2024 Edexcel IAS-Level Chemistry (XCH11) 模擬試題連答案詳解

There is a very large increase between the third and fourth ionization energies (from \(2745\) to \(11577\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from a shell closer to the nucleus, meaning the element has three valence electrons and forms a stable \(X^{3+}\) ion. The oxide of a Group 3 element is \(X_2O_3\).

1 mark for correct option (B).

1. Find the molar mass of the hydrocarbon: \(M = \frac{5.60\text{ g}}{0.100\text{ mol}} = 56.0\text{ g mol}^{-1}\). 2. Calculate the mass of carbon in one mole: \(0.857 \times 56.0 = 48.0\text{ g}\), which corresponds to \(\frac{48.0}{12.0} = 4\) carbon atoms. 3. Calculate the mass of hydrogen in one mole: \(56.0 - 48.0 = 8.0\text{ g}\), which corresponds to \(\frac{8.0}{1.0} = 8\) hydrogen atoms. 4. The molecular formula is therefore \(C_4H_8\).

1 mark for correct option (C).

The amide ion, \(NH_2^-\), has 2 bonding pairs and 2 lone pairs on the nitrogen atom (total of 8 valence electrons). This results in a tetrahedral arrangement of electron pairs, but a bent shape. The repulsion from the two lone pairs reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\).

1 mark for correct option (B).

The structural isomers for \(C_5H_{12}\) are: 1) pentane (straight chain), 2) 2-methylbutane, and 3) 2,2-dimethylpropane. There are exactly 3 structural isomers.

1 mark for correct option (B).

The electrophilic addition of HBr to but-1-ene can form either a primary carbocation (\(CH_3CH_2CH_2CH_2^+\)) or a secondary carbocation (\(CH_3CH_2CH^+CH_3\)). The secondary carbocation is more stable due to the electron-donating inductive effect of two alkyl groups compared to one in the primary carbocation. Thus, the major product is formed via the more stable secondary carbocation.

1 mark for correct option (B).

Using the ideal gas equation: \(PV = nRT\). The molar mass of \(N_2\) is \(28.0\text{ g mol}^{-1}\). The number of moles is \(n = \frac{1.40}{28.0} = 0.0500\text{ mol}\). Temperature \(T = 20.0 + 273.15 = 293.15\text{ K}\). Pressure \(P = 100\text{ kPa} = 100,000\text{ Pa}\). Rearranging for volume: \(V = \frac{nRT}{P} = \frac{0.0500 \times 8.31 \times 293.15}{100,000} = 0.001218\text{ m}^3\). Converting to \(dm^3\) by multiplying by 1000 gives \(1.22\text{ dm}^3\).

1 mark for correct option (B).

Manganese has atomic number 25, with electronic configuration \([Ar] 4s^2 3d^5\). When forming the \(Mn^{2+}\) ion, it loses the two \(4s\) electrons, resulting in the configuration \([Ar] 3d^5\), which is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\).

1 mark for correct option (B).

According to Fajan's rules, covalent character increases with high polarizing power of the cation and high polarizability of the anion. The \(Li^+\) cation is smaller than the \(Na^+\) cation, so it has greater polarizing power. The \(I^-\) anion is larger than the \(F^-\) anion, so it is more easily polarized. Therefore, lithium iodide (\(LiI\)) has the greatest covalent character.

1 mark for correct option (A).

The successive ionization energies show a very large increase (or 'jump') between the third and fourth ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner electron shell closer to the nucleus, meaning there are three electrons in the outermost shell. Therefore, element \( X \) belongs to Group 3 (or Group 13).

1 mark: correct answer is C.

1. Calculate moles of \(\text{HCl}\):
\(\text{n(HCl)} = \text{concentration} \times \text{volume} = 2.00\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0500\text{ mol}\).

2. From the equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\):
\(\text{n(Na}_2\text{CO}_3) = \frac{0.0500}{2} = 0.0250\text{ mol}\).

3. Calculate mass of \(\text{Na}_2\text{CO}_3\):
\(\text{mass} = 0.0250\text{ mol} \times 106.0\text{ g mol}^{-1} = 2.65\text{ g}\).

1 mark: correct answer is B.

\(\text{H}_3\text{O}^+\) (the hydronium ion) has three bonding pairs of electrons and one lone pair of electrons around the central oxygen atom. According to VSEPR theory, these four electron pairs arrange themselves in a tetrahedral geometry to minimize repulsion, resulting in a trigonal pyramidal molecular geometry which is non-planar. All other species (\(\text{BF}_3\), \(\text{CO}_3^{2-}\), and \(\text{C}_2\text{H}_4\)) are planar.

1 mark: correct answer is C.

A termination step is a step in a free-radical mechanism where two free radicals combine to form a stable covalent molecule, ending the chain reaction. In option C, two methyl radicals (\(\text{CH}_3^\bullet\)) combine to form ethane (\(\text{C}_2\text{H}_6\)). Option A and B are propagation steps, and option D is an initiation step.

1 mark: correct answer is C.

For an alkene to exhibit stereoisomerism (cis/trans or E/Z isomerism), each carbon atom of the double bond must be attached to two different groups. In \(\text{CH}_3\text{CH}=\text{CHCH}_3\) (but-2-ene), both double-bonded carbons are attached to a hydrogen atom (\(-\text{H}\)) and a methyl group (\(-\text{CH}_3\)). In all the other options, at least one of the double-bonded carbon atoms is attached to two identical groups (either two \(-\text{H}\) atoms or two \(-\text{CH}_3\) groups).

1 mark: correct answer is A.

The atomic number of iron (\(\text{Fe}\)) is 26, so its ground state atom 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 cations, they lose their outermost \(4\text{s}\) electrons first before the \(3\text{d}\) electrons. Therefore, to form \(\text{Fe}^{3+}\), two electrons are lost from the \(4\text{s}\) subshell and one from the \(3\text{d}\) subshell, leaving \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5\).

1 mark: correct answer is A.

Assume \(100\text{ g}\) of the compound:
- Mass of \(\text{Cu} = 88.8\text{ g}\)
- Mass of \(\text{O} = 100 - 88.8 = 11.2\text{ g}\)

Moles of \(\text{Cu} = \frac{88.8}{63.5} = 1.398\text{ mol}\)
Moles of \(\text{O} = \frac{11.2}{16.0} = 0.700\text{ mol}\)

Divide by the smallest value (\(0.700\)):
- \(\text{Cu} = \frac{1.398}{0.700} = 1.997 \approx 2\)
- \(\text{O} = \frac{0.700}{0.700} = 1\)

Therefore, the empirical formula is \(\text{Cu}_2\text{O}\).

1 mark: correct answer is B.

According to Fajan's rules, covalent character in ionic compounds increases with high positive charge on the cation and small ionic radius (high charge density of the cation). Among the cations \(\text{Na}^+\), \(\text{Mg}^{2+}\), \(\text{Al}^{3+}\), and \(\text{Ca}^{2+}\), the \(\text{Al}^{3+}\) ion has the highest charge (+3) and the smallest ionic radius, giving it the highest charge density. Therefore, \(\text{AlCl}_3\) exhibits the greatest degree of covalent character.

1 mark: correct answer is C.

