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1. Assume a \(100\text{ g}\) sample. This gives \(30.4\text{ g}\) of \(N\) and \(69.6\text{ g}\) of \(O\).
2. Calculate the number of moles of each element:
- \(n(\text{N}) = \frac{30.4\text{ g}}{14.0\text{ g mol}^{-1}} \approx 2.17\text{ mol}\)
- \(n(\text{O}) = \frac{69.6\text{ g}}{16.0\text{ g mol}^{-1}} \approx 4.35\text{ mol}\)
3. Find the simplest whole-number ratio by dividing by the smallest value:
- \(\text{N}: \frac{2.17}{2.17} = 1\)
- \(\text{O}: \frac{4.35}{2.17} \approx 2\)
4. The empirical formula is \(\text{NO}_2\).

Award 1 mark for the correct answer: B.
- 1 mark for calculating correct mole ratio (1 : 2) and identifying NO2.
- Award 0 marks for incorrect formula options.

Across Period 3, the electronegativity increases. This is because the nuclear charge increases (more protons are added to the nucleus) while the shielding remains approximately constant (electrons are added to the same main energy level). As a result, there is a stronger attraction between the nucleus and the bonding electrons. Atomic radius decreases across the period, which also enhances this attraction, rather than increasing.

Award 1 mark for the correct answer: B.
- Reject responses stating electronegativity decreases.
- Reject responses attributing the increase to an increase in atomic radius.

1. Identify the reducing agent: The reducing agent is the species that undergoes oxidation (loses electrons / experiences an increase in oxidation state).
2. Determine the oxidation state of sulfur in \(\text{SO}_3^{2-}\):
\(x + 3(-2) = -2 \Rightarrow x = +4\).
3. Determine the oxidation state of sulfur in \(\text{SO}_4^{2-}\):
\(x + 4(-2) = -2 \Rightarrow x = +6\).
4. Since sulfur increases in oxidation state from +4 to +6, \(\text{SO}_3^{2-}\) is oxidized and is therefore the reducing agent.

Award 1 mark for the correct answer: B.
- 1 mark for identifying SO3(2-) as the reducing agent and +4 to +6 as the oxidation state change.

- \(\text{NH}_4^+\) has a tetrahedral geometry with a bond angle of \(109.5^\circ\).
- \(\text{H}_3\text{O}^+\) has a trigonal pyramidal geometry with a bond angle of approximately \(107^\circ\).
- \(\text{CO}_3^{2-}\) has 3 charge centers around the central carbon atom and no lone pairs, resulting in a trigonal planar geometry with a bond angle of exactly \(120^\circ\).
- \(\text{SF}_6\) has an octahedral geometry with bond angles of \(90^\circ\) and \(180^\circ\).

Award 1 mark for the correct answer: C.
- Reject A, B, and D based on VSEPR theory shapes and bond angles.

A conjugate acid-base pair consists of two species that differ by a single proton (\(\text{H}^+\)). Here, \(\text{HCO}_3^-\)(acid) and \(\text{CO}_3^{2-}\)(base) differ by exactly one proton: \(\text{HCO}_3^- \rightarrow \text{CO}_3^{2-} + \text{H}^+\). Hence, they represent a conjugate acid-base pair.

Award 1 mark for the correct answer: C.
- 1 mark for identifying the correct pair differing by only one proton.

1. The compound contains the carboxylate ester linkage group \(-\text{COO}-\), indicating it belongs to the ester family.
2. In naming an ester, the alkyl group attached directly to the single-bonded oxygen is named first: \(-\text{CH}_3\) is "methyl".
3. The acyl group containing the carbonyl carbon is named second, based on the total number of carbon atoms in that chain: \(\text{CH}_3\text{CH}_2\text{CO}-\) has 3 carbons, so it is "propanoate".
4. Therefore, the IUPAC name is methyl propanoate.

Award 1 mark for the correct answer: A.
- Reject B (incorrect naming order/parent chain selection).
- Reject C and D (incorrect functional groups).

1. Doubling the volumes of both reactant solutions doubles the amounts of moles of reactants reacting, which doubles the total heat energy released (\(q_{new} = 2q\)).
2. However, doubling the volumes also doubles the total mass of the aqueous solution being heated (\(m_{new} = 2m\)).
3. Using the equation \(q = m \cdot c \cdot \Delta T\):
\(\Delta T_{new} = \frac{q_{new}}{m_{new} \cdot c} = \frac{2q}{2m \cdot c} = \frac{q}{m \cdot c} = \Delta T\).
4. Therefore, the temperature change remains exactly the same, \(\Delta T\).

Award 1 mark for the correct answer: B.
- 1 mark for correctly realizing that heat released and mass heated scale proportionally, leaving temperature change unchanged.

Adding a catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). On a Maxwell-Boltzmann distribution graph, this is represented by shifting the vertical line for the activation energy to the left. A catalyst does not affect the actual kinetic energy of the gas molecules, the shape of the distribution curve, or the area under the curve (which represents the total number of molecules).

Award 1 mark for the correct answer: C.
- Reject A and D as temperature is constant so the curve shape and average kinetic energy are unchanged.
- Reject B as the total number of molecules (area) remains constant.

The oxidizing agent is the species that undergoes reduction (a decrease in its oxidation state). In the reactant MnO4-, the oxidation state of Mn is +7. In the product Mn2+, the oxidation state of Mn is +2. Since the oxidation state of manganese decreases, it is reduced, meaning MnO4- acts as the oxidizing agent.

[1 mark] Award for choosing option a. Reject all other options.

First ionization energy generally increases across a period due to increasing nuclear charge and constant shielding. However, there is a deviation between Group 2 (Mg) and Group 13 (Al). Magnesium has the electron configuration [Ne] 3s^2, while aluminium has [Ne] 3s^2 3p^1. The outermost electron of Al is in the 3p subshell, which is higher in energy and more shielded from the nucleus than the 3s electron in Mg, making it easier to remove. Therefore, the first ionization energy of Al is lower than that of Mg, resulting in the order: Na < Al < Mg < Si.

