2025 IB DP Chemistry 模擬試題連答案詳解

Aluminium sulfate dissociates according to the equation: \(\text{Al}_2(\text{SO}_4)_3(\text{s}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq})\). This means that 1 mole of \(\text{Al}_2(\text{SO}_4)_3\) dissociates to yield 5 moles of ions. For \(0.20\text{ mol}\) of the compound, the total amount of ions is \(0.20\text{ mol} \times 5 = 1.0\text{ mol}\). The total number of ions is therefore \(1.0\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 6.0 \times 10^{23}\).

[1 mark] for selecting the correct option C. Award 0 marks for any incorrect option.

The nitrite ion, \(\text{NO}_2^-\), has 18 valence electrons. The central nitrogen atom is bonded to two oxygen atoms and has one lone pair, giving it 3 electron domains. The electron domain geometry is trigonal planar. The presence of the lone pair repels the bonding pairs, reducing the bond angle from the ideal \(120^\circ\) to approximately \(115^\circ\). The resulting molecular geometry is bent (or V-shaped).

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

According to the Br%C3%B8nsted-Lowry acid-base theory, a conjugate acid is formed when a base accepts a proton (\(\text{H}^+\)). Adding \(\text{H}^+\) to \(\text{HPO}_4^{2-}\) results in \(\text{H}_2\text{PO}_4^-\).

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

A reducing agent undergoes oxidation by losing electrons. The strongest reducing agent is the species that is most easily oxidized, which corresponds to the most negative standard reduction potential. Zinc has the lowest standard reduction potential (\(-0.76\text{ V}\)), so \(\text{Zn}(\text{s})\) has the greatest tendency to lose electrons and act as a reducing agent.

[1 mark] for selecting the correct option A. Award 0 marks for any incorrect option.

Since the forward reaction is exothermic (\(\Delta H < 0\)), increasing the temperature favors the reverse, endothermic reaction to absorb the added thermal energy. This shifts the position of equilibrium to the left. Because the concentration of products decreases and the concentration of reactants increases, the value of the equilibrium constant \(K_c\) decreases.

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

First, calculate the heat energy absorbed by the water: \(q = m c \Delta T = 100\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 20\text{ K} = 8360\text{ J} = 8.36\text{ kJ}\). Next, calculate the chemical amount of ethanol burned: \(n = \frac{0.92\text{ g}}{46.0\text{ g mol}^{-1}} = 0.020\text{ mol}\). Finally, calculate the enthalpy of combustion: \(\Delta H_c = -\frac{q}{n} = -\frac{8.36\text{ kJ}}{0.020\text{ mol}} = -418\text{ kJ mol}^{-1}\). The negative sign indicates that the combustion reaction is exothermic.

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

A tertiary alcohol is characterized by having the hydroxyl (\(-\text{OH}\)) group attached to a carbon atom that is directly bonded to three other carbon atoms. In 2-methylpropan-2-ol, the central carbon is bonded to three methyl groups and the hydroxyl group, making it a tertiary alcohol. Butan-2-ol and pentan-3-ol are secondary alcohols, and 2-methylpropan-1-ol is a primary alcohol.

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

Using the combined gas law, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Rearranging the formula to solve for the final volume \(V_2\) gives: \(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}\). Substituting the given values: \(V_2 = 2.0\text{ dm}^3 \times \frac{100\text{ kPa}}{200\text{ kPa}} \times \frac{600\text{ K}}{300\text{ K}} = 2.0 \times 0.5 \times 2 = 2.0\text{ dm}^3\).

[1 mark] for selecting the correct option B. Award 0 marks for any incorrect option.

Using the combined gas law, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\).

Rearranging for \(V_2\):
\(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}\)

Given:
- \(V_1 = 2.00\text{ dm}^3\)
- \(\frac{P_1}{P_2} = \frac{1}{2}\) (since the pressure is doubled, \(P_2 = 2 P_1\))
- \(T_1 = 300\text{ K}\)
- \(T_2 = 450\text{ K}\)

Substituting these values:
\(V_2 = 2.00 \times \frac{1}{2} \times \frac{450}{300} = 1.00 \times 1.5 = 1.50\text{ dm}^3\).

[1] Award 1 mark for the correct answer (B).
[0] Other options are incorrect.

Assume a \(100\text{ g}\) sample of the oxide.
- Mass of \(\text{P} = 43.6\text{ g}\)
- Mass of \(\text{O} = 100.0 - 43.6 = 56.4\text{ g}\)

Convert masses to moles:
- Moles of \(\text{P} = \frac{43.6\text{ g}}{31.0\text{ g mol}^{-1}} \approx 1.41\text{ mol}\)
- Moles of \(\text{O} = \frac{56.4\text{ g}}{16.0\text{ g mol}^{-1}} \approx 3.53\text{ mol}\)

Divide by the smaller value to get the molar ratio:
- \(\text{P} = \frac{1.41}{1.41} = 1\)
- \(\text{O} = \frac{3.53}{1.41} \approx 2.5\)

Multiply by 2 to obtain whole numbers:
- \(\text{P} = 2\)
- \(\text{O} = 5\)

Therefore, the empirical formula is \(\text{P}_2\text{O}_5\).

[1] Award 1 mark for the correct answer (B).
[0] Other options are incorrect.

- \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair (tetrahedral domain geometry, trigonal pyramidal molecular geometry) with a bond angle of approximately \(107^\circ\).
- \(\text{SO}_2\) has 2 bonding domains and 1 lone pair on the sulfur atom (trigonal planar domain geometry, bent molecular geometry) with a bond angle of approximately \(119^\circ\), which is closest to \(120^\circ\).
- \(\text{H}_2\text{O}\) has 2 bonding pairs and 2 lone pairs (tetrahedral domain geometry, bent molecular geometry) with a bond angle of approximately \(104.5^\circ\).
- \(\text{CH}_4\) has 4 bonding pairs and 0 lone pairs (tetrahedral molecular geometry) with a bond angle of \(109.5^\circ\).

[1] Award 1 mark for the correct answer (B).
[0] Other options are incorrect.

A dissociation of \(1.00\%\) means that the fraction of acid dissociated is \(0.0100\).

