MetadataPastPaper.sampleTitle

First, find the volume of \(CO_2(g)\) produced. Gaseous \(CO_2\) is absorbed by concentrated sodium hydroxide solution, so the reduction in volume is equal to the volume of \(CO_2\):

$$\text{Volume of } CO_2 = 110 - 30 = 80\text{ cm}^3$$

Since \(20\text{ cm}^3\) of the hydrocarbon \(C_xH_y\) produces \(80\text{ cm}^3\) of \(CO_2\), we can determine the number of carbon atoms, \(x\):

$$x = \frac{80}{20} = 4$$

Next, determine the volume of oxygen that reacted. The remaining gas after absorption is \(30\text{ cm}^3\) of excess \(O_2\). Therefore, the volume of oxygen reacted is:

$$\text{Volume of } O_2 \text{ reacted} = 150 - 30 = 120\text{ cm}^3$$

The ratio of reacting volumes of hydrocarbon to oxygen is:

$$1 : \frac{120}{20} = 1 : 6$$

The general equation for combustion of a hydrocarbon is:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

Using \(x = 4\) and the reacting ratio of oxygen:

$$4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8$$

Therefore, the molecular formula of the hydrocarbon is \(C_4H_8\).

1 mark for the correct option (B).

The rate of hydrolysis of halogenoalkanes is determined by two main factors:

1. **The strength of the carbon-halogen bond**: Carbon-iodine (\(C-I\)) bonds are the weakest, followed by \(C-Br\) and \(C-Cl\). Weak bonds are broken much more easily, making iodoalkanes more reactive than bromoalkanes and chloroalkanes.

2. **The mechanism of the reaction**: Tertiary halogenoalkanes undergo nucleophilic substitution via the \(S_N1\) mechanism, which forms a highly stable tertiary carbocation intermediate. This pathway has a lower activation energy and is much faster than the \(S_N2\) pathway followed by primary halogenoalkanes.

Combining these two factors, 2-iodo-2-methylpropane (a tertiary iodoalkane) is the most reactive and will produce a yellow precipitate of silver iodide (\(AgI\)) almost instantly, which is the fastest among the options given.

1 mark for the correct option (C).

1. The thermal decomposition of \(X\) yields a metal oxide, a brown gas (\(NO_2\)), and a gas that relights a glowing splint (\(O_2\)). This shows that \(X\) is a Group 2 nitrate.

2. When an aqueous solution of this Group 2 nitrate is treated with dilute sulfuric acid, it forms a thick white precipitate of the metal sulfate:

$$M(NO_3)_2(aq) + H_2SO_4(aq) \rightarrow MSO_4(s) + 2HNO_3(aq)$$

Group 2 sulfates become less soluble down the group. Magnesium sulfate is soluble, calcium sulfate is slightly soluble, while barium sulfate is highly insoluble and forms a dense white precipitate. Therefore, the metal cation must be barium (\(Ba^{2+}\)), and compound \(X\) is barium nitrate, \(Ba(NO_3)_2\).

1 mark for the correct option (B).

- **Immediate effect**: Decreasing the volume increases the concentration of all gas particles inside the syringe. Because the concentration of brown \(NO_2\) molecules is suddenly increased, the mixture immediately becomes darker brown.

- **Subsequent effect**: According to Le Chatelier's principle, the system will oppose the increased pressure (caused by decreasing the volume) by shifting the equilibrium to the side with fewer gas molecules. In this reaction, there is 1 mole of gas on the reactant side (\(N_2O_4\)) and 2 moles of gas on the product side (\(NO_2\)). The equilibrium will shift to the left, converting dark brown \(NO_2\) into colorless \(N_2O_4\). As a result, the mixture subsequently becomes lighter brown.

1 mark for the correct option (B).

- \(BF_3\): The boron atom has 3 valence electrons and forms 3 bonding pairs with no lone pairs. According to VSEPR theory, it has a trigonal planar shape.

- \(CO_3^{2-}\): The central carbon atom has 3 areas of electron density (two single bonds, one double bond) and no lone pairs. This gives it a trigonal planar shape.

- \(H_3O^+\): The central oxygen atom has 6 valence electrons, minus 1 for the positive charge, leaving 5. It forms 3 single covalent bonds with hydrogen atoms, leaving 1 lone pair of electrons. The 3 bonding pairs and 1 lone pair arrange themselves in a tetrahedral geometry, resulting in a **trigonal pyramidal** shape, which is non-planar.

- \(C_2H_4\): Each carbon atom is \(sp^2\) hybridized and is bonded to two hydrogens and one carbon in a trigonal planar arrangement. The entire molecule is planar.

Thus, only \(H_3O^+\) is non-planar.

1 mark for the correct option (C).

The reactant is 2-methylbut-2-ene, \((CH_3)_2C=CHCH_3\).

When \(H^+\) adds to the double bond, it can form either:
1. A secondary carbocation: \((CH_3)_2CH-CH^+-CH_3\)
2. A tertiary carbocation: \((CH_3)_2C^+-CH_2CH_3\)

The tertiary carbocation is more stable than the secondary carbocation due to the greater electron-donating inductive effect of three alkyl groups attached to the positively charged carbon. Therefore, the major intermediate is the tertiary carbocation.

In the subsequent step, the bromide ion (\(Br^-\)) attacks the tertiary carbocation to form the major product, 2-bromo-2-methylbutane, which has the structural formula \((CH_3)_2C(Br)CH_2CH_3\).

1 mark for the correct option (C).

The equation for the formation of propane is:

$$3C(s) + 4H_2(g) \rightarrow C_3H_8(g)$$

Using Hess's Law and enthalpy changes of combustion:

$$\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})$$

$$\Delta H_f^\ominus = \left[ 3 \times \Delta H_c^\ominus[C(s)] + 4 \times \Delta H_c^\ominus[H_2(g)] \right] - \Delta H_c^\ominus[C_3H_8(g)]$$

$$\Delta H_f^\ominus = \left[ 3(-394) + 4(-286) \right] - (-2220)$$

$$\Delta H_f^\ominus = