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Carbon dioxide (\(\text{CO}_2\)) is linear, non-polar, and has \(sp\) hybridization on the carbon atom. Methanal (\(\text{HCHO}\)) is trigonal planar around the carbon atom, meaning the carbon is \(sp^2\) hybridized. Due to the electronegativity difference between carbon and oxygen, and the asymmetric shape, the molecule is polar. Methane (\(\text{CH}_4\)) is tetrahedral, non-polar, and has \(sp^3\) hybridization. Ethyne (\(\text{C}_2\text{H}_2\)) is linear, non-polar, and has \(sp\) hybridization.

Award [1] for correct option B. Award [0] for incorrect options.

Phosphorus (\([\text{Ne}] 3s^2 3p^3\)) has three singly occupied 3p orbitals. Sulfur (\([\text{Ne}] 3s^2 3p^4\)) has one doubly occupied 3p orbital and two singly occupied ones. The repulsion between the two paired electrons in the same orbital of sulfur decreases the attraction to the nucleus, making it easier to remove one of these electrons compared to a singly occupied orbital in phosphorus. Therefore, sulfur has a lower first ionization energy despite its higher nuclear charge.

Award [1] for correct option C. Award [0] for incorrect options.

Assume a \(100\text{ g}\) sample of the oxide: Mass of nitrogen is \(30.4\text{ g}\) and mass of oxygen is \(69.6\text{ g}\). Calculate the moles: \(n(\text{N}) = 30.4 / 14.01 = 2.17\text{ mol}\), and \(n(\text{O}) = 69.6 / 16.00 = 4.35\text{ mol}\). The mole ratio of nitrogen to oxygen is \(2.17 : 4.35 = 1 : 2.0\). Therefore, the empirical formula is \(\text{NO}_2\).

Award [1] for correct option B. Award [0] for incorrect options.

The Maxwell-Boltzmann distribution plots the number of particles (or fraction of particles) against their kinetic energy. The total area under this curve represents the total number of particles in the sample, which remains constant regardless of temperature changes (though the shape of the curve changes).

Award [1] for correct option D. Award [0] for incorrect options.

The equilibrium constant, \(K_c\), is only affected by changes in temperature. Because the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium to the right to release heat according to Le Chatelier's principle. This increases the concentration of products relative to reactants, thereby increasing the value of \(K_c\).

Award [1] for correct option B. Award [0] for incorrect options.

According to the Bronsted-Lowry theory, a conjugate base is formed when an acid donates a proton. Therefore, \(\text{HPO}_4^{2-} \rightarrow \text{H}^+ + \text{PO}_4^{3-}\). The conjugate base of \(\text{HPO}_4^{2-}\) is the phosphate ion, \(\text{PO}_4^{3-}\).

Award [1] for correct option B. Award [0] for incorrect options.

We determine the oxidation state of chlorine in each species, assuming oxygen has an oxidation state of \(-2\): In \(\text{ClO}^-\), the oxidation state of Cl is \(+1\). In \(\text{ClO}_3^-\), the oxidation state of Cl is \(+5\). In \(\text{ClO}_2\), the oxidation state of Cl is \(+4\). In \(\text{Cl}_2\text{O}_7\), the oxidation state of Cl is \(+7\). Thus, chlorine has the highest oxidation state (\(+7\)) in \(\text{Cl}_2\text{O}_7\).

Award [1] for correct option D. Award [0] for incorrect options.

Methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)) is an ester. The characteristic functional group is the ester group (\(-\text{COO}-\)), which is formed by the reaction of a carboxylic acid (propanoic acid) and an alcohol (methanol).

Award [1] for correct option B. Award [0] for incorrect options.

First, calculate the number of moles of carbon dioxide gas produced:
\(n(\text{CO}_2) = \frac{V}{V_{\text{m}}} = \frac{360\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0150\text{ mol}\)

Since the reaction of a group 2 metal carbonate with hydrochloric acid proceeds according to the equation:
\(\text{MCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{MCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)

The mole ratio between \(\text{MCO}_3\) and \(\text{CO}_2\) is \(1:1\). Thus:
\(n(\text{MCO}_3) = 0.0150\text{ mol}\)

The molar mass of \(\text{MCO}_3\) is:
\(M(\text{MCO}_3) = \frac{\text{mass}}{n} = \frac{1.50\text{ g}}{0.0150\text{ mol}} = 100\text{ g mol}^{-1}\)

To find the relative atomic mass of metal \(\text{M}\):
\(M(\text{M}) = M(\text{MCO}_3) - M(\text{CO}_3^{2-}) = 100 - (12.01 + 3 \times 16.00) = 100 - 60.01 = 39.99\text{ g mol}^{-1}\)

This molar mass corresponds to calcium (\(\text{Ca}\), \(M_{\text{r}} = 40.08\)).

