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First, calculate heat energy released using \( q = m c \Delta T \): \( q = 250.0 \times 4.18 \times (71.5 - 20.5) = 53310\text{ J} = 53.31\text{ kJ} \). Next, calculate the moles of pentan-1-ol burned: \( M_r(\text{C}_5\text{H}_{11}\text{OH}) = (5 \times 12.0) + (12 \times 1.0) + 16.0 = 88.0\text{ g mol}^{-1} \). \( \text{moles} = 1.76 / 88.0 = 0.0200\text{ mol} \). The enthalpy change of combustion is \( \Delta H_c = -q / n = -53.31 / 0.0200 = -2665.5\text{ kJ mol}^{-1} \), which rounds to \( -2670\text{ kJ mol}^{-1} \).

1 mark for the correct option B.

Using the ideal gas equation \( pV = nRT \), we rearrange to find \( n = \frac{pV}{RT} \). Convert units: \( p = 102000\text{ Pa} \), \( V = 380 \times 10^{-6}\text{ m}^3 \), \( T = 350\text{ K} \). \( n = \frac{102000 \times 380 \times 10^{-6}}{8.314 \times 350} = 0.01332\text{ mol} \). Calculate molar mass: \( M_r = \frac{\text{mass}}{n} = \frac{0.932}{0.01332} = 70.0\text{ g mol}^{-1} \). The molecular formula with a molar mass of \( 70.0\text{ g mol}^{-1} \) is \( \text{C}_5\text{H}_{10} \) (\( 5 \times 12.0 + 10 \times 1.0 = 70.0 \)).

1 mark for the correct option B.

The amide ion, \( \text{NH}_2^- \), has 2 bonding pairs and 2 lone pairs on the central nitrogen atom (total of 8 valence electrons around N). This results in a non-linear (bent) shape based on a tetrahedral arrangement, with a bond angle of approximately \( 104.5^\circ \) due to lone pair-lone pair repulsion. \( \text{NH}_4^+ \) is tetrahedral (\( 109.5^\circ \)), \( \text{BF}_3 \) is trigonal planar (\( 120^\circ \)), and \( \text{CO}_2 \) is linear (\( 180^\circ \)).

1 mark for the correct option A.

3-methylhex-2-ene, \( \text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3 \), has different groups on each carbon of the double bond (C2 has H and \( \text{CH}_3 \); C3 has \( \text{CH}_3 \) and \( \text{C}_3\text{H}_7 \)), meaning it exhibits \( E/Z \) stereoisomerism. Addition of \( \text{H}^+ \) to C2 forms a stable tertiary carbocation at C3, which reacts with \( \text{Br}^- \) to form 3-bromo-3-methylhexane as the major product. 2-methylhex-2-ene and 2,3-dimethylbut-2-ene do not exhibit \( E/Z \) stereoisomerism because one of the double-bonded carbons has two identical methyl groups. 3-methylbut-1-ene also lacks \( E/Z \) stereoisomerism.

1 mark for the correct option B.

In option C, nitrogen changes from an oxidation state of \( -3 \) in \( \text{NH}_3 \) to \( +2 \) in \( \text{NO} \), which is an overall increase of 5 units. In option A, nitrogen in \( \text{NO}_2 \) (\( +4 \)) disproportionates to \( +3 \) and \( +5 \) (change of 1). In option B, nitrogen in \( \text{NH}_4^+ \) (\( -3 \)) and \( \text{NO}_3^- \) (\( +5 \)) goes to \( \text{N}_2\text{O} \) (\( +1 \)) (changes of 4). In option D, nitrogen in \( \text{HNO}_3 \) (\( +5 \)) goes to \( \text{NO} \) (\( +2 \)) (change of 3).

1 mark for the correct option C.

Chlorine is more reactive than bromine and displaces bromide ions from \( \text{NaBr} \): \( \text{Cl}_2(\text{aq}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq}) \). The displaced bromine (\( \text{Br}_2 \)) is non-polar and preferentially dissolves in the top organic cyclohexane layer, giving it a characteristic orange-brown colour. Iodine would give a violet/purple layer.

1 mark for the correct option C.

The rate of nucleophilic substitution (hydrolysis) of haloalkanes is determined by the bond enthalpy of the carbon-halogen bond, not its polarity. The carbon-iodine bond is the weakest bond (lowest bond enthalpy) among the three, so it requires the least energy to break. This allows 1-iodobutane to react the fastest. Conversely, the carbon-chlorine bond is the strongest, making 1-chlorobutane react the slowest.

1 mark for the correct option B.

The numerical value of the equilibrium constant \( K_c \) is dependent solely on temperature. Changes in pressure, concentration, or the addition of a catalyst do not alter the value of \( K_c \). Since the forward reaction is exothermic (\( \Delta H < 0 \)), decreasing the temperature shifts the equilibrium position to the right to oppose the cooling, increasing the product concentration and decreasing the reactant concentrations at equilibrium, which increases \( K_c \).

1 mark for the correct option D.

Selenium has an atomic number of 34, which represents the number of protons. The mass number is 79, so the number of neutrons is calculated as \(79 - 34 = 45\). Since the ion has a charge of \(2-\), it has two extra electrons compared to a neutral selenium atom, resulting in \(34 + 2 = 36\) electrons.

1 mark for the correct option (A).

First, convert the gas volume into \(dm^3\):
\(240\text{ cm}^3 = 0.240\text{ dm}^3\).

Next, calculate the amount in moles of the alkane:
\(n = \frac{V}{V_m} = \frac{0.240\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.010\text{ mol}\).

Now, calculate the molar mass (\(M\)) of the alkane:
\(M = \frac{m}{n} = \frac{0.58\text{ g}}{0.010\text{ mol}} = 58\text{ g mol}^{-1}\).

The general formula for an alkane is \(\text{C}_n\text{H}_{2n+2}\).
\(12n + (2n + 2) = 58 \Rightarrow 14n + 2 = 58 \Rightarrow n = 4\).
Therefore, the molecular formula is \(\text{C}_4\text{H}_{10}\).

1 mark for the correct option (C).

To find the bond angle, determine the number of bonding pairs and lone pairs around the central atom:
- \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs. It is tetrahedral with a bond angle of \(109.5^\circ\).
- \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair. It is trigonal pyramidal with a bond angle of \(107^\circ\).
- \(\text{NH}_2^-\): Nitrogen has 5 valence electrons, plus 1 from the negative charge, giving 6. It shares 2 electrons with two hydrogens, giving 2 bonding pairs and 2 lone pairs. The arrangement of electron pairs is tetrahedral, but the shape is non-linear (bent) with a bond angle of approximately \(104.5^\circ\).
- \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs. It is trigonal planar with a bond angle of \(120^\circ\).

1 mark for the correct option (C).

Calculate the total energy required to break the reactant bonds:
- \(4 \times (\text{N}-\text{H}) = 4 \times 391 = 1564\text{ kJ}\)
- \(1 \times (\text{N}-\text{N}) = 158\text{ kJ}\)
- \(1 \times (\text{O}=\text{O}) = 498\text{ kJ}\)
Total energy for bond breaking = \(1564 + 158 + 498 = 2220\text{ kJ}\)

Calculate the total energy released when product bonds are formed:
- \(1 \times (\text{N}\equiv\text{N}) = 945\text{ kJ}\)
- \(4 \times (\text{O}-\text{H}) = 4 \times 463 = 1852\text{ kJ}\)
Total energy for bond making = \(945 + 1852 = 2797\text{ kJ}\)

Enthalpy change of reaction (\(\Delta H\)):
\(\Delta H = \text{energy of bonds broken} - \text{energy of bonds made} = 2220 - 2797 = -577\text{ kJ mol}^{-1}\).

1 mark for the correct option (A).