[1 mark] Correct choice is b.

The carbonate ion has three electron domains around the central carbon atom and no lone pairs. This results in a trigonal planar electron domain geometry and molecular geometry. Due to resonance, the three carbon-oxygen bonds are identical, resulting in bond angles of exactly 120 degrees. NH3 and H2O have lone pairs on the central atom that reduce their bond angles below the tetrahedral angle, and CH4 is tetrahedral with 109.5 degree angles.

[1 mark] Correct choice is b.

Assume a 100 g sample of the compound. Mass of N is 30.4 g, and mass of O is 100 - 30.4 = 69.6 g. Moles of N = 30.4 / 14.0 = 2.17 mol. Moles of O = 69.6 / 16.0 = 4.35 mol. Dividing both values by the smaller number (2.17) gives a ratio of N : O of approximately 1 : 2. Therefore, the empirical formula is NO2.

[1 mark] Correct choice is b.

A conjugate acid-base pair consists of two species that differ by exactly one proton (H+). In this reaction, H2PO4- acts as a Bronsted-Lowry acid by donating a proton to form its conjugate base, HPO42-. Therefore, H2PO4- and HPO42- represent a conjugate acid-base pair.

[1 mark] Correct choice is c.

The compound HOCH2CH2COOCH3 contains a hydroxyl group (-OH) which characterizes the alcohol functional group, and an ester group (-COO-) which characterizes the ester functional group.

[1 mark] Correct choice is a.

The heat released (q) in Joules is calculated using q = m * c * dT = 100 * 4.18 * 15.0 J. Converting this value to kilojoules requires dividing by 1000, giving q = (100 * 4.18 * 15.0) / 1000 kJ. Since the temperature increased, the dissolution is exothermic and the enthalpy change (dH) must be negative. To find dH per mole of solute, we divide the heat value by the number of moles (0.10 mol), resulting in dH = - (100 * 4.18 * 15.0) / (0.10 * 1000) kJ/mol.

[1 mark] Correct choice is a.

A catalyst increases the reaction rate by providing an alternative pathway with a lower activation energy. However, it does not alter the potential energy of the reactants or the products, meaning the enthalpy change (dH) of the reaction remains unchanged. Therefore, Reaction 1 has a lower activation energy and the same enthalpy change as Reaction 2.

[1 mark] Correct choice is b.

First, calculate the moles of chloride ions in each option: Option A: \(0.10\text{ mol AlCl}_3\) contains \(0.10 \times 3 = 0.30\text{ mol Cl}^-\). Option B: \(10.0\text{ g} / 95.21\text{ g mol}^{-1} \approx 0.105\text{ mol MgCl}_2\), which contains \(0.105 \times 2 = 0.210\text{ mol Cl}^-\). Option C: \(10.0\text{ g} / 42.39\text{ g mol}^{-1} \approx 0.236\text{ mol LiCl}\), which contains \(0.236 \times 1 = 0.236\text{ mol Cl}^-\). Option D: Moles of \(\text{BaCl}_2 = 0.100\text{ dm}^3 \times 2.0\text{ mol dm}^{-3} = 0.20\text{ mol}\). This contains \(0.20 \times 2 = 0.40\text{ mol Cl}^-\). Therefore, Option D contains the greatest amount of chloride ions.

Award 1 mark for correct choice D. No partial marks.

\(\text{P}_4\text{O}_{10}\) is an acidic molecular oxide that reacts with water to form phosphoric acid (\(\text{H}_3\text{PO}_4\)), lowering the pH below 7. \(\text{SO}_3\) is also an acidic molecular oxide that reacts with water to form sulfuric acid (\(\text{H}_2\text{SO}_4\)), lowering the pH below 7. \(\text{SiO}_2\) has a giant covalent structure and is completely insoluble in water, so it does not affect the pH of water. Thus, only I and III form acidic solutions.

Award 1 mark for correct choice B.

\(\text{CO}_3^{2-}\) has 3 bonding domains and 0 lone pairs on the central carbon atom, resulting in a trigonal planar electron-domain geometry and molecular geometry with bond angles of exactly \(120^\circ\). \(\text{NH}_3\) (trigonal pyramidal) has a bond angle of approximately \(107^\circ\) due to the lone pair. \(\text{H}_3\text{O}^+\) (trigonal pyramidal) also has a bond angle of approximately \(107^\circ\). \(\text{OF}_2\) (bent) has 2 bonding domains and 2 lone pairs on the oxygen atom, resulting in a bond angle of approximately \(103^\circ\).

Award 1 mark for correct choice C.

First, calculate the heat released (\(q\)) during dissolution: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\ \text{K}^{-1} \times 12.0\text{ K} = 5040\text{ J} = 5.04\text{ kJ}\). Since the temperature increased, the reaction is exothermic, meaning \(\Delta H\) is negative. Thus, \(\Delta H = -5.04\text{ kJ}\). Next, calculate the molar enthalpy of solution: \(\Delta H_{\text{sol}} = \frac{-5.04\text{ kJ}}{0.10\text{ mol}} = -50.4\text{ kJ mol}^{-1}\).

Award 1 mark for correct choice B.

In the permanganate ion (\(\text{MnO}_4^-\)), the oxidation state of Mn is \(+7\) (since \(x + 4(-2) = -1\) leads to \(x = +7\)). In \(\text{Mn}^{2+}\), the oxidation state of Mn is \(+2\). Therefore, manganese undergoes a change in oxidation state from \(+7\) to \(+2\), which represents a reduction. \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\) (oxidation state increases from \(+2\) to \(+3\)), so \(\text{Fe}^{2+}\) acts as the reducing agent. The oxidation states of hydrogen and oxygen remain unchanged at \(+1\) and \(-2\) respectively.

Award 1 mark for correct choice B.

A catalyst provides an alternative pathway for the reaction with a lower activation energy (\(E_a\)). This does not change the temperature or the average kinetic energy of the particles, nor does it affect collision frequency. Instead, because \(E_a\) is lowered, a greater fraction of the existing collisions have energy equal to or greater than the new activation energy, which increases the rate of both the forward and reverse reactions equally.