[1 mark] for correct selection of Option B.
- Correctly determines moles of carbon dioxide (0.0150 mol).
- Correctly determines molar mass of the metal carbonate (100 g/mol).
- Correctly identifies calcium (Ca).

- In \(\text{SO}_2\), the sulfur atom has 6 valence electrons. It forms one double bond and one coordinate (or double) bond with oxygen, leaving one lone pair on the sulfur atom. The steric number (number of bonding domains + lone pairs) is \(2 + 1 = 3\), which corresponds to \(\text{sp}^2\) hybridization with exactly one lone pair on the central sulfur atom.
- In \(\text{CO}_2\), carbon is \(\text{sp}\) hybridized with 0 lone pairs.
- In \(\text{NH}_3\), nitrogen is \(\text{sp}^3\) hybridized with 1 lone pair.
- In \(\text{H}_2\text{O}\), oxygen is \(\text{sp}^3\) hybridized with 2 lone pairs.

[1 mark] for correct selection of Option B.

The rate of the overall reaction is determined by the slow rate-determining step (Step 2):
\(\text{Rate} = k_2 [\text{N}_2\text{O}_2][\text{H}_2]\)

Since \(\text{N}_2\text{O}_2\) is an intermediate, its concentration must be expressed in terms of the reactants using the fast equilibrium of Step 1:
\(K_{\text{eq}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \Rightarrow [\text{N}_2\text{O}_2] = K_{\text{eq}} [\text{NO}]^2\)

Substitute this expression back into the rate equation:
\(\text{Rate} = k_2 K_{\text{eq}} [\text{NO}]^2 [\text{H}_2]\)

Setting \(k = k_2 K_{\text{eq}}\), the rate expression is:
\(\text{Rate} = k [\text{NO}]^2 [\text{H}_2]\)

[1 mark] for correct selection of Option B.

For a weak acid:
\(K_{\text{a}} \approx \frac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}}\)

\([\text{H}^+] = \sqrt{K_{\text{a}} \times [\text{HA}]} = \sqrt{(1.0 \times 10^{-5}) \times (0.10)} = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\)

Calculating the pH:
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.0\)

[1 mark] for correct selection of Option B.

To be thermodynamically spontaneous under standard conditions, the standard cell potential (\(E^\ominus_{\text{cell}}\)) must be positive.

Let us evaluate the options:
- For Option B: \(\text{Cu}(s) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Fe}^{2+}(aq)\)
- Reduction: \(\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\ominus = +0.77\text{ V}\)
- Oxidation: \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^- \quad E^\ominus = +0.34\text{ V}\)
- \(E^\ominus_{\text{cell}} = E^\ominus_{\text{red}} - E^\ominus_{\text{ox}} = +0.77\text{ V} - 0.34\text{ V} = +0.43\text{ V}\) (Spontaneous)

Checking other options:
- Option A: \(E^\ominus_{\text{cell}} = 0.54 - 0.77 = -0.23\text{ V}\) (Non-spontaneous)
- Option C: \(E^\ominus_{\text{cell}} = 0.15 - 0.34 = -0.19\text{ V}\) (Non-spontaneous)
- Option D: \(E^\ominus_{\text{cell}} = 0.54 - 0.77 = -0.23\text{ V}\) (Non-spontaneous)

[1 mark] for correct selection of Option B.

An alcohol is classified as tertiary if the carbon atom bonded to the hydroxyl group (\(\text{-OH}\)) is bonded to three other carbon atoms.
- In 2-methylbutan-2-ol, the carbon at position 2 is bonded to: a methyl group at position 1, a methyl group at position 2, and an ethyl group (carbon atoms 3 and 4). Since it is bonded to three other carbons, it is a tertiary alcohol.
- 3-methylbutan-2-ol is a secondary alcohol.
- 2-methylbutan-1-ol is a primary alcohol.
- Pentan-3-ol is a secondary alcohol.

[1 mark] for correct selection of Option A.

- \(\text{Na}_2\text{O}\) is a basic oxide and forms strongly alkaline \(\text{NaOH}\) in water.
- \(\text{MgO}\) is a basic oxide and forms weakly alkaline \(\text{Mg(OH)}_2\) in water.
- \(\text{SiO}_2\) is an acidic non-metal oxide but is insoluble in water due to its giant covalent structure, so it does not lower the pH of water.
- \(\text{P}_4\text{O}_{10}\) is a simple molecular non-metal oxide that reacts vigorously with water to form phosphoric acid (\(\text{H}_3\text{PO}_4\)), producing a highly acidic solution.