Calculate the equilibrium amounts (moles) of reactants and products:
- Initial: \(\text{SO}_2 = 2.0\text{ mol}\), \(\text{O}_2 = 1.0\text{ mol}\), \(\text{SO}_3 = 0\text{ mol}\)
- Change: \(\text{SO}_3 = +1.6\text{ mol}\), \(\text{SO}_2 = -1.6\text{ mol}\), \(\text{O}_2 = -0.8\text{ mol}\)
- Equilibrium: \(\text{SO}_2 = 0.4\text{ mol}\), \(\text{O}_2 = 0.2\text{ mol}\), \(\text{SO}_3 = 1.6\text{ mol}\)

Now, convert these molar amounts to equilibrium concentrations by dividing by the volume of \(2.0\text{ dm}^3\):
- \([\text{SO}_2]_{eq} = \frac{0.4\text{ mol}}{2.0\text{ dm}^3} = 0.2\text{ mol dm}^{-3}\)
- \([\text{O}_2]_{eq} = \frac{0.2\text{ mol}}{2.0\text{ dm}^3} = 0.1\text{ mol dm}^{-3}\)
- \([\text{SO}_3]_{eq} = \frac{1.6\text{ mol}}{2.0\text{ dm}^3} = 0.8\text{ mol dm}^{-3}\)

Write the expression for \(K_c\) and calculate its value:
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.8)^2}{(0.2)^2 \times 0.1} = \frac{0.64}{0.04 \times 0.1} = 160\text{ dm}^3\text{ mol}^{-1}\).

1 mark for the correct option (C).

To exhibit E/Z isomerism, an alkene must have two different groups attached to each of the two double-bonded carbon atoms.
- In 2-methylbut-2-ene, one of the carbon atoms in the double bond is attached to two identical methyl groups: \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\).
- In 1,1-dichloroprop-1-ene, one of the carbon atoms in the double bond is attached to two identical chlorine atoms: \(\text{Cl}_2\text{C}=\text{CHCH}_3\).
- In 2-methylpent-1-ene, the terminal carbon has two identical hydrogen atoms: \(\text{H}_2\text{C}=\text{C(CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\).
- In 2-chlorobut-2-ene, \(\text{CH}_3\text{C(Cl)}=\text{CHCH}_3\), the left carbon is bonded to \(-\text{Cl}\) and \(-\text{CH}_3\) (different), and the right carbon is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (different). Thus, it has E/Z isomers.

1 mark for the correct option (C).

The rate of hydrolysis of haloalkanes depends on two factors: the strength of the C-Halogen bond and the classification of the haloalkane (primary, secondary, or tertiary):
1. Carbon-halogen bond strength decreases in the order: \(\text{C-Cl} > \text{C-Br} > \text{C-I}\). This means iodoalkanes have the weakest bonds and undergo hydrolysis fastest.
2. Tertiary haloalkanes hydrolyze significantly faster than secondary and primary haloalkanes because they react via the SN1 mechanism, which involves forming a stable tertiary carbocation intermediate.

Comparing the options, 2-iodo-2-methylpropane is a tertiary iodoalkane, so it has both the weakest carbon-halogen bond and a tertiary structure, leading to the fastest rate of hydrolysis.

1 mark for the correct option (D).

An alcohol that is oxidized under reflux to a product that is not a carboxylic acid and cannot be further oxidized must be a secondary alcohol (which oxidizes to a ketone).
- Butan-1-ol and 2-methylpropan-1-ol are primary alcohols and would oxidize to carboxylic acids under reflux.
- 2-methylpropan-2-ol is a tertiary alcohol and does not oxidize.
- Butan-2-ol is a secondary alcohol and is oxidized to butanone (a ketone). Ketones cannot be oxidized further using acidified potassium dichromate(VI).

1 mark for the correct option (B).

1. Calculate the moles of H2 gas collected: n(H2) = volume in cm3 / 24000 cm3 mol-1 = 120 / 24000 = 0.00500 mol. 2. From the stoichiometry of the equation, 1 mole of M reacts to produce 1 mole of H2. Therefore, n(M) = 0.00500 mol. 3. Calculate the molar mass of M: Molar mass = mass / moles = 0.1215 g / 0.00500 mol = 24.3 g mol-1. 4. Checking the periodic table, the metal with a relative atomic mass of 24.3 is Magnesium.

1 mark for the correct option B. Award 1 mark for calculating n(H2) = 0.00500 mol, determining the molar mass of M is 24.3 g/mol, and correctly identifying the metal as Magnesium.

- NH4+ has 4 bonding pairs and 0 lone pairs around the central nitrogen, leading to a tetrahedral shape with a bond angle of 109.5 degrees. - NH2- has 2 bonding pairs and 2 lone pairs around the central nitrogen, leading to a non-linear shape with a bond angle of approximately 104.5 degrees. - H3O+ has 3 bonding pairs and 1 lone pair around the central oxygen, leading to a trigonal pyramidal shape with a bond angle of approximately 107 degrees. - CO2 has 2 double bonding regions and 0 lone pairs around the central carbon, leading to a linear shape with a bond angle of 180 degrees.

1 mark for the correct option C. Award 1 mark for identifying that the H3O+ ion has a trigonal pyramidal geometry due to 3 bonding pairs and 1 lone pair.

The rate of hydrolysis of haloalkanes is determined by the bond enthalpy of the carbon-halogen bond, not its polarity. The C-I bond is weaker and has a lower bond enthalpy than the C-Cl bond, meaning it is broken much more easily during the nucleophilic substitution reaction. Consequently, 1-iodobutane reacts significantly faster than 1-chlorobutane.

1 mark for the correct option B. Award 1 mark for identifying that reaction rate depends on the lower bond enthalpy of the C-I bond compared to the C-Cl bond.

By applying Hess's Law using combustion values: D_f H = Sum[D_c H(reactants)] - Sum[D_c H(products)]. For propene: D_f H = [3 * D_c H(C) + 3 * D_c H(H2)] - [D_c H(C3H6)] = [3 * (-394) + 3 * (-286)] - (-2058) = [-1182 - 858] + 2058 = -2040 + 2058 = +18 kJ mol-1.

1 mark for the correct option C. Award 1 mark for applying the Hess's Law cycle correctly to calculate +18 kJ mol-1.

Step 1: Write the equation for the reaction: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\). Step 2: Calculate the moles of \(\text{HCl}\) used: \(n(\text{HCl}) = 0.100 \text{ mol dm}^{-3} \times 0.02120 \text{ dm}^3 = 2.12 \times 10^{-3} \text{ mol}\). Step 3: Calculate the moles of sodium carbonate in the titrated \(25.0\text{ cm}^3\) portion: \(n(\text{Na}_2\text{CO}_3) = 0.5 \times 2.12 \times 10^{-3} = 1.06 \times 10^{-3} \text{ mol}\). Step 4: Scale up to find the moles of sodium carbonate in the original \(250.0\text{ cm}^3\) flask: \(1.06 \times 10^{-3} \times 10 = 0.0106 \text{ mol}\). Step 5: Find the molar mass of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\): \(M = 3.03 \text{ g} / 0.0106 \text{ mol} = 285.85 \text{ g mol}^{-1}\). Step 6: Determine \(x\) using the molar mass of anhydrous sodium carbonate (\(106.0 \text{ g mol}^{-1}\)): \(x \times 18.0 = 285.85 - 106.0 = 179.85\), therefore \(x = 9.99\), which rounds to 10.

Mark 1 (Method): Calculation of the moles of \(\text{HCl}\) (\(2.12 \times 10^{-3} \text{ mol}\)) and the corresponding moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3\) (\(1.06 \times 10^{-3} \text{ mol}\)). Mark 2 (Method): Conversion of moles to find the molar mass of the hydrated salt (\(285.85 \text{ g mol}^{-1}\)). Mark 3 (Accuracy): Correct integer calculation of \(x = 10\).

