

(c)(i)
- M1: Correct expression: \(K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}\) [1]
- M2: State units as 'none' or 'no units'. [1]

(c)(ii)
- M1: Correct equilibrium moles of \(\text{H}_2\) (\(0.088\text{ mol}\)) AND \(\text{HI}\) (\(0.624\text{ mol}\)). [1]
- M2: Divide equilibrium moles by \(2.00\text{ dm}^3\) to obtain correct equilibrium concentrations: \([\text{HI}] = 0.312\text{ mol dm}^{-3}\) and \([\text{H}_2] = [\text{I}_2] = 0.044\text{ mol dm}^{-3}\). [1]
- M3: Correct substitution of concentrations (or moles, as volume cancels) into the \(K_c\) expression. [1]
- M4: Correct final calculated value of \(0.0199\) (accept \(0.02\) or \(0.020\) if correctly rounded). [1]

**(a)**
- Compound **W** is formed by the oxidation of alcohol **Z**. Since **W** does not react with Tollens' reagent, it cannot be an aldehyde, meaning it must be a ketone.
- Since **W** forms a yellow precipitate with alkaline aqueous iodine, it must contain a methyl ketone group (\(\text{CH}_3\text{CO}-\)). Therefore, the simplest possible ketone that fits this description is propanone, \(\text{CH}_3\text{COCH}_3\).
- Since **W** is propanone, the parent alcohol **Z** must be the secondary alcohol propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\).
- The ester **X** contains 5 carbon atoms. Since alcohol **Z** contains 3 carbon atoms, the carboxylic acid **Y** must contain \(5 - 3 = 2\) carbon atoms, which is ethanoic acid, \(\text{CH}_3\text{COOH}\).
- Recombining **Y** and **Z** gives the ester **X**, which is 1-methylethyl ethanoate (isopropyl ethanoate), \(\text{CH}_3\text{COOCH}(\text{CH}_3)_2\).

**(b)**
The hydrolysis is reversible and catalysed by dilute acid:
\(\text{CH}_3\text{COOCH}(\text{CH}_3)_2\text{(l)} + \text{H}_2\text{O}\text{(l)} \rightleftharpoons \text{CH}_3\text{COOH}\text{(aq)} + \text{CH}_3\text{CH(OH)CH}_3\text{(aq)}\)
(State symbols of (l) or (aq) are acceptable for the organic compounds depending on whether the system is treated as a pure liquid mixture or in aqueous solution.)

**(c) (i)**
- Reagents: Acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\) / \(\text{H}^+\)(aq).
- Conditions: Heat under reflux (or distillation, as ketones do not undergo further oxidation).

**(c) (ii)**
- Reagents: Aqueous iodine and aqueous sodium hydroxide, \(\text{I}_2\text{(aq)} / \text{NaOH(aq)}\) (alkaline aqueous iodine).
- Yellow precipitate formula: \(\text{CHI}_3\) (triiodomethane).

**(d)**
Propan-2-ol is produced industrially by the hydration of propene:
- Reagents: Steam (\(\text{H}_2\text{O(g)}\)) in the presence of a phosphoric acid catalyst (\(\text{H}_3\text{PO}_4\)).
- Conditions: \(300^\circ\text{C}\) and \(60\text{ atm}\) pressure.

**(a)** [4 marks]
- Award [1] for the correct formula of **X**: \(\text{CH}_3\text{COOCH}(\text{CH}_3)_2\) (or skeletal equivalent).
- Award [1] for the correct formula of **Y**: \(\text{CH}_3\text{COOH}\) (or skeletal equivalent).
- Award [1] for the correct formula of **Z**: \(\text{CH}_3\text{CH(OH)CH}_3\) or \(\text{(CH}_3)_2\text{CHOH}\) (or skeletal equivalent).
- Award [1] for the correct formula of **W**: \(\text{CH}_3\text{COCH}_3\) (or skeletal equivalent).

**(b)** [2 marks]
- Award [1] for the correct balanced equation: \(\text{CH}_3\text{COOCH}(\text{CH}_3)_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{CH}_3\text{CH(OH)CH}_3\) (allow single arrow).
- Award [1] for correct state symbols on all species: (l) or (aq) for reactants and products.

