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The successive ionization energies show a massive increase between the third and fourth values (from 2745 to 11578 kJ mol\(^{-1}\)). This indicates that the fourth electron is removed from an inner quantum shell closer to the nucleus, which experiences a much greater effective nuclear charge. Therefore, the element has three electrons in its outermost shell, placing it in Group 3 (or Group 13 in modern IUPAC notation).

1 Mark: Correct option C. [Accept: identifying that the largest jump is between the third and fourth ionization energies, which corresponds to 3 valence electrons.]

Using the ideal gas equation, \(pV = nRT\):

1. Convert all parameters to SI units:
- Temperature, \(T = 127 + 273 = 400\text{ K}\)
- Pressure, \(p = 100\text{ kPa} = 1.00 \times 10^5\text{ Pa}\)
- Volume, \(V = 166.2\text{ cm}^3 = 1.662 \times 10^{-4}\text{ m}^3\)

2. Calculate the number of moles (\(n\)):
\(n = \frac{pV}{RT} = \frac{1.00 \times 10^5 \times 1.662 \times 10^{-4}}{8.31 \times 400} = \frac{16.62}{3324} = 0.00500\text{ mol}\)

3. Calculate the relative molecular mass (\(M_r\)):
\(M_r = \frac{\text{mass}}{n} = \frac{0.230}{0.00500} = 46.0\text{ g mol}^{-1}\)

1 Mark: Correct option A. [Accept: standard calculations showing correct conversions and yielding \(M_r = 46.0\).]

Let's analyze the shapes of each choice:
- \(\text{CO}_2\): Linear shape with 2 bonding regions and 0 lone pairs on carbon, bond angle is \(180^\circ\).
- \(\text{NH}_2^-\): The nitrogen atom has 5 valence electrons, plus 1 extra electron from the negative charge, giving 6. Two are used in N-H bonds, leaving 4 non-bonding electrons (2 lone pairs). Under VSEPR theory, these 4 electron pairs arrange tetrahedrally. The presence of two lone pairs repels the bonding pairs more than single bonds would, reducing the bond angle from \(109.5^\circ\) to approximately \(104.5^\circ\) in a non-linear (bent) shape.
- \(\text{BF}_3\): Trigonal planar shape with 3 bonding pairs and 0 lone pairs, bond angle is \(120^\circ\).
- \(\text{NH}_4^+\): Tetrahedral shape with 4 bonding pairs and 0 lone pairs, bond angle is \(109.5^\circ\).

1 Mark: Correct option B. [Accept: correct identification of the presence of two bonding pairs and two lone pairs on the central nitrogen atom.]

Thermal stability of Group 2 carbonates increases down the group. The carbonate ion (\(\text{CO}_3^{2-}\)) is a large anion that is easily polarized. Down Group 2, the cation charge remains \(+2\), but the ionic radius increases (from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\)). This decrease in charge density of the cation down the group means that the larger \(\text{Ba}^{2+}\) ion has less polarizing power and distorts the electron cloud of the carbonate ion to a lesser extent. Thus, the carbon-oxygen bonds within the carbonate ion are weakened less, requiring more thermal energy to decompose.

1 Mark: Correct option A. [Reject: answers implying that barium has a higher charge density or that ionic bonding in the lattice itself directly dictates the thermal stability without referring to polarization.]

In this reaction:
- The reactant iodine, \(\text{I}_2\), has an oxidation state of 0.
- In the product sodium iodide, \(\text{NaI}\), the oxidation state of iodine is \(-1\) (reduction: \(0 \rightarrow -1\)).
- In the product sodium iodate, \(\text{NaIO}_3\), the oxidation state of iodine is \(+5\) (oxidation: \(0 \rightarrow +5\)).

Because the same element is simultaneously oxidized and reduced, this is a disproportionation reaction.

1 Mark: Correct option C. [Accept: identifying that disproportionation involves simultaneous oxidation and reduction of the same element, and confirming the correct oxidation states of \(0\), \(-1\), and \(+5\).]

Iron (\(\text{Fe}\)) has an atomic number of 26. Its ground-state atomic configuration is:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2\)

When transition metals form ions, the \(4\text{s}\) electrons are lost first, before any \(3\text{d}\) electrons are removed. Therefore, the \(\text{Fe}^{2+}\) ion loses its two \(4\text{s}\) electrons, leaving the configuration:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6\)

1 Mark: Correct option A. [Reject: configurations where the \(4\text{s}\) electrons are retained over the \(3\text{d}\) electrons, such as option B or C.]

Atom economy is defined as:

\(\text{Atom Economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all products}} \times 100\%\)

- The desired product is ethanol: \(2 \times 46.0 = 92.0\text{ g mol}^{-1}\).
- The total mass of products is: \(2 \times 46.0\text{ (ethanol)} + 2 \times 44.0\text{ (carbon dioxide)} = 180.0\text{ g mol}^{-1}\).

\(\text{Atom Economy} = \frac{92.0}{180.0} \times 100\% = 51.11\% \approx 51.1\%\)

1 Mark: Correct option B. [Accept: method calculating the ratio of the mass of desired ethanol over the total mass of the products.]

When solid potassium bromide reacts with concentrated sulfuric acid:
1. An acid-base reaction occurs to yield hydrogen bromide gas:
\(\text{KBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{KHSO}_4\text{(s)} + \text{HBr(g)}\)
2. Because bromide ions are strong enough reducing agents, some of the hydrogen bromide is oxidized to bromine gas (seen as brown fumes), reducing the sulfuric acid to sulfur dioxide gas:
\(2\text{HBr(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)

Thus, the gaseous products present in the mixture of gases evolved are \(\text{HBr}\), \(\text{Br}_2\), and \(\text{SO}_2\). Note that bromide ions are not strong enough reducing agents to reduce sulfuric acid to elemental sulfur or hydrogen sulfide (which only occurs with iodide ions).

1 Mark: Correct option C. [Accept: lists of all three gases that are produced by both the acid-base and subsequent redox steps.]

1. Analyze the successive ionization energies to identify the group of element \( \text{X} \). There is a very large jump between the 3rd and 4th ionization energies (from 2745 to 11577 kJ/mol). This shows that the fourth electron is removed from an inner quantum shell, indicating that \( \text{X} \) has 3 valence electrons and belongs to Group 13. 2. Since \( \text{X} \) is in Period 3, it is aluminium (\( \text{Al} \)). 3. The neutral aluminium atom has 13 electrons with the electronic configuration: \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^1 \). 4. The \( \text{Al}^{2+} \) ion is formed by removing two electrons from the outer shell, giving the configuration: \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^1 \).

1 Mark: Identify the element as Aluminium (or symbol \( \text{Al} \)) and explain that the large increase is between the 3rd and 4th ionization energies. 1 Mark: Correctly write the electronic configuration of the \( 2+ \) ion as \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^1 \). 0.5 Mark: Correctly relate the number of valence electrons to the element's group.

1. Calculate the amount in moles of hydrogen gas collected: \( n(\text{H}_2) = \frac{75.0}{24000} = 0.003125\text{ mol} \). 2. From the balanced chemical equation, the mole ratio of \( \text{M} : \text{H}_2 \) is 1:1. Therefore: \( n(\text{M}) = 0.003125\text{ mol} \). 3. Calculate the relative atomic mass of the metal \( \text{M} \): \( A_{\text{r}} = \frac{0.125\text{ g}}{0.003125\text{ mol}} = 40.0 \). 4. Comparing this value to the periodic table, the Group 2 metal with an atomic mass close to 40.0 is Calcium (\( \text{Ca} \)).

