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The second electron of calcium is removed from the 4s subshell, which is in the fourth main energy level (outermost shell). The third electron must be removed from the 3p subshell, which is in the third main energy level (an inner shell closer to the nucleus). Electrons in the third shell experience significantly less shielding from the positive nuclear charge and are closer to the nucleus. This results in a much stronger electrostatic force of attraction, requiring much more energy to remove the third electron.

M1 (1.0 mark): State that the second electron is removed from the 4th shell / 4s orbital, whereas the third electron is removed from the 3rd shell / 3p orbital. M2 (1.0 mark): Explain that the 3rd shell is closer to the nucleus and experiences less shielding. M3 (0.5 mark): State that there is a stronger electrostatic attraction between the nucleus and the third electron.

Chlorine has 7 outer shell electrons. The negative charge adds 1 electron, giving 8 electrons. Chlorine forms 2 single bonds with the 2 fluorine atoms, which uses 2 electrons, leaving 6 non-bonding electrons (3 lone pairs). There are 5 electron pairs in total around the central chlorine atom, giving a trigonal bipyramidal electron pair geometry. To minimize repulsion, the 3 lone pairs occupy the equatorial positions at 120 degrees to each other. This forces the 2 bonding pairs into axial positions, resulting in a linear molecular shape and a bond angle of \(180^\circ\).

M1 (1.0 mark): Correctly identify that there are 2 bonding pairs and 3 lone pairs (5 electron pairs in total) around the central chlorine atom. M2 (1.0 mark): State that the molecular shape is linear and the bond angle is \(180^\circ\). M3 (0.5 mark): Explain that lone pairs repel more than bonding pairs and occupy equatorial positions to minimize repulsion.

1. Convert units to SI: \(P = 100\text{ kPa} = 100,000\text{ Pa}\), \(V = 340\text{ cm}^3 = 3.40 \times 10^{-4}\text{ m}^3\), \(T = 293\text{ K}\). 2. Rearrange the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT}\). 3. Calculate the number of moles: \(n = \frac{100,000 \times 3.40 \times 10^{-4}}{8.31 \times 293} = \frac{34.0}{2434.83} = 0.013964\text{ mol}\). 4. Calculate molar mass: \(M_r = \frac{\text{mass}}{n} = \frac{0.557}{0.013964} = 39.9\text{ g mol}^{-1}\). 5. Identify the noble gas with relative atomic mass of approximately 39.9 as Argon (Ar).

M1 (1.0 mark): Correctly convert units and rearrange equation to calculate moles: \(n = 0.0140\text{ mol}\) (or \(0.01396\text{ mol}\)). M2 (1.0 mark): Calculate the relative atomic mass as \(39.9\) (allow \(39.8 - 40.1\)). M3 (0.5 mark): Correctly identify the gas as Argon (Ar).

Silicon has a giant covalent macromolecular structure in which every silicon atom is tetrahedrally bonded to four other silicon atoms by strong covalent bonds. Breaking these strong covalent bonds throughout the giant lattice requires a very large amount of thermal energy. In contrast, phosphorus exists as simple molecular \(\text{P}_4\) molecules held together only by weak intermolecular van der Waals forces. Overcoming these weak van der Waals forces requires significantly less thermal energy than breaking covalent bonds, resulting in a much lower melting point for phosphorus.

M1 (1.0 mark): Identify that silicon has a giant covalent/macromolecular structure and explain that strong covalent bonds must be broken. M2 (1.0 mark): Identify that phosphorus has a simple molecular structure (or exists as \(\text{P}_4\) molecules) with weak van der Waals forces between molecules. M3 (0.5 mark): Link the difference in energy required to overcome these forces/bonds to the relative melting points (breaking covalent bonds requires much more energy than overcoming van der Waals forces).

1. Calculate temperature change: \(\Delta T = 58.2 - 21.5 = 36.7\text{ K}\). 2. Calculate heat energy transferred: \(q = m c \Delta T = 150\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 36.7\text{ K} = 23010.9\text{ J} = 23.011\text{ kJ}\). 3. Calculate moles of propan-1-ol combusted: \(n = \frac{\text{mass}}{M_r} = \frac{1.50}{60.0} = 0.0250\text{ mol}\). 4. Calculate molar enthalpy change: \(\Delta H_c = -\frac{q}{n} = -\frac{23.011}{0.0250} = -920.44\text{ kJ mol}^{-1}\). 5. Expressing to 3 significant figures gives \(-920\text{ kJ mol}^{-1}\).

M1 (1.0 mark): Calculate heat energy transferred \(q = 23.0\text{ kJ}\) (or \(23011\text{ J}\)). M2 (1.0 mark): Calculate moles of propan-1-ol = \(0.0250\text{ mol}\) and divide heat by moles to obtain \(920.4\text{ kJ mol}^{-1}\). M3 (0.5 mark): State the final answer with a negative sign and to 3 significant figures: \(-920\text{ kJ mol}^{-1}\) (allow \(-9.20 \times 10^2\)).

When sodium bromide reacts with concentrated sulfuric acid, the bromide ions are strong enough reducing agents to reduce sulfur in sulfuric acid from an oxidation state of +6 to +4 in sulfur dioxide (\(\text{SO}_2\)). The bromide ions are oxidized to bromine gas (\(\text{Br}_2\)), which is the orange-brown vapor. The balanced chemical equation is: \( 2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \) or ionic: \( 2\text{Br}^- + 2\text{H}^+ + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \). Because sulfuric acid oxidizes bromide ions to bromine, its role is an oxidizing agent.

M1 (1.0 mark): Provide a balanced overall molecular or ionic equation: \( 2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \) (or \( 2\text{Br}^- + 2\text{H}^+ + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \)). M2 (1.0 mark): Correctly state that sulfuric acid acts as an oxidizing agent (or is reduced). M3 (0.5 mark): Identify that the orange-brown vapor is bromine (\(\text{Br}_2\)) and the choking gas is sulfur dioxide (\(\text{SO}_2\)).

1. The solubility of Group 2 hydroxides increases down the group from Mg(OH)2 (sparingly soluble) to Ba(OH)2 (highly soluble). 2. To test for sulfate ions, add barium chloride solution (\(\text{BaCl}_2\)) acidified with hydrochloric acid (\(\text{HCl}\)) or nitric acid (\(\text{HNO}_3\)). 3. The observation for a positive result is the formation of a white precipitate of barium sulfate (\(\text{BaSO}_4\)).

M1 (1.0 mark): State that the solubility of Group 2 hydroxides increases down the group. M2 (1.0 mark): Reagent: Acidified barium chloride solution (accept barium chloride acidified with HCl or HNO3; reject sulfuric acid). M3 (0.5 mark): Observation: White precipitate.

1. Write the reduction half-equation for dichromate(VI): \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \). 2. Write the oxidation half-equation for iron(II): \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \). 3. Multiply the oxidation half-equation by 6 to balance electrons: \( 6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^- \). 4. Combine the two half-equations: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \).

M1 (1.0 mark): Write the correct reduction half-equation: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \). M2 (1.0 mark): Write the correct overall balanced ionic equation: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \). M3 (0.5 mark): Correctly identify that the oxidation state of chromium changes from +6 to +3.

1. Write the electron configurations:
Phosphorus is \(1s^2 2s^2 2p^6 3s^2 3p^3\) and sulfur is \(1s^2 2s^2 2p^6 3s^2 3p^4\).
2. Contrast the 3p subshells:
Phosphorus has three singly occupied 3p orbitals (no pairing).
Sulfur has four electrons in the 3p subshell, meaning one 3p orbital contains a pair of electrons.
3. Identify the effect of repulsion:
The mutual repulsion between the two paired electrons in this 3p orbital of sulfur makes it easier to remove one of them than from the half-filled subshell of phosphorus.

M1 (1 mark): Identifies that sulfur has a paired electron in a 3p orbital, whereas phosphorus has only unpaired electrons in its 3p orbitals (or compares configurations \(3p^3\) vs \(3p^4\)).
M2 (1 mark): Explains that there is mutual repulsion between the paired electrons in the same orbital in sulfur.
M3 (0.5 marks): Concludes that this repulsion makes it require less energy to remove the electron from sulfur.

1. Determine the number of valence electrons around the central chlorine atom: Chlorine is in Group 17, so it has 7 valence electrons. The positive charge (\(+\)) means we subtract 1 electron, leaving 6 valence electrons.
2. Count bonding and non-bonding electrons: Each of the 2 fluorine atoms forms a single covalent bond with chlorine. This uses 2 electrons, leaving \(6 - 2 = 4\) non-bonding electrons.
3. Calculate the number of lone pairs: 4 non-bonding electrons correspond to 2 lone pairs.
4. Determine shape and angle: With 2 bonding pairs and 2 lone pairs, the electron pair geometry is based on a tetrahedron. The shape of the ion is bent (or V-shaped). Due to the greater repulsion of the 2 lone pairs compared to the bonding pairs, the bond angle is reduced from the tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\).

M1 (1 mark): Correctly identifies 2 bonding pairs and 2 lone pairs on the central chlorine atom.
M2 (1 mark): Identifies the shape as bent / V-shaped / non-linear.
M3 (0.5 marks): States a bond angle in the range \(104^\circ\) to \(105^\circ\) (accept any value in this range or significantly less than \(109.5^\circ\) with correct reasoning).

1. Convert pressure to Pascals: \(P = 101\text{ kPa} = 101,000\text{ Pa}\).
2. Use the ideal gas equation \(PV = nRT\) to find the moles (\(n\)) of gas X:
\(n = \frac{PV}{RT} = \frac{101000 \times 1.15 \times 10^{-4}}{8.31 \times 298}\)
\(n = \frac{11.615}{2476.38} = 4.6903 \times 10^{-3}\text{ mol}\).
3. Calculate the relative molecular mass (\(M_r\)):
\(M_r = \frac{\text{mass}}{n} = \frac{0.284}{4.6903 \times 10^{-3}} = 60.55\text{ g mol}^{-1}\).
Rounding to 3 significant figures gives \(60.6\).