Step 1: Write the balanced chemical equation for the reaction:
\(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)

Step 2: Calculate the number of moles of \(\text{CO}_2\) gas collected:
\(n(\text{CO}_2) = \frac{504 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.0210 \text{ mol}\)

Step 3: Determine the moles of pure \(\text{CaCO}_3\) reacted:
Since the molar ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, \(n(\text{CaCO}_3) = 0.0210 \text{ mol}\).

Step 4: Calculate the mass of pure \(\text{CaCO}_3\):
\(\text{mass} = 0.0210 \text{ mol} \times 100.1 \text{ g mol}^{-1} = 2.1021 \text{ g}\)

Step 5: Calculate the percentage purity of the sample:
\(\text{Percentage purity} = \frac{2.1021 \text{ g}}{2.40 \text{ g}} \times 100 = 87.5875\% \approx 87.6\%\)

Award 1 mark for correct option B.
- Correct conversion of gas volume to moles (0.0210 mol).
- Correct determination of mass of pure calcium carbonate (2.10 g).
- Correct evaluation of percentage purity (87.6%).

Step 1: Analyze the successive ionization energies to find the largest increase (jump):
- Between 1st and 2nd: 1239 kJ mol\(^{-1}\)
- Between 2nd and 3rd: 928 kJ mol\(^{-1}\)
- Between 3rd and 4th: 8833 kJ mol\(^{-1}\) (very large jump)

Step 2: The large jump between the 3rd and 4th ionization energies indicates that the fourth electron is being removed from a lower energy level closer to the nucleus. Therefore, element X has three valence electrons and belongs to Group 3 (Group 13).

Step 3: Since X is in Period 3 and Group 3, it is Aluminium (Al).

Step 4: The stable oxide of Aluminium consists of Al\(^{3+}\) and O\(^{2-}\) ions, giving the chemical formula \(\text{Al}_2\text{O}_3\).

Award 1 mark for correct option C.
- Identification of the main jump between the 3rd and 4th ionization energies.
- Deduction that the element is in Group 3 (Aluminium) and forms the stable oxide \(\text{Al}_2\text{O}_3\).

- \(\text{CO}_2\): The central C atom has 4 valence electrons, forms 2 double bonds, and has 0 lone pairs. Shape is linear.
- \(\text{SO}_2\): The central S atom has 6 valence electrons, forms 2 double bonds (using 4 electrons), leaving 1 lone pair. Its shape is non-linear (bent).
- \(\text{BeCl}_2\): The central Be atom has 2 valence electrons, forms 2 single bonds, and has 0 lone pairs. Shape is linear.
- \(\text{BF}_3\): The central B atom has 3 valence electrons, forms 3 single bonds, and has 0 lone pairs. Shape is trigonal planar.

Award 1 mark for correct option B.
- Correctly identifying the molecular geometries and number of lone pairs on each central atom to conclude that \(\text{SO}_2\) is non-linear with a lone pair.

For an alkene to exhibit stereoisomerism (\(E/Z\) isomerism), each carbon atom in the double bond must be bonded to two different groups.
- 2-methylbut-2-ene: \((\text{CH}_3)_2\text{C=CHCH}_3\). The first carbon atom of the C=C bond is bonded to two identical methyl groups, so it cannot show stereoisomerism.
- 2-methylbut-1-ene: \(\text{CH}_2\text{=C(CH}_3)\text{CH}_2\text{CH}_3\). The terminal carbon atom is bonded to two identical hydrogen atoms, so it cannot show stereoisomerism.
- pent-1-ene: \(\text{CH}_2\text{=CHCH}_2\text{CH}_2\text{CH}_3\). The terminal carbon atom is bonded to two identical hydrogen atoms, so it cannot show stereoisomerism.
- pent-2-ene: \(\text{CH}_3\text{CH=CHCH}_2\text{CH}_3\). One carbon of the C=C bond is attached to \(-\text{H}\) and \(-\text{CH}_3\); the other is attached to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Both double-bonded carbons are attached to two different groups, so it exhibits stereoisomerism.

Award 1 mark for correct option D.
- Identification that pent-2-ene satisfies the structural requirements for geometric (\(E/Z\)) isomerism.

(a)
1. Calculate moles of \(\text{HCl}\) used in the titration:
\[n(\text{HCl}) = C \times V = 0.100\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\]

2. Determine moles of \(M_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) sample:
From the stoichiometry of the equation, \(1\text{ mol}\) of \(M_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(HCl\):
\[n(M_2\text{CO}_3) \text{ in } 25.0\text{ cm}^3 = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\]

3. Determine moles of \(M_2\text{CO}_3\) in the original \(250.0\text{ cm}^3\) solution:
\[n(M_2\text{CO}_3) \text{ in } 250.0\text{ cm}^3 = 1.25 \times 10^{-3}\text{ mol} \times 10 = 1.25 \times 10^{-2}\text{ mol}\]

4. Calculate the molar mass of \(M_2\text{CO}_3 \cdot x\text{H}_2\text{O}\):
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{3.575\text{ g}}{1.25 \times 10^{-2}\text{ mol}} = 286\text{ g mol}^{-1}\]

(b)
(i)
1. Mass of water lost on heating:
\[\text{Mass of } \text{H}_2\text{O} = 3.575\text{ g} - 1.325\text{ g} = 2.250\text{ g}\]

2. Moles of water lost:
\[n(\text{H}_2\text{O}) = \frac{2.250\text{ g}}{18.0\text{ g mol}^{-1}} = 0.125\text{ mol}\]

3. Determine the ratio of moles of water to moles of the metal carbonate:
\[x = \frac{n(\text{H}_2\text{O})}{n(M_2\text{CO}_3)} = \frac{0.125\text{ mol}}{1.25 \times 10^{-2}\text{ mol}} = 10\]

(ii)
1. Molar mass of anhydrous \(M_2\text{CO}_3\):
\[M_r(M_2\text{CO}_3) = M_r(\text{hydrated}) - 10 \times M_r(\text{H}_2\text{O}) = 286.0 - 180.0 = 106.0\text{ g mol}^{-1}\]

2. Determine \(A_r(M)\):
\[2 \times A_r(M) + A_r(\text{C}) + 3 \times A_r(\text{O}) = 106.0\]
\[2 \times A_r(M) + 12.0 + 48.0 = 106.0\]
\[2 \times A_r(M) = 46.0 \implies A_r(M) = 23.0\]

3. Identify metal \(M\):
Metal \(M\) is sodium (\(\text{Na}\)).

Part (a) [6 marks]:
- M1: Calculation of moles of \(\text{HCl}\) (\(2.50 \times 10^{-3}\text{ mol}\)) (1)
- M2: Division of moles of \(\text{HCl}\) by 2 to get moles of \(M_2\text{CO}_3\) in \(25.0\text{ cm}^3\) (\(1.25 \times 10^{-3}\text{ mol}\)) (1)
- M3: Multiplication by 10 to get moles of \(M_2\text{CO}_3\) in \(250.0\text{ cm}^3\) (\(1.25 \times 10^{-2}\text{ mol}\)) (1)
- M4: Correct formula linking molar mass, mass, and moles (1)
- M5: Correct calculation to give \(286\) (1)
- M6: Correct unit: \(\text{g mol}^{-1}\) (dependent on correct value of 286 or via ECF) (1)