Therefore, the concentration of hydrogen ions, \([\text{H}^+]\), is:
\([\text{H}^+] = 0.100\text{ mol dm}^{-3} \times 0.0100 = 1.00 \times 10^{-3}\text{ mol dm}^{-3}\)

Using the definition of pH:
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.00 \times 10^{-3}) = 3.00\).

[1] Award 1 mark for the correct answer (C).
[0] Other options are incorrect.

Oxidation is the increase in oxidation state (or loss of electrons).
- In A, \(\underline{\text{Fe}^{2+}}\) is oxidized to \(\text{Fe}^{3+}\) (oxidation state increases from +2 to +3).
- In B, \(\text{H}^+\) in \(\underline{\text{H}^+}\) is reduced to \(\text{H}_2\) (oxidation state decreases from +1 to 0).
- In C, \(\text{Cu}^{2+}\) in \(\underline{\text{Cu}^{2+}}\) is reduced to \(\text{Cu}\) (oxidation state decreases from +2 to 0).
- In D, \(\text{Cl}_2\) in \(\underline{\text{Cl}_2}\) is reduced to \(\text{Cl}^-\) (oxidation state decreases from 0 to -1).

[1] Award 1 mark for the correct answer (A).
[0] Other options are incorrect.

- Total mass of mixture, \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\).
- Temperature change, \(\Delta T = 26.8 - 20.0 = 6.8\text{ K}\).
- Heat released in joules, \(q = 100.0 \times 4.18 \times 6.8\text{ J}\).
- To convert to kilojoules, divide by 1000: \(q = \frac{100.0 \times 4.18 \times 6.8}{1000}\text{ kJ}\).
- Moles of water formed, \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).
- Since the temperature increased, the reaction is exothermic, so \(\Delta H\) must be negative:
\(\Delta H = -\frac{q}{n} = -\frac{100.0 \times 4.18 \times 6.8}{1000 \times 0.0500}\text{ kJ mol}^{-1}\).

[1] Award 1 mark for the correct answer (A).
[0] Other options are incorrect.

- The chromium atom (\(\text{Cr}\)) has an atomic number of 24. Its ground-state electron configuration is \([\text{Ar}] 4\text{s}^1 3\text{d}^5\) (due to the extra stability of a half-filled d-subshell).
- To form the \(\text{Cr}^{3+}\) ion, three electrons are removed. Electrons are removed from the outer \(4\text{s}\) subshell first, then from the \(3\text{d}\) subshell.
- Removing one electron from \(4\text{s}\) and two electrons from \(3\text{d}\) leaves \([\text{Ar}] 3\text{d}^3\).

[1] Award 1 mark for the correct answer (A).
[0] Other options are incorrect.

- The structural segment \(-\text{CH(OH)}-\) contains a hydroxyl group (characteristic of an alcohol).
- The structural segment \(-\text{COOCH}_3\) is an ester group (carbonyl adjacent to an ether linkage).
- Therefore, the functional groups present are hydroxyl and ester.

[1] Award 1 mark for the correct answer (B).
[0] Other options are incorrect.

From the stoichiometry of the reaction, \(0.10\text{ mol}\) of \(\text{C}_x\text{H}_y\) produces \(0.40\text{ mol}\) of \(\text{CO}_2\) and \(0.50\text{ mol}\) of \(\text{H}_2\text{O}\). Scaling this to \(1.0\text{ mol}\) of the hydrocarbon: \(1.0\text{ mol}\) of \(\text{C}_x\text{H}_y\) produces \(4.0\text{ mol}\) of \(\text{CO}_2\) and \(5.0\text{ mol}\) of \(\text{H}_2\text{O}\). Thus, \(x = 4\) and \(y = 5 \times 2 = 10\). The molecular formula of the hydrocarbon is \(\text{C}_4\text{H}_{10}\).

Award [1] for the correct answer B. Award [0] for other options.

The carbonate ion, \(\text{CO}_3^{2-}\), has three bonding domains and zero lone pairs around the central carbon atom. This gives a trigonal planar geometry with a bond angle of \(120^\circ\). \(\text{SF}_4\) has five electron domains, leading to a seesaw molecular geometry. \(\text{OF}_2\) has four electron domains (two bonding, two lone pairs), leading to a bent molecular geometry. \(\text{PCl}_3\) has four electron domains (three bonding, one lone pair), leading to a trigonal pyramidal molecular geometry.

Award [1] for the correct answer B. Award [0] for other options.

A Br\u00f8nsted\u2013Lowry conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). Here, \(\text{H}_2\text{PO}_4^-\text{(aq)}\) acts as an acid by donating a proton to become its conjugate base, \(\text{HPO}_4^{2-}\text{(aq)}\). These two species constitute a conjugate acid-base pair.

Award [1] for the correct answer C. Award [0] for other options.

In the nitrate ion (\(\text{NO}_3^-\)), the oxidation state of oxygen is \(-2\). Let the oxidation state of nitrogen be \(x\): \(x + 3(-2) = -1 \Rightarrow x = +5\). In the ammonium ion (\(\text{NH}_4^+\)), the oxidation state of hydrogen is \(+1\). Let the oxidation state of nitrogen be \(y\): \(y + 4(+1) = +1 \Rightarrow y = -3\). Therefore, the oxidation state changes from \(+5\) to \(-3\).

Award [1] for the correct answer A. Award [0] for other options.

Applying the combined gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Convert temperatures to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Rearranging the equation for \(V_2\): \(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.00 \times \frac{100}{50} \times \frac{600}{300} = 2.00 \times 2 \times 2 = 8.00\text{ dm}^3\).

Award [1] for the correct answer D. Award [0] for other options.

The heat energy absorbed during dissolution is given by \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 4.0\text{ K}\) (in Joules). Dividing by \(1000\) converts this to kilojoules: \(\frac{100.0 \times 4.18 \times 4.0}{1000}\text{ kJ}\). Since the temperature of the water decreases, the process is endothermic, so \(\Delta H_{\text{sol}}\) has a positive sign. Dividing the energy change by the number of moles of the salt gives the molar enthalpy change: \(\Delta H_{\text{sol}} = +\frac{100.0 \times 4.18 \times 4.0}{0.050 \times 1000}\text{ kJ mol}^{-1}\).

Award [1] for the correct answer B. Award [0] for other options.