Award 1 mark for correct choice C.

Methyl 2-hydroxybenzoate contains: an ester group (\(-\text{COO}-\)), a hydroxyl group (\(-\text{OH}\)) directly attached to the benzene ring, and an arene group (the benzene ring itself). It does not contain an ether group (\(\text{R}-\text{O}-\text{R}\)).

Award 1 mark for correct choice C.

The ground-state electron configuration of a neutral copper atom (Cu, Z = 29) is \([\text{Ar}] 4s^1 3d^{10}\) due to the extra stability of a fully filled d-subshell. When forming transition metal cations, electrons are lost from the s-orbital first. To form \(\text{Cu}^{2+}\), two electrons are removed: one from the \(4s\) orbital and one from the \(3d\) orbital. This yields \([\text{Ar}] 3d^9\).

Award 1 mark for correct choice A.

One formula unit of \(\text{Fe}_2(\text{SO}_4)_3\) dissociates into two \(\text{Fe}^{3+}\) ions and three \(\text{SO}_4^{2-}\) ions, giving a total of 5 ions per formula unit. Therefore, \(0.10\text{ mol}\) of \(\text{Fe}_2(\text{SO}_4)_3\) contains \(0.10 \times 5 = 0.50\text{ mol}\) of ions. The number of ions is \(0.50 L\).

[1 mark] for identifying that there are 5 ions per formula unit and correctly multiplying to get \(0.50 L\). Award 0 marks for other options.

When the volumes of both reactant solutions are halved, the number of moles of reactants is halved, meaning the heat energy released (\(q\)) is also halved. However, the total mass/volume of the solution being heated is also halved (from \(100.0\text{ cm}^3\) to \(50.0\text{ cm}^3\)). Since \(\Delta T = \frac{q}{m \cdot c}\), dividing both \(q\) and \(m\) by 2 results in the same temperature change, \(\Delta T\).

[1 mark] for recognizing that both heat released and solution mass are halved, keeping \(\Delta T\) constant.

A catalyst provides an alternative reaction pathway with a lower activation energy. This allows a greater fraction of reactant molecules to have energy equal to or greater than the activation energy, increasing the frequency of successful collisions. It does not affect average kinetic energy (only temperature does), total collision frequency, or the enthalpy change of the reaction.

[1 mark] for selecting the option stating that a catalyst provides an alternative pathway with a lower activation energy.

The ground-state electron configuration of a neutral iron atom (\(\text{Fe}\), atomic number 26) is \([\text{Ar}] 4s^2 3d^6\). When transition metals form cations, they lose the outer-shell \(s\) electrons first. Thus, the two electrons are removed from the \(4s\) subshell, resulting in \([\text{Ar}] 3d^6\).

[1 mark] for identifying the correct electron configuration of the transition metal cation, recognizing that \(4s\) electrons are lost first.

According to the Brønsted-Lowry theory, an acid becomes its conjugate base after losing a proton (\(\text{H}^+\)). Thus, removing one hydrogen ion from \(\text{H}_2\text{PO}_4^-\): \(\text{H}_2\text{PO}_4^- \rightarrow \text{HPO}_4^{2-} + \text{H}^+\). Therefore, \(\text{HPO}_4^{2-}\) is the conjugate base.

[1 mark] for identifying \(\text{HPO}_4^{2-}\) as the conjugate base of \(\text{H}_2\text{PO}_4^-\).

Let \(x\) be the oxidation state of sulfur. Oxygen has a standard oxidation state of \(-2\). The sum of the oxidation states in the polyatomic ion must equal its overall charge of \(-2\): \(2(x) + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).

[1 mark] for correctly calculating the oxidation state of sulfur as \(+2\).

First, write the oxidation half-equation: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\). Second, write the reduction half-equation: \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\). Multiplying the oxidation half-equation by 6 to balance the electrons and combining both half-equations yields the balanced net ionic equation: \(6\text{Fe}^{2+}(\text{aq}) + \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\).

Award 1 mark for the correct oxidation half-equation (or showing the 6:1 ratio of iron to dichromate). Award 1 mark for the correct reduction half-equation with balanced \(\text{H}^+\) and \(\text{H}_2\text{O}\). Award 1 mark for the final fully balanced net ionic equation (state symbols are not required for marks).

Across Period 3, the first ionization energy generally increases because nuclear charge increases (more protons) while shielding remains approximately constant (electrons are added to the same valence shell), leading to a stronger electrostatic attraction on the valence electrons. Aluminum (\(\text{Al}\)) is an exception because its valence electron is in a \(3\text{p}\) orbital, which is higher in energy and better shielded than the \(3\text{s}\) orbital of magnesium (\(\text{Mg}\)). Alternatively, sulfur (\(\text{S}\)) is an exception because its outer electron is paired in a \(3\text{p}\) orbital, resulting in spin-pair repulsion which makes the electron easier to remove compared to phosphorus (\(\text{P}\)).

Award 1 mark for explaining the general increase across Period 3 (increased nuclear charge / effective nuclear charge and similar shielding). Award 1 mark for identifying a correct exception (aluminum or sulfur). Award 1 mark for explaining the chosen exception correctly (orbital energy difference/shielding for Al, or spin-pair repulsion in a p-orbital for S).

The Lewis structure of \(\text{SF}_4\) shows sulfur as the central atom with 10 valence electrons (an expanded octet): 4 bonding pairs with fluorine atoms and 1 lone pair. Based on 5 electron domains, the electron-domain geometry is trigonal bipyramidal. To minimize repulsion, the lone pair occupies an equatorial position, resulting in a see-saw molecular geometry. The lone pair repels the bonding pairs, reducing the bond angles to less than \(120^\circ\) (specifically around \(102^\circ\) - \(117^\circ\) for the equatorial F-S-F angle) and less than \(90^\circ\) (around \(87^\circ\) for the axial-equatorial F-S-F angle).

