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The ozone molecule, \(\text{O}_3\), has three electron domains around the central oxygen atom (one single bond, one double bond, and one lone pair). This results in a trigonal planar electron domain geometry. Due to the extra repulsion from the lone pair, the bond angle is slightly less than \(120^\circ\), measuring approximately \(117^\circ\). In contrast, \(\text{NH}_3\) is trigonal pyramidal with a bond angle of \(107^\circ\), \(\text{CO}_2\) is linear with a bond angle of \(180^\circ\), and \(\text{H}_2\text{S}\) is bent with a bond angle of approximately \(92^\circ\).

Award 1 mark for the correct option A. No partial marks.

The value of the equilibrium constant \(K_c\) is only affected by temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature will shift the equilibrium position to the right to release heat (according to Le Chatelier's principle). This increases the concentration of the products and decreases the concentration of the reactants, thereby increasing the value of \(K_c\) and the yield of \(\text{C}(g)\).

Award 1 mark for the correct option A. No partial marks.

A conjugate acid-base pair consists of two species that differ by a single proton (\(\text{H}^+\)). The conjugate acid has one more proton than its conjugate base. In this reaction, \(\text{H}_2\text{PO}_4^-\) acts as an acid by donating a proton to form its conjugate base, \(\text{HPO}_4^{2-}\). Thus, \(\text{H}_2\text{PO}_4^-\) (acid) and \(\text{HPO}_4^{2-}\) (base) form a conjugate pair.

Award 1 mark for the correct option A. No partial marks.

Assume a 100 g sample of the oxide. The mass of oxygen is 60 g and the mass of sulfur is 100 - 60 = 40 g. Convert mass to moles: n(S) = 40 / 32 = 1.25 mol; n(O) = 60 / 16 = 3.75 mol. Determine the simplest whole-number ratio by dividing by the smallest value: S = 1.25 / 1.25 = 1; O = 3.75 / 1.25 = 3. Therefore, the empirical formula is \(\text{SO}_3\).

Award 1 mark for the correct option C. No partial marks.

Using Hess's Law, we can manipulate the given equations to obtain the target equation: \(\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)\). Keep equation (1) as is: \(\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)\) with \(\Delta H = -394\text{ kJ}\). Reverse equation (2) and divide it by 2: \(\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g)\) with \(\Delta H = -0.5 * (-566\text{ kJ}) = +283\text{ kJ}\). Summing these two modified equations gives: \(\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)\). The total enthalpy change is -394 kJ + 283 kJ = -111 kJ.

Award 1 mark for the correct option A. No partial marks.

The neutral iron atom (\(\text{Fe}\)) has the electron configuration \([\text{Ar}] 4s^2 3d^6\). When transition metals form positive ions, they lose electrons from the outermost s orbital (the 4s orbital) first, before losing any from the d orbital. Removing two electrons to form the \(\text{Fe}^{2+}\) ion results in the configuration \([\text{Ar}] 3d^6\).

Award 1 mark for the correct option B. No partial marks.

Since zinc has a more negative standard electrode potential (\(E^\theta = -0.76\text{ V}\)) compared to copper (\(E^\theta = +0.34\text{ V}\)), zinc is more easily oxidized and acts as the anode. The half-reaction at the anode is \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\), meaning the zinc electrode loses mass as solid zinc is oxidized into aqueous ions. Therefore, option C is correct.

Award 1 mark for the correct option C. No partial marks.

The rate-determining step is the slowest step in the reaction mechanism (Step 1). The rate expression is determined solely by the reactants of this rate-determining step. Since two molecules of \(\text{NO}_2\) react in Step 1, the rate expression is \(\text{Rate} = k[\text{NO}_2]^2\).

Award 1 mark for the correct option A. No partial marks.

To calculate the formal charge (FC) of an atom:
\(\text{FC} = V - (N + B)\)
where \(V\) is the number of valence electrons of the free atom, \(N\) is the number of non-bonding valence electrons, and \(B\) is the number of bonding pairs (or single bonds).

- **Sulfur (S):** Group 16, so \(V = 6\). In this structure, S has 3 lone pairs (6 non-bonding electrons) and 1 single bond.
\(\text{FC} = 6 - (6 + 1) = -1\)
- **Carbon (C):** Group 14, so \(V = 4\). Carbon has 0 lone pairs and 4 bonds (one single, one triple).
\(\text{FC} = 4 - (0 + 4) = 0\)
- **Nitrogen (N):** Group 15, so \(V = 5\). Nitrogen has 1 lone pair (2 non-bonding electrons) and a triple bond (3 bonds).
\(\text{FC} = 5 - (2 + 3) = 0\)

Thus, the formal charges are \(-1\), \(0\), and \(0\) respectively.

[1 mark] awarded for selecting option A.
[0 marks] for any other option.

In the reverse reaction:
\(\text{HPO}_4^{2-}(\text{aq}) + \text{H}_2\text{CO}_3(\text{aq}) \rightarrow \text{H}_2\text{PO}_4^-(\text{aq}) + \text{HCO}_3^-(\text{aq})\)

According to the Brønsted-Lowry theory, an acid is a proton (\(\text{H}^+\)) donor. Here, \(\text{H}_2\text{CO}_3\) donates a proton to \(\text{HPO}_4^{2-}\) to form \(\text{HCO}_3^-\) and \(\text{H}_2\text{PO}_4^-\). Thus, \(\text{H}_2\text{CO}_3\) acts as the Brønsted-Lowry acid in the reverse process.

[1 mark] awarded for selecting option D.
[0 marks] for any other option.

1. Determine the anode and cathode:
- The half-cell with the more positive reduction potential undergoes reduction at the cathode: \(\text{Fe}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Fe}(\text{s})\) (\(E^\theta = -0.44\text{ V}\)).
- The half-cell with the more negative reduction potential undergoes oxidation at the anode: \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\) (\(E^\theta = -0.74\text{ V}\)).

2. Calculate the cell potential:
\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = -0.44\text{ V} - (-0.74\text{ V}) = +0.30\text{ V}\).

3. Identify the direction of electron flow:
Electrons always flow from the anode (where oxidation occurs and electrons are released) to the cathode (where reduction occurs and electrons are consumed) in the external circuit. Thus, electrons flow from the chromium electrode to the iron electrode.

[1 mark] awarded for selecting option A.
[0 marks] for any other option.

The rate-determining step is the slowest step in the mechanism (Step 1). Therefore, the rate expression is determined directly by the reactants involved in this step:
\(\text{Rate} = k[\text{A}][\text{B}]\).

Reaction intermediates are species that are produced in an earlier step and consumed in a subsequent step, and thus do not appear in the overall balanced equation. Looking at the steps, \(\text{AB}\) is produced in Step 1 and consumed in Step 2; \(\text{AB}_2\) is produced in Step 2 and consumed in Step 3. Therefore, both \(\text{AB}\) and \(\text{AB}_2\) are intermediates.

[1 mark] awarded for selecting option A.
[0 marks] for any other option.

1. Calculate heat released, \(q\):
\(q = m \cdot c \cdot \Delta T\)
\(m = 100\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100\text{ g}\)
\(q = 100\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.0\text{ K} = 4180\text{ J} = 4.18\text{ kJ}\)

2. Calculate moles of limiting reactant, \(n\):
\(n(\text{CuSO}_4) = 0.10\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.010\text{ mol}\)

3. Calculate molar enthalpy change, \(\Delta H\):
Since the temperature increases, the reaction is exothermic, making \(\Delta H\) negative:
\(\Delta H = -\frac{q}{n} = -\frac{4.18\text{ kJ}}{0.010\text{ mol}} = -418\text{ kJ mol}^{-1}\).