Part (b)(i) [3 marks]:
- M1: Mass of water lost = \(2.250\text{ g}\) (1)
- M2: Moles of water lost = \(0.125\text{ mol}\) (1)
- M3: Correct calculation showing ratio of \(0.125 : 0.0125 = 10\) (1)

Part (b)(ii) [3 marks]:
- M1: Calculation of \(M_r(M_2\text{CO}_3) = 106.0\) (1)
- M2: Correct calculation of \(A_r(M) = 23.0\) (1)
- M3: Identification of metal as sodium / \(\text{Na}\) (1)

(a)
Let the percentage abundance of \({}^{69}\text{Ga}\) be \(y\%\).
Therefore, the abundance of \({}^{71}\text{Ga}\) is \((100 - y)\%\).
\[A_r(\text{Ga}) = \frac{69y + 71(100-y)}{100} = 69.723\]
\[69y + 7100 - 71y = 6972.3\]
\[-2y = -127.7\]
\[y = 63.850\%\]
Abundance of \({}^{69}\text{Ga} = 63.850\%\)
Abundance of \({}^{71}\text{Ga} = 100 - 63.850 = 36.150\%\)

(b)
(i) High-energy electrons from an electron gun are fired at the vaporised sample. This knocks off one electron from each atom/molecule to form positive (\(1+\)) ions.
(ii) An electric field applies an equal force to accelerate all the positive ions, giving them all the exact same kinetic energy.

(c)
\[\text{Na}^{2+}(\text{g}) \rightarrow \text{Na}^{3+}(\text{g}) + \text{e}^-\]

(d)
The second electron is removed from the second shell (\(2p\) subshell), whereas the third electron is removed from a shell closer to the nucleus / different principal quantum energy level / closer to the positive nucleus, experiencing significantly less shielding, which results in a much stronger electrostatic force of attraction between the nucleus and the remaining outer electron.

Part (a) [4 marks]:
- M1: Setting up a correct algebraic equation for the weighted average (e.g., \(\frac{69y + 71(100-y)}{100} = 69.723\)) (1)
- M2: Correct rearrangement to find \(y\) (1)
- M3: Correct abundance of \({}^{69}\text{Ga} = 63.850\%\) (to 3 d.p.) (1)
- M4: Correct abundance of \({}^{71}\text{Ga} = 36.150\%\) (to 3 d.p.) (1)

Part (b) [4 marks]:
- (i) M1: High-energy electrons fired from electron gun (1); M2: knocks out an electron from the sample to form \(1+\) ions (1)
- (ii) M1: Accelerates positive ions using an electric field (1); M2: to give all ions the same kinetic energy (1)

Part (c) [2 marks]:
- M1: Correct species: \(\text{Na}^{2+}\) and \(\text{Na}^{3+}\) and \(\text{e}^-\) (1)
- M2: Correct state symbols \((\text{g})\) on both sodium species (1)

Part (d) [2 marks]:
- M1: Electron is removed from an inner shell/energy level (closer to the nucleus) (1)
- M2: Experiences less shielding, leading to a much stronger attraction to the nucleus (1)

(a)
(i)
1. Phosphorus has 5 outer shell electrons, sharing 3 with chlorine atoms. This leaves 3 bonding pairs and 1 lone pair of electrons around the central phosphorus atom.
2. Electron pairs repel to find a position of maximum separation (minimum repulsion).
3. Lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, which pushes the bonds closer together.
4. Shape: Trigonal pyramidal. Suggested bond angle: \(107^\circ\) (accept range \(100^\circ - 108^\circ\)).

(ii)
Shape: Trigonal bipyramidal. Bond angles: \(90^\circ\) and \(120^\circ\) (also accept \(180^\circ\)).

(b)
(i) Phosphorous originally has 5 valence electrons. The \(+\) charge means it has lost 1 electron, leaving 4 valence electrons. Each of these 4 electrons is shared with 1 electron from each of the 4 chlorine atoms to form 4 single covalent bonds.
Central P atom has 8 electrons in its outer shell (4 dots from P, 4 crosses from Cl). Each Cl atom has 8 outer electrons (1 cross, 7 dots/crosses of its own).
Draw the structure inside square brackets with a \(+\) sign outside.

(ii) \([\text{PCl}_6]^-\) has 6 bonding pairs and 0 lone pairs around the central phosphorus atom. Its shape is octahedral.

(c)
Ammonia (\(\text{NH}_3\)) molecules form hydrogen bonds between them because nitrogen is highly electronegative and has a lone pair of electrons. Phosphine (\(\text{PH}_3\)) cannot form hydrogen bonds as phosphorus is not electronegative enough; it only forms weaker London forces and permanent dipole-dipole interactions. More energy is required to overcome the stronger hydrogen bonds in ammonia, resulting in a higher boiling temperature.

Part (a)(i) [4 marks]:
- M1: State that there are 3 bonding pairs and 1 lone pair around phosphorus (1)
- M2: State that electron pairs repel to positions of minimum repulsion (or lone pairs repel more than bonding pairs) (1)
- M3: Shape is trigonal pyramidal (1)
- M4: Bond angle is \(107^\circ\) (accept \(100^\circ - 108^\circ\)) (1)

Part (a)(ii) [2 marks]:
- M1: Trigonal bipyramidal (1)
- M2: \(90^\circ\) and \(120^\circ\) (1)

Part (b)(i) [3 marks]:
- M1: Central P shown with 4 single bonds (pairs of dots and crosses) to 4 Cl atoms (1)
- M2: Remaining 6 non-bonding valence electrons shown around each Cl atom (1)
- M3: Square brackets with a \(+\) charge outside (1)

Part (b)(ii) [1 mark]:
- M1: Octahedral (1)

Part (c) [2 marks]:
- M1: Identify that \(\text{NH}_3\) has hydrogen bonding while \(\text{PH}_3\) has London forces / permanent dipole-dipole forces (1)
- M2: Hydrogen bonds are stronger and require more energy to break (1)

(a)
(i)
1. Molar mass of \(\text{CO}_2 = 12.0 + 2(16.0) = 44.0\text{ g mol}^{-1}\)
2. Calculate the moles of \(\text{CO}_2\) produced:
\[n(\text{CO}_2) = \frac{33.0\text{ g}}{44.0\text{ g mol}^{-1}} = 0.750\text{ mol}\]
3. Determine the ratio of moles of \(\text{CO}_2\) to moles of alkane \(Y\):
\[n = \frac{n(\text{CO}_2)}{n(Y)} = \frac{0.750\text{ mol}}{0.150\text{ mol}} = 5\]
4. Since \(Y\) is an alkane with the general formula \(\text{C}_n\text{H}_{2n+2}\), the molecular formula when \(n = 5\) is \(\text{C}_5\text{H}_{12}\).