M1 (1 mark): Correct rearrangement of \(PV = nRT\) and substitution, ensuring pressure is converted to \(101,000\text{ Pa}\).
M2 (0.5 marks): Correct calculation of moles: \(n = 4.69 \times 10^{-3}\text{ mol}\) (allow consequential error from conversion).
M3 (1 mark): Correct calculation of \(M_r = 60.6\) (accept range \(60.4\) to \(60.8\) based on rounding).

1. Write the equation for the standard enthalpy of formation of liquid ethanol:
\(2\text{C(s)} + 3\text{H}_2\text{(g)} + 0.5\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\)
2. Construct the Hess's Law expression using enthalpies of combustion (where reactants and products both combust to form \(\text{CO}_2\) and \(\text{H}_2\text{O}\)):
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [2 \times \Delta_c H^\theta(\text{C(s)}) + 3 \times \Delta_c H^\theta(\text{H}_2\text{(g)})] - [\Delta_c H^\theta(\text{C}_2\text{H}_5\text{OH(l)})]\)
3. Substitute the values:
\(\Delta_f H^\theta = [2(-393.5) + 3(-285.8)] - [-1367.3]\)
\(\Delta_f H^\theta = [-787.0 - 857.4] + 1367.3\)
\(\Delta_f H^\theta = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

M1 (1 mark): States a correct Hess's Law cycle or algebraic equation representing the formation of ethanol from its combustion products: \(\Delta_f H = 2\Delta_c H(\text{C}) + 3\Delta_c H(\text{H}_2) - \Delta_c H(\text{ethanol})\).
M2 (1 mark): Correct substitution of the data into the equation: \(2(-393.5) + 3(-285.8) - (-1367.3)\).
M3 (0.5 marks): Obtains the final value of \(-277.1\text{ kJ mol}^{-1}\) with the correct negative sign (reject if sign is positive).

To distinguish between Group 2 cations (\(\text{Mg}^{2+}\) and \(\text{Ba}^{2+}\)), we can exploit the trend in the solubility of Group 2 sulfates or hydroxides.

Method 1: Add a soluble sulfate, such as aqueous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) or sulfuric acid (\(\text{H}_2\text{SO}_4\)).
- Barium sulfate is highly insoluble, so a white precipitate of \(\text{BaSO}_4\) will form in the barium chloride tube.
- Magnesium sulfate is soluble, so no visible reaction occurs (solution remains colourless/clear).

Method 2: Add a soluble hydroxide, such as aqueous sodium hydroxide (\(\text{NaOH}\)).
- Magnesium hydroxide is insoluble, so a white precipitate of \(\text{Mg(OH)}_2\) will form in the magnesium chloride tube.
- Barium hydroxide is soluble, so no visible reaction occurs.

M1 (1 mark): Identifies a suitable reagent (aqueous sodium sulfate, sulfuric acid, or sodium hydroxide).
M2 (0.5 marks): Describes the correct negative observation (no visible change/remains colourless) for the correct chloride (e.g., no change with magnesium chloride if sulfate is used, or no change with barium chloride if hydroxide is used).
M3 (1 mark): Describes the correct positive observation (white precipitate formed) for the correct chloride (e.g., white precipitate with barium chloride if sulfate is used, or white precipitate with magnesium chloride if hydroxide is used).

1. Write the reduction half-equation for the dichromate(VI) ion in acidic medium:
Balance Cr atoms: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}\)
Balance oxygen with water: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Balance hydrogen with H⁺: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Balance charges with electrons: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

2. Write the oxidation half-equation for iron(II):
\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)

3. Multiply the iron half-equation by 6 so that the number of electrons transferred matches:
\(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-\)

4. Combine the two half-equations (the electrons cancel out):
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}\)

M1 (1 mark): Correctly identifies or constructs the reduction half-equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\).
M2 (0.5 marks): Correctly states the oxidation half-equation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\).
M3 (1 mark): Combines both equations correctly to produce the final overall equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}\) (fully balanced with respect to both atoms and charges).

1. Convert the gas data into standard SI units:
\(P = 100\text{ kPa} = 1.00 \times 10^5\text{ Pa}\)
\(V = 436\text{ cm}^3 = 4.36 \times 10^{-4}\text{ m}^3\)
\(T = 300\text{ K}\)

2. Calculate the total moles of gas produced using the ideal gas equation:
\(n_{\text{gas}} = \frac{PV}{RT} = \frac{1.00 \times 10^5 \times 4.36 \times 10^{-4}}{8.31 \times 300} = 0.017489\text{ mol}\)

3. Use the stoichiometric ratio from the balanced equation to find the moles of \(Ca(NO_3)_2\):
\(2\text{ mol}\) of \(Ca(NO_3)_2\) produces \(5\text{ mol}\) of gas (\(4\text{ mol } NO_2 + 1\text{ mol } O_2\)).
\(n(Ca(NO_3)_2) = 0.017489 \times \frac{2}{5} = 0.0069956\text{ mol}\)

4. Calculate the mass of pure \(Ca(NO_3)_2\):
\(M_r(Ca(NO_3)_2) = 40.1 + (2 \times 14.0) + (6 \times 16.0) = 164.1\)
\(m = 0.0069956 \times 164.1 = 1.148\text{ g}\)

5. Calculate percentage purity:
\(\text{Percentage purity} = \frac{1.148}{1.30} \times 100 = 88.3\%\) (3 significant figures)

- **1 Mark**: Correct calculation of the total moles of gas (\(0.0175\text{ mol}\)) using \(PV=nRT\) with appropriate unit conversions.
- **1 Mark**: Correct application of the stoichiometry ratio (\(2:5\)) to find moles of \(Ca(NO_3)_2\) (\(0.00700\text{ mol}\)).
- **1 Mark**: Correct calculation of the mass of pure \(Ca(NO_3)_2\) (\(1.15\text{ g}\) or \(1.148\text{ g}\)).
- **1 Mark**: Correct percentage purity calculation leading to \(88.3\%\).
- **0.5 Marks**: Final answer expressed to 3 significant figures.

1. Calculate heat energy transferred (\(q\)):
\(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)
\(\Delta T = 27.3 - 19.2 = 8.1\text{ K}\)
\(q = mc\Delta T = 100.0 \times 4.18 \times 8.1 = 3385.8\text{ J} = 3.3858\text{ kJ}\)

2. Determine the limiting reactant and moles of water formed:
\(n(HNO_3) = 0.0500 \times 1.20 = 0.0600\text{ mol}\)
\(n(KOH) = 0.0500 \times 1.40 = 0.0700\text{ mol}\)
Since they react in a 1:1 ratio, \(HNO_3\) is the limiting reactant.
\(n(H_2O\text{ formed}) = 0.0600\text{ mol}\)

3. Calculate the molar enthalpy of neutralization (\(\Delta H_{\text{neut}}\)):
\(\Delta H_{\text{neut}} = -\frac{q}{n(H_2O)} = -\frac{3.3858}{0.0600} = -56.43\text{ kJ mol}^{-1}\)
To 3 significant figures, this is \(-56.4\text{ kJ mol}^{-1}\).

- **1 Mark**: Calculation of heat transferred \(q = 3385.8\text{ J}\) (or \(3.39\text{ kJ}\)).
- **1 Mark**: Correct identification of limiting reactant and calculation of moles of water formed (\(0.0600\text{ mol}\)).
- **1 Mark**: Dividing the calculated heat energy (in \(\text{kJ}\)) by the correct moles of water formed.
- **1 Mark**: Correct sign (negative, for exothermic reaction) and value of \(-56.4\).
- **0.5 Marks**: Correct units of \(\text{kJ mol}^{-1}\) included in final working/answer.

1. Let the percentage abundance of \(^{26}\text{X}\) be \(y\%\).
2. Therefore, the abundance of \(^{25}\text{X}\) is \(2y\%\).
3. Since the sum of all abundances is \(100\%\), the abundance of \(^{24}\text{X}\) is \((100 - 3y)\%\).

4. Set up the weighted average relative atomic mass equation:
\[A_r = \frac{24(100 - 3y) + 25(2y) + 26y}{100} = 24.32\]

5. Solve for \(y\):
\[2400 - 72y + 50y + 26y = 2432\]
\[2400 + 4y = 2432\]
\[4y = 32\]
\[y = 8.0\]

6. Determine percentage abundances:
- \(^{26}\text{X} = 8.0\%\)
- \(^{25}\text{X} = 2 \times 8.0 = 16.0\%\)
- \(^{24}\text{X} = 100 - 3(8.0) = 76.0\%\)

- **1 Mark**: Setting up correct algebraic terms for the three isotope abundances (e.g., \(y\), \(2y\), and \(100-3y\)).
- **1 Mark**: Setting up the correct relative atomic mass equation using the algebraic terms.
- **1 Mark**: Correctly solving the algebraic equation to find the value of \(y = 8.0\).
- **1 Mark**: Finding all three correct final percentage abundances (\(76.0\%\), \(16.0\%\), \(8.0\%\)).
- **0.5 Marks**: Clear presentation of final abundances with appropriate decimal precision.

