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The largest jump in ionization energy occurs between the 5th and the 6th ionization energies (from 6273 to 22233 kJ mol^{-1}). This indicates that the 6th electron is removed from an inner shell closer to the nucleus, which is more strongly attracted by the nuclear charge. Therefore, element X has 5 valence electrons in its outer shell. Since X is in Period 3 and has 5 valence electrons, it belongs to Group 15 (Group 5), which is Phosphorus.

1 mark: Correctly identifies B as the answer based on the jump between the 5th and 6th ionization energies.

The volume decrease when shaken with sodium hydroxide is due to the absorption of carbon dioxide gas: Volume of CO_2 = 55 cm^3 - 25 cm^3 = 30 cm^3. Since 10 cm^3 of hydrocarbon produced 30 cm^3 of CO_2, 1 molecule of hydrocarbon contains 3 carbon atoms (x = 3). The remaining gas is the unreacted oxygen, which is 25 cm^3. Therefore, the volume of oxygen reacted is: Volume of O_2 reacted = 70 cm^3 - 25 cm^3 = 45 cm^3. The molar ratio of hydrocarbon to oxygen reacted is 10 : 45 = 1 : 4.5. The general equation for the complete combustion of a hydrocarbon is C_3H_y + (3 + y/4)O_2 -> 3CO_2 + (y/2)H_2O. Thus, we have 3 + y/4 = 4.5, which gives y = 6. So the molecular formula is C_3H_6.

1 mark: Correctly deduces C as the formula by calculating the volume of CO_2 and reacted O_2.

The geometry of electron pairs around a central atom depends on the total number of bonding regions and lone pairs: In NH_2^-, nitrogen has 2 bonding pairs and 2 lone pairs (total of 4 electron pairs), which are arranged in a tetrahedral geometry. In PH_3, phosphorus has 3 bonding pairs and 1 lone pair (total of 4 electron pairs), which are arranged in a tetrahedral geometry. In H_3O^+, oxygen has 3 bonding pairs and 1 lone pair (total of 4 electron pairs), which are arranged in a tetrahedral geometry. In CO_3^{2-}, the carbon atom forms one double bond and two single bonds with three oxygen atoms, with no lone pairs on the carbon. This results in 3 electron-dense regions, which are arranged in a trigonal planar geometry, not tetrahedral.

1 mark: Selects B as the correct species that does not have a tetrahedral electron-pair geometry.

Let the relative abundance of ^{25}Mg be x%. Since the sum of the abundances of all isotopes is 100%, the abundance of ^{26}Mg is (21.0 - x)%. Using the relative atomic mass formula: A_r = [ (79.0 * 24.0) + (x * 25.0) + ((21.0 - x) * 26.0) ] / 100 = 24.32. Multiply by 100: 1896.0 + 25.0x + 546.0 - 26.0x = 2432. This simplifies to 2442.0 - x = 2432, which gives x = 10.0%. Therefore, the relative abundance of ^{25}Mg is 10.0%.

1 mark: Correctly calculates the relative abundance of ^{25}Mg to be 10.0% (Option A).

1. Calculate the molar mass of anhydrous iron(II) sulfate, FeSO_4: M(FeSO_4) = 55.8 + 32.1 + (4 * 16.0) = 151.9 g mol^{-1}. 2. Calculate the number of moles of anhydrous FeSO_4 produced: n(FeSO_4) = 3.04 g / 151.9 g mol^{-1} = 0.0200 mol. 3. Calculate the mass of water lost during heating: Mass of H_2O = 5.56 g - 3.04 g = 2.52 g. 4. Calculate the number of moles of water lost: n(H_2O) = 2.52 g / 18.0 g mol^{-1} = 0.140 mol. 5. Determine the value of x: x = n(H_2O) / n(FeSO_4) = 0.140 / 0.0200 = 7. Thus, the formula is FeSO_4 . 7H_2O.

1 mark: Correctly calculates x = 7 (Option C).

In a propagation step, a free radical reacts with a stable molecule to produce a new free radical and a new stable molecule. Option A and Option D represent termination steps, because two free radicals combine to form a stable molecule. Option B is incorrect because hydrogen radicals (H^\bullet) are highly unstable and do not form in this reaction; instead, Cl^\bullet abstracts a hydrogen atom to form HCl and a methyl radical. Option C represents a valid propagation step where a methyl radical reacts with a chlorine molecule to form chloromethane and regenerate a chlorine radical.

1 mark: Correctly identifies C as the propagation step.

Bond polarity is determined by the difference in electronegativity between the two bonded atoms. The greater the difference in electronegativity, the more polar the covalent bond. Fluorine is the most electronegative element in the periodic table (electronegativity of 4.0), whereas carbon has an electronegativity of 2.5. The electronegativity differences are: C-H: 2.5 - 2.1 = 0.4; C-Cl: 3.0 - 2.5 = 0.5; C-O: 3.5 - 2.5 = 1.0; C-F: 4.0 - 2.5 = 1.5. Therefore, the C-F bond is the most polar.

1 mark: Correctly selects B as the most polar bond.

For an alkene to exhibit E/Z stereoisomerism, both carbon atoms of the double bond must be attached to two different groups. In 2-methylbut-2-ene, the carbon on the left is bonded to two identical methyl groups, so no E/Z stereoisomerism is possible. In but-1-ene, the terminal carbon of the double bond is bonded to two identical hydrogen atoms, so no E/Z stereoisomerism is possible. In 2,3-dimethylbut-2-ene, both double-bonded carbons are attached to two identical methyl groups, so no E/Z stereoisomerism is possible. In pent-2-ene, one double-bonded carbon is bonded to -H and -CH_3 (two different groups), and the other double-bonded carbon is bonded to -H and -CH_2CH_3 (two different groups). Therefore, it exhibits E/Z stereoisomerism.

1 mark: Correctly identifies D as the alkene exhibiting E/Z stereoisomerism.

Looking at the successive ionization energies of \(X\), 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 being removed from an inner quantum shell, meaning there are three electrons in the outer shell of \(X\). Thus, \(X\) forms a stable \(X^{3+}\) ion. Since the oxide ion is \(O^{2-}\), the compound formed between them is \(X_2O_3\).

1 mark: Correctly identifies that the large jump between the 3rd and 4th ionization energies indicates a 3+ ion and matches it to the formula \(X_2O_3\).

The molar mass of anhydrous \(\text{MgCl}_2 = 24.3 + 2 \times 35.5 = 95.3\text{ g mol}^{-1}\). The moles of anhydrous \(\text{MgCl}_2 = 1.90 / 95.3 = 0.01994\text{ mol}\). The mass of water lost = \(4.06 - 1.90 = 2.16\text{ g}\). The moles of \(\text{H}_2\text{O} = 2.16 / 18.0 = 0.120\text{ mol}\). The mole ratio of water to anhydrous salt is \(x = 0.120 / 0.01994 = 6.02\), which rounds to the nearest whole number 6.

1 mark: Correctly calculates moles of anhydrous salt and water to deduce the value of \(x = 6\).

We determine the number of bonding pairs of electrons around the central atom in each molecule: \(\text{SF}_4\) has 4 bonding pairs (and 1 lone pair on S); \(\text{BF}_3\) has 3 bonding pairs; \(\text{H}_2\text{S}\) has 2 bonding pairs (and 2 lone pairs on S); \(\text{PCl}_5\) has 5 bonding pairs (with 0 lone pairs on P). Therefore, \(\text{PCl}_5\) contains the greatest number of bonding pairs.

1 mark: Correctly identifies that \(\text{PCl}_5\) has 5 bonding pairs, which is the largest among the options.

In the free radical chlorination of methane, the propagation steps must involve a radical reactant and a radical product. The two main propagation steps are: 1) \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) and 2) \(\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\). Therefore, option C is a valid propagation step. Option A is initiation, option B does not occur because the formation of a hydrogen radical is energetically unfavorable, and option D is a termination step.

1 mark: Correctly identifies option C as a propagation step and rejects initiation, termination, or non-viable steps.