Step 1: Calculate the heat energy transferred to the water using \(q = m c \Delta T\). Here, \(m = 150.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T = 58.5 - 21.5 = 37.0\text{ K}\). \(q = 150.0 \times 4.18 \times 37.0 = 23201\text{ J} = 23.201\text{ kJ}\). Step 2: Calculate the amount, in mol, of methanol burned. Molar mass of \(\text{CH}_3\text{OH} = 12.0 + 3(1.0) + 16.0 + 1.0 = 32.0\text{ g mol}^{-1}\). \(n(\text{CH}_3\text{OH}) = 1.28\text{ g} / 32.0\text{ g mol}^{-1} = 0.0400\text{ mol}\). Step 3: Calculate the enthalpy change per mole and include a negative sign since the reaction is exothermic: \(\Delta H_c = -23.201\text{ kJ} / 0.0400\text{ mol} = -580.025\text{ kJ mol}^{-1}\). To 3 significant figures, this is \(-580\text{ kJ mol}^{-1}\).

Mark 1 (Method): Correct calculation of heat energy, \(q\), as \(23.2\text{ kJ}\) or \(23201\text{ J}\). Mark 2 (Method): Correct calculation of the amount of methanol (\(0.0400\text{ mol}\)) and dividing the heat energy by the number of moles. Mark 3 (Accuracy): Correct final answer of \(-580\) with the negative sign, to 3 significant figures.

Boron is in Group 13 and has 3 electrons in its outer valence shell. In \(\text{BF}_3\), boron forms 3 single covalent bonds with fluorine atoms, meaning there are 3 bonding electron pairs and no lone pairs surrounding the central boron atom. According to the valence shell electron-pair repulsion (VSEPR) theory, these 3 bonding pairs repel each other equally and move as far apart as possible to minimise repulsion. This results in a trigonal planar arrangement with a bond angle of \(120^\circ\).

Mark 1 (Accuracy): Identifies the shape as trigonal planar and the bond angle as \(120^\circ\). Mark 2 (Method): States that there are three bonding electron pairs and zero lone pairs around the central boron atom. Mark 3 (Method): Explains that the bonding pairs repel each other equally to reach a position of maximum separation or minimum repulsion.

First, establish the equilibrium moles of each species. Reaction: \(\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}\). Initial moles: \(\text{H}_2 = 1.50\), \(\text{I}_2 = 1.50\), \(\text{HI} = 0\). Change in moles: Since \(2.40\text{ mol}\) of \(\text{HI}\) is formed, the change in \(\text{HI}\) is \(+2.40\text{ mol}\). Correspondingly, the changes in reactants are \(-1.20\text{ mol}\) for each of \(\text{H}_2\) and \(\text{I}_2\) (due to the 1:1:2 stoichiometry). Equilibrium moles: \(n(\text{H}_2) = 1.50 - 1.20 = 0.30\text{ mol}\); \(n(\text{I}_2) = 1.50 - 1.20 = 0.30\text{ mol}\); \(n(\text{HI}) = 2.40\text{ mol}\). The expression for \(K_c\) is \(K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(2.40/V)^2}{(0.30/V)(0.30/V)}\). The volume terms cancel out completely: \(K_c = \frac{(2.40)^2}{(0.30) \times (0.30)} = \frac{5.76}{0.09} = 64\).

Mark 1 (Method): For calculating the equilibrium amount of reactants: \(n(\text{H}_2) = 0.30\text{ mol}\) and \(n(\text{I}_2) = 0.30\text{ mol}\). Mark 2 (Method): Writing the correct expression for \(K_c\) and substituting the equilibrium values correctly. Mark 3 (Accuracy): Correct calculated value of 64 with no units.

The electrophilic addition of \(\text{HBr}\) to propene can proceed via two different pathways. If the hydrogen atom adds to carbon-1, a secondary carbocation (\(\text{CH}_3\text{CH}^{+}\text{CH}_3\)) is formed. If the hydrogen atom adds to carbon-2, a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^{+}\)) is formed. The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups that help to disperse the positive charge (known as the inductive effect). The major product, 2-bromopropane, is therefore formed preferentially from this more stable intermediate.

Mark 1 (Accuracy): Correct name of major product as 2-bromopropane. Mark 2 (Method): Explanation that the reaction proceeds via a secondary carbocation intermediate (compared to a primary carbocation for the minor product). Mark 3 (Method): Explanation that secondary carbocations are more stable than primary carbocations due to the inductive/electron-donating effect of the adjacent alkyl groups.

Let the fractional abundance of \({}^{63}\text{Cu}\) be \(y\). The fractional abundance of \({}^{65}\text{Cu}\) is therefore \((1 - y)\). The relative atomic mass is given by: \(A_r = 63y + 65(1 - y) = 63.55\). Expand and simplify the equation: \(63y + 65 - 65y = 63.55 \Rightarrow -2y = -1.45 \Rightarrow y = 0.725\). Convert the fractional abundance to a percentage: \(0.725 \times 100 = 72.5\%\).

Mark 1 (Method): Formulates a correct algebraic expression relating the isotopic masses and fractional (or percentage) abundances to the relative atomic mass. Mark 2 (Method): Shows correct algebraic rearrangement to isolate the variable representing the abundance of \({}^{63}\text{Cu}\). Mark 3 (Accuracy): Gives the final percentage abundance as \(72.5\%\) (or 72.5) to 3 significant figures.

First, the student should add dilute nitric acid to the solution (to remove any carbonate ions), followed by aqueous silver nitrate. A cream-coloured precipitate of silver bromide, \(\text{AgBr(s)}\), will form. To distinguish this from silver chloride (white) and silver iodide (yellow), the student can test solubility in ammonia. Add dilute aqueous ammonia; the cream silver bromide precipitate will remain insoluble. Then add concentrated aqueous ammonia; the silver bromide precipitate will dissolve to form a colourless solution.

Mark 1 (Accuracy): Adds aqueous silver nitrate (acidified with dilute nitric acid) to obtain a cream precipitate. Mark 2 (Method): Shows that dilute aqueous ammonia does not dissolve the cream precipitate. Mark 3 (Method): Shows that concentrated aqueous ammonia dissolves the cream precipitate to give a colourless solution.

Since alcohol A is oxidized to a carboxylic acid under reflux, A must be a primary alcohol. The product carboxylic acid B has the formula \(\text{C}_4\text{H}_8\text{O}_2\), meaning there are 4 carbon atoms. There are two possible primary alcohols with 4 carbon atoms: (1) Butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)), which oxidizes to butanoic acid, and (2) 2-methylpropan-1-ol (\((\text{CH}_3)_2\text{CHCH}_2\text{OH}\)), which oxidizes to 2-methylpropanoic acid. Drawing the skeletal structure of either of these alcohols and writing the matching carboxylic acid name completes the answer.

Mark 1 (Method): Identifies that alcohol A must be a primary alcohol. Mark 2 (Accuracy): Draws a correct skeletal formula for either butan-1-ol or 2-methylpropan-1-ol. Mark 3 (Accuracy): Gives the correct matching systematic name of carboxylic acid B (either 'butanoic acid' or '2-methylpropanoic acid').

1. Express the weighted average calculation using the isotopic abundances: \(\text{RAM} = \frac{(69 \times 60.11) + (71 \times 39.89)}{100}\)
2. Calculate the intermediate values: \(69 \times 60.11 = 4147.59\) and \(71 \times 39.89 = 2832.19\)
3. Sum these values: \(4147.59 + 2832.19 = 6979.78\)
4. Divide by 100: \(\text{RAM} = 69.7978\)
5. Round to 2 decimal places: \(69.80\).