1. Write the balanced ionic equation for the redox reaction:
\[MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\]

2. Calculate the moles of manganate(VII) ions used:
\[n(MnO_4^-) = C \times V = 0.0200 \times \frac{16.80}{1000} = 3.36 \times 10^{-4}\text{ mol}\]

3. Calculate the moles of iron(II) ions using the 1:5 stoichiometric ratio:
\[n(Fe^{2+}) = 5 \times n(MnO_4^-) = 5 \times 3.36 \times 10^{-4} = 1.68 \times 10^{-3}\text{ mol}\]

4. Calculate the mass of iron(II) ions (\(A_r(Fe) = 55.8\)):
\[m(Fe^{2+}) = 1.68 \times 10^{-3} \times 55.8 = 0.093744\text{ g}\]

5. Calculate the percentage by mass of \(Fe^{2+}\) in the tablet:
\[\%\text{ by mass} = \frac{0.093744\text{ g}}{0.242\text{ g}} \times 100 = 38.737\% \approx 38.7\%\]

- **1 Mark**: Correct calculation of moles of \(MnO_4^-\) (\(3.36 \times 10^{-4}\text{ mol}\)).
- **1 Mark**: Correctly using the 1:5 stoichiometry to find moles of \(Fe^{2+}\) (\(1.68 \times 10^{-3}\text{ mol}\)).
- **1 Mark**: Correct calculation of the mass of iron (\(0.0937\text{ g}\) or \(0.09374\text{ g}\)).
- **1 Mark**: Correct percentage by mass calculation leading to \(38.7\%\).
- **0.5 Marks**: Reporting the final answer to 3 significant figures.

1. Express the reaction enthalpy in terms of bond breaking and bond making:
\[\Delta H = \sum (\text{bond enthalpies of reactants}) - \sum (\text{bond enthalpies of products})\]

2. Bonds broken in reactants:
- \(1 \times (N-N)\) bond
- \(4 \times (N-H)\) bonds \(= 4 \times 391 = 1564\text{ kJ mol}^{-1}\)
- \(1 \times (O=O)\) bond \(= 496\text{ kJ mol}^{-1}\)
\[\text{Total broken} = E(N-N) + 1564 + 496 = E(N-N) + 2060\text{ kJ mol}^{-1}\]

3. Bonds made in products:
- \(1 \times (N\equiv N)\) bond \(= 944\text{ kJ mol}^{-1}\)
- \(4 \times (O-H)\) bonds (in \(2H_2O\)) \(= 4 \times 463 = 1852\text{ kJ mol}^{-1}\)
\[\text{Total made} = 944 + 1852 = 2796\text{ kJ mol}^{-1}\]

4. Set up the equation and solve for \(E(N-N)\):
\[-534 = [E(N-N) + 2060] - 2796\]
\[-534 = E(N-N) - 736\]
\[E(N-N) = -534 + 736 = +202\text{ kJ mol}^{-1}\]

- **1 Mark**: Correct determination of reactant bonds broken excluding \(N-N\) (\(2060\text{ kJ mol}^{-1}\)).
- **1 Mark**: Correct determination of product bonds made (\(2796\text{ kJ mol}^{-1}\)).
- **1 Mark**: Correct algebraic expression relating the reactants, products, and overall reaction enthalpy.
- **1 Mark**: Correct final calculated value of \(202\).
- **0.5 Marks**: Correct sign (positive/\(+\)) and unit (\(\text{kJ mol}^{-1}\)) provided.

The 3D diagram should show sulfur (S) as the central atom. Two fluorine (F) atoms are in axial positions (forming a linear or near-linear F-S-F backbone). Two F atoms are in equatorial positions shown with a wedge and a dash (or simple lines) at an angle of less than 120 degrees due to repulsion. One equatorial position is occupied by a lone pair (represented as two dots or a lobe). The shape name is 'seesaw'. According to VSEPR theory, electron pairs repel each other to be as far apart as possible. Lone pairs repel more than bonding pairs, forcing the equatorial bonding pairs closer together (angle < 120 degrees) and axial bonding pairs slightly away from the lone pair (angle < 180 degrees).

1 mark: Correct 3D diagram showing seesaw geometry with the central sulfur atom, four fluorine atoms, and one equatorial lone pair clearly indicated. 1.1 marks: Correctly naming the shape as 'seesaw' and explaining that the 5 electron pairs minimise repulsion by adopting a trigonal bipyramidal orientation with the lone pair in the equatorial position.

The diagram should depict a regular, closely packed grid or array of circles representing sodium cations, each clearly labeled with a '+' sign or '\(\text{Na}^+\)'. Surrounding these cations, small dots or circles labeled '-' or '\(\text{e}^-\)' represent the delocalised sea of electrons. To explain ductility: the metallic bond is non-directional. When stress is applied, layers of metal ions slide past one another. The delocalised electrons are mobile and continue to hold the sliding layers together, preventing the lattice from shattering.

1 mark: Drawing a regular lattice containing labelled sodium cations (\(\text{Na}^+\)) and a sea of delocalised electrons. 1.1 marks: Explaining ductility in terms of layers of cations sliding past one another while the delocalised electrons maintain the non-directional electrostatic attractions.

The schematic diagram should be a linear flowchart or block diagram representing the TOF mass spectrometer: Sample introduction leading into an ionisation chamber, followed by negatively/positively charged acceleration plates, a long field-free drift tube (flight tube), and ending at the ion detector. The two main ionization techniques are: 1. Electron impact (vaporised sample bombarded with high-energy electrons from an electron gun to knock off an electron, forming 1+ ions). 2. Electrospray ionisation (sample dissolved in a volatile solvent and injected through a fine hypodermic needle at high voltage, gaining a proton, \(\text{H}^+\)).

1 mark: Correctly sequenced diagram showing the stages: Ionisation, Acceleration, Drift Tube, and Detector in order. 1.1 marks: Correctly naming both 'electron impact' (or electron ionisation) and 'electrospray ionisation' as the two methods of ionisation.

The flowchart should show two branches: one for \(\text{Cl}^-\text{(aq)}\) and one for \(\text{I}^-\text{(aq)}\). First step: Add \(\text{AgNO}_3\text{(aq)}\). \(\text{Cl}^-\) branch shows 'White precipitate of \(\text{AgCl}\)'. \(\text{I}^-\) branch shows 'Yellow precipitate of \(\text{AgI}\)'. Second step: Add concentrated \(\text{NH}_3\text{(aq)}\). \(\text{Cl}^-\) branch shows 'Precipitate dissolves to form a colourless solution'. \(\text{I}^-\) branch shows 'Precipitate remains / insoluble'.

1 mark: Correctly matching the precipitate colours (white for \(\text{AgCl}\), yellow for \(\text{AgI}\)). 1.1 marks: Correctly identifying the solubility behaviour in concentrated ammonia (\(\text{AgCl}\) dissolves, \(\text{AgI}\) remains insoluble).

In the enthalpy profile diagram: 1. The vertical axis is labeled 'Enthalpy' (or H) and the horizontal axis is 'Progress of reaction' (or Reaction coordinate). 2. Reactants are at a higher level than the products. 3. A curve rises from the reactants to a peak (transition state) and then drops down to the products' level. 4. The activation energy (\(E_{\text{a}}\)) is shown with an arrow pointing upwards from the reactant level to the peak. 5. The enthalpy change (\(\Delta H\)) is shown with an arrow pointing downwards from the reactant level to the product level, indicating an exothermic process.

1 mark: Correctly drawing the exothermic profile with reactants higher than products and labeled axes. 1.1 marks: Correctly indicating \(E_{\text{a}}\) as an upward arrow from reactants to the peak, and \(\Delta H\) as a downward arrow from reactants to products.

The drawing must depict a volumetric flask, which has a wide flat bottom and a very narrow neck. A single calibration line (graduation mark) must be drawn horizontally across the neck. The liquid inside must have a concave meniscus. The bottom of the curved meniscus must rest exactly on the graduation line. To prepare the solution correctly: the solid solute is dissolved first in a beaker, transferred with washings to the flask, and water is added until near the line. The final drops are added slowly with a teat pipette at eye level to avoid overshoot.

1 mark: Drawing a flask with a narrow neck, showing a clear graduation line and a concave meniscus with the bottom of the curve touching the line. 1.1 marks: Stating that the final volume must be adjusted dropwise using a pipette and read at eye level to avoid parallax error.

Mechanism steps:
1. The C-Br bond in 1-bromobutane is polar. Label the carbon atom as \(\text{C}^{\delta+}\) and the bromine atom as \(\text{Br}^{\delta-}\).
2. Draw a curly arrow starting from a lone pair on the oxygen atom of the hydroxide ion (\(\text{OH}^-\)) to the \(\text{C}^{\delta+}\) carbon atom.
3. Draw a curly arrow from the C-Br bond to the Br atom.
4. The products are butan-1-ol and a bromide ion (\(\text{Br}^-\)).

Percentage yield calculation:
- Moles of 1-bromobutane = \(13.7 \text{ g} / 137.0 \text{ g/mol} = 0.100 \text{ mol}\)
- Theoretical yield of butan-1-ol = \(0.100 \text{ mol} \times 74.0 \text{ g/mol} = 7.40 \text{ g}\)
- \(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)
- \(\text{Percentage Yield} = \frac{5.18 \text{ g}}{7.40 \text{ g}} \times 100 = 70.0\%\)

- M1: For showing correct \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) dipole on 1-bromobutane (1 mark)
- M2: For drawing a curly arrow from a lone pair on the oxygen of \(\text{OH}^-\) to the \(\text{C}^{\delta+}\) carbon atom (1 mark)
- M3: For drawing a curly arrow from the C-Br bond to the Br atom (1 mark)
- M4 (1.2 marks): For the percentage yield formula and correct calculation of \(70.0\%\) (1 mark for correct numbers, 0.2 marks for correct formula/working).

Mechanism steps:
1. Label the hydrogen bromide dipole as \(\text{H}^{\delta+}-\text{Br}^{\delta-}\).
2. Draw a curly arrow from the double bond of propene to the electrophilic hydrogen of \(\text{H-Br}\).
3. Draw a curly arrow from the H-Br bond to the bromine atom.
4. Show the secondary carbocation intermediate: \(\text{CH}_3\text{CH}^+\text{CH}_3\), and a bromide ion \(\text{Br}^-\).
5. Draw a curly arrow from the lone pair on the bromide ion (\(\text{Br}^-\)) to the carbon containing the positive charge.