The structural isomers of alkene with the molecular formula \(\text{C}_5\text{H}_{10}\) are: 1) pent-1-ene, 2) pent-2-ene, 3) 2-methylbut-1-ene, 4) 3-methylbut-1-ene, and 5) 2-methylbut-2-ene. Cycloalkanes are excluded because the question specifies alkenes, and stereoisomers (cis/trans) are excluded by the question instruction.

1 mark: Correctly identifies all 5 unique structural alkene isomers.

Let \(y\) be the fractional abundance of \({^7}\text{Li}\). The fractional abundance of \({^6}\text{Li}\) is \(1 - y\). Setting up the equation for relative atomic mass: \(6.015(1 - y) + 7.016y = 6.941\). Expanding and solving gives \(6.015 + 1.001y = 6.941\), so \(1.001y = 0.926\) and \(y = 0.926 / 1.001 \approx 0.925\). This corresponds to a percentage abundance of 92.5%.

1 mark: Correctly sets up the isotopic abundance equation and solves for the percentage of \({^7}\text{Li}\).

Covalent character in an ionic compound is maximized when there is high polarization of the anion by the cation. According to Fajan's rules, this occurs with a small, highly charging cation and a large, polarizable anion. \(\text{Li}^+\) is much smaller than \(\text{Cs}^+\), so it has greater polarizing power. \(\text{I}^-\) is much larger than \(\text{F}^-\), so it is much more polarizable. Thus, \(\text{LiI}\) has the greatest degree of covalent character.

1 mark: Correctly applies Fajan's rules to identify \(\text{LiI}\) as having the greatest covalent character.

The formation of a mixture of major and minor products during the addition of \(\text{HBr}\) indicates that the reactant alkene is unsymmetrical. For but-1-ene (\(\text{CH}_2=\text{CHCH}_2\text{CH}_3\)), addition of \(\text{H}^+\) to C1 forms a secondary carbocation, leading to 2-bromobutane as the major product, whereas addition to C2 forms a primary carbocation, leading to 1-bromobutane as the minor product. But-2-ene is a symmetrical alkene and only yields 2-bromobutane (with no minor product). 2-Methylpropene would yield 2-bromo-2-methylpropane as the major product. Cyclobutene would yield bromocyclobutane.

1 mark: Correctly identifies that but-1-ene is unsymmetrical and yields both 2-bromobutane and 1-bromobutane in major/minor proportions.

The largest jump in successive ionization energies occurs between the \(5\text{th}\) and the \(6\text{th}\) ionization energies (from \(6274\) to \(21268 \text{ kJ mol}^{-1}\)). This indicates that the \(6\text{th}\) electron is being removed from an inner quantum shell, which is closer to the nucleus and less shielded. Therefore, element \(X\) has 5 valence electrons and belongs to Group 15 (Group 5). In Period 3, this element is phosphorus, which forms the stable hydride \(\text{PH}_3\). This corresponds to the general formula \(\text{XH}_3\).

Correct option selected: B (1 mark)

1. Calculate the number of moles of \(\text{MnO}_2\):
\(\text{Moles of MnO}_2 = \frac{4.35\text{ g}}{87.0\text{ g mol}^{-1}} = 0.050\text{ mol}\).

2. Determine moles of \(\text{Cl}_2\) using the stoichiometry from the equation:
The molar ratio of \(\text{MnO}_2 : \text{Cl}_2\) is \(1:1\).
Therefore, \(\text{moles of Cl}_2 = 0.050\text{ mol}\).

3. Calculate the volume of \(\text{Cl}_2\) gas at r.t.p.:
\(\text{Volume} = 0.050\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\).

Correct option selected: C (1 mark)

All three species have four electron pairs around the central nitrogen atom, arranged in a tetrahedral-like geometry:
- \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs, giving a regular tetrahedral shape with a bond angle of \(109.5^\circ\).
- \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair. The greater repulsion from the lone pair reduces the bond angle to approximately \(107^\circ\).
- \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs. The strong repulsion between the two lone pairs reduces the bond angle further to approximately \(104.5^\circ\).

Therefore, the order of decreasing bond angle is \(\text{NH}_4^+ > \text{NH}_3 > \text{NH}_2^-\).

Correct option selected: A (1 mark)

The reaction proceeds via electrophilic addition. Addition of \(\text{H}^+\) to the double bond of 2-methylbut-2-ene can form two possible carbocation intermediates:
1. A secondary carbocation: \(\text{(CH}_3)_2\text{CH}-\text{C}^+\text{HCH}_3\)
2. A tertiary carbocation: \(\text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\)

A tertiary carbocation is more stable than a secondary carbocation due to the electron-donating inductive effect of the three alkyl groups attached to the positively charged carbon atom. The reaction pathway via the more stable tertiary carbocation intermediate is favoured, resulting in the major product being 2-bromo-2-methylbutane.

Correct option selected: A (1 mark)

Detailed step-by-step solution:

(a) Experimental procedure:
1. Weigh the empty, clean crucible with its lid.
2. Add the hydrated iron(II) sulfate and reweigh the crucible, lid, and salt.
3. Heat the crucible gently at first (with the lid slightly ajar to allow water vapour to escape), then strongly for about 5 minutes.
4. Allow to cool, reweigh, and repeat the heating, cooling, and weighing process until a constant mass is obtained.

(b) Calculations:
- Mass of hydrated salt = \(23.41 - 18.45 = 4.96\text{ g}\)
- Mass of anhydrous residue (\(\text{FeSO}_4\)) = \(21.16 - 18.45 = 2.71\text{ g}\)
- Mass of water lost = \(4.96 - 2.71 = 2.25\text{ g}\)
- Moles of water = \(2.25 / 18.0 = 0.125\text{ mol}\)
- Molar mass of \(\text{FeSO}_4 = 55.8 + 32.1 + (16.0 \times 4) = 151.9\text{ g mol}^{-1}\)
- Moles of \(\text{FeSO}_4 = 2.71 / 151.9 = 0.01784\text{ mol}\)
- Ratio \(x = 0.125 / 0.01784 = 7.007\)
- Since \(x\) must be an integer, \(x = 7\).

(c) Heating to constant mass ensures that all of the water of crystallisation has been completely evaporated. If not all water is removed, the mass loss is smaller than it should be, resulting in an underestimated value of \(x\).

(d) The calculated value of \(x\) would be higher than the true value. This is because the loss of gaseous sulfur oxides would increase the apparent mass loss, making it appear that a larger mass of water had been lost from the crystal lattice.

(e) Use crucible tongs to handle the hot crucible to avoid severe burns, or place the hot crucible on a heatproof mat to prevent damage to the bench surface.

**Part (a) [4 marks]**
- **M1**: Weigh the empty crucible + lid and then with the hydrated salt. (1)
- **M2**: Heat gently at first, then strongly (with lid slightly open). (1)
- **M3**: Allow to cool before weighing. (1)
- **M4**: Repeat heating, cooling, and weighing until the mass remains constant. (1)

**Part (b) [5 marks]**
- **M1**: Calculate mass of \(\text{FeSO}_4\) (\(2.71\text{ g}\)) and water lost (\(2.25\text{ g}\)). (1)
- **M2**: Calculate moles of water = \(0.125\text{ mol}\). (1)
- **M3**: Calculate molar mass of \(\text{FeSO}_4 = 151.9\text{ g mol}^{-1}\) and moles of \(\text{FeSO}_4 = 0.01784\text{ mol}\). (1)
- **M4**: State the ratio \(n(\text{H}_2\text{O}) / n(\text{FeSO}_4) = 0.125 / 0.01784 = 7.007\). (1)
- **M5**: Give the final answer \(x = 7\) (must be an integer). (1)

**Part (c) [2 marks]**
- **M1**: To ensure all the water of crystallisation has been removed. (1)
- **M2**: To prevent underestimating the mass of water lost / value of \(x\). (1)

**Part (d) [2 marks]**
- **M1**: Calculated value of \(x\) would be higher / larger. (1)
- **M2**: Because gaseous sulfur oxides escape, increasing the total mass lost, which is assumed to be only water. (1)

**Part (e) [2 marks]**
- **M1**: Use crucible tongs (or equivalent suitable safety measure like heatproof mat). (1)
- **M2**: Because the crucible is extremely hot and can cause burns (or to protect the laboratory bench/balance). (1)
*Reject: Wearing lab coat / gloves without specifying heatproof or tongs.*

Detailed step-by-step solution:

(a) Procedure:
1. Place \(50.0\text{ cm}^3\) of the copper(II) sulfate solution in a polystyrene cup placed inside a beaker for stability.
2. Measure and record the temperature of the solution every minute for 3 minutes.
3. At the 4th minute, add the weighed sample of zinc powder and stir thoroughly, but do not record the temperature at this minute.
4. Continue recording the temperature every minute from the 5th minute to about 10 minutes, then plot a graph of temperature against time to find the maximum temperature change by extrapolation.