[1 mark] for correct selection of Option C.

A real gas behaves most like an ideal gas when intermolecular attractions and the volume of the gas particles themselves are minimized:
1. At high temperatures, the thermal kinetic energy of the particles is very high, making intermolecular attractive forces negligible.
2. At low pressures, the gas particles are highly spread out, making the volume of individual molecules negligible compared to the total volume of the container.

[1 mark] for correct selection of Option C.

First, calculate the theoretical yield of \(\text{CO}_2\) in moles: since the mole ratio of \(\text{M}_2\text{CO}_3\) to \(\text{CO}_2\) is 1:1, decomposing \(0.10\text{ mol}\) of \(\text{M}_2\text{CO}_3\) should theoretically yield \(0.10\text{ mol}\) of \(\text{CO}_2\). Next, calculate the theoretical volume of \(\text{CO}_2\) at STP: \(\text{Volume} = 0.10\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 2.27\text{ dm}^3\). Finally, calculate the percentage yield: \(\text{Percentage yield} = \frac{\text{Actual volume}}{\text{Theoretical volume}} \times 100\% = \frac{1.79}{2.27} \times 100\% \approx 79\%\).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

\(\text{XeF}_4\) has 8 valence electrons from xenon and 7 from each of the four fluorine atoms, giving a total of 36 valence electrons. The central xenon atom has 4 single bonding pairs and 2 lone pairs. This corresponds to an octahedral electron domain geometry and a square planar molecular geometry to minimize electron-pair repulsion. \(\text{SF}_4\) is seesaw-shaped, while \(\text{CF}_4\) and \(\text{PCl}_4^+\) are both tetrahedral.

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

Phosphorus has a valence electron configuration of \(3\text{s}^2 3\text{p}^3\) with three singly occupied \(3\text{p}\) orbitals. Sulfur has a valence configuration of \(3\text{s}^2 3\text{p}^4\), meaning it has one doubly occupied \(3\text{p}\) orbital. The mutual repulsion between the two electrons in the same \(3\text{p}\) orbital in sulfur makes it easier to remove one of these electrons, resulting in a lower first ionization energy than phosphorus despite sulfur's higher nuclear charge.

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

The half-cell with the more positive standard reduction potential (\(\text{Ag}^+/\text{Ag}\)) undergoes reduction at the cathode: \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\). The half-cell with the less positive potential (\(\text{Cr}^{3+}/\text{Cr}\)) undergoes oxidation at the anode: \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\). The standard cell potential is \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.80\text{ V} - (-0.74\text{ V}) = +1.54\text{ V}\). Multiplying the reduction half-equation by 3 to balance the electrons gives the overall cell reaction: \(3\text{Ag}^+(\text{aq}) + \text{Cr}(\text{s}) \rightarrow 3\text{Ag}(\text{s}) + \text{Cr}^{3+}(\text{aq})\).

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

The rate constant \(k\) is independent of the reactant concentration or pressure changes. It only depends on temperature and the activation energy (which can be lowered by a catalyst). According to the Arrhenius equation, increasing the temperature increases the value of the rate constant \(k\).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

A basic buffer solution (pH > 7) consists of a weak base and its conjugate acid. In mixture A, \(50\text{ cm}^3\) of \(0.10\text{ mol dm}^{-3}\ \text{NH}_3\) (\(0.0050\text{ mol}\)) reacts with \(25\text{ cm}^3\) of \(0.10\text{ mol dm}^{-3}\ \text{HCl}\) (\(0.0025\text{ mol}\)). This neutralizes half of the ammonia, leaving \(0.0025\text{ mol}\) of unreacted \(\text{NH}_3\) and forming \(0.0025\text{ mol}\) of \(\text{NH}_4^+\). Because the weak base and its conjugate acid are present in comparable amounts, it forms a buffer with pH > 7. Mixture B forms an acidic buffer (pH < 7). Mixture C results in complete neutralization forming the salt \(\text{NH}_4\text{Cl}\) (pH < 7). Mixture D contains excess strong acid (pH < 7).

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

The benzene ring represents an arene functional group. The \(-\text{COOCH}_3\) substituent is an ester group. The \(-\text{OH}\) group is attached directly to the aromatic benzene ring, making it a phenol group (rather than an aliphatic alcohol). Therefore, the correct functional groups are ester, phenol, and arene.

