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First, count the valence electrons around the central chlorine atom: Chlorine has 7 valence electrons, plus 1 extra electron from the overall negative charge, giving 8 electrons. Two oxygen atoms are bonded to the chlorine, which uses 4 electrons for bonding (forming 2 bonding regions), leaving 4 non-bonding valence electrons as 2 lone pairs. According to VSEPR theory, these 4 electron pairs repel each other to adopt a tetrahedral spatial arrangement. The shape of the molecule, determined by the positions of the atoms only, is bent/non-linear. Because lone pairs exert greater repulsive forces than bonding pairs, the \( \text{O}-\text{Cl}-\text{O} \) bond angle is reduced from the perfect tetrahedral angle of \( 109.5^\circ \) to around \( 104.5^\circ \).

[0.4 marks] Correctly stating there are 2 bonding pairs/regions and 2 lone pairs around the central chlorine atom. [0.4 marks] Identifying the shape as bent / non-linear / V-shaped. [0.4 marks] Giving a bond angle between \( 103^\circ \) and \( 106^\circ \). [0.4 marks] Explaining that lone pairs repel more than bonding pairs, causing compression of the bond angle.

Both \( \text{ICl} \) and \( \text{Br}_2 \) are diatomic molecules containing 70 electrons in total (I = 53, Cl = 17; Br = 35 + 35 = 70). This means the strength of the London dispersion (temporary/induced dipole-dipole) forces is very similar in both substances. However, because chlorine is significantly more electronegative than iodine, the covalent bond in \( \text{ICl} \) is polar, establishing a permanent dipole across the molecule. In contrast, bromine is a homonuclear diatomic molecule and is entirely non-polar. Consequently, \( \text{ICl} \) experiences additional permanent dipole-dipole intermolecular attractions, which are stronger than London dispersion forces alone. More thermal energy is needed to break these stronger intermolecular forces in \( \text{ICl} \), resulting in a higher boiling point.

[0.4 marks] Stating that both molecules have similar London (induced dipole-dipole) forces because they have the same number of electrons. [0.4 marks] Stating that \( \text{ICl} \) is polar / has permanent dipole-dipole forces. [0.4 marks] Stating that \( \text{Br}_2 \) is non-polar / only has London forces. [0.4 marks] Concluding that permanent dipole-dipole forces are stronger and require more thermal energy to overcome.

Sulfur is in Group 16 and has 6 valence electrons. In \( \text{SF}_4 \), it forms 4 single covalent bonds with fluorine atoms, leaving 2 non-bonding valence electrons (1 lone pair). This gives a total of 5 electron pairs (steric number = 5) around the central S atom. To minimize repulsion, these pairs adopt a trigonal bipyramidal arrangement. The lone pair preferentially occupies one of the equatorial positions to minimize its \( 90^\circ \) interactions. The resulting molecular shape is seesaw (or sawhorse). The lone pair is more diffuse and exerts stronger repulsive forces than the bonding pairs, which forces the axial fluorine atoms slightly away from the lone pair, reducing the ideal axial-equatorial bond angle from \( 90^\circ \) to around \( 86^\circ - 89^\circ \) (experimentally about \( 87.3^\circ \)).

[0.4 marks] Correctly identifying the molecular shape as seesaw / sawhorse. [0.4 marks] Stating that the axial-equatorial bond angle is compressed below \( 90^\circ \) (accept range \( 86^\circ \) to \( 89^\circ \)). [0.4 marks] Identifying that the central sulfur atom has 4 bonding pairs and 1 lone pair. [0.4 marks] Explaining that the lone pair exerts greater repulsion than the bonding pairs, compressing the angles.

Both compounds have a giant ionic lattice, but the strength of the electrostatic forces of attraction depends on the charges of the ions and their sizes (charge density). In \( \text{MgO} \), the ions are doubly charged (\( \text{Mg}^{2+} \) and \( \text{O}^{2-} \)), whereas in \( \text{NaF} \), the ions are singly charged (\( \text{Na}^+ \) and \( \text{F}^- \)). Because electrostatic attraction is directly proportional to the product of the ionic charges, the attractive forces in \( \text{MgO} \) are much stronger. Furthermore, \( \text{Mg}^{2+} \) has a smaller ionic radius than \( \text{Na}^+ \) because it has more protons attracting the same number of electrons (isoelectronic). The smaller ionic size allows the ions to pack closer together, further strengthening the electrostatic attraction. Consequently, substantially more thermal energy is needed to disrupt the lattice of \( \text{MgO} \), resulting in its much higher melting point.

[0.4 marks] Identifying that \( \text{Mg}^{2+} \) and \( \text{O}^{2-} \) have higher charges than \( \text{Na}^+ \) and \( \text{F}^- \). [0.4 marks] Explaining that \( \text{Mg}^{2+} \) is smaller than \( \text{Na}^+ \) (or referencing higher charge density). [0.4 marks] Describing the attraction as electrostatic forces of attraction between oppositely charged ions in a giant ionic lattice. [0.4 marks] Concluding that significantly more energy is required to overcome these stronger forces in \( \text{MgO} \).

The hydronium ion, \( \text{H}_3\text{O}^+ \), has a central oxygen atom. Oxygen starts with 6 valence electrons, but losing 1 electron (due to the positive charge) leaves 5. Three of these electrons are shared with three hydrogen atoms in single covalent bonds, leaving 2 non-bonding electrons as 1 lone pair. This gives 4 electron pairs in total (3 bonding pairs, 1 lone pair). The shape is therefore trigonal pyramidal (similar to ammonia, \( \text{NH}_3 \)). The lone pair compresses the ideal tetrahedral bond angle from \( 109.5^\circ \) to approximately \( 107^\circ \). The third hydrogen is attached via a coordinate (dative covalent) bond, which is formed when the neutral water molecule (specifically a lone pair on the oxygen atom) donates both electrons to the empty orbital of an incoming \( \text{H}^+ \) ion.

[0.4 marks] Identifying the shape as trigonal pyramidal. [0.4 marks] Stating the bond angle is between \( 106^\circ \) and \( 108^\circ \). [0.4 marks] Explaining that the coordinate bond is formed by the donation of a lone pair from the oxygen atom of water. [0.4 marks] Stating that this lone pair is donated to a hydrogen ion (\( \text{H}^+ \)) / proton / species with an empty valence orbital.

When concentrated sulfuric acid reacts with solid sodium chloride, hydrogen chloride gas is produced, which is observed as misty white fumes. The chemical equation is: \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\). Alternatively, the reaction can be written as: \(2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}\).

M1: Misty/white fumes observed (0.5 marks). M2: Balanced equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) (or 2:1 ratio to form \(\text{Na}_2\text{SO}_4\)) (1.0 mark). Allow state symbols but not required.

The cream precipitate is silver bromide, \(\text{AgBr}\). Silver chloride is white and dissolves in dilute ammonia; silver iodide is yellow and is insoluble in both dilute and concentrated ammonia. Silver bromide is insoluble in dilute ammonia but dissolves in concentrated ammonia to form a colorless complex.

M1: Silver bromide / \(\text{AgBr}\) (0.5 marks). M2: Precipitate is insoluble in dilute ammonia AND dissolves in concentrated ammonia (1.0 mark).

Sodium iodide is a strong reducing agent and reduces sulfuric acid to hydrogen sulfide (\(\text{H}_2\text{S}\)), which has a characteristic rotten egg smell. The iodide ions are oxidized to iodine (\(\text{I}_2\)), which is observed as a dark grey solid.

M1: Dark grey solid is Iodine / \(\text{I}_2\) (0.75 marks). M2: Gas is Hydrogen sulfide / \(\text{H}_2\text{S}\) (0.75 marks).

The reaction of chlorine with cold, dilute aqueous sodium hydroxide is: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). Chlorine is converted into sodium chloride (\(\text{NaCl}\)), where its oxidation state is -1, and sodium chlorate(I) (\(\text{NaClO}\)), where its oxidation state is +1.

M1: Balanced equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (0.5 marks). M2: Oxidation state of Cl in \(\text{NaCl}\) is -1 AND in \(\text{NaClO}\) is +1 (1.0 mark, 0.5 marks each).

The equation is \(\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}\). In this reaction, the oxidation state of chlorine in \(\text{Cl}_2\) is 0. It is reduced to -1 in \(\text{HCl}\) and oxidized to +1 in \(\text{HClO}\). Since the same element is simultaneously oxidized and reduced, it is a disproportionation reaction.

M1: Balanced equation: \(\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}\) (0.5 marks). M2: Explanation that chlorine is simultaneously oxidized (from 0 to +1 in \(\text{HClO}\)) and reduced (from 0 to -1 in \(\text{HCl}\)) (1.0 mark).

Down Group 7, the reducing ability of halide ions increases. As you go down the group, the halide ions have more electron shells, so the ionic radius increases and there is more shielding. Consequently, the attraction between the nucleus and the outermost electrons becomes weaker, allowing the outer electrons to be lost more easily.

M1: State that reducing ability increases down the group (0.5 marks). M2: Explain that ionic radius increases / there is more shielding (0.5 marks). M3: State that outer electrons are less strongly attracted to the nucleus and therefore more easily lost (0.5 marks).

Let the percentage abundance of \(^{24}\text{Mg}\) be \(x\%\).
Since the total abundance of all isotopes must equal \(100\%\), the percentage abundance of \(^{26}\text{Mg}\) is:
\((100 - 10.0 - x)\% = (90.0 - x)\%\)

The relative atomic mass \(A_r\) is calculated as follows:
\(A_r = \frac{(24 \times x) + (25 \times 10.0) + (26 \times (90.0 - x))}{100} = 24.31\)

Multiply both sides by 100 to clear the fraction:
\(24x + 250 + 26(90.0 - x) = 2431\)

Expand the brackets:
\(24x + 250 + 2340 - 26x = 2431\)

Combine like terms:
\(-2x + 2590 = 2431\)
\(2x = 159\)
\(x = 79.5\%\)

- 1 mark: Correct algebraic setup representing the relative atomic mass expression, e.g., \(\frac{24x + 250 + 26(90-x)}{100} = 24.31\).
- 1 mark: Correct rearrangement and simplification to solve for \(x\) (e.g., \(2x = 159\) or equivalent step).
- 0.25 marks: Correct final answer of \(79.5\%\) (accept 79.5 without % sign; reject 80 or other incorrect rounding).

1. Convert the mass of a single \(^{48}\text{Ti}^{+}\) ion to kg:
\(m = \frac{48 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 7.9708 \times 10^{-26}\text{ kg}\)

2. Calculate the velocity of the ion using \(KE = \frac{1}{2}mv^2\):
\(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 2.40 \times 10^{-15}}{7.9708 \times 10^{-26}}} = 2.4540 \times 10^{5}\text{ m s}^{-1}\)

3. Calculate the time of flight using \(t = \frac{d}{v}\):
\(t = \frac{1.50}{2.4540 \times 10^{5}} = 6.1125 \times 10^{-6}\text{ s}\)

Rounded to 3 significant figures: \(6.11 \times 10^{-6}\text{ s}\).

- 1 mark: Correct mass of a single \(^{48}\text{Ti}^{+}\) ion calculated in kilograms (\(7.97 \times 10^{-26}\text{ kg}\)).
- 1 mark: Correct velocity calculation (\(2.45 \times 10^5\text{ m s}^{-1}\)) OR correct combined algebraic expression for \(t\).
- 0.25 marks: Correct final time of flight (\(6.11 \times 10^{-6}\text{ s}\); allow range of \(6.11 \times 10^{-6}\) to \(6.12 \times 10^{-6}\) depending on rounding).

Since both ions have the same kinetic energy (\(KE\)) and travel the same distance (\(d\)):
\(KE = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_2 v_2^2\)

Since \(v = \frac{d}{t}\):
\(m_1 \left(\frac{d}{t_1}\right)^2 = m_2 \left(\frac{d}{t_2}\right)^2 \implies \frac{m_1}{t_1^2} = \frac{m_2}{t_2^2}\)

Rearranging to find \(t_2\):
\(t_2 = t_1 \sqrt{\frac{m_2}{m_1}}\)

Substitute the isotopic masses and the given flight time:
\(t_2 = (1.25 \times 10^{-5}\text{ s}) \times \sqrt{\frac{37}{35}}\)
\(t_2 = (1.25 \times 10^{-5}) \times 1.02817 = 1.2852 \times 10^{-5}\text{ s}\)

Rounded to 3 significant figures: \(1.29 \times 10^{-5}\text{ s}\).