Atom economy calculation:
- \(\text{Atom Economy} = \frac{\text{Molecular mass of desired product}}{\text{Total molecular mass of all reactants}} \times 100\)
- Since this is an addition reaction where all reactants combine to form a single product, the atom economy is \(100\%\).

- M1: For drawing a curly arrow from the \(\text{C=C}\) double bond to the \(\text{H}\) of \(\text{H-Br}\) and showing the \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) dipole (1 mark)
- M2: For drawing a curly arrow from the H-Br bond to the Br atom and showing the correct structure of the secondary carbocation intermediate (1 mark)
- M3: For drawing a curly arrow from a lone pair on the \(\text{Br}^-\) ion to the positively charged carbon of the carbocation (1 mark)
- M4 (1.2 marks): For stating the correct atom economy formula and calculating the value as \(100\%\) (1.2 marks).

Mechanism steps:
1. Protonation: Draw a curly arrow from a lone pair on the oxygen atom of the alcohol group to \(\text{H}^+\) from the acid catalyst, forming the protonated intermediate \(\text{CH}_3\text{CH}_2\text{CH}(\text{OH}_2^+)\text{CH}_3\).
2. Loss of water: Draw a curly arrow from the C-O bond to the oxygen atom to form water and the carbocation intermediate \(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\).
3. Elimination: Draw a curly arrow from a C-H bond of the adjacent carbon (C3) to the adjacent C-C bond to regenerate the catalyst and form the double bond of but-2-ene.

Percentage yield calculation:
- Moles of butan-2-ol = \(14.8 \text{ g} / 74.0 \text{ g/mol} = 0.200 \text{ mol}\)
- Theoretical yield of but-2-ene = \(0.200 \text{ mol} \times 56.0 \text{ g/mol} = 11.2 \text{ g}\)
- \(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)
- \(\text{Percentage Yield} = \frac{6.72 \text{ g}}{11.2 \text{ g}} \times 100 = 60.0\%\)

- M1: For drawing a curly arrow from the lone pair of the oxygen on butan-2-ol to the proton (\(\text{H}^+\)) (1 mark)
- M2: For drawing a curly arrow from the C-O bond of the protonated intermediate to the oxygen atom to release water, and showing the carbocation intermediate (1 mark)
- M3: For drawing a curly arrow from the C-H bond on the adjacent carbon (C3) to the C-C bond next to the positive charge to form the C=C bond (1 mark)
- M4 (1.2 marks): For stating the formula for percentage yield and correctly calculating the percentage yield as \(60.0\%\) (1.2 marks).

First, calculate the heat energy released using \(q = m c \Delta T\):
\(q = 200.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 19.5\text{ K} = 16302\text{ J} = 16.302\text{ kJ}\).

Next, calculate the amount of methanol combusted in moles:
\(\text{moles of methanol} = \frac{0.800\text{ g}}{32.0\text{ g mol}^{-1}} = 0.0250\text{ mol}\).

Finally, calculate the enthalpy change of combustion per mole:
\(\Delta H_c = -\frac{q}{n} = -\frac{16.302\text{ kJ}}{0.0250\text{ mol}} = -652.08\text{ kJ mol}^{-1}\).

Rounding to 3 significant figures gives \(-652\text{ kJ mol}^{-1}\).

M1: Correct calculation of heat energy transferred, \(q = 16.3\text{ kJ}\) (allow \(16300\text{ J}\)) [1 mark]
M2: Correct calculation of moles of methanol, \(n = 0.0250\text{ mol}\) [1 mark]
M3: Correct calculation of \(\Delta H_c = -652\text{ kJ mol}^{-1}\) (must have negative sign and be to 3 significant figures) [1 mark]

First, calculate the theoretical maximum moles of cyclohexene that can be produced:
\(\text{moles of cyclohexanol starting material} = \frac{12.5\text{ g}}{100.0\text{ g mol}^{-1}} = 0.125\text{ mol}\).
Since the reaction stoichiometry is 1:1, the theoretical moles of cyclohexene = \(0.125\text{ mol}\).

Next, calculate the theoretical mass of cyclohexene:
\(\text{theoretical mass} = 0.125\text{ mol} \times 82.0\text{ g mol}^{-1} = 10.25\text{ g}\).

Finally, calculate the percentage yield:
\(\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100 = \frac{6.15}{10.25} \times 100 = 60.0\%\).

M1: Correct calculation of theoretical moles of cyclohexene (\(0.125\text{ mol}\)) [1 mark]
M2: Correct calculation of theoretical mass of cyclohexene (\(10.25\text{ g}\)) [1 mark]
M3: Correct final percentage yield calculation (\(60.0\%\)) to 3 significant figures [1 mark]

First, determine the equilibrium amounts of all species in moles:
Initial \(\text{PCl}_5 = 0.800\text{ mol}\), \(\text{PCl}_3 = 0\text{ mol}\), \(\text{Cl}_2 = 0\text{ mol}\).
Equilibrium \(\text{PCl}_5 = 0.240\text{ mol}\).
Moles of \(\text{PCl}_5\) reacted = \(0.800 - 0.240 = 0.560\text{ mol}\).
By stoichiometry, equilibrium moles of \(\text{PCl}_3 = 0.560\text{ mol}\) and \(\text{Cl}_2 = 0.560\text{ mol}\).

Next, calculate the equilibrium concentrations in \(\text{mol dm}^{-3}\) (Volume = \(2.00\text{ dm}^3\)):
\([\text{PCl}_5] = \frac{0.240}{2.00} = 0.120\text{ mol dm}^{-3}\)
\([\text{PCl}_3] = \frac{0.560}{2.00} = 0.280\text{ mol dm}^{-3}\)
\([\text{Cl}_2] = \frac{0.560}{2.00} = 0.280\text{ mol dm}^{-3}\)

Finally, calculate \(K_c\):
\(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.280 \times 0.280}{0.120} = 0.65333...\text{ mol dm}^{-3}\).
To 3 significant figures, \(K_c = 0.653\text{ mol dm}^{-3}\).

M1: Correctly determining equilibrium moles of \(\text{PCl}_3\) and \(\text{Cl}_2\) as \(0.560\text{ mol}\) [1 mark]
M2: Correctly dividing by \(2.00\text{ dm}^3\) to find all equilibrium concentrations [1 mark]
M3: Correct calculation of \(K_c = 0.653\) with units \(\text{mol dm}^{-3}\) [1 mark]

First, calculate the number of moles of chloride ions precipitated as \(\text{AgCl}\):
\(\text{moles of AgCl} = \frac{2.87\text{ g}}{143.4\text{ g mol}^{-1}} = 0.0200\text{ mol}\).
Since each mole of a monochloroalkane contains one mole of chlorine, the moles of chloroalkane = \(0.0200\text{ mol}\).

Next, calculate the molar mass of the chloroalkane:
\(\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1.85\text{ g}}{0.0200\text{ mol}} = 92.5\text{ g mol}^{-1}\).

Finally, deduce the molecular formula:
Subtracting the atomic mass of chlorine (\(35.5\)) gives the mass of the alkyl group \(\text{R}\):
\(M_r(\text{R}) = 92.5 - 35.5 = 57.0\text{ g mol}^{-1}\).
An alkyl group has the general formula \(\text{C}_n\text{H}_{2n+1}\):
\(12n + (2n + 1) = 57 \Rightarrow 14n = 56 \Rightarrow n = 4\).
Therefore, the molecular formula is \(\text{C}_4\text{H}_9\text{Cl}\).

M1: Calculate moles of \(\text{AgCl} = 0.0200\text{ mol}\) [1 mark]
M2: Calculate molar mass of chloroalkane = \(92.5\text{ g mol}^{-1}\) (allow \(92.3 - 92.7\)) [1 mark]
M3: Deduces the molecular formula as \(\text{C}_4\text{H}_9\text{Cl}\) [1 mark]

First, convert all values to SI units:
\(T = 87.0 + 273.15 = 360\text{ K}\) (or \(360\text{ K}\) using \(273\))
\(p = 100\text{ kPa} = 1.00 \times 10^5\text{ Pa}\)
\(V = 449\text{ cm}^3 = 449 \times 10^{-6}\text{ m}^3 = 4.49 \times 10^{-4}\text{ m}^3\)

Next, use the ideal gas equation (\(pV = nRT\)) to find the number of moles (\(n\)):
\(n = \frac{pV}{RT} = \frac{1.00 \times 10^5\text{ Pa} \times 4.49 \times 10^{-4}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 360\text{ K}} = \frac{44.9}{2991.6} = 0.0150\text{ mol}\).

Finally, calculate the molar mass (\(M_r\)) of the liquid:
\(M_r = \frac{\text{mass}}{n} = \frac{1.27\text{ g}}{0.0150\text{ mol}} = 84.7\text{ g mol}^{-1}\).

M1: Convert temperature to \(360\text{ K}\) and volume to \(4.49 \times 10^{-4}\text{ m}^3\) [1 mark]
M2: Correct rearrangement and calculation of \(n = 0.0150\text{ mol}\) (allow \(0.015\)) [1 mark]
M3: Calculate molar mass = \(84.7\) (allow \(84.6 - 84.8\)) to 3 significant figures [1 mark]

1. 2-methylbut-2-ene reacts with hydrogen bromide in an electrophilic addition. The electrophilic addition begins with the addition of a proton (\(\text{H}^+\)) to the double bond. 2. This can form two different carbocation intermediates: a tertiary carbocation, \(\text{(CH}_3\text{)}_2\text{C}^+\text{CH}_2\text{CH}_3\), or a secondary carbocation, \(\text{(CH}_3\text{)}_2\text{CHCH}^+\text{CH}_3\). 3. The tertiary carbocation is more stable because it has three electron-releasing alkyl groups that decrease the positive charge on the carbon (positive inductive effect), whereas the secondary carbocation has only two. 4. Therefore, the major product (2-bromo-2-methylbutane) is formed via the more stable tertiary carbocation intermediate.