1 mark for the correct mathematical expression showing calculation of weighted mean. 1 mark for correct evaluation to 69.7978. 0.77 mark for final rounding to 69.80 (must be 2 decimal places).

1. Divide the mass percentage of each element by its relative atomic mass:
C: \(52.17 / 12.0 = 4.348\text{ mol}\)
H: \(13.04 / 1.0 = 13.04\text{ mol}\)
O: \(34.79 / 16.0 = 2.174\text{ mol}\)
2. Divide each value by the smallest calculated value (2.174) to obtain the simplest ratio:
C: \(4.348 / 2.174 = 2.00\)
H: \(13.04 / 2.174 = 6.00\)
O: \(2.174 / 2.174 = 1.00\)
3. The simplest whole-number ratio is 2 : 6 : 1, giving the empirical formula \(\text{C}_2\text{H}_6\text{O}\).

1 mark for dividing percentages by correct atomic masses to find moles. 1 mark for obtaining the correct simplest ratio of 2 : 6 : 1. 0.77 mark for writing the correct empirical formula C2H6O.

1. Phosphorus has 5 valence electrons. It forms 3 single covalent bonds with 3 fluorine atoms, leaving 1 lone pair.
2. There are 4 areas of electron density (3 bonding pairs and 1 lone pair) around the central phosphorus atom.
3. These electron pairs repel each other to get as far apart as possible.
4. Lone pairs repel more than bonding pairs, reducing the standard tetrahedral bond angle from \(109.5^\circ\) to approximately \(107^\circ\).
5. The molecular shape is trigonal pyramidal.

1 mark for identifying 3 bonding pairs and 1 lone pair of electrons around the P atom. 1 mark for predicting the trigonal pyramidal shape and an angle of 107 degrees (allow 106-108 degrees). 0.77 mark for explaining that electron pairs repel to be as far apart as possible, and lone pairs repel more than bonding pairs.

1. Calculate energy transferred to water: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 22.5\text{ K} = 9405\text{ J} = 9.405\text{ kJ}\)
2. Calculate moles of propan-1-ol burned: \(n = \frac{0.600\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0100\text{ mol}\)
3. Calculate enthalpy change of combustion: \(\Delta H_c = -\frac{q}{n} = -\frac{9.405\text{ kJ}}{0.0100\text{ mol}} = -940.5\text{ kJ mol}^{-1}\)
4. Rounding to 3 significant figures gives \(-941\text{ kJ mol}^{-1}\). (Negative sign is required as combustion is exothermic).

1 mark for calculating energy transferred, q = 9405 J (or 9.405 kJ). 1 mark for calculating moles of propan-1-ol = 0.0100 mol. 0.77 mark for the final enthalpy value of -941 kJ mol-1 with negative sign and to 3 significant figures.

1. Write the equilibrium constant expression: \(K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\)
2. Calculate equilibrium concentrations by dividing moles by volume (\(2.00\text{ dm}^3\)): \([\text{NO}_2] = \frac{0.400}{2.00} = 0.200\text{ mol dm}^{-3}\), \([\text{N}_2\text{O}_4] = \frac{1.20}{2.00} = 0.600\text{ mol dm}^{-3}\)
3. Substitute values into the expression: \(K_c = \frac{0.600}{(0.200)^2} = \frac{0.600}{0.0400} = 15.0\)
4. Work out units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\).

1 mark for dividing moles by volume to find equilibrium concentrations. 1 mark for correct calculation of Kc value as 15 or 15.0. 0.77 mark for correct units of dm3 mol-1.

1. Stereoisomerism (specifically E/Z isomerism) requires restricted rotation about the \(\text{C}=\text{C}\) double bond, which both molecules have due to the pi-bond.
2. It also requires that both carbon atoms in the double bond are attached to two different groups.
3. In pent-2-ene, carbon-2 is attached to \(-\text{H}\) and \(-\text{CH}_3\), and carbon-3 is attached to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Since both carbons have two different groups, it shows stereoisomerism.
4. In pent-1-ene, carbon-1 is attached to two identical hydrogen atoms (\(-\text{H}\) and \(-\text{H}\)), meaning it cannot show stereoisomerism.

1 mark for identifying that stereoisomerism is due to restricted rotation about the C=C double bond. 1 mark for stating that both carbon atoms involved in the double bond must be attached to two different groups. 0.77 mark for pointing out that pent-1-ene has one carbon attached to two identical hydrogen atoms, while pent-2-ene has different groups on both carbons.

1. Write the balanced equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (or ionic form).
2. State the initial oxidation state of chlorine in \(\text{Cl}_2\), which is 0.
3. State the oxidation state of chlorine in \(\text{NaCl}\), which is -1 (reduction from 0 to -1).
4. State the oxidation state of chlorine in \(\text{NaClO}\), which is +1 (oxidation from 0 to +1).
5. Conclude that disproportionation is the simultaneous oxidation and reduction of the same element in a single reaction.

1 mark for the correct, balanced equation: Cl2 + 2NaOH -> NaCl + NaClO + H2O. 1 mark for correctly assigning oxidation numbers of Cl as 0 in Cl2, -1 in NaCl, and +1 in NaClO. 0.77 mark for defining disproportionation as the simultaneous oxidation and reduction of the same element (chlorine).

1. Write the balanced chemical equation: \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\).
2. Calculate the moles of \(\text{NaOH}\) used: \(n(\text{NaOH}) = C \times V = 0.120\text{ mol dm}^{-3} \times \frac{22.50}{1000}\text{ dm}^3 = 2.70 \times 10^{-3}\text{ mol}\).
3. Determine moles of \(\text{H}_2\text{SO}_4\) using the 1:2 reaction ratio: \(n(\text{H}_2\text{SO}_4) = \frac{2.70 \times 10^{-3}}{2} = 1.35 \times 10^{-3}\text{ mol}\).
4. Calculate the concentration of \(\text{H}_2\text{SO}_4\): \(C(\text{H}_2\text{SO}_4) = \frac{n}{V} = \frac{1.35 \times 10^{-3}\text{ mol}}{25.00/1000\text{ dm}^3} = 0.0540\text{ mol dm}^{-3}\).

1 mark for calculating the correct number of moles of NaOH as 2.70 x 10-3 mol. 1 mark for using the correct 1:2 reaction stoichiometry to calculate moles of H2SO4 as 1.35 x 10-3 mol. 0.77 mark for the correct final concentration of 0.0540 mol dm-3, reported to 3 significant figures.

1. Calculate the energy transferred to the water, \(q\):
\(q = m c \Delta T\)
\(\Delta T = 68.5 - 19.5 = 49.0\ ^\circ\text{C}\) (or \(49.0\text{ K}\))
\(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 49.0\text{ K} = 30723\text{ J} = 30.723\text{ kJ}\)

2. Calculate the amount, in moles, of pentan-1-ol burned:
\(M_r(\text{C}_5\text{H}_{11}\text{OH}) = (5 \times 12.0) + (12 \times 1.0) + 16.0 = 88.0\text{ g mol}^{-1}\)
\(n = \frac{1.10\text{ g}}{88.0\text{ g mol}^{-1}} = 0.0125\text{ mol}\)

3. Calculate the enthalpy change of combustion, \(\Delta_c H\):
\(\Delta_c H = -\frac{30.723\text{ kJ}}{0.0125\text{ mol}} = -2457.84\text{ kJ mol}^{-1}\)

Rounding to 3 significant figures gives \(-2460\text{ kJ mol}^{-1}\).

• Mark 1: Correct calculation of heat energy change \(q = 30.723\text{ kJ}\) (or \(30723\text{ J}\)).
• Mark 2: Correct calculation of amount of pentan-1-ol burned \(n = 0.0125\text{ mol}\) (from using \(M_r = 88.0\) or \(88.15\)).
• Mark 3: Correct final value of \(-2460\text{ kJ mol}^{-1}\) (must have negative sign, 3 significant figures, and allow error carried forward from previous steps).