[1 mark] awarded for selecting option B.
[0 marks] for any other option.

Phosphorus has the electron configuration \([\text{Ne}] 3s^2 3p^3\) with three unpaired electrons in three separate, degenerate \(3p\) orbitals. Sulfur has the electron configuration \([\text{Ne}] 3s^2 3p^4\). In sulfur, one of the \(3p\) orbitals contains a pair of electrons. The mutual electrostatic repulsion between these two paired electrons in the same orbital makes it easier to remove one of them, resulting in a lower first ionization energy for sulfur compared to phosphorus.

[1 mark] awarded for selecting option C.
[0 marks] for any other option.

The Gibbs free energy change is given by the relation:
\(\Delta G = \Delta H - T\Delta S\)

For a reaction to be spontaneous, \(\Delta G\) must be negative (\(\Delta G < 0\)).
- If \(\Delta H > 0\) (endothermic) and \(\Delta S > 0\) (increase in entropy):
- At low temperatures, \(T\Delta S\) is small, so \(\Delta H > T\Delta S\), making \(\Delta G > 0\) (non-spontaneous).
- At high temperatures, \(T\Delta S\) becomes larger than \(\Delta H\), making \(\Delta G < 0\) (spontaneous).

Thus, option A is the correct set of conditions.

[1 mark] awarded for selecting option A.
[0 marks] for any other option.

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

Substitute the given values into the equation:
\(\frac{P \cdot V}{T} = \frac{0.5P \cdot V_2}{2T}\)

Simplify the expression by cancelling \(P\) and \(T\) from both sides:
\(V = \frac{0.5 \cdot V_2}{2}\)
\(V = 0.25 \cdot V_2\)
\(V_2 = 4V\)

Therefore, the new volume of the gas is \(4V\).

[1 mark] awarded for selecting option D.
[0 marks] for any other option.

1. From the balanced chemical equation, 1 mole of \(\text{C}_3\text{H}_8\) reacts to produce 4 moles of \(\text{H}_2\text{O}\).
2. Therefore, \(0.20\text{ mol}\) of \(\text{C}_3\text{H}_8\) produces \(0.20 \times 4 = 0.80\text{ mol}\) of \(\text{H}_2\text{O}\).
3. The molar mass of water, \(\text{H}_2\text{O}\), is approximately \(18.0\text{ g mol}^{-1}\).
4. Mass of water = \(0.80\text{ mol} \times 18.0\text{ g mol}^{-1} = 14.4\text{ g}\).

[1 mark] for correct choice C.
Award 0 marks for any other option.

Sulfur tetrafluoride (\(\text{SF}_4\)) has 5 electron domains around the central sulfur atom (4 bonding pairs and 1 lone pair). This arrangement is based on a trigonal bipyramidal electron domain geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a see-saw molecular geometry.

- \(\text{CF}_4\) has 4 electron domains (tetrahedral).
- \(\text{XeF}_4\) has 6 electron domains (square planar with 2 lone pairs).
- \(\text{ClF}_4^-\) also has 6 electron domains (square planar with 2 lone pairs).

[1 mark] for correct choice A.

1. Compare Experiment 1 and 2: \([\text{B}]\) is kept constant while \([\text{A}]\) is doubled. The rate doubles (from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\)). This indicates a first-order dependence on \([\text{A}]\).
2. Compare Experiment 1 and 3: \([\text{A}]\) is kept constant while \([\text{B}]\) is doubled. The rate quadruples (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). This indicates a second-order dependence on \([\text{B}]\).
3. The rate expression is: Rate = \(k[\text{A}]^1[\text{B}]^2\).
4. The overall order of reaction is the sum of individual orders: \(1 + 2 = 3\).

[1 mark] for correct choice C.

In the electrolysis of concentrated aqueous \(\text{NaCl}\):
- At the anode (positive electrode), both \(\text{Cl}^-\text{(aq)}\) and \(\text{H}_2\text{O(l)}\) are present. Due to high concentration, chloride ions are preferentially oxidized to form chlorine gas, \(\text{Cl}_2\text{(g)}\).
- At the cathode (negative electrode), both \(\text{Na}^+\text{(aq)}\) and \(\text{H}_2\text{O(l)}\) (or \(\text{H}^+\)) are present. Water is more easily reduced than sodium ions due to its higher reduction potential, preferentially forming hydrogen gas, \(\text{H}_2\text{(g)}\).

[1 mark] for correct choice B.

1. Initially, \(\text{pH} = 2.0\), which means \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2}\text{ mol dm}^{-3} = 0.01\text{ mol dm}^{-3}\).
2. The solution is diluted from \(10.0\text{ cm}^3\) to \(1000\text{ cm}^3\), representing a dilution factor of: \(\frac{1000}{10.0} = 100\).
3. After dilution, the concentration of hydrogen ions becomes: \([\text{H}^+] = \frac{10^{-2}}{100} = 10^{-4}\text{ mol dm}^{-3}\).
4. The new pH is: \(\text{pH} = -\log_{10}(10^{-4}) = 4.0\).

[1 mark] for correct choice C.

To find the standard enthalpy of formation of carbon monoxide, we need the enthalpy change for the reaction:
\(\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}\)

Using Hess's Law:
- Keep the first equation as is: \(\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H_1^\theta = -394\text{ kJ mol}^{-1}\)
- Reverse the second equation: \(\text{CO}_2\text{(g)} \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad \Delta H_2^\theta = +283\text{ kJ mol}^{-1}\)
- Add the two equations together:
\(\text{C(s)} + \text{O}_2\text{(g)} + \text{CO}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)}\)
Simplifies to:
\(\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}\)

Calculate the net enthalpy:
\(\Delta H^\theta = -394 + 283 = -111\text{ kJ mol}^{-1}\).

[1 mark] for correct choice B.

- Phosphorus has the valence electron configuration \(3\text{s}^2 3\text{p}^3\) where the three \(3\text{p}\) orbitals are each singly occupied (half-filled stability).
- Sulfur has the valence electron configuration \(3\text{s}^2 3\text{p}^4\) where one of the \(3\text{p}\) orbitals is doubly occupied.
- The pairing of electrons in this orbital results in mutual electrostatic repulsion, making it easier to remove one of these paired electrons from sulfur compared to a single electron from phosphorus, despite sulfur's higher nuclear charge.

[1 mark] for correct choice C.

- The only factor that changes the numerical value of the equilibrium constant, \(K_c\), is temperature.
- Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right to favor the exothermic direction (Le Chatelier's principle), which increases the yield of \(\text{SO}_3\) and consequently increases the value of \(K_c\).
- Increasing the pressure or concentration of reactants shifts the equilibrium to the right (increasing yield) but does not affect the value of \(K_c\).

[1 mark] for correct choice B.

One formula unit of aluminium sulfate, \(\text{Al}_2(\text{SO}_4)_3\), contains 12 oxygen atoms. Therefore, 1 mole of \(\text{Al}_2(\text{SO}_4)_3\) contains 12 moles of oxygen atoms. For \(0.10\text{ mol}\) of \(\text{Al}_2(\text{SO}_4)_3\), the amount of oxygen atoms is \(0.10 \times 12 = 1.2\text{ mol}\). The total number of oxygen atoms is \(1.2 \times L\).

[1 mark] for selecting the correct option C.

Bond length decreases as bond order increases. In \(\text{N}_2\text{H}_4\), there is a single nitrogen-to-nitrogen bond (bond order 1). In \(\text{N}_2\text{F}_2\) and \(\text{N}_2\text{H}_2\), there is a double bond (bond order 2). In molecular nitrogen, \(\text{N}_2\), there is a triple bond (bond order 3). Therefore, \(\text{N}_2\) has the shortest bond length.