- 1 mark: Recognition and use of the proportionality \(t \propto \sqrt{m}\) or relationship \(\frac{t_1^2}{m_1} = \frac{t_2^2}{m_2}\).
- 1 mark: Correct substitution of values, e.g., \(t_2 = 1.25 \times 10^{-5} \times \sqrt{\frac{37}{35}}\).
- 0.25 marks: Correct final answer of \(1.29 \times 10^{-5}\text{ s}\) (allow \(1.285 \times 10^{-5}\) to \(1.29 \times 10^{-5}\)).

1. Calculate the mass of one \(^{84}\text{Kr}^{+}\) ion in kg:
\(m = \frac{84 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.3949 \times 10^{-25}\text{ kg}\)

2. Use \(KE = \frac{1}{2}mv^2\) to find the velocity of the ion:
\(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.85 \times 10^{-15}}{1.3949 \times 10^{-25}}} = 1.6287 \times 10^{5}\text{ m s}^{-1}\)

3. Calculate the distance (drift tube length) using \(d = v \times t\):
\(d = (1.6287 \times 10^{5}\text{ m s}^{-1}) \times (8.20 \times 10^{-6}\text{ s}) = 1.3355\text{ m}\)

Rounded to 3 significant figures: \(1.34\text{ m}\).

- 1 mark: Correct mass of a single \(^{84}\text{Kr}^{+}\) ion in kilograms (\(1.39 \times 10^{-25}\text{ kg}\) or \(1.40 \times 10^{-25}\text{ kg}\)).
- 1 mark: Correct calculation of velocity (\(1.63 \times 10^5\text{ m s}^{-1}\)).
- 0.25 marks: Correct final distance (\(1.34\text{ m}\); allow \(1.33\text{ m}\) to \(1.34\text{ m}\) depending on intermediate rounding).

First, consider boron trifluoride: boron has three valence electrons and forms three bonding pairs with no lone pairs. This results in a symmetrical trigonal planar geometry. Even though the B-F bonds are highly polar, the individual dipole moments are equal and act in opposite directions, cancelling each other out completely. Thus, \(\text{BF}_3\) is non-polar. Next, consider nitrogen trifluoride: nitrogen has five valence electrons, forming three bonding pairs and retaining one lone pair. This gives \(\text{NF}_3\) an asymmetrical trigonal pyramidal shape. The highly polar N-F bond dipoles do not cancel out, and the lone pair also contributes to the overall molecular dipole. Consequently, \(\text{NF}_3\) has a permanent net dipole moment, making it polar.

1 mark: State that \(\text{BF}_3\) has a symmetrical trigonal planar geometry/shape, causing the bond dipoles to cancel. 1 mark: State that \(\text{NF}_3\) has an asymmetrical trigonal pyramidal geometry/shape (due to a lone pair), meaning bond dipoles do not cancel. 0.5 mark: Identify that fluorine is more electronegative than boron or nitrogen, which causes polar bonds.

The standard enthalpy of formation (\(\Delta_fH^\theta\)) is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 100 kPa). For liquid ethanol, the elements are solid carbon (graphite), hydrogen gas, and oxygen gas. The balanced equation forming 1 mole of ethanol is: \(2\text{C(s, graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(l)}\). The standard enthalpy of formation of any element in its standard state (its most stable physical state under standard conditions) is defined as zero by convention. Since graphite is the reference standard state for carbon, its \(\Delta_fH^\theta\) is zero.

1 mark: Correct formulas and balancing for the equation to form 1 mole of liquid ethanol. 0.5 mark: Correct state symbols for all species in the equation: C(s) / C(s, graphite), H2(g), O2(g), and CH3CH2OH(l). 1 mark: Explaining that graphite is the standard state (or most stable allotrope/state under standard conditions) of carbon, and elements in their standard states have an enthalpy of formation of zero by definition.

To illustrate hydrogen bonding, the partial charges on the highly polar oxygen-hydrogen bonds must be marked: \(\delta-\)\ on the oxygen atoms and \(\delta+\)\ on the hydrogen atoms. Oxygen has two lone pairs of electrons in both water and methanol. The hydrogen bond is drawn as a dashed or dotted line from a lone pair on the oxygen atom of one molecule (e.g., water) to the \(\delta+\)\ hydrogen atom of the polar O-H bond of the other molecule (e.g., methanol). To show the correct geometry, the covalent O-H bond, the hydrogen bond, and the accepting O atom must form a straight line (bond angle of approximately 180 degrees) due to the alignment of orbital overlap.

1 mark: Correctly identified partial charges (\(\delta-\)\ on oxygen and \(\delta+\)\ on hydrogen) on the polar O-H groups of both molecules. 1 mark: The hydrogen bond shown as a dotted/dashed line connecting a lone pair on an oxygen atom to the \(\delta+\)\ hydrogen atom of the other molecule. 0.5 mark: Stating or showing that the O-H...O atoms are linear (180 degree bond angle around the hydrogen atom).

Using Hess's Law, the enthalpy change of the reaction is given by: \(\Delta H^\theta = \sum \Delta_fH^\theta\text{(products)} - \sum \Delta_fH^\theta\text{(reactants)}\). First, calculate the sum of the enthalpies of formation for the products: \(\sum \Delta_fH^\theta\text{(products)} = [3 \times \Delta_fH^\theta\text{(Cu(s))}] + \Delta_fH^\theta\text{(N}_2\text{(g))} + [3 \times \Delta_fH^\theta\text{(H}_2\text{O(l))}]\). Since \(\text{Cu(s)}\) and \(\text{N}_2\text{(g)}\) are elements in their standard states, their enthalpies of formation are zero. Therefore: \(\sum \Delta_fH^\theta\text{(products)} = 0 + 0 + [3 \times (-286.0)] = -858.0\text{ kJ mol}^{-1}\). Next, calculate the sum of the enthalpies of formation for the reactants: \(\sum \Delta_fH^\theta\text{(reactants)} = [2 \times \Delta_fH^\theta\text{(NH}_3\text{(g))}] + [3 \times \Delta_fH^\theta\text{(CuO(s))}] = [2 \times (-46.0)] + [3 \times (-155.0)] = -92.0 + (-465.0) = -557.0\text{ kJ mol}^{-1}\). Finally, calculate \(\Delta H^\theta\): \(\Delta H^\theta = -858.0 - (-557.0) = -301.0\text{ kJ mol}^{-1}\).

1 mark: Correct substitution of stoichiometric coefficients and standard values to calculate total reactants enthalpy (\(-557.0\text{ kJ}\)) and products enthalpy (\(-858.0\text{ kJ}\)). 1 mark: Correctly setting up the Hess's Law expression: \(\Delta H^\theta = \sum \Delta_fH^\theta\text{(products)} - \sum \Delta_fH^\theta\text{(reactants)}\) (\(-858.0 - (-557.0)\)). 0.5 mark: For the final correct value of \(-301.0\text{ kJ mol}^{-1}\) (or \(-301\text{ kJ mol}^{-1}\)). Deduct 0.5 overall if standard units are missing or the sign is positive.

First, calculate the amount, in moles, of hydrochloric acid used: \(n(\text{HCl}) = \frac{18.20}{1000} \times 0.0500 = 9.10 \times 10^{-4}\text{ mol}\). Since hydrochloric acid is monoprotic, the moles of \(\text{H}^+\) ions added are \(9.10 \times 10^{-4}\text{ mol}\). Since neutralisation occurs when \(n(\text{H}^+) = n(\text{OH}^-)\), the amount of hydroxide ions in the 25.0 \(\text{cm}^3\) sample is also \(9.10 \times 10^{-4}\text{ mol}\). Finally, calculate the concentration of hydroxide ions: \([\text{OH}^-] = \frac{9.10 \times 10^{-4}}{0.0250} = 0.0364\text{ mol dm}^{-3}\).

M1: Calculates the moles of hydrochloric acid correctly: \(9.10 \times 10^{-4}\text{ mol}\) (1 mark). M2: Calculates the concentration of hydroxide ions correctly: \(0.0364\text{ mol dm}^{-3}\) (allow \(3.64 \times 10^{-2}\)) (1 mark).

First, calculate the volume of the titre delivered: \(\text{Titre} = 24.35 - 0.15 = 24.20\text{ cm}^3\). Because a titre requires two readings (initial and final), the total uncertainty is \(2 \times 0.05 = 0.10\text{ cm}^3\). Calculate the percentage uncertainty: \(\text{Percentage uncertainty} = \left(\frac{0.10}{24.20}\right) \times 100 = 0.413\%\).

M1: Identifies the total uncertainty as \(\pm 0.10\text{ cm}^3\) and the titre as \(24.20\text{ cm}^3\) (1 mark). M2: Calculates the percentage uncertainty as 0.41% or 0.413% (1 mark).

Since this is a titration between a strong base and a strong acid, either phenolphthalein or methyl orange can be used. If phenolphthalein is used, the initial solution is alkaline, so it starts pink and turns colourless at the end-point. If methyl orange is used, it starts yellow and turns orange (or red/orange) at the end-point.

M1: States a suitable indicator: Phenolphthalein OR Methyl orange (1 mark). M2: Describes the correct colour change: pink to colourless (for phenolphthalein) OR yellow to orange/red (for methyl orange) (1 mark). Reject 'clear' for colourless.

Calculate the moles of \(\text{HCl}\) used: \(n(\text{HCl}) = 5.60 \times 10^{-3} \times 1.50 \times 10^{-3} = 8.40 \times 10^{-6}\text{ mol}\). The reaction equation is: \(\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}\). Thus, the moles of \(\text{Mg(OH)}_2\) in the sample are \(4.20 \times 10^{-6}\text{ mol}\). Calculate the concentration of \(\text{Mg(OH)}_2\): \([\text{Mg(OH)}_2] = \frac{4.20 \times 10^{-6}}{0.0500} = 8.40 \times 10^{-5}\text{ mol dm}^{-3}\). Multiply by \(M_r\) to find the solubility: \(\text{Solubility} = 8.40 \times 10^{-5} \times 58.3 = 4.90 \times 10^{-3}\text{ g dm}^{-3}\).

M1: Calculates the moles of \(\text{Mg(OH)}_2\) in the sample correctly: \(4.20 \times 10^{-6}\text{ mol}\) (1 mark). M2: Calculates the solubility in \(\text{g dm}^{-3}\): \(4.90 \times 10^{-3}\text{ g dm}^{-3}\) (allow \(4.9 \times 10^{-3}\)) (1 mark).

Calculate the moles of \(\text{HCl}\) used: \(n(\text{HCl}) = 0.03370 \times 0.200 = 6.74 \times 10^{-3}\text{ mol}\). The reaction is: \(\text{M(OH)}_2 + 2\text{HCl} \rightarrow \text{MCl}_2 + 2\text{H}_2\text{O}\). Thus, the moles of \(\text{M(OH)}_2\) are \(\frac{6.74 \times 10^{-3}}{2} = 3.37 \times 10^{-3}\text{ mol}\). Calculate the molar mass of \(\text{M(OH)}_2\): \(M_r = \frac{0.250}{3.37 \times 10^{-3}} = 74.2\text{ g mol}^{-1}\). Subtract the mass of the two hydroxide groups: \(A_r(\text{M}) = 74.2 - 2(17.0) = 40.2\text{ g mol}^{-1}\). This corresponds to calcium (\(A_r\) of \(\text{Ca} = 40.1\)).

M1: Calculates the molar mass of \(\text{M(OH)}_2\) as approximately \(74\text{ g mol}^{-1}\) (or \(A_r(\text{M}) \approx 40\)) (1 mark). M2: Identifies the metal M as calcium / Ca (1 mark).