Esters have the general formula \(\text{R}^1\text{-COO-R}^2\), where \(\text{R}^2\) must be an alkyl group. The ester isomers of \(\text{C}_4\text{H}_8\text{O}_2\) are: (1) Methyl propanoate: \(\text{CH}_3\text{CH}_2\text{COOCH}_3\), (2) Ethyl ethanoate: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\), (3) Propyl methanoate: \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\), and (4) Isopropyl methanoate: \(\text{HCOOCH(CH}_3)_2\). There are exactly 4 ester isomers.

Award [1] for the correct answer C. Award [0] for other options.

The value of the equilibrium constant, \(K_c\), is dependent only on temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature (decreasing the temperature) shifts the equilibrium position to the right to favor the exothermic process. This increases the concentration of products relative to reactants at equilibrium, thereby increasing the value of \(K_c\).

Award [1] for the correct answer B. Award [0] for other options.

In carbon dioxide (I), the carbon-oxygen bonds are double bonds with a bond order of 2.0. In the methanoate ion (II), resonance occurs between two carbon-oxygen bonds (one single and one double), giving a bond order of 1.5. In the carbonate ion (III), resonance occurs over three carbon-oxygen bonds (one double and two single), giving a bond order of 1.33. Higher bond orders correspond to shorter, stronger bonds. Thus, the bond lengths increase in the order: \(\text{CO}_2\) (shortest) < \(\text{HCOO}^-\) < \(\text{CO}_3^{2-}\) (longest).

[1] Award mark for selecting option A. Option B represents the reverse order (decreasing bond length). Option C and D are incorrect because they do not correctly identify the bond orders or their relationship with bond length.

The amount of propanoic acid (HA) initially is \(0.0500\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.0050\text{ mol}\). The amount of NaOH added is \(0.0250\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.0025\text{ mol}\). NaOH reacts completely with the acid to form the conjugate base (\(\text{A}^-\)): \(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\). After reaction, \(0.0025\text{ mol}\) of HA remains, and \(0.0025\text{ mol}\) of \(\text{A}^-\) is formed. Because \([\text{HA}] = [\text{A}^-]\), the Henderson-Hasselbalch equation \(\text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}])\) simplifies to \(\text{pH} = \text{p}K_a = 4.87\).

[1] Correct option is B. Option A is \(\text{p}K_a / 2\). Option C is neutral pH. Option D is \(14.00 - \text{p}K_a\).

At the cathode, reduction occurs. The species with the most positive standard reduction potential is reduced first. Comparing \(\text{Cu}^{2+}\) (\(+0.34\text{ V}\)), \(\text{Ni}^{2+}\) (\(-0.26\text{ V}\)), and water (\(-0.83\text{ V}\)), \(\text{Cu}^{2+}\) is reduced first to form \(\text{Cu}(s)\). At the anode, oxidation occurs. Water is oxidized in preference to sulfate ions (which are extremely stable and difficult to oxidize under standard conditions) to form \(\text{O}_2(g)\) according to the reaction \(2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\).

[1] Award mark for option A. Option B is incorrect because \(\text{Ni}^{2+}\) is less readily reduced than \(\text{Cu}^{2+}\). Option C is incorrect because water is less readily reduced than copper(II) ions. Option D is incorrect because sulfate is not oxidized to sulfur dioxide at the anode.

Assume a \(100\text{ g}\) sample. Mass of nitrogen = \(30.4\text{ g}\), mass of oxygen = \(100 - 30.4 = 69.6\text{ g}\). Convert masses to moles: Moles of \(\text{N} = 30.4 / 14.0 = 2.17\text{ mol}\); Moles of \(\text{O} = 69.6 / 16.0 = 4.35\text{ mol}\). Divide by the smallest number of moles to find the simplest ratio: \(\text{N} = 2.17 / 2.17 = 1\); \(\text{O} = 4.35 / 2.17 = 2.00\). The empirical formula is \(\text{NO}_2\).

[1] Award mark for option B. Option A, C, and D are incorrect ratios.

Energy emitted is given by \(E = h\nu\), where \(\nu\) is frequency. Therefore, the transition with the largest energy change (\(\Delta E\)) emits radiation with the highest frequency. Transition down to the ground state (\(n = 1\)) involves the largest energy difference because the energy levels converge as \(n\) increases. Thus, the transition \(n = 2 \rightarrow n = 1\) (UV region) has a much larger energy difference than transitions between higher levels (e.g. \(n = 3 \rightarrow n = 2\) which is in the visible region, or \(n = 4 \rightarrow n = 3\) in the infrared region). Option D represents absorption, not emission.

[1] Award mark for option C. Option A and B have lower energy changes and therefore lower frequencies. Option D represents absorption, not emission.

A carbonyl compound contains a \(\text{C}=\text{O}\) group. This includes aldehydes and ketones. Let's find all possible structural isomers for \(\text{C}_4\text{H}_8\text{O}\) containing a carbonyl group. Aldehydes: (1) butanal: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\), (2) 2-methylpropanal: \((\text{CH}_3)_2\text{CHCHO}\). Ketones: (3) butanone: \(\text{CH}_3\text{COCH}_2\text{CH}_3\). There are no other ways to arrange the 4 carbons with a carbonyl group. Thus, there are exactly 3 structural isomers that are carbonyl compounds.

[1] Award mark for option B. Option A is incorrect as it misses one isomer. Option C and D are incorrect as they count other functional group isomers (e.g. cyclic ethers or unsaturated alcohols) which are not carbonyl compounds.

1. Calculate heat absorbed (q): \(q = m \cdot c \cdot \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{ K}^{-1} \times 3.20 \text{ K} = 1337.6 \text{ J} = 1.338 \text{ kJ}\). 2. Calculate moles of \(\text{NH}_4\text{Cl}\) (Molar mass = \(53.50 \text{ g mol}^{-1}\)): \(n = \frac{5.00 \text{ g}}{53.50 \text{ g mol}^{-1}} = 0.0935 \text{ mol}\). 3. Calculate enthalpy change of solution: \(\Delta H_{\text{sol}} = +\frac{1.338 \text{ kJ}}{0.0935 \text{ mol}} = +14.3 \text{ kJ mol}^{-1}\). 4. A systematic error that makes the result less endothermic is heat transfer from the surroundings into the calorimeter due to a lack of lid or poor insulation, which prevents the temperature from dropping as low as it should.