[1 mark] for selecting the correct option C.

A conjugate acid-base pair differs by only a single proton (\(\text{H}^+\)). Here, \(\text{H}_2\text{PO}_4^-\) is the acid and \(\text{HPO}_4^{2-}\) is its conjugate base, because \(\text{H}_2\text{PO}_4^-\) can donate a proton to form \(\text{HPO}_4^{2-}\).

[1 mark] for selecting the correct option B.

The total area under the Maxwell-Boltzmann curve represents the total number of particles, which remains constant. As temperature increases, the average kinetic energy increases, shifting the peak to the right. Because the total area must remain constant, the peak must also shift down (be lower).

[1 mark] for selecting the correct option B.

Let the oxidation state of sulfur be \(x\). Oxygen has an oxidation state of \(-2\). The sum of oxidation states is equal to the overall charge of the ion: \(2x + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).

[1 mark] for selecting the correct option A.

The heat released by the reaction, \(q\), is proportional to the moles of reactants, which is halved when volumes are halved. However, the total mass (and volume) of the solution being heated is also halved. Since \(\Delta T = \frac{q}{m \times c}\), halving both \(q\) and \(m\) results in the same temperature change, \(\Delta T\).

[1 mark] for selecting the correct option B.

Iron (\(\text{Fe}\)) has atomic number 26, with ground-state electron configuration \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 4\text{s}^2 3\text{d}^6\). When forming a transition metal cation, electrons are lost first from the outermost \(4\text{s}\) orbital before the \(3\text{d}\) orbital. Thus, the \(\text{Fe}^{2+}\) ion loses the two \(4\text{s}\) electrons to give \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6\).

[1 mark] for selecting the correct option B.

Down Group 17, the molecular size increases, which increases the strength of London dispersion forces between the diatomic molecules. This results in an increase in boiling point. First ionization energy, electronegativity, and reactivity with metals all decrease down the group.

[1 mark] for selecting the correct option C.

A conjugate acid-base pair consists of two chemical species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, \(\text{HPO}_4^{2-}\) acts as an acid by donating a proton to form its conjugate base, \(\text{PO}_4^{3-}\). Therefore, \(\text{HPO}_4^{2-}\) and \(\text{PO}_4^{3-}\) are a conjugate acid-base pair.

Award [1] for correct answer C. [0] for any other option.

Let the initial rate be \(R_1 = k[\text{X}][\text{Y}]^2\). If the concentration of \(\text{X}\) is doubled to \(2[\text{X}]\) and the concentration of \(\text{Y}\) is halved to \(0.5[\text{Y}]\), the new rate \(R_2\) becomes: \(R_2 = k(2[\text{X}])(0.5[\text{Y}])^2 = k \cdot 2[\text{X}] \cdot 0.25[\text{Y}]^2 = 0.5 \cdot k[\text{X}][\text{Y}]^2 = 0.5 R_1\). Thus, the rate decreases by a factor of 2.

Award [1] for correct answer A. [0] for any other option.

Iron (\(\text{Fe}\), Z = 26) has the ground-state electron configuration \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). When forming the transition metal cation \(\text{Fe}^{3+}\), three electrons are lost. Transition metals always lose their outer-shell s electrons first before losing d electrons. Therefore, two electrons are removed from the \(4\text{s}\) subshell, and one electron is removed from the \(3\text{d}\) subshell, leaving \([\text{Ar}] 3\text{d}^5\).

Award [1] for correct answer A. [0] for any other option.

Use the equation \(q = m \cdot c \cdot \Delta T\). Here, the mass of water \(m = 50.0\text{ g}\), the specific heat capacity \(c = 4.18\text{ J g}^{-1}\ \text{K}^{-1}\), and the temperature change \(\Delta T = 40.0 - 20.0 = 20.0\ ^\circ\text{C}\) (or \(20.0\text{ K}\)). Calculation: \(q = 50.0 \times 4.18 \times 20.0 = 4180\text{ J} = 4.18\text{ kJ}\).

Award [1] for correct answer A. [0] for any other option.

\(\text{NH}_3\) has 5 valence electrons on nitrogen, bonding to 3 hydrogens, leaving 1 lone pair. This gives a steric number of 4 (tetrahedral electron-pair geometry) and a trigonal pyramidal molecular geometry. \(\text{BF}_3\) is trigonal planar, \(\text{CH}_4\) is tetrahedral, and \(\text{H}_2\text{O}\) is bent.

Award [1] for correct answer B. [0] for any other option.

For a reaction to be spontaneous, the Gibbs free energy change must be negative (\(\Delta G^\ominus < 0\)). Under standard conditions, \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus < 0\), which rearranges to \(T > \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting \(\Delta H^\ominus\) to Joules: \(\Delta H^\ominus = 80 \times 10^3\text{ J mol}^{-1} = 80000\text{ J mol}^{-1}\). Therefore, \(T > \frac{80000}{200} = 400\text{ K}\). The reaction becomes spontaneous at temperatures above \(400\text{ K}\).

Award [1] for correct answer A. [0] for any other option.

In the thiosulfate ion (\(\text{S}_2\text{O}_3^{2-}\)), let \(x\) be the average oxidation state of sulfur. Oxygen typically has an oxidation state of \(-2\). The sum of the oxidation states must equal the overall charge of the ion: \(2(x) + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).

Award [1] for correct answer A. [0] for any other option.

One molecule of ethanol, \(\text{C}_2\text{H}_5\text{OH}\), contains \(2 \text{ (carbon)} + 5 \text{ (hydrogen)} + 1 \text{ (oxygen)} + 1 \text{ (hydrogen)} = 9\) atoms. To find the total number of atoms in \(0.20\text{ mol}\) of ethanol: \(\text{Number of atoms} = 0.20\text{ mol} \times 9\text{ atoms/molecule} \times 6.0 \times 10^{23}\text{ molecules/mol} = 1.8\text{ mol of atoms} \times 6.0 \times 10^{23}\text{ atoms/mol} = 1.08 \times 10^{24}\) atoms. Let's re-verify the chemical formula atoms count: \(\text{C}_2\text{H}_5\text{OH}\) has \(2 + 5 + 1 + 1 = 9\) atoms. Wait, \(2 + 6 + 1 = 9\) atoms. \(0.20 \times 9 \times 6.0 \times 10^{23} = 1.08 \times 10^{24}\) atoms. Wait, let's recalculate if the option says \(9.6 \times 10^{23}\). If we counted 8 atoms (like \(\text{C}_2\text{H}_5\text{O}\) by mistake), it would be \(9.6 \times 10^{23}\). Let's check: \(\text{C}_2\text{H}_5\text{OH}\) has 2 carbons, 6 hydrogens, and 1 oxygen = 9 atoms. Thus, \(0.20 \text{ mol} \times 9 = 1.8 \text{ mol}\) of atoms, which is \(1.8 \times 6.0 \times 10^{23} = 1.08 \times 10^{24}\). If we round to \(1.1 \times 10^{24}\) atoms, that is option C. Let's use \(1.1 \times 10^{24}\) as the correct answer (C).

Award [1] for correct answer C. [0] for any other option.

(a) For \(\text{SO}_2\): Total valence electrons = 18. Sulfur can expand its octet. The most stable Lewis structure minimizing formal charges (all formal charges = 0) has sulfur double-bonded to both oxygens with one lone pair on sulfur: \(\text{O}=\ddot{\text{S}}=\text{O}\).
For \(\text{SO}_3\): Total valence electrons = 24. The most stable structure with formal charges of zero has sulfur double-bonded to all three oxygens: three double bonds, zero lone pairs on sulfur.