Award 1 mark for a correct Lewis structure of \(\text{SF}_4\) (showing 4 single S-F bonds, 1 lone pair on S, and 3 lone pairs on each F atom). Award 1 mark for predicting 'see-saw' (or 'seesaw') molecular geometry. Award 1 mark for estimating bond angles as less than \(120^\circ\) and/or less than \(90^\circ\) (or specifying values like \(102^\circ-117^\circ\) and \(87^\circ-89^\circ\)).

First, find the mass percentage of oxygen: \(100.00\% - 40.00\% - 6.71\% = 53.29\%\). Next, calculate the mole amounts in a \(100\text{ g}\) sample: \(n(\text{C}) = 40.00 / 12.01 = 3.33\text{ mol}\), \(n(\text{H}) = 6.71 / 1.01 = 6.64\text{ mol}\), \(n(\text{O}) = 53.29 / 16.00 = 3.33\text{ mol}\). Dividing each value by the smallest value (3.33) yields a ratio of \(\text{C} : \text{H} : \text{O} = 1 : 2 : 1\). Thus, the empirical formula is \(\text{CH}_2\text{O}\).

Award 1 mark for calculating correct amount in moles of each element. Award 1 mark for finding the simplest integer ratio of 1:2:1. Award 1 mark for stating the correct final empirical formula of \(\text{CH}_2\text{O}\).

Nitric acid (\(\text{HNO}_3\)) is a strong acid that dissociates completely in water, whereas nitrous acid (\(\text{HNO}_2\)) is a weak acid that dissociates only partially. Method 1: Measure pH using a pH meter or universal indicator. Nitric acid will have a lower pH (pH = 1.0) than nitrous acid (pH > 1.0). Method 2: Measure electrical conductivity. Nitric acid will have a higher electrical conductivity than nitrous acid due to a higher concentration of mobile ions. Method 3: React equal volumes of both acids with a reactive metal (e.g., magnesium ribbon). Nitric acid will react more vigorously, producing bubbles of hydrogen gas at a faster rate.

Award 1 mark for identifying two valid methods (e.g., pH measurement and electrical conductivity, or rate of gas evolution with a reactive metal/carbonate). Award 1 mark for describing the correct observation for Method 1 (e.g., lower pH / faster bubbling for nitric acid). Award 1 mark for describing the correct observation for Method 2 (e.g., higher conductivity for nitric acid / higher pH for nitrous acid).

For the carboxylic acid class, a structural isomer is butanoic acid (\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{COOH}\)). Its full structural formula must show all bonds, including the double-bonded oxygen and the hydroxyl group on the terminal carbon. For the ester class, one structural isomer is ethyl ethanoate (\(\text{CH}_3-\text{COO}-\text{CH}_2-\text{CH}_3\)). Its full structural formula shows a carbonyl group on a two-carbon chain bonded to an oxygen connected to an ethyl group. (Other valid esters like methyl propanoate or propyl methanoate are also acceptable).

Award 1 mark for drawing the correct full structural formula and giving the correct IUPAC name of a valid carboxylic acid (e.g., butanoic acid or 2-methylpropanoic acid). Award 1 mark for drawing the correct full structural formula and giving the correct IUPAC name of a valid ester (e.g., ethyl ethanoate, methyl propanoate, or propyl methanoate). Award 1 mark for drawing correct structures showing all atoms and all bonds clearly (ensuring carbon has 4 bonds, oxygen 2, hydrogen 1).

First, calculate the heat energy absorbed by the water: \(q = m \cdot c \cdot \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (44.8 - 20.2)\text{ K} = 15424.2\text{ J} = 15.42\text{ kJ}\). Next, calculate the amount of methanol burned in moles: \(n(\text{CH}_3\text{OH}) = 0.800\text{ g} / 32.05\text{ g mol}^{-1} = 0.02496\text{ mol}\). Finally, compute the enthalpy change of combustion (with a negative sign for an exothermic reaction): \(\Delta H_\text{c} = -q / n = -15.42\text{ kJ} / 0.02496\text{ mol} = -618\text{ kJ mol}^{-1}\).

Award 1 mark for calculating the heat absorbed by water (\(15.42\text{ kJ}\) or \(15424.2\text{ J}\)). Award 1 mark for calculating the amount of methanol in moles (\(0.0250\text{ mol}\)). Award 1 mark for the correct final enthalpy value of \(-618\text{ kJ mol}^{-1}\) (must include the negative sign; accept range \(-617\) to \(-620\)).

(i) The electronic configuration of \(\text{Cu}\) is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\). (ii) The electronic configuration of \(\text{Cu}^{2+}\) is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\) (as electrons are lost from the \(4\text{s}\) orbital first). The expected filling pattern would predict \([\text{Ar}] 3\text{d}^9 4\text{s}^2\) for a copper atom. However, a fully filled \(3\text{d}\) subshell (\(3\text{d}^{10}\)) is particularly stable due to symmetrical electron distribution and lower inter-electronic repulsion, making the \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\) configuration lower in energy.

Award 1 mark for the correct full electron configuration of \(\text{Cu}\). Award 1 mark for the correct full electron configuration of \(\text{Cu}^{2+}\). Award 1 mark for explaining that a full \(3\text{d}^{10}\) subshell offers extra energetic stability compared to \(3\text{d}^9 4\text{s}^2\).

1. Write the balanced redox equation to determine the molar ratio:
\(5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
The mole ratio of \(\text{MnO}_4^-\rightleftharpoons \text{Fe}^{2+}\) is \(1:5\).

2. Calculate the amount (in moles) of \(\text{MnO}_4^-\rightleftharpoons\) used:
\(n(\text{MnO}_4^-) = c \times V = 0.0200\text{ mol dm}^{-3} \times \frac{18.50}{1000}\text{ dm}^3 = 3.70 \times 10^{-4}\text{ mol}\).

3. Calculate the amount (in moles) of \(\text{Fe}^{2+}\):
\(n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-) = 5 \times 3.70 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\).

4. Calculate the concentration of \(\text{Fe}^{2+}\):
\(c(\text{Fe}^{2+}) = \frac{n}{V} = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0740\text{ mol dm}^{-3}\).