[1 mark] for the correct calculation of the enthalpy change (+14.3 kJ mol-1). The positive sign must be present. [0.56 marks] for identifying heat transfer from the surroundings / poor insulation of the calorimeter.

1. Convert units to SI: \(P = 101.3 \times 10^3 \text{ Pa}\), \(V = 82.0 \times 10^{-6} \text{ m}^3\), \(T = 95.0 + 273.15 = 368.15 \text{ K}\). 2. Calculate moles using the ideal gas law: \(n = \frac{PV}{RT} = \frac{101.3 \times 10^3 \text{ Pa} \times 82.0 \times 10^{-6} \text{ m}^3}{8.314 \text{ J K}^{-1}\text{ mol}^{-1} \times 368.15 \text{ K}} = 0.002714 \text{ mol}\). 3. Calculate molar mass (M): \(M = \frac{m}{n} = \frac{0.245 \text{ g}}{0.002714 \text{ mol}} = 90.3 \text{ g mol}^{-1}\). 4. Precaution: Keep the syringe in a temperature-controlled oven or vapor jacket heated well above the boiling point of the volatile liquid to prevent condensation.

[1 mark] for the correct calculation of molar mass (90.3 g mol-1). [0.56 marks] for identifying an appropriate precaution, such as heating well above the boiling point of the liquid or waiting for the syringe volume to stabilize.

1. The initial rate is the rate of reaction at the very start (\(t = 0\)). On a plot of product volume against time, this is determined by drawing a tangent to the curve at the origin (\(t = 0\)) and calculating the gradient (slope) of this tangent line. 2. The independent variable in this graph is time, which is plotted on the horizontal (x) axis.

[1 mark] for describing drawing a tangent at t = 0 and calculating its gradient/slope. [0.56 marks] for identifying time as the variable on the horizontal axis.

1. Calculate moles of \(\text{Na}_2\text{CO}_3\): \(n = 0.02500 \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 0.00125 \text{ mol}\). 2. Find moles of \(\text{HCl}\) using stoichiometry (1:2 ratio): \(n(\text{HCl}) = 2 \times 0.00125 \text{ mol} = 0.00250 \text{ mol}\). 3. Calculate concentration of \(\text{HCl}\): \(C = \frac{0.00250 \text{ mol}}{0.02145 \text{ dm}^3} = 0.117 \text{ mol dm}^{-3}\) (to 3 sig figs). 4. A suitable indicator for an equivalence point at pH 4.0 is methyl orange (which changes color in the range 3.1-4.4).

[1 mark] for the correct concentration (0.117 mol dm-3). [0.56 marks] for naming a suitable indicator (accept methyl orange or bromophenol blue).

1. Moles of \(\text{MnO}_4^-\) used: \(n = 0.01850 \text{ dm}^3 \times 0.0200 \text{ mol dm}^{-3} = 3.70 \times 10^{-4} \text{ mol}\). 2. Moles of \(\text{Fe}^{2+}\) in the \(25.00 \text{ cm}^3\) aliquot: \(5 \times 3.70 \times 10^{-4} \text{ mol} = 1.85 \times 10^{-3} \text{ mol}\) (due to 1:5 stoichiometry). 3. Moles of \(\text{Fe}^{2+}\) in the entire \(250.0 \text{ cm}^3\) flask: \(1.85 \times 10^{-3} \text{ mol} \times \frac{250.0}{25.00} = 1.85 \times 10^{-2} \text{ mol}\) of \(\text{Fe}^{2+}\). 4. Mass of \(\text{Fe}^{2+}\) in the sample: \(m = 1.85 \times 10^{-2} \text{ mol} \times 55.85 \text{ g mol}^{-1} = 1.03 \text{ g}\). 5. The end-point is observed when a single drop of excess purple \(\text{MnO}_4^-\) turns the colorless solution in the conical flask to a permanent pale pink color.

[1 mark] for the correct mass calculation (1.03 g). [0.56 marks] for the correct description of the end-point (colorless to permanent pale pink).

1. Mass of anhydrous salt (\(\text{MgSO}_4\)): \(26.76 \text{ g} - 24.35 \text{ g} = 2.41 \text{ g}\). Moles of \(\text{MgSO}_4\) (Molar mass = \(120.37 \text{ g mol}^{-1}\)): \(n = \frac{2.41 \text{ g}}{120.37 \text{ g mol}^{-1}} = 0.0200 \text{ mol}\). 2. Mass of water lost: \(29.28 \text{ g} - 26.76 \text{ g} = 2.52 \text{ g}\). Moles of \(\text{H}_2\text{O}\) (Molar mass = \(18.02 \text{ g mol}^{-1}\)): \(n = \frac{2.52 \text{ g}}{18.02 \text{ g mol}^{-1}} = 0.140 \text{ mol}\). 3. Calculate ratio \(x = \frac{0.140}{0.0200} = 7\). 4. Heating to constant mass ensures that all water of crystallization has been fully released from the hydrate, leaving only the anhydrous compound behind.

[1 mark] for calculating x = 7. [0.56 marks] for explaining that heating to constant mass ensures complete dehydration of the salt.

1. Build an ICE table: Initial ethanoic acid = 1.00 mol, ethanol = 1.00 mol, ester = 0 mol, water = 0 mol. 2. Change: Since 0.33 mol of ethanoic acid remains, \(1.00 - 0.33 = 0.67 \text{ mol}\) reacted. 3. Equilibrium: ethanoic acid = 0.33 mol, ethanol = 0.33 mol, ethyl ethanoate = 0.67 mol, water = 0.67 mol. 4. Since the volume terms cancel out in the expression: \(K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} = \frac{0.67 \times 0.67}{0.33 \times 0.33} = 4.12 \approx 4.1\). 5. The strong acid catalyst increases the rates of both the forward and reverse reactions, reducing the time required for the system to reach equilibrium.

[1 mark] for the correct calculation of Kc (4.1 or 4.12). [0.56 marks] for explaining that the catalyst increases the rate of reaction to reach equilibrium faster.