1 Mark: Calculate the moles of hydrogen gas (0.003125 mol). 1 Mark: Calculate the relative atomic mass of \( \text{M} \) as 40.0. 0.5 Mark: Correctly identify the metal as calcium / \( \text{Ca} \).

1. Determine the shapes: \( \text{PF}_3 \) has 3 bonding pairs and 1 lone pair of electrons around the central phosphorus atom, which results in a trigonal pyramidal shape. \( \text{BF}_3 \) has 3 bonding pairs and 0 lone pairs of electrons around the central boron atom, resulting in a trigonal planar shape. 2. Explain the polarity: \( \text{BF}_3 \) is a symmetrical molecule. The individual polar \( \text{B-F} \) bond dipoles cancel each other out perfectly, resulting in no overall net molecular dipole moment. \( \text{PF}_3 \) is asymmetrical due to the presence of the lone pair of electrons on the phosphorus atom. The polar \( \text{P-F} \) bond dipoles do not cancel each other out, resulting in a net molecular dipole moment (polar).

1 Mark: Correctly state the shape of \( \text{PF}_3 \) (trigonal pyramidal) and \( \text{BF}_3 \) (trigonal planar). 1 Mark: Explain that \( \text{BF}_3 \) is symmetrical and its bond dipoles cancel out. 0.5 Mark: Explain that \( \text{PF}_3 \) is asymmetrical and its bond dipoles do not cancel out.

1. Cation Charge and Size: Both magnesium (\( \text{Mg}^{2+} \)) and barium (\( \text{Ba}^{2+} \)) ions have the same ionic charge (2+). However, the \( \text{Mg}^{2+} \) ion has a much smaller ionic radius than the \( \text{Ba}^{2+} \) ion. 2. Charge Density: Due to its smaller size, the \( \text{Mg}^{2+} \) ion has a much higher charge density than the \( \text{Ba}^{2+} \) ion. 3. Polarization: The highly charge-dense \( \text{Mg}^{2+} \) ion is more effective at polarizing (distorting the electron cloud of) the neighboring carbonate (\( \text{CO}_3^{2-} \)) ion. 4. Bond Weakening: This polarization weakens one of the carbon-oxygen bonds in the carbonate ion, making it easier to break and form carbon dioxide. Therefore, less thermal energy is required for the thermal decomposition of \( \text{MgCO}_3 \) compared to \( \text{BaCO}_3 \).

1 Mark: Identify that the magnesium ion is smaller than the barium ion (while both have a 2+ charge) and therefore has a higher charge density. 1 Mark: Explain that the higher charge density of \( \text{Mg}^{2+} \) results in stronger polarization/distortion of the carbonate ion's electron cloud. 0.5 Mark: Conclude that this polarization weakens the C-O bond, lowering the thermal energy required for decomposition.

1. Oxidation Numbers: In \( \text{Cl}_2 \), the oxidation state of chlorine is 0. In potassium chlorate(V), \( \text{KClO}_3 \), (or the chlorate(V) ion, \( \text{ClO}_3^- \)), the oxidation state of chlorine is +5. In potassium chloride, \( \text{KCl} \), (or the chloride ion, \( \text{Cl}^- \)), the oxidation state of chlorine is -1. 2. Disproportionation: Chlorine is simultaneously oxidized (from 0 to +5) and reduced (from 0 to -1) during this reaction. This meets the definition of disproportionation. 3. Balanced Ionic Equation: Using oxidation states to balance: Oxidation: \( \text{Cl}_2 \rightarrow 2\text{ClO}_3^- \) (10 electrons lost per \( \text{Cl}_2 \)); Reduction: \( \text{Cl}_2 \rightarrow 2\text{Cl}^- \) (2 electrons gained per \( \text{Cl}_2 \)). Multiplying the reduction process by 5 to balance electrons gives: \( 3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O} \).

1 Mark: Identify the oxidation states of Cl in \( \text{Cl}_2 \) (0), \( \text{ClO}_3^- \) (+5), and \( \text{Cl}^- \) (-1), and state that chlorine is both oxidized and reduced. 1 Mark: Write the correctly balanced ionic equation: \( 3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O} \). 0.5 Mark: Ensure overall charge and mass balance are correct.

(i) Determine the mass percentage of oxygen: \( 100\% - 30.4\% = 69.6\% \). Calculate the relative number of moles: For Nitrogen: \( \frac{30.4\text{ g}}{14.0\text{ g mol}^{-1}} = 2.17\text{ mol} \); For Oxygen: \( \frac{69.6\text{ g}}{16.0\text{ g mol}^{-1}} = 4.35\text{ mol} \). Divide by the smallest value (2.17): Nitrogen: 1, Oxygen: 2. The empirical formula is \( \text{NO}_2 \). (ii) Determine the formula mass of the empirical unit: \( M(\text{NO}_2) = 14.0 + (2 \times 16.0) = 46.0\text{ g mol}^{-1} \). Compare this with the given molar mass: \( \text{Ratio} = \frac{92.0}{46.0} = 2 \). Multiply the empirical formula subscripts by 2 to find the molecular formula: \( \text{N}_2\text{O}_4 \).

1 Mark: Calculate correct molar ratio of nitrogen and oxygen (approx. 1 : 2) showing the working. 0.5 Mark: Deduce the empirical formula as \( \text{NO}_2 \). 1 Mark: Show comparison of molar mass to empirical formula mass (92.0 / 46.0 = 2) and deduce the molecular formula as \( \text{N}_2\text{O}_4 \).

1. Intermolecular Forces: Halogens exist as simple diatomic covalent molecules. The intermolecular forces between these molecules are London forces (instantaneous dipole-induced dipole forces). 2. Trend down Group 7: As you go down the group from chlorine to iodine, the size of the halogen atoms increases, and the number of electrons in the molecules increases. 3. Strength of London Forces: The larger electron cloud is more easily polarized, which results in stronger London forces between molecules. 4. Energy requirement: More thermal energy is needed to overcome these stronger London forces, leading to an increase in boiling points.

1 Mark: Identify that the intermolecular forces between halogen molecules are London forces (instantaneous dipole-induced dipole forces). 1 Mark: State that the number of electrons increases down the group, which increases the strength of the London forces. 0.5 Mark: State that more thermal energy is needed to overcome these stronger forces, resulting in higher boiling points.

1. Calculate the moles of ethanol used: \( n(\text{ethanol}) = \frac{11.5\text{ g}}{46.0\text{ g mol}^{-1}} = 0.250\text{ mol} \). 2. Determine the theoretical yield: The stoichiometry of the reaction is 1:1, so 0.250 moles of ethanol theoretically produces 0.250 moles of ethyl ethanoate. \( \text{Theoretical mass} = 0.250\text{ mol} \times 88.0\text{ g mol}^{-1} = 22.0\text{ g} \). 3. Calculate percentage yield: \( \text{Percentage yield} = \frac{14.3\text{ g}}{22.0\text{ g}} \times 100 = 65.0\% \).

1 Mark: Calculate the moles of ethanol (0.250 mol) and state the theoretical yield of ethyl ethanoate is 22.0 g. 1 Mark: Correctly calculate the percentage yield as 65.0%. 0.5 Mark: Show clear working with correct formula for percentage yield.