(b) \(\text{SO}_2\):
- Electron domains around sulfur = 3 (two double bonds, one lone pair).
- Electron domain geometry = trigonal planar.
- Molecular geometry = bent (V-shaped).
- Bond angle = slightly less than \(120^\circ\) (typically \(119^\circ\)) due to lone pair-bonding pair repulsion.
\(\text{SO}_3\):
- Electron domains around sulfur = 3 (three double bonds, zero lone pairs).
- Electron domain geometry = trigonal planar.
- Molecular geometry = trigonal planar.
- Bond angle = \(120^\circ\).

(c) In both molecules, the actual bonds are resonance hybrids of the contributing Lewis structures, meaning all S-O bonds within each molecule are identical in length and intermediate between single and double bonds. In \(\text{SO}_2\), the average bond order is 1.5; in \(\text{SO}_3\), the average bond order is 1.33. Comparing the actual bond lengths: the S-O bonds in \(\text{SO}_3\) (bond order 1.33) are slightly longer than those in \(\text{SO}_2\) (bond order 1.5) due to the lower bond order.

- (a) [4 marks]:
- Correct Lewis structure for \(\text{SO}_2\) with 2 double bonds and 1 lone pair on S [1 mark].
- All non-bonding electrons shown on O atoms [1 mark].
- Correct Lewis structure for \(\text{SO}_3\) with 3 double bonds and 0 lone pairs on S [1 mark].
- All non-bonding electrons shown on O atoms [1 mark].
- (b) [4 marks]:
- \(\text{SO}_2\): trigonal planar domain geometry, bent molecular shape [1 mark]; bond angle < \(120^\circ\) (accept \(115^\circ - 119^\circ\)) [1 mark].
- \(\text{SO}_3\): trigonal planar domain geometry, trigonal planar shape [1 mark]; bond angle \(120^\circ\) [1 mark].
- (c) [2 marks]:
- Statement that resonance results in equal S-O bond lengths within each molecule [1 mark].
- Deducting that \(\text{SO}_2\) has a higher bond order (1.5) than \(\text{SO}_3\) (1.33), hence \(\text{SO}_2\) has shorter S-O bonds [1 mark].

(a) Let \(\text{HA}\) represent propanoic acid. \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\).
Assuming \([\text{H}^+] \approx [\text{A}^-]\) and \([\text{HA}]_{eq} \approx [\text{HA}]_{initial} = 0.100\text{ mol dm}^{-3}\):
\(1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.100}\)
\([\text{H}^+] = \sqrt{1.35 \times 10^{-6}} = 1.162 \times 10^{-3}\text{ mol dm}^{-3}\n\text{pH} = -\log(1.162 \times 10^{-3}) = 2.93\).

(b) Adding \(12.50\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \(\text{NaOH}\) to \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid represents the half-equivalence point because exactly half of the acid has been neutralized.
At the half-equivalence point, \([\text{HA}] = [\text{A}^-]\).
Therefore, \(K_a = [\text{H}^+]\), and so \(\text{pH} = \text{p}K_a\).
\(\text{pH} = -\log(1.35 \times 10^{-5}) = 4.87\).

(c) At the equivalence point, all propanoic acid has reacted with hydroxide ions to form propanoate ions, \(\text{CH}_3\text{CH}_2\text{COO}^-\), which is a weak conjugate base.
The propanoate ion undergoes hydrolysis in water to produce hydroxide ions, making the solution basic (pH > 7).
Equation: \(\text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq})\).

- (a) [3 marks]:
- Set up equation: \(K_a = \frac{[\text{H}^+]^2}{[\text{HA}]}\) [1 mark].
- Calculate \([\text{H}^+] = 1.16 \times 10^{-3}\text{ mol dm}^{-3}\) [1 mark].
- Calculate \(\text{pH} = 2.93\) [1 mark]. Deduct 1 mark if no assumption is stated.
- (b) [3 marks]:
- Identify this as the half-equivalence point [1 mark].
- State that at the half-equivalence point, \(\text{pH} = \text{p}K_a\) [1 mark].
- Calculate \(\text{pH} = 4.87\) [1 mark].
- (c) [4 marks]:
- State that the equivalence point pH is greater than 7 [1 mark].
- Explain that the conjugate base (propanoate ion, \(\text{CH}_3\text{CH}_2\text{COO}^-\)) is formed [1 mark].
- Write the correct equilibrium reaction: \(\text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq})\) [1 mark].
- State that production of \(\text{OH}^-\)\ yields an alkaline solution [1 mark].

(a) The anode is chromium and the cathode is copper.
Cell notation: \(\text{Cr}(\text{s}) \mid \text{Cr}^{3+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})\).

(b) Oxidation half-reaction: \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\)
Reduction half-reaction: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)
Balanced overall reaction: \(2\text{Cr}(\text{s}) + 3\text{Cu}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cu}(\text{s})\).
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.34\text{ V} - (-0.74\text{ V}) = +1.08\text{ V}\).

(c) Electrons flow through the external wire from the chromium anode to the copper cathode. The salt bridge completes the circuit and maintains electrical neutrality. Anions in the salt bridge migrate towards the anode compartment (chromium half-cell) to balance the positive \(\text{Cr}^{3+}\) ion build-up, while cations migrate towards the cathode compartment (copper half-cell) to replace consumed \(\text{Cu}^{2+}\) ions.

(d) \(\Delta G^\theta = -nFE^\theta_{\text{cell}}\)
With \(n = 6\) moles of electrons:
\(\Delta G^\theta = -6 \times 96,500\text{ C mol}^{-1} \times 1.08\text{ V} = -625,320\text{ J mol}^{-1} = -625\text{ kJ mol}^{-1}\).

- (a) [2 marks]:
- Correct representation of anode on left and cathode on right: \(\text{Cr}(\text{s}) \mid \text{Cr}^{3+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})\) [1 mark].
- Correct state symbols and phase boundaries [1 mark].
- (b) [3 marks]:
- Correctly balanced overall equation [2 marks].
- Calculation of standard cell potential: \(E^\theta_{\text{cell}} = +1.08\text{ V}\) [1 mark].
- (c) [3 marks]:
- Electron flow from Cr electrode to Cu electrode in external circuit [1 mark].
- Salt bridge maintains electrical neutrality [1 mark].
- Anions migrate to Cr half-cell AND cations migrate to Cu half-cell [1 mark].
- (d) [2 marks]:
- Use of \(\Delta G^\theta = -nFE^\theta_{\text{cell}}\) with \(n = 6\) [1 mark].
- Value of \(-625\text{ kJ mol}^{-1}\) [1 mark]. Deduct 1 mark if negative sign is missing or units are incorrect.

(a) Comparing Exp 1 and Exp 2: \([\text{CO}]\) is kept constant. \([\text{NO}_2]\) doubles and the rate increases by a factor of \(4 = 2^2\). Thus, the reaction is second order with respect to \(\text{NO}_2\).
Comparing Exp 2 and Exp 3: \([\text{NO}_2]\) is kept constant. \([\text{CO}]\) doubles and the rate remains unchanged. Thus, the reaction is zero order with respect to \(\text{CO}\).
Rate expression: \(\text{Rate} = k[\text{NO}_2]^2\).

(b) Using Exp 1: \(2.1 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2\).
\(k = 2.1 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) Proposed Mechanism:
Step 1 (slow, rate-determining): \(\text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{NO}_3(\text{g}) + \text{NO}(\text{g})\)
Step 2 (fast): \(\text{NO}_3(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{NO}_2(\text{g}) + \text{CO}_2(\text{g})\)
Sum: \(\text{NO}_2(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{NO}(\text{g}) + \text{CO}_2(\text{g})\).
This matches the rate-determining step reactant stoichiometry.