1. Test 1: Add sodium hydrogen carbonate (\(\text{NaHCO}_3\)) or sodium carbonate (\(\text{Na}_2\text{CO}_3\)) to separate samples. Propanoic acid, being a carboxylic acid, will react to produce carbon dioxide gas (bubbles/effervescence). Propan-1-ol and propanone will not react. 2. Test 2: Warm the remaining two liquids with acidified potassium dichromate(VI) solution, \(\text{K}_2\text{Cr}_2\text{O}_7\text{/H}^+\). Propan-1-ol (a primary alcohol) will be oxidized to propanal/propanoic acid, and the chromium is reduced, changing the solution's color from orange to green. Propanone (a ketone) is resistant to oxidation and the solution remains orange.

[0.78 marks] for Test 1: suitable carbonate reagent (e.g., NaHCO3/Na2CO3) or blue litmus paper, and positive observation of bubbles or color change to red for propanoic acid. [0.78 marks] for Test 2: suitable oxidizing agent (acidified potassium dichromate(VI) or potassium manganate(VII)) and warming, and positive observation of orange to green or purple to colorless for propan-1-ol.

When calculating the heat released using \( q = mc\Delta T \), only the heat absorbed by the solution is accounted for. Since the calorimeter itself also absorbs some of the heat released by the reaction, the actual heat released is greater than the calculated \( q \). Neglecting this heat capacity results in an underestimate of the magnitude of the enthalpy of neutralization, making the value less negative (smaller in magnitude).

Award 0.78 marks for stating that the calculated magnitude is smaller or less exothermic. Award 0.78 marks for explaining that some heat is absorbed by the calorimeter container itself, meaning the temperature rise of the water alone underestimates the total heat released.

If some water of crystallization remains, the measured mass of the anhydrous residue will be too high. This leads to a calculated mass of water lost that is systematically too low. Consequently, the calculated number of moles of water per mole of magnesium sulfate (\( x \)) will be lower than the true value.

Award 0.78 marks for stating that the calculated value of \( x \) will be lower than the true value. Award 0.78 marks for explaining that the calculated mass of water lost is too low due to the higher mass of the remaining residue.

From the ideal gas equation rearranged for molar mass, \( M = \frac{mRT}{PV} \), the calculated molar mass is directly proportional to the temperature \( T \) (in Kelvin). If the temperature is recorded as higher than the true value, the calculated value of \( M \) will be systematically higher than the actual molar mass.

Award 0.78 marks for identifying that the calculated molar mass is systematically too high. Award 0.78 marks for explaining that molar mass is directly proportional to the recorded temperature in the ideal gas equation, so an overestimate of temperature increases the calculated mass.

The pH meter reads lower than the actual pH because it was calibrated to a lower nominal value than the actual buffer pH. Consequently, the measured pH at the half-equivalence point (which equals the recorded \( pK_a \)) is systematically lower than the true value. Since \( K_a = 10^{-pK_a} \), a lower recorded \( pK_a \) yields a systematically higher calculated \( K_a \).

Award 0.78 marks for identifying that the calculated \( K_a \) will be systematically higher. Award 0.78 marks for explaining that the calibration causes the meter to read lower pH values, leading to a lower recorded \( pK_a \) and therefore a higher calculated \( K_a \).

Because the human reaction-time delay of \( 0.25\text{ s} \) occurs both when starting and when stopping the stopwatch, the two delays act in the same direction and cancel each other out, leaving the measured time interval unchanged. Thus, there is no significant systematic error introduced, and the accuracy of the calculated rate is unaffected.

Award 0.78 marks for stating that the delay has a negligible effect on the accuracy of the time interval. Award 0.78 marks for explaining that the delay at the start and the delay at the stop cancel each other out.

A non-volatile impurity such as dissolved salts or a more polar solvent like water can form stronger intermolecular forces (such as hydrogen bonds) with ethanol. This decreases the overall vapor pressure of the solution at any given temperature. Consequently, a higher temperature is required for the vapor pressure to equal atmospheric pressure, elevating the boiling point.

Award 0.78 marks for identifying a suitable impurity (e.g., water, a dissolved salt, or a less volatile compound). Award 0.78 marks for explaining that the impurity lowers the vapor pressure of the liquid mixture, thereby requiring more thermal energy (higher temperature) to reach boiling point.

The esterification and reverse hydrolysis reactions are slow but ongoing at room temperature. Titrating the acid shifts the equilibrium by removing one of the reactants. Performing the titration rapidly or in iced water quenches (slows down to a negligible rate) the reaction, ensuring that the concentration of ethanoic acid measured corresponds to its true value at the equilibrium state.

Award 0.78 marks for stating that the rapid titration or cold temperature freezes/quenches the reaction. Award 0.78 marks for explaining that this prevents the equilibrium from shifting (or prevents further reaction of reactants/products) during the analytical measurement.

A beaker has a high volume uncertainty (typically \( \pm 5\% \) or more) compared to a volumetric flask (typically \( \pm 0.1\% \)). This introduces a large systematic and random error in the concentration of the primary standard, which directly reduces both the accuracy and precision of the standardized potassium manganate(VII) solution.

Award 0.78 marks for stating that both accuracy and precision will be significantly reduced. Award 0.78 marks for explaining that beakers have a very high volume uncertainty compared to volumetric flasks, leading to a large uncertainty in the concentration of the standard solution.

First, calculate the initial concentrations of the reactants: \([\text{CO}]_i = 0.200\text{ mol} / 2.00\text{ dm}^3 = 0.100\text{ mol dm}^{-3}\) and \([\text{H}_2\text{O}]_i = 0.300\text{ mol} / 2.00\text{ dm}^3 = 0.150\text{ mol dm}^{-3}\). Let \(x\) be the change in concentration of reactants at equilibrium. Since the initial concentration of \(\text{CO}_2\) is zero and its equilibrium concentration is \(0.040\text{ mol dm}^{-3}\), \(x = 0.040\text{ mol dm}^{-3}\). The equilibrium concentrations are: \([\text{CO}] = 0.100 - 0.040 = 0.060\text{ mol dm}^{-3}\), \([\text{H}_2\text{O}] = 0.150 - 0.040 = 0.110\text{ mol dm}^{-3}\), \([\text{CO}_2] = 0.040\text{ mol dm}^{-3}\), and \([\text{H}_2] = 0.040\text{ mol dm}^{-3}\). The expression for \(K_c\) is: \(K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}\). Substituting the equilibrium values gives: \(K_c = \frac{0.040 \times 0.040}{0.060 \times 0.110} = \frac{0.0016}{0.0066} = 0.24\) (or \(0.242\) with 3 significant figures).