**(c) (i)** [2 marks]
- Award [1] for reagent: \(\text{K}_2\text{Cr}_2\text{O}_7\) and \(\text{H}_2\text{SO}_4\) / acidified dichromate(VI) (Reject: Tollens' or Fehling's, or dichromate without acid).
- Award [1] for conditions: Heat under reflux or heat and distil.

**(c) (ii)** [2 marks]
- Award [1] for reagents: Aqueous iodine in sodium hydroxide / \(\text{I}_2\) and \(\text{NaOH}\)(aq).
- Award [1] for the formula of the yellow precipitate: \(\text{CHI}_3\).

**(d)** [1 mark]
- Award [1] for: Steam / \(\text{H}_2\text{O(g)}\) with phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst (accept high temperature and pressure, e.g., \(300^\circ\text{C}\) and \(60\text{ atm}\)).

(a)(i) The stereoisomerism exhibited by X is E-Z (or geometrical / cis-trans) isomerism. This arises because rotation around the C=C bond is restricted by the presence of the \( \pi \) bond. For E-Z isomerism to occur, both carbon atoms involved in the double bond must be attached to two different atoms or groups of atoms. In 3-methylpent-2-ene, C2 is attached to -H and -\( \text{CH}_3 \), while C3 is attached to -\( \text{CH}_3 \) and -\( \text{CH}_2\text{CH}_3 \).
(ii) The IUPAC name for X is 3-methylpent-2-ene.
For the E-isomer: The high priority groups on each carbon of the C=C bond are on opposite sides. On C2, the high priority group is -\( \text{CH}_3 \) (vs -H). On C3, the high priority group is -\( \text{CH}_2\text{CH}_3 \) (vs -\( \text{CH}_3 \)). Thus, in the E-isomer, the C2 methyl group and the C3 ethyl group are on opposite sides of the double bond axis.
For the Z-isomer: These high priority groups are on the same side of the double bond axis.
(b)(i) When 3-methylpent-2-ene is cleaved by hot, concentrated, acidified \( \text{KMnO}_4 \):
- The C3 side containing the -\( \text{CH}_3 \) and -\( \text{CH}_2\text{CH}_3 \) groups forms the ketone, butanone: \( \text{CH}_3\text{COCH}_2\text{CH}_3 \) (compound Y).
- The C2 side containing the -H and -\( \text{CH}_3 \) groups initially forms ethanal, which is rapidly oxidized to ethanoic acid: \( \text{CH}_3\text{COOH} \) (compound Z).
(ii) Compound Z is ethanoic acid.

Part (a)(i):
- 1 Mark: Identifies the stereoisomerism as geometrical / E-Z / cis-trans isomerism.
- 1 Mark: Explains that there is restricted rotation about the C=C double bond (or due to the \( \pi \) bond) AND that each carbon of the double bond is attached to two different groups.

Part (a)(ii):
- 1 Mark: Correct skeletal structure of E-3-methylpent-2-ene clearly labelled.
- 1 Mark: Correct skeletal structure of Z-3-methylpent-2-ene clearly labelled.

Part (b)(i):
- 1 Mark: Correct structural/skeletal formula of Y as \( \text{CH}_3\text{COCH}_2\text{CH}_3 \) (or \( \text{CH}_3\text{CH}_2\text{COCH}_3 \)).
- 1 Mark: Correct structural/skeletal formula of Z as \( \text{CH}_3\text{COOH} \) (reject ethanal).

Part (b)(ii):
- 1 Mark: Correctly names Z as ethanoic acid (accept acetic acid).

(a) (i)
- Determine mass % of oxygen: \(100 - (52.17 + 13.04) = 34.79\%\)
- Convert percentages to moles:
- \(n(\text{C}) = \frac{52.17}{12.0} = 4.35\text{ mol}\)
- \(n(\text{H}) = \frac{13.04}{1.0} = 13.04\text{ mol}\)
- \(n(\text{O}) = \frac{34.79}{16.0} = 2.17\text{ mol}\)
- Divide by the smallest mole value (2.17):
- \(\text{C} = \frac{4.35}{2.17} = 2.0\)
- \(\text{H} = \frac{13.04}{2.17} = 6.0\)
- \(\text{O} = \frac{2.17}{2.17} = 1.0\)
- Empirical formula is \(\text{C}_2\text{H}_6\text{O}\).