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

Using the combined gas law, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Temperatures must be converted to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Rearranging the formula for the new volume: \(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 3.0\text{ dm}^3 \times \frac{100\text{ kPa}}{200\text{ kPa}} \times \frac{600\text{ K}}{300\text{ K}} = 3.0 \times 0.5 \times 2 = 3.0\text{ dm}^3\).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

To find the new volume of the gas, we use the combined gas law:

\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]

First, convert temperatures from Celsius to Kelvin:
\[T_1 = 27 + 273.15 = 300.15\text{ K} \approx 300\text{ K}\]
\[T_2 = 327 + 273.15 = 600.15\text{ K} \approx 600\text{ K}\]

Rearrange the equation to solve for the final volume \(V_2\):

\[V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}\]

Substitute the given values into the equation:

\[V_2 = 2.0\text{ dm}^3 \times \frac{100\text{ kPa}}{200\text{ kPa}} \times \frac{600\text{ K}}{300\text{ K}}\]
\[V_2 = 2.0 \times 0.5 \times 2 = 2.0\text{ dm}^3\]

[1 mark] Award for correct option B.
- Award 0 marks for other options.

- Option A is incorrect: \(\text{Al}_2\text{O}_3\) is ionic with significant covalent character, not a giant covalent structure.
- Option B is correct: \(\text{SiO}_2\) (silicon dioxide) has a macromolecular (giant covalent) structure and behaves as an acidic oxide, reacting with strong bases.
- Option C is incorrect: \(\text{SO}_3\) has a molecular covalent structure but behaves as an acidic oxide, not basic.
- Option D is incorrect: \(\text{Na}_2\text{O}\) has an ionic structure and is a basic oxide, not acidic.

[1 mark] Award for correct option B.

First, identify the anode and cathode:
- The half-cell with the higher standard reduction potential undergoes reduction. Here, \(E^\theta(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}\), which is higher than \(+0.34\text{ V}\). Thus, silver ions are reduced at the silver electrode (cathode): \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\). Since silver ions are consumed, their concentration decreases.
- The copper electrode undergoes oxidation (anode): \(\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\). This causes the mass of the copper electrode to decrease.
- Electrons flow through the external circuit from the negative anode (copper) to the positive cathode (silver).

Therefore, option C is the correct statement.

[1 mark] Award for correct option C.

- Option A is incorrect: Doubling the concentration of both reactants increases the rate by a factor of \(2^2 \times 2^1 = 8\).
- Option B is incorrect: Elementary termolecular reactions are extremely rare, and since the stoichiometry indicates a total of 4 reactant molecules, this reaction must proceed via a multi-step mechanism.
- Option C is correct: The overall reaction order is \(2 + 1 = 3\).
\[\text{Units of } k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\]
- Option D is incorrect: Since the order with respect to \(\text{H}_2\) is 1, doubling its concentration will only double the initial rate.

[1 mark] Award for correct option C.

A basic buffer (pH > 7) consists of a weak base and its conjugate acid. It can be prepared by reacting excess weak base with a smaller amount of strong acid.

- In option A: We mix \(50.0\text{ cm}^3\) of \(0.10\text{ mol dm}^{-3}\text{ NH}_3\) (\(5.0\text{ mmol}\)) with \(25.0\text{ cm}^3\) of \(0.10\text{ mol dm}^{-3}\text{ HCl}\) (\(2.5\text{ mmol}\)).
\(\text{NH}_3\) is a weak base, and \(\text{HCl}\) is a strong acid.
They react in a 1:1 ratio:
\[\text{NH}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NH}_4\text{Cl}(\text{aq})\]
Since \(\text{NH}_3\) is in excess (\(5.0\text{ mmol} > 2.5\text{ mmol}\)), all \(\text{HCl}\) is consumed, leaving \(2.5\text{ mmol}\) of unreacted \(\text{NH}_3\) and producing \(2.5\text{ mmol}\) of \(\text{NH}_4^+\). This mixture of a weak base and its conjugate acid forms a buffer with pH > 7.
- In option B: This results in a mixture of excess \(\text{CH}_3\text{COOH}\) and \(\text{CH}_3\text{COO}^-\), which is an acidic buffer (pH < 7).
- In option C: Equimolar amounts of \(\text{NH}_3\) and \(\text{HCl}\) react completely to form \(\text{NH}_4\text{Cl}\), which undergoes salt hydrolysis to form an acidic solution (pH < 7) and is not a buffer.
- In option D: This creates a solution containing excess strong base (\(\text{NaOH}\)) and neutral salt (\(\text{NaCl}\)), which is not a buffer system.