(d) The potential energy profile shows a two-peak reaction pathway where the first peak representing Step 1 (slow) is higher in energy than the second peak representing Step 2 (fast), with products lower in energy than reactants.

- (a) [3 marks]:
- Deduce second-order with respect to \(\text{NO}_2\) [1 mark].
- Deduce zero-order with respect to \(\text{CO}\) [1 mark].
- Correct rate expression: \(\text{Rate} = k[\text{NO}_2]^2\) [1 mark].
- (b) [2 marks]:
- Value of \(k = 2.1 \times 10^{-2}\) [1 mark].
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) [1 mark].
- (c) [3 marks]:
- Step 1: \(2\text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}\) (slow) [1 mark].
- Step 2: \(\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2\) (fast) [1 mark].
- Verify sum matches overall reaction [1 mark].
- (d) [2 marks]:
- Exothermic profile with correct axis labels [1 mark].
- Two peaks, with the first peak higher than the second peak [1 mark].

(a) First electron affinity of chlorine is the enthalpy change when one mole of gaseous chlorine atoms gains one mole of electrons to form one mole of gaseous chloride ions: \(\text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g})\).

(b) The Born-Haber cycle should show a closed loop or vertical energy diagram with levels:
- Baseline: \(\text{Ca}(\text{s}) + \text{Cl}_2(\text{g})\)
- Downward to \(\text{CaCl}_2(\text{s})\) representing \(\Delta H_{\text{f}}^\theta = -796\text{ kJ mol}^{-1}\)
- Upward steps: atomization of Ca \((+178)\), ionization of Ca \((+590 + 1145)\), dissociation of \(\text{Cl}_2\) \((+242)\)
- Downward step: electron affinity of 2 moles of \(\text{Cl}\) atoms \((2 \times -349 = -698)\)
- Downward/upward path connecting gaseous ions \(\text{Ca}^{2+}(\text{g}) + 2\text{Cl}^-(\text{g})\) to the solid lattice, representing lattice enthalpy.

(c) Traveling the cycle using Hess's Law:
\(\Delta H_{\text{f}}^0 = \Delta H_{\text{at}}^\theta(\text{Ca}) + \Delta H_{\text{IE1}}^\theta + \Delta H_{\text{IE2}}^\theta + E(\text{Cl}-\text{Cl}) + 2\Delta H_{\text{EA}}^\theta(\text{Cl}) - \Delta H_{\text{latt}}^\theta\)
\(-796 = 178 + 590 + 1145 + 242 + 2(-349) - \Delta H_{\text{latt}}^\theta\)
\(-796 = 1457 - \Delta H_{\text{latt}}^\theta\)
\(\Delta H_{\text{latt}}^\theta = 1457 - (-796) = +2253\text{ kJ mol}^{-1}\).

(d) Calcium iodide contains the highly polarizable iodide ion (\(\text{I}^-\)). The small, highly charged calcium ion (\(\text{Ca}^{2+}\)) polarizes the electron cloud of the iodide ion, introducing covalent character into the bonding which stabilizes the lattice and increases experimental lattice energy.

- (a) [2 marks]:
- Enthalpy change for 1 mole of gaseous atoms gaining 1 mole of electrons [1 mark].
- Equation with state symbols: \(\text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g})\) [1 mark].
- (b) [4 marks]:
- Correct representation of elements in standard states and enthalpy of formation [1 mark].
- Correct atomization of Ca and dissociation of Cl2 (using +242) [1 mark].
- Correct first and second ionization steps of Ca [1 mark].
- Correct electron affinity (multiplied by 2) and lattice enthalpy transition [1 mark].
- (c) [3 marks]:
- Correctly set up the cycle equation [1 mark].
- Correct substitution of values, factoring in 2 moles of Cl EA [1 mark].
- Calculated value: \(+2253\text{ kJ mol}^{-1}\) [1 mark].
- (d) [1 mark]:
- Suggests polarization of iodide ion / covalent character in calcium iodide [1 mark].

(a) Total mass = \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\).
\(\Delta T = 26.1 - 19.5 = 6.6\text{ K}\).
\(q = mc\Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.6\text{ K} = 2758.8\text{ J}\) (or \(2.76\text{ kJ}\)).

(b) Moles of \(\text{H}^+\) reacting = \(0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
Moles of water formed = \(0.0500\text{ mol}\).
\(\Delta H_{\text{neut}} = -\frac{q}{\text{moles of water}} = -\frac{2758.8\text{ J}}{0.0500\text{ mol}} = -55,176\text{ J mol}^{-1} = -55.2\text{ kJ mol}^{-1}\).

(c) Systematic errors:
1. Heat loss to surroundings: Lower measured maximum temperature, leading to a less exothermic (less negative) calculated \(\Delta H_{\text{neut}}\).
2. Specific heat capacity of solution assumed to be same as water (\(4.18\)): Ions in solution decrease the actual specific heat capacity slightly. Using \(4.18\) overestimates heat released, making the value appear more exothermic.

(d) Ethanoic acid is a weak acid and only partially dissociates in aqueous solution. Energy is absorbed in the endothermic dissociation of undissociated ethanoic acid molecules, reducing the net heat energy released during neutralization.

- (a) [2 marks]:
- Correct mass of \(100.0\text{ g}\) and temperature difference \(\Delta T = 6.6\text{ K}\) [1 mark].
- Correctly calculated heat energy released (\(q = 2758.8\text{ J}\)) [1 mark].
- (b) [3 marks]:
- Moles of reactants/water calculated as \(0.0500\text{ mol}\) [1 mark].
- Division of heat energy by moles [1 mark].
- Correct negative sign and final value \(-55.2\text{ kJ mol}^{-1}\) [1 mark].
- (c) [3 marks]:
- Identify heat loss to surroundings [1 mark] and explain its less-exothermic impact on the calculated value [1 mark].
- Identify another valid source of error (e.g., specific heat capacity assumption) [1 mark].
- (d) [2 marks]:
- State that ethanoic acid is a weak/partially dissociated acid [1 mark].
- Explain that dissociation is endothermic, absorbing some of the neutralization energy [1 mark].

(a) The trend in the acid-base character of Period 3 oxides changes from basic (sodium and magnesium oxides) to amphoteric (aluminum oxide) to acidic (silicon, phosphorus, sulfur, and chlorine oxides).
- Reaction of sodium oxide with water:
\(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq})\)
- Reaction of tetraphosphorus decaoxide with water:
\(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\)
- Reaction of aluminum oxide with hydrochloric acid:
\(\text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)

(b) Liquid sodium oxide (\(\text{Na}_2\text{O}\)) is an electrical conductor because it has a giant ionic structure, and melting allows the \(\text{Na}^+\) and \(\text{O}^{2-}\) ions to move freely. Liquid silicon dioxide (\(\text{SiO}_2\)) is a non-conductor because it has a giant covalent macromolecular structure where atoms are linked by strong covalent bonds; there are no mobile ions or free electrons.

(c) First ionization energy increases across Period 3 due to increasing nuclear charge and decreasing atomic radius. Exceptions occur at Al (outermost electron in higher-energy \(3\text{p}\) orbital, which is more shielded than \(3\text{s}\)) and S (electron-electron pairing repulsion in a \(3\text{p}\) orbital, making it easier to remove).