Award [1] for the correct mole ratio of 1:5 (explicitly stated or shown in calculations).
Award [1] for calculating \(n(\text{Fe}^{2+}) = 1.85 \times 10^{-3}\text{ mol}\).
Award [1] for the final concentration of \(0.0740\text{ mol dm}^{-3}\) (accept \(0.074\)).

(i) Each iodine atom has 7 valence electrons, and the negative charge adds 1 electron: \(3 \times 7 + 1 = 22\) valence electrons.

(ii) The central iodine atom is bonded to two other iodine atoms and has three lone pairs of electrons (total of 5 electron domains). According to VSEPR theory, these 5 domains adopt a trigonal bipyramidal arrangement, with the three lone pairs in the equatorial positions to minimize repulsion. The remaining two bonding domains are in axial positions, leading to a linear molecular geometry.

(iii) For a linear molecular geometry with the axial positions occupied, the bond angle is exactly \(180^\circ\).

(i) Award [1] for 22 (valence electrons).
(ii) Award [1] for linear.
(iii) Award [1] for \(180^\circ\) (accept values in the range \(175^\circ\) to \(180^\circ\)).

1. Calculate the heat released by the reaction (\(q\)):
\(q = m \cdot c \cdot \Delta T\)
\(m = 105.00\text{ g}\) (mass of the solution)
\(\Delta T = 28.8^\circ\text{C} - 20.2^\circ\text{C} = 8.6\text{ K}\)
\(q = 105.00\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 8.6\text{ K} = 3774.54\text{ J} = 3.775\text{ kJ}\)

2. Calculate the amount (in moles) of \(\text{CaCl}_2\) dissolved:
\(n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00\text{ g}}{110.98\text{ g mol}^{-1}} = 0.04505\text{ mol}\)

3. Calculate \(\Delta H_{\text{sol}}\):
\(\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{3.77454\text{ kJ}}{0.04505\text{ mol}} = -83.785\text{ kJ mol}^{-1} \approx -83.8\text{ kJ mol}^{-1}\).

Award [1] for calculating the heat change \(q = 3.77\text{ kJ}\) (or \(3770\text{ J}\); accept \(3.59\text{ kJ}\) if water mass of \(100.0\text{ g}\) is used instead of solution mass).
Award [1] for calculating the moles of \(\text{CaCl}_2 = 0.0451\text{ mol}\).
Award [1] for the final value of \(-83.8\text{ kJ mol}^{-1}\) (accept range \(-83.7\) to \(-84.0\text{ kJ mol}^{-1}\), or \(-79.7\) to \(-80.0\text{ kJ mol}^{-1}\) if water mass was used). Negative sign is required for the mark.

(i) The ionization equation is:
\(\text{C}_2\text{H}_5\text{COOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq})\)
(Accept \(\text{C}_2\text{H}_5\text{COOH}(\text{aq}) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq})\))

(ii) Since propanoic acid is a weak acid, we can use the approximation:
\(K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}\)
\(1.34 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.125}\)
\([\text{H}^+]^2 = 1.34 \times 10^{-5} \times 0.125 = 1.675 \times 10^{-6}\)
\([\text{H}^+] = \sqrt{1.675 \times 10^{-6}} = 1.294 \times 10^{-3}\text{ mol dm}^{-3}\)

\(\text{pH} = -\log_{10}(1.294 \times 10^{-3}) = 2.89\)

(i) Award [1] for a correct ionization equation with equilibrium arrows.
(ii) Award [1] for calculating \([\text{H}^+] = 1.29 \times 10^{-3}\text{ mol dm}^{-3}\).
Award [1] for \(\text{pH} = 2.89\) (accept \(2.9\)).

1. Molar mass of \(\text{Na}_2\text{SO}_4 = 2(22.99) + 32.07 + 4(16.00) = 142.05\text{ g mol}^{-1}\). 2. Moles of \(\text{Na}_2\text{SO}_4 = \frac{3.55\text{ g}}{142.05\text{ g mol}^{-1}} = 0.0250\text{ mol}\). 3. Concentration of \(\text{Na}_2\text{SO}_4 = \frac{0.0250\text{ mol}}{0.2500\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\). 4. Since each unit of \(\text{Na}_2\text{SO}_4\) dissociates into two \(\text{Na}^+\) ions, the concentration of \(\text{Na}^+ = 2 \times 0.100 = 0.200\text{ mol dm}^{-3}\).

[1 mark]: Calculating moles of sodium sulfate correctly as \(0.0250\text{ mol}\) (or obtaining the correct molar mass of \(142.05\text{ g mol}^{-1}\)). [1 mark]: Multiplying the concentration or moles by 2 to account for the \(2:1\) mole ratio of \(\text{Na}^+\) to \(\text{Na}_2\text{SO}_4\). [0.5 marks]: Providing the final answer of \(0.200\text{ mol dm}^{-3}\) to 3 significant figures.

1. Mass of solution = \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). 2. Temperature change, \(\Delta T = 27.80\ ^\circ\text{C} - 21.20\ ^\circ\text{C} = 6.60\text{ K}\). 3. Heat released, \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{K}^{-1} \times 6.60\text{ K} = 2758.8\text{ J} = 2.759\text{ kJ}\). 4. Moles of reaction, \(n = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). 5. Enthalpy change of neutralization, \(\Delta H_{\text{n}} = -\frac{q}{n} = -\frac{2.759\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).

[1 mark]: Calculating the heat change \(q = 2.76\text{ kJ}\) (or \(2760\text{ J}\)) correctly. [1 mark]: Calculating the amount of reactant in moles (\(0.0500\text{ mol}\)). [0.5 marks]: Giving the final answer with a negative sign and appropriate units (\(-55.2\text{ kJ mol}^{-1}\)).