(ii)
1. Write the balanced equation for the complete combustion of pentane:
\[\text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O}\]
2. Moles of \(\text{H}_2\text{O}\) produced = \(6 \times n(\text{C}_5\text{H}_{12}) = 6 \times 0.150\text{ mol} = 0.900\text{ mol}\)
3. Calculate mass of \(\text{H}_2\text{O}\):
\[\text{Mass} = 0.900\text{ mol} \times 18.0\text{ g mol}^{-1} = 16.2\text{ g}\]

(b)
(i) UV radiation provides the energy to cause homolytic fission of the \(\text{Cl-Cl}\) covalent bond (creating chlorine free radicals).
(ii)
Initiation:
\[\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\]
Propagation step 1:
\[\text{C}_5\text{H}_{12} + \text{Cl}^\bullet \rightarrow \text{C}_5\text{H}_{11}^\bullet + \text{HCl}\]
Propagation step 2:
\[\text{C}_5\text{H}_{11}^\bullet + \text{Cl}_2 \rightarrow \text{C}_5\text{H}_{11}\text{Cl} + \text{Cl}^\bullet\]

(iii)
During the termination stage of the reaction, two pentyl free radicals can collide and combine to form a single decane molecule:
\[2\text{C}_5\text{H}_{11}^\bullet \rightarrow \text{C}_{10}\text{H}_{22}\]

Part (a)(i) [4 marks]:
- M1: Calculation of moles of \(\text{CO}_2\) = \(0.750\text{ mol}\) (1)
- M2: Identifying that \(0.750\text{ mol}\) of \(\text{CO}_2\) contains \(0.750\text{ mol}\) of carbon atoms (1)
- M3: Calculation of ratio \(0.750 / 0.150 = 5\) carbon atoms per molecule (1)
- M4: Correct application of the general formula \(\text{C}_n\text{H}_{2n+2}\) to arrive at \(\text{C}_5\text{H}_{12}\) (1)

Part (a)(ii) [2 marks]:
- M1: Calculation of moles of water produced = \(0.900\text{ mol}\) (1)
- M2: Mass of water = \(16.2\text{ g}\) (1)

Part (b)(i) [1 mark]:
- M1: Provides energy to break the \(\text{Cl-Cl}\) bond homolytically (accept: provides activation energy for homolytic fission of chlorine) (1)

Part (b)(ii) [3 marks]:
- M1: Correct initiation equation with radical symbol on Cl (1)
- M2: Correct first propagation equation with radical on pentyl (1)
- M3: Correct second propagation equation showing regenerated chlorine radical (1)

Part (b)(iii) [2 marks]:
- M1: Collision/combination of two pentyl radicals (1)
- M2: Termination step equation: \(2\text{C}_5\text{H}_{11}^\bullet \rightarrow \text{C}_{10}\text{H}_{22}\) (1)

(a)
(i)
1. Calculate moles of alkene \(Z\) reacted (\(M_r(\text{C}_5\text{H}_{10}) = 5 \times 12.0 + 10 \times 1.0 = 70.0\text{ g mol}^{-1}\)):
\[n(Z) = \frac{1.40\text{ g}}{70.0\text{ g mol}^{-1}} = 0.020\text{ mol}\]
2. Molar mass of the dibromoalkane, \(\text{C}_5\text{H}_{10}\text{Br}_2 = 70.0 + 2(79.9) = 229.8\text{ g mol}^{-1}\).
3. Calculate the expected mass of the product if a 1:1 reaction occurs:
\[\text{Expected mass} = 0.020\text{ mol} \times 229.8\text{ g mol}^{-1} = 4.596\text{ g} \approx 4.60\text{ g}\]
Since the expected mass of the addition product matches the experimental mass (\(4.60\text{ g}\)), the reaction occurs in a 1:1 ratio, which is consistent with the presence of one double bond (a monoalkene).

(ii)
Because \(Z\) exhibits stereoisomerism, it must be pent-2-ene (pent-1-ene does not have stereoisomers as C1 has two hydrogens).
Skeletal structures should show:
- (E)-pent-2-ene (or trans-pent-2-ene): the methyl and ethyl groups are on opposite sides of the double bond.
- (Z)-pent-2-ene (or cis-pent-2-ene): the methyl and ethyl groups are on the same side of the double bond.
Names: (E)-pent-2-ene / trans-pent-2-ene AND (Z)-pent-2-ene / cis-pent-2-ene.

(b)
(i)
1. Draw pent-1-ene: \(\text{CH}_2\text{=CHCH}_2\text{CH}_2\text{CH}_3\).
2. Draw \(\text{H-Cl}\) with dipoles: \(\text{H}^{\delta+}\text{-Cl}^{\delta-}\).
3. Curly arrow starts from the double bond of pent-1-ene pointing to the \(\text{H}^{\delta+}\).
4. Curly arrow from the \(\text{H-Cl}\) bond pointing to \(\text{Cl}\).
5. Draw the intermediate carbocation: secondary carbocation on C2 (\(\text{CH}_3\text{-CH}^+\text{-CH}_2\text{CH}_2\text{CH}_3\)) and a chloride ion \(\text{Cl}^-\) with a lone pair and negative charge.
6. Curly arrow from the lone pair on the chloride ion pointing to the carbocation carbon \(\text{C}^+\).
7. Product: 2-chloropentane.

(ii)
The reaction proceeds via a secondary carbocation intermediate rather than a primary carbocation. The secondary carbocation is more stable than the primary carbocation due to the electron-releasing inductive effect of the two alkyl groups compared to only one in the primary carbocation.

Part (a)(i) [3 marks]:
- M1: Moles of \(Z\) = \(0.020\text{ mol}\) (1)
- M2: Calculation of \(M_r\) of dibromoalkane = \(229.8\) (1)
- M3: Multiplication of moles by molar mass to get \(4.60\text{ g}\) and stating this confirms 1:1 addition (1)

Part (a)(ii) [4 marks]:
- M1: Correct skeletal structure of (E)-pent-2-ene (1)
- M2: Correct skeletal structure of (Z)-pent-2-ene (1)
- M3: Correct IUPAC name: (E)-pent-2-ene / trans-pent-2-ene (1)
- M4: Correct IUPAC name: (Z)-pent-2-ene / cis-pent-2-ene (1)

Part (b)(i) [4 marks]:
- M1: Curly arrow from double bond to \(\text{H}^{\delta+}\) of \(\text{H-Cl}\), showing correct dipole on \(\text{H-Cl}\) (1)
- M2: Curly arrow from \(\text{H-Cl}\) bond to \(\text{Cl}\) (1)
- M3: Drawing correct secondary carbocation intermediate (1)
- M4: Curly arrow from lone pair on \(\text{Cl}^-\) to the carbon with positive charge, yielding 2-chloropentane (1)

Part (b)(ii) [1 mark]:
- M1: The secondary carbocation intermediate is more stable than the primary carbocation because of the electron-releasing/inductive effect of two alkyl groups (1)

Hydrogen fluoride (HF) has the highest boiling temperature because of the presence of strong intermolecular hydrogen bonds between its molecules. The other hydrogen halides (HCl, HBr, HI) only have weaker permanent dipole-dipole and London forces. Although HI has larger London forces than HCl and HBr due to more electrons, these are still weaker than the hydrogen bonding present in HF.

1 mark for selecting A.

Going down Group 2, the solubility of the sulfates decreases (magnesium sulfate is highly soluble, whereas barium sulfate is virtually insoluble). Conversely, the solubility of the hydroxides increases down the group (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble).

1 mark for selecting D.