(ii)
- The empirical formula mass \(M_r(\text{C}_2\text{H}_6\text{O}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\).
- Since the empirical formula mass equals the molecular ion peak value (\(m/z = 46\)), the molecular formula is \(\text{C}_2\text{H}_6\text{O}\).

(b) (i)
- Chlorine exists naturally as two main isotopes: \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\).
- The natural abundance ratio of \(^{35}\text{Cl}\) to \(^{37}\text{Cl}\) is approximately \(3:1\), which results in two molecular ion peaks with a height ratio of \(3:1\).

(ii)
- The peak at \(m/z = 15\) is due to the methyl cation fragment: \(\text{CH}_3^+\).
- Equation for its formation from the molecular ion:
\([\text{C}_3\text{H}_7^{35}\text{Cl}]^{\bullet+} \rightarrow \text{CH}_3^+ + \cdot\text{C}_2\text{H}_4^{35}\text{Cl}\) (or \(\text{C}_3\text{H}_7\text{Cl}^+ \rightarrow \text{CH}_3^+ + \text{C}_2\text{H}_4\text{Cl}\))

(iii)
- Using the formula:
\(\text{Number of C atoms } (n) = \frac{100}{1.1} \times \frac{\text{abundance of } M+1}{\text{abundance of } M}\)
- \(n = \frac{100}{1.1} \times \frac{3.3}{100} = 3\).
- Therefore, there are 3 carbon atoms in one molecule of **Y**.

**Part (a)(i) [3 marks]**
- **M1**: Calculates % of O as 34.79% and calculates moles of each element: C = 4.35, H = 13.04, O = 2.17 (allow minor rounding errors).
- **M2**: Correct mole ratio shown (C : H : O = 2 : 6 : 1) by dividing by the smallest number of moles.
- **M3**: States correct empirical formula: \(\text{C}_2\text{H}_6\text{O}\).

**Part (a)(ii) [1 mark]**
- **M1**: States that the empirical formula mass (46) is equal to the molecular ion mass (46), hence molecular formula is \(\text{C}_2\text{H}_6\text{O}\).

**Part (b)(i) [2 marks]**
- **M1**: Explains that chlorine has two isotopes, \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\).
- **M2**: States that the natural abundance ratio of these isotopes is \(3:1\), leading to a \(3:1\) height ratio of the \(M\) and \(M+2\) peaks.

**Part (b)(ii) [2 marks]**
- **M1**: Identifies species as \(\text{CH}_3^+\). *Do not accept species without a positive charge.*
- **M2**: Writes correct equation, e.g., \([\text{C}_3\text{H}_7\text{Cl}]^{\bullet+} \rightarrow \text{CH}_3^+ + \cdot\text{C}_2\text{H}_4\text{Cl}\) or \(\text{C}_3\text{H}_7\text{Cl}^+ \rightarrow \text{CH}_3^+ + \text{C}_2\text{H}_4\text{Cl}\).

**Part (b)(iii) [2 marks]**
- **M1**: Shows correct working using the formula \(n = \frac{\text{abundance of } M+1}{1.1 \% \times \text{abundance of } M} \times 100\) (or equivalent ratio method).
- **M2**: Correctly calculates the number of carbon atoms as 3.

### **Part (a)**
Concordant titres are those within \(0.10\text{ cm}^3\) of each other.
* Concordant titres are Titration 1 (\(24.95\text{ cm}^3\)) and Titration 2 (\(25.05\text{ cm}^3\)).
* \(\text{Average titre} = \frac{24.95 + 25.05}{2} = 25.00\text{ cm}^3\).