[1.0 Mark] Correctly identifying that the reaction proceeds via the more stable tertiary carbocation intermediate (compared to the secondary carbocation). [1.0 Mark] Explaining carbocation stability in terms of the positive inductive effect of the three electron-releasing alkyl groups. [0.7 Marks] Linking the major product directly to the bromide ion nucleophilic attack on this tertiary carbocation intermediate. Reject simple statements of Markovnikov's rule without carbocation stability explanation.

1. Mean bond enthalpies apply strictly to substances in the gaseous state. 2. Ethanol is a liquid under standard conditions, so the experimental value includes the enthalpy change of vaporization (an endothermic process that requires energy input), making the overall experimental combustion value less exothermic. 3. Furthermore, practical experimental setups suffer from heat loss to the calorimeter/surroundings and the potential for incomplete combustion of the ethanol fuel, both of which also contribute to a lower experimental heat release than the theoretical value.

[1.0 Mark] Explaining that mean bond enthalpies refer to molecules in the gaseous state, whereas ethanol is a liquid in its standard state. [1.0 Mark] Mentioning that heat is required to vaporize liquid ethanol (or that vaporization is endothermic). [0.7 Marks] Identifying experimental errors such as heat loss to the surroundings or incomplete combustion of ethanol. Reject general errors like 'human error'.

1. The reaction is a nucleophilic substitution where ammonia acts as a nucleophile. 2. The initial product, butan-1-amine, also contains a lone pair of electrons on the nitrogen atom, making it a nucleophile. 3. If 1-bromobutane is in excess or in equal amounts, the newly formed butan-1-amine will react with remaining 1-bromobutane to form dibutylamine (a secondary amine), and eventually tributylamine and quaternary ammonium salts. 4. Using a large excess of ammonia ensures that any 1-bromobutane molecule is much more likely to collide with, and react with, an ammonia molecule rather than a butan-1-amine molecule, minimizing further substitution.

[1.0 Mark] Correct IUPAC name: butan-1-amine (accept 1-aminobutane). [1.0 Mark] Explanation that the product (butan-1-amine) is also a nucleophile and can undergo further substitution. [0.7 Marks] Explaining that excess ammonia ensures collision/reaction occurs preferentially with ammonia, maximizing primary amine yield. Reject 'to ensure complete reaction' alone.

1. Stereoisomerism in alkenes (E/Z isomerism) arises from two main criteria: restricted rotation about the \(\text{C}=\text{C}\) double bond due to the presence of a \(\pi\)-bond, and having two different groups attached to each of the carbon atoms in the double bond. 2. In 1,2-dichloroethene, each carbon in the double bond is bonded to one hydrogen atom and one chlorine atom. Since both carbons are bonded to two different groups, it exists as (E)-1,2-dichloroethene and (Z)-1,2-dichloroethene. 3. In 1,1-dichloroethene, one carbon is bonded to two chlorine atoms and the other is bonded to two hydrogen atoms. Because at least one carbon is bonded to two identical groups, swapping the groups does not result in a new stereoisomer.

[1.0 Mark] Mentioning restricted rotation about the carbon-carbon double bond (due to the pi-bond). [1.0 Mark] Explaining that stereoisomerism requires each carbon atom in the double bond to be attached to two different groups. [0.7 Marks] Showing that in 1,1-dichloroethene, one carbon is attached to two identical chlorine atoms (or two identical hydrogen atoms), which prevents stereoisomerism.

1. Dehydration of butan-2-ol can yield but-1-ene, (E)-but-2-ene, and (Z)-but-2-ene. The two stereoisomers are (E)-but-2-ene and (Z)-but-2-ene. 2. These two isomers are diastereomers (geometric isomers) and have different arrangements of groups in space, resulting in different overall molecular dipoles and packing. Thus, they have different boiling points. 3. Fractional distillation separates compounds based on differences in boiling points. Enantiomers cannot be separated this way because they have identical physical properties (including boiling points) due to their mirror-image symmetry in an achiral environment.

[1.0 Mark] Correctly naming the two stereoisomers: (E)-but-2-ene and (Z)-but-2-ene (accept trans-but-2-ene and cis-but-2-ene). [1.0 Mark] Explaining that diastereomers/geometric isomers have different boiling points due to different polarities/intermolecular forces. [0.7 Marks] Explaining that enantiomers have identical physical properties (including boiling points) and thus cannot be separated by fractional distillation.

1. Write the expression for \(K_c\) using the standard equilibrium law: \(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\). Square brackets denote equilibrium concentrations. 2. The forward reaction is exothermic (\(\Delta H < 0\)). 3. According to Le Chatelier's principle, an increase in temperature will cause the equilibrium position to shift in the endothermic direction (to the left) to minimize the temperature increase. 4. As the equilibrium shifts to the left, the concentration of the product \(\text{NH}_3\) decreases, and the concentrations of the reactants \(\text{N}_2\) and \(\text{H}_2\) increase. 5. Therefore, the ratio represented by \(K_c\) decreases.

[1.0 Mark] Correct expression for \(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\) (must have square brackets; round brackets are rejected). [1.0 Mark] Stating that the value of \(K_c\) decreases with an increase in temperature. [0.7 Marks] Explaining the shift in terms of Le Chatelier's principle: the endothermic direction is favoured to oppose the increase in temperature, leading to fewer products and more reactants.

1. Propanal is an aldehyde and can be easily oxidized to a carboxylic acid. Propanone is a ketone and cannot be easily oxidized because it lacks a hydrogen atom directly bonded to the carbonyl carbon. 2. Tollens' reagent contains \([\text{Ag(NH}_3\text{)}_2]^+\) ions. When warmed with propanal, the aldehyde reduces the silver ions to metallic silver, forming a mirror. Ketones do not react. 3. Alternatively, Fehling's solution (containing \(\text{Cu}^{2+}\) ions) can be used. When warmed with propanal, the blue solution forms a red precipitate of copper(I) oxide (\(\text{Cu}_2\text{O}\)). Propanone remains blue.

[1.0 Mark] Name a suitable reagent: Tollens' reagent (accept Fehling's solution or acidified potassium dichromate(VI)). [1.0 Mark] Correct observation for propanal: silver mirror (or red precipitate for Fehling's, or orange-to-green colour change for dichromate). [0.7 Marks] Correct observation for propanone: no change / remains colourless (or remains blue for Fehling's, or remains orange for dichromate).

1. The reaction between methane and chlorine proceeds via a free-radical substitution mechanism. 2. The reaction is initiated by ultraviolet (UV) light, which provides the energy to homolytically cleave the \(\text{Cl}-\text{Cl}\) bond. 3. In the first propagation step, a chlorine radical abstracts a hydrogen atom from methane, producing a methyl radical and hydrogen chloride: \(\text{CH}_4 + \cdot\text{Cl} \rightarrow \cdot\text{CH}_3 + \text{HCl}\). 4. In the second propagation step, the methyl radical reacts with a chlorine molecule, forming the chloromethane product and regenerating the chlorine radical: \(\cdot\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot\text{Cl}\).

[1.0 Mark] Correct first propagation step with single electron radicals correctly placed: \(\text{CH}_4 + \cdot\text{Cl} \rightarrow \cdot\text{CH}_3 + \text{HCl}\). [1.0 Mark] Correct second propagation step: \(\cdot\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot\text{Cl}\). [0.7 Marks] Identifying Ultraviolet (UV) light (or high temperature \(\ge 400\,^\circ\text{C}\)) as the necessary condition. Reject 'sunlight' without mentioning UV.

1. Nucleophilic attack: The first NH3 molecule acts as a nucleophile, donating its lone pair of electrons to the carbocationic center of the C-Br polar bond. 2. Intermediate deprotonation: The intermediate [CH3CH2CH2NH3]+ is deprotonated by a second NH3 molecule acting as a base, yielding the primary amine product and ammonium bromide. 3. Yield optimization: Excess ammonia ensures that the concentration of ammonia is significantly higher than that of propylamine, preventing further nucleophilic attack by the product on unreacted 1-bromopropane.

1 mark: Identifies the roles of the two ammonia molecules (first is nucleophile, second is base). 1 mark: Explains the formation of the intermediate [CH3CH2CH2NH3]+ and the deprotonation step. 0.7 marks: Explains that excess ammonia prevents further substitution of propylamine to form secondary/tertiary amines.

1. Addition of H+ to 2-methylbut-2-ene can form a tertiary carbocation or a secondary carbocation. 2. The tertiary carbocation is stabilized by the positive inductive effect of three electron-donating alkyl groups, making it more stable than the secondary carbocation, which is only stabilized by two. 3. Subsequent attack of the bromide ion on the more stable tertiary carbocation yields 2-bromo-2-methylbutane as the major product.

1 mark: Correct IUPAC name: 2-bromo-2-methylbutane. 1 mark: Identifies the major product intermediate as a tertiary carbocation and the minor as a secondary carbocation. 0.7 marks: Explains that the tertiary carbocation is more stable due to the positive inductive effect of three electron-donating alkyl groups compared to two.

1. Definition: Enthalpy change when 1 mole of substance is burned completely in oxygen with all reactants and products in their standard states under standard conditions of 100 kPa and 298 K. 2. Equation: \(C_3H_7OH(l) + 4.5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\) (or with whole numbers: \(2C_3H_7OH(l) + 9O_2(g) \rightarrow 6CO_2(g) + 8H_2O(l)\), but the definition of enthalpy of combustion refers to 1 mole of fuel, so 1 mole of propan-1-ol equation is preferred).