Calculate the initial moles of \(\text{HCl}\): \(0.0500 \times 0.500 = 0.0250\text{ mol}\). Calculate the moles of excess \(\text{HCl}\) by finding moles of \(\text{NaOH}\) needed: \(n(\text{NaOH}) = 0.01420 \times 0.250 = 3.55 \times 10^{-3}\text{ mol}\). Moles of \(\text{HCl}\) that reacted with \(\text{Ca(OH)}_2\) are \(0.0250 - 0.00355 = 0.02145\text{ mol}\). The reaction equation is \(\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}\), so moles of \(\text{Ca(OH)}_2\) reacted are \(\frac{0.02145}{2} = 0.010725\text{ mol}\). Calculate the mass: \(\text{mass} = 0.010725 \times 74.1 = 0.795\text{ g}\).

M1: Calculates the moles of hydrochloric acid that reacted with calcium hydroxide: \(0.02145\text{ mol}\) (1 mark). M2: Calculates the mass of \(\text{Ca(OH)}_2\) that reacted: \(0.795\text{ g}\) (allow 0.79 to 0.80) (1 mark).

Magnesium sulfate is soluble, while barium sulfate is highly insoluble. Adding a source of sulfate ions, such as sodium sulfate or sulfuric acid, allows for distinction. Magnesium chloride remains clear and colourless because magnesium sulfate does not precipitate. Barium chloride reacts to form a thick white precipitate of barium sulfate. The ionic equation represents the formation of this solid precipitate from its aqueous ions: \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}.

M1: Identify a suitable reagent: aqueous sodium sulfate / sulfuric acid (accept any soluble sulfate) [1 mark]. M2: Both observations correct: no visible change/remains colourless with MgCl2 AND white precipitate with BaCl2 [1 mark]. M3: Correct ionic equation with state symbols: \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} [0.8 marks].

First, convert all units to SI: P = 101,000 Pa, V = 89.0 x 10^-6 m^3, T = 373 K. Use the ideal gas equation PV = nRT to find the number of moles (n): n = PV / RT = (101,000 x 89.0 x 10^-6) / (8.31 x 373) = 8.989 / 3099.63 = 0.002900 mol. Now, calculate Mr: Mr = mass / n = 0.285 / 0.002900 = 98.275, which rounds to 98.3 to 3 significant figures.

M1: Correct conversion of volume to m3 (89.0 x 10^-6 m3) and pressure to Pa (101,000 Pa) [1 mark]. M2: Correct calculation of moles of gas (n = 0.002900 mol) [1 mark]. M3: Correct calculation of Mr to 3 significant figures (98.3) [0.8 marks]. (Allow ECF if conversion step is incorrect but calculation is mathematically sound)

Magnesium reacts with steam (gaseous water) to produce solid magnesium oxide and hydrogen gas: \text{Mg(s)} + \text{H}_2\text{O(g)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)}. The reaction is highly exothermic, producing a bright white light and a white ash (MgO). It is much faster than the reaction with cold water because steam is at a much higher temperature, providing higher kinetic energy so that more colliding molecules exceed the activation energy threshold.

M1: Correct balanced chemical equation with state symbols: \text{Mg(s)} + \text{H}_2\text{O(g)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)} [1 mark]. M2: State one visual observation: bright white light or white solid/ash [1 mark]. M3: Kinetic explanation: Higher temperature of steam means particles have higher kinetic energy, so a larger fraction of collisions have energy \ge E_a [0.8 marks].

Using a Born-Haber cycle / Hess's Law cycle for dissolution: \Delta H_{sol} = \Delta H_{L,diss} + \Delta H_{hyd}(\text{Ca}^{2+}) + 2 \times \Delta H_{hyd}(\text{Cl}^-). Substitute the values: \Delta H_{sol} = 2258 + (-1579) + 2 \times (-378) = 2258 - 1579 - 756 = -77\text{ kJ mol}^{-1}.

M1: Show correct cycle or expression: \Delta H_{sol} = \Delta H_{L,diss} + \Delta H_{hyd}(\text{Ca}^{2+}) + 2 \times \Delta H_{hyd}(\text{Cl}^-) [1 mark]. M2: Correct multiplication of chloride hydration enthalpy by 2 (-756) [1 mark]. M3: Final correct value with correct sign (-77 kJ mol^-1) [0.8 marks].

We can rearrange the ideal gas equation: PV = nRT. Since n = m/M, we can write PV = mRT/M. Rearranging for density (\rho = m/V) gives \rho = PM / RT. Make sure to use consistent units: P = 100,000 Pa, M = 0.0280 kg mol^-1 (since N2 has Mr = 28.0), T = 298 K. \rho = (100,000 x 0.0280) / (8.31 x 298) = 2800 / 2476.38 = 1.1307 kg m^-3. Since 1 kg m^-3 is equivalent to 1 g dm^-3, the density is 1.13 g dm^-3.

M1: Correct algebraic rearrangement to relate density to pressure, molar mass, and temperature: \rho = PM / RT [1 mark]. M2: Correct values substituted with molar mass of N2 as 28.0 g mol^-1 (or 0.0280 kg mol^-1) [1 mark]. M3: Final answer of 1.13 to 3 significant figures [0.8 marks].

The general trend in first ionisation energy across Period 3 is an increase. This is because from sodium to argon, the number of protons in the nucleus increases (increasing nuclear charge), while the shielding effect remains relatively constant as outer electrons are added to the same main energy level. Consequently, the outer electrons experience a stronger electrostatic attraction to the nucleus, requiring more energy to be removed.

1.33 marks total: 0.44 marks for identifying the general increasing trend; 0.44 marks for stating the increase in nuclear charge with similar shielding; 0.45 marks for explaining the stronger attraction of the nucleus on outer electrons.

Phosphorus has the outer electronic configuration \( 3s^2 3p^3 \) with three unpaired electrons in separate 3p orbitals. Sulfur has the configuration \( 3s^2 3p^4 \) with one pair of electrons in a 3p orbital. The repulsion between the two paired electrons in sulfur's 3p orbital makes it easier to remove an electron from sulfur than from phosphorus, resulting in a lower first ionisation energy.

1.33 marks total: 0.66 marks for identifying that sulfur has a paired electron in a 3p orbital whereas phosphorus has only unpaired 3p electrons; 0.67 marks for explaining that electron-pair repulsion in sulfur's 3p orbital makes it easier to remove an electron.

Magnesium has the electronic configuration \( 1s^2 2s^2 2p^6 3s^2 \), so its outer electron is in a 3s orbital. Aluminum has the configuration \( 1s^2 2s^2 2p^6 3s^2 3p^1 \), so its outer electron is in a 3p orbital. The 3p orbital is higher in energy and experiences shielding from the 3s electrons, making the outer electron in aluminum easier to remove than that in magnesium.

1.33 marks total: 0.66 marks for identifying that aluminum's outer electron is in a 3p subshell whereas magnesium's is in a 3s subshell; 0.67 marks for explaining that the 3p subshell is higher in energy and experiences shielding from the 3s subshell, making its electron easier to remove.

Sodium (Na) has the highest second ionisation energy. Its first electron is removed from the \( 3s^1 \) subshell. The second electron must be removed from the \( 2p^6 \) subshell, which is in a lower main energy level (closer to the nucleus) and experiences significantly less shielding. This results in a massive increase in electrostatic attraction, requiring much more energy.

1.33 marks total: 0.44 marks for identifying Sodium (Na); 0.44 marks for stating the second electron is removed from an inner shell (2p subshell); 0.45 marks for explaining that this inner shell experiences much less shielding and stronger nuclear attraction.

Across Period 3 from sodium to chlorine, the number of protons in the nucleus increases, resulting in an increased nuclear charge. However, the outer electrons are added to the same main energy level (shell 3), meaning they experience a similar level of shielding from inner shells. The stronger electrostatic attraction from the increasing nuclear charge pulls the outer electron shell closer to the nucleus, decreasing the atomic radius.

1.33 marks total: 0.44 marks for stating nuclear charge increases; 0.44 marks for stating shielding remains similar; 0.45 marks for explaining that outer electrons are pulled closer to the nucleus.

The third ionisation energy represents the energy required to remove one mole of electrons from one mole of gaseous \( \text{Mg}^{2+} \) ions to form one mole of gaseous \( \text{Mg}^{3+} \) ions. The balanced equation is: \( \text{Mg}^{2+}(g) \rightarrow \text{Mg}^{3+}(g) + e^- \).

1.33 marks total: 0.66 marks for correct species and state symbols (\( \text{Mg}^{2+}(g) \) and \( \text{Mg}^{3+}(g) \)); 0.67 marks for the correct inclusion of the electron (\( e^- \)) on the product side.

When temperature decreases, the average kinetic energy of the molecules decreases. This causes the Maxwell-Boltzmann curve to shift: 1. The peak moves to the left (lower energy) and becomes taller because the total number of molecules (area under the curve) must remain constant. 2. The area under the curve to the right of the activation energy \(E_a\) decreases, meaning a smaller fraction of molecules possess energy greater than or equal to \(E_a\).

1 mark: Peak shifts to the left and is higher. 0.38 marks: Fraction of molecules with energy greater than or equal to \(E_a\) decreases (or the area to the right of \(E_a\) decreases).

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. On a Maxwell-Boltzmann distribution, this is represented by shifting the activation energy line to the left. Consequently, a larger fraction of molecules have energy greater than or equal to this new lower activation energy, leading to a higher frequency of successful collisions.

1 mark: Catalyst provides an alternative pathway with a lower activation energy. 0.38 marks: A larger fraction of molecules now have energy greater than or equal to the activation energy (or there is an increased frequency of successful collisions).

The balanced chemical equation for the incomplete combustion of octane is: \(\text{C}_8\text{H}_{18}(\text{g}) + 8.5\text{O}_2(\text{g}) \rightarrow 8\text{CO}(\text{g}) + 9\text{H}_2\text{O}(\text{g})\) (or \(2\text{C}_8\text{H}_{18}(\text{g}) + 17\text{O}_2(\text{g}) \rightarrow 16\text{CO}(\text{g}) + 18\text{H}_2\text{O}(\text{g})\)). Carbon monoxide is toxic/poisonous because it binds strongly to hemoglobin, preventing the transport of oxygen around the body.

1 mark: Correct balanced equation (allow fractions, ignore state symbols). 0.38 marks: Toxic / poisonous / reduces the oxygen capacity of blood (do not accept 'causes global warming' or other general environmental issues).

Carbon monoxide and nitrogen monoxide react over a precious metal catalyst to produce carbon dioxide and nitrogen. The balanced equation is: \(2\text{CO}(\text{g}) + 2\text{NO}(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + \text{N}_2(\text{g})\) (or \(\text{CO} + \text{NO} \rightarrow \text{CO}_2 + 0.5\text{N}_2\)). Common catalysts are transition metals such as platinum, palladium, or rhodium.

1 mark: Correct balanced equation (allow fractions, ignore state symbols). 0.38 marks: Named catalyst: Platinum / Pt / Palladium / Pd / Rhodium / Rh.

The total area under the Maxwell-Boltzmann distribution curve represents the total number of molecules in the sample. Although increasing the temperature changes the distribution of molecular energies (making the curve flatter and shifting the peak to the right), the total number of molecules in the closed system remains constant, so the area under the curve does not change.

1 mark: Total area represents the total number of particles/molecules. 0.38 marks: The total number of molecules/particles remains constant (or system is closed/no molecules escape).

The standard enthalpy of combustion is the enthalpy change measured when 1 mole of a substance is completely burned in excess oxygen under standard conditions (100 kPa, specified temperature usually 298 K), with all reactants and products in their standard states.

1 mark: Enthalpy change when 1 mole of a substance is burned completely in oxygen. 0.38 marks: Under standard conditions, with all reactants and products in their standard states.

A heterogeneous catalyst operates in a different physical state or phase from the reactants (e.g., a solid transition metal catalyst reacting with gaseous reactants). The initial step is adsorption, where the reactant molecules form weak bonds with active sites on the catalyst surface, which weakens the bonds within the reactant molecules and lowers the activation energy.

1 mark: A catalyst in a different physical state/phase from the reactants. 0.38 marks: Adsorption of reactants onto the catalyst surface / active sites.