---

### **Part (b)**
* **Step 1**:
\(\text{Moles of } \text{KMnO}_4 = \text{concentration} \times \text{volume (dm}^3\)\)
\(\text{Moles of } \text{MnO}_4^- = 0.0200\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 5.00 \times 10^{-4}\text{ mol}\)

* **Step 2**:
From the ionic equation, \(1\text{ mol of } \text{MnO}_4^-\) reacts with \(5\text{ mol of } \text{Fe}^{2+}\).
\(\text{Moles of } \text{Fe}^{2+} \text{ in } 25.00\text{ cm}^3 = 5.00 \times 10^{-4} \times 5 = 2.50 \times 10^{-3}\text{ mol}\)

* **Step 3**:
\(\text{Moles of } \text{Fe}^{2+} \text{ in } 250.0\text{ cm}^3 = 2.50 \times 10^{-3} \times \frac{250.0}{25.00} = 2.50 \times 10^{-2}\text{ mol}\)

* **Step 4**:
\(\text{Molar mass } (M_r) \text{ of } \text{FeSO}_4 \cdot x\text{H}_2\text{O} = \frac{\text{Mass}}{\text{Moles}}\)
\(M_r = \frac{6.95\text{ g}}{2.50 \times 10^{-2}\text{ mol}} = 278.0\text{ g mol}^{-1}\)

* **Step 5**:
\(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\)
\(M_r(x\text{H}_2\text{O}) = 278.0 - 151.9 = 126.1\text{ g mol}^{-1}\)
\(x = \frac{126.1}{18.0} = 7.01 \approx 7\)

---

### **Part (c)**
(i) \(\text{Percentage uncertainty} = \frac{\text{Uncertainty}}{\text{Volume measured}} \times 100\)
\(\text{Percentage uncertainty} = \frac{0.06}{25.00} \times 100 = 0.24\%\)

(ii) If prepared in water without acid, some of the \(\text{Fe}^{2+}\) ions would be oxidized by dissolved oxygen in the air to \(\text{Fe}^{3+}\) (or would hydrolyze to form an iron oxide/hydroxide precipitate). As a result, the concentration of active \(\text{Fe}^{2+}\) ions in **FA 1** would decrease, leading to a smaller average titre value.

---

### **Part (d)**
* No indicator is needed because potassium manganate(VII), \(\text{KMnO}_4\), acts as a self-indicator. The purple \(\text{MnO}_4^-\) ions are reduced to colorless (or virtually colorless) \(\text{Mn}^{2+}\) ions during the titration.
* At the end-point, when all \(\text{Fe}^{2+}\) has reacted, the first excess drop of \(\text{MnO}_4^-\) turns the solution from colorless (or very pale green) to a permanent pale pink.

### **Part (a) [3 Marks]**
* **M1**: Identifies Titrations 1 and 2 as the only concordant titres (within \(\pm 0.10\text{ cm}^3\)). (1)
* **M2**: Calculates the correct average titre value of \(25.00\text{ cm}^3\) (must be given to 2 decimal places with units). (1)
* **M3**: Shows clear working for the average titration calculation. (1)

### **Part (b) [6 Marks]**
* **M4 (Step 1)**: Correctly calculates the amount of \(\text{MnO}_4^-\) as \(5.00 \times 10^{-4}\text{ mol}\). (1)
* **M5 (Step 2)**: Multiplies moles of \(\text{MnO}_4^-\) by 5 to find moles of \(\text{Fe}^{2+}\) in \(25.00\text{ cm}^3\) (\(2.50 \times 10^{-3}\text{ mol}\)). (1)
* **M6 (Step 3)**: Correctly scales up by 10 to find moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) (\(2.50 \times 10^{-2}\text{ mol}\)). (1)
* **M7 (Step 4)**: Shows correct method to calculate \(M_r\) of hydrated salt (\(\frac{\text{mass}}{\text{moles}}\)). (1)
* **M8 (Step 4)**: Obtains \(278\text{ g mol}^{-1}\) (or \(278.0\)) based on previous steps. (1)
* **M9 (Step 5)**: Correctly calculates \(x = 7\) as an integer. (1)

### **Part (c) [3 Marks]**
* **M10**: Calculates the percentage uncertainty of the class B pipette: \(\frac{0.06}{25.00} \times 100 = 0.24\%\). (1)
* **M11**: States that the titre would decrease/be smaller. (1)
* **M12**: Explains that \(\text{Fe}^{2+}\) ions would oxidize to \(\text{Fe}^{3+}\) (by air/oxygen) in the absence of acid, lowering the concentration of \(\text{Fe}^{2+}\). (1)

### **Part (d) [2 Marks]**
* **M13**: Explains that \(\text{KMnO}_4\) acts as a self-indicator because the purple \(\text{MnO}_4^-\) reactant turns into colorless \(\text{Mn}^{2+}\) product. (1)
* **M14**: States the correct end-point colour change: colorless (or pale green) to permanent pale pink. (1) *(Do not accept purple to colorless as the end-point is reached by adding manganate to iron(II))*.