There is a very large increase (jump) between the third and fourth ionization energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is removed from an inner shell closer to the nucleus, which experiences a much greater effective nuclear charge. Therefore, the element has 3 valence electrons and belongs to Group 3 (or Group 13). Its stable ion is \(X^{3+}\) and its oxide has the formula \(X_2\text{O}_3\).

1 mark: Identify Group 3 / 13 and state that the large jump is between the 3rd and 4th ionization energies. 1 mark: Explain that this indicates 3 valence electrons in the outer shell. 0.5 marks: Deduce the correct formula of the oxide as \(X_2\text{O}_3\) (allow \(\text{Al}_2\text{O}_3\)).

Phosphorus is in Group 5 and has 5 valence electrons. The positive charge means it has lost 1 electron, leaving 4 valence electrons. These 4 electrons form 4 single covalent bonds with 4 chlorine atoms, resulting in 4 bonding pairs and 0 lone pairs. According to electron pair repulsion theory, these 4 bonding pairs repel each other equally and arrange themselves as far apart as possible to minimize repulsion, yielding a tetrahedral shape with a bond angle of \(109.5^\circ\).

1 mark: Stating tetrahedral shape. 1 mark: Correct bond angle of \(109.5^\circ\). 0.5 marks: Stating there are 4 bonding pairs and 0 lone pairs around the central phosphorus atom.

Convert units to SI units: \(V = 380 \times 10^{-6}\text{ m}^3\), \(T = 25 + 273 = 298\text{ K}\), \(p = 101 \times 10^3\text{ Pa}\). Use the ideal gas equation: \(pV = nRT \Rightarrow n = \frac{pV}{RT}\). \(n = \frac{101 \times 10^3 \times 380 \times 10^{-6}}{8.31 \times 298} = 0.0155\text{ mol}\). Molar mass \(M = \frac{\text{mass}}{n} = \frac{0.455\text{ g}}{0.0155\text{ mol}} = 29.4\text{ g mol}^{-1}\).

1 mark: Correct unit conversions (volume to \(3.80 \times 10^{-4}\text{ m}^3\), temperature to \(298\text{ K}\), and pressure to \(101000\text{ Pa}\)). 1 mark: Calculation of moles of gas as \(0.0155\text{ mol}\). 0.5 marks: Correct calculation of molar mass as \(29.4\text{ g mol}^{-1}\) (accept range \(29.3\) to \(29.5\)).

Mass of water lost = \(4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\). Moles of \(\text{CuSO}_4 = \frac{3.19}{159.6} = 0.0200\text{ mol}\). Moles of water lost = \(\frac{1.80}{18.0} = 0.100\text{ mol}\). The mole ratio of water to anhydrous salt is \(\frac{0.100}{0.0200} = 5\). Therefore, \(x = 5\).

1 mark: Calculate mass of water lost (\(1.80\text{ g}\)) and moles of water (\(0.100\text{ mol}\)). 1 mark: Calculate moles of anhydrous copper(II) sulfate (\(0.0200\text{ mol}\)). 0.5 marks: Determine the simple whole-number ratio of 5.

Going down Group 2, the ionic radius of the Group 2 cation increases (from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\)) while the ionic charge remains \(2+\). This results in a decrease in charge density down the group. The larger cations have a lower polarizing power, so they distort the electron cloud of the carbonate ion to a lesser extent. Less distortion means the covalent bonds within the carbonate ion are weakened less, requiring a higher temperature to decompose the carbonate.

1 mark: State that thermal stability increases down the group. 1 mark: Explain that cationic radius increases down the group, resulting in lower charge density and weaker polarizing power. 0.5 marks: Explain that this leads to less distortion / polarization of the carbonate ion, meaning more heat is needed to break the bonds.

A cream precipitate with silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\), forming silver bromide (\(\text{AgBr}\)). To confirm this, aqueous ammonia is added. Silver bromide is insoluble in dilute aqueous ammonia but dissolves upon addition of concentrated aqueous ammonia to form a clear, colorless solution.

1 mark: Identify the halide ion as bromide (\(\text{Br}^-\)) or silver bromide precipitate. 1 mark: State that the cream precipitate does not dissolve in dilute ammonia. 0.5 marks: State that the precipitate dissolves in concentrated ammonia.

In hot alkali, chlorine gas disproportionates into chloride and chlorate(V) ions. The half-equations are: \(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\)- (reduction) and \(\text{Cl}_2 + 12\text{OH}^- \rightarrow 2\text{ClO}_3^- + 6\text{H}_2\text{O} + 10e^-\)- (oxidation). Multiplying the reduction half-equation by 5 and combining gives: 6\(\text{Cl}_2\) + 12\(\text{OH}^-\)- \(\rightarrow\) 10\(\text{Cl}^-\)- + 2\(\text{ClO}_3^-\)- + 6\(\text{H}_2\text{O}\). Simplifying the coefficients gives: 3\(\text{Cl}_2\) + 6\(\text{OH}^-\)- \(\rightarrow\) 5\(\text{Cl}^-\)- + \(\text{ClO}_3^-\)- + 3\(\text{H}_2\text{O}\). Thus, the balanced molecular equation is: 3\(\text{Cl}_2\) + 6\(\text{NaOH}\) \(\rightarrow\) 5\(\text{NaCl}\) + \(\text{NaClO}_3\) + 3\(\text{H}_2\text{O}\). The oxidation state of chlorine in \(\text{NaCl}\) is -1 and in \(\text{NaClO}_3\) is +5.

1 mark: Correctly balanced equation with correct states or formulas (3\(\text{Cl}_2\) + 6\(\text{NaOH}\) \(\rightarrow\) 5\(\text{NaCl}\) + \(\text{NaClO}_3\) + 3\(\text{H}_2\text{O}\)). 1 mark: Identify the oxidation states of chlorine in the products as -1 and +5. 0.5 marks: Correctly define or explain that chlorine is both oxidized and reduced (disproportionation).

The electronic configuration of an iron atom (Fe, Z = 26) is 1\(\text{s}^2\) 2\(\text{s}^2\) 2\(\text{p}^6\) 3\(\text{s}^2\) 3\(\text{p}^6\) 3\(\text{d}^6\) 4\(\text{s}^2\). When forming the \(\text{Fe}^{2+}\) ion, the two 4s electrons are lost first, resulting in 1\(\text{s}^2\) 2\(\text{s}^2\) 2\(\text{p}^6\) 3\(\text{s}^2\) 3\(\text{p}^6\) 3\(\text{d}^6\) (or [Ar] 3\(\text{d}^6\)). For the ionization energy trend: Phosphorus has three unpaired electrons in its 3p subshell (\(3\text{p}_x^1 3\text{p}_y^1 3\text{p}_z^1\)), while sulfur has four 3p electrons, meaning one of the 3p orbitals contains a pair of electrons (\(3\text{p}_x^2 3\text{p}_y^1 3\text{p}_z^1\)). The spin-pair repulsion between these two electrons in the same orbital makes it easier to remove one of them from sulfur, overcoming the increase in nuclear charge.

1 mark: Correct full electronic configuration for \(\text{Fe}^{2+}\) (must be 1\(\text{s}^2\) 2\(\text{s}^2\) 2\(\text{p}^6\) 3\(\text{s}^2\) 3\(\text{p}^6\) 3\(\text{d}^6\)). 1 mark: Explain that sulfur has paired electrons in a 3p orbital while phosphorus has only singly occupied 3p orbitals. 0.5 marks: Identify that spin-pair repulsion in sulfur's orbital makes the electron easier to remove.