Sulfur dioxide is oxidized in the atmosphere by oxygen to sulfur trioxide: \(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g})\) (or \(\text{SO}_2 + 0.5\text{O}_2 \rightarrow \text{SO}_3\)). Sulfur trioxide then reacts with atmospheric water droplets to form sulfuric acid, which causes acid rain: \(\text{SO}_3(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{aq})\).

1 mark: Correct balanced equation for the oxidation of \(\text{SO}_2\) to \(\text{SO}_3\) (allow fractions). 0.38 marks: Correct balanced equation for the reaction of \(\text{SO}_3\) with water to form \(\text{H}_2\text{SO}_4\).

1. Decane is a larger molecule with a larger surface area and a greater number of electrons compared to pentane.
2. This results in stronger van der Waals (London dispersion) forces between the decane molecules, which require more thermal energy to overcome, resulting in a higher boiling point.

M1: State that decane has a larger surface area / more electrons / larger molecular size than pentane (1 mark).
M2: State that decane has stronger van der Waals forces (or London dispersion forces) between molecules that require more energy to overcome (1 mark). Reject references to breaking covalent bonds.

Thermal cracking requires high temperatures (typically 450-900 C) and high pressure (up to 7000 kPa). This homolytic fission process primarily produces a high proportion of alkenes, which are highly useful for polymer synthesis.

M1: High temperature (accept any temperature value in the range 400-900 C / 700-1200 K) AND high pressure (accept 7000 kPa / 70 atm) (1 mark).
M2: Alkenes (1 mark). Accept specific alkene examples like 'ethene'.

Pentane (\(\text{C}_5\text{H}_{12}\)) is a straight-chain alkane. Its two branched chain isomers are:
1. A four-carbon chain with a methyl branch: 2-methylbutane.
2. A three-carbon chain with two methyl branches on the middle carbon: 2,2-dimethylpropane.

M1: 2-methylbutane (1 mark). Accept methylbutane.
M2: 2,2-dimethylpropane (1 mark). Accept dimethylpropane. Reject any other structures or non-IUPAC names.

First, determine the remaining atoms from the reactant:
Carbon atoms left: \(14 - 8 = 6\)
Hydrogen atoms left: \(30 - 18 = 12\)
Since three identical alkene molecules are formed, divide these atoms by 3:
Carbon per alkene molecule: \(6 / 3 = 2\)
Hydrogen per alkene molecule: \(12 / 3 = 4\)
Thus, the alkene is ethene with the formula \(\text{C}_2\text{H}_4\).
The balanced equation is: \(\text{C}_{14}\text{H}_{30} \rightarrow \text{C}_8\text{H}_{18} + 3\text{C}_2\text{H}_4\).

M1: Identify the formula of the alkene as \(\text{C}_2\text{H}_4\) (1 mark).
M2: Provide the fully balanced chemical equation: \(\text{C}_{14}\text{H}_{30} \rightarrow \text{C}_8\text{H}_{18} + 3\text{C}_2\text{H}_4\) (1 mark).

1. Identify the longest continuous carbon chain containing the double bond: 5 carbons (pentene).
2. Number the carbon chain starting from the end closer to the double bond to give the alkene functional group the lowest possible locant. This makes it a pent-1-ene.
3. Identify the substituents and their positions on the numbered chain:
- Chlorine atom at position 4 (4-chloro)
- Methyl group at position 3 (3-methyl)
4. List substituents in alphabetical order: 'chloro' before 'methyl'.
5. Combine these elements into the final name: 4-chloro-3-methylpent-1-ene.

1 mark: 4-chloro-3-methylpent-1-ene.

[Accept: 4-chloro-3-methylpentene]
[Reject: 2-chloro-3-methylpent-4-ene, 4-chloro-3-methylpent-4-ene]

1. To find the monomer, replace the single carbon-carbon bond in the backbone of the repeating unit with a double bond:
\(\text{CH}_2=\text{C(CH}_3\text{)(Cl)}\)
2. Find the longest continuous carbon chain containing the double bond: 3 carbons (propene).
3. Number the carbon chain starting at the end with the double bond (C1 = \(\text{CH}_2\), C2 = \(\text{C}\), C3 = \(\text{CH}_3\)).
4. Identify the substituent: a chlorine atom is attached to carbon 2.
5. Combining these gives 2-chloropropene (or 2-chloroprop-1-ene).

1 mark: 2-chloropropene OR 2-chloroprop-1-ene.

[Reject: 2-chloroprop-2-ene, chloropropene (the locant '2' is mandatory to distinguish it from 1-chloropropene)]

1. Write the structural formula of 3-methylbut-1-ene: \(\text{CH}_2=\text{CH-CH(CH}_3\text{)}_2\).
2. A position isomer must keep the same carbon skeleton (methylbutane skeleton) but have the functional group (double bond) in a different position.
3. Moving the double bond to the middle of the chain yields \(\text{(CH}_3\text{)}_2\text{C=CH-CH}_3\), which is 2-methylbut-2-ene.
4. Note: 2-methylbut-1-ene is a chain isomer, not a position isomer, because the methyl substituent is attached to a different carbon on the chain.

1 mark: 2-methylbut-2-ene.

[Reject: 2-methylbut-1-ene (this is a chain isomer)]

During nucleophilic substitution, the carbon-halogen bond must break. The carbon-chlorine bond is more polar than the carbon-iodine bond because chlorine is more electronegative than iodine. This polarity would theoretically make the carbon in 1-chlorobutane more susceptible to nucleophilic attack. However, the rate of reaction depends primarily on the strength of the bond being broken (bond enthalpy) rather than its polarity. The C-I bond is longer and weaker (lower bond enthalpy of approximately \(238 \text{ kJ mol}^{-1}\)) than the C-Cl bond (approximately \(338 \text{ kJ mol}^{-1}\)), making it much easier to break. Therefore, 1-iodobutane hydrolyses significantly faster than 1-chlorobutane.

1 mark: 1-iodobutane is faster because the C-I bond is weaker / has a lower bond enthalpy than the C-Cl bond (allow reverse argument for C-Cl).
0.67 marks: Bond enthalpy / bond strength is the dominant factor in determining the rate of substitution, outweighing the effect of bond polarity.

In this reaction, ammonia acts as a nucleophile to substitute the bromine atom in bromoethane, forming ethylamine. However, the primary amine product, ethylamine, contains a nitrogen atom with a lone pair and is also a nucleophile. If bromoethane is readily available, the ethylamine will react with it to form diethylamine (a secondary amine), which can further react to form triethylamine and eventually tetraethylammonium bromide. Using a large excess of ammonia ensures that any bromoethane molecule is far more likely to collide with an ammonia molecule than with an ethylamine molecule, thereby maximizing the yield of the primary amine.

1 mark: Excess ammonia is used to prevent further substitution / to ensure the primary amine (ethylamine) is the major product.
0.67 marks: Diethylamine / secondary amine / tertiary amine / quaternary ammonium salt (or formula like \((\text{C}_2\text{H}_5)_2\text{NH}\)) would increase in yield if excess bromoethane is used.

The hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile, attacking the electron-deficient carbon atom (\(\text{C}^{\delta+}\)) bonded to the bromine atom. The bromine atom leaves as a bromide ion (\(\text{Br}^-\)), taking both bonding electrons from the C-Br bond with it. This process where both bonding electrons go to one of the atoms is called heterolytic fission. The resulting organic product is a four-carbon alcohol with the hydroxyl group on the second carbon, named butan-2-ol.

1 mark: butan-2-ol (accept 2-butanol; reject butanol / butan-1-ol).
0.67 marks: heterolytic (fission) (reject homolytic).

1. Identify the major product: When H+ adds to the double bond, it can add to carbon-1 to form a tertiary carbocation at carbon-2, or add to carbon-2 to form a primary carbocation at carbon-1. Addition of bromide to the tertiary carbocation yields the major product, 2-bromo-2-methylbutane.
2. Compare intermediate carbocation stability: The intermediate tertiary carbocation is more stable than the alternative primary carbocation.
3. Explain in terms of the inductive effect: Alkyl groups are electron-donating (positive inductive effect). The tertiary carbocation has three alkyl groups attached to the positive carbon, which disperse the positive charge more effectively than the single alkyl group attached to the positive carbon in the primary carbocation.

M1: State that the major product is 2-bromo-2-methylbutane (1 mark)
M2: State that the reaction proceeds via a tertiary carbocation intermediate which is more stable than the primary carbocation (1 mark)
M3: Explain that the tertiary carbocation is stabilized by the electron-releasing / positive inductive effect of three alkyl groups, which disperses the positive charge (1.33 marks)
[Accept 'electron-donating' for 'electron-releasing']

Propagation Step 1: A chlorine radical abstract a hydrogen atom from ethane, generating an ethyl radical and hydrogen chloride molecule.
Propagation Step 2: The ethyl radical reacts with a chlorine molecule to produce chloroethane and regenerate the chlorine radical.
Preventing further substitution: If chlorine is in excess or equimolar amounts, the chloroethane produced will continue to react with chlorine radicals to form di-, tri-, and polychloroethanes. An excess of ethane ensures that chlorine radicals are far more likely to collide with unreacted ethane molecules than with chloroethane molecules, maximizing the yield of the desired monochloroalkane.

M1: Correct equation for Propagation Step 1: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\) (1 mark)
M2: Correct equation for Propagation Step 2: \(\text{CH}_3\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}^\bullet\) (1 mark)
M3: Explanation of excess ethane: To limit further substitution / chlorination reactions and maximize the yield of mono-substituted chloroethane (1.33 marks)

1. Identify the key factor: The rate of electrophilic addition with hydrogen halides depends primarily on the bond enthalpy of the H–X bond, not the polarity of the bond.
2. Bond strength comparison: The H–I bond is much weaker than the H–Cl bond. This is due to the larger size of the iodine atom, which leads to longer bond length and less efficient overlap of the bonding orbitals.
3. Link to reaction rate: Since the H–I bond is weaker, it breaks much more readily during electrophilic attack on the carbon-carbon double bond, lowering the activation energy of the rate-determining step.

M1: States that the H–I bond is weaker than the H–Cl bond / has a lower bond enthalpy (1 mark)
M2: Explains that iodine has a larger atomic radius / more electron shielding, leading to longer bonds and poorer orbital overlap (1 mark)
M3: Links this to easier heterolytic fission / lower activation energy for the rate-determining step (1.33 marks)
[Reject any argument that suggests HCl reacts faster due to being more polar]

Butan-2-ol is a secondary alcohol and is oxidized to butanone, reducing the orange dichromate(VI) ions \(\text{Cr}_2\text{O}_7^{2-}\) to green chromium(III) ions \(\text{Cr}^{3+}\). 2-methylpropan-2-ol is a tertiary alcohol and is resistant to oxidation under these conditions, so the solution remains orange.

1 mark: Correct observation for butan-2-ol (orange to green / green solution formed).
1 mark: Correct observation for 2-methylpropan-2-ol (remains orange / no change).

Tollens' reagent contains the diamminesilver(I) complex, which is reduced to metallic silver by aldehydes like propanal, forming a silver mirror. Ketones like propanone cannot be easily oxidized and show no reaction. Alternatively, Fehling's solution can be used, which forms a red precipitate with propanal and remains blue with propanone.

1 mark: Suitable reagent identified (Tollens' reagent OR Fehling's solution).
1 mark: Correct corresponding observations for both compounds (silver mirror with propanal and no change with propanone OR red precipitate with propanal and remains blue with propanone).

Ethene undergoes partial oxidation in the presence of a silver catalyst at elevated temperatures to form epoxyethane: \(2\text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{C}_2\text{H}_4\text{O}\).

1 mark: Balanced equation for the oxidation of ethene to epoxyethane.
1 mark: Silver (\(\text{Ag}\)) catalyst.

The acid-catalyzed ring opening of epoxyethane with water yields ethane-1,2-diol: \(\text{C}_2\text{H}_4\text{O} + \text{H}_2\text{O} \rightarrow \text{HOCH}_2\text{CH}_2\text{OH}\). Ethane-1,2-diol is widely used as an engine coolant/antifreeze and in the manufacture of polyesters like polyethylene terephthalate (PET).