1 mark: Full definition including 'one mole', 'completely burned in oxygen', and 'standard conditions/standard states'. 1 mark: Correct balanced equation with propan-1-ol coefficient as 1 (or balanced equivalent). 0.7 marks: All state symbols correct: (l) for propan-1-ol and water, (g) for oxygen and carbon dioxide.

1. Propan-1-ol is a primary alcohol, so it can be oxidized by acidified potassium dichromate(VI). The chromium(VI) is reduced to green chromium(III). 2. 2-methylpropan-2-ol is a tertiary alcohol, which cannot be oxidized easily because there is no hydrogen atom on the carbon holding the -OH group. Therefore, no reaction occurs, and the solution remains orange.

1 mark: Acidified potassium dichromate(VI) and heating/refluxing (accept acidified sodium dichromate or H+/Cr2O7^2-). 1 mark: Propan-1-ol observation: Orange to green. 0.7 marks: 2-methylpropan-2-ol observation: No change / remains orange.

1. The forward reaction is exothermic (\(\Delta H < 0\)). 2. According to Le Chatelier's principle, raising the temperature shifts the equilibrium to the left (the endothermic direction) to absorb the heat. 3. Since \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\), shifting the equilibrium to the left decreases the numerator and increases the denominator, leading to a smaller value of \(K_c\).

1 mark: States that Kc decreases. 1 mark: Explains that the reaction shifts in the endothermic direction / to the left. 0.7 marks: Connects the shift to opposing the temperature increase (Le Chatelier's principle) and states how this changes the ratio of products to reactants.

1. Propagation 1: A chlorine free radical abstracts a hydrogen atom from ethane, producing an ethyl radical and hydrogen chloride. 2. Propagation 2: The ethyl radical reacts with a chlorine molecule, forming chloroethane and regenerating the chlorine radical. 3. Initiation: Ultraviolet light is required to provide the energy to homolytically split the Cl-Cl bond.

1 mark: Correct equation for Propagation 1 (must show radical dots clearly on Cl and ethyl radical, e.g., \(\cdot CH_2CH_3\) or \(C_2H_5\cdot\)). 1 mark: Correct equation for Propagation 2. 0.7 marks: UV light / ultraviolet light.

1. Carbonyl group identification: The band at \(1715\text{ cm}^{-1}\) corresponds to a carbonyl (C=O) group. 2. Exclusion of alcohol and aldehyde: The lack of a broad O-H absorption band above \(3200\text{ cm}^{-1}\) means it is not an alcohol (e.g., prop-2-en-1-ol). The failure to reduce Tollens' reagent means it is not an aldehyde. 3. Name deduction: A ketone with three carbon atoms must be propanone.

1 mark: Correct IUPAC name: Propanone. 1 mark: Links the \(1715\text{ cm}^{-1}\) band to C=O (carbonyl) and explains the absence of O-H. 0.7 marks: Explains that the negative Tollens' test confirms it is a ketone and not an aldehyde.

1. Definition: Stereoisomers have the same structural formula but different spatial arrangement of atoms. 2. Requirements for E/Z: Restricted rotation around C=C (due to the pi bond) and each of the double-bonded carbons must be attached to two different atoms or groups. 3. Comparison: In but-2-ene, both C-2 and C-3 are bonded to -H and -CH3 (different). In but-1-ene, C-1 is bonded to two identical -H atoms, so no stereoisomerism is possible.

1 mark: Complete definition of stereoisomerism (same structural formula, different spatial arrangement). 1 mark: Explains that but-2-ene has restricted C=C rotation and both double-bonded carbons have two different groups attached. 0.7 marks: Explains that but-1-ene has one double-bonded carbon with two identical groups (H atoms).

1. Determine equilibrium moles: Initial moles: SO2 = 3.00, O2 = 2.00, SO3 = 0.00. Equilibrium moles of SO3 = 2.20. Moles of SO2 reacted = 2.20, so equilibrium moles of SO2 = 3.00 - 2.20 = 0.80 mol. Moles of O2 reacted = 1.10, so equilibrium moles of O2 = 2.00 - 1.10 = 0.90 mol. 2. Determine equilibrium concentrations (Volume = 4.00 dm^3): [SO3] = 2.20 / 4.00 = 0.55 mol dm^-3. [SO2] = 0.80 / 4.00 = 0.20 mol dm^-3. [O2] = 0.90 / 4.00 = 0.225 mol dm^-3. 3. Calculate Kc: Kc = [SO3]^2 / ([SO2]^2 * [O2]) = (0.55)^2 / ((0.20)^2 * 0.225) = 0.3025 / (0.04 * 0.225) = 33.61 dm^3 mol^-1. Rounded to 3 significant figures, Kc = 33.6 dm^3 mol^-1.

1 mark for calculating correct equilibrium moles of SO2 (0.80 mol) and O2 (0.90 mol). 1 mark for dividing moles by volume (4.00 dm^3) to obtain concentrations. 1 mark for the correct Kc expression. 1 mark for the correct numerical value of 33.6. 0.5 marks for correct units (dm3 mol-1).

1. Calculate the rate of oxygen gas production in moles per second: Rate of O2 production = 1.80 cm^3 s^-1 / 24000 cm^3 mol^-1 = 7.50 x 10^-5 mol s^-1. 2. Use stoichiometry to find the rate of H2O2 consumption: According to the balanced equation, 2 moles of H2O2 decompose for every 1 mole of O2 produced. Rate of H2O2 consumption = 2 * 7.50 x 10^-5 = 1.50 x 10^-4 mol s^-1. 3. Calculate the rate of concentration decrease: Volume of reaction solution = 50.0 cm^3 = 0.0500 dm^3. Rate of decrease in concentration = 1.50 x 10^-4 mol s^-1 / 0.0500 dm^3 = 3.00 x 10^-3 mol dm^-3 s^-1.

1 mark for converting volume rate of oxygen to mole rate (7.50 x 10^-5 mol s^-1). 1.5 marks for multiplying the mole rate by 2 to account for reaction stoichiometry (1.50 x 10^-4 mol s^-1). 1 mark for dividing the mole rate of H2O2 by the volume of solution in dm^3 (0.0500 dm^3). 1 mark for the correct calculation and units of the final answer (3.00 x 10^-3 mol dm^-3 s^-1).

1. Calculate moles of NaOH used in titration: n(NaOH) = (14.00 / 1000) * 0.150 = 2.10 x 10^-3 mol. 2. Moles of excess HCl in 25.0 cm^3 aliquot: n(HCl) = 2.10 x 10^-3 mol. 3. Moles of excess HCl in the total 250.0 cm^3 solution: n(HCl) excess = 2.10 x 10^-3 * (250.0 / 25.0) = 2.10 x 10^-2 mol. 4. Calculate initial moles of HCl added: n(HCl) initial = (50.0 / 1000) * 1.00 = 0.0500 mol. 5. Calculate moles of HCl that reacted with MgCO3: n(HCl) reacted = 0.0500 - 0.0210 = 0.0290 mol. 6. Moles of MgCO3 reacted: The reaction equation is MgCO3 + 2HCl -> MgCl2 + CO2 + H2O. Therefore, n(MgCO3) = 0.0290 / 2 = 0.0145 mol. 7. Mass of MgCO3: m(MgCO3) = 0.0145 * 84.3 = 1.222 g. 8. Percentage by mass: Percentage = (1.222 / 1.30) * 100 = 94.0%.

1 mark for calculating moles of NaOH and scaling to find the total excess HCl (2.10 x 10^-2 mol). 1 mark for calculating reacted moles of HCl (0.0290 mol). 1 mark for halving the moles of HCl to find moles of MgCO3 (0.0145 mol). 1 mark for calculating mass of MgCO3 (1.222 g). 0.5 marks for correct final percentage to 3 significant figures (94.0%).

1. Calculate heat changes (q = m c dT): For anhydrous CaCl2: q1 = 100.0 * 4.18 * 10.4 = 4347.2 J = 4.3472 kJ. Moles of CaCl2 = 5.55 / 111.0 = 0.0500 mol. dH_sol(anhydrous) = -(4.3472 / 0.0500) = -86.94 kJ mol^-1. For hydrated CaCl2.6H2O: q2 = 100.0 * 4.18 * (-2.1) = -877.8 J = -0.8778 kJ. Moles of CaCl2.6H2O = 10.95 / 219.0 = 0.0500 mol. dH_sol(hydrated) = -(-0.8778 / 0.0500) = +17.56 kJ mol^-1. 2. Construct Hess's cycle: CaCl2(s) + 6H2O(l) -> CaCl2.6H2O(s). By Hess's law: dH_hyd = dH_sol(anhydrous) - dH_sol(hydrated) = -86.94 - (+17.56) = -104.5 kJ mol^-1. Rounded to 3 significant figures: -105 kJ mol^-1.

1 mark for calculating dH_sol(anhydrous) = -86.94 kJ mol^-1 (must include correct sign). 1 mark for calculating dH_sol(hydrated) = +17.56 kJ mol^-1 (must include correct sign). 1.5 marks for showing a correct Hess's Law relationship or cycle. 1 mark for the correct calculation and sign of final answer (-105 kJ mol-1).

1. Identify reaction stoichiometry: MnO4^- + 8H^+ + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O. Therefore, moles of MnO4^- react with Fe^2+ in a 1:5 ratio. 2. Calculate moles of MnO4^- used: n(MnO4^-) = 0.02380 dm^3 * 0.0200 mol dm^-3 = 4.76 x 10^-4 mol. 3. Calculate moles of Fe^2+ in the 25.0 cm^3 sample: n(Fe^2+) = 5 * 4.76 x 10^-4 = 2.38 x 10^-3 mol. 4. Calculate total moles of Fe^2+ in 250.0 cm^3: n(Fe^2+) total = 2.38 x 10^-3 * 10 = 2.38 x 10^-2 mol. 5. Calculate mass of iron: m(Fe) = 2.38 x 10^-2 mol * 55.8 g mol^-1 = 1.328 g. 6. Calculate percentage of iron: Percentage = (1.328 / 1.45) * 100 = 91.6% (to 3 sig figs).