1. Determine the equilibrium moles of each species using stoichiometry:
• \(n(\text{C}) = 0.150\text{ mol}\) (given)
• \(n(\text{D}) = 0.150\text{ mol}\) (since 1:1 ratio with C)
• \(n(\text{A}) = 0.400 - 0.150 = 0.250\text{ mol}\)
• \(n(\text{B}) = 0.400 - (2 \times 0.150) = 0.100\text{ mol}\)

2. Divide moles by volume (\(2.00\text{ dm}^3\)) to find equilibrium concentrations:
• \([\text{A}] = \frac{0.250}{2.00} = 0.125\text{ mol dm}^{-3}\)
• \([\text{B}] = \frac{0.100}{2.00} = 0.0500\text{ mol dm}^{-3}\)
• \([\text{C}] = \frac{0.150}{2.00} = 0.0750\text{ mol dm}^{-3}\)
• \([\text{D}] = \frac{0.150}{2.00} = 0.0750\text{ mol dm}^{-3}\)

3. Set up the \(K_c\) expression:
\[K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]^2}\]

4. Substitute concentrations into the \(K_c\) expression:
\[K_c = \frac{0.0750 \times 0.0750}{0.125 \times (0.0500)^2} = \frac{0.005625}{0.0003125} = 18.0\]

5. Determine the units:
\[\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\]

• Mark 1: Deduces equilibrium moles of all species: \(n(\text{A}) = 0.250\text{ mol}\), \(n(\text{B}) = 0.100\text{ mol}\), and \(n(\text{D}) = 0.150\text{ mol}\).
• Mark 2: Calculates concentrations by dividing moles by \(2.00\) AND provides the correct \(K_c\) expression: \(K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]^2}\).
• Mark 3: Correctly calculates \(K_c = 18.0\) (or \(18\)) with units: \(\text{dm}^3\text{ mol}^{-1}\) (allow ecf on calculated concentrations).

1. Calculate the mass of water lost during heating:
\( 4.29\text{ g} - 1.59\text{ g} = 2.70\text{ g} \)

2. Calculate the amount, in moles, of anhydrous \( \text{Na}_2\text{CO}_3 \):
\( M_r(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1} \)
\( n(\text{Na}_2\text{CO}_3) = \frac{1.59\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0150\text{ mol} \)

3. Calculate the amount, in moles, of water lost:
\( M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1} \)
\( n(\text{H}_2\text{O}) = \frac{2.70\text{ g}}{18.0\text{ g mol}^{-1}} = 0.150\text{ mol} \)

4. Determine the molar ratio of water to anhydrous salt:
\( x = \frac{0.150}{0.0150} = 10 \)

Therefore, the formula is \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \).

Mark 1: Award 1 mark for calculating the mass of water lost (2.70 g) and calculating the moles of anhydrous salt (0.0150 mol).
Mark 2: Award 1 mark for calculating the moles of water (0.150 mol).
Mark 3: Award 1 mark for calculating the correct whole-number ratio value of x = 10.

1. Write the equation for the standard enthalpy change of formation of propene:
\( 3\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_3\text{H}_6(g) \)

2. Set up the Hess's Law cycle using combustion products (carbon dioxide and water):
\( \Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) \)

3. Substitute the combustion values into the expression:
\( \Delta_f H^\theta = [3 \times (-394) + 3 \times (-286)] - [-2058] \)
\( \Delta_f H^\theta = [-1182 - 858] + 2058 \)
\( \Delta_f H^\theta = -2040 + 2058 = +18\text{ kJ mol}^{-1} \).

Mark 1: Award 1 mark for showing a correct Hess's cycle or expression linking the formation reaction to the combustion reactions (e.g., \( 3 \times \Delta_c H^\theta[\text{C}] + 3 \times \Delta_c H^\theta[\text{H}_2] - \Delta_c H^\theta[\text{C}_3\text{H}_6] \)).
Mark 2: Award 1 mark for the correct calculation of the intermediate combustion sum (\( -2040\text{ kJ mol}^{-1} \)).
Mark 3: Award 1 mark for the correct final answer of +18 (or 18) with positive sign (accept without sign but preferred).

1. Determine the number of electron pairs around the central oxygen atom in \( \text{H}_3\text{O}^+ \):
- Oxygen has 6 valence electrons.
- The positive charge means 1 electron is lost, leaving 5 electrons.
- There are 3 shared pairs bonded to hydrogen atoms, which leaves 1 lone pair remaining.
- Total: 4 electron pairs (3 bonding pairs, 1 lone pair).

2. Determine molecular shape and bond angle:
- The base geometry for 4 electron pairs is tetrahedral, but with 1 lone pair, the shape is trigonal pyramidal (or pyramidal).
- The lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, reducing the bond angle from the tetrahedral value of 109.5° to approximately 107°.

3. Explanation:
- Electron pairs repel each other to get as far apart as possible.
- Lone pairs repel more strongly than bonding pairs, which reduces the bond angles.

Mark 1: Award 1 mark for identifying the shape as trigonal pyramidal (or pyramidal) and stating the approximate bond angle is 107° (accept any value in the range 106° to 108°).
Mark 2: Award 1 mark for stating that there are 3 bonding pairs and 1 lone pair around the central oxygen atom.
Mark 3: Award 1 mark for explaining that electron pairs repel to minimize repulsion (or move as far apart as possible) and that lone pairs repel more than bonding pairs.

1. The reaction of 1-bromobutane with silver nitrate produces silver bromide, \( \text{AgBr} \), which is a cream-coloured precipitate.

2. The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane.

3. Trend explanation: The rate depends on the bond enthalpy (strength) of the carbon-halogen (C-X) bond. Down Group 7, the atomic radius of the halogen increases, resulting in longer, weaker C-X bonds. The weaker C-I bond is broken more easily than the C-Br and C-Cl bonds, making 1-iodobutane the fastest to hydrolyse.

Mark 1: Award 1 mark for stating the precipitate with 1-bromobutane is cream (or cream-coloured).
Mark 2: Award 1 mark for stating that the rate of hydrolysis increases from chlorine to iodine (or 1-iodobutane reacts fastest and 1-chlorobutane reacts slowest).
Mark 3: Award 1 mark for explaining that C-X bond strength / bond enthalpy decreases down the group (C-Cl > C-Br > C-I), meaning the C-I bond is the easiest to break (Reject explanations based on electronegativity or bond polarity).

1. Determine initial concentrations:
\( [\text{SO}_3]_{\text{initial}} = \frac{0.400\text{ mol}}{2.00\text{ dm}^3} = 0.200\text{ mol dm}^{-3} \)
\( [\text{SO}_2]_{\text{initial}} = 0\text{ mol dm}^{-3} \)
\( [\text{O}_2]_{\text{initial}} = 0\text{ mol dm}^{-3} \)

2. Determine equilibrium concentrations from stoichiometry:
At equilibrium, \( [\text{O}_2]_{\text{eq}} = 0.050\text{ mol dm}^{-3} \).
This requires the formation of \( 2 \times 0.050 = 0.100\text{ mol dm}^{-3} \) of \( \text{SO}_2 \) and the consumption of \( 2 \times 0.050 = 0.100\text{ mol dm}^{-3} \) of \( \text{SO}_3 \).