1 mark: Correct balanced chemical equation (accept molecular formula \(\text{C}_2\text{H}_6\text{O}_2\) for the product).
1 mark: Valid industrial use (e.g., antifreeze, engine coolant, polyester manufacture).

Dehydration of butan-2-ol can eliminate a hydrogen atom from either carbon-1 or carbon-3. Elimination from carbon-3 yields but-2-ene, which exists as a pair of stereoisomers due to restricted rotation about the double bond: (E)-but-2-ene (trans-but-2-ene) and (Z)-but-2-ene (cis-but-2-ene). These are geometric or E-Z isomers.

1 mark: Both IUPAC names correct ((E)-but-2-ene and (Z)-but-2-ene, or cis-but-2-ene and trans-but-2-ene).
1 mark: Type of stereoisomerism identified as E-Z isomerism or geometric isomerism.

The only isomer of \(\text{C}_4\text{H}_{10}\text{O}\) that cannot be oxidized by acidified potassium dichromate(VI) is the tertiary alcohol, 2-methylpropan-2-ol. Tertiary alcohols cannot be oxidized because the carbon atom bonded to the \(\text{-OH}\) group is bonded to three other carbon atoms and has no C-H bonds to break during oxidation.

1 mark: Correct IUPAC name (2-methylpropan-2-ol).
1 mark: Correct explanation (it is a tertiary alcohol / there is no hydrogen atom on the carbon carrying the \(\text{-OH}\) group).

First, calculate the moles of cyclohexanol used: \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{15.0}{100.0} = 0.150\text{ mol}\). Since the stoichiometry of the dehydration of cyclohexanol to cyclohexene is 1:1, the theoretical yield of cyclohexene is also \(0.150\text{ mol}\). Next, apply the percentage yield to find the actual moles of cyclohexene obtained: \(\text{actual moles} = 0.150 \times 0.680 = 0.102\text{ mol}\). Finally, calculate the actual mass of cyclohexene: \(\text{mass} = 0.102 \times 82.0 = 8.364\text{ g}\). To 3 significant figures, this is \(8.36\text{ g}\).

M1: Moles of cyclohexanol = \(\frac{15.0}{100.0} = 0.150\text{ mol}\) (1.0 mark). M2: Actual moles of cyclohexene = \(0.150 \times 0.680 = 0.102\text{ mol}\) (1.0 mark). M3: Mass of cyclohexene = \(0.102 \times 82.0 = 8.36\text{ g}\) (0.4 mark). Accept range 8.36 to 8.37 g.

First, calculate the mass of cyclohexanol used: \(\text{mass} = \text{density} \times \text{volume} = 0.962\text{ g cm}^{-3} \times 12.0\text{ cm}^3 = 11.544\text{ g}\). Next, calculate the moles of cyclohexanol: \(\text{moles} = \frac{11.544}{100.0} = 0.11544\text{ mol}\). The theoretical yield of cyclohexene is therefore \(0.11544\text{ mol}\). Calculate the theoretical mass of cyclohexene: \(\text{theoretical mass} = 0.11544\text{ mol} \times 82.0\text{ g mol}^{-1} = 9.466\text{ g}\). Finally, calculate the percentage yield: \(\text{yield} = \frac{5.50}{9.466} \times 100 = 58.102\%\). To 3 significant figures, this is \(58.1\%\).

M1: Mass of cyclohexanol = 11.544 g and Moles of cyclohexanol = 0.1154 mol (1.0 mark). M2: Theoretical mass of cyclohexene = \(0.11544 \times 82.0 = 9.47\text{ g}\) (1.0 mark). M3: Percentage yield = \(\frac{5.50}{9.466} \times 100 = 58.1\%\) (0.4 mark). Accept 58.0% to 58.2%.

First, find the moles of cyclohexene needed: \(\text{moles} = \frac{10.0}{82.0} = 0.12195\text{ mol}\). Since the reaction has a 65.0% yield, the theoretical moles of cyclohexanol required is \(\frac{0.12195}{0.650} = 0.18762\text{ mol}\). Convert this to the mass of cyclohexanol required: \(\text{mass} = 0.18762\text{ mol} \times 100.0\text{ g mol}^{-1} = 18.762\text{ g}\). Finally, find the volume of cyclohexanol using its density: \(\text{volume} = \frac{\text{mass}}{\text{density}} = \frac{18.762}{0.962} = 19.503\text{ cm}^3\). To 3 significant figures, this is \(19.5\text{ cm}^3\).

M1: Moles of cyclohexene needed = 0.122 mol and theoretical moles of cyclohexanol required = 0.188 mol (1.0 mark). M2: Mass of cyclohexanol required = 18.8 g (1.0 mark). M3: Volume of cyclohexanol = \(\frac{18.762}{0.962} = 19.5\text{ cm}^3\) (0.4 mark). Accept 19.4 to 19.6 cm3.

First, calculate the moles of cyclohexanol: \(\text{moles} = \frac{18.0}{100.0} = 0.180\text{ mol}\). The theoretical yield of cyclohexene is also \(0.180\text{ mol}\), which corresponds to a theoretical mass of \(0.180\text{ mol} \times 82.0\text{ g mol}^{-1} = 14.76\text{ g}\). Next, calculate the actual mass of cyclohexene collected: \(\text{actual mass} = \text{volume} \times \text{density} = 11.2\text{ cm}^3 \times 0.811\text{ g cm}^{-3} = 9.0832\text{ g}\). Finally, calculate the percentage yield: \(\text{yield} = \frac{9.0832}{14.76} \times 100 = 61.539\%\). To 3 significant figures, this is \(61.5\%\).

M1: Theoretical mass of cyclohexene = \(0.180 \times 82.0 = 14.76\text{ g}\) (1.0 mark). M2: Actual mass of cyclohexene collected = \(11.2 \times 0.811 = 9.083\text{ g}\) (1.0 mark). M3: Percentage yield = \(\frac{9.0832}{14.76} \times 100 = 61.5\%\) (0.4 mark). Accept 61.4% to 61.6%.

First, calculate the theoretical mass of cyclohexene expected from 0.220 mol of cyclohexanol: \(\text{theoretical mass} = 0.220\text{ mol} \times 82.0\text{ g mol}^{-1} = 18.04\text{ g}\). Next, calculate the percentage yield: \(\text{yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100 = \frac{13.1}{18.04} \times 100 = 72.616\%\). To 3 significant figures, this is \(72.6\%\).

M1: Theoretical mass of cyclohexene = \(0.220 \times 82.0 = 18.04\text{ g}\) (1.0 mark). M2: Expression or working for percentage yield showing actual over theoretical (1.0 mark). M3: Percentage yield = 72.6% (0.4 mark). Accept 72.5% to 72.7%.

The equilibrium constant expression is written with the concentration of products in the numerator and the concentration of reactants in the denominator, each raised to the power of their stoichiometric coefficients:
\[K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\]

To find the units:
\[\text{Units} = \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{1}{\text{mol}^2 \text{dm}^{-6}} = \text{dm}^6 \text{mol}^{-2}\]

1 mark for the correct expression for \(K_c\) (must include square brackets; state symbols not required; penalize round brackets).
0.8 mark for the correct units: \(\text{dm}^6 \text{mol}^{-2}\) (or \(\text{mol}^{-2} \text{dm}^6\)).

1. Determine equilibrium moles:
- Initial moles of \(\text{PCl}_5 = 0.200\text{ mol}\)
- Equilibrium moles of \(\text{PCl}_5 = 0.120\text{ mol}\)
- Change in moles = \(-0.080\text{ mol}\)
- Equilibrium moles of \(\text{PCl}_3 = +0.080\text{ mol}\)
- Equilibrium moles of \(\text{Cl}_2 = +0.080\text{ mol}\)

2. Calculate equilibrium concentrations (divide by volume \(2.50\text{ dm}^3\)):
- \([\text{PCl}_5] = \frac{0.120}{2.50} = 0.0480\text{ mol dm}^{-3}\)
- \([\text{PCl}_3] = \frac{0.080}{2.50} = 0.0320\text{ mol dm}^{-3}\)
- \([\text{Cl}_2] = \frac{0.080}{2.50} = 0.0320\text{ mol dm}^{-3}\)

3. Calculate \(K_c\):
\[K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.0320 \times 0.0320}{0.0480} = 0.0213\text{ mol dm}^{-3}\]

0.5 marks: Deduces equilibrium moles of \(\text{PCl}_3\) and \(\text{Cl}_2\) are both \(0.080\text{ mol}\).
0.5 marks: Correctly divides moles by volume (\(2.50\text{ dm}^3\)) to find concentrations.
0.5 marks: Correct calculation of the numerical value \(0.0213\) (accept \(0.021\) or \(2.13 \times 10^{-2}\)).
0.3 marks: Correct units: \(\text{mol dm}^{-3}\).

1. Effect on \(K_c\): Because the forward reaction is exothermic, an increase in temperature shifts the position of equilibrium to the left (the endothermic direction) to absorb the added thermal energy. As a result, the concentrations of reactants increase relative to products, and the value of \(K_c\) decreases.
2. Effect on yield: The position of equilibrium shifts to the left, which decreases the equilibrium yield of ammonia.

1.0 mark: States that the value of \(K_c\) decreases AND explains this is because the forward reaction is exothermic, so the equilibrium shifts in the endothermic direction (to the left).
0.8 mark: States that the equilibrium yield of ammonia decreases because the system shifts to the left to oppose the increase in temperature.

(i) The position of equilibrium is unchanged. This is because there are equal moles of gaseous reactants (2 moles) and gaseous products (2 moles) on both sides of the equation. Thus, pressure changes do not favor either side.
(ii) The value of \(K_c\) is unchanged. The equilibrium constant \(K_c\) is temperature-dependent and is unaffected by changes in pressure or concentration.

0.5 marks: States that the position of equilibrium is unchanged/unaffected.
0.5 marks: Explains that there is an equal number of moles of gas on both sides of the reaction equation.
0.5 marks: States that the value of \(K_c\) is unchanged.
0.3 marks: Explains that \(K_c\) only changes with temperature.

Let \(x\) be the number of moles of ethyl ethanoate formed at equilibrium.
- Moles of ethanoic acid at equilibrium = \(1.0 - x\)
- Moles of ethanol at equilibrium = \(1.0 - x\)
- Moles of ethyl ethanoate at equilibrium = \(x\)
- Moles of water at equilibrium = \(x\)

Since total volume \(V\) cancels out in the \(K_c\) expression:
\[K_c = \frac{[\text{ester}][\text{water}]}{[\text{acid}][\text{alcohol}]} = \frac{(x/V)(x/V)}{((1.0-x)/V)((1.0-x)/V)} = \frac{x^2}{(1.0 - x)^2}\]

Given \(K_c = 4.0\):
\[\frac{x^2}{(1.0 - x)^2} = 4.0\]
Taking the square root of both sides:
\[\frac{x}{1.0 - x} = 2.0\]
\[x = 2.0(1.0 - x)\]
\[x = 2.0 - 2.0x\]
\[3.0x = 2.0\]
\[x = \frac{2}{3} \approx 0.67\text{ mol}\]

0.4 marks: Expresses the equilibrium moles of reactants as \((1.0 - x)\) and products as \(x\).
0.4 marks: Correctly sets up the \(K_c\) equation, showing that volume cancels out.
0.5 marks: Solves the mathematical equation by taking the square root to get \(x / (1.0 - x) = 2.0\).
0.5 marks: Calculates the final answer as \(0.67\text{ mol}\) (accept \(0.667\) or \(2/3\)).

(a) Silicon has 4 valence electrons, all of which are bonded to chlorine atoms, giving 4 bonding pairs and 0 lone pairs. According to VSEPR theory, 4 electron pairs adopt a tetrahedral geometry with a bond angle of \(109.5^\circ\).
(b) Silicon tetrachloride undergoes rapid hydrolysis with water to form silicon dioxide (or orthosilicic acid, \(\text{H}_4\text{SiO}_4\)) and hydrogen chloride: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\).