1. Set up equilibrium expression: \(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\). Let \([\text{H}^+] = [\text{A}^-] = x\). 2. \(1.34 \times 10^{-5} = \frac{x^2}{0.150} \implies x^2 = 2.01 \times 10^{-6}\). 3. \(x = [\text{H}^+] = \sqrt{2.01 \times 10^{-6}} = 1.418 \times 10^{-3}\text{ mol dm}^{-3}\). 4. \(\text{pH} = -\log_{10}(1.418 \times 10^{-3}) = 2.85\).

[1 mark]: Setting up the correct expression and solving for \(x^2\) (or \([\text{H}^+]^2 = 2.01 \times 10^{-6}\)). [1 mark]: Correctly solving for \([\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}\). [0.5 marks]: Finding the final \(\text{pH}\) value of 2.85 (must be given to 2 decimal places).

1. Balance the ionic equation: \(5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}\). This gives a mole ratio of \(n(\text{Fe}^{2+}) : n(\text{MnO}_4^-) = 5 : 1\). 2. Moles of \(\text{MnO}_4^-\text{ used} = 0.0200\text{ mol dm}^{-3} \times 0.01850\text{ dm}^3 = 3.70 \times 10^{-4}\text{ mol}\). 3. Moles of \(\text{Fe}^{2+} = 5 \times 3.70 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\). 4. Concentration of \(\text{Fe}^{2+} = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0740\text{ mol dm}^{-3}\).

[1 mark]: Finding the correct 5:1 stoichiometric ratio from balancing the half-reactions or full equation. [1 mark]: Correctly calculating the moles of iron(II) (\(1.85 \times 10^{-3}\text{ mol}\)). [0.5 marks]: Correctly determining the concentration of \(\text{Fe}^{2+}\) with 3 significant figures and units (\(0.0740\text{ mol dm}^{-3}\)).

To draw the Lewis structure of \(\text{NO}_2^+\):
1. Calculate the total number of valence electrons: \(5\text{ (from N)} + 2 \times 6\text{ (from O)} - 1\text{ (for the positive charge)} = 16\) electrons (8 pairs).
2. Nitrogen is the central atom. Connect the oxygen atoms to nitrogen with double bonds: \(\text{O}=\text{N}=\text{O}\). This uses 8 electrons (4 pairs).
3. Complete the octets of the outer oxygen atoms by adding two lone pairs (4 non-bonding electrons) to each oxygen. This uses the remaining 8 electrons.
4. Nitrogen has a formal charge of \(+1\) and now has a complete octet (4 bonding pairs). The overall ion carries a \(+1\) charge, which can be represented by placing brackets around the structure with a superscript \(+\).

Award [1] for a linear structure showing double bonds between the central nitrogen atom and both oxygen atoms, with no lone pairs on the nitrogen atom.
Award [1] for showing two lone pairs on each oxygen atom AND indicating the positive charge (either as a '+' charge on the nitrogen atom or by enclosing the entire structure in square brackets with a '+' superscript).

[Total: 2 marks]

Methyl ethanoate is an ester with the molecular formula \(\text{C}_3\text{H}_6\text{O}_2\).
A full structural formula must show all atoms and all bonds explicitly:
- A central carbon atom double-bonded to an oxygen atom (carbonyl group) and single-bonded to another oxygen atom.
- This central carbon is bonded to a methyl group (\(\text{-CH}_3\)), showing all three \(\text{C}-\text{H}\) bonds.
- The single-bonded oxygen is bonded to another methyl group (\(\text{-CH}_3\)), showing all three \(\text{C}-\text{H}\) bonds.

Structure representation:
H O H
| // |
H--C--C--O--C--H
| |
H H

Award [1] for the correct ester linkage, showing the \(\text{C}=\text{O}\) double bond and \(\text{C}-\text{O}\) single bond.
Award [1] for the complete structure showing all carbon and hydrogen atoms, with all \(\text{C}-\text{H}\) and \(\text{C}-\text{C}\) bonds fully drawn out (no condensed representations such as \(\text{-CH}_3\) or \(\text{-COOCH}_3\) accepted).

[Total: 2 marks]

1. Incomplete combustion of ethanol: This produces carbon monoxide and/or soot instead of carbon dioxide, which releases less energy per mole of ethanol burned than complete combustion.
2. Evaporation of ethanol from the wick: Some ethanol evaporates from the wick when the burner is extinguished or during weighing, leading to an overestimation of the mass of ethanol burned (which makes the calculated enthalpy change per mole less exothermic).
Alternatively, heat capacity of the copper beaker: Some heat released is used to heat the copper container itself and is not accounted for in the calculation using only the heat capacity of water, resulting in a lower calculated enthalpy change.

[1 mark] for identifying incomplete combustion / formation of CO or soot.
[1 mark] for identifying either: evaporation of ethanol from the wick before/after weighing OR heat absorbed by the copper beaker/calorimeter not being accounted for.

1. As the reaction proceeds, the concentration of hydrogen peroxide reactant decreases. According to collision theory, this reduces the frequency of successful collisions between reactant particles per unit time, thereby decreasing the reaction rate.
2. The initial rate of reaction is found by drawing a tangent to the curve on the volume-time graph at \(t = 0\text{ s}\) and calculating the gradient of this tangent (change in volume divided by change in time).

[1 mark]: for explaining that concentration of reactants decreases, leading to a lower frequency of successful collisions (or fewer successful collisions per unit time).
[1 mark]: for stating that the initial rate is determined by drawing a tangent at \(t = 0\) and calculating its gradient.

1. Determine the mass of water lost:
\(\text{Mass of H}_2\text{O} = 5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}\)

2. Calculate the amount in moles of anhydrous \(\text{CuSO}_4\) and water:
\(\text{Moles of CuSO}_4 = \frac{3.20\text{ g}}{159.62\text{ g mol}^{-1}} = 0.0200\text{ mol}\)
\(\text{Moles of H}_2\text{O} = \frac{1.80\text{ g}}{18.02\text{ g mol}^{-1}} = 0.100\text{ mol}\)

3. Determine the mole ratio of water to anhydrous salt:
\(x = \frac{0.100\text{ mol}}{0.0200\text{ mol}} = 5\)

Therefore, \(x = 5\).