Award [1] for calculating correct equilibrium concentrations of all species: \([\text{CO}] = 0.060\text{ M}\), \([\text{H}_2\text{O}] = 0.110\text{ M}\), and \([\text{H}_2] = 0.040\text{ M}\). Award [1] for the final correct calculation of \(K_c = 0.24\) (accept 0.242).

For a weak acid, the dissociation can be represented as \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\). Using the approximation that \([\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}}\), we have: \(K_a = \frac{[\text{H}^+]^2}{[\text{HA}]}\). Rearranging to solve for hydrogen ion concentration: \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.34 \times 10^{-5} \times 0.125} = \sqrt{1.675 \times 10^{-6}} = 1.294 \times 10^{-3}\text{ mol dm}^{-3}\). Calculating the pH: \(\text{pH} = -\log_{10}(1.294 \times 10^{-3}) = 2.89\).

Award [1] for calculating \([\text{H}^+] = 1.29 \times 10^{-3}\text{ mol dm}^{-3}\). Award [1] for calculating the correct pH of 2.89.

Convert mass percentages to moles assuming a \(100\text{ g}\) sample: moles of C = \(40.0\text{ g} / 12.01\text{ g mol}^{-1} = 3.33\text{ mol}\); moles of H = \(6.7\text{ g} / 1.01\text{ g mol}^{-1} = 6.63\text{ mol}\); moles of O = \(53.3\text{ g} / 16.00\text{ g mol}^{-1} = 3.33\text{ mol}\). Divide each mole value by the smallest value (3.33): ratio for C = \(3.33/3.33 = 1\); ratio for H = \(6.63/3.33 \approx 2\); ratio for O = \(3.33/3.33 = 1\). Therefore, the empirical formula is \(\text{CH}_2\text{O}\).

Award [1] for correctly determining the molar ratios of the elements (3.33 mol C : 6.63 mol H : 3.33 mol O). Award [1] for writing the correct empirical formula \(\text{CH}_2\text{O}\).

Formal charge is calculated as: \(\text{FC} = V - N - \frac{1}{2}B\), where \(V\) is valence electrons, \(N\) is non-bonding valence electrons, and \(B\) is bonding electrons. For the Carbon atom: \(V = 4\), \(N = 0\), \(B = 8\) (from 4 bonds), so \(\text{FC} = 4 - 0 - \frac{1}{2}(8) = 0\). For the single-bonded Oxygen atom: \(V = 6\), \(N = 6\) (3 lone pairs), \(B = 2\) (1 bond), so \(\text{FC} = 6 - 6 - \frac{1}{2}(2) = -1\).

Award [1] for the correct formal charge on Carbon (0). Award [1] for the correct formal charge on the single-bonded Oxygen (-1).

The reduction half-reaction is: \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). The oxidation half-reaction is: \(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^-\). To balance electrons, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5: \(2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}\) and \(5\text{H}_2\text{C}_2\text{O}_4 \rightarrow 10\text{CO}_2 + 10\text{H}^+ + 10e^-\). Adding the reactions and simplifying \(\text{H}^+\) species: \(2\text{MnO}_4^-(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l)\). The coefficient of \(\text{H}^+(aq)\) is 6.

Award [1] for successfully balancing both half-reactions including electrons. Award [1] for identifying the correct final coefficient of \(\text{H}^+\) as 6.

The structural formula of 3-hydroxybutanal contains a hydroxyl group (\(-\text{OH}\)) on the third carbon, which belongs to the alcohol family, and a carbonyl group at the end of the chain (\(-\text{CHO}\)), which belongs to the aldehyde family.

Award [1] for identifying 'alcohol' (or 'hydroxyl group'). Award [1] for identifying 'aldehyde' (or 'carbonyl group' / 'alkanal').

Calculate the total mass of the mixture: \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). The temperature change: \(\Delta T = 26.1 - 19.5 = 6.6\text{ K}\). The heat absorbed by the solution: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.6\text{ K} = 2758.8\text{ J} = 2.759\text{ kJ}\). The amount of reactants used: \(n = C \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). The enthalpy of neutralization per mole: \(\Delta H = -\frac{q}{n} = -\frac{2.759\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\) (or \(-55\text{ kJ mol}^{-1}\) to two significant figures based on temperature change).

Award [1] for calculating heat released \(q = 2.76\text{ kJ}\). Award [1] for dividing by \(0.0500\text{ mol}\) and reporting a negative value of \(-55\) or \(-55.2\text{ kJ mol}^{-1}\).

Convert all values to SI units: \(P = 1.013 \times 10^5\text{ Pa}\), \(V = 2.500 \times 10^{-4}\text{ m}^3\), \(T = 98.0 + 273.15 = 371.15\text{ K}\). Use the ideal gas law \(PV = nRT\) to find moles \(n = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 2.500 \times 10^{-4}}{8.314 \times 371.15} = 0.008207\text{ mol}\). Calculate the molar mass: \(M = \frac{\text{mass}}{n} = \frac{0.582\text{ g}}{0.008207\text{ mol}} = 70.9\text{ g mol}^{-1}\).

Award [1] for calculating the correct number of moles, \(n = 0.00821\text{ mol}\) (or showing correct unit conversions). Award [1] for finding the final molar mass of \(70.9\text{ g mol}^{-1}\) (accept range 70.8 - 71.1).

Carbon dioxide (\(\text{CO}_2\)) has a simple molecular structure with weak intermolecular (London dispersion) forces between individual molecules. Relatively little thermal energy is needed to overcome these weak forces, making it a gas at room temperature. In contrast, silicon dioxide (\(\text{SiO}_2\)) has a giant covalent lattice structure in which each silicon atom is covalently bonded to four oxygen atoms, and each oxygen to two silicon atoms. Breaking these strong covalent bonds throughout the giant structure requires a large amount of energy, resulting in a very high melting point.

Award [1 mark] for identifying that \(\text{CO}_2\) is a simple molecular structure with weak intermolecular forces (or London dispersion forces). Award [1 mark] for identifying that \(\text{SiO}_2\) is a giant covalent structure with strong covalent bonds that must be broken.