Total: 2.2 marks
- 1.1 marks: Correctly identifying the shape as tetrahedral and the bond angle as 109.5° (accept 109° to 110°).
- 1.1 marks: Correct balanced equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) (allow \(\text{H}_4\text{SiO}_4\) instead of \(\text{SiO}_2 + 2\text{H}_2\text{O}\); state symbols not strictly required but equation must be balanced).

An amphoteric substance can react with both acids and bases.
(a) Reaction with acid (acting as a base):
\(\text{Al}_2\text{O}_3(\text{s}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)

(b) Reaction with base (acting as an acid):
\(\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al}(\text{OH})_4]^-(\text{aq})\)
Alternatively, reaction to form aluminate is accepted: \(\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) \rightarrow 2\text{AlO}_2^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\).

Total: 2.2 marks
- 1.1 marks: Correct balanced ionic equation for the reaction with acid (state symbols not required).
- 1.1 marks: Correct balanced ionic equation for the reaction with base (accept either the tetrahydroxoaluminate complex product or the aluminate \(\text{AlO}_2^-\)/\([\text{Al(OH)}_4\cdot\text{H}_2\text{O}]^-\)/similar variations, properly balanced).

(a) Gaseous \(\text{PCl}_5\) has 5 bonding pairs and no lone pairs on the central phosphorus atom. This results in a trigonal bipyramidal shape. The bond angles are \(90^\circ\) (axial-equatorial) and \(120^\circ\) (equatorial-equatorial). The axial-axial bond angle is \(180^\circ\).
(b) In excess water, \(\text{PCl}_5\) undergoes complete hydrolysis to form phosphoric acid and hydrogen chloride:
\(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\).

Total: 2.2 marks
- 1.1 marks: Correct molecular shape (trigonal bipyramidal) and at least two correct angles (90° and 120°, 180° also accepted).
- 1.1 marks: Correct balanced chemical equation for complete hydrolysis (state symbols not required): \(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\). (Reject partial hydrolysis equation forming \(\text{POCl}_3\) as the question specifies excess water).

(a) When \(\text{MgCl}_2\) dissolves, it undergoes very slight hydrolysis, giving a pH of approximately 6.5 (accept 6 to 7). When \(\text{AlCl}_3\) is added to water, it undergoes vigorous hydrolysis to yield an acidic solution with a pH of approximately 2 to 3.
(b) The aluminium ion, \(\text{Al}^{3+}\), has a smaller ionic radius and a higher positive charge compared to the magnesium ion, \(\text{Mg}^{2+}\). This gives the \(\text{Al}^{3+}\) ion a very high charge density. When hydrated, \(\text{Al}^{3+}\) strongly polarizes the electron density in the \(\text{O}-\text{H}\) bonds of the coordinated water molecules in \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\), facilitating the release of hydrogen ions (protons, \(\text{H}^+\)) into the solution, making it highly acidic.

Total: 2.2 marks
- 1.1 marks: Both pH values correct: \(\text{MgCl}_2\) pH 6.0 - 7.0 AND \(\text{AlCl}_3\) pH 2.0 - 3.0.
- 1.1 marks: Clear explanation based on: (1) \(\text{Al}^{3+}\) having a higher charge and smaller size (or higher charge density) than \(\text{Mg}^{2+}\), and (2) this high charge density polarizes/weakens the \(\text{O}-\text{H}\) bond in water molecules, releasing \(\text{H}^+\)/hydroxonium ions.

(a) When white crystalline \(\text{NaCl}\) is added to water, it dissolves readily to form a clear, colorless solution. No vigorous chemical reaction occurs, and the pH of the solution remains neutral (pH = 7).
(b) When \(\text{S}_2\text{Cl}_2\) (a simple molecular covalent compound) reacts with water, it undergoes hydrolysis in which covalent bonds (both \(\text{S}-\text{Cl}\) and possibly \(\text{S}-\text{S}\)) are broken to form new chemical products (such as \(\text{HCl}\), \(\text{S}\), \(\text{SO}_2\)). This is a chemical reaction (hydrolysis). In contrast, dissolving \(\text{NaCl}\) is a physical process where ionic bonds in the giant ionic lattice are broken by hydration, but no covalent bonds are broken, and no hydrolysis occurs.

Total: 2.2 marks
- 1.1 marks: Correct description of \(\text{NaCl}\) dissolving to form a neutral solution (pH = 7) without chemical reaction.
- 1.1 marks: Explaining that covalent bonds in \(\text{S}_2\text{Cl}_2\) are broken through a chemical hydrolysis reaction with water, whereas for \(\text{NaCl}\) only ionic bonds are broken/solvated (or no chemical reaction occurs).

We use a Born-Haber cycle where:
\(\Delta H_f^{\theta}(\text{MgCl}_2) = \Delta H_{at}^{\theta}(\text{Mg}) + IE_1(\text{Mg}) + IE_2(\text{Mg}) + 2 \times \Delta H_{at}^{\theta}(\text{Cl}) + 2 \times EA_1(\text{Cl}) - \Delta H_{L,diss}^{\theta}(\text{MgCl}_2)\)

Substituting the values:
\(-641 = 148 + 738 + 1451 + 2(121) + 2(-349) - \Delta H_{L,diss}^{\theta}\)
\(-641 = 148 + 738 + 1451 + 242 - 698 - \Delta H_{L,diss}^{\theta}\)
\(-641 = 1881 - \Delta H_{L,diss}^{\theta}\)

Rearranging for \(\Delta H_{L,diss}^{\theta}\):
\(\Delta H_{L,diss}^{\theta} = 1881 - (-641) = +2522 \text{ kJ mol}^{-1}\)

M1 (1.33 marks): Correctly setting up the algebraic expression for the Born-Haber cycle or showing correct substitution of values (allowing for one transcription/arithmetic error or missing factor of 2).
M2 (1.00 mark): Correct final value of \(+2522 \text{ kJ mol}^{-1}\) (must have positive sign or unit).

The relationship between enthalpy of solution, lattice dissociation enthalpy, and hydration enthalpies is:
\(\Delta H_{sol}^{\theta}(\text{CaCl}_2) = \Delta H_{L,diss}^{\theta}(\text{CaCl}_2) + \Delta H_{hyd}^{\theta}(\text{Ca}^{2+}) + 2 \times \Delta H_{hyd}^{\theta}(\text{Cl}^{-})\)

Substituting the known values:
\(-83 = 2258 + \Delta H_{hyd}^{\theta}(\text{Ca}^{2+}) + 2(-378)\)
\(-83 = 2258 + \Delta H_{hyd}^{\theta}(\text{Ca}^{2+}) - 756\)
\(-83 = 1502 + \Delta H_{hyd}^{\theta}(\text{Ca}^{2+})\)

Solving for \(\Delta H_{hyd}^{\theta}(\text{Ca}^{2+})\):
\(\Delta H_{hyd}^{\theta}(\text{Ca}^{2+}) = -83 - 1502 = -1585 \text{ kJ mol}^{-1}\)

M1 (1.33 marks): Correctly constructing the energy cycle or algebraic equation, including the factor of 2 for \(\Delta H_{hyd}^{\theta}(\text{Cl}^{-})\).
M2 (1.00 mark): Correct calculation to yield \(-1585 \text{ kJ mol}^{-1}\) (accept without units, but minus sign is essential).

The equation for the formation of butane is:
\(4\text{C}(s) + 5\text{H}_2(g) \rightarrow \text{C}_4\text{H}_{10}(g)\)

Using the Hess cycle based on combustion products:
\(\Delta H_f^{\theta}(\text{C}_4\text{H}_{10}) = \sum \Delta H_c^{\theta}(\text{reactants}) - \sum \Delta H_c^{\theta}(\text{products})\)
\(\Delta H_f^{\theta}(\text{C}_4\text{H}_{10}) = [4 \times \Delta H_c^{\theta}(\text{C}) + 5 \times \Delta H_c^{\theta}(\text{H}_2)] - \Delta H_c^{\theta}(\text{C}_4\text{H}_{10})\)

Substitute values:
\(\Delta H_f^{\theta}(\text{C}_4\text{H}_{10}) = [4(-394) + 5(-286)] - (-2877)\)
\(\Delta H_f^{\theta}(\text{C}_4\text{H}_{10}) = [-1576 - 1430] + 2877\)
\(\Delta H_f^{\theta}(\text{C}_4\text{H}_{10}) = -3006 + 2877 = -129 \text{ kJ mol}^{-1}\)

M1 (1.33 marks): Correct expression or cycle showing the multiplication of Carbon by 4 and Hydrogen by 5, and subtracting the combustion of butane.
M2 (1.00 mark): Correct evaluation to \(-129 \text{ kJ mol}^{-1}\) (with negative sign).

Using the Born-Haber cycle equation:
\(\Delta H_f^{\theta}(\text{NaF}) = \Delta H_{at}^{\theta}(\text{Na}) + IE_1(\text{Na}) + \Delta H_{at}^{\theta}(\text{F}) + EA_1(\text{F}) + \Delta H_{L,form}^{\theta}(\text{NaF})\)

Substitute the known values:
\(-569 = 109 + 496 + 79 + EA_1(\text{F}) + (-918)\)
\(-569 = 684 - 918 + EA_1(\text{F})\)
\(-569 = -234 + EA_1(\text{F})\)

Rearranging for \(EA_1(\text{F})\):
\(EA_1(\text{F}) = -569 - (-234) = -335 \text{ kJ mol}^{-1}\)

M1 (1.33 marks): Correct algebraic setup or intermediate sum shown (e.g. \(684\) or \(-234\)).
M2 (1.00 mark): Correct final answer of \(-335 \text{ kJ mol}^{-1}\) (negative sign required).

1. The calcium ion (\(\text{Ca}^{2+}\)) and oxide ion (\(\text{O}^{2-}\)) have higher charges (\(+2\) and \(-2\) respectively) compared to the potassium (\(\text{K}^{+}\), \(+1\)) and chloride (\(\text{Cl}^{-}\), \(-1\)) ions.
2. The ions in \(\text{CaO}\) (specifically \(\text{O}^{2-}\) and \(\text{Ca}^{2+}\)) also have smaller ionic radii than \(\text{Cl}^{-}\) and \(\text{K}^{+}\).
3. Therefore, the electrostatic attraction between the oppositely charged ions in the lattice of \(\text{CaO}\) is much stronger than in \(\text{KCl}\), which requires significantly more energy to overcome.

M1 (1.33 marks): Mentions higher ionic charges on \(\text{Ca}^{2+}\) and \(\text{O}^{2-}\) (or 2+ and 2- vs 1+ and 1-), OR smaller sizes of the ions (either \(\text{Ca}^{2+}\) vs \(\text{K}^{+}\) or \(\text{O}^{2-}\) vs \(\text{Cl}^{-}\)).
M2 (1.00 mark): Connects this to much stronger electrostatic forces of attraction between ions in \(\text{CaO}\) than in \(\text{KCl}\), leading to a more endothermic lattice dissociation enthalpy.

Enthalpy change can be calculated as:
\(\Delta H = \sum \text{Bond enthalpies of reactants (broken)} - \sum \text{Bond enthalpies of products (formed)}\)

Bonds broken:
- \(1 \times \text{C}=\text{C} = 612 \text{ kJ mol}^{-1}\)
- \(4 \times \text{C}-\text{H} = 4 \times 412 = 1648 \text{ kJ mol}^{-1}\)
- \(1 \times \text{H}-\text{H} = 436 \text{ kJ mol}^{-1}\)
Total bonds broken = \(612 + 1648 + 436 = 2696 \text{ kJ mol}^{-1}\)

Bonds formed:
- \(1 \times \text{C}-\text{C} = 348 \text{ kJ mol}^{-1}\)
- \(6 \times \text{C}-\text{H} = 6 \times 412 = 2472 \text{ kJ mol}^{-1}\)
Total bonds formed = \(348 + 2472 = 2820 \text{ kJ mol}^{-1}\)

\(\Delta H = 2696 - 2820 = -124 \text{ kJ mol}^{-1}\)

M1 (1.33 marks): Correct sum of bond enthalpies for bonds broken (\(2696\)) and bonds formed (\(2820\)), OR correctly simplifying by only considering bonds that change (broken: \(1 \times \text{C}=\text{C}\) and \(1 \times \text{H}-\text{H} = 1048\); formed: \(1 \times \text{C}-\text{C}\) and \(2 \times \text{C}-\text{H} = 1172\)).
M2 (1.00 mark): Correct calculation of final answer as \(-124 \text{ kJ mol}^{-1}\) (with negative sign).