(b) Calculation:
- \(\Delta T = 31.8 - 19.5 = 12.3\text{ }^\circ\text{C}\) (or \(12.3\text{ K}\))
- Mass of solution \(m = 50.0\text{ g}\)
- Heat transferred \(q = m c \Delta T = 50.0 \times 4.18 \times 12.3 = 2570.7\text{ J} = 2.5707\text{ kJ}\)
- Moles of \(\text{CuSO}_4 = \text{volume} \times \text{concentration} = 0.0500 \times 0.200 = 0.0100\text{ mol}\)
- Zinc is in excess, so copper(II) sulfate is the limiting reagent.
- \(\Delta H = -\frac{q}{n} = -\frac{2.5707\text{ kJ}}{0.0100\text{ mol}} = -257.07\text{ kJ mol}^{-1}\)
- Rounded to 3 significant figures: \(-257\text{ kJ mol}^{-1}\).

(c) Improvements to reduce heat loss:
1. Use a lid on the polystyrene cup.
2. Use a double-walled polystyrene cup (nest one cup inside another) to increase insulation.

(d) Prediction and justification:
- Temperature rise \(\Delta T\) will be \(24.6\text{ }^\circ\text{C}\) (i.e. double the original temperature rise).
- The concentration of \(\text{CuSO}_4\) is doubled, so the number of moles of \(\text{CuSO}_4\) reacting is doubled (\(0.0200\text{ mol}\)).
- Since \(\Delta H\) is constant, the energy released (\(q\)) will also double.
- The volume (and thus the mass) of the solution remains constant at \(50.0\text{ g}\), so the temperature rise must double because \(\Delta T = \frac{q}{m c}\).

**Part (a) [4 marks]**
- **M1**: Place \(50.0\text{ cm}^3\) of copper(II) sulfate solution into a polystyrene cup. (1)
- **M2**: Measure and record the temperature of the solution at regular intervals (e.g. every minute) for several minutes before adding the zinc. (1)
- **M3**: Add the zinc at a specific minute (e.g. 4th minute), stir continuously. (1)
- **M4**: Measure temperature at regular intervals after addition and plot temperature against time (to extrapolate back to the mixing time to find the maximum temperature rise). (1)

**Part (b) [5 marks]**
- **M1**: Calculate temperature change \(\Delta T = 12.3\text{ }^\circ\text{C}\) (or \(\text{K}\)). (1)
- **M2**: Calculate heat released \(q = 50.0 \times 4.18 \times 12.3 = 2570.7\text{ J}\) (or \(2.57\text{ kJ}\)). (1)
- **M3**: Calculate moles of copper(II) sulfate = \(0.0100\text{ mol}\). (1)
- **M4**: Divide \(q\) by moles of copper(II) sulfate (e.g. \(2570.7 / 0.0100 = 257070\text{ J mol}^{-1}\)). (1)
- **M5**: Correct final answer of \(-257\text{ kJ mol}^{-1}\) (with negative sign and 3 significant figures). (1)

**Part (c) [2 marks]**
- **M1**: Add a lid to the polystyrene cup. (1)
- **M2**: Use a beaker/cotton wool or a second cup (double-cup) to provide extra insulation. (1)

**Part (d) [4 marks]**
- **M1**: State that \(\Delta T\) will double / be \(24.6\text{ }^\circ\text{C}\). (1)
- **M2**: Explain that the moles of reactants/copper(II) ions are doubled. (1)
- **M3**: State that the heat energy released (\(q\)) is doubled. (1)
- **M4**: Explain that the mass of solution remains the same, so \(\Delta T\) doubles (from \(\Delta T = q / mc\)). (1)

Detailed step-by-step solution:

(a) Preparation and Separation:
1. Place \(10.0\text{ g}\) of cyclohexanol in a round-bottomed flask and carefully add concentrated phosphoric acid (catalyst) and some anti-bumping granules.
2. Set up the flask for distillation. Heat the mixture gently using a heating mantle (or water bath/sand bath) because cyclohexene is highly flammable.
3. Collect the distillate boiling below \(95\text{ }^\circ\text{C}\) in a receiver cooled in an ice bath.
4. Transfer the distillate (which contains both aqueous and organic layers) into a separating funnel.
5. Allow the layers to separate, then run off and discard the lower aqueous layer, retaining the upper organic layer of crude cyclohexene.

(b) Calculation:
- Moles of cyclohexanol = \(10.0 / 100.1 = 0.0999\text{ mol}\)
- Since the stoichiometry is 1:1, theoretical moles of cyclohexene = \(0.0999\text{ mol}\)
- Theoretical mass of cyclohexene = \(0.0999 \times 82.1 = 8.202\text{ g}\)
- Actual mass of cyclohexene = \(5.40\text{ g}\)
- Percentage yield = \((5.40 / 8.202) \times 100 = 65.84\%\)
- Rounded to 3 significant figures: \(65.8\%\) (or \(66\%\)).

(c) Test for unsaturation:
- Reagent: Bromine water (or bromine in an organic solvent).
- Observation before: Orange/yellow/brown solution.
- Observation after: Decolourises (becomes colourless).

(d) Purification:
1. To remove acid: Transfer the crude cyclohexene to a separating funnel, add sodium hydrogencarbonate solution (or sodium carbonate solution), shake, and release pressure/gas regularly.
2. Separate the aqueous layer and discard it.
3. To remove water: Transfer the organic layer to a conical flask, add an anhydrous inorganic salt (such as anhydrous calcium chloride or anhydrous sodium sulfate), and swirl.
4. Leave until the liquid becomes completely clear, then filter or decant the dry cyclohexene into a clean flask and perform a final distillation to obtain pure cyclohexene.

**Part (a) [5 marks]**
- **M1**: Mix cyclohexanol and concentrated phosphoric acid with anti-bumping granules in a flask. (1)
- **M2**: Heat the mixture using a heating mantle / sand bath (safe heating source). (1)
- **M3**: Distil the mixture and collect the distillate (in a cooled receiver). (1)
- **M4**: Transfer distillate to a separating funnel. (1)
- **M5**: Separate the layers and discard the aqueous layer / retain the organic layer. (1)

**Part (b) [3 marks]**
- **M1**: Calculate moles of cyclohexanol = \(10.0 / 100.1 = 0.0999\text{ mol}\). (1)
- **M2**: Calculate theoretical mass of cyclohexene = \(0.0999 \times 82.1 = 8.202\text{ g}\) (or actual moles = \(5.40 / 82.1 = 0.06577\text{ mol}\)). (1)
- **M3**: Calculate percentage yield = \((5.40 / 8.202) \times 100 = 65.8\%\) (accept \(66\%\)). (1)

**Part (c) [3 marks]**
- **M1**: Reagent: bromine water / bromine. (1)
- **M2**: Observation: decolourises / orange-brown to colourless. (1)
- **M3**: Explains this shows the presence of \(\text{C}=\text{C}\) double bond / unsaturation. (1)
*(Alternatively, use acidified potassium manganate(VII), purple to colourless)*

**Part (d) [4 marks]**
- **M1**: Wash the organic layer with sodium hydrogencarbonate (or sodium carbonate) solution to neutralise/remove phosphoric acid. (1)
- **M2**: Release pressure periodically when using the separating funnel. (1)
- **M3**: Add an anhydrous salt (e.g. \(\text{CaCl}_2\) or \(\text{Na}_2\text{SO}_4\)) to dry the product/remove water. (1)
- **M4**: Decant/filter the liquid and re-distil to obtain pure dry cyclohexene. (1)

Detailed step-by-step solution:

(a) Preparation of diluted solution:
1. Pipette exactly \(10.0\text{ cm}^3\) of the commercial bleach using a volumetric pipette and transfer it to a clean \(250.0\text{ cm}^3\) volumetric flask.
2. Add deionised water to the flask, swirling to mix, until the liquid is close to the graduation mark.
3. Use a teat pipette to add deionised water dropwise until the bottom of the meniscus is exactly on the graduation line.
4. Invert the flask several times to ensure a completely homogeneous solution.