1 mark for stating or using the 1:5 mole ratio of manganate(VII) to iron(II). 1 mark for calculating moles of MnO4^- (4.76 x 10^-4 mol) and multiplying by 5 to find moles of Fe^2+ in the aliquot. 1 mark for scaling up by 10 to find total moles of Fe^2+ (2.38 x 10^-2 mol). 1 mark for calculating mass of iron (1.328 g). 0.5 marks for correct percentage to 3 significant figures (91.6%).

Let x be the mass of KCl and y be the mass of KBr in grams. 1. Establish simultaneous equations: Equation 1: x + y = 2.500 => y = 2.500 - x. Moles of KCl = x / 74.6, which forms (x / 74.6) mol of AgCl. Mass of AgCl = (143.4 / 74.6) * x = 1.9222 x. Moles of KBr = y / 119.0, which forms (y / 119.0) mol of AgBr. Mass of AgBr = (187.8 / 119.0) * y = 1.5781 y. Equation 2: 1.9222 x + 1.5781 y = 4.250. 2. Solve for x: Substitute y: 1.9222 x + 1.5781 (2.500 - x) = 4.250. 1.9222 x + 3.9453 - 1.5781 x = 4.250. 0.3441 x = 0.3047 => x = 0.8855 g. 3. Calculate percentage of KCl: Percentage = (0.8855 / 2.500) * 100 = 35.4%.

1.5 marks for expressing the masses of silver halides in terms of algebraic variables. 1.5 marks for setting up and solving the simultaneous equations to find the mass of KCl (0.8855 g). 1.5 marks for calculating the correct percentage of 35.4%.

1. Calculate moles of reactants: n(Ba^2+) = 0.1500 dm^3 * 0.0400 mol dm^-3 = 6.00 x 10^-3 mol. n(SO4^2-) = 0.2500 dm^3 * 0.0300 mol dm^-3 = 7.50 x 10^-3 mol. 2. Determine limiting reactant: The precipitation reaction is Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s). Since they react in a 1:1 mole ratio, Ba^2+ is the limiting reactant (6.00 x 10^-3 < 7.50 x 10^-3). 3. Calculate mass of precipitate: Moles of BaSO4 formed = 6.00 x 10^-3 mol. Mass of BaSO4 = 6.00 x 10^-3 mol * 233.4 g mol^-1 = 1.4004 g. Rounded to 3 significant figures: 1.40 g.

1 mark for calculating moles of Ba2+ (6.00 x 10^-3 mol). 1 mark for calculating moles of SO42- (7.50 x 10^-3 mol). 1.5 marks for identifying Ba2+ as the limiting reactant with explanation. 1 mark for calculating the correct mass of barium sulfate formed (1.40 g).

1. Calculate moles of chloride ions in the sample: n(AgCl) = 4.302 g / 143.4 g mol^-1 = 0.0300 mol. Since each mole of AgCl contains 1 mole of Cl^-, n(Cl^-) = 0.0300 mol. 2. Calculate the mass of chlorine in the sample: Mass of Cl = 0.0300 mol * 35.5 g mol^-1 = 1.065 g. 3. Calculate the mass of element X in the sample: Mass of X = 1.335 g - 1.065 g = 0.270 g. 4. Identify element X: Test possible ratios: If formula is XCl3 (n=3), moles of X = 0.0300 / 3 = 0.0100 mol. Molar mass of X = 0.270 g / 0.0100 mol = 27.0 g mol^-1. This matches Aluminium (Al), which is in Period 3. 5. Calculate percentage by mass of X (Aluminium): Percentage = (0.270 / 1.335) * 100 = 20.22%.

1 mark for calculating moles of Cl- (0.0300 mol). 1 mark for calculating mass of Cl (1.065 g) and mass of X (0.270 g). 1.5 marks for showing calculations that identify X as Aluminium (Ar = 27.0) with formula AlCl3. 1 mark for calculating the correct percentage of X in the chloride (20.2%).

The lattice enthalpy of formation is defined as the enthalpy change when one mole of an solid ionic crystal lattice is formed from its constituent gaseous ions under standard conditions. Thus, gaseous calcium ions and gaseous chloride ions react to form solid calcium chloride: \(\text{Ca}^{2+}(\text{g}) + 2\text{Cl}^{-}(\text{g}) \rightarrow \text{CaCl}_2(\text{s})\).

M1: Correct balanced chemical equation with correct species: \(\text{Ca}^{2+} + 2\text{Cl}^{-} \rightarrow \text{CaCl}_2\) (1 mark). M2: All state symbols correct: \(\text{g}\) for reactants and \(\text{s}\) for the product (1 mark). Note: Reject equations starting from elements or with incorrect charges on ions.

Sodium reacts vigorously with cold water to form sodium hydroxide and hydrogen gas: \(2\text{Na}(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq}) + \text{H}_2(\text{g})\). Sodium hydroxide is a strong base, which dissociates fully in water to yield a strongly alkaline solution with a pH between 12 and 14.

M1: Correct balanced equation with state symbols: \(2\text{Na}(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq}) + \text{H}_2(\text{g})\) (or ionic equivalent: \(2\text{Na}(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Na}^{+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) + \text{H}_2(\text{g})\)) (1 mark). M2: State pH in the range 12 to 14 (1 mark).

When temperature decreases, the average kinetic energy of the molecules decreases. This causes the peak of the curve (most probable energy) to shift to the left (lower energy) and become taller (higher fraction of molecules at this lower energy). The curve is lower at higher energy values, meaning a smaller fraction of molecules possess energy equal to or greater than the activation energy.

M1: Peak shifts to the left and is higher/taller (1 mark). M2: Curve is lower on the right side / fewer molecules have energy greater than or equal to Ea (1 mark).

Chlorine is more reactive than bromine, so it displaces bromide ions from solution: \(\text{Cl}_2(\text{g}) + 2\text{Br}^{-}(\text{aq}) \rightarrow 2\text{Cl}^{-}(\text{aq}) + \text{Br}_2(\text{aq})\). The formation of aqueous bromine causes the colourless solution to turn orange or yellow-brown.

M1: Correct ionic equation with correct state symbols: \(\text{Cl}_2(\text{g}) + 2\text{Br}^{-}(\text{aq}) \rightarrow 2\text{Cl}^{-}(\text{aq}) + \text{Br}_2(\text{aq})\) (accept \(\text{Cl}_2(\text{aq})\)) (1 mark). M2: Observation: (Colourless solution turns) orange / yellow-brown / brown (1 mark) [Reject: red, red-brown gas, or just 'turns yellow'].

Since the forward reaction is exothermic, an increase in temperature will cause the system to shift to oppose the change by favouring the endothermic reverse reaction. This moves the equilibrium position to the left, decreasing the concentrations of products and increasing the concentrations of reactants, which decreases the value of Kc.

M1: State that Kc decreases (1 mark). M2: Explain that the forward reaction is exothermic, so an increase in temperature shifts the position of equilibrium to the left to oppose the temperature increase (1 mark).

The d-orbitals of a transition metal split into different energy levels when ligands coordinate. The size of this energy gap depends on the oxidation state of the transition metal ion, the identity of the ligands, and the coordination number or geometry of the complex.

M1: Any one correct factor such as oxidation state of the metal or identity of ligand (1 mark). M2: A second distinct correct factor such as coordination number or shape of the complex (1 mark).

Using Hess's Law for the Born-Haber cycle: Enthalpy of formation = Enthalpy of atomisation of Na + First ionisation energy of Na + Enthalpy of atomisation of Cl + First electron affinity of Cl - Lattice enthalpy of dissociation. Substituting the values: -411 = +107 + 496 + 122 - 349 - L.D. This simplifies to -411 = +376 - L.D., which gives L.D. = +787 kJ mol^-1.

M1: Correct expression or substitute values correctly: -411 = 107 + 496 + 122 - 349 - L.D. (1 mark). M2: Correct final answer: +787 kJ mol^-1 (1 mark). Reject -787.

The solubility of Group 2 sulfates decreases down the group. Barium sulfate is highly insoluble, and adding barium ions to sulfate ions results in a white precipitate: Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s).

M1: State that solubility decreases down the group (1 mark). M2: Correct ionic equation with state symbols: Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s) (1 mark).

In a Hess's Law cycle for the hydration of an anhydrous salt, both the anhydrous salt, \( \text{CaCl}_2(s) \), and the hydrated salt, \( \text{CaCl}_2 \cdot 2\text{H}_2\text{O}(s) \), are dissolved in excess water to form the exact same final aqueous solution.

Reaction 1: Dissolution of the anhydrous salt:
\( \text{CaCl}_2(s) \xrightarrow{\text{aq}} \text{CaCl}_2(aq) \) (or separated into ions: \( \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) \))

Reaction 2: Dissolution of the dihydrate:
\( \text{CaCl}_2 \cdot 2\text{H}_2\text{O}(s) \xrightarrow{\text{aq}} \text{CaCl}_2(aq) + 2\text{H}_2\text{O}(l) \) (or separated into ions: \( \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{H}_2\text{O}(l) \))

[1 mark] For the balanced equation of the anhydrous salt dissolving, including state symbols: \( \text{CaCl}_2(s) \rightarrow \text{CaCl}_2(aq) \) (or separated ions).
[1 mark] For the balanced equation of the dihydrate dissolving, including state symbols: \( \text{CaCl}_2 \cdot 2\text{H}_2\text{O}(s) \rightarrow \text{CaCl}_2(aq) + 2\text{H}_2\text{O}(l) \) (or separated ions). Allow 'aq' or '+ aq' as reactant, but state symbols for reactants and products must be correct.