- \( [\text{SO}_3]_{\text{eq}} = 0.200 - 0.100 = 0.100\text{ mol dm}^{-3} \)
- \( [\text{SO}_2]_{\text{eq}} = 0.100\text{ mol dm}^{-3} \)
- \( [\text{O}_2]_{\text{eq}} = 0.050\text{ mol dm}^{-3} \)

3. Write the expression for \( K_c \):
\( K_c = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2} \)

4. Substitute values and calculate \( K_c \):
\( K_c = \frac{(0.100)^2 \times 0.050}{(0.100)^2} = 0.050 \)

5. Determine units:
\( \text{Units} = \frac{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})}{(\text{mol dm}^{-3})^2} = \text{mol dm}^{-3} \)

Mark 1: Award 1 mark for finding the correct equilibrium concentrations of \( [\text{SO}_3]_{\text{eq}} = 0.100\text{ mol dm}^{-3} \) and \( [\text{SO}_2]_{\text{eq}} = 0.100\text{ mol dm}^{-3} \).
Mark 2: Award 1 mark for writing the correct \( K_c \) expression and substituting the calculated concentrations.
Mark 3: Award 1 mark for the correct numerical value of 0.050 (or \( 5.0 \times 10^{-2} \)) with the units \( \text{mol dm}^{-3} \).

1. 2-methylbut-2-ene, \( (\text{CH}_3)_2\text{C}=\text{CHCH}_3 \), reacts with \( \text{H}^+ \) from \( \text{HBr} \) to form carbocation intermediates.

2. Protonation can occur at two different positions:
- Proton addition to C3 forms a tertiary carbocation at C2: \( (\text{CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3 \).
- Proton addition to C2 forms a secondary carbocation at C3: \( (\text{CH}_3)_2\text{CH}-\text{CH}^+-\text{CH}_3 \).

3. The tertiary carbocation is more stable than the secondary carbocation because it has three electron-donating alkyl groups attached to the positively charged carbon atom compared to two. This inductive effect disperses the positive charge, lowering the activation energy for this pathway.

4. Subsequent nucleophilic attack by \( \text{Br}^- \) on the more stable tertiary carbocation produces 2-bromo-2-methylbutane as the major product.

Mark 1: Award 1 mark for naming the major product as 2-bromo-2-methylbutane (or drawing its structure / skeletal formula).
Mark 2: Award 1 mark for stating that the reaction proceeds via a more stable tertiary carbocation intermediate compared to a secondary carbocation intermediate.
Mark 3: Award 1 mark for explaining that alkyl groups are electron-donating (or have a positive inductive effect) which stabilises the positive charge on the carbocation.

Step 1: Hydration of propene to form the intermediate alcohol.
- Reagents and conditions: Steam (\( \text{H}_2\text{O}(g) \)) and a concentrated phosphoric acid catalyst (\( \text{H}_3\text{PO}_4 \)), under high temperature and pressure (or concentrated \( \text{H}_2\text{SO}_4 \) followed by water).
- Intermediate compound: Propan-2-ol.

Step 2: Oxidation of the secondary alcohol to a ketone.
- Reagents and conditions: Acidified potassium dichromate(VI) (\( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4 \)) heated under reflux.

Mark 1: Award 1 mark for correctly identifying the intermediate as propan-2-ol (by name or formula).
Mark 2: Award 1 mark for Step 1 reagents and conditions: Steam and concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) catalyst (allow concentrated sulfuric acid then water).
Mark 3: Award 1 mark for Step 2 reagents and conditions: Acidified potassium dichromate(VI) (or \( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4 \)) heated under reflux.

1. Determine molecular formula:
- Empirical mass of \( \text{C}_2\text{H}_4\text{O} = (2 \times 12.0) + (4 \times 1.0) + 16.0 = 44.0 \).
- Since the molecular ion peak \( m/z = 88.0 \), the molecular formula must be twice the empirical formula:
\( \text{C}_4\text{H}_8\text{O}_2 \).

2. Analyze infrared spectrum:
- The very broad absorption between \( 2500 \text{ and } 3300\text{ cm}^{-1} \) is characteristic of an O-H stretch in a carboxylic acid.
- The strong, sharp absorption at \( 1715\text{ cm}^{-1} \) is characteristic of a C=O carbonyl stretch.
- Together, these peaks confirm the presence of a carboxylic acid functional group.

3. Draw/write a possible structure:
- A 4-carbon carboxylic acid is butanoic acid: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \) (or methylpropanoic acid: \( (\text{CH}_3)_2\text{CHCOOH} \)).

Mark 1: Award 1 mark for deducing the correct molecular formula as \( \text{C}_4\text{H}_8\text{O}2 \).
Mark 2: Award 1 mark for identifying the functional group as a carboxylic acid, referencing the broad O-H stretch at 2500-3300 \( \text{cm}^{-1} \) and the C=O stretch at 1715 \( \text{cm}^{-1} \).
Mark 3: Award 1 mark for drawing/writing a correct structural, displayed, or skeletal formula of butanoic acid or methylpropanoic acid.

1. Identify the bonds broken in the reactants:
- Propene has 1 \(C=C\) bond, 1 \(C-C\) bond, and 6 \(C-H\) bonds.
- Hydrogen has 1 \(H-H\) bond.
- Total energy required to break bonds = \(611 + 347 + (6 \times 413) + 436 = 3872\text{ kJ mol}^{-1}\).

2. Identify the bonds formed in the product:
- Propane has 2 \(C-C\) bonds and 8 \(C-H\) bonds.
- Total energy released when bonds are formed = \((2 \times 347) + (8 \times 413) = 694 + 3304 = 3998\text{ kJ mol}^{-1}\).

3. Calculate \(\Delta_r H\):
- \(\Delta_r H = \text{Energy of bonds broken} - \text{Energy of bonds formed}\)
- \(\Delta_r H = 3872 - 3998 = -126\text{ kJ mol}^{-1}\).

Award 1 mark for calculating correct total bond enthalpy of reactants (3872).
Award 1 mark for calculating correct total bond enthalpy of products (3998).
Award 0.9 marks for the final calculated value of -126 with correct sign.

1. Determine the mass of water lost:
- Mass of \(H_2O = 4.31\text{ g} - 3.41\text{ g} = 0.90\text{ g}\).

2. Calculate the amount in moles of anhydrous \(CaSO_4\) (\(M_r = 136.2\text{ g mol}^{-1}\)):
- \(n(CaSO_4) = \frac{3.41\text{ g}}{136.2\text{ g mol}^{-1}} = 0.0250\text{ mol}\).

3. Calculate the amount in moles of water lost (\(M_r = 18.0\text{ g mol}^{-1}\)):
- \(n(H_2O) = \frac{0.90\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\).

4. Determine the simplest molar ratio:
- \(\text{Ratio} = \frac{0.0500}{0.0250} = 2\).
- Therefore, \(x = 2\).

Award 1 mark for calculating moles of CaSO4 (0.0250) and water (0.0500).
Award 1.9 marks for calculating the correct molar ratio, giving x = 2.

1. Sulfur has 6 valence electrons and forms 6 single covalent bonding pairs of electrons with fluorine atoms, leaving no lone pairs on the sulfur atom.
2. According to electron-pair repulsion theory, electron pairs repel each other and move as far apart as possible to minimise repulsion.
3. The 6 bonding pairs repel equally, adopting a symmetrical octahedral geometry with bond angles of \(90^\circ\).

Award 1 mark for stating there are 6 bonding pairs and 0 lone pairs around the central S atom.
Award 1 mark for stating that electron pairs repel each other to minimise repulsion.
Award 0.9 marks for linking the equal repulsion of 6 pairs to the symmetrical octahedral shape and 90 degree bond angles.

1. Chlorine is a stronger oxidising agent than bromine and displaces bromide ions from solution.
2. Observation: The colorless solution turns orange/yellow due to the formation of aqueous bromine (\(Br_2(aq)\)).
3. Ionic equation: \(Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)\).

Award 1 mark for stating that the colorless solution turns orange or yellow.
Award 1 mark for the correct ionic species in the equation: Cl2 + 2Br- -> 2Cl- + Br2.
Award 0.9 marks for correct state symbols (aq) on all reactants and products.