(b) Calculations:
- Moles of thiosulfate used in titration:
\(n(\text{S}_2\text{O}_3^{2-}) = \text{volume} \times \text{concentration} = 0.02145\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.145 \times 10^{-3}\text{ mol}\)

- From the second equation:
\(n(\text{I}_2) = \frac{1}{2} n(\text{S}_2\text{O}_3^{2-}) = \frac{2.145 \times 10^{-3}}{2} = 1.0725 \times 10^{-3}\text{ mol}\)

- From the first equation, the mole ratio of \(\text{ClO}^-\text{ to I}_2\) is 1:1, so:
\(n(\text{ClO}^-) \text{ in } 25.0\text{ cm}^3 = 1.0725 \times 10^{-3}\text{ mol}\)

- Scale up to the total volume of diluted bleach (\(250.0\text{ cm}^3\)):
\(n(\text{ClO}^-) \text{ in } 250.0\text{ cm}^3 = 1.0725 \times 10^{-3} \times \frac{250.0}{25.0} = 1.0725 \times 10^{-2}\text{ mol}\)

- Since the dilution was \(10.0\text{ cm}^3\) to \(250.0\text{ cm}^3\), this mole amount was originally in the \(10.0\text{ cm}^3\) of commercial bleach:
\(c(\text{NaClO}) = \frac{1.0725 \times 10^{-2}\text{ mol}}{0.0100\text{ dm}^3} = 1.0725\text{ mol dm}^{-3}\)

- To 3 significant figures, the concentration is \(1.07\text{ mol dm}^{-3}\).

(c) Indicator and end-point:
- Indicator: Starch solution.
- Timing: Added near the end-point when the solution becomes pale straw-yellow.
- Colour change: Blue-black to colourless.

(d) Why starch is added late:
If starch is added too early (when the iodine concentration is high), it forms an insoluble or very stable starch-iodine complex. This prevents the iodine from reacting readily with the thiosulfate, leading to an inaccurate/sluggish end-point or underestimation of the titre.

**Part (a) [4 marks]**
- **M1**: Use a volumetric pipette to measure \(10.0\text{ cm}^3\) of commercial bleach. (1)
- **M2**: Transfer to a \(250.0\text{ cm}^3\) volumetric flask. (1)
- **M3**: Add deionised water up to the mark (meniscus on the line). (1)
- **M4**: Invert several times to mix thoroughly. (1)

**Part (b) [6 marks]**
- **M1**: Calculate moles of thiosulfate: \(0.02145 \times 0.100 = 2.145 \times 10^{-3}\text{ mol}\). (1)
- **M2**: Relate moles of thiosulfate to moles of iodine (1:2 ratio): \(1.0725 \times 10^{-3}\text{ mol}\). (1)
- **M3**: Relate moles of iodine to moles of hypochlorite in \(25.0\text{ cm}^3\) (1:1 ratio): \(1.0725 \times 10^{-3}\text{ mol}\). (1)
- **M4**: Multiply by 10 to find moles of hypochlorite in \(250.0\text{ cm}^3\): \(1.0725 \times 10^{-2}\text{ mol}\). (1)
- **M5**: Divide by \(0.0100\text{ dm}^3\) (the original volume of bleach). (1)
- **M6**: Correct final concentration of \(1.07\text{ mol dm}^{-3}\) (3 sig fig). (1)

**Part (c) [3 marks]**
- **M1**: State indicator is starch solution. (1)
- **M2**: Added when the solution is pale yellow / straw-coloured. (1)
- **M3**: Colour change from blue-black to colourless. (1)
*Reject: Clear instead of colourless.*

**Part (d) [2 marks]**
- **M1**: Starch forms an insoluble / stable complex with iodine at high concentration. (1)
- **M2**: This leads to a sluggish/inaccurate end-point (or some iodine remains bound). (1)

Carbon dioxide (\(CO_2\)), methane (\(CH_4\)), and tetrachloromethane (\(CCl_4\)) are highly symmetrical molecules, which means their individual bond dipoles cancel each other out, resulting in a net dipole moment of zero. Chloromethane (\(CH_3Cl\)) is asymmetrical and polar, so it experiences significant permanent dipole-dipole intermolecular attractions.

Award 1.18 marks for the correct answer. Reject all other options.

Using Hess's Law based on enthalpy changes of combustion:
\(\Delta H^\ominus_f = 2 \times \Delta H^\ominus_c[C(s)] + 3 \times \Delta H^\ominus_c[H_2(g)] - \Delta H^\ominus_c[C_2H_6(g)]\)

\(\Delta H^\ominus_f = 2(-394) + 3(-286) - (-1560)\)
\(\Delta H^\ominus_f = -788 - 858 + 1560 = -86 \text{ kJ mol}^{-1}\)

Award 1.18 marks for the correct answer. Reject all other options.

The solubility of Group 2 hydroxides increases down the group from magnesium hydroxide to barium hydroxide. Therefore, barium hydroxide is the most soluble among the given options.

Award 1.18 marks for the correct answer. Reject all other options.

The peak at \(1720\text{ cm}^{-1}\) indicates a carbonyl group (\(C=O\)), which rules out propan-1-ol and prop-2-en-1-ol (which have \(O-H\) groups). The peak at \(m/z = 29\) in the mass spectrum corresponds to the \(CHO^+\) or \(CH_3CH_2^+\) fragment ion, which can be formed by the fragmentation of propanal (\(CH_3CH_2CHO\)). Propanone (\(CH_3COCH_3\)) does not readily fragment to give an intensive peak at \(m/z = 29\); its major acylium fragment peak would appear at \(m/z = 43\) (\(CH_3CO^+\)).

Award 1.18 marks for the correct answer. Reject all other options.

As the temperature increases, the particles gain kinetic energy, shifting the overall distribution curve to the right (the position of the peak moves to a higher energy). To keep the total area under the curve constant (since the total number of particles remains unchanged), the curve must broaden and flatten, meaning the maximum peak height decreases.

Award 1.18 marks for the correct answer. Reject all other options.

Potassium iodide is oxidized by concentrated sulfuric acid, and the hydrogen sulfate/sulfuric acid is reduced to several products. The reduction products of sulfuric acid include sulfur dioxide (\(SO_2\), a colourless acidic gas), sulfur (\(S\), a yellow solid), and hydrogen sulfide (\(H_2S\), a toxic gas with a rotten-egg odour). Iodine (\(I_2\)) is the oxidation product of iodide ions, forming a grey-black solid or purple vapour. Hence, the yellow solid reduction product is sulfur (\(S\)).

Award 1.18 marks for the correct answer. Reject all other options.

The rate of hydrolysis of halogenoalkanes depends on the carbon-halogen bond strength and the mechanism. The \(C-I\) bond is weaker than the \(C-Br\) and \(C-Cl\) bonds, making iodoalkanes more reactive. Furthermore, tertiary halogenoalkanes (like 2-iodo-2-methylpropane) react significantly faster than primary halogenoalkanes (like 1-iodobutane) because they undergo \(S_N1\) substitution via a highly stable tertiary carbocation intermediate.

Award 1.18 marks for the correct answer. Reject all other options.

1. Find temperature change: \(\Delta T = 26.3 - 19.5 = 6.8\text{ K}\).
2. Find total mass of solution: \(m = 50.0 + 50.0 = 100.0\text{ g}\).
3. Calculate energy released: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.8424\text{ kJ}\).
4. Calculate the amount of water formed: \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).
5. Calculate standard enthalpy of neutralisation: \(\Delta H = -\frac{q}{n} = -\frac{2.8424\text{ kJ}}{0.0500\text{ mol}} = -56.8\text{ kJ mol}^{-1}\).