Using Hess's Law: \(2\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{l})\). The standard enthalpy change of formation is calculated by: \(\Delta H_{\text{f}} = \sum \Delta H_{\text{c}}(\text{reactants}) - \sum \Delta H_{\text{c}}(\text{products})\), which gives \(\Delta H_{\text{f}} = [2 \times \Delta H_{\text{c}}(\text{C}) + 3 \times \Delta H_{\text{c}}(\text{H}_2)] - \Delta H_{\text{c}}(\text{C}_2\text{H}_5\text{OH}) = [2(-394) + 3(-286)] - (-1367) = [-788 - 858] + 1367 = -1646 + 1367 = -279\text{ kJ mol}^{-1}\).

1 mark for selecting A.

The rate of hydrolysis of halogenoalkanes depends on two factors: the carbon-halogen bond strength (C-I is weaker than C-Cl, so iodoalkanes react faster than chloroalkanes) and the mechanism pathway (tertiary halogenoalkanes react much faster via the stable tertiary carbocation intermediate in the \(S_N1\) mechanism compared to primary halogenoalkanes via \(S_N2\)). Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.

1 mark for selecting D.

A catalyst does not alter the energy of the reactant molecules, so the Maxwell-Boltzmann distribution curve (which describes molecular energies at a constant temperature) remains completely unchanged. Instead, the catalyst provides an alternative reaction pathway with a lower activation energy, meaning the line indicating \(E_a\) shifts to the left on the energy axis.

1 mark for selecting C.

The disproportionation reaction of chlorine in hot, concentrated sodium hydroxide is: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\). The products containing chlorine are sodium chloride (\(\text{NaCl}\)), where Cl has an oxidation state of \(-1\), and sodium chlorate(V) (\(\text{NaClO}_3\)), where Cl has an oxidation state of \(+5\).

1 mark for selecting C.

An absorption peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad peak at \(3200-3750\text{ cm}^{-1}\) means there is no hydroxyl (\(\text{O-H}\)) group. Propanal contains a carbonyl group and no alcohol or carboxylic acid hydroxyl group, and matches the molecular formula \(\text{C}_3\text{H}_6\text{O}\).

1 mark for selecting C.

By definition, dynamic chemical equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction, and both continue to occur (dynamic) so that the macroscopic concentrations of reactants and products remain constant, but not necessarily equal.

1 mark for selecting B.

First, calculate the energy transferred to the water using \(q = m c \Delta T\). Here, \(m = 100\text{ g}\), \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T = 10.5\text{ K}\). Thus, \(q = 100 \times 4.18 \times 10.5 = 4389\text{ J} = 4.389\text{ kJ}\). Next, calculate the chemical amount of methanol burned: \(n = \frac{0.400}{32.0} = 0.0125\text{ mol}\). Finally, the enthalpy change of combustion is \(\Delta H_c^\ominus = -\frac{q}{n} = -\frac{4.389}{0.0125} = -351.12\text{ kJ mol}^{-1}\), which rounds to \(-351\text{ kJ mol}^{-1}\).

Award 1 mark for correct answer B. Option A is calculated by dividing temperature change by mass; Option C has the incorrect sign; Option D has a decimal place error.

Propan-1-ol has the molecular formula \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) and contains an \(\text{-OH}\) group, allowing it to form strong intermolecular hydrogen bonds. The other compounds (propanal, propanone, and butane) only have weaker intermolecular forces (London forces and/or permanent dipole-dipole forces) and thus require less thermal energy to vaporise.

Award 1 mark for correct answer C.

Down Group 2, the cationic radius increases (from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\)) while the ionic charge remains constant. Therefore, the \(\text{Ba}^{2+}\) ion has a smaller charge density compared to the \(\text{Mg}^{2+}\) ion. The lower charge density means the barium ion polarises the carbonate ion less, keeping the carbonate ion more stable and raising the thermal decomposition temperature.

Award 1 mark for correct answer A.

The rate of nucleophilic substitution in halogenoalkanes depends on the strength of the carbon-halogen bond. Down Group 7, bond enthalpy decreases as bond length increases. Therefore, the C-I bond is the weakest and breaks most readily, yielding iodide ions fastest to react with silver ions and form a yellow precipitate. The C-Cl bond is the strongest and breaks slowest. The order of precipitate appearance from fastest to slowest is 1-iodobutane > 1-bromobutane > 1-chlorobutane.

Award 1 mark for correct answer D.

Adding a catalyst provides an alternative reaction pathway with a lower activation energy, meaning the minimum energy required for reaction shifts to a lower value (to the left) on the energy axis. Because the temperature of the system is unchanged, the actual distribution of molecular kinetic energies described by the Maxwell-Boltzmann distribution curve remains exactly the same.

Award 1 mark for correct answer B.

In cold, dilute aqueous sodium hydroxide, chlorine gas undergoes a disproportionation reaction: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). The two chlorine-containing species formed in solution are sodium chloride (\(\text{NaCl}\)) and sodium chlorate(I) (\(\text{NaClO}\)).

Award 1 mark for correct answer C.

The strong, sharp absorption peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C=O}\)). The absence of a broad absorption in the range \(3200-3600\text{ cm}^{-1}\) indicates there is no alcohol (\(\text{O-H}\)) group. Propanone is a ketone (\(\text{C}_3\text{H}_6\text{O}\)) containing a carbonyl group and no alcohol group, matching the spectral characteristics.

Award 1 mark for correct answer A.

Only a change in temperature can alter the numerical value of the equilibrium constant, \(K_c\). Because the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the position of equilibrium to the right to produce heat (opposing the change), which increases both the yield of \(\text{SO}_3\) and the value of \(K_c\).

Award 1 mark for correct answer D.

As the halogen group is descended from fluorine to iodine, the size of the halogen atom increases, resulting in a larger number of electrons in the molecule. This leads to stronger London (dispersion) forces between molecules, which require more thermal energy to overcome, thereby increasing the boiling temperature. Even though the carbon-halogen bond becomes less polar, the increase in London forces is the dominant factor.

Select Option D. [1 mark] for selecting D. Reject all other options.

The \( \text{Mg}^{2+} \) ion has a smaller ionic radius than the \( \text{Ba}^{2+} \) ion but carries the same 2+ charge, resulting in a higher charge density. This higher charge density allows the \( \text{Mg}^{2+} \) ion to polarise the electron cloud of the nitrate anion more effectively, weakening the covalent bonds within the nitrate ion and making it easier to decompose thermally. Therefore, magnesium nitrate decomposes at a lower temperature than barium nitrate. The decomposition of Group 2 nitrates produces the metal oxide, nitrogen dioxide (a brown gas), and oxygen.

Select Option B. [1 mark] for selecting B. Reject all other options.

Tertiary halogenoalkanes react with nucleophiles predominantly via an \( \text{S}_{\text{N}}1 \) mechanism because they form a relatively stable tertiary carbocation intermediate. This intermediate is stabilised by the electron-donating inductive effect of the three alkyl groups attached to the positive carbon. 2-chloro-2-methylpropane is a tertiary halogenoalkane, whereas 1-chlorobutane and 1-chloro-2-methylpropane are primary, and 2-chlorobutane is secondary.

Select Option D. [1 mark] for selecting D. Reject all other options.