1. Major product: 2-bromopropane.
2. Explanation: The electrophilic addition mechanism proceeds via a carbocation intermediate. The reaction to form 2-bromopropane goes through a secondary carbocation intermediate (\(CH_3C^+HCH_3\)), whereas 1-bromopropane goes through a primary carbocation intermediate (\(CH_3CH_2C^+H_2\)).
3. The secondary carbocation is more stable than the primary carbocation because the two electron-releasing alkyl groups reduce the positive charge density on the carbon atom (inductive effect).

Award 1 mark for identifying 2-bromopropane as the major product.
Award 1 mark for stating that the reaction proceeds via a secondary carbocation intermediate.
Award 0.9 marks for explaining that the secondary carbocation is more stable due to the inductive effect of more alkyl groups.

1. 1-iodobutane reacts the fastest (hydrolyses first).
2. Hydrolysis involves breaking the carbon-halogen bond (\(C-X\)).
3. The C-I bond is the weakest bond (has the lowest bond enthalpy) because the iodine atom is larger than chlorine and bromine, leading to a longer bond and weaker electrostatic attraction between the bonded pair of electrons and the nuclei. Therefore, the C-I bond is broken most easily, resulting in the fastest rate.

Award 1 mark for identifying 1-iodobutane as the fastest-reacting haloalkane.
Award 1 mark for stating that bond enthalpy / bond strength determines the rate of hydrolysis (not bond polarity).
Award 0.9 marks for explaining that the C-I bond has the lowest bond enthalpy / is the weakest, and therefore is broken most easily.

1. Establish the equilibrium amounts:
- Initial moles: \(PCl_5 = 2.00\text{ mol}\); \(PCl_3 = 0\text{ mol}\); \(Cl_2 = 0\text{ mol}\).
- Equilibrium moles of \(PCl_3 = 0.60\text{ mol}\).
- Since the stoichiometry is 1:1:1, equilibrium moles of \(Cl_2 = 0.60\text{ mol}\).
- Equilibrium moles of \(PCl_5 = 2.00 - 0.60 = 1.40\text{ mol}\).

2. Calculate the equilibrium concentrations (using flask volume \(V = 5.00\text{ dm}^3\)):
- \([PCl_5]_{eq} = \frac{1.40}{5.00} = 0.28\text{ mol dm}^{-3}\)
- \([PCl_3]_{eq} = \frac{0.60}{5.00} = 0.12\text{ mol dm}^{-3}\)
- \([Cl_2]_{eq} = \frac{0.60}{5.00} = 0.12\text{ mol dm}^{-3}\)

3. Calculate \(K_c\):
- \(K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.12 \times 0.12}{0.28} = \frac{0.0144}{0.28} = 0.0514\text{ (3 sig figs)}\).

4. Determine the units:
- \(\text{Units} = \frac{\text{mol dm}^{-3} \times \text{mol dm}^{-3}}{\text{mol dm}^{-3}} = \text{mol dm}^{-3}\).

Award 1 mark for finding correct equilibrium amounts (1.40 mol PCl5, 0.60 mol Cl2).
Award 1 mark for dividing moles by 5.00 dm^3 to find equilibrium concentrations.
Award 0.9 marks for the correct value of 0.0514 with correct units of mol dm^-3.

1. Step 1: Free radical substitution of propane to form 2-bromopropane (or 2-chloropropane).
- Reagents: Bromine (\(Br_2\)) or chlorine (\(Cl_2\))
- Conditions: UV light / radiation
- Intermediate compound: 2-bromopropane (or 2-chloropropane)

2. Step 2: Nucleophilic substitution (hydrolysis) of the halogenoalkane intermediate.
- Reagents: Aqueous sodium hydroxide (\(NaOH(aq)\)) or aqueous potassium hydroxide (\(KOH(aq)\))
- Conditions: Heat under reflux

Award 1 mark for Step 1 reagents (Br2/Cl2) and conditions (UV light) to form 2-bromopropane or 2-chloropropane as intermediate.
Award 1 mark for Step 2 reagents (NaOH(aq) / KOH(aq)) and conditions (heat under reflux / warm).
Award 0.9 marks for correctly naming the intermediate as 2-bromopropane or 2-chloropropane.

Step 1: Calculate the mass of water lost: 2.78 g - 1.52 g = 1.26 g. Step 2: Calculate the amount in moles of anhydrous FeSO4 (Mr = 151.9 g mol-1): Moles of FeSO4 = 1.52 / 151.9 = 0.0100 mol. Step 3: Calculate the amount in moles of H2O (Mr = 18.0 g mol-1): Moles of H2O = 1.26 / 18.0 = 0.0700 mol. Step 4: Find the simplest molar ratio: Ratio = 0.0700 / 0.0100 = 7. Therefore, x = 7.

Award 1 mark for calculating the mass of water lost as 1.26 g and the moles of water as 0.0700 mol. Award 1 mark for calculating the moles of FeSO4 as 0.0100 mol. Award 1 mark for the correct final whole number value of x = 7, supported by working.

The equation for the formation of butane-1,3-diene is: 4C(s) + 3H2(g) -> C4H6(l). According to Hess's Law, the enthalpy change of formation is equal to the sum of the combustion enthalpies of the reactants minus the combustion enthalpy of the product: \(\Delta_f H^\theta\) = [4 * (-394) + 3 * (-286)] - [-2540] = [-1576 - 858] + 2540 = -2434 + 2540 = +106 kJ mol-1.

Award 1 mark for correctly showing the sum of combustion enthalpies of reactants: 4 * (-394) + 3 * (-286) = -2434 kJ mol-1. Award 1 mark for setting up the subtraction of the product: -2434 - (-2540). Award 1 mark for the correct final answer of +106 kJ mol-1 (accept 106; do not accept -106).

Step 1: Calculate the equilibrium concentrations of each species by dividing the moles by the volume (2.00 dm3): [SO2] = 0.400 / 2.00 = 0.200 mol dm-3, [O2] = 0.300 / 2.00 = 0.150 mol dm-3, [SO3] = 0.800 / 2.00 = 0.400 mol dm-3. Step 2: Write the expression for Kc: Kc = [SO3]^2 / ([SO2]^2 * [O2]). Step 3: Calculate the value of Kc: Kc = 0.400^2 / (0.200^2 * 0.150) = 0.160 / (0.0400 * 0.150) = 26.7. Step 4: Determine the units: Units = (mol dm-3)^2 / ((mol dm-3)^2 * (mol dm-3)) = dm3 mol-1.

Award 1 mark for calculating all three equilibrium concentrations correctly: [SO2] = 0.200, [O2] = 0.150, and [SO3] = 0.400 mol dm-3. Award 1 mark for a correct Kc expression and calculation showing the value is 26.7 (or 26.67). Award 1 mark for the correct units of dm3 mol-1 (or mol-1 dm3).

1. Order of rate of hydrolysis: 1-iodobutane is the fastest, followed by 1-bromobutane, and 1-chlorobutane is the slowest. 2. Explanation: The rate of hydrolysis depends on the bond enthalpy (strength) of the carbon-halogen bond rather than its polarity. The C-I bond is the weakest bond (has the lowest bond enthalpy), which means it requires the least energy to break. Consequently, 1-iodobutane reacts the fastest. The C-Cl bond is the strongest bond (has the highest bond enthalpy), so it requires the most energy to break, making 1-chlorobutane react the slowest.

Award 1 mark for stating the correct order: 1-iodobutane > 1-bromobutane > 1-chlorobutane (or equivalent wording). Award 1 mark for stating that the rate of reaction depends on the strength or bond enthalpy of the C-X bond (rather than bond polarity). Award 1 mark for explaining that the C-I bond has the lowest bond enthalpy (or is the weakest) and thus breaks most easily.