1. Identify the ionic sizes and charges: Both \(\text{Mg}^{2+}\) and \(\text{Ba}^{2+}\) have a \(2+\) charge, but the \(\text{Mg}^{2+}\) ion is significantly smaller in ionic radius than the \(\text{Ba}^{2+}\) ion.
2. Connect to charge density and polarising power: Because of its smaller size and identical charge, the \(\text{Mg}^{2+}\) ion has a higher charge density and thus a much greater polarising power than the \(\text{Ba}^{2+}\) ion.
3. Connect to the carbonate ion: The highly polarising \(\text{Mg}^{2+}\) ion distorts the electron cloud of the neighbouring carbonate (\(\text{CO}_3^{2-}\)) ion more effectively. This polarising action weakens the carbon–oxygen (\(\text{C–O}\)) covalent bonds within the carbonate group, decreasing the activation energy and thermal stability, thus allowing decomposition at a lower temperature.

* **M1 (1 mark):** State that the magnesium ion is smaller / has a smaller ionic radius than the barium ion (or that they both have a \(2+\) charge).
* **M2 (1 mark):** State that the magnesium ion has a higher charge density or greater polarising power than the barium ion.
* **M3 (0.5 marks):** State that the magnesium ion polarises / distorts the electron cloud of the carbonate ion, which weakens the carbon–oxygen (\(\text{C–O}\)) bond.

*Accept:* Correct reverse arguments for the barium ion.
*Reject:* References to magnesium/barium atoms instead of ions.

1. Find the mass of water lost during heating:
\(\text{Mass of water} = 4.41\text{ g} - 3.33\text{ g} = 1.08\text{ g}\)

2. Calculate the amount in moles of anhydrous \(\text{CaCl}_2\) and water (using molar masses: \(M_r(\text{CaCl}_2) = 111.1\text{ g mol}^{-1}\) and \(M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1}\)):
\(\text{Moles of CaCl}_2 = \frac{3.33}{111.1} = 0.0300\text{ mol}\)
\(\text{Moles of H}_2\text{O} = \frac{1.08}{18.0} = 0.0600\text{ mol}\)

3. Determine the ratio of moles:
\(\text{Ratio} = \frac{0.0600}{0.0300} = 2\)

Therefore, \(x = 2\).

* **M1 (1 mark):** Correctly calculate the mass of water lost (\(1.08\text{ g}\)) and find the amount of both substances in moles (\(\text{CaCl}_2 = 0.0300\text{ mol}\) and \(\text{H}_2\text{O} = 0.0600\text{ mol}\)).
* **M2 (1 mark):** Show a clear method of dividing the moles of water by the moles of anhydrous salt to find the mole ratio.
* **M3 (0.5 marks):** State the correct final value of \(x = 2\).

*Accept:* Alternative correct calculation methods (e.g., using percentage composition).
*Note:* Allow full Error Carried Forward (ECF) in M2 and M3 for minor mathematical errors made in M1.

1. Write the full electron configuration for phosphorus: \(1s^2 2s^2 2p^6 3s^2 3p^3\).
2. Write the full electron configuration for sulfur: \(1s^2 2s^2 2p^6 3s^2 3p^4\).
3. In phosphorus, the three electrons in the 3p sub-shell occupy different orbitals singly (parallel spins), following Hund's rule. This is a relatively stable, half-filled configuration.
4. In sulfur, the fourth 3p electron must pair up with another electron in one of the 3p orbitals.
5. The repulsion between these two paired electrons in the same orbital in sulfur lowers the energy required to remove one of them, making the first ionization energy of sulfur slightly lower than that of phosphorus.

M1: Correct electron configuration of P: \(1s^2 2s^2 2p^6 3s^2 3p^3\) (allow shorthand \([\text{Ne}] 3s^2 3p^3\)) (1)
M2: Correct electron configuration of S: \(1s^2 2s^2 2p^6 3s^2 3p^4\) (allow shorthand \([\text{Ne}] 3s^2 3p^4\)) (1)
M3: State that phosphorus has a half-filled 3p sub-shell / single electrons in each 3p orbital (1)
M4: State that sulfur has a paired set of electrons in one of the 3p orbitals (1)
M5: Explain that the mutual repulsion between the paired electrons in sulfur makes the electron easier to remove (1)

(a) Calculation of \(x\):
- Mass of hydrated salt = \(25.09\text{ g} - 22.31\text{ g} = 2.78\text{ g}\)
- Mass of anhydrous residue (\(\text{FeSO}_4\)) = \(23.83\text{ g} - 22.31\text{ g} = 1.52\text{ g}\)
- Mass of water lost = \(25.09\text{ g} - 23.83\text{ g} = 1.26\text{ g}\)
- Molar mass of \(\text{FeSO}_4\) = \(55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\)
- Moles of \(\text{FeSO}_4\) = \(1.52 / 151.9 = 0.01001\text{ mol}\)
- Molar mass of \(\text{H}_2\text{O}\) = \(18.0\text{ g mol}^{-1}\)
- Moles of \(\text{H}_2\text{O}\) = \(1.26 / 18.0 = 0.07000\text{ mol}\)
- Ratio of \(\text{H}_2\text{O} : \text{FeSO}_4\) = \(0.07000 / 0.01001 = 6.99\), which rounds to 7.
- Therefore, \(x = 7\).

(b) Without a lid, the highly hygroscopic anhydrous \(\text{FeSO}_4\) absorbs water vapour from the air during the cooling process. This increases the mass of the final residue, making the calculated mass loss of water smaller than it actually was, leading to a lower calculated value of \(x\).

M1: Calculate mass of anhydrous salt (\(1.52\text{ g}\)) and water lost (\(1.26\text{ g}\)) (1)
M2: Calculate moles of anhydrous \(\text{FeSO}_4\) = \(0.0100\text{ mol}\) (using \(M_r = 151.9\text{ g mol}^{-1}\)) (1)
M3: Calculate moles of \(\text{H}_2\text{O}\) = \(0.0700\text{ mol}\) (using \(M_r = 18.0\text{ g mol}^{-1}\)) (1)
M4: Differentiate ratio to find \(x = 7\) (must be an integer) (1)
M5: Suggestion for (b): Anhydrous salt reabsorbs water vapour/moisture from the air during cooling (1)

The thermal stability of Group 2 carbonates increases down the group because:
1. Cation ionic radius increases down the group (from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\)).
2. Cation charge remains constant at \(+2\), which means the charge density decreases down the group.
3. A cation with lower charge density is less polarising and causes less distortion of the electron cloud of the adjacent carbonate (\(\text{CO}_3^{2-}\)) ion.
4. Less polarisation means the C-O covalent bonds within the carbonate ion are weakened to a lesser extent.
5. Consequently, more thermal energy is required to break these bonds and decompose the carbonate into the metal oxide and carbon dioxide.