**(a)**
**(i)** The completed table with \( \Delta T \) values is:
- Exp 1: \( 26.0 - 20.5 = 5.5 ^\circ\text{C} \)
- Exp 2: \( 31.4 - 20.5 = 10.9 ^\circ\text{C} \)
- Exp 3: \( 36.9 - 20.5 = 16.4 ^\circ\text{C} \)
- Exp 4: \( 32.8 - 20.5 = 12.3 ^\circ\text{C} \)
- Exp 5: \( 28.7 - 20.5 = 8.2 ^\circ\text{C} \)
- Exp 6: \( 24.6 - 20.5 = 4.1 ^\circ\text{C} \)

**(ii)**
- Uncertainty of a single thermometer reading is \( \pm 0.05 ^\circ\text{C} \).
- Since \( \Delta T \) is a difference between two readings (final - initial), the uncertainty in \( \Delta T \) is \( 2 \times 0.05 = \pm 0.10 ^\circ\text{C} \) (or \( \pm 0.1 ^\circ\text{C} \)).
- For Experiment 6: \( \Delta T = 4.1 ^\circ\text{C} \).
- \( \text{Percentage uncertainty} = \frac{0.10}{4.1} \times 100\\% = 2.44\\% \) (or \( 2.4\\% \)).
*(Note: If a student uses \( \pm 0.1 ^\circ\text{C} \) for a single reading and \( \pm 0.2 ^\circ\text{C} \) for the difference, then the percentage uncertainty is \( \frac{0.2}{4.1} \times 100\\% = 4.88\\% \).)*

**(b)**
**(i)** The graph is plotted with \( \Delta T \) on the y-axis and the volume of FB 1 on the x-axis. A straight line of best fit is drawn through the first three points (Experiments 1, 2, 3), and another straight line of best fit is drawn through the remaining points (Experiments 3, 4, 5, 6). The two straight lines intersect at the point of maximum temperature rise.

**(ii)** Reading from the intersection point of the two lines of best fit:
- Volume of FB 1 at maximum temperature rise = \( 30.0\text{ cm}^3 \)
- Volume of FB 2 at maximum temperature rise = \( 50.0 - 30.0 = 20.0\text{ cm}^3 \)

**(c)**
**(i)** \( n(\text{NaOH}) = C \times V = 2.00\text{ mol dm}^{-3} \times \frac{30.0}{1000}\text{ dm}^3 = 0.0600\text{ mol} \)

**(ii)** \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
From the balanced equation, \( 2 \text{ moles of NaOH} \) react with \( 1 \text{ mole of H}_2\text{SO}_4 \).
\( n(\text{H}_2\text{SO}_4) = \frac{1}{2} \times n(\text{NaOH}) = \frac{0.0600}{2} = 0.0300\text{ mol} \)
Volume of FB 2 = \( 20.0\text{ cm}^3 = 0.0200\text{ dm}^3 \)
\( [\text{H}_2\text{SO}_4] = \frac{n(\text{H}_2\text{SO}_4)}{V(\text{H}_2\text{SO}_4)} = \frac{0.0300}{0.0200} = 1.50\text{ mol dm}^{-3} \)

**(d)**
**(i)** \( q = m c \Delta T \)
Total volume of solution = \( 50.0\text{ cm}^3 \), mass \( m = 50.0\text{ g} \).
Using the maximum temperature rise \( \Delta T = 16.4 ^\circ\text{C} \) (or \( 16.4\text{ K} \)):
\( q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 16.4\text{ K} = 3427.6\text{ J} \) (or \( 3.43\text{ kJ} \) or \( 3430\text{ J} \) to 3 sig fig).