When the temperature is increased, the Maxwell-Boltzmann distribution curve flattens and shifts to the right (higher energies). This shift causes the peak of the curve, representing the most probable molecular energy (\( E_{\text{mp}} \)), to move to a higher energy value. Additionally, the area under the curve to the right of the activation energy (\( E_{\text{a}} \)) increases significantly, meaning that a larger fraction of molecules possess energy greater than or equal to \( E_{\text{a}} \).

Select Option A. [1 mark] for selecting A. Reject all other options.

(a)(i) Observations confirming redox: Red-brown fumes/vapor of bromine are produced. A choking gas (sulfur dioxide) is formed. (ii) Acid-base equation: \(\text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)}\). Redox equation: \(2\text{HBr(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\). (iii) In the first reaction, sulfuric acid acts as an acid (proton donor). In the second, it acts as an oxidizing agent. (b)(i) Oxidation state of S in \(\text{H}_2\text{SO}_4\) is +6. Oxidation state of S in \(\text{H}_2\text{S}\) is -2. Iodide ions are stronger reducing agents than bromide ions because the ionic radius of iodide is larger. The outer electrons are further from the nucleus and experience more shielding, so they are lost more easily. (ii) Balanced ionic equation: \(8\text{I}^- + \text{H}_2\text{SO}_4 + 8\text{H}^+ \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}\).

(a)(i) 1 mark for identifying red-brown vapor/liquid of bromine. 1 mark for identifying a choking/pungent gas of sulfur dioxide. (ii) 1 mark for correct acid-base equation (accept products as \(\text{Na}_2\text{SO}_4\)). 1 mark for correct balanced redox equation. (iii) 1 mark for stating both roles correctly (acid/proton donor and oxidizing agent). (b)(i) 1 mark for correct oxidation states (+6 and -2). 1 mark for stating iodide has a larger ionic radius/more shielding. 1 mark for explaining that outer electrons are less strongly held and lost more easily. (ii) 2 marks for fully correct balanced ionic equation. Allow 1 mark if species are correct but balancing is incorrect.

(a)(i) Since Y has molecular formula \(C_4H_8O\) and does not react with Fehling's solution, it is a ketone. Thus, Y is butanone, and X is butan-2-ol. Structures: X is \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\), Y is \(\text{CH}_3\text{COCH}_2\text{CH}_3\). (ii) Reaction type: Oxidation. Color change: Orange to green. (b) X contains an O-H bond (alcohol) which gives a broad, strong peak at \(3200 - 3750\ \text{cm}^{-1}\). Y contains a C=O bond (carbonyl) which gives a sharp, strong peak at \(1675 - 1750\ \text{cm}^{-1}\). Distinguishing is achieved by monitoring the disappearance of the O-H peak and the appearance of the C=O peak. (c) In an \(S_N1\) mechanism, loss of the leaving group (chloride) forms a planar carbocation intermediate: \(\text{[CH}_3\text{CHCH}_2\text{CH}_3\text{]}^+\). The nucleophile (\(\text{OH}^-\)) can attack this planar intermediate with equal probability from either side, producing an equal 50:50 mixture of the two optical isomers (enantiomers), which is a racemic mixture.

(a)(i) 1 mark for naming X as butan-2-ol. 1 mark for naming Y as butanone. 1 mark for drawing correct structures for both. (ii) 1 mark for oxidation and color change from orange to green. (b) 1 mark for stating X shows broad O-H absorption in the range \(3200 - 3750\ \text{cm}^{-1}\). 1 mark for stating Y shows sharp C=O absorption in the range \(1675 - 1750\ \text{cm}^{-1}\). 1 mark for explaining that comparing the presence/absence of these characteristic peaks distinguishes the compounds. (c) 1 mark for stating that the carbocation intermediate is planar. 1 mark for explaining that the nucleophile can attack with equal probability from either side, resulting in a 50:50 mixture of enantiomers.

(a)(i) \(\Delta T = 48.5 - 18.5 = 30.0\ \text{K}\). \(q = m c \Delta T = 150.0 \times 4.18 \times 30.0 = 18810\ \text{J}\) (or 18.81 kJ). (ii) \(\text{moles} = 0.600 / 60.0 = 0.0100\ \text{mol}\). (iii) \(\Delta_c H = -18.81\ \text{kJ} / 0.0100\ \text{mol} = -1880\ \text{kJ mol}^{-1}\) (to 3 significant figures). (b) Reasons: 1. Heat loss to the surroundings/calorimeter. 2. Incomplete combustion of propan-1-ol (forming carbon monoxide or carbon soot). 3. Evaporation of propan-1-ol from the wick after extinguishing. (c) Formation equation: \(3\text{C(s)} + 4\text{H}_2\text{(g)} + 0.5\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\). Using Hess's law: \(\Delta_f H = 3\Delta_c H[\text{C}] + 4\Delta_c H[\text{H}_2] - \Delta_c H[\text{C}_3\text{H}_7\text{OH}]\). \(\Delta_f H = 3(-394) + 4(-286) - (-2021) = -1182 - 1144 + 2021 = -305\ \text{kJ mol}^{-1}\).

(a)(i) 1 mark for correct calculation of heat energy: \(18810\ \text{J}\) (or \(18.8\ \text{kJ}\)). (ii) 1 mark for correct moles: \(0.0100\ \text{mol}\). (iii) 1 mark for dividing heat by moles. 1 mark for final value of \(-1880\ \text{kJ mol}^{-1}\) with negative sign and 3 sig figs. (b) 1 mark for each valid reason up to a maximum of 3 marks (e.g. heat loss to surroundings, incomplete combustion, evaporation of fuel, non-standard conditions). (c) 1 mark for showing a correct Hess cycle diagram or formula. 1 mark for correct calculation setup: \(3(-394) + 4(-286) - (-2021)\). 1 mark for correct final value of \(-305\ \text{kJ mol}^{-1}\).

(a)(i) The diagram should show 'Number of molecules' on the y-axis and 'Energy' on the x-axis. The \(T_1\) curve starts at the origin, rises to a peak, and asymptotes to the x-axis. The \(T_2\) curve has a lower peak shifted to the right, crossing \(T_1\) once, and lies above \(T_1\) at higher energy values. (ii) \(E_{cat}\) is drawn to the left of \(E_a\) on the energy axis. A catalyst provides an alternative pathway with a lower activation energy. Therefore, a larger fraction/proportion of molecules have energy \(\ge E_{cat}\), resulting in a higher frequency of successful collisions. (b)(i) Increasing pressure increases the equilibrium yield of methanol because the system shifts to the side with fewer moles of gas (1 mole on products vs 3 moles on reactants) to oppose the increase in pressure. (ii) Increasing temperature decreases the yield because the forward reaction is exothermic, so the equilibrium shifts to the left (endothermic direction). However, it increases the reaction rate because molecules have more kinetic energy, leading to more frequent and energetic collisions.

(a)(i) 1 mark for correct axes labels. 1 mark for correct \(T_1\) shape. 1 mark for correct \(T_2\) shape showing lower peak shifted to the right. (ii) 1 mark for marking \(E_a\) and \(E_{cat}\) on the x-axis with \(E_{cat} < E_a\). 1 mark for stating that a catalyst lowers the activation energy by providing an alternative pathway. 1 mark for stating that more molecules have energy greater than the activation energy, increasing the rate of successful collisions. (b)(i) 1 mark for stating yield increases. 1 mark for explaining that equilibrium shifts to the side with fewer gas molecules (1 vs 3) to reduce pressure. (ii) 1 mark for stating yield decreases because forward reaction is exothermic. 1 mark for stating rate increases because of more frequent and successful collisions.