[1 mark] Award for correct option A.

A trigonal pyramidal molecular geometry occurs when the central atom has 3 bonding pairs and 1 lone pair of electrons (steric number = 4).

- In option A:
- \(\text{NH}_3\): Nitrogen has 5 valence electrons. It forms 3 single bonds with H and has 1 lone pair. (Trigonal pyramidal)
- \(\text{H}_3\text{O}^+\): Oxygen has 6 valence electrons, minus 1 for the positive charge = 5. It forms 3 single bonds with H and has 1 lone pair. (Trigonal pyramidal)
- \(\text{PF}_3\): Phosphorus has 5 valence electrons. It forms 3 single bonds with F and has 1 lone pair. (Trigonal pyramidal)
Therefore, all species in A have a trigonal pyramidal geometry.

- In option B:
- \(\text{BF}_3\) and \(\text{SO}_3\) are trigonal planar.
- In option C:
- \(\text{CO}_3^{2-}\) is trigonal planar.
- In option D:
- \(\text{CH}_4\) is tetrahedral.

[1 mark] Award for correct option A.

(a) To establish a steady initial temperature baseline and ensure the solution is at thermal equilibrium with the surroundings before the reaction begins.

(b) To compensate for heat loss to the surroundings that occurs during the course of the reaction before the maximum temperature is reached. Extrapolating back to the exact time of mixing (\(t = 3.5\text{ min}\)) estimates the temperature change that would have occurred if the reaction had been instantaneous with zero heat loss.

(c) First, calculate the heat released (\(q\)):
\(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 12.4\text{ K} = 2591.6\text{ J} = 2.59\text{ kJ}\)

Next, calculate the amount (in moles) of \(\text{Cu}^{2+}\) reacted:
\(n(\text{Cu}^{2+}) = 0.0500\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol}\)

Calculate \(\Delta H\):
\(\Delta H = -\frac{q}{n} = -\frac{2.5916\text{ kJ}}{0.0100\text{ mol}} = -259\text{ kJ mol}^{-1}\)

(d) Glass is a much better thermal conductor than polystyrene. More heat is lost to the surroundings during the reaction, which results in a smaller measured temperature change (\(\Delta T\)). Consequently, the calculated heat of reaction (\(q\)) and the calculated value of \(\Delta H\) will be less negative (lower in magnitude).

(a) Award [1] for establishing a steady baseline temperature/determining the initial rate of heat exchange with surroundings.

(b) Award [1] for stating that extrapolation corrects for heat loss to the surroundings during the reaction; and [0.25] for stating that it estimates the temperature change if the reaction had been instantaneous.

(c) Award [1] for calculating the heat released: \(q = 2.59\text{ kJ}\) (or \(2592\text{ J}\)); Award [1] for dividing by \(0.0100\text{ mol}\) to obtain \(-259\text{ kJ mol}^{-1}\) (must have negative sign and correct unit). Accept range \(-259\) to \(-260\).

(d) Award [1] for identifying that more heat is lost because glass is a poorer insulator/better thermal conductor; Award [1] for concluding that \(\Delta T\) is smaller, so the calculated \(\Delta H\) is less negative / lower in magnitude.

(a) Independent variable: Time. Dependent variable: Volume of carbon dioxide gas collected.

(b) Percentage uncertainty = \(\frac{\text{Uncertainty}}{\text{Measured Value}} \times 100\% = \frac{0.5}{15.0} \times 100\% = 3.33\%\) (or \(3.3\%\)).

(c) (i) Powdered calcium carbonate has a larger total surface area than large marble chips. This increases the frequency of collisions between reactant particles (specifically, \(\text{H}^+\) ions and the solid \(\text{CaCO}_3\) surface), resulting in a higher rate of successful collisions per unit time and thus an increased initial rate of reaction.
(ii) The final volume of gas collected will be the same in both experiments. This is because the mass and moles of the limiting reactant (\(\text{CaCO}_3\), since \(\text{HCl}\) is in excess) are identical, which means the same stoichiometry yields the same total number of moles of carbon dioxide gas.

(a) Award [0.5] for identifying time as the independent variable and [0.5] for volume of gas as the dependent variable.

(b) Award [1] for correct formula/substitution: \(\frac{0.5}{15.0} \times 100\); and [0.25] for correct numerical calculation: \(3.33\%\) (or \(3.3\%\)).