Since the reaction is endothermic, increasing the temperature drives the system to absorb heat. According to Le Chatelier's principle, the equilibrium will shift in the endothermic direction (to the right, favoring the forward reaction) to counteract this temperature increase. Because the concentration of products (\(\text{NO}_2\)) increases and the concentration of reactants (\(\text{N}_2\text{O}_4\)) decreases, the value of the equilibrium constant \(K_{\text{c}} = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}\) increases.

Award [1 mark] for stating that \(K_{\text{c}}\) increases. Award [1 mark] for stating that the position of equilibrium shifts to the right.

First, determine initial moles: \(n(\text{CH}_3\text{COOH}) = 0.0500 \times 0.10 = 5.00 \times 10^{-3}\text{ mol}\) and \(n(\text{NaOH}) = 0.0250 \times 0.10 = 2.50 \times 10^{-3}\text{ mol}\). Since NaOH is limiting, \(2.50 \times 10^{-3}\text{ mol}\) of acid reacts to form \(2.50 \times 10^{-3}\text{ mol}\) of ethanoate conjugate base, leaving \(2.50 \times 10^{-3}\text{ mol}\) of unreacted ethanoic acid. Since \([\text{acid}] = [\text{conjugate base}]\), \([\text{H}^+] = K_{\text{a}} = 1.8 \times 10^{-5}\). Thus, \(\text{pH} = -\log_{10}(1.8 \times 10^{-5}) = 4.74\).

Award [1 mark] for showing that \([\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]\) in the buffer solution. Award [1 mark] for calculating \(\text{pH} = 4.74\) (accept 4.7).

During the electrolysis of molten lead(II) bromide, bromide ions migrate to the positive electrode (anode) and lose electrons: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\). Lead ions migrate to the negative electrode (cathode) and gain electrons: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\). Oxidation is the loss of electrons, which occurs at the anode.

Award [1 mark] for both correct, balanced half-equations. Award [1 mark] for identifying the anode as the site of oxidation.

Propan-1-ol is a primary alcohol. Under distillation conditions with an oxidizing agent, it is oxidized to the aldehyde propanal. Under reflux conditions, it undergoes complete oxidation to form the carboxylic acid propanoic acid.

Award [1 mark] for (i) propanal. Award [1 mark] for (ii) propanoic acid.

Using the ideal gas law \(PV = nRT\), we convert the pressure to \(1.013 \times 10^5\text{ Pa}\), volume to \(2.800 \times 10^{-4}\text{ m}^3\), and temperature to \(298.15\text{ K}\). Calculating the number of moles: \(n = \frac{1.013 \times 10^5 \times 2.800 \times 10^{-4}}{8.31 \times 298.15} = 0.01145\text{ mol}\). Molar mass \(M = \frac{m}{n} = \frac{0.584}{0.01145} = 51.0\text{ g mol}^{-1}\).

Award [1 mark] for calculating the moles of gas as \(n \approx 0.0114\text{ mol}\). Award [1 mark] for calculating the correct molar mass \(51.0\text{ g mol}^{-1}\) (accept range \(50.8 - 51.2\)).

First, find empirical formula ratio by dividing mass percentages by molar masses: \(n(\text{C}) = \frac{40.0}{12.01} = 3.33\), \(n(\text{H}) = \frac{6.7}{1.01} = 6.63\), \(n(\text{O}) = \frac{53.3}{16.00} = 3.33\). This yields a ratio of \(1 : 2 : 1\), so the empirical formula is \(\text{CH}_2\text{O}\). The empirical formula mass is \(30.03\text{ g mol}^{-1}\). Since the molecular mass is \(180.18\text{ g mol}^{-1}\), the scale factor is \(\frac{180.18}{30.03} = 6\). Thus, the molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_6\).

Award [1 mark] for the correct empirical formula \(\text{CH}_2\text{O}\). Award [1 mark] for the correct molecular formula \(\text{C}_6\text{H}_{12}\text{O}_6\).

First, calculate heat released: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.60\text{ K} = 2758.8\text{ J} = 2.7588\text{ kJ}\). Find moles of water produced: \(n = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). Finally, calculate enthalpy of neutralization: \(\Delta H = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).

Award [1 mark] for calculating the correct heat energy change (\(q = 2.76\text{ kJ}\)). Award [1 mark] for dividing by \(0.0500\text{ mol}\) and including a negative sign to get \(-55.2\text{ kJ mol}^{-1}\) (accept range \(-55.0\) to \(-55.3\)).

Mass of water lost = \(2.497\text{ g} - 1.597\text{ g} = 0.900\text{ g}\). Molar mass of \(\text{H}_2\text{O} = 18.02\text{ g mol}^{-1}\). Moles of water = \(0.900\text{ g} / 18.02\text{ g mol}^{-1} = 0.0499\text{ mol}\). Molar mass of \(\text{CuSO}_4 = 159.62\text{ g mol}^{-1}\). Moles of anhydrous residue = \(1.597\text{ g} / 159.62\text{ g mol}^{-1} = 0.0100\text{ mol}\). Ratio \(x = 0.0499 / 0.0100 = 4.99\), which rounds to the nearest integer 5.

Award [1] for calculating the correct moles of water (0.050 mol) and anhydrous copper(II) sulfate (0.010 mol). Award [1] for correctly determining the ratio x = 5.

Sulfur has 6 valence electrons and each fluorine has 7, giving a total of \(6 + (4 \times 7) = 34\) valence electrons. This translates to 4 single bonds with fluorine and 1 lone pair on the sulfur atom (10 electrons around S). 5 electron domains around the central sulfur atom give a trigonal bipyramidal electron domain geometry. The molecular geometry, which considers only the positions of the atoms, is seesaw (or sawhorse).

Award [1] for describing or drawing the correct Lewis structure with 4 single S-F bonds and 1 lone pair on S. Award [1] for identifying trigonal bipyramidal electron domain geometry and seesaw/sawhorse molecular geometry.

\(\text{pH} = 3.00\) means \([\text{H}^+] = 10^{-3.00} = 1.00 \times 10^{-3}\text{ mol dm}^{-3}\). For weak acid dissociation, \([\text{H}^+] = [\text{A}^-] = 1.00 \times 10^{-3}\text{ mol dm}^{-3}\). Using the equilibrium expression \(K_a = [\text{H}^+][\text{A}^-] / [\text{HA}]\) and the approximation that \([\text{HA}] \approx 0.100\text{ mol dm}^{-3}\), we get \(K_a = (1.00 \times 10^{-3})^2 / 0.100 = 1.0 \times 10^{-5}\text{ mol dm}^{-3}\). If no approximation is made, \(K_a = (1.00 \times 10^{-3})^2 / 0.099 = 1.01 \times 10^{-5}\text{ mol dm}^{-3}\).