Award 1.18 marks for the correct answer. Reject all other options.

Butane is a non-polar hydrocarbon and only has London forces. Propanal is a polar molecule with permanent dipole-dipole forces in addition to London forces. Propan-1-ol has hydrogen bonding between its molecules, which is the strongest type of intermolecular force. Therefore, the order of increasing boiling temperature is butane < propanal < propan-1-ol.

1 mark for the correct option selection B. Reject other options because they do not correctly rank the intermolecular force strengths of London forces < permanent dipole-dipole < hydrogen bonding.

Going down Group 2 from beryllium to barium, the solubility of their hydroxides increases (magnesium hydroxide is sparingly soluble, barium hydroxide is highly soluble), whereas the solubility of their sulfates decreases (magnesium sulfate is highly soluble, barium sulfate is insoluble).

1 mark for the correct option B. Reject options A, C, and D as they display incorrect directions for the solubility trends of either the hydroxides or sulfates.

The molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) represents a saturated acyclic alcohol. Compound \(\text{X}\) forms a precipitate with 2,4-DNPH, which means it must be a carbonyl compound (either an aldehyde or a ketone). However, compound \(\text{X}\) does not react with sodium carbonate solution, indicating that it is not a carboxylic acid. If the starting material were a primary alcohol (options A or D), heating under reflux with acidified potassium dichromate(VI) would oxidize it to a carboxylic acid. If the starting material was a tertiary alcohol (option C), it would not undergo oxidation. Only the secondary alcohol, butan-2-ol (option B), is oxidized under reflux to a ketone (butanone), which gives a positive 2,4-DNPH test and is resistant to further oxidation to a carboxylic acid.

1 mark for identifying B as the correct secondary alcohol. Reject options A, C, and D as they would yield carboxylic acids (which would react with sodium carbonate) or show no oxidation at all.

Total mass of solution \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\).
Heat energy released \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 6.50\ ^\circ\text{C} = 2717\text{ J} = 2.717\text{ kJ}\).
Moles of reactants: \(n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\), \(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
Moles of water formed in neutralization = \(0.0500\text{ mol}\).
\(\Delta H_{\text{neut}} = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.34\text{ kJ mol}^{-1} \approx -54.3\text{ kJ mol}^{-1}\).

1 mark for the correct answer A. Reject B, C, D which result from incorrect arithmetic operations (e.g., failing to sum the volumes to find the mass of the solution, or incorrect mole calculations).

Decreasing the volume of the container at constant temperature increases the total pressure. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium towards the side with fewer gas moles to reduce the pressure. There are 3 moles of gas on the left-hand side and 2 moles of gas on the right-hand side, so the position of equilibrium shifts to the right, increasing the yield of \(\text{NO}_2\).

1 mark for selecting B. Reject A as increasing temperature shifts the equilibrium to the endothermic side (left). Reject C as catalysts do not affect equilibrium position. Reject D as removing reactants shifts equilibrium to the left.

Viscosity is determined by the strength and number of intermolecular forces. Propane-1,2,3-triol (glycerol) has three hydroxyl (\(\text{-OH}\)) groups per molecule, which allows it to form an extensive network of intermolecular hydrogen bonds. This strong network of hydrogen bonds resists flow, making it significantly more viscous than propan-1-ol, propan-2-ol, or propane-1,2-diol.

1 mark for selecting the correct compound C. Reject other choices because they contain fewer hydroxyl groups and therefore exhibit weaker overall intermolecular hydrogen bonding.

When concentrated sulfuric acid reacts with solid potassium iodide, iodide ions act as strong reducing agents. Hydrogen iodide gas is produced first. Iodide then reduces sulfuric acid to sulfur dioxide, sulfur, and hydrogen sulfide, and is itself oxidized to iodine. Since there is no source of chlorine in the reactants, chlorine gas (\(\text{Cl}_2\)) cannot be formed.

1 mark for option D. Reject A, B, and C as these are all known products formed from the reduction of sulfuric acid by iodide ions and the associated acid-base reactions.

Tertiary halogenoalkanes react predominantly via the \(\text{S}_\text{N}1\) mechanism. The only tertiary halogenoalkane in the options is 2-bromo-2-methylpropane. When 2-bromo-2-methylpropane is heated with ethanolic potassium hydroxide, elimination of HBr occurs. Because all three methyl groups are equivalent, elimination can only yield one alkene product, 2-methylpropene. Thus, both conditions are satisfied.

1 mark for selecting C. Reject A and D because they are primary halogenoalkanes and react predominantly via \(\text{S}_\text{N}2\). Reject B because its elimination reaction gives a mixture of alkene isomers (but-1-ene, cis-but-2-ene, and trans-but-2-ene).

1. **Identify the class of alcohol \(Y\)**: Since alcohol \(Y\) is resistant to oxidation by acidified potassium dichromate(VI) (no color change is observed), it must be a tertiary (\(3^\circ\)) alcohol.

2. **Determine the structure of \(Y\)**: The only tertiary alcohol with the molecular formula \(\text{C}_4\text{H}_9\text{OH}\) is 2-methylpropan-2-ol, \(\text{(CH}_3\text{)}_3\text{COH}\).

3. **Determine the structure of \(X\)**: Since \(X\) undergoes nucleophilic substitution with aqueous hydroxide ions to form \(Y\), the halogen atom must be on the same tertiary carbon. Therefore, \(X\) is 2-bromo-2-methylpropane, \(\text{(CH}_3\text{)}_3\text{CBr}\).

Correct Answer: D (1.18 marks)

Distractor Analysis:
- **A (1-bromobutane)**: Incorrect. This is a primary halogenoalkane which hydrolyzes to butan-1-ol (a primary alcohol). Primary alcohols are easily oxidized to aldehydes and carboxylic acids, which would turn the green-orange acidified dichromate green.
- **B (2-bromobutane)**: Incorrect. This is a secondary halogenoalkane which hydrolyzes to butan-2-ol (a secondary alcohol). Secondary alcohols are easily oxidized to ketones, which would turn the acidified dichromate green.
- **C (1-bromo-2-methylpropane)**: Incorrect. This is a primary halogenoalkane which hydrolyzes to 2-methylpropan-1-ol (a primary alcohol). This would be oxidized, changing the color of the dichromate solution.

1. **Identify the class of alcohol \(Y\)**: Since alcohol \(Y\) is resistant to oxidation by acidified potassium dichromate(VI) (no color change is observed), it must be a tertiary (\(3^\circ\)) alcohol.

2. **Determine the structure of \(Y\)**: The only tertiary alcohol with the molecular formula \(\text{C}_4\text{H}_9\text{OH}\) is 2-methylpropan-2-ol, \(\text{(CH}_3\text{)}_3\text{COH}\).

3. **Determine the structure of \(X\)**: Since \(X\) undergoes nucleophilic substitution with aqueous hydroxide ions to form \(Y\), the halogen atom must be on the same tertiary carbon. Therefore, \(X\) is 2-bromo-2-methylpropane, \(\text{(CH}_3\text{)}_3\text{CBr}\).

Correct Answer: D (1.18 marks)

Distractor Analysis:
- **A (1-bromobutane)**: Incorrect. This is a primary halogenoalkane which hydrolyzes to butan-1-ol (a primary alcohol). Primary alcohols are easily oxidized to aldehydes and carboxylic acids, which would turn the green-orange acidified dichromate green.
- **B (2-bromobutane)**: Incorrect. This is a secondary halogenoalkane which hydrolyzes to butan-2-ol (a secondary alcohol). Secondary alcohols are easily oxidized to ketones, which would turn the acidified dichromate green.
- **C (1-bromo-2-methylpropane)**: Incorrect. This is a primary halogenoalkane which hydrolyzes to 2-methylpropan-1-ol (a primary alcohol). This would be oxidized, changing the color of the dichromate solution.

Detailed working:

(a) Excess NaOH ensures that methanoic acid (the limiting reagent) reacts completely.