(a) HF has hydrogen bonding and London forces. HCl, HBr, and HI have permanent dipole-dipole forces and London forces. HF has an anomalously high boiling point because fluorine is highly electronegative, making the H-F bond highly polar, leading to strong intermolecular hydrogen bonds which require significant energy to break. From HCl to HI, boiling points increase because the halogen atom size increases down the group, increasing the number of electrons. This increases the strength of London forces (instantaneous dipole-induced dipole forces), which outweigh the decrease in permanent dipoles. (b) \(\text{CH}_4\) cannot form hydrogen bonds because carbon is not electronegative enough; thus it only has weak London forces, explaining why its boiling point is low with no anomaly. The boiling points of Group 4 hydrides increase steadily from \(\text{CH}_4\) to \(\text{SnH}_4\) because the number of electrons per molecule increases down the group, leading to stronger London forces between the non-polar molecules.

(a) 1 mark for stating HF has hydrogen bonds and London forces. 1 mark for stating others have permanent dipole-dipole and London forces. 1 mark for explaining HF's high boiling point due to strong hydrogen bonding from highly electronegative F. 1 mark for stating hydrogen bonds are much stronger than London/dipole forces. 1 mark for explaining that from HCl to HI, the number of electrons increases. 1 mark for concluding this leads to stronger London forces. (b) 1 mark for stating carbon has low electronegativity/the C-H bond is non-polar. 1 mark for concluding \(\text{CH}_4\) cannot form hydrogen bonds. 1 mark for stating Group 4 hydride boiling points increase down the group. 1 mark for explaining this is due to increasing electron count causing stronger London forces.

(a) Thermal stability of Group 2 carbonates increases down the group. Down the group, the cationic radius of \(\text{M}^{2+}\) increases while the charge remains +2. This decreases the charge density of the cation. Consequently, the cation has a weaker polarizing effect on the large carbonate (\(\text{CO}_3^{2-}\)) ion, leading to less distortion/polarization of the C-O covalent bonds. This makes the carbonate ion more stable, requiring a higher temperature to decompose. (b)(i) Solubility of Group 2 hydroxides increases down the group, whereas the solubility of Group 2 sulfates decreases down the group. (ii) To distinguish: Add \(\text{Na}_2\text{SO}_4\text{(aq)}\) to both. \(\text{Ba}^{2+}\) forms a thick white precipitate (\(\text{BaSO}_4\)), whereas \(\text{Mg}^{2+}\) shows no visible change. Add \(\text{NaOH(aq)}\) to both. \(\text{Mg}^{2+}\) forms a white precipitate (\(\text{Mg(OH)}_2\)), whereas \(\text{Ba}^{2+}\) shows no visible change (as barium hydroxide is highly soluble).

(a) 1 mark for stating thermal stability increases down the group. 1 mark for stating that the cationic radius of \(\text{M}^{2+}\) increases down the group. 1 mark for stating that the charge density of the cation decreases. 1 mark for explaining that this causes less polarization/distortion of the carbonate ion. 1 mark for relating this to less weakening of the C-O bonds, making the compound more stable to heat. (b)(i) 1 mark for stating hydroxides become more soluble down the group. 1 mark for stating sulfates become less soluble down the group. (ii) 1 mark for using sulfate ions to get a white precipitate with \(\text{Ba}^{2+}\) and no change with \(\text{Mg}^{2+}\). 1 mark for using hydroxide ions to get a white precipitate with \(\text{Mg}^{2+}\) and no change with \(\text{Ba}^{2+}\). 1 mark for clearly identifying the formulas of the precipitates formed (\(\text{BaSO}_4\) and \(\text{Mg(OH)}_2\)).

(a) Concordant titres are within \(0.10\text{ cm}^3\) of each other. These are Titre 2 (\(25.10\text{ cm}^3\)), Titre 3 (\(24.90\text{ cm}^3\)), and Titre 4 (\(25.00\text{ cm}^3\)). Mean titre = \(\frac{25.10 + 24.90 + 25.00}{3} = 25.00\text{ cm}^3\).

(b) Yellow to orange (or peach/pink-orange).

(c) \(\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\)

(d) Moles of \(\text{HCl}\) in mean titre = \(0.100\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\).
From the chemical equation, \(2\text{ mol}\) of \(\text{HCl}\) reacts with \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 1.25 \times 10^{-3} \times 10 = 0.0125\text{ mol}\).

(e) Molar mass of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} = \frac{\text{mass}}{\text{moles}} = \frac{3.58}{0.0125} = 286.4\text{ g mol}^{-1}\).
Molar mass of anhydrous \(\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\).
Mass of \(x\text{H}_2\text{O} = 286.4 - 106.0 = 180.4\text{ g mol}^{-1}\).
\(x = \frac{180.4}{18.0} = 10.02 \approx 10\).

(f) A damp weighing bottle contains water, meaning some of the recorded mass of the solid is actually water from the bottle. Therefore, the actual mass of hydrated sodium carbonate transferred to the volumetric flask is less than \(3.58\text{ g}\). This means there are fewer moles of \(\text{Na}_2\text{CO}_3\) in the flask than calculated. This results in a smaller calculated value for the moles of anhydrous salt, which makes the calculated molar mass of the hydrated salt appear larger than it actually is. Consequently, the calculated mass of water of crystallisation and thus the value of \(x\) will be too high.

(a)
- Selects Titres 2, 3, and 4 (concordant within \(\pm 0.10\text{ cm}^3\)) (1)
- Correctly calculates mean = \(25.00\text{ cm}^3\) (1)

(b)
- Yellow to orange / peach (do not accept pink or red as final color) (1)

(c)
- Balanced equation with correct formulas: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\) (1)

(d)
- Moles of \(\text{HCl} = 2.50 \times 10^{-3}\text{ mol}\) (1)
- Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = 1.25 \times 10^{-3}\text{ mol}\) (1)
- Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 0.0125\text{ mol}\) (1)

(e)
- Molar mass of hydrated salt = \(286.4\text{ g mol}^{-1}\) (1)
- Mass of water = \(180.4\text{ g mol}^{-1}\) (1)
- Value of \(x = 10\) (must be a whole number) (1)

(f)
- Water on the bottle increases the measured mass of the solid / less actual solid is added than weighed (1)
- Calculated moles of \(\text{Na}_2\text{CO}_3\) based on titration is unaffected, but the apparent mass of solid is higher, leading to a larger calculated molar mass (1)
- Therefore, the calculated \(x\) is too high (1)

(a) To ensure that all of the copper(II) sulfate reacts completely, making it the limiting reactant.

(b) The reaction is not instantaneous, so some heat loss to the surroundings occurs while the reaction is still taking place. Extrapolating the cooling curve back to the time of mixing (the 4th minute) estimates the temperature rise that would have occurred if the reaction had been instantaneous and no heat was lost.