M1: State that thermal stability increases down the group (1)
M2: State that the ionic radius of the cation increases (down the group) (1)
M3: State that the charge remains the same (\(+2\)), so charge density of the cation decreases (1)
M4: Explain that the larger cation has a weaker polarising effect on / distorts the electron cloud of the carbonate ion less (1)
M5: Conclude that the C-O bond within the carbonate ion is weakened less, requiring more energy to break (1)

1. Under electron pair repulsion theory, both \(\text{NH}_3\) and \(\text{NF}_3\) have 4 pairs of electrons around the central N atom: 3 bonding pairs and 1 lone pair.
2. Electron pairs repel to be as far apart as possible. Since lone pairs repel more than bonding pairs, the bond angles are compressed from the ideal tetrahedral angle of \(109.5^\circ\) to a trigonal pyramidal shape.
3. Fluorine is highly electronegative (the most electronegative element) and is more electronegative than nitrogen. Nitrogen is more electronegative than hydrogen.
4. In \(\text{NF}_3\), the bonding electron pairs in the N-F bonds are pulled away from the central nitrogen atom towards the fluorine atoms.
5. This decreases the electron density of the bonding pairs near the nitrogen atom, reducing the repulsion between the N-F bonding pairs. The lone pair on nitrogen can therefore push the bonding pairs closer together, resulting in a smaller bond angle of \(102.5^\circ\) compared to \(107^\circ\) in \(\text{NH}_3\).

M1: State that both molecules have 3 bonding pairs and 1 lone pair of electrons around the central nitrogen atom (1)
M2: State that lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, leading to a trigonal pyramidal shape (1)
M3: State that fluorine is more electronegative than nitrogen (or hydrogen), pulling the bonding electron pairs closer to fluorine / away from nitrogen (1)
M4: State that in \(\text{NH}_3\), the bonding pairs are closer to the nitrogen atom (1)
M5: Explain that reduced repulsion between the N-F bonding pairs near nitrogen allows the lone pair to compress the bond angle further in \(\text{NF}_3\) (1)

(a)
- In \(\text{K}_2\text{Cr}_2\text{O}_7\), potassium has an oxidation number of \(+1\) and oxygen has \(-2\). Thus, \(2(+1) + 2(\text{Cr}) + 7(-2) = 0 \implies 2\text{Cr} = +12 \implies \text{Cr} = +6\).
- In the simple ion \(\text{Cr}^{3+}\), the oxidation number is \(+3\).

(b)
- Oxidation: Chloride ions are oxidized to chlorine gas: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\).
- Reduction: Dichromate(VI) ions are reduced to chromium(III) ions in acid: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\).

(c)
- To combine the half-equations, the number of electrons must be balanced. Multiply the oxidation half-equation by 3:
\(3 \times (2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-) \implies 6\text{Cl}^- \rightarrow 3\text{Cl}_2 + 6\text{e}^-\)
- Add the two half-equations together, cancelling the 6 electrons on both sides:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Cl}^- \rightarrow 2\text{Cr}^{3+} + 3\text{Cl}_2 + 7\text{H}_2\text{O}\).

M1: State correct oxidation states: Cr is +6 in \(\text{K}_2\text{Cr}_2\text{O}_7\) and +3 in \(\text{Cr}^{3+}\) (1)
M2: Provide correct oxidation half-equation: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\)(or with state symbols) (1)
M3: Provide correct reduction half-equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) (1)
M4: Show scaling of the oxidation half-equation by 3 to balance electrons (6 electrons in total) (1)
M5: Write the fully balanced overall ionic equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Cl}^- \rightarrow 2\text{Cr}^{3+} + 3\text{Cl}_2 + 7\text{H}_2\text{O}\) (1)

Step 1: Calculate heat energy transferred using \( q = mc\Delta T \). \( q = 100.0 \times 4.18 \times 42.0 = 17556 \text{ J} = 17.556 \text{ kJ} \). Step 2: Calculate moles of methanol burned: \( n = \frac{1.60}{32.0} = 0.050 \text{ mol} \). Step 3: Calculate the molar enthalpy of combustion: \( \Delta H_c = -\frac{q}{n} = -\frac{17.556}{0.050} = -351.12 \text{ kJ mol}^{-1} \approx -351 \text{ kJ mol}^{-1} \).

1 mark for correct selection of option A.

During electrophilic addition to 2-methylbut-2-ene (\( \text{(CH}_3)_2\text{C=CHCH}_3 \)), protonation occurs at carbon-3 to form the more stable tertiary carbocation, \( \text{(CH}_3)_2\text{C}^+\text{-CH}_2\text{CH}_3 \). This carbocation is stabilized by the electron-donating inductive effect of three alkyl groups.

1 mark for correct selection of option A.

In nucleophilic substitution, a curly arrow starts from a lone pair on the nucleophile's oxygen (\( \text{OH}^- \)) and points to the partially positive, halogen-bonded carbon. A second curly arrow starts from the C-Br bond and points to the bromine atom, showing heterolytic bond fission.

1 mark for correct selection of option B.

The strong, sharp peak at \( 1715 \text{ cm}^{-1} \) indicates a carbonyl group (\( \text{C=O} \)). The lack of an O-H stretch at \( 3200\text{--}3600 \text{ cm}^{-1} \) rules out alcohols. The ketone propanone (\( \text{C}_3\text{H}_6\text{O} \)) fits these characteristics perfectly.

1 mark for correct selection of option B.

To synthesize an aldehyde from a primary alcohol, a distillation setup is used with a limited amount of the oxidizing agent. Distillation allows the low-boiling aldehyde (propanal) to be vaporized and collected immediately, preventing further oxidation to a carboxylic acid.

1 mark for correct selection of option D.

A catalyst lowers the activation energy (\( E_a \)) by providing an alternative pathway, but it has no effect on the temperature or kinetic energy of the molecules. Therefore, the shape and peak of the Maxwell-Boltzmann distribution curve remain unchanged.

1 mark for correct selection of option B.

The forward reaction is exothermic (\( \Delta H < 0 \)), so lowering the temperature shifts the equilibrium to the right to favor the forward reaction. There are 3 moles of gas on the left and 1 mole on the right; increasing the pressure shifts the equilibrium to the side with fewer moles of gas (the right). Thus, a combination of decreased temperature and increased pressure yields the highest amount of methanol.

1 mark for correct selection of option D.

The percentage atom economy = \( \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100 \). Here, desired product is bromoethane (\( 108.9 \text{ g mol}^{-1} \)). Sum of reactant masses = \( 46.0 + 102.9 + 98.1 = 247.0 \text{ g mol}^{-1} \). Therefore, Atom Economy = \( \frac{108.9}{247.0} \times 100 = 44.089\% \approx 44.1\% \).

1 mark for correct selection of option A.

First, calculate the heat energy transferred to the water using q = m * c * delta T: q = 100.0 g * 4.18 J g^-1 K^-1 * 22.5 K = 9405 J = 9.405 kJ. Next, calculate the number of moles of propan-1-ol burned: moles = mass / molar mass = 0.620 g / 60.1 g mol^-1 = 0.010316 mol. Finally, calculate the enthalpy change per mole: delta H = -q / moles = -9.405 kJ / 0.010316 mol = -911.68 kJ mol^-1. Rounding to 3 significant figures gives -912 kJ mol^-1.

1 mark for calculating correct heat energy transferred (9.405 kJ). 1 mark for calculating correct moles of propan-1-ol (0.0103 mol). 0.6 mark for correct final value with negative sign (-912 kJ mol^-1). Reject positive values.

The conversion of a halogenoalkane (2-bromopropane) to an alcohol (propan-2-ol) is carried out using aqueous sodium hydroxide (NaOH) or potassium hydroxide (KOH) as the reagent. The mixture is heated under reflux to prevent the loss of volatile compounds. This reaction proceeds via a nucleophilic substitution mechanism, where the hydroxide ion acts as the nucleophile to replace the bromide ion.