1. **Initial slow rate:** The reaction is between two negatively charged ions: permanganate (\(\text{MnO}_4^-\) ) and oxalate (\(\text{C}_2\text{O}_4^{2-}\)). Because they are both anions, they repel each other strongly. This repulsion results in a very high activation energy, meaning very few collisions have sufficient energy to react at room temperature.
2. **Rapid increase in rate:** As the reaction proceeds, manganese(II) ions (\(\text{Mn}^{2+}\)) are produced. These \(\text{Mn}^{2+}\) ions act as an autocatalyst, lowering the activation energy for the reaction by providing an alternative reaction pathway.
3. **Eventual decrease to zero:** As reactants are consumed, their concentrations decrease, leading to fewer collisions per unit time, and eventually the reaction stops when one of the limiting reactants is entirely used up.

- **1 mark** for stating that both reacting ions (\(\text{MnO}_4^-\) and \(\text{C}_2\text{O}_4^{2-}\)) are negatively charged, resulting in mutual repulsion and high activation energy.
- **1 mark** for identifying that \(\text{Mn}^{2+}\) is produced in the reaction and acts as an autocatalyst.
- **0.75 marks** for stating that the rate eventually decreases to zero because the reactants are consumed / their concentrations decrease.

1. **Calculate the moles of \(\text{MnO}_4^-\) used:**\
\(n(\text{MnO}_4^-) = C \times V = 0.0200\text{ mol dm}^{-3} \times \frac{18.40}{1000}\text{ dm}^3 = 3.68 \times 10^{-4}\text{ mol}\)\
\
2. **Determine the moles of \(\text{C}_2\text{O}_4^{2-}\) in \(25.0\text{ cm}^3\):**\
The redox ratio is \(2\text{ MnO}_4^- : 5\text{ C}_2\text{O}_4^{2-}\).\
\(n(\text{C}_2\text{O}_4^{2-}) = 3.68 \times 10^{-4}\text{ mol} \times \frac{5}{2} = 9.20 \times 10^{-4}\text{ mol}\)\
\
3. **Calculate the concentration of \(\text{Na}_2\text{C}_2\text{O}_4\) in \(\text{mol dm}^{-3}\):**\
\(C(\text{Na}_2\text{C}_2\text{O}_4) = \frac{9.20 \times 10^{-4}\text{ mol}}{0.0250\text{ dm}^3} = 0.0368\text{ mol dm}^{-3}\)\
\
4. **Convert concentration to \(\text{g dm}^{-3}\):**\
\(\text{Concentration in g dm}^{-3} = 0.0368\text{ mol dm}^{-3} \times 134.0\text{ g mol}^{-1} = 4.93\text{ g dm}^{-3}\) (to 3 significant figures).

- **1 mark** for calculating moles of manganate(VII): \(3.68 \times 10^{-4}\text{ mol}\).\
- **1 mark** for calculating concentration of sodium ethanedioate in \(\text{mol dm}^{-3}\): \(0.0368\text{ mol dm}^{-3}\) (using correct 5:2 ratio).\
- **0.75 marks** for converting to mass concentration: \(4.93\text{ g dm}^{-3}\) (accept \(4.93\) to \(4.94\), must be 3 significant figures).

1. **Step 1:** The catalyst \(\text{Mn}^{2+}\) is oxidized by \(\text{MnO}_4^-\). The half-equations are combined to give:\
\(4\text{Mn}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Mn}^{3+} + 4\text{H}_2\text{O}\)\
\
2. **Step 2:** The intermediate \(\text{Mn}^{3+}\) then oxidizes the ethanedioate ions (\(\text{C}_2\text{O}_4^{2-}\)) to carbon dioxide, regenerating the catalyst:\
\(2\text{Mn}^{3+} + \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 2\text{CO}_2\)

- **1 mark** for a balanced ionic equation for Step 1: \(4\text{Mn}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Mn}^{3+} + 4\text{H}_2\text{O}\) (allow correct species unbalanced for 0.5 mark).\
- **1 mark** for a balanced ionic equation for Step 2: \(2\text{Mn}^{3+} + \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 2\text{CO}_2\) (allow correct species unbalanced for 0.5 mark).\
- **0.75 marks** for explaining that \(\text{Mn}^{3+}\) acts as an intermediate that easily reacts with negatively charged ethanedioate due to opposite charges reducing the activation energy.

At room temperature, the rate of the uncatalyzed reaction is extremely slow because both reactant anions repel each other. Warming the mixture to around \(60\text{ }^\circ\text{C}\) provides enough thermal energy to overcome this initial high activation energy, starting the reaction and producing the \(\text{Mn}^{2+}\) catalyst. Once the catalyst is present, the reaction can proceed rapidly even at lower temperatures.\
\
The end-point is self-indicating. The purple manganate(VII) ions are reduced to colorless manganese(II) ions until all ethanedioate has reacted. At the end-point, the addition of one extra drop of manganate(VII) results in a permanent pale pink color in the conical flask.

- **1 mark** for stating that heating is needed to provide the high activation energy / overcome repulsion of negative reactant ions to initiate the reaction.\
- **1 mark** for stating that the reaction is self-indicating / potassium manganate(VII) acts as its own indicator.\
- **0.75 marks** for describing the end-point color change: from colorless to a permanent pale pink.

When three bidentate ethane-1,2-diamine ligands replace six monodentate water ligands, the equation is: [Cu(H2O)6]2+ + 3en -> [Cu(en)3]2+ + 6H2O.
This reaction starts with 4 reactant particles and produces 7 product particles. This increase in the number of particles in solution leads to a significant increase in disorder, resulting in a highly positive entropy change (\Delta S^\ominus > 0). Because the number and strength of coordinate bonds broken and formed are very similar, the enthalpy change (\Delta H^\ominus) is close to zero. Thus, according to \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus, the free energy change is negative, making the product thermodynamically very stable.

M1: State that the reaction results in an increase in the number of particles/species in solution (from 4 to 7). [1 mark]
M2: State that this leads to a significant increase in entropy / positive entropy change (\Delta S > 0). [1 mark]
M3: State that because \Delta H is approximately zero, the Gibbs free energy change (\Delta G) is negative, making the reaction thermodynamically feasible/stable. [1 mark]

Square planar complexes have coordinate bond angles of 90 degrees and 180 degrees. This allows identical ligands to be either adjacent to each other (cis-isomer, 90-degree angle) or opposite each other (trans-isomer, 180-degree angle). Conversely, in a tetrahedral complex, the geometry is fully symmetrical with all bond angles equal to 109.5 degrees. Every position is adjacent and equidistant to every other position, meaning that changing the positions of ligands does not create a non-superimposable isomer.

M1: Explain that in the square planar geometry, different bond angles (90 degrees and 180 degrees) allow cis (adjacent) and trans (opposite) spatial arrangements of ligands. [1 mark]
M2: Explain that in a tetrahedral complex, all bond angles are identical (109.5 degrees) and all ligand positions are equidistant. [1 mark]
M3: Conclude that any alternative arrangement of ligands in the tetrahedral geometry can be rotated to superimpose on the original, meaning no geometric isomers can exist. [1 mark]

The reaction represents the replacement of six monodentate water ligands by one hexadentate EDTA4- ligand:
[Co(H2O)6]2+(aq) + EDTA4-(aq) -> [Co(EDTA)]2-(aq) + 6H2O(l).
Because EDTA4- is hexadentate, it forms six coordinate bonds to the central cobalt(II) ion. Therefore, the cobalt ion is octahedral both before and after the reaction. The coordination number remains 6 (no change). Consequently, the coordinate bond angles remain approximately 90 degrees (octahedral geometry).

M1: Balanced equation with correct formulas and charges: [Co(H2O)6]2+ + EDTA4- -> [Co(EDTA)]2- + 6H2O. [1 mark]
M2: State that the coordination number remains 6 (or change is 0). [1 mark]
M3: State that the coordinate bond angle remains approximately 90 degrees (accept octahedral angle). [1 mark]

To calculate the energy difference, we use the Planck-Einstein relation:

\(\Delta E = \frac{hc}{\lambda}\)

First, convert the wavelength from nanometers to meters:
\(\lambda = 540 \text{ nm} = 540 \times 10^{-9} \text{ m} = 5.40 \times 10^{-7} \text{ m}\)

Next, substitute the values into the equation:
\(\Delta E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{5.40 \times 10^{-7} \text{ m}}\)
\(\Delta E = \frac{1.989 \times 10^{-25} \text{ J m}}{5.40 \times 10^{-7} \text{ m}}\)
\(\Delta E = 3.6833... \times 10^{-19} \text{ J}\)

Rounding to 3 significant figures gives:
\(\Delta E = 3.68 \times 10^{-19} \text{ J}\)

• [1 mark] for the correct formula application and conversion of wavelength to meters: \(\Delta E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{540 \times 10^{-9}}\).
• [0.67 marks] for the final correct answer to 3 significant figures: \(3.68 \times 10^{-19} \text{ J}\) (accept standard scientific notation, units must be correct if given).

When excess concentrated hydrochloric acid is added to aqueous copper(II) ions, a ligand substitution reaction takes place. The neutral water ligands are replaced by negative chloride ligands:

\([\text{Cu}(\text{H}_2\text{O)}_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)

The starting hexaaquacopper(II) complex is pale blue, and the resulting tetrachlorocuprate(II) complex is yellow. The color transition often goes through a green phase due to a mixture of the blue reactant and yellow product.

• [1 mark] for the correct formula of the product complex, including the correct charge: \([\text{CuCl}_4]^{2-}\).
• [0.67 marks] for stating the correct color change from pale blue to yellow (accept yellow-green or green).

The transition metal ions in both complexes have different oxidation states (\(\text{Fe}^{2+}\) vs \(\text{Fe}^{3+}\)). The greater positive charge on the \(\text{Fe}^{3+}\) ion polarises and pulls the water ligands closer than in the \(\text{Fe}^{2+}\) ion. This stronger interaction causes a larger splitting of the d-orbitals (a larger d-d energy gap, \(\Delta E\)). Because \(\Delta E\) is different, the ions absorb different frequencies/wavelengths of visible light, causing different complementary colors to be transmitted and observed.

• [1 mark] for identifying that the iron ions have different oxidation states (or different charges/charge densities).
• [0.67 marks] for linking the different charges to a different size of the d-orbital splitting (\(\Delta E\)) and thus different frequencies/wavelengths of light absorbed (or different complementary colors transmitted).

In the absence of oxygen (which would otherwise oxidize cobalt(II) to cobalt(III)), concentrated ammonia replaces all six water ligands in the hexaaquacobalt(II) ion to form the hexaamminecobalt(II) ion:

\([\text{Co}(\text{H}_2\text{O)}_6]^{2+} + 6\text{NH}_3 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 6\text{H}_2\text{O}\)

Since the coordination number of the cobalt ion remains 6, the shape of the complex is octahedral.

• [1 mark] for the correct balanced equation with all formulas and charges correct.
• [0.67 marks] for stating the shape is octahedral.

1. In the hexaaquatitanium(III) complex, the surrounding water ligands create an electric field that splits the five degenerate 3d orbitals into two distinct energy levels separated by an energy gap, \(\Delta E\).
2. Titanium(III) has a \(3\text{d}^1\) configuration, meaning its single d-electron occupies one of the lower energy d-orbitals.
3. When white light passes through, the electron absorbs a photon of light equivalent in energy to \(\Delta E\) (which corresponds to green light), and is promoted (excited) to a higher energy d-orbital (a d-d transition).
4. The remaining unabsorbed wavelengths (purple) are transmitted.

• [1 mark] for stating that the presence of ligands causes the d-orbitals to split into two different energy levels (or ground and excited states).
• [0.67 marks] for explaining that a d-electron is promoted/excited from the lower energy level to the higher energy level by absorbing a photon of energy equal to \(\Delta E\).