(b) \(q = m c \Delta T\)
\(m = 50.0 + 50.0 = 100.0\text{ g}\)
\(\Delta T = 25.8 - 19.5 = 6.3\text{ }^\circ\text{C} = 6.3\text{ K}\)
\(q = 100.0 \times 4.18 \times 6.3 = 2633.4\text{ J}\)

(c) Moles of \(\text{HCOOH} = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
Moles of \(\text{NaOH} = 0.0500\text{ dm}^3 \times 1.05\text{ mol dm}^{-3} = 0.0525\text{ mol}\).
Since \(\text{HCOOH}\) is limiting, the moles of water formed is \(0.0500\text{ mol}\).

(d) \(\Delta H_{\text{neu}} = -\frac{q}{n} = -\frac{2.6334\text{ kJ}}{0.0500\text{ mol}} = -52.668\text{ kJ mol}^{-1}\). Rounded to 3 s.f., this is \(-52.7\text{ kJ mol}^{-1}\).

(e) Methanoic acid is a weak acid and is only partially ionised in aqueous solution. Energy is absorbed/needed to fully ionise the remaining methanoic acid molecules (break O-H bonds) before neutralisation can occur, which reduces the net energy released.

(f) Absolute uncertainty in temperature difference = \(2 \times 0.1 = \pm 0.2\text{ }^\circ\text{C}\).
Percentage uncertainty = \(\frac{0.2}{6.3} \times 100\% = 3.17\%\) (or 3.2\%).

(g) Put a lid on the cup to reduce heat loss by evaporation, and place the cup inside a larger beaker insulated with cotton wool to reduce heat loss by conduction.

(a) 1 mark: to ensure all methanoic acid is neutralised / complete reaction of the weak acid.

(b) 2 marks:
- 1 mark for correct calculation of temperature rise (6.3 K) and mass (100.0 g).
- 1 mark for calculating energy as 2633.4 J (or 2.63 kJ).

(c) 2 marks:
- 1 mark for showing moles of HCOOH = 0.0500 mol and NaOH = 0.0525 mol.
- 1 mark for identifying that 0.0500 mol of water is formed (since HCOOH is limiting).

(d) 3 marks:
- 1 mark for converting energy to kJ (2.6334 kJ).
- 1 mark for dividing energy by moles of water (2.6334 / 0.0500).
- 1 mark for correct negative sign and units (kJ mol-1) to 3 s.f. (-52.7 kJ mol-1).

(e) 3 marks:
- 1 mark for stating methanoic acid is a weak acid / partially ionised.
- 1 mark for stating energy is needed to ionise the acid / break O-H bonds.
- 1 mark for stating this absorption of energy makes the reaction less exothermic overall.

(f) 2 marks:
- 1 mark for doubling the uncertainty to +/- 0.2 C.
- 1 mark for calculating percentage uncertainty as 3.17% (or 3.2%).

(g) 2 marks:
- 1 mark for using a lid (to prevent heat loss by evaporation/convection).
- 1 mark for wrapping the cup in insulating material / placing in a beaker with cotton wool.

Detailed working:

(a) Starch indicator is used. The colour change at the end-point is blue-black to colourless. It must be added close to the end-point when the solution is pale straw-yellow, because adding it too early causes the starch-iodine complex to precipitate and fail to react reversibly.

(b) In \(\text{ClO}^-\), Cl has an oxidation state of +1. In \(\text{Cl}^-\), Cl is -1. This is a decrease in oxidation number, so reduction. In \(\text{I}^-\), I has an oxidation state of -1. In \(\text{I}_2\), I is 0. This is an increase in oxidation number, so oxidation. Since both oxidation and reduction occur simultaneously, it is a redox reaction.

(c) Moles of \(\text{Na}_2\text{S}_2\text{O}_3 = 0.100\text{ mol dm}^{-3} \times \frac{24.30}{1000}\text{ dm}^3 = 2.43 \times 10^{-3}\text{ mol}\).

(d) From Reaction 2, moles of \(\text{I}_2 = \frac{1}{2} \times \text{moles of }\text{S}_2\text{O}_3^{2-} = 1.215 \times 10^{-3}\text{ mol}\).
From Reaction 1, moles of \(\text{ClO}^- = \text{moles of }\text{I}_2 = 1.215 \times 10^{-3}\text{ mol}\) in \(25.0\text{ cm}^3\).
Moles of \(\text{ClO}^-\) in the \(250.0\text{ cm}^3\) volumetric flask = \(1.215 \times 10^{-3} \times 10 = 1.215 \times 10^{-2}\text{ mol}\).
This amount came from the \(10.0\text{ cm}^3\) sample of undiluted bleach.
Concentration of \(\text{ClO}^-\)
in undiluted bleach = \(\frac{1.215 \times 10^{-2}\text{ mol}}{0.0100\text{ dm}^3} = 1.215\text{ mol dm}^{-3}\).

(e) In \(1.00\text{ dm}^3\) of bleach, moles of \(\text{ClO}^- = 1.215\text{ mol}\).
Moles of equivalent \(\text{Cl}_2 = 1.215\text{ mol}\).
Mass of equivalent \(\text{Cl}_2 = 1.215\text{ mol} \times 70.9\text{ g mol}^{-1} = 86.14\text{ g}\).
Mass of \(1.00\text{ dm}^3\) of bleach = \(1000\text{ cm}^3 \times 1.08\text{ g cm}^{-3} = 1080\text{ g}\).
Percentage of active chlorine = \(\frac{86.14}{1080} \times 100\% = 7.98\%\).
Since \(7.98\% > 4.5\%\), the bleach has a higher active chlorine concentration than the minimum claimed, so the claim is accurate (or conservative).

(a) 3 marks:
- 1 mark for starch.
- 1 mark for blue-black to colourless (reject 'clear').
- 1 mark for adding when the solution is pale yellow / straw-coloured (near the end-point).

(b) 3 marks:
- 1 mark for identifying oxidation numbers of chlorine (from +1 to -1) and stating it is reduced.
- 1 mark for identifying oxidation numbers of iodine (from -1 to 0) and stating it is oxidised.
- 1 mark for stating that redox involves simultaneous reduction and oxidation.

(c) 1 mark: 2.43 x 10^-3 mol (or 0.00243 mol).

(d) 4 marks:
- 1 mark for calculating moles of I2 as 1.215 x 10^-3 mol.
- 1 mark for setting moles of ClO- in 25 cm3 equal to 1.215 x 10^-3 mol.
- 1 mark for scaling up to 250 cm3 to get 1.215 x 10^-2 mol of ClO-.
- 1 mark for dividing by 0.0100 dm3 to show 1.215 mol dm-3.

(e) 4 marks:
- 1 mark for calculating mass of Cl2 in 1 dm3 = 86.14 g (or 86.1 g).
- 1 mark for calculating total mass of 1 dm3 bleach = 1080 g.
- 1 mark for calculating mass percentage of active chlorine = 7.98% (or 8.0%).
- 1 mark for comparing 7.98% with 4.5% and concluding that the claim is correct/exceeded.

Detailed working:

(a) Compound X must be 2-bromo-2-methylpropane. The molecular formula \(\text{C}_4\text{H}_9\text{Br}\) has several isomers. The rate of hydrolysis of halogenoalkanes is in the order tertiary > secondary > primary. The 'almost instant' cream precipitate indicates a tertiary halogenoalkane. Furthermore, the hydrolysis of a tertiary halogenoalkane produces a tertiary alcohol. Alcohol Y (2-methylpropan-2-ol) is a tertiary alcohol. Tertiary alcohols cannot be oxidised by acidified potassium dichromate(VI) because there is no hydrogen atom on the carbon bonded to the hydroxyl group; hence, the solution remains orange. IUPAC name is 2-bromo-2-methylpropane.

(b) \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\).

(c) Halogenoalkanes are insoluble in water, but soluble in ethanol. Silver nitrate is soluble in water. Ethanol acts as a co-solvent, allowing the reactants to mix into a single phase so they can react.

(d) Down Group 7, the size of the halogen atom increases. The C-I bond length is longer than the C-Br bond length, which makes the C-I bond weaker (lower bond enthalpy). Since bond breaking is the rate-determining step in hydrolysis, the weaker C-I bond breaks more quickly than the C-Br bond, resulting in a faster rate of reaction despite the C-I bond being less polar.