[1 mark]: for calculating the correct number of moles of anhydrous \(\text{CuSO}_4\) (0.0200 mol) and water (0.100 mol) (or showing a correct ratio of 1 to 5 from steps).
[1 mark]: for the correct final integer value of \(x = 5\).

1. Convert values to correct SI units:
\(P = 1.00 \times 10^5\text{ Pa}\)
\(V = 83.1\text{ cm}^3 = 83.1 \times 10^{-6}\text{ m}^3\)
\(T = 100 + 273 = 373\text{ K}\) (accept \(373.15\text{ K}\))

2. Apply the ideal gas equation to find moles (\(n\)):
\(PV = nRT \implies n = \frac{PV}{RT}\)
\(n = \frac{1.00 \times 10^5\text{ Pa} \times 83.1 \times 10^{-6}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 373\text{ K}} = \frac{8.31}{3099.63} = 0.00268\text{ mol}\)

3. Calculate molar mass (\(M\)):
\(M = \frac{m}{n} = \frac{0.200\text{ g}}{0.00268\text{ mol}} = 74.6\text{ g mol}^{-1}\)

[1 mark]: for calculating the correct number of moles \(n = 0.00268\text{ mol}\) (or converting all variables correctly to SI units and substituting them into the ideal gas equation).
[1 mark]: for calculating the final molar mass of \(74.6\text{ g mol}^{-1}\) (accept values in the range 74.5 to 74.7 depending on rounding used for temperature).

First, calculate the temperature change of the water: \(\Delta T = 54.7 - 21.2 = 33.5\text{ K}\). Next, calculate the heat energy absorbed by the water: \(q = m c \Delta T = 200.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 33.5\text{ K} = 28006\text{ J} = 28.01\text{ kJ}\). Calculate the mass of ethanol burned: \(120.50\text{ g} - 119.12\text{ g} = 1.38\text{ g}\). Convert mass of ethanol to moles using its molar mass (46.08 g/mol): \(n = 1.38\text{ g} / 46.08\text{ g mol}^{-1} = 0.02995\text{ mol}\). Finally, calculate the molar enthalpy of combustion: \(\Delta H_c = -q / n = -28.01\text{ kJ} / 0.02995\text{ mol} = -935\text{ kJ mol}^{-1}\).

Award [1] for calculating the correct heat absorbed: \(q = 28.0\text{ kJ}\) (or \(28000\text{ J}\)). Award [1] for calculating the correct amount of ethanol burned: \(n = 0.0299\text{ mol}\). Award [1.5] for calculating the correct enthalpy of combustion: \(-935\text{ kJ mol}^{-1}\) (award [1] for the value 935, and [0.5] for the negative sign and correct units). Accept answers in the range of -934 to -936.

First, calculate the mass of anhydrous sodium carbonate: \(26.28\text{ g} - 24.35\text{ g} = 1.93\text{ g}\). Calculate the mass of water lost: \(29.56\text{ g} - 26.28\text{ g} = 3.28\text{ g}\). Next, calculate the chemical amount of anhydrous sodium carbonate: \(n(\text{Na}_2\text{CO}_3) = 1.93\text{ g} / 105.99\text{ g mol}^{-1} = 0.0182\text{ mol}\). Calculate the chemical amount of water: \(n(\text{H}_2\text{O}) = 3.28\text{ g} / 18.02\text{ g mol}^{-1} = 0.182\text{ mol}\). Determine the ratio of \(x\): \(x = n(\text{H}_2\text{O}) / n(\text{Na}_2\text{CO}_3) = 0.182 / 0.0182 = 10\). Thus, \(x = 10\).

Award [1] for determining the mass of anhydrous sodium carbonate (1.93 g) and mass of water (3.28 g). Award [1.5] for determining both chemical amounts: \(n(\text{Na}_2\text{CO}_3) = 0.0182\text{ mol}\) and \(n(\text{H}_2\text{O}) = 0.182\text{ mol}\) (award [0.75] for each). Award [1] for determining the simplest whole number ratio and stating \(x = 10\).

Dissolving crude aspirin in a minimum volume of hot solvent ensures that the solution becomes saturated with respect to aspirin. This maximizes the yield of recrystallized aspirin when the solution cools, as a larger volume of solvent would keep more of the aspirin dissolved at lower temperatures. Upon cooling, pure aspirin crystallizes out because its solubility decreases significantly, while the soluble impurities remain in the cold solvent.

Award [1] for stating that a minimum volume is used to ensure the solution is saturated / to maximize the precipitation of pure aspirin on cooling (as more solvent would keep aspirin dissolved). Award [1] for explaining that upon cooling, pure aspirin crystallizes out while soluble impurities remain dissolved in the cold solvent (or insoluble impurities are filtered off beforehand).

The four-membered beta-lactam ring has bond angles of approximately \(90^\circ\), which are significantly smaller than the normal tetrahedral (\(109.5^\circ\)) or trigonal planar (\(120^\circ\)) bond angles expected for \(sp^3\) and \(sp^2\) hybridized atoms. This causes significant ring strain, making the amide carbonyl group highly reactive. The ring easily opens to form a covalent bond with the active site of the transpeptidase enzyme, irreversibly inhibiting cell wall cross-linking in bacteria.

Award [1] for identifying that the ring contains highly strained bond angles of \(90^\circ\) (which makes it unstable/highly reactive). Award [1] for explaining that the ring opens and covalently binds to/inhibits the transpeptidase enzyme, stopping the synthesis of the bacterial cell wall.

Morphine has two polar hydroxyl (\(-\text{OH}\)) groups that form hydrogen bonds with water, making it relatively hydrophilic. In contrast, diamorphine has these groups converted to less polar ester/acetyl (\(-\text{OCOCH}_3\)) groups, which increases its lipid solubility (lipophilicity). Since the blood-brain barrier consists of a lipid bilayer, the lipid-soluble diamorphine can cross it much more efficiently and quickly than morphine, resulting in a higher concentration reaching the brain.