Award [1] for calculating \([\text{H}^+] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\). Award [1] for the correct value of \(K_a\) of \(1.0 \times 10^{-5}\) (or \(1.01 \times 10^{-5}\)) with the units \(\text{mol dm}^{-3}\).

At the anode, oxidation occurs: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\). At the cathode, reduction occurs: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). The phase boundary is shown as a single vertical line, and the salt bridge is a double vertical line. Thus, the cell notation is \(\text{Zn(s)} \mid \text{Zn}^{2+}\text{(aq)} \parallel \text{Cu}^{2+}\text{(aq)} \mid \text{Cu(s)}\).

Award [1] for representing oxidation on the left and reduction on the right. Award [1] for correct phase boundaries (|) and salt bridge (||).

Using the combined gas law: \(P_1 V_1 / T_1 = P_2 V_2 / T_2\). We can rearrange this to find \(V_2\): \(V_2 = V_1 \times (P_1 / P_2) \times (T_2 / T_1)\). Since absolute temperature is doubled, \(T_2 / T_1 = 2\). Thus, \(V_2 = 4.00 \times (101.3 / 150.0) \times 2 = 5.40\text{ dm}^3\).

Award [1] for correctly rearranging the combined gas law. Award [1] for the final volume of 5.40 dm^3.

Heat absorbed by the water: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 22.5\text{ K} = 9405\text{ J} = 9.405\text{ kJ}\). Moles of methanol burned: \(n = 0.800\text{ g} / 32.04\text{ g mol}^{-1} = 0.02497\text{ mol}\). Experimental enthalpy of combustion: \(\Delta H_c = -q / n = -9.405\text{ kJ} / 0.02497\text{ mol} = -377\text{ kJ mol}^{-1}\).

Award [1] for calculating the correct heat released (9.41 kJ) and moles of methanol (0.0250 mol). Award [1] for calculating \(\Delta H_c = -377\text{ kJ mol}^{-1}\) (negative sign is required).

Ethyl ethanoate is an ester (functional group: ester). It is formed through the esterification reaction of an alcohol (ethanol) and a carboxylic acid (ethanoic acid) in the presence of an acid catalyst such as concentrated sulfuric acid.

Award [1] for identifying the functional group as an ester. Award [1] for identifying ethanol and ethanoic acid as the reagents.

The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant of the forward reaction: \(K_{c,\text{reverse}} = 1 / K_{c,\text{forward}} = 1 / (4.0 \times 10^{-2}) = 25\).

Award [1] for stating the reciprocal relationship. Award [1] for the final answer of 25.

1. Calculate heat released, \(q\): \(q = m \cdot c \cdot \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 27.0\text{ K} = 16929\text{ J} = 16.929\text{ kJ}\). 2. Calculate amount of methanol, \(n\): \(n = 1.28\text{ g} / 32.05\text{ g mol}^{-1} = 0.039938\text{ mol}\). 3. Calculate \(\Delta H_c\): \(\Delta H_c = -q / n = -16.929\text{ kJ} / 0.039938\text{ mol} = -423.89\text{ kJ mol}^{-1}\), which rounds to \(-424\text{ kJ mol}^{-1}\).

Award 1 mark for calculating the heat absorbed by water: \(16.9\text{ kJ}\) (or \(16929\text{ J}\)). Award 1 mark for the correct calculation of moles of methanol (\(0.0399\text{ mol}\)) and final value of \(-424\text{ kJ mol}^{-1}\) (must include negative sign and correct unit/3 sig figs). Accept \(-424\).

1. Set up the weak acid approximation: \(K_a \approx [\text{H}^+]^2 / [\text{HA}]\). 2. Rearrange to find \([\text{H}^+]\): \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.34 \times 10^{-5} \times 0.150} = \sqrt{2.01 \times 10^{-6}} = 1.418 \times 10^{-3}\text{ mol dm}^{-3}\). 3. Calculate pH: \(\text{pH} = -\log_{10}(1.418 \times 10^{-3}) = 2.848\). Rounded to 2 decimal places, \(\text{pH} = 2.85\).

Award 1 mark for setting up the calculation and determining \([\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}\) (or equivalent). Award 1 mark for the final pH of \(2.85\) (accept range \(2.84 - 2.86\) depending on intermediate rounding).

1. Convert percentages to moles (assuming \(100\text{ g}\) of substance): \(n(\text{C}) = 40.0 / 12.01 = 3.33\text{ mol}\); \(n(\text{H}) = 6.7 / 1.01 = 6.63\text{ mol}\); \(n(\text{O}) = 53.3 / 16.00 = 3.33\text{ mol}\). 2. Divide each by the smallest number of moles (\(3.33\)): Ratio of \(\text{C} : \text{H} : \text{O} = 3.33/3.33 : 6.63/3.33 : 3.33/3.33 = 1 : 1.99 : 1 \approx 1 : 2 : 1\). Thus, the empirical formula is \(\text{CH}_2\text{O}\).

Award 1 mark for finding the correct mole ratio of \(1 : 2 : 1\) (or showing the mole values: C: 3.33, H: 6.63, O: 3.33). Award 1 mark for stating the correct empirical formula as \(\text{CH}_2\text{O}\).

1. Identify cathode and anode half-reactions: Silver is reduced (cathode): \(E^\theta_{\text{red}} = +0.80\text{ V}\). Nickel is oxidized (anode): \(E^\theta_{\text{red}} = -0.26\text{ V}\). 2. Calculate cell potential: \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (-0.26\text{ V}) = +1.06\text{ V}\). 3. Determine spontaneity: Since the cell potential is positive (\(E^\theta_{\text{cell}} > 0\)), the reaction is spontaneous.

Award 1 mark for correct calculation of \(E^\theta_{\text{cell}} = +1.06\text{ V}\) (accept \(1.06\text{ V}\)). Award 1 mark for stating that the reaction is spontaneous because \(E^\theta_{\text{cell}}\) is positive.