(e)(i) Alkene Z is 2-methylpropene. Its skeletal structure consists of a central carbon atom with a double bond to one CH2 group, and two single bonds to methyl groups. (Looking like a 'Y' shape with a double line on one of the branches).

(ii) Bromine has two stable isotopes: \(^{79}\text{Br}\) and \(^{81}\text{Br}\), which occur naturally in an approximately 1:1 ratio. The molecular ion peak at \(m/z = 136\) corresponds to \([\text{C}_4\text{H}_9^{79}\text{Br}]^+\), and the peak at \(m/z = 138\) corresponds to \([\text{C}_4\text{H}_9^{81}\text{Br}]^+\). Since they are present in equal abundance, the heights of these two peaks are equal.

(a) 5 marks:
- 1 mark for stating X is a tertiary halogenoalkane / Y is a tertiary alcohol.
- 1 mark for linking rapid hydrolysis to the stable tertiary carbocation intermediate / tertiary structure.
- 1 mark for linking the orange colour remaining (no oxidation) to Y being a tertiary alcohol.
- 1 mark for drawing/describing the correct structure of 2-bromo-2-methylpropane.
- 1 mark for IUPAC name: 2-bromo-2-methylpropane.

(b) 2 marks:
- 1 mark for correct formulae: Ag+ and Br- to form AgBr.
- 1 mark for correct state symbols: (aq), (aq) and (s).

(c) 1 mark: acts as a mutual / common solvent / allows the immiscible reactants to mix.

(d) 3 marks:
- 1 mark for stating that the C-I bond is weaker / has lower bond enthalpy than the C-Br bond.
- 1 mark for explaining that iodine is a larger atom, resulting in a longer, weaker bond.
- 1 mark for stating that bond enthalpy / bond strength is the overriding factor (not bond polarity) in determining rate.

(e)(i) 1 mark for correct skeletal structure of 2-methylpropene.

(ii) 3 marks:
- 1 mark for stating bromine exists as 79Br and 81Br isotopes.
- 1 mark for stating these isotopes have a 1:1 / equal ratio of abundance.
- 1 mark for matching m/z = 136 to C4H9(79)Br+ and m/z = 138 to C4H9(81)Br+.

Detailed working:

(a) Moles of cyclohexanol reacted = \(12.0\text{ g} / 100.2\text{ g mol}^{-1} = 0.11976\text{ mol}\).
Since the reaction stoichiometry is 1:1, the theoretical yield of cyclohexene is \(0.11976\text{ mol}\).
Theoretical mass of cyclohexene = \(0.11976\text{ mol} \times 82.1\text{ g mol}^{-1} = 9.832\text{ g}\).
Percentage yield = \(\frac{6.20}{9.832} \times 100\% = 63.06\% \approx 63.1\%\).

(b) Atom economy = \(\frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100 = \frac{82.1}{100.2} \times 100 = 81.94\% \approx 81.9\%\).

(c)(i) The bromine water changes colour from orange / yellow / brown to colourless.

(ii) In bromine water, the major product is 2-bromocyclohexan-1-ol. Its structure consists of a six-membered carbon ring (cyclohexane) with a hydroxyl group (\(-\text{OH}\)) and a bromine atom (\(-\text{Br}\)) attached to adjacent carbon atoms. (Accept 1,2-dibromocyclohexane if the student assumes simple bromination).

(iii) Mechanism description:
1. The \(\pi\) electrons in the \(\text{C=C}\) double bond repel the electrons in the \(\text{Br-Br}\) bond, inducing a dipole: \(\text{Br}^{\delta+}-\text{Br}^{\delta-}\).
2. A curly arrow is drawn from the double bond to the \(\text{Br}^{\delta+}\) atom.
3. A concurrent curly arrow is drawn from the \(\text{Br-Br}\) bond to the \(\text{Br}^{\delta-}\) atom, leading to the cleavage of the bond and formation of a bromide ion (\(\text{Br}^-\)).
4. This forms a carbocation intermediate (or a cyclic bromonium ion) and a bromide ion.
5. A curly arrow is drawn from the lone pair of the bromide ion to the positively charged carbon of the carbocation, forming 1,2-dibromocyclohexane.

(d) During addition polymerisation, the double bond of cyclohexene opens up. The repeating unit is a cyclohexane ring where the two carbons that originally shared the double bond are now single-bonded and have single bonds extending outwards to show connection to adjacent monomer units. This can be drawn as a six-membered ring with horizontal bonds extending from carbons 1 and 2, enclosed in square brackets with 'n'.

(a) 3 marks:
- 1 mark for calculating moles of cyclohexanol = 0.120 mol.
- 1 mark for calculating theoretical mass of cyclohexene = 9.83 g.
- 1 mark for calculating percentage yield = 63.1% (accept 63%).

(b) 2 marks:
- 1 mark for setting up ratio: 82.1 / 100.2.
- 1 mark for correct calculation of atom economy = 81.9% (or 82% if using integer mass 82/100).

(c)(i) 1 mark: orange/brown/yellow to colourless (reject 'clear').

(ii) 2 marks:
- 1 mark for correct cyclohexane ring.
- 1 mark for adjacent -OH and -Br groups (accept adjacent -Br and -Br groups for 1,2-dibromocyclohexane).

(iii) 5 marks:
- 1 mark for explaining that the high electron density of the double bond induces a dipole on the Br2 molecule.
- 1 mark for a curly arrow from the double bond to the Br(d+) atom.
- 1 mark for a curly arrow from the Br-Br bond to the Br(d-) atom.
- 1 mark for drawing the correct intermediate (either a carbocation with a + charge on carbon and Br on the adjacent carbon, or a cyclic bromonium ion) AND a bromide ion with a lone pair and negative charge.
- 1 mark for a curly arrow from the lone pair of Br- to the carbocation carbon (or to one of the ring carbons of the bromonium ion).

(d) 2 marks:
- 1 mark for showing a cyclohexane ring with the double bond replaced by a single bond.
- 1 mark for drawing single bonds extending outwards from the two adjacent carbons, showing continuation of the polymer chain, with brackets.

(a) Mean initial temperature = \((19.5 + 19.7) / 2 = 19.6^\circ\text{C}\). (b) Temperature change, \(\Delta T = 26.2 - 19.6 = 6.6^\circ\text{C}\). Total volume of solution = \(25.0 + 25.0 = 50.0\text{ cm}^3\). Mass, \(m = 50.0\text{ g}\). Heat energy released, \(q = m \times c \times \Delta T = 50.0 \times 4.18 \times 6.6 = 1379.4\text{ J}\). (c) \(\text{Moles of HCl} = 0.0250 \times 1.00 = 0.0250\text{ mol}\); \(\text{Moles of NaOH} = 0.0250 \times 1.00 = 0.0250\text{ mol}\). Since the reaction ratio is 1:1, moles of water formed = \(0.0250\text{ mol}\). (d) \(\Delta H_{\text{neut}} = -q / n = -1.3794\text{ kJ} / 0.0250\text{ mol} = -55.176\text{ kJ mol}^{-1}\). To 3 significant figures, this is \(-55.2\text{ kJ mol}^{-1}\). (e) Glass is a better thermal conductor than polystyrene, resulting in greater heat loss to the surroundings. This decreases the maximum temperature reached, leading to a smaller calculated temperature rise and a less exothermic (less negative) calculated enthalpy value. (f) Add a lid to the beaker, or wrap the outer walls of the glass beaker in insulating material such as cotton wool or bubble wrap.

(a) 1 Mark: Correct mean initial temperature of 19.6 [\(^\circ\text{C}\)]. (b) 2 Marks: 1 mark for calculating temperature change \(\Delta T = 6.6\ [^\circ\text{C}]\) and 1 mark for calculating \(q = 1379.4\ [\text{J}]\) (accept 1380 or 1.38 kJ). (c) 1 Mark: Moles of water = 0.0250 [mol]. (d) 2 Marks: 1 mark for correct division of heat by moles, 1 mark for correct value to 3 significant figures with a negative sign (-55.2 [\(\text{kJ mol}^{-1}\)]). (e) 2 Marks: 1 mark for stating that glass allows more heat loss to the surroundings than polystyrene, 1 mark for concluding that the calculated enthalpy value becomes less negative/exothermic. (f) 2 Marks: 1 mark for suggesting an insulating modification (e.g., adding a lid or wrapping with insulating material) and 1 mark for explaining that this decreases heat loss.