**(ii)** Suitable improvements include:
- Nesting the plastic cup inside another plastic cup to create an air insulation layer.
- Placing the plastic cup inside a beaker and filling the surrounding gap with cotton wool, vermiculite, or polyurethane foam.
- Using a vacuum-jacketed vessel.

**(a)(i)** [1 mark]
- Award 1 mark for all six temperature rises correctly calculated:
Exp 1: 5.5, Exp 2: 10.9, Exp 3: 16.4, Exp 4: 12.3, Exp 5: 8.2, Exp 6: 4.1

**(a)(ii)** [2 marks]
- 1 mark: Uncertainty in a single reading is \( \pm 0.05 ^\circ\text{C} \) OR uncertainty in the temperature rise (difference) is \( \pm 0.10 ^\circ\text{C} \).
- 1 mark: Correct calculation of percentage uncertainty for Experiment 6, showing working:
\( \frac{0.10}{4.1} \times 100\\% = 2.44\\% \) (or \( 2.4\\% \)).
*Accept \( \frac{0.2}{4.1} \times 100\\% = 4.88\\% \) if a single reading uncertainty is taken as \( \pm 0.1 ^\circ\text{C} \).*

**(b)(i)** [3 marks]
- 1 mark: Axes labeled correctly (y-axis: \( \Delta T / ^\circ\text{C} \), x-axis: volume of FB 1 / \( \text{cm}^3 \)) with linear scales where the plotted points occupy more than half of the grid area.
- 1 mark: All points plotted accurately within half a small square.
- 1 mark: Two straight lines of best fit drawn using a ruler, intersecting cleanly at a single maximum point.

**(b)(ii)** [1 mark]
- 1 mark: Volumes correctly read from the intersection of the two lines of best fit. Based on the data:
Volume of FB 1 = \( 30.0\text{ cm}^3 \) (accept \( 29.5 - 30.5\text{ cm}^3 \))
Volume of FB 2 = \( 20.0\text{ cm}^3 \) (calculated as \( 50.0 - \text{Volume of FB 1} \))

**(c)(i)** [1 mark]
- 1 mark: Correct calculation of moles of \( \text{NaOH} \) based on the student's volume from (b)(ii):
\( n(\text{NaOH}) = 2.00 \times \frac{30.0}{1000} = 0.0600\text{ mol} \)

**(c)(ii)** [2 marks]
- 1 mark: Correct balanced equation: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \) (accept correct ionic equation showing 2:1 stoichiometry).
- 1 mark: Correct calculation of \( [\text{H}_2\text{SO}_4] \) using the 2:1 stoichiometry and the student's volume of FB 2 from (b)(ii):
\( [\text{H}_2\text{SO}_4] = \frac{0.0600 / 2}{0.0200} = 1.50\text{ mol dm}^{-3} \) (allow error carried forward from (c)(i) and (b)(ii)).

**(d)(i)** [1 mark]
- 1 mark: Correct calculation of \( q \) using \( q = m c \Delta T \):
\( q = 50.0 \times 4.18 \times 16.4 = 3427.6\text{ J} \) (or \( 3.43\text{ kJ} \) or \( 3430\text{ J} \)).
*Accept calculation using the student's maximum temperature rise.*

**(d)(ii)** [1 mark]
- 1 mark: Accept any valid insulation method that does not involve a lid, such as: nesting the cup in another plastic cup (to create a dead-air space), packing wool or foam around the cup inside a beaker, or using a vacuum jacket.

### Part (a) Expected Observations:

* **FA 5 (containing \(\text{Mg}^{2+}\) and \(\text{Cl}^-\))**:
* **Test 1 (NaOH)**: White precipitate formed, which is insoluble in excess.
* **Test 3 (NH3)**: White precipitate formed, which is insoluble in excess.
* **Test 4 (AgNO3 + NH3)**: White precipitate formed with silver nitrate, which completely dissolves in aqueous ammonia to give a colorless solution.