**1. Formula Determination:**
- Carbon moles: \(64.86 / 12.0 = 5.405\)
- Hydrogen moles: \(13.51 / 1.0 = 13.51\)
- Oxygen moles: \(21.62 / 16.0 = 1.351\)
- Dividing by the smallest number: \(C = 4\), \(H = 10\), \(O = 1\). Empirical formula is \(C_4H_{10}O\).
- The molecular ion peak \(M^+\) is at \(m/z = 74.0\). Since the formula mass of \(C_4H_{10}O\) is \((4 \times 12.0) + (10 \times 1.0) + 16.0 = 74.0\), the molecular formula is also \(C_4H_{10}O\).

**2. IR Analysis:**
- Peak at \(3200\text{--}3600\text{ cm}^{-1}\) indicates the presence of an \(O\text{-H}\) bond in an alcohol.
- Absence of any peak in the range \(1630\text{--}1820\text{ cm}^{-1}\) indicates that no carbonyl (\(C\text{=O}\)) group is present, ruling out aldehydes, ketones, or carboxylic acids.

**3. Chemical Test & Isomer Deduction:**
- There are four isomers of alcohol with formula \(C_4H_{10}O\): butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol.
- Heating under reflux with acidified potassium dichromate(VI) results in the solution remaining orange. This means the alcohol is not oxidized (tertiary alcohol).
- The only tertiary alcohol among the isomers is 2-methylpropan-2-ol.
- This is also a branched-chain alcohol, matching the initial description.

**4. Mass Spectrum Fragment:**
- The peak at \(m/z = 59.0\) represents the loss of a methyl group (\(M - 15\)) from the molecular ion: \([(CH_3)_3COH]^{+\bullet} \rightarrow [(CH_3)_2COH]^+ + \cdot CH_3\).
- The ion responsible for the peak at \(m/z = 59.0\) is \([(CH_3)_2COH]^+\).

**Level 3 (5–6 marks):**
- Calculates both the correct empirical and molecular formulas (\(C_4H_{10}O\)) with working.
- Correctly assigns the IR band to an alcohol \(O\text{-H}\) group and states the absence of a \(C\text{=O}\) carbonyl group.
- Deduces that the alcohol must be tertiary because it is resistant to oxidation by acidified potassium dichromate(VI).
- Gives the correct IUPAC name (2-methylpropan-2-ol) and draws the fully displayed formula showing all bonds (including the \(O\text{-H}\) bond).
- Identifies the fragment ion at \(m/z = 59.0\) as \([(CH_3)_2COH]^+\) (formula must include the positive charge).
- Explanation is highly structured, logical, and uses appropriate scientific terminology throughout.

**Level 2 (3–4 marks):**
- Calculates the empirical formula correctly and identifies at least two other aspects (e.g., identifies \(O\text{-H}\) band, deduces tertiary alcohol, or names the structure).
- Attempts a displayed formula but may have minor errors (e.g., omitting the bond between \(O\) and \(H\)).
- Fragment ion may be incorrectly identified or missing the positive charge.
- There is a line of reasoning with some structure and chemistry terminology.

**Level 1 (1–2 marks):**
- Attempts empirical formula calculation or identifies the presence of an alcohol group from the IR peak.
- Structure may be incorrect or incomplete.
- Answer contains some scientific terminology but lacks structure or clarity.

**Level 0 (0 marks):**
- No response or no relevant chemical information.

**1. Direct Measurement Limitation:**
- High temperature is required to decompose calcium carbonate, making it impossible to measure heat changes in a simple laboratory calorimeter. Heat loss to the surroundings would be too significant, and the reaction cannot be easily contained.

**2. Hess's Law Cycle:**
- Top of cycle: \(\text{CaCO}_3(\text{s}) \xrightarrow{\Delta H_{\text{r}}} \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)
- Bottom of cycle: \(\text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)
- Arrow from reactant \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq})\) pointing down to the bottom products represents \(\Delta H_1\).
- Arrow from product \(\text{CaO}(\text{s}) + 2\text{HCl}(\text{aq}) + \text{CO}_2(\text{g})\) pointing down to the bottom products represents \(\Delta H_2\).
- By Hess's Law: \(\Delta H_1 = \Delta H_{\text{r}} + \Delta H_2\), therefore \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\).

**3. Calculations:**
- **Reaction 1 (\(\Delta H_1\)):**
- \(q_1 = m c \Delta T_1 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 3.3\text{ K} = 689.7\text{ J} = 0.6897\text{ kJ}\)
- \(n(\text{CaCO}_3) = 3.00\text{ g} / 100.1\text{ g mol}^{-1} = 0.02997\text{ mol}\) (or \(3.00 / 100 = 0.0300\text{ mol}\))
- Since temperature rose, reaction is exothermic: \(\Delta H_1 = -0.6897 / 0.02997 = -23.01\text{ kJ mol}^{-1}\) (or \(-23.0\text{ kJ mol}^{-1}\))

- **Reaction 2 (\(\Delta H_2\)):**
- \(q_2 = m c \Delta T_2 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 22.6\text{ K} = 4723.4\text{ J} = 4.7234\text{ kJ}\)
- \(n(\text{CaO}) = 1.68\text{ g} / 56.1\text{ g mol}^{-1} = 0.02995\text{ mol}\) (or \(1.68 / 56 = 0.0300\text{ mol}\))
- Since temperature rose, reaction is exothermic: \(\Delta H_2 = -4.7234 / 0.02995 = -157.71\text{ kJ mol}^{-1}\) (or \(-157.7\text{ kJ mol}^{-1}\))

- **Determining \(\Delta H_{\text{r}}\):**
- \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\)
- Using exact values: \(\Delta H_{\text{r}} = -23.01 - (-157.71) = +134.7\text{ kJ mol}^{-1}\)
- Using rounded values (with \(M_r\) of \(100\) and \(56\)): \(\Delta H_{\text{r}} = -23.0 - (-157.4) = +134.4\text{ kJ mol}^{-1}\) (Accept any value between \(+134.0\) and \(+135.0\text{ kJ mol}^{-1}\) depending on rounding, provided the sign is positive and the working is shown).

**Level 3 (5–6 marks):**
- Explains clearly that the reaction requires high temperatures, making direct calorimeter measurements impossible due to massive heat loss.
- Draws a fully labeled Hess's Law cycle showing all chemical species at each stage, with arrows pointing in the correct direction.
- Calculates \(q_1\), \(q_2\) and the respective moles of \(\text{CaCO}_3\) and \(\text{CaO}\) correctly.
- Obtains correct values for \(\Delta H_1\) (approx. \(-23.0\text{ kJ mol}^{-1}\)) and \(\Delta H_2\) (approx. \(-157.7\text{ kJ mol}^{-1}\)), ensuring negative signs are included.
- Calculates the final \(\Delta H_{\text{r}}\) correctly (approx. \(+134.7\text{ kJ mol}^{-1}\)), ensuring the sign is positive and the units are correct.
- Structure is highly logical and step-by-step.

**Level 2 (3–4 marks):**
- Explains the limitation OR draws a mostly correct Hess's Law cycle.
- Calculates \(q\) and moles correctly for both reactions, but may have made a sign error or failed to convert to \(\text{kJ mol}^{-1}\).
- Calculates a final value for \(\Delta H_{\text{r}}\) but with carried-forward calculation errors.
- The explanation of steps is reasonably clear and follows a structured approach.

**Level 1 (1–2 marks):**
- Identifies the basic concept of Hess's Law (e.g., writing the equation \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\)).
- Attempts heat energy calculations (\(q = m c \Delta T\)) but contains major errors.
- Explanation is fragmented or poorly structured.

**Level 0 (0 marks):**
- No response or no relevant chemical information.