(a) Hydrochloric acid must be in excess to ensure that all of the metal carbonate reacts completely. (b) Moles of \(\text{CO}_2 = \text{volume} / \text{molar volume} = 32.0\text{ cm}^3 / 24000\text{ cm}^3\text{ mol}^{-1} = 1.333 \times 10^{-3}\text{ mol}\). (c) The reaction stoichiometry between \(\text{M}_2\text{CO}_3\) and \(\text{CO}_2\) is 1:1, so moles of \(\text{M}_2\text{CO}_3 = 1.333 \times 10^{-3}\text{ mol}\). (d) Molar mass \(M_r = \text{mass} / \text{moles} = 0.185\text{ g} / 1.333 \times 10^{-3}\text{ mol} = 138.8\text{ g mol}^{-1}\) (accept 139). (e) \(M_r(\text{M}_2\text{CO}_3) = 2 \times A_r(\text{M}) + 12.0 + 3(16.0) = 138.8 \Rightarrow 2 \times A_r(\text{M}) + 60.0 = 138.8 \Rightarrow 2 \times A_r(\text{M}) = 78.8 \Rightarrow A_r(\text{M}) = 39.4\text{ g mol}^{-1}\). Comparing this with the periodic table, the metal M is potassium (K, \(A_r = 39.1\)). (f) Some carbon dioxide dissolved in the aqueous acid solution; the reaction did not go to completion.

(a) 1 Mark: To ensure complete reaction of the metal carbonate. (b) 2 Marks: 1 mark for correct formula usage (32 / 24000), 1 mark for correct calculation (1.33 x 10^-3 [mol]). (c) 1 Mark: 1:1 ratio used to state 1.33 x 10^-3 [mol] (accept follow-through). (d) 2 Marks: 1 mark for setting up formula (0.185 / moles), 1 mark for correct calculation of 138.8 [g/mol] (accept 139). (e) 2 Marks: 1 mark for setting up equation \(2 \times A_r(\text{M}) + 60 = 138.8\), 1 mark for calculating \(A_r \approx 39.4\) and identifying the metal as Potassium / K. (f) 2 Marks: 1 mark each for any two valid reasons: carbon dioxide dissolves in water/acid; reaction is incomplete; temperature was lower than RTP / pressure was higher than RTP.

(a) Heating to constant mass ensures that all the water of crystallization has been completely driven off from the hydrated salt. (b) Mass of anhydrous barium chloride = final mass - empty crucible mass = \(20.81 - 18.42 = 2.39\text{ g}\). Mass of water lost = initial hydrated mass - final mass = \(21.23 - 20.81 = 0.42\text{ g}\). (c) Moles of \(\text{BaCl}_2 = 2.39 / 208.2 = 0.0115\text{ mol}\). Moles of \(\text{H}_2\text{O} = 0.42 / 18.0 = 0.0233\text{ mol}\). (d) Mole ratio of \(\text{H}_2\text{O}\) to \(\text{BaCl}_2 = 0.0233 / 0.0115 = 2.03 \approx 2\). Therefore, \(x = 2\) and the formula is \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\). (e) If cooled in open air, the anhydrous barium chloride would reabsorb moisture from the atmosphere. This would increase the measured mass of the anhydrous residue, making the calculated mass of water lost appear smaller than it is, which would result in a lower calculated value of \(x\).

(a) 2 Marks: 1 mark for stating that it ensures all water is removed / fully dehydrated, 1 mark for explaining that constant mass indicates no further mass loss (water loss) is occurring. (b) 2 Marks: 1 mark for mass of anhydrous salt = 2.39 [g], 1 mark for mass of water lost = 0.42 [g]. (c) 2 Marks: 1 mark for moles of \(\text{BaCl}_2 = 0.0115\text{ mol}\), 1 mark for moles of \(\text{H}_2\text{O} = 0.0233\text{ mol}\). (d) 2 Marks: 1 mark for ratio calculation leading to \(x \approx 2\), 1 mark for stating the formula as \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\). (e) 2 Marks: 1 mark for stating that the anhydrous residue absorbs water from the air, 1 mark for concluding that the measured mass of anhydrous residue is higher, leading to a lower calculated mass of water lost and a lower calculated value of \(x\).

(a) Titres 1 and 2 are concordant (identical, within \(0.1\text{ cm}^3\)). The rough titre is excluded. Mean titre = \((24.00 + 24.00) / 2 = 24.00\text{ cm}^3\). (b) Moles of \(\text{NaOH} = (24.00 \times 0.100) / 1000 = 2.40 \times 10^{-3}\text{ mol}\). (c) According to the equation, \(1\text{ mol}\) of \(\text{H}_2\text{A}\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Moles of \(\text{H}_2\text{A}\) in \(25.0\text{ cm}^3 = 2.40 \times 10^{-3} / 2 = 1.20 \times 10^{-3}\text{ mol}\). (d) Moles of \(\text{H}_2\text{A}\) in \(250.0\text{ cm}^3 = 1.20 \times 10^{-3} \times 10 = 0.0120\text{ mol}\). Molar mass \(M_r = \text{mass} / \text{moles} = 1.08\text{ g} / 0.0120\text{ mol} = 90.0\text{ g mol}^{-1}\). (e) Dilution occurs because of water remaining in the burette, which decreases the concentration of the \(\text{NaOH}\) solution. As a result, a larger volume of the diluted \(\text{NaOH}\) solution is required to neutralize the acid. This increases the calculated moles of \(\text{NaOH}\) and thus the calculated moles of \(\text{H}_2\text{A}\), leading to a lower calculated molar mass (since \(M_r = \text{mass} / \text{moles}\)).

(a) 2 Marks: 1 mark for choosing Titres 1 and 2 and stating they are concordant (within 0.1 cm3), 1 mark for correct mean calculation of 24.00 [cm3]. (b) 1 Mark: Correct calculation of moles of NaOH = 2.40 x 10^-3 [mol]. (c) 2 Marks: 1 mark for acknowledging 1:2 ratio, 1 mark for calculating moles of acid in 25 cm3 = 1.20 x 10^-3 [mol]. (d) 3 Marks: 1 mark for calculating moles in 250 cm3 = 0.0120 [mol], 1 mark for using Mr = mass / moles, 1 mark for correct calculation of 90.0 [g/mol]. (e) 2 Marks: 1 mark for stating that water in the burette dilutes the NaOH, increasing the titre volume, 1 mark for explaining that this leads to a higher calculated number of moles of acid and hence a lower calculated molar mass.

(a) Ethanol acts as a mutual solvent to allow the halogenoalkane (immiscible in water) and aqueous silver nitrate to mix and react in a single phase. (b) Halide ion: bromide (\(\text{Br}^-\)). Precipitate: silver bromide, \(\text{AgBr}\). (c) The rate of hydrolysis increases down Group 7 (1-chlorobutane < 1-bromobutane < 1-iodobutane). This is because the carbon-halogen bond strength (bond enthalpy) decreases down the group (C-Cl > C-Br > C-I). The weaker the bond, the less energy is required to break it, making the reaction faster. (d) \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\). (e) Use a fume cupboard because halogenoalkanes are toxic or volatile, OR keep away from naked flames because halogenoalkanes are flammable.

(a) 1 Mark: Acts as a mutual/common solvent. (b) 2 Marks: 1 mark for identifying bromide ion (\(\text{Br}^-\)), 1 mark for formula of precipitate (\(\text{AgBr}\)). (c) 3 Marks: 1 mark for stating that rate of hydrolysis increases down the group, 1 mark for stating that bond strength / bond enthalpy decreases down the group, 1 mark for relating weaker bond to easier breaking / lower activation energy. (d) 2 Marks: 1 mark for correct formulas of reactants and products, 1 mark for correct state symbols. (e) 2 Marks: 1 mark for a correct safety precaution (e.g., use a fume cupboard, wear gloves, keep away from ignition sources), 1 mark for a matching reason (e.g., toxic/irritant vapors, skin absorption, flammable).