(c) (i) Award [1] for stating that powder increases the total surface area; and [1] for stating that this increases collision frequency / number of collisions per unit time (leading to more successful collisions per unit time).
(ii) Award [1] for stating that the final volume of gas is unchanged/the same; and [1] for explaining that the amount/moles of the limiting reactant (\(\text{CaCO}_3\)) is the same in both runs.

(a) Mass of volatile liquid = \(10.456\text{ g} - 10.224\text{ g} = 0.232\text{ g}\).

(b) First, convert experimental values to standard SI units:
\(P = 101.3\text{ kPa} = 1.013 \times 10^5\text{ Pa}\)
\(V = 84.0\text{ cm}^3 = 84.0 \times 10^{-6}\text{ m}^3\)
\(T = 98.0 + 273.15 = 371.15\text{ K}\)

Now use \(PV = nRT\):
\(n = \frac{PV}{RT} = \frac{1.013 \times 10^5\text{ Pa} \times 84.0 \times 10^{-6}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 371.15\text{ K}} = 2.76 \times 10^{-3}\text{ mol}\) (or \(0.00276\text{ mol}\)).

(c) Molar mass (\(M\)):
\(M = \frac{m}{n} = \frac{0.232\text{ g}}{2.759 \times 10^{-3}\text{ mol}} = 84.1\text{ g mol}^{-1}\).

(d) If some vapor condenses in the needle, the measured volume of gas (\(V\)) in the syringe will be lower than the true volume of the vaporized sample. Consequently, the calculated number of moles (\(n\)) will be underestimated. Because molar mass is inversely proportional to the number of moles (\(M = \frac{m}{n}\)), a lower value of \(n\) leads to a higher calculated molar mass.

(a) Award [1] for \(0.232\text{ g}\).

(b) Award [1] for correct conversion of units (specifically \(V = 8.40 \times 10^{-5}\text{ m}^3\) and \(T = 371\text{ K}\)). Award [1] for calculating moles correctly: \(2.76 \times 10^{-3}\text{ mol}\) (accept \(2.75 \times 10^{-3}\) to \(2.80 \times 10^{-3}\)).

(c) Award [1] for calculated value of \(84.1\) (accept range \(82.9\) to \(84.4\) depending on intermediate rounding). Award [0.25] for correct unit \(\text{g mol}^{-1}\) or \(\text{g/mol}\).

(d) Award [1] for stating that the volume of gas read would be too low, hence the calculated number of moles is too low. Award [1] for concluding that the calculated molar mass is too high / overestimated.

(a) Colorless to permanent pale pink. (Do not accept 'clear' instead of 'colorless').

(b) Concordant titres are within \(\pm 0.10\text{ cm}^3\) of each other. Titrations 1, 2, and 3 are concordant (\(14.65\), \(14.55\), and \(14.60\text{ cm}^3\)), whereas the rough titration is not.
Mean volume = \(\frac{14.65 + 14.55 + 14.60}{3} = 14.60\text{ cm}^3\).

(c) Equation: \(\text{CH}_3\text{COOH}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{CH}_3\text{COONa}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

Moles of \(\text{NaOH}\) used in titration = \(0.100\text{ mol dm}^{-3} \times 0.01460\text{ dm}^3 = 0.00146\text{ mol}\).

Since stoichiometry is 1:1, moles of \(\text{CH}_3\text{COOH}\) in the \(25.00\text{ cm}^3\) aliquot = \(0.00146\text{ mol}\).

Concentration of diluted vinegar = \(\frac{0.00146\text{ mol}}{0.02500\text{ dm}^3} = 0.0584\text{ mol dm}^{-3}\).

Since the vinegar was diluted by a factor of 10 (\(25.00\text{ cm}^3\) to \(250.0\text{ cm}^3\)), the concentration of the original, undiluted commercial vinegar is:
\(0.0584\text{ mol dm}^{-3} \times 10 = 0.584\text{ mol dm}^{-3}\).

(d) Distilled water remaining in the burette dilutes the sodium hydroxide solution, lowering its actual concentration below \(0.100\text{ mol dm}^{-3}\). Consequently, a larger volume of this dilute \(\text{NaOH}\) solution will be required to reach the end point. This larger titre volume leads to a higher calculated number of moles of acid, and therefore an overestimate (higher calculated value) of the vinegar's concentration.

(a) Award [1] for colorless to (pale) pink. (Reject 'clear' as the initial state).

(b) Award [0.5] for selecting and averageing only Titrations 1, 2, and 3 (concordant titres) and excluding the rough titration; Award [0.75] for correct calculation of the mean: \(14.60\text{ cm}^3\).

(c) Award [1] for calculating the moles of NaOH (\(1.46 \times 10^{-3}\text{ mol}\)) and concentration of diluted vinegar (\(0.0584\text{ mol dm}^{-3}\)); Award [1] for correctly multiplying by the dilution factor of 10 to obtain \(0.584\text{ mol dm}^{-3}\) (accept \(0.58\) to \(0.59\)).

(d) Award [1] for stating that the water dilutes the NaOH solution, so a larger volume of NaOH is needed to reach the end point; Award [1] for concluding that the calculated concentration of ethanoic acid in vinegar is too high / overestimated.

(a) Iodine is a colored product (yellow/brown in solution), while all reactants are colorless. The increase in absorbance over time can be measured using a spectrophotometer to determine the reaction rate. (b)(i) \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\). (b)(ii) The overall order is 2. (c) \(\text{Rate} = (1.15 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}) \times (0.045\text{ mol dm}^{-3}) \times (0.080\text{ mol dm}^{-3}) = 4.14 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). (d) Increasing the temperature increases the average kinetic energy of the particles. This leads to more frequent collisions per unit time and, more importantly, a significantly larger fraction of collisions possessing energy greater than or equal to the activation energy. (e) Step 1 (slow/rate-determining step): \(\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{IO}^- + \text{H}_2\text{O}\). Step 2 (fast): \(\text{IO}^- + \text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + \text{H}_2\text{O}\).

(a) Award [1] for noting that iodine is colored / yellow-brown and reactants are colorless. Award [1] for stating that the rate of increase in color intensity/absorbance is measured over time. (b)(i) Award [1] for \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\). (b)(ii) Award [1] for overall order of 2. (c) Award [1] for \(4.14 \times 10^{-5}\) and [1] for correct units \(\text{mol dm}^{-3}\text{ s}^{-1}\). (d) Award [1] for stating that particles have higher kinetic energy leading to more frequent collisions. Award [1] for stating that a greater fraction of particles have energy \(\ge E_a\). (e) Award [1] for a correct rate-determining step involving one molecule of \(\text{H}_2\text{O}_2\) and one ion of \(\text{I}^-\). Award [1] for a fast step that, when added to the slow step, yields the correct overall equation.

(a) The copper electrode is the anode. Half-equation: \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-\). (b) \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 0.80\text{ V} - 0.34\text{ V} = 0.46\text{ V}\). (c) \(\text{Cu}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)\). (d) Electrons flow from the copper anode to the silver cathode through the external wire. The salt bridge completes the electrical circuit and maintains charge neutrality by allowing anions to migrate toward the anode and cations to migrate toward the cathode. (e) \(Q = I \times t = 1.50\text{ A} \times (45.0 \times 60\text{ s}) = 4050\text{ C}\). \(n(e^-) = \frac{4050\text{ C}}{96500\text{ C mol}^{-1}} = 0.04197\text{ mol}\). Since \(\text{Ag}^{+}(aq) + e^- \rightarrow \text{Ag}(s)\), \(n(\text{Ag}) = 0.04197\text{ mol}\). \(\text{Mass} = 0.04197\text{ mol} \times 107.87\text{ g mol}^{-1} = 4.53\text{ g}\).

(a) Award [1] for identifying copper/Cu as the anode. Award [1] for the oxidation half-equation: \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-\). (b) Award [1] for \(0.46\text{ V}\). (c) Award [1] for correct reactants and products, and [1] for correct balancing and state symbols. (d) Award [1] for stating electrons flow from the copper anode to the silver cathode. Award [1] for stating that the salt bridge completes the circuit / maintains electrical neutrality by ion migration. (e) Award [1] for calculating charge as \(4050\text{ C}\). Award [1] for finding moles of electrons / silver as \(0.042\text{ mol}\). Award [1] for final mass of \(4.53\text{ g}\) (accept 4.50 g to 4.53 g).

(a) The Lewis structure of \(\text{NO}_3^-\space\) shows a central nitrogen double-bonded to one oxygen and single-bonded to two other oxygens. The single-bonded oxygens have formal charges of \(-1\), the nitrogen has a formal charge of \(+1\), and the double-bonded oxygen has a formal charge of \(0\). (b) Electron domain geometry is trigonal planar; molecular geometry is trigonal planar. (c) Experimental evidence shows that all three nitrogen-oxygen bonds are of equal length (intermediate between a single and a double bond). This is due to the delocalization of the \(\pi\) electrons across the three resonance structures, making each bond equivalent. (d) Carbonate also has three resonance structures, yielding a carbon-oxygen bond order of \(1.33\) (or \(4/3\)). Nitrate has a nitrogen-oxygen bond order of \(1.33\). The bond orders are equal because both polyatomic ions have 4 bonding pairs distributed over 3 terminal oxygen atoms.

(a) Award [1] for correct connectivity and octets. Award [1] for correct formal charges (\(-1\) on two oxygens, \(+1\) on nitrogen, \(0\) on one oxygen) or brackets showing a charge of \(-1\). (b) Award [1] for trigonal planar electron domain geometry. Award [1] for trigonal planar molecular geometry. (c) Award [1] for stating that all N-O bond lengths are equal. Award [1] for resonance/delocalization of \(\pi\) electrons. Award [1] for mentioning that the bond order is intermediate (1.33). (d) Award [1] for stating both bond orders are equal (1.33 or 4/3). Award [1] for stating carbonate has 3 resonance structures / 4 bonding pairs over 3 bonds. Award [1] for stating nitrate also has 3 resonance structures / 4 bonding pairs over 3 bonds.

(a) Atomic radius decreases across Period 3. This occurs because the nuclear charge increases as protons are added to the nucleus, while shielding remains approximately constant because electrons are added to the same main energy level. This increases the effective nuclear charge, pulling the outer electrons closer. (b) Phosphorus has a \(3\text{p}^3\) configuration where all three 3p orbitals are singly occupied. Sulfur has a \(3\text{p}^4\) configuration, meaning one 3p orbital contains a pair of electrons. The electrostatic repulsion between these paired electrons (spin-pair repulsion) makes it easier to remove the first electron from sulfur. (c)(i) \(\text{MgO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Mg(OH)}_2(aq)\). The product is basic. (c)(ii) Amphoteric oxide: \(\text{Al}_2\text{O}_3\). Reaction with strong acid (e.g., \(\text{HCl}\)): \(\text{Al}_2\text{O}_3(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2\text{O}(l)\).

(a) Award [1] for stating that atomic radius decreases. Award [1] for stating that nuclear charge / number of protons increases. Award [1] for stating that shielding remains constant / outer electrons are in the same shell, leading to a stronger pull. (b) Award [1] for identifying that sulfur has a paired electron in its 3p orbital (or \(3\text{p}^4\) vs \(3\text{p}^3\)). Award [1] for stating that spin-pair repulsion makes this electron easier to remove. (c)(i) Award [1] for balanced equation: \(\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2\). Award [1] for basic nature of the product. (c)(ii) Award [1] for \(\text{Al}_2\text{O}_3\). Award [2] for a fully balanced equation with any common strong acid (e.g. \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)), deduct [1] if unbalanced or wrong products.

(a)(i) \(\Delta T = 48.8 - 18.4 = 30.4\text{ K}\). \(q = m \cdot c \cdot \Delta T = 150.0 \times 4.18 \times 30.4 = 19060.8\text{ J} = 19.1\text{ kJ}\). (a)(ii) \(M(\text{CH}_3\text{OH}) = 32.05\text{ g mol}^{-1}\). \(n = 1.28 / 32.05 = 0.0399\text{ mol}\). (a)(iii) \(\Delta H_c = -q/n = -19.0608 / 0.0399 = -478\text{ kJ mol}^{-1}\). (b)(i) \(\text{Percentage error} = |(-478 - (-726)) / -726| \times 100\% = 34.2\%\). (b)(ii) Heat lost to the surroundings/beaker; incomplete combustion of methanol. (c) The lid prevents heat loss via evaporation and convection. Copper has a higher thermal conductivity than glass, ensuring faster heat transfer and reducing the time window for heat loss to the air.

(a)(i) Award [1] for calculating \(\Delta T = 30.4\text{ K}\). Award [1] for \(q = 19.1\text{ kJ}\) (or \(19.06\text{ kJ}\)). (a)(ii) Award [1] for \(0.0399\text{ mol}\) (accept \(0.040\text{ mol}\)). (a)(iii) Award [1] for correct division \(19.06 / 0.0399 = 478\text{ kJ mol}^{-1}\). Award [1] for correct negative sign and unit. (b)(i) Award [1] for \(34.2\%\) (accept \(34\%\)). (b)(ii) Award [1] each for any two of: heat loss to surroundings, incomplete combustion (soot formation), evaporation of methanol from the wick before/after weighing. (c) Award [1] for explaining that a lid reduces heat loss via evaporation/convection. Award [1] for explaining that copper conducts heat faster than glass, which limits the time available for heat loss.