(c) \(q = m \times c \times \Delta T\)
Mass of solution, \(m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\)
\(q = 50.0 \times 4.18 \times 12.5 = 2612.5\text{ J}\)
To 3 significant figures, \(q = 2610\text{ J}\) (or \(2.61\text{ kJ}\)).

(d) Moles of \(\text{CuSO}_4 = \text{concentration} \times \text{volume} = 0.200 \times 0.0500 = 0.0100\text{ mol}\).
Since zinc is in excess, moles of reaction = \(0.0100\text{ mol}\).
\(\Delta H = -\frac{q}{\text{moles}} = -\frac{2.6125\text{ kJ}}{0.0100\text{ mol}} = -261.25\text{ kJ mol}^{-1}\).
To 3 significant figures, \(\Delta H = -261\text{ kJ mol}^{-1}\).

(e) The zinc powder might have an oxide layer (zinc oxide), meaning less active zinc is available to react, slowing down the reaction and reducing the maximum temperature rise. Alternatively, the reaction might not have gone to completion, reducing the heat released.

(f) Since the temperature rise is calculated from two temperature readings (initial and final temperature), the total uncertainty is \(2 \times 0.1 = 0.2\text{ }^{\circ}\text{C}\).
\(Percentage\ uncertainty = \frac{\text{total uncertainty}}{\text{temperature rise}} \times 100\% = \frac{0.2}{12.5} \times 100\% = 1.6\%\).

(a)
- To ensure all copper(II) ions / CuSO4 react completely (1)

(b)
- Reaction is not instantaneous / heat loss occurs during the reaction (1)
- Extrapolation corrects for heat loss to estimate the temperature rise if reaction was instantaneous (1)

(c)
- Calculation of mass of solution = 50.0 g (1)
- Calculation of q = 2612.5 J / 2610 J / 2.61 kJ (1)

(d)
- Moles of copper(II) sulfate = 0.0100 mol (1)
- Division of heat by moles (e.g. 2.6125 / 0.0100) (1)
- Correct final value with negative sign and units: -261 kJ mol-1 (accept answers in range -261 to -261.3) (1)

(e)
- Zinc is coated with an oxide layer / zinc has reacted with oxygen (1)
- This reduces the rate of reaction / means less heat is produced per second, leading to more heat loss and a lower temperature rise (1)

(f)
- Total uncertainty in temperature difference = 0.2 °C (1)
- Percentage uncertainty = (0.2 / 12.5) x 100 = 1.6% (1)

(a) Brick-red flame indicates the presence of calcium ions, \(\text{Ca}^{2+}\).

(b) Formation of a cream precipitate with silver nitrate that is insoluble in dilute ammonia but soluble in concentrated ammonia indicates bromide ions, \(\text{Br}^-\).

(c) Silver ions react with bromide ions to form silver bromide:
\(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)

(d) Calcium ions react with sulfate ions from sulfuric acid to form a white precipitate of calcium sulfate:
\(\text{Ca}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{CaSO}_4\text{(s)}\)

(e) The chemical formula of Compound **Y** is \(\text{CaBr}_2\).

(f) Clean a platinum (or nichrome) wire by dipping it into concentrated hydrochloric acid and then holding it in a hot Bunsen burner flame. Repeat this cleaning process until the wire does not produce any color in the flame. Dip the clean wire back into concentrated hydrochloric acid and then into the solid sample of **Y** so that some solid adheres to the wire. Place the wire in the non-luminous (blue) Bunsen burner flame and observe the flame color.

(g) Nitric acid is added to react with and remove any carbonate or hydrogencarbonate impurities that might be present in the sample. If these impurities were not removed, they would react with silver nitrate to form a white precipitate of silver carbonate, giving a false positive result.

(a)
- Calcium / Ca2+ (1)

(b)
- Bromide / Br- (do not accept bromine) (1)

(c)
- Correct species: Ag+ + Br- -> AgBr (1)
- Correct state symbols: (aq) + (aq) -> (s) (1)

(d)
- Correct species: Ca2+ + SO42- -> CaSO4 (1)
- Correct state symbols: (aq) + (aq) -> (s) (1)

(e)
- CaBr2 (1)

(f)
- Dip platinum / nichrome wire into concentrated HCl and heat in flame (to clean it) (1)
- Repeat until no color is produced in flame (1)
- Dip wire in concentrated HCl and then into the solid sample (1)
- Place in a non-luminous / blue / roaring Bunsen flame (1)

(g)
- To react with / remove carbonate / hydrogencarbonate / hydroxide ions (1)
- Which would otherwise form a precipitate with silver ions (e.g. silver carbonate) (1)

(a) Reflux allows the mixture to be heated to its boiling point for a prolonged period without the volatile reactants or products escaping from the flask, as vapor condenses and flows back into the reaction flask.

(b)
(i) The organic layer is the lower layer because its density (\(1.27\text{ g cm}^{-3}\)) is greater than that of the aqueous layer (approximately \(1.00\text{ g cm}^{-3}\)). Alternatively, add a small volume of deionised water to the funnel and observe which layer increases in volume (the upper aqueous layer, confirming the lower is organic).
(ii) Aqueous sodium hydrogencarbonate, \(\text{NaHCO}_3\) (or sodium carbonate, \(\text{Na}_2\text{CO}_3\)).
(iii) Insert the stopper, invert the funnel, and shake gently. Open the tap periodically while inverted to release built-up pressure from carbon dioxide gas.

(c)
(i) Anhydrous calcium chloride, \(\text{CaCl}_2\) (or anhydrous sodium sulfate, \(\text{Na}_2\text{SO}_4\), or anhydrous magnesium sulfate, \(\text{MgSO}_4\)).
(ii) The liquid changes from cloudy to completely clear.

(d) Around \(101 - 103\text{ }^{\circ}\text{C}\) (or at \(102\text{ }^{\circ}\text{C}\)).

(e)
Moles of 1-butanol = \(\frac{7.40}{74.0} = 0.100\text{ mol}\).
Since the reaction stoichiometry is 1:1, theoretical moles of 1-bromobutane = \(0.100\text{ mol}\).
Theoretical mass of 1-bromobutane = \(0.100 \times 137.0 = 13.7\text{ g}\).
Percentage yield = \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% = \frac{8.22}{13.7} \times 100\% = 60.0\%\).

(a)
- Vapor condenses and returns to the flask / prevents volatile substances escaping (1)
- Allows prolonged heating / reaction to reach completion (1)

(b)
- (i) Lower layer because its density is greater / add water and see which layer increases in volume (which is the aqueous layer, so other is organic) (1)
- (ii) Sodium hydrogencarbonate / NaHCO3 / sodium carbonate / Na2CO3 (1)
- (iii) Invert and shake (with stopper in place) (1) and open tap/stopcock regularly to release pressure (1)

(c)
- (i) Anhydrous calcium chloride / CaCl2 / anhydrous sodium sulfate / Na2SO4 / anhydrous magnesium sulfate / MgSO4 (1)
- (ii) Liquid goes from cloudy to clear / transparent (1)

(d)
- 101 – 103 °C (accept any range containing 102 °C within ± 2 °C) (1)

(e)
- Moles of 1-butanol = 0.100 mol (1)
- Theoretical mass of 1-bromobutane = 13.7 g (1)
- Percentage yield = (8.22 / 13.7) x 100 = 60.0% (1)