Award [1] for comparing structures: diamorphine has ester groups while morphine has hydroxyl groups, making diamorphine less polar / more lipid-soluble / more lipophilic. Award [1] for explaining that lipid solubility allows diamorphine to penetrate the non-polar blood-brain barrier much more easily/rapidly.

Ranitidine acts as an \(\text{H}_2\)-receptor antagonist. It binds to histamine receptors on the parietal cells in the gastric glands, blocking histamine from stimulating the release of stomach acid. Omeprazole acts as a proton pump inhibitor. It undergoes a chemical transformation in the acidic environment of the parietal cells to its active form, which then covalently binds to and irreversibly inhibits the \(\text{H}^+/\text{K}^+\)-ATPase enzyme (the proton pump), directly preventing the active transport of \(\text{H}^+\) ions into the stomach lumen.

Award [1] for ranitidine: acts as an \(\text{H}_2\)-receptor antagonist / blocks histamine receptors to prevent acid secretion. Award [1] for omeprazole: acts as a proton pump inhibitor / inhibits the \(\text{H}^+/\text{K}^+\)-ATPase enzyme to stop \(\text{H}^+\) transport.

The antiviral drugs oseltamivir and zanamivir target the viral enzyme neuraminidase. Neuraminidase normally functions by cleaving the sialic acid receptor on the host cell surface, allowing newly synthesized influenza virions to bud off and escape from the infected host cell. By inhibiting neuraminidase, the new virus particles remain anchored to the host cell and cannot escape to infect surrounding healthy host cells.

Award [1] for identifying the enzyme: neuraminidase. Award [1] for explaining the mechanism: prevents the newly replicated virus particles from escaping/budding from the host cell surface / prevents them from spreading to and infecting other cells.

To mitigate the environmental impact of solvent waste in pharmaceutical manufacturing, green chemistry principles can be applied: 1. Solvent recovery and reuse: Designing systems that capture, purify (e.g., via distillation), and recycle organic solvents within the manufacturing process. 2. Use of safer solvents: Replacing volatile, hazardous, or chlorinated organic solvents with environmentally benign alternatives, such as water, ionic liquids, or supercritical carbon dioxide (\(\text{CO}_2\)).

Award [1] for each of two valid suggestions: Solvent recovery/recycling/reuse; Use of safer/alternative solvents (such as water, supercritical \(\text{CO}_2\), or bio-derived solvents); Designing solvent-free reaction pathways; Minimizing the number of reaction steps to reduce total solvent consumption.

1. Find the concentration of hydrogen ions, \([\text{H}^+]\), from the pH:
\([\text{H}^+] = 10^{-\text{pH}} = 10^{-7.40} = 3.98 \times 10^{-8}\text{ mol dm}^{-3}\)

2. Write the acid dissociation constant expression for the weak acid \(\text{H}_2\text{PO}_4^-\):
\(K_a = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\)

3. Substitute the known values to find the required ratio or concentration of \(\text{HPO}_4^{2-}\):
\(6.20 \times 10^{-8} = \frac{(3.98 \times 10^{-8}) \times [\text{HPO}_4^{2-}]}{0.100}\)

\([\text{HPO}_4^{2-}] = 0.1558\text{ mol dm}^{-3}\) (or \(0.1556\text{ mol dm}^{-3}\) if using \(\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}\))

4. Calculate the amount of moles of \(\text{Na}_2\text{HPO}_4\) in \(500.0\text{ cm}^3\) (\(0.5000\text{ dm}^3\)):
\(n(\text{Na}_2\text{HPO}_4) = 0.1558\text{ mol dm}^{-3} \times 0.5000\text{ dm}^3 = 0.0779\text{ mol}\) (or \(0.0778\text{ mol}\))

5. Calculate the mass of \(\text{Na}_2\text{HPO}_4\):
\(m = n \times M = 0.0779\text{ mol} \times 141.96\text{ g mol}^{-1} = 11.06\text{ g}\) (or \(11.05\text{ g}\))

To three significant figures, this rounds to \(11.1\text{ g}\) (or \(11.0\text{ g}\) depending on rounding).

[1 mark]: Correctly calculates \([\text{H}^+] = 3.98 \times 10^{-8}\text{ mol dm}^{-3}\) OR \(\text{p}K_a = 7.21\).
[1 mark]: Correctly sets up the equilibrium expression or Henderson-Hasselbalch equation and solves for \([\text{HPO}_4^{2-}] = 0.156\text{ mol dm}^{-3}\) (accept values between \(0.155\) and \(0.156\)).
[1 mark]: Calculates the amount of moles: \(n = 0.0778\text{ mol}\) to \(0.0779\text{ mol}\).
[1 mark]: Calculates the mass: \(11.1\text{ g}\) (accept \(11.0\text{ g}\) to \(11.1\text{ g}\)).

1. Convert the time from days to hours:
\(t = 7.00\text{ days} \times 24\text{ hours/day} = 168.0\text{ hours}\)

2. Determine the number of half-lives, \(n\):
\(n = \frac{t}{t_{1/2}} = \frac{168.0}{64.1} = 2.621\)

3. Calculate the remaining activity, \(A\):
\(A = A_0 \times (0.5)^n\)

\(A = 3.50 \times 10^8 \times (0.5)^{2.621} = 5.69 \times 10^7\text{ Bq}\)

Alternative method using decay constant, \(\lambda\):
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{64.1\text{ h}} = 0.01081\text{ h}^{-1}\)

\(A = A_0 e^{-\lambda t} = 3.50 \times 10^8 \times e^{-(0.01081 \times 168.0)} = 5.69 \times 10^7\text{ Bq}\)

[1 mark]: Converts 7.00 days to 168 hours.
[1 mark]: Correctly calculates the decay constant \(\lambda = 0.01081\text{ h}^{-1}\) OR the number of half-lives \(n = 2.621\).
[1 mark]: Shows correct substitution into the radioactive decay equation (exponential or half-life base form).
[1 mark]: Arrives at final activity of \(5.69 \times 10^7\text{ Bq}\) (accept values from \(5.68 \times 10^7\) to \(5.71 \times 10^7\)).