The first ionisation energy of sodium is the energy required to remove one mole of electrons from one mole of gaseous sodium atoms to form one mole of gaseous sodium ions:
\( \text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^- \)

The 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:
\( \text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g) \)

[1 mark] \( \text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^- \) (accept \( \text{Na}(g) - \text{e}^- \rightarrow \text{Na}^+(g) \)). State symbols must be gaseous (g).
[1 mark] \( \text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g) \). State symbols must be gaseous (g). Reject equations involving diatomic chlorine, e.g., \( \frac{1}{2}\text{Cl}_2 \).

When chlorine reacts with cold, dilute sodium hydroxide, sodium chloride (\( \text{NaCl} \)), sodium chlorate(I) (\( \text{NaClO} \)), and water are formed. The ionic equation for this reaction is:
\( \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O} \)

In this reaction, the oxidation state of chlorine in molecular chlorine (\( \text{Cl}_2 \)) is \( 0 \). It is reduced to \( -1 \) in the chloride ion (\( \text{Cl}^- \)) and oxidised to \( +1 \) in the chlorate(I) ion (\( \text{ClO}^- \)). Since the same element is simultaneously oxidised and reduced, this is a disproportionation reaction.

[1 mark] For the balanced ionic equation: \( \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O} \) (state symbols not required, but must be correct if used).
[1 mark] For stating that the oxidation state of chlorine changes from \( 0 \) to \( -1 \) and \( +1 \) (both product oxidation states must be correctly stated).

Magnesium reacts rapidly with steam (gaseous water) to form solid white magnesium oxide and hydrogen gas:
\( \text{Mg}(s) + \text{H}_2\text{O}(g) \rightarrow \text{MgO}(s) + \text{H}_2(g) \)

During this reaction, the magnesium burns with a bright white flame and a white solid (magnesium oxide) is produced.

[1 mark] For the correct equation with state symbols: \( \text{Mg}(s) + \text{H}_2\text{O}(g) \rightarrow \text{MgO}(s) + \text{H}_2(g) \). Reject if state symbol of steam is written as (l).
[1 mark] For a correct observation: Accept 'bright white light', 'bright white flame', 'white solid formed', or 'white ash formed'. Reject: 'fizzing' or 'bubbles' (as steam is already gaseous, bubbling is not a clear observation).

1. At \(T_2\), the peak of the curve shifts to the right and is lower, meaning the molecules have a higher average kinetic energy (1 mark).
There is a larger area under the curve to the right of the activation energy \(E_a\) at \(T_2\) than at \(T_1\), which represents a larger fraction of molecules with energy equal to or greater than the activation energy (1 mark).
This results in a significantly higher frequency of successful collisions per unit time, which increases the rate of reaction (1 mark).

2. The total area under each curve represents the total number of gaseous particles/molecules in the sample, which remains constant (1 mark).

M1 (1 mark): State that at \(T_2\), the curve shifts to the right / has a lower peak / molecules have higher average energy.
M2 (1 mark): State that a greater proportion/fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)) at \(T_2\) (or reference to larger area under the curve to the right of \(E_a\)).
M3 (1 mark): State that there is a higher frequency of successful collisions (or more successful collisions per unit time). Reject 'more collisions' without a rate/time element or the word 'successful'.
M4 (1 mark): Identify that the total area represents the total number of molecules/particles in the gas sample.

1. According to Hess's Law, the enthalpy change for the direct route equals the sum of the enthalpy changes for the indirect route:
\(\Delta H_{\text{r}} + \Delta H_2 = \Delta H_1\)
Rearranging gives:
\(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\) (1 mark)

2. Substitute the given values into the equation:
\(\Delta H_{\text{r}} = -66.5 - (+11.7)\) (1 mark for working)
\(\Delta H_{\text{r}} = -78.2\text{ kJ mol}^{-1}\) (1 mark for correct answer with sign and units)

3. It is not possible to measure the enthalpy of hydration directly because water reacts with solid anhydrous copper(II) sulfate to form hydrated copper(II) sulfate, but some of the solid will inevitably dissolve in the water, and it is impossible to add exactly the right stoichiometric ratio of water to solid without dissolving any of it. Additionally, solid-state reactions are too slow to allow accurate temperature change measurements due to heat loss to the surroundings. (1 mark for any valid reason)

M1 (1 mark): \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\) (or \(\Delta H_{\text{r}} + \Delta H_2 = \Delta H_1\))
M2 (1 mark): Correct substitution: \(-66.5 - 11.7\)
M3 (1 mark): Correct answer: \(-78.2\text{ kJ mol}^{-1}\) (allow consequential error on M2 but must have correct units and negative sign)
M4 (1 mark): Explanation: Difficult to add exact amount of water without dissolving the solid / reaction is too slow to measure temperature changes accurately / water dissolves the solid.

1. Determine the change in concentration for each substance from \(t = 0\) to equilibrium:
Change in \([\text{X}] = 0.80 - 0.20 = 0.60\text{ mol dm}^{-3}\) (decrease) (1 mark)
Change in \([\text{Y}] = 1.20 - 0.00 = 1.20\text{ mol dm}^{-3}\) (increase)
The ratio of changes is:
\(\text{X} : \text{Y} = 0.60 : 1.20 = 1 : 2\)
Thus, \(a = 1\) and \(b = 2\), giving the balanced equation: \(\text{X}(\text{g}) \rightleftharpoons 2\text{Y}(\text{g})\) (1 mark).

2. The equilibrium constant expression is:
\(K_c = \frac{[\text{Y}]^2}{[\text{X}]}\) (1 mark)
Substitute the equilibrium concentrations:
\(K_c = \frac{(1.20)^2}{0.20} = \frac{1.44}{0.20} = 7.20\text{ mol dm}^{-3}\) (1 mark for correct value and units).

M1 (1 mark): Calculate the change in concentration: \(\Delta [\text{X}] = -0.60\text{ mol dm}^{-3}\) and \(\Delta [\text{Y}] = +1.20\text{ mol dm}^{-3}\).
M2 (1 mark): Deduce the ratio \(1 : 2\) to give \(a = 1\) and \(b = 2\).
M3 (1 mark): Write the correct expression for \(K_c = \frac{[\text{Y}]^2}{[\text{X}]}\) (allow consequential expression based on their coefficients in part 1).
M4 (1 mark): Calculate the value \(7.2\) (or \(7.20\)) and correct units: \(\text{mol dm}^{-3}\) (allow consequential calculation based on their \(K_c\) expression).

1.
- \(\text{NaX}\) produces only misty fumes (hydrogen halide, \(\text{HCl}\)), indicating no redox reaction has taken place because chloride is a weak reducing agent. Thus, \(\text{X}^- = \text{Cl}^-\) (1 mark).
- \(\text{NaY}\) produces misty fumes (\(\text{HBr}\)), brown fumes (\(\text{Br}_2\)), and a choking gas (\(\text{SO}_2\)), which shows bromide is strong enough to reduce sulfuric acid to sulfur dioxide. Thus, \(\text{Y}^- = \text{Br}^-\) (1 mark).
- \(\text{NaZ}\) produces misty fumes (\(\text{HI}\)), a purple vapor (\(\text{I}_2\)), and a gas with a smell of rotten eggs (\(\text{H}_2\text{S}\)), which shows iodide is a very strong reducing agent capable of reducing sulfuric acid to hydrogen sulfide. Thus, \(\text{Z}^- = \text{I}^-\) (1 mark).

2. The equation for the reaction of \(\text{NaCl}\) with concentrated sulfuric acid is:
\(\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g})\) (1 mark for balanced equation with correct state symbols).
Note: \(2\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{Na}_2\text{SO}_4(\text{s}) + 2\text{HCl}(\text{g})\) is also accepted.

M1 (1 mark): Identify \(\text{X}^- = \text{Cl}^-\) (or chloride).
M2 (1 mark): Identify \(\text{Y}^- = \text{Br}^-\) (or bromide).
M3 (1 mark): Identify \(\text{Z}^- = \text{I}^-\) (or iodide).
M4 (1 mark): \(\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g})\) (accept state symbols \((\text{aq})\) for sulfuric acid or sodium bisulfate, but concentrated acid is strictly liquid. Accept balanced equation forming \(\text{Na}_2\text{SO}_4\)). Both formulas and state symbols must be correct.

1. To distinguish between \(\text{MgCl}_2(\text{aq})\) and \(\text{BaCl}_2(\text{aq})\):
- Reagent: Add aqueous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) or sulfuric acid (\(\text{H}_2\text{SO}_4\)) (1 mark).
- Observation with \(\text{MgCl}_2\): No visible change / remains colorless (since magnesium sulfate is highly soluble) (0.5 marks).
- Observation with \(\text{BaCl}_2\): A white precipitate forms (since barium sulfate is highly insoluble) (0.5 marks).

Alternative Reagent: Aqueous sodium hydroxide (\(\text{NaOH}\)) (1 mark).
- Observation with \(\text{MgCl}_2\): A white precipitate forms (since magnesium hydroxide is insoluble) (0.5 marks).
- Observation with \(\text{BaCl}_2\): No visible change / remains colorless (since barium hydroxide is soluble) (0.5 marks).

2. The ionic equation for the precipitation of barium sulfate (if sulfate reagent is used):
\(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)
(1 mark for correct species, 1 mark for correct state symbols)

Alternative ionic equation (if hydroxide reagent is used):
\(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s})\)
(1 mark for correct species, 1 mark for correct state symbols)

M1 (1 mark): Identify a suitable reagent: sodium sulfate / sulfuric acid / any soluble sulfate (or sodium hydroxide / any soluble hydroxide).
M2 (1 mark): State correct observations for both: Barium chloride forms a white precipitate and magnesium chloride has no change (or vice versa if hydroxide is used).
M3 (1 mark): Write correct ionic equation: e.g., \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\).
M4 (1 mark): Correct state symbols for their ionic equation: \((\text{aq}) + (\text{aq}) \rightarrow (\text{s})\).