1 mark for reagent (aqueous sodium hydroxide / KOH, reject solid or ethanolic). 1 mark for mechanism type (nucleophilic substitution). 0.6 mark for conditions (heating under reflux).

Adding a catalyst provides an alternative reaction pathway with a lower activation energy (Ea). On a Maxwell-Boltzmann distribution curve, the activation energy threshold is shifted to the left (a lower energy value). This means that a significantly larger proportion of molecules now possess kinetic energy greater than or equal to this new activation energy, resulting in a higher frequency of successful collisions and thus an increased reaction rate.

1 mark for stating that a catalyst provides an alternative pathway with a lower activation energy. 1 mark for explaining that this shifts the activation energy line to the left on the Maxwell-Boltzmann distribution. 0.6 mark for concluding that a greater fraction of molecules have energy greater than or equal to the activation energy, increasing successful collision frequency.

Propan-1-ol contains an alcohol functional group, which exhibits a broad, characteristic O-H stretching absorption band in the range of 3200-3750 cm^-1, but lacks a carbonyl (C=O) band. Propanone, a ketone, does not have an O-H group but exhibits a strong, sharp carbonyl (C=O) stretching absorption band in the range of 1675-1750 cm^-1. Comparing these distinctive peaks allows for clear identification of each substance.

1 mark for identifying the O-H absorption band in propan-1-ol with correct wavenumber range (3200-3750 cm^-1) and broad shape. 1 mark for identifying the C=O absorption band in propanone with correct wavenumber range (1675-1750 cm^-1) and sharp shape. 0.6 mark for stating that the absence of these respective peaks in the other molecule allows them to be distinguished.

In the first step of the electrophilic addition, a curly arrow starts from the electron-rich C=C double bond of propene and points to the partially positive hydrogen atom of HBr. Concurrently, a second curly arrow starts from the H-Br covalent bond and points to the bromine atom. 2-bromopropane is the major product because the addition of H+ to the terminal carbon forms a secondary carbocation (CH3-CH+-CH3). This secondary carbocation is more stable than the alternative primary carbocation (CH3-CH2-CH2+) due to the electron-donating inductive effect of two methyl groups, lowering the activation energy for this pathway.

1 mark for describing the correct direction of both curly arrows in the first step. 1 mark for identifying that the major pathway involves a more stable secondary carbocation intermediate compared to a primary carbocation. 0.6 mark for explaining stability in terms of the inductive effect of the two alkyl groups.

According to Le Chatelier's principle, if the temperature is increased, the equilibrium will shift in the endothermic direction (to the left) to absorb the added heat, thereby decreasing the yield of methanol. If the pressure is increased, the system shifts to favor the side with fewer gas molecules (the right side, where there is 1 mole of gas compared to 3 moles on the left) to reduce pressure, thereby increasing the yield of methanol.

1 mark for predicting and explaining the effect of increasing temperature (shifts left/endothermic direction, decreases yield). 1 mark for predicting and explaining the effect of increasing pressure (shifts right/fewer gas moles direction, increases yield). 0.6 mark for citing correct numbers of gaseous moles (3 on left vs 1 on right) and exothermic nature of the forward reaction.

First, find the moles of MgCO3: moles = 2.10 g / 84.3 g mol^-1 = 0.024911 mol. Since the stoichiometry of MgCO3 to CO2 is 1:1, moles of CO2 = 0.024911 mol. Convert pressure to Pa: P = 101,000 Pa. Use the ideal gas equation V = nRT / P: V = (0.024911 mol * 8.31 J mol^-1 K^-1 * 353 K) / 101,000 Pa = 0.0007235 m^3. Convert m^3 to dm^3 by multiplying by 1000: V = 0.7235 dm^3, which rounds to 0.724 dm^3.

1 mark for calculating the correct moles of CO2 (0.0249 mol). 1 mark for correct use of the ideal gas equation with consistent units (P in Pa and T in K). 0.6 mark for calculating the final volume as 0.724 dm^3 (accept 0.720 to 0.730 depending on intermediate rounding).

1. Placing a rubber bung in the top of the condenser seals the apparatus, creating a closed system. Heating a closed system causes gas pressure to build up rapidly, which can lead to the glass apparatus shattering or exploding violently. 2. Connecting the water cooling hoses so water enters at the top means the condenser jacket will not fill completely because of gravity. This results in highly inefficient cooling, meaning volatile organic vapours will not condense properly and will escape into the laboratory, representing a fire and inhalation hazard.

1 mark for identifying that sealing the top creates a closed system and leads to dangerous pressure build-up/explosion. 1 mark for identifying that water entering from the top results in an incomplete fill of the cooling jacket. 0.6 mark for linking the poor water flow to inefficient cooling and escape of volatile reaction components.

During the electrophilic addition of HBr to propene, the electrophile H+ can add to either carbon-1 or carbon-2. Adding to carbon-1 generates a secondary carbocation intermediate, CH3CH(+)CH3. Adding to carbon-2 generates a primary carbocation intermediate, CH3CH2CH2(+). Secondary carbocations are more stable than primary carbocations because they have two electron-releasing alkyl (methyl) groups that stabilize the positive charge through the inductive effect, lowering the activation energy for this pathway and resulting in 2-bromopropane as the major product.

M1: Identify that the major product goes through a secondary carbocation intermediate while the minor goes through a primary carbocation (1 mark). M2: State that the secondary carbocation is more stable than the primary carbocation (1 mark). M3: Explain the stability in terms of the electron-releasing inductive effect of alkyl groups (0.6 marks).

Halogenation of secondary alcohols like butan-2-ol to form chloroalkanes is best achieved using phosphorus pentachloride (PCl5) at room temperature. The reaction proceeds as: CH3CH(OH)CH2CH3 + PCl5 -> CH3CHClCH2CH3 + POCl3 + HCl. Alternatively, concentrated hydrochloric acid (HCl) with anhydrous zinc chloride catalyst can be accepted.

M1: State a suitable reagent: phosphorus pentachloride / PCl5 (1 mark). M2: Give correct organic product 2-chlorobutane (1 mark). M3: Give correct inorganic side products POCl3 and HCl and write a balanced equation (0.6 marks). Accept HCl/ZnCl2 reagent with equation producing water.

First, calculate energy required to break bonds in reactants: 3 x C-H = 3 x 413 = 1239 kJ mol^-1; 1 x C-O = 358 kJ mol^-1; 1 x O-H = 463 kJ mol^-1; 1.5 x O=O = 1.5 x 498 = 747 kJ mol^-1. Total energy broken = 1239 + 358 + 463 + 747 = 2807 kJ mol^-1. Second, calculate energy released from making bonds in products: 2 x C=O = 2 x 805 = 1610 kJ mol^-1; 4 x O-H (in 2 H2O) = 4 x 463 = 1852 kJ mol^-1. Total energy made = 1610 + 1852 = 3462 kJ mol^-1. Enthalpy change = Total energy broken - Total energy made = 2807 - 3462 = -655 kJ mol^-1.

M1: Sum of bond enthalpies of reactants broken calculated as +2807 kJ mol^-1 (1 mark). M2: Sum of bond enthalpies of products formed calculated as -3462 kJ mol^-1 (1 mark). M3: Enthalpy change calculated as -655 kJ mol^-1 with correct negative sign (0.6 marks).

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy (Ea) than the uncatalyzed route. As a result of this lower activation barrier, a larger proportion of reactant molecules possess energy equal to or greater than the activation energy (E >= Ea) at the same temperature. This leads to a significantly higher frequency of successful collisions per unit time, thereby increasing the reaction rate.