- (a) [5 marks]:
- Describe trend from basic to amphoteric to acidic [1 mark].
- Correct balanced equation for \(\text{Na}_2\text{O}\) + water [1 mark].
- Correct balanced equation for \(\text{P}_4\text{O}_{10}\) + water [1 mark].
- Correct balanced equation for \(\text{Al}_2\text{O}_3\) + acid [1 mark].
- State that \(\text{Al}_2\text{O}_3\) is amphoteric [1 mark].
- (b) [3 marks]:
- Liquid \(\text{Na}_2\text{O}\) conducts due to free-moving ions in the molten state [1 mark].
- Liquid \(\text{SiO}_2\) does not conduct because it is giant covalent and has no free ions/electrons [1 mark].
- Relate directly to bonding types [1 mark].
- (c) [2 marks]:
- State that ionization energy generally increases due to increasing nuclear charge [1 mark].
- Explain either the Al or S exception with orbital configurations [1 mark].

(a) 1-chlorobutane structural formula:
\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{Cl}\)
Classification: Primary halogenoalkane.

2-chloro-2-methylpropane structural formula:
\((\text{CH}_3)_3\text{C}-\text{Cl}\)
Classification: Tertiary halogenoalkane.

(b) Mechanism of \(\text{S}_\text{N}2\) substitution:
- \(\text{OH}^-\)\ nucleophile performs a backside attack on the partially positive carbon (\(\text{C}^{\delta+}\)).
- A curly arrow goes from the lone pair on the oxygen of \(\text{OH}^-\)\ to the \(\text{C}^{\delta+}\).
- A concurrent curly arrow goes from the C-Cl bond to the Cl atom.
- The transition state shows the pentacoordinate central carbon with partial bonds to OH and Cl, partial negative charges (\(\delta-\)) on OH and Cl, and brackets with a negative sign.
- Products are butan-1-ol and chloride ion.

(c) 2-chloro-2-methylpropane undergoes nucleophilic substitution via the \(\text{S}_\text{N}1\) mechanism. It cannot undergo \(\text{S}_\text{N}2\) due to steric hindrance from the three bulky methyl groups which block the nucleophile's backside attack. Furthermore, the intermediate tertiary carbocation is highly stable because the three electron-donating methyl groups disperse the positive charge through the inductive effect.

- (a) [4 marks]:
- Correct structure of 1-chlorobutane [1 mark] and primary classification [1 mark].
- Correct structure of 2-chloro-2-methylpropane [1 mark] and tertiary classification [1 mark].
- (b) [4 marks]:
- Curly arrow from \(\text{OH}^-\)\ lone pair to carbon [1 mark].
- Correct dipoles on C-Cl and arrow from C-Cl bond to Cl [1 mark].
- Correct transition state (pentacoordinate carbon, correct partial charges and overall charge) [1 mark].
- Correct products [1 mark].
- (c) [2 marks]:
- Identify \(\text{S}_\text{N}1\) mechanism for 2-chloro-2-methylpropane [1 mark].
- Explain why in terms of steric hindrance or stability of the tertiary carbocation [1 mark].

(a) Cobalt metal, \(\text{Co}\), has the lower standard reduction potential (\(-0.28\text{ V}\)) compared to silver (\(+0.80\text{ V}\)). Thus, cobalt is oxidized at the anode: \(\text{Co(s)} \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{e}^-\).

(b) Multiplying the silver reduction half-equation by 2 and combining it with the cobalt oxidation half-equation yields: \(\text{Co(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{Ag(s)}\). The cell potential is \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V}\).

(c) Electrons flow through the external circuit from the negative electrode (anode, cobalt) to the positive electrode (cathode, silver). The salt bridge completes the electrical circuit and maintains electrical neutrality within both half-cells by allowing anions to migrate to the anode and cations to migrate to the cathode.

(d) Increasing the concentration of the product ion, \(\text{Co}^{2+}(\text{aq})\), shifts the position of the cell equilibrium to the left (reactants side) according to Le Chatelier's principle. This decreases the tendency of the forward reaction to occur, thus decreasing the overall cell potential.

(e) The standard conditions for the standard hydrogen electrode (SHE) are:
1. Temperature of \(298\text{ K}\) / \(25\text{ }^\circ\text{C}\)
2. Acid concentration of \([\text{H}^+] = 1.0\text{ mol dm}^{-3}\)
3. Hydrogen gas pressure of \(100\text{ kPa}\) / \(1\text{ bar}\)

(a)
- Cobalt / \(\text{Co}\) is oxidized [1]
- \(\text{Co(s)} \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{e}^-\)
(Accept reversible arrows; ignore state symbols) [1]

(b)
- \(\text{Co(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{Ag(s)}\) (Ignore state symbols) [1]
- \(E^\theta_{\text{cell}} = +1.08\text{ V}\) (Accept 1.08 V) [1]

(c)
- Electrons flow from the cobalt electrode to the silver electrode (or anode to cathode) [1]
- Salt bridge completes the circuit AND maintains electrical neutrality / balance of charge [1]

(d)
- Cell potential decreases / is less positive [1]
- Increasing product concentration / \([\text{Co}^{2+}]\) shifts the position of equilibrium to the left / towards the reactants [1]

(e)
Any two of:
- Temperature of \(298\text{ K}\) / \(25\text{ }^\circ\text{C}\) [1]
- \([\text{H}^+] = 1.0\text{ mol dm}^{-3}\) / acid concentration of \(1.0\text{ mol dm}^{-3}\) (do not accept "concentration of acid is \(1\text{ M}\)" without reference to \(\text{H}^+\) concentration if diprotic) [1]
- Hydrogen gas pressure of \(100\text{ kPa}\) / \(1\text{ bar}\) / \(1\text{ atm}\) [1]

**(a)** The variable controlled by the investigator is the independent variable, which is Time (\(t\)). The variable that is measured is the dependent variable, which is the Volume of \(\text{CO}_2\) (\(V\)).

**(b)** Collision theory dictates that reaction rate depends on the frequency of successful collisions. As the reaction progresses, \(\text{HCl}\) is consumed, which decreases its concentration. This reduces the number of reactant particles per unit volume, lowering collision frequency and thus the rate.

**(c)** Average rate between 10 s and 40 s:
$$\text{Average Rate} = \frac{V_{40} - V_{10}}{t_{40} - t_{10}} = \frac{47.0\text{ cm}^3 - 15.0\text{ cm}^3}{40\text{ s} - 10\text{ s}} = \frac{32.0\text{ cm}^3}{30\text{ s}} = 1.07\text{ cm}^3\text{ s}^{-1}$$

**(d)** The actual volume (\(58.0\text{ cm}^3\)) is lower than the theoretical yield (\(62.0\text{ cm}^3\)). Since leaks are ruled out, this is explained by:
- \(\text{CO}_2\) being slightly soluble in water/aqueous solutions, so some gas remained dissolved in the reaction mixture.
- Deviation from standard conditions (e.g., temperature was lower than \(298\text{ K}\) or pressure was higher than \(100\text{ kPa}\)), causing a lower gas volume than predicted by the ideal gas molar volume.
- Impurities in the reactant, meaning fewer moles of gas were produced.

**(e)** Percentage uncertainty is calculated as:
$$\text{Percentage Uncertainty} = \frac{\text{Absolute Uncertainty}}{\text{Measured Value}} \times 100\%$$
Since absolute uncertainty is constant (\(\pm 0.5\text{ cm}^3\)) and the measured volume increases, the percentage uncertainty decreases.

**(a)** [1 mark]
Award [1] for correctly identifying both independent (Time) and dependent (Volume of \(\text{CO}_2\)) variables.

**(b)** [2 marks]
Award [1] for stating concentration of reactant/\(\text{HCl}\) decreases over time.
Award [1] for linking this to lower frequency of collisions (or fewer successful collisions per unit time).