* **FA 6 (containing \(\text{Zn}^{2+}\) and \(\text{I}^-\))**:
* **Test 1 (NaOH)**: White precipitate formed, which dissolves in excess to give a colorless solution.
* **Test 3 (NH3)**: White precipitate formed, which dissolves in excess to give a colorless solution.
* **Test 4 (AgNO3 + NH3)**: Yellow precipitate formed with silver nitrate, which remains insoluble on addition of aqueous ammonia.

* **FA 7 (containing \(\text{NH}_4^+\) and \(\text{Br}^-\))**:
* **Test 1 (NaOH)**: No precipitate / colorless solution remains.
* **Test 2 (Warm with NaOH)**: Gas is evolved which turns damp red litmus paper blue (ammonia gas).
* **Test 3 (NH3)**: No precipitate / colorless solution remains.
* **Test 4 (AgNO3 + NH3)**: Cream precipitate formed with silver nitrate, which is insoluble/partially soluble in dilute aqueous ammonia.

### Part (b) Identifications:

* **FA 5**: Cation is \(\text{Mg}^{2+}\); Anion is \(\text{Cl}^-\).
* *Evidence*: The white precipitates formed with both \(\text{NaOH}\) and \(\text{NH}_3\) which are insoluble in excess indicate \(\text{Mg}^{2+}\). The white precipitate with \(\text{AgNO}_3\) that dissolves in \(\text{NH}_3\) confirms \(\text{Cl}^-\).
* **FA 6**: Cation is \(\text{Zn}^{2+}\); Anion is \(\text{I}^-\).
* *Evidence*: The white precipitates formed with \(\text{NaOH}\) and \(\text{NH}_3\) which both dissolve in excess to form colorless solutions confirm \(\text{Zn}^{2+}\). The yellow precipitate with \(\text{AgNO}_3\) that is insoluble in \(\text{NH}_3\) confirms \(\text{I}^-\).
* **FA 7**: Cation is \(\text{NH}_4^+\); Anion is \(\text{Br}^-\).
* *Evidence*: Production of an alkaline gas (ammonia) on warming with alkali confirms \(\text{NH}_4^+\). The cream precipitate with \(\text{AgNO}_3\) that is insoluble/partially soluble in dilute \(\text{NH}_3\) confirms \(\text{Br}^-\).

### Part (c) Ionic Equations:

(i) \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\)
(ii) \(\text{Zn}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Zn(OH)}_2(\text{s})\)

**Part (a): Observation Table (8 Marks)**

* **M1**: FA 5 with NaOH(aq) AND NH3(aq) both give a white precipitate, insoluble in excess [1]
* **M2**: FA 5 with AgNO3(aq) gives a white precipitate, which dissolves in NH3(aq) [1]
* **M3**: FA 6 with NaOH(aq) AND NH3(aq) both give a white precipitate, soluble in excess [1]
* **M4**: FA 6 with AgNO3(aq) gives a yellow precipitate, insoluble in NH3(aq) [1]
* **M5**: FA 7 with NaOH(aq) AND NH3(aq) both give no precipitate / remain colorless [1]
* **M6**: FA 7 on warming with NaOH(aq) produces a gas that turns damp red litmus paper blue [1]
* **M7**: FA 7 with AgNO3(aq) gives a cream precipitate, insoluble/partially soluble in dilute NH3(aq) [1]
* **M8**: Recording of observations is systematic and uses correct chemical terminology (e.g. 'precipitate', 'insoluble in excess', 'turns damp red litmus paper blue', 'no precipitate') [1]

**Part (b): Identifications and Evidence (4 Marks)**

* **M9**: Correctly identifies FA 5 as magnesium chloride / containing Mg2+ and Cl- [1]
* **M10**: Correctly identifies FA 6 as zinc iodide / containing Zn2+ and I- [1]
* **M11**: Correctly identifies FA 7 as ammonium bromide / containing NH4+ and Br- [1]
* **M12**: Gives convincing supporting evidence for all three solutions based on observations [1]

**Part (c): Ionic Equations (2 Marks)**

* **M13**: Correct ionic equation for AgCl precipitation with state symbols: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\) [1]
* **M14**: Correct ionic equation for Zn(OH)2 precipitation with state symbols: \(\text{Zn}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Zn(OH)}_2(\text{s})\) [1]
*(Note: State symbols must be correct and present for both equations)*