M1: State that the catalyst provides an alternative pathway with a lower activation energy (1 mark). M2: Explain that a greater proportion/fraction of molecules now have energy equal to or greater than the activation energy (1 mark). M3: Link to an increased frequency of successful collisions (0.6 marks).

The peak at m/z = 31 in primary alcohols like propan-1-ol corresponds to the stable oxonium-like ion [CH2OH]+, which is formed via alpha-cleavage of the C-C bond adjacent to the oxygen atom. The equation for the fragmentation of the molecular ion [CH3CH2CH2OH]+ is: [CH3CH2CH2OH]+ -> .CH2CH3 + [CH2OH]+, producing an ethyl radical and the charged fragment.

M1: Correctly identify the species as [CH2OH]+ or CH2OH+ (1 mark, reject if charge is missing or negative). M2: Correctly identify the neutral fragment as the ethyl radical, .CH2CH3 or C2H5 (1 mark). M3: Write a balanced equation showing the molecular ion fragmenting into the species and radical (0.6 marks).

Use the ideal gas equation: PV = nRT to find n = PV / RT. First convert all terms to SI units: P = 101 kPa = 1.01 x 10^5 Pa; V = 278 cm³ = 2.78 x 10^-4 m³; T = 100 °C = 373 K. n = (1.01 x 10^5 * 2.78 x 10^-4) / (8.31 * 373) = 28.078 / 3099.63 = 0.0090585 mol. Molar mass (M) = mass / moles = 0.850 g / 0.0090585 mol = 93.837 g mol^-1. To 3 significant figures, this is 93.8 g mol^-1.

M1: Convert temperature to 373 K and volume to 2.78 x 10^-4 m³ (1 mark). M2: Correct calculation of moles n = 0.00906 mol (1 mark). M3: Calculate molar mass as 93.8 g mol^-1 (0.6 marks, allow 94 or 93.8 to 94.0).

According to Le Chatelier's principle, if temperature is increased, the system will shift to oppose this change by favoring the endothermic direction. Because the forward reaction is exothermic (dH is negative), the reverse reaction is endothermic. Therefore, the equilibrium shifts to the left, decreasing the yield of ammonia. Since the position of equilibrium shifts to favor the reactants at a higher temperature, the ratio of product concentration to reactant concentrations decreases, meaning the value of Kc decreases.

M1: State that the yield of ammonia decreases (1 mark). M2: Explain that the forward reaction is exothermic, so the equilibrium shifts to the left in the endothermic direction to absorb added heat (1 mark). M3: State that the value of Kc decreases (0.6 marks).

Heating under reflux allows prolonged heating of the reaction mixture without losing volatile organic reactants and products (such as ethanol and the intermediate ethanal). Vapours rise, condense in the vertical condenser, and drip back down into the reaction flask to undergo complete oxidation to ethanoic acid. If distillation were used, the volatile intermediate ethanal would distil off first and prevent the formation of ethanoic acid. The acidified potassium dichromate(VI) acts as the oxidizing agent and is reduced from orange Cr2O7^2- to green Cr^3+.

M1: Explain that reflux prevents the escape of volatile components/ethanol/ethanal, ensuring complete reaction/oxidation (1 mark). M2: Explain that distillation would remove the intermediate ethanal before it can be fully oxidized (1 mark). M3: State the correct colour change from orange to green (0.6 marks).

1. **Mechanism Identification**: The reaction is a nucleophilic substitution. Under aqueous conditions, the hydroxide ion behaves as a nucleophile.
2. **Mechanism Details**:
- The polar carbon-bromine bond has a dipole: \(\text{C}^{\delta+}-\text{Br}^{\delta-}\).
- The nucleophilic hydroxide ion (\(\text{OH}^-\)) has at least one lone pair of electrons on the oxygen atom.
- A curly arrow is drawn starting from a lone pair on the oxygen of the \(\text{OH}^-\)\ ion to the carbon atom of the C-Br bond, representing the formation of a coordinate covalent bond.
- A second curly arrow is drawn starting from the C-Br bond to the bromine atom, representing the heterolytic cleavage of the bond.
- The products of this step are propan-2-ol and a bromide ion (\(\text{Br}^-\)).

1 mark: Correctly identifying the mechanism as nucleophilic substitution and the role of the hydroxide ion as a nucleophile (or lone pair donor).
1 mark: Explaining the initiation of the mechanism, including the correct polar dipole representation (\(\text{C}^{\delta+}-\text{Br}^{\delta-}\)) and a curly arrow starting from a lone pair on the oxygen of the hydroxide ion to the carbon atom.
0.6 marks: Explaining the bond breaking step with a curly arrow starting from the C-Br bond to the bromine atom, and correctly identifying the products as propan-2-ol and a bromide ion.

1. **Calculate the heat energy transferred (q)** using \(q = m c \Delta T\):
\(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 36.0\text{ K} = 15048\text{ J} = 15.048\text{ kJ}\)

2. **Calculate the number of moles of methanol burned**:
\(n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.025\text{ mol}\)

3. **Calculate the enthalpy change of combustion (\(\Delta_c H^\ominus\))**:
\(\Delta_c H^\ominus = -\frac{q}{n} = -\frac{15.048\text{ kJ}}{0.025\text{ mol}} = -601.92\text{ kJ mol}^{-1}\)

Rounding to 3 significant figures gives \(-602\text{ kJ mol}^{-1}\).

1 mark: Correct calculation of heat energy transferred, \(q = 15.05\text{ kJ}\) (or \(15048\text{ J}\)).
1 mark: Correct calculation of moles of methanol, \(n = 0.025\text{ mol}\).
0.6 marks: Correct final enthalpy change calculation (\(-602\text{ kJ mol}^{-1}\)), including the negative sign and 3 significant figures. (Accept \(-601.9\) to \(-602\)).

1. **Purpose of Reflux**: Heating volatile organic compounds in an open system leads to evaporation and loss of materials, which reduces yield and presents safety hazards (many organic vapours are flammable/toxic). Reflux allows the mixture to be boiled continuously; vapours rise, condense on the cool surfaces of the vertical condenser, and drip back down into the reaction flask.
2. **Apparatus Setup**: The condenser must be positioned vertically above the round-bottom or pear-shaped flask. The cooling water must enter the condenser jacket from the lower inlet and flow out of the upper outlet to ensure the jacket is fully filled with cold water. The system must not be sealed (must remain open at the top of the condenser) to prevent pressure build-up and potential explosion.

1 mark: Stating that reflux prevents the loss of volatile components (solvent/reactants/products) during prolonged heating.
1 mark: Stating that the condenser must be positioned vertically above the reaction flask.
0.6 marks: Specifying that water must flow in at the bottom and out at the top of the condenser, and that the top of the apparatus must remain open.

To verify that propan-1-ol has been completely oxidised to propanoic acid:
1. Propan-1-ol contains an alcohol functional group with a broad O-H bond absorption located at \(3200-3600\text{ cm}^{-1}\). If conversion is complete, this characteristic alcohol O-H peak will be completely absent from the spectrum of the product.
2. Propanoic acid contains a carboxylic acid functional group, which has two diagnostic peaks:
- A sharp, intense peak due to the carbonyl group (\(\text{C}=\text{O}\)) in the region \(1680-1750\text{ cm}^{-1}\).
- A very broad absorption band due to the acidic O-H bond in the region \(2500-3300\text{ cm}^{-1}\), which typically overlaps with the C-H stretching absorptions.
Therefore, the spectrum of the successfully converted product will show both the \(\text{C}=\text{O}\) and acidic \(\text{O}-\text{H}\) peaks, and have no trace of the alcohol \(\text{O}-\text{H}\) peak.