**(c)** [2 marks]
Award [1] for correct numerical setup: \(\frac{47.0 - 15.0}{40 - 10}\).
Award [1] for final answer of \(1.07\text{ cm}^3\text{ s}^{-1}\) (or \(1.1\text{ cm}^3\text{ s}^{-1}\)) with correct units.

**(d)** [2 marks]
Award [1] each for any two of the following:
- Carbon dioxide gas dissolved in the reaction solution.
- Actual temperature was lower than \(298\text{ K}\) / actual pressure was higher than \(100\text{ kPa}\).
- Impurities in the calcium carbonate (resulting in fewer moles of reactants reacting).
- Accept incomplete reaction due to passivation or evaporation of acid.
*Do NOT accept "gas escaped" as the question rules out gas leaks.*

**(e)** [1 mark]
Award [1] for stating that the percentage uncertainty decreases.

**(a)** First, determine the mass of the reaction mixture:
$$\text{Total volume} = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3$$
$$\text{Mass (m)} = 100.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100.0\text{ g}$$
Determine the temperature change:
$$\Delta T = 28.2^\circ\text{C} - 21.5^\circ\text{C} = 6.7\text{ K}$$
Calculate the heat energy released, \(q\):
$$q = mc\Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ}$$

**(b)** Calculate the moles of water formed in this neutralization:
$$n(\text{H}_2\text{O}) = n(\text{HCl}) = n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}$$
Calculate \(\Delta H_n\) with correct negative sign (exothermic reaction):
$$\Delta H_n = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.012\text{ kJ mol}^{-1} \approx -56.0\text{ kJ mol}^{-1}$$

**(c)** Calculate the percentage error:
$$\text{Percentage error} = \frac{|\text{Experimental Value} - \text{Literature Value}|}{|\text{Literature Value}|} \times 100\%$$
$$\text{Percentage error} = \frac{|-56.0 - (-57.1)|}{57.1} \times 100\% = \frac{1.1}{57.1} \times 100\% = 1.93\%$$

**(d)** A systematic error is heat loss to the surroundings (air, cup, thermometer) resulting in a lower recorded maximum temperature. This can be mitigated by using a lid on the cup, nesting the polystyrene cup inside another cup to increase insulation, or using a vacuum flask/bomb calorimeter.

**(a)** [2 marks]
Award [1] for calculating mass of 100.0 g and temperature difference of 6.7 K (or 6.7 °C).
Award [1] for the correct final heat value of 2.80 kJ (or 2.8 kJ).

**(b)** [2 marks]
Award [1] for calculating moles of water formed as 0.0500 mol.
Award [1] for dividing heat by moles with the correct negative sign (\(-56.0\text{ kJ mol}^{-1}\) or \(-56\text{ kJ mol}^{-1}\)).

**(c)** [1 mark]
Award [1] for calculating percentage error as 1.93% (accept 1.9% or 2% if consistent with rounding in previous steps).

**(d)** [2 marks]
Award [1] for identifying heat loss to the surroundings / absorption of heat by the cup/thermometer.
Award [1] for suggesting a logical improvement (e.g. use a lid, use a double-nested polystyrene cup, use a vacuum flask, or perform a temperature-time extrapolation correction).

(a) At pH 7.0, which is higher than the two acidic \(pK_{\text{a}}\) values of 1.88 and 3.65 but lower than the basic \(pK_{\text{a}}\) of 9.60, both carboxylic acid groups will be deprotonated to form carboxylate anions (\(-\text{COO}^-\)), while the amino group remains protonated as an ammonium ion (\(-\text{NH}_3^+\)). The overall charge of the molecule is \(-1\). (b) For acidic amino acids, the isoelectric point (\(pI\)) lies between the \(pK_{\text{a}}\) values of the two carboxylic acid groups. It is calculated as the arithmetic mean of these two values: \(pI = \frac{1.88 + 3.65}{2} = 2.765 \approx 2.77\).

Award [1 mark] for correctly describing the protonation state of aspartic acid at pH 7.0: both carboxyl groups are deprotonated as \(-\text{COO}^-\) and the amino group is protonated as \(-\text{NH}_3^+\). Award [1 mark] for showing that the calculation of \(pI\) requires averaging the two lowest \(pK_{\text{a}}\) values (associated with the acidic carboxyl groups). Award [1 mark] for the correct calculation: \(pI = 2.77\) (accept 2.77 or 2.8).

(a) The Michaelis constant, \(K_{\text{m}}\), is defined as the substrate concentration at which the reaction velocity is half of its maximum rate (\(\frac{1}{2}V_{\max}\)). It is inversely proportional to enzyme affinity: a lower \(K_{\text{m}}\) value means the enzyme has a higher affinity for its substrate, requiring less substrate to reach half-maximal speed. (b) A competitive inhibitor structurally resembles the substrate and binds reversibly to the active site. Because it blocks the active site, more substrate is needed to achieve the same rate of reaction, which increases the apparent \(K_{\text{m}}\). At infinitely high substrate concentrations, the substrate completely outcompetes the inhibitor, allowing the enzyme to still reach its original maximum velocity (\(V_{\max}\)).

Award [1 mark] for defining \(K_{\text{m}}\) as the substrate concentration at half \(V_{\max}\) AND stating that a lower \(K_{\text{m}}\) corresponds to a higher affinity. Award [1 mark] for identifying the inhibition as competitive. Award [1 mark] for explaining that competitive inhibitors bind reversibly to the active site, which can be overcome at very high substrate concentrations, leaving \(V_{\max}\) unaffected.

(a) A fully saturated 18-carbon carboxylic acid (stearic acid) has the formula \(\text{C}_{18}\text{H}_{36}\text{O}_2\). Since linolenic acid has the formula \(\text{C}_{18}\text{H}_{30}\text{O}_2\), it is deficient by 6 hydrogen atoms. Each double bond represents a deficiency of 2 hydrogen atoms compared to the saturated compound. Therefore, there are \(\frac{6}{2} = 3\) carbon-carbon double bonds in one molecule of linolenic acid. (b) Each \(\text{C}=\text{C}\) double bond reacts with one molecule of iodine (\(\text{I}_2\)). Thus, 1 mole of linolenic acid reacts with 3 moles of \(\text{I}_2\). Number of moles of linolenic acid in 100 g: \(n(\text{acid}) = \frac{100\text{ g}}{278.44\text{ g mol}^{-1}} = 0.35915\text{ mol}\). Number of moles of \(\text{I}_2\) reacting: \(n(\text{I}_2) = 3 \times 0.35915\text{ mol} = 1.0775\text{ mol}\). Mass of \(\text{I}_2\): \(m(\text{I}_2) = 1.0775\text{ mol} \times 253.80\text{ g mol}^{-1} = 273.46\text{ g}\). Therefore, the iodine number is 273.

Award [1 mark] for correctly determining 3 \(\text{C}=\text{C}\) double bonds based on the molecular formula comparison with saturated fatty acids. Award [1 mark] for establishing the stoichiometric ratio of \(1\text{ mol acid} : 3\text{ mol }\text{I}_2\). Award [1 mark] for the correct final iodine number of 273 (accept values from 273 to 274).

(a) Starch (specifically amylose) consists of \(\alpha\)-glucose monomers connected by \(\alpha\)-1,4-glycosidic linkages. This leads to a coiled/helical polymeric structure. In contrast, cellulose consists of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic linkages, where alternating glucose units are flipped \(180^\circ\). This allows cellulose chains to remain linear, pack closely together, and form extensive intermolecular hydrogen bonds, leading to tough microfibrils. (b) Human digestive systems produce enzymes such as amylase, which specifically recognize and catalyze the hydrolysis of \(\alpha\)-1,4-glycosidic bonds in starch. However, humans do not produce the enzyme cellulase, which is required to hydrolyze the highly stable \(\beta\)-1,4-glycosidic bonds of cellulose. Therefore, cellulose passes through the human digestive tract undigested as dietary fiber.