The reaction replaces 6 monodentate water ligands with 3 bidentate ethane-1,2-diamine ligands. Comparing the stoichiometry of the reactants and products:
- Reactants: 4 particles (1 hydrated metal complex and 3 ligand molecules)
- Products: 7 particles (1 chelated metal complex and 6 water molecules)

This represents a significant increase in the number of free particles in solution, resulting in a large positive entropy change (\(\Delta S^\theta > 0\)) due to increased disorder. Since the strength of the Ni-N coordination bonds is similar to the Ni-O bonds broken, the enthalpy change (\(\Delta H^\theta\)) is close to zero. Consequently, the Gibbs free energy change (\(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)) becomes highly negative, making the reaction thermodynamically very favorable (the chelate effect).

• [1 mark] for explaining that there is an increase in the number of particles (from 4 on the left to 7 on the right), which leads to a positive entropy change (\(\Delta S^\theta > 0\) / more disorder).
• [0.67 marks] for linking this positive \(\Delta S^\theta\) (with small/negligible \(\Delta H^\theta\)) to a negative Gibbs free energy change (\(\Delta G^\theta < 0\)), making the process highly favorable.

First, identify the anode and cathode by comparing standard electrode potentials. The silver electrode has the more positive potential (\(+0.80\text{ V} > +0.77\text{ V}\)), so it undergoes reduction (cathode) and is placed on the right. The iron half-cell undergoes oxidation (anode) and is placed on the left. Since both species in the iron half-cell are in solution, an inert platinum electrode is used. Left-hand side (oxidation): \(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq})\). Right-hand side (reduction): \(\text{Ag}^{+}(\text{aq}) \mid \text{Ag}(\text{s})\). Combining these with a double vertical line (salt bridge) gives: \(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Ag}^{+}(\text{aq}) \mid \text{Ag}(\text{s})\).

[0.8 marks] For writing the correct anode half-cell on the left with Pt electrode and correct phase boundaries: Pt(s) | Fe2+(aq), Fe3+(aq). [0.8 marks] For writing the correct cathode half-cell on the right with correct phase boundaries and salt bridge: || Ag+(aq) | Ag(s).

Using the standard formula: \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). The reduction reaction occurs at the cathode (oxygen electrode): \(E^\theta = +1.23\text{ V}\). The oxidation reaction occurs at the anode (methanol electrode): \(E^\theta = +0.02\text{ V}\). Therefore, \(E^\theta_{\text{cell}} = +1.23\text{ V} - (+0.02\text{ V}) = +1.21\text{ V}\).

[0.8 marks] For identifying the correct anode and cathode potentials or using the correct expression. [0.8 marks] For calculating the final answer of +1.21 V (accept 1.21 V, penalize incorrect units).

In a hydrogen-oxygen fuel cell, hydrogen reacts with oxygen to form water as the only product. The balanced equation is \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\) (or \(\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}\)).

[0.8 marks] For identifying correct reactants (H2 and O2) and product (H2O). [0.8 marks] For correctly balancing the equation.

According to Le Chatelier's principle, decreasing the concentration of \(\text{Cu}^{2+}(\text{aq})\) shifts the position of the equilibrium \(\text{Cu}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Cu}(\text{s})\) to the left. This shifts the system towards the release of electrons, making the electrode potential less positive (it decreases).

[0.8 marks] For stating that the electrode potential decreases / becomes less positive / becomes more negative. [0.8 marks] For explaining that the equilibrium shifts to the left to oppose the decrease in concentration of Cu2+.

By convention, the left-hand side represents oxidation (the anode) and the right-hand side represents reduction (the cathode). The oxidation half-reaction is: \(\text{H}_2(\text{g}) \rightarrow 2\text{H}^{+}(\text{aq}) + 2e^-\). The species undergoing oxidation is \(\text{H}_2\), meaning it acts as the reducing agent.

[0.8 marks] For identifying that oxidation occurs in the left-hand half-cell. [0.8 marks] For identifying H2 (hydrogen gas) as the reducing agent (reject H+ or Pt).

1. Calculate initial moles of \(\text{Ba(OH)}_2\):
\(n(\text{Ba(OH)}_2) = 0.0100\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 1.50 \times 10^{-3}\text{ mol}\)

2. Since each mole of \(\text{Ba(OH)}_2\) dissociates to release two moles of \(\text{OH}^-\):
\(n(\text{OH}^-) = 2 \times 1.50 \times 10^{-3}\text{ mol} = 3.00 \times 10^{-3}\text{ mol}\)

3. Calculate the concentration of hydroxide ions after dilution to \(250\text{ cm}^3\) (\(0.250\text{ dm}^3\)):
\([\text{OH}^-] = \frac{3.00 \times 10^{-3}\text{ mol}}{0.250\text{ dm}^3} = 0.0120\text{ mol dm}^{-3}\)

4. Use \(K_w\) to find hydrogen ion concentration:
\([\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.0120} = 8.33 \times 10^{-13}\text{ mol dm}^{-3}\)

5. Calculate pH:
\(\text{pH} = -\log_{10}(8.33 \times 10^{-13}) = 12.08\)

M1: Calculates correct moles of hydroxide ions: \(3.00 \times 10^{-3}\text{ mol}\) (1 mark)
M2: Calculates correct diluted concentration of hydroxide ions: \(0.0120\text{ mol dm}^{-3}\) (0.71 marks)
M3: Correctly calculates pH as 12.08 (accept 12.1) (0.72 marks)

1. Determine the hydroxide ion concentration in the saturated solution. Barium hydroxide dissociates fully in aqueous solution according to:
\(\text{Ba(OH)}_2(aq) \rightarrow \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq)\)
Therefore, \([\text{OH}^-] = 2 \times [\text{Ba(OH)}_2] = 2 \times 0.115\text{ mol dm}^{-3} = 0.230\text{ mol dm}^{-3}\).

2. Use the ionic product of water, \(K_w\), to calculate \([\text{H}^+]\):
\([\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.230} = 4.35 \times 10^{-14}\text{ mol dm}^{-3}\).

3. Calculate the pH:
\(\text{pH} = -\log_{10}(4.35 \times 10^{-14}) = 13.36\).

M1: Deduces \([\text{OH}^-] = 0.230\text{ mol dm}^{-3}\) (1 mark)
M2: Utilizes \(K_w\) or \(\text{pOH}\) relationship correctly (0.71 marks)
M3: Obtains final pH of 13.36 (0.72 marks)

1. Find the concentration (or moles) of propanoic acid and sodium propanoate after mixing. Because equal volumes of both are mixed, the total volume doubles, halving the concentration of each:
\([\text{CH}_3\text{CH}_2\text{COOH}] = 0.0500\text{ mol dm}^{-3}\)
\([\text{CH}_3\text{CH}_2\text{COO}^-] = 0.0400\text{ mol dm}^{-3}\)

2. Use the weak acid buffer expression:
\([\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{CH}_2\text{COOH}]}{[\text{CH}_3\text{CH}_2\text{COO}^-]}\)
\([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0500}{0.0400} = 1.6875 \times 10^{-5}\text{ mol dm}^{-3}\)

3. Calculate the pH:
\(\text{pH} = -\log_{10}(1.6875 \times 10^{-5}) = 4.77\).

M1: Identifies the correct ratio of acid to salt conjugate base (1 mark)
M2: Calculates \([\text{H}^+] = 1.69 \times 10^{-5}\text{ mol dm}^{-3}\) (0.71 marks)
M3: Obtains pH of 4.77 (accept 4.8) (0.72 marks)

1. Calculate \([\text{H}^+]\) from the given pH:
\([\text{H}^+] = 10^{-12.50} = 3.162 \times 10^{-13}\text{ mol dm}^{-3}\)

2. Find the hydroxide ion concentration using \(K_w\):
\([\text{OH}^-] = \frac{1.00 \times 10^{-14}}{3.162 \times 10^{-13}} = 0.03162\text{ mol dm}^{-3}\)

3. Relate hydroxide concentration to barium hydroxide concentration:
\([\text{Ba(OH)}_2] = \frac{[\text{OH}^-]}{2} = 0.01581\text{ mol dm}^{-3}\)

4. Determine the number of moles required for \(500\text{ cm}^3\) (\(0.500\text{ dm}^3\)):
\(n = 0.01581 \times 0.500 = 7.906 \times 10^{-3}\text{ mol}\)

5. Calculate the mass:
\(\text{mass} = 7.906 \times 10^{-3}\text{ mol} \times 171.3\text{ g mol}^{-1} = 1.354\text{ g} \approx 1.35\text{ g}\).

M1: Correctly determines \([\text{OH}^-] = 0.0316\text{ mol dm}^{-3}\) (1 mark)
M2: Deduces moles of \(\text{Ba(OH)}_2\) required in \(500\text{ cm}^3\) is \(7.91 \times 10^{-3}\text{ mol}\) (0.71 marks)
M3: Calculates final mass of 1.35 g (accept 1.35 to 1.36) (0.72 marks)

1. Calculate the initial moles of hydrogen ions added:
\(n(\text{H}^+) = 0.0400\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 8.00 \times 10^{-3}\text{ mol}\)

2. Calculate the initial moles of hydroxide ions added:
\(n(\text{OH}^-) = 2 \times (0.0600\text{ dm}^3 \times 0.100\text{ mol dm}^{-3}) = 1.20 \times 10^{-2}\text{ mol}\)

3. Determine excess moles of hydroxide ions:
\(n(\text{OH}^-)_{\text{excess}} = 1.20 \times 10^{-2} - 8.00 \times 10^{-3} = 4.00 \times 10^{-3}\text{ mol}\)

4. Calculate the concentration of excess hydroxide ions in the total volume (\(40.0 + 60.0 = 100.0\text{ cm}^3 = 0.100\text{ dm}^3\)):
\([\text{OH}^-] = \frac{4.00 \times 10^{-3}\text{ mol}}{0.100\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\)

5. Calculate hydrogen ion concentration and pH:
\([\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0400} = 2.50 \times 10^{-13}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(2.50 \times 10^{-13}) = 12.60\).

M1: Correctly calculates excess moles of \(\text{OH}^-\text{ as } 4.00 \times 10^{-3}\text{ mol}\) (1 mark)
M2: Divides by total volume to find \([\text{OH}^-] = 0.0400\text{ mol dm}^{-3}\) (0.71 marks)
M3: Correctly calculates pH as 12.60 (0.72 marks)

1. Determine the concentration of methanoic acid (\(\text{HCOOH}\)) after dilution:
\([\text{HCOOH}]_{\text{diluted}} = \frac{25.0}{250} \times 0.500\text{ mol dm}^{-3} = 0.0500\text{ mol dm}^{-3}\)

2. For a weak acid, use the approximation:
\([\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]}\)
\([\text{H}^+] = \sqrt{1.78 \times 10^{-4} \times 0.0500} = \sqrt{8.90 \times 10^{-6}} = 2.983 \times 10^{-3}\text{ mol dm}^{-3}\)

3. Calculate the pH:
\(\text{pH} = -\log_{10}(2.983 \times 10^{-3}) = 2.53\).

M1: Calculates correct diluted acid concentration of \(0.0500\text{ mol dm}^{-3}\) (1 mark)
M2: Applies weak acid approximation to calculate \([\text{H}^+] = 2.98 \times 10^{-3}\text{ mol dm}^{-3}\) (0.71 marks)
M3: Obtains final pH of 2.53 (0.72 marks)

1. Trend: The solubility of Group 2 hydroxides increases down the group from \(\text{Mg(OH)}_2\) to \(\text{Ba(OH)}_2\).

2. Effect: As a result, when descending the group, a saturated solution of the hydroxide will contain a higher concentration of dissolved hydroxide ions (\(\text{OH}^-\)).

3. pH: Because the concentration of hydroxide ions increases, the saturated solutions become more alkaline, resulting in a higher pH.

M1: States that solubility of Group 2 hydroxides increases down the group (1 mark)
M2: Explains that this leads to a higher concentration of hydroxide ions in saturated solutions (0.71 marks)
M3: Concludes that the pH of the saturated solutions increases down the group (0.72 marks)