1 mark: Identifying the absence/disappearance of the alcohol O-H absorption in the range \(3200-3600\text{ cm}^{-1}\).
1 mark: Identifying the appearance of the carbonyl (C=O) absorption peak in the range \(1680-1750\text{ cm}^{-1}\).
0.6 marks: Identifying the appearance of the very broad carboxylic acid O-H absorption peak in the range \(2500-3300\text{ cm}^{-1}\).

A logical synthetic pathway to convert propene (an alkene) into propan-2-ol (a secondary alcohol) is:
- **Step 1: Preparation of intermediate halogenoalkane**
- Reagent: Hydrogen bromide (\(\text{HBr}\)) or hydrogen chloride (\(\text{HCl}\)).
- Conditions: Room temperature.
- Reaction Type: Electrophilic addition.
- Intermediate Product: 2-bromopropane (or 2-chloropropane) as the major product (following Markovnikov's rule).
- **Step 2: Conversion of halogenoalkane to alcohol**
- Reagent: Aqueous potassium hydroxide (\(\text{KOH}\)) or aqueous sodium hydroxide (\(\text{NaOH}\)).
- Conditions: Heat under reflux.
- Reaction Type: Nucleophilic substitution.
- Final Product: Propan-2-ol.

1 mark: Step 1: Correctly identifying the reagent (\(\text{HBr}\) or \(\text{HCl}\)) at room temperature and the reaction type as electrophilic addition.
1 mark: Step 2: Correctly identifying the reagent (aqueous \(\text{KOH}\) or aqueous \(\text{NaOH}\)) heated under reflux and the reaction type as nucleophilic substitution.
0.6 marks: Identifying 2-bromopropane (or 2-chloropropane) as the intermediate product.

(a) Step 1: Calculate temperature change: \(\Delta T = 48.5 - 20.2 = 28.3\text{ K}\). Step 2: Calculate heat energy absorbed by water: \(q = m c \Delta T = 100.0 \times 4.18 \times 28.3 = 11829.4\text{ J} = 11.8294\text{ kJ}\). Step 3: Calculate mass of propan-1-ol burned: \(124.55 - 124.12 = 0.43\text{ g}\). Step 4: Calculate moles of propan-1-ol burned: \(n = 0.43 / 60.0 = 0.007167\text{ mol}\). Step 5: Calculate enthalpy change: \(\Delta_c H = -q / n = -11.8294 / 0.007167 = -1650.6\text{ kJ mol}^{-1}\), which rounds to \(-1650\text{ kJ mol}^{-1}\) (3 s.f.). (b) Reason 1: Incomplete combustion of propan-1-ol (which releases less heat energy). Reason 2: Evaporation of propan-1-ol from the wick after weighing / Heat capacity of the copper calorimeter itself was neglected / Heat lost to the air and copper can rather than solely the water.

Part (a): [1 M] for calculating temperature change and heat energy transferred (\(q = 11.83\text{ kJ}\)). [1 M] for calculating moles of fuel burned (\(0.00717\text{ mol}\)). [1 M] for dividing heat energy by moles to get final value, including the negative sign. [1 M] for correct 3 s.f. value of \(-1650\text{ kJ mol}^{-1}\) (allow range \(-1650\) to \(-1660\) depending on intermediate rounding). Part (b): [1 M] for stating 'incomplete combustion' of the fuel. [1 M] for stating either 'evaporation of fuel from the wick', 'heat capacity of the copper calorimeter neglected' or 'non-standard conditions'.

(a) Step-by-step procedure: 1. Transfer the crude mixture to a separating funnel, add sodium hydrogencarbonate solution to neutralise any acid, shake carefully and release the pressure by opening the tap. 2. Allow the mixture to separate into two layers, then run off and discard the lower aqueous layer, retaining the upper organic layer of cyclohexene. 3. Transfer the cyclohexene to a conical flask, add anhydrous calcium chloride, and swirl the mixture until the liquid changes from cloudy to completely clear. 4. Decant or filter the liquid into a distillation flask, then perform a final fractional distillation, collecting the distillate boiling at the boiling point of cyclohexene (approximately \(83 ^\circ\text{C}\)). (b) Role of anhydrous calcium chloride: It acts as a drying agent to remove trace water from the organic product. Reason for choice: Concentrated sulfuric acid cannot be used because it is an dehydrating agent/strong acid that would react with the cyclohexene alkene double bond, causing polymerization or addition reactions.

Part (a): [1 M] for using a separating funnel and washing with sodium hydrogencarbonate (or sodium chloride) solution, including releasing pressure. [1 M] for separating the layers and discarding the aqueous layer. [1 M] for adding anhydrous calcium chloride (or sodium sulfate) to dry the organic layer until clear. [1 M] for performing a second distillation to collect the fraction at the boiling point of cyclohexene. Part (b): [1 M] for identifying its role as a drying agent / removing water. [1 M] for explaining that concentrated sulfuric acid would react with/hydrate/polymerize the cyclohexene alkene double bond.

Step 1: Calculate the amount, in moles, of \(HCl\) used in the titration: \(n(HCl) = c \times V = 0.100\text{ mol dm}^{-3} \times 0.02500\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\). Step 2: Use the reaction stoichiometry to find the moles of \(Na_2CO_3\) in the \(25.0\text{ cm}^3\) sample: \(n(Na_2CO_3) = \frac{1}{2} \times n(HCl) = 1.25 \times 10^{-3}\text{ mol}\). Step 3: Calculate the moles of \(Na_2CO_3\) in the original \(250.0\text{ cm}^3\) volumetric flask: \(n_{\text{total}} = 1.25 \times 10^{-3}\text{ mol} \times 10 = 0.0125\text{ mol}\). Step 4: Calculate the molar mass of \(Na_2CO_3 \cdot xH_2O\): \(M = \frac{\text{mass}}{\text{moles}} = \frac{3.58\text{ g}}{0.0125\text{ mol}} = 286.4\text{ g mol}^{-1}\). Step 5: Calculate the molar mass of anhydrous sodium carbonate, \(Na_2CO_3\): \(M(Na_2CO_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\). Step 6: Find the value of \(x\): \(M(xH_2O) = 286.4 - 106.0 = 180.4\text{ g mol}^{-1}\). Since \(M(H_2O) = 18.0\text{ g mol}^{-1}\), \(x = \frac{180.4}{18.0} = 10.02\), which rounds to the nearest whole number of 10.

[1 M] for calculating moles of \(HCl\) as \(2.50 \times 10^{-3}\text{ mol}\). [1 M] for using the 1:2 ratio to get moles of \(Na_2CO_3\) in sample as \(1.25 \times 10^{-3}\text{ mol}\). [1 M] for multiplying by 10 to find moles in \(250.0\text{ cm}^3\) as \(0.0125\text{ mol}\). [1 M] for calculating the relative formula mass of the hydrated salt as \(286.4\text{ g mol}^{-1}\) (allow ECF from incorrect moles). [1 M] for calculating the relative formula mass of anhydrous \(Na_2CO_3\) as \(106.0\text{ g mol}^{-1}\). [1 M] for calculating \(x = 10\) (must be a whole number, allow 10 from correct calculation).