Award [1 mark] for contrasting monomers and linkages: starch has \(\alpha\)-glucose with \(\alpha\)-1,4 linkages, while cellulose has \(\beta\)-glucose with \(\beta\)-1,4 linkages. Award [1 mark] for contrasting physical structures: starch has a helical/coiled structure, whereas cellulose is linear and stabilized by intermolecular hydrogen bonding. Award [1 mark] for explaining that humans produce amylase (which hydrolyzes \(\alpha\)-linkages) but lack cellulase (required to hydrolyze \(\beta\)-linkages).

(a) Vitamin C (ascorbic acid) contains four polar hydroxyl (\(-\text{OH}\)) groups and a lactone ring, enabling it to form extensive hydrogen bonds with water molecules. This makes it highly water-soluble. On the other hand, Vitamin A (retinol) consists of a long, non-polar, hydrophobic conjugated hydrocarbon chain with only one terminal hydroxyl group. The non-polar region dominates its properties, so it primarily participates in London dispersion forces, making it insoluble in water but highly soluble in lipids (fat-soluble). (b) Because Vitamin A is lipid-soluble, it can accumulate and be stored for long periods in the liver and adipose (fat) tissues. Over-consumption can lead to toxicity (hypervitaminosis A). In contrast, water-soluble Vitamin C cannot be stored efficiently in the body; any excess in the bloodstream is filtered by the kidneys and excreted in urine, making toxicity highly unlikely but requiring regular dietary intake.

Award [1 mark] for explaining that Vitamin C's multiple hydroxyl groups allow it to hydrogen-bond with water, making it water-soluble. Award [1 mark] for explaining that Vitamin A's dominant non-polar hydrocarbon chain restricts it to London dispersion forces, making it fat-soluble (insoluble in water). Award [1 mark] for stating that fat-soluble Vitamin A accumulates in adipose/liver tissue (posing a toxicity risk), whereas water-soluble Vitamin C is excreted in urine (low toxicity risk).

(a) Both chlorophyll and hemoglobin contain an extensive conjugated system of alternating single and double carbon-carbon bonds (with delocalized \(\pi\) electrons) in the porphyrin ring. This extensive conjugation lowers the energy gap (\(\Delta E\)) between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). The transition energy corresponds to wavelengths in the visible light spectrum (\(E = \frac{hc}{\lambda}\)). When light of specific wavelengths is absorbed, electrons are excited, and the complementary wavelengths are transmitted or reflected, producing the pigment's observed color. (b) Chlorophyll contains magnesium (\(\text{Mg}^{2+}\)) coordinated in its porphyrin ring. Its primary function is to absorb light energy to drive the light-dependent reactions of photosynthesis. Hemoglobin consists of four polypeptide chains, each containing a heme group with an iron (\(\text{Fe}^{2+}\)) ion at its center. Its primary function is the reversible binding and transport of oxygen molecules from the lungs to tissues.

Award [1 mark] for explaining that the extensive conjugated system (alternating single and double bonds/delocalized \(\pi\) electrons) reduces the energy gap (\(\Delta E\)) between HOMO and LUMO. Award [1 mark] for explaining that this smaller energy transition allows the absorption of visible light, with the unabsorbed wavelengths giving the pigment its color. Award [1 mark] for correctly identifying that chlorophyll contains \(\text{Mg}^{2+}\) (for photosynthesis/light absorption) and hemoglobin contains \(\text{Fe}^{2+}\) (for oxygen transport).

(a) Individual nucleotides within a single DNA strand are joined together covalently by a phosphodiester linkage. This bond is formed between the phosphate group attached to the 5' carbon of one deoxyribose sugar and the hydroxyl group on the 3' carbon of the adjacent deoxyribose sugar. (b) The double-helix structure is stabilized by hydrogen bonding between the nitrogenous bases on opposite, antiparallel strands. Complementary base-pairing occurs between a purine and a pyrimidine: Adenine (A) pairs with Thymine (T) via two hydrogen bonds, and Guanine (G) pairs with Cytosine (C) via three hydrogen bonds. These highly specific hydrogen bonds hold the two strands together securely while maintaining a uniform distance between the backbones.

Award [1 mark] for identifying covalent phosphodiester bonds (or linkages between the sugar and phosphate groups) as the bond within a single strand. Award [1 mark] for explaining that hydrogen bonding between complementary base pairs on opposite strands holds the double-stranded structure together. Award [1 mark] for specifying that Adenine-Thymine (A-T) forms 2 hydrogen bonds, and Guanine-Cytosine (G-C) forms 3 hydrogen bonds.

(a) Traditional addition polymers like polyethylene (PE) have a backbone consisting of strong, non-polar carbon-carbon (\(\text{C}-\text{C}\)) single bonds. Microorganisms do not possess enzymes capable of breaking these stable bonds, making PE extremely persistent in the environment. Condensation polymers like polyethylene terephthalate (PET) contain ester bonds (polar linkages) in their backbone. These bonds are susceptible to slow chemical or enzymatic hydrolysis, rendering PET more biodegradable/degradable than PE, although still somewhat persistent. (b) Structural features that promote biodegradation include having polar, hydrolyzable linkages in the polymer backbone, such as ester (\(-\text{COO}-\)) or amide (\(-\text{CONH}-\)) linkages (or incorporating natural, enzyme-recognizable monomers like starch or lactic acid). Explanation: These polar bonds can be easily targeted by extracellular enzymes (like esterases or peptidases) produced by microorganisms. The enzymes catalyze the hydrolysis (breakdown) of these linkages, splitting the long polymer chain into small, water-soluble, non-toxic monomers or oligomers that the microorganisms can then metabolize.

Award [1 mark] for explaining that addition polymers (like PE) are persistent due to stable, non-polar carbon-carbon (\(\text{C}-\text{C}\)) backbones that are resistant to cleavage. Award [1 mark] for explaining that condensation polymers (like PET) contain polar linkages (ester bonds) that can undergo slow hydrolysis. Award [1 mark] for identifying a hydrolyzable structural feature (e.g., ester or amide bonds, or natural polymer backbones) and explaining that microbial enzymes can hydrolyze these linkages to break down the polymer into small, non-toxic molecules.

At the optimum pH of 7.4, the tertiary structure of the enzyme is stabilized by various interactions between amino acid residue side chains, including ionic bonds (salt bridges) and hydrogen bonds.

When the pH is decreased to 3.0:
1. The high concentration of \(H^+\) ions leads to the protonation of negatively charged side chains, such as carboxylate groups (\(-COO^-\) becomes \(-COOH\)).
2. This protonation neutralizes the negative charges, disrupting the electrostatic attractions (ionic bonds or salt bridges) and hydrogen bonds that maintain the specific three-dimensional conformation of the protein.
3. The enzyme undergoes denaturation, altering the shape of the active site so that the substrate can no longer fit, leading to a loss of catalytic activity.

Award [1] mark for each of the following up to a maximum of [3.33] marks:
- State that activity decreases/ceases because the enzyme is denatured OR the conformation of the active site is altered. [1]
- Explain that decreasing pH (increasing \(H^+\) concentration) results in the protonation of side-chain groups (e.g., carboxylate/\(-COO^-\) to carboxylic acid/\(-COOH\) OR amino/\(-NH_2\) to ammonium/\(-NH_3^+\)). [1]
- Explain that this protonation disrupts ionic bonds (salt bridges) OR hydrogen bonds that stabilize the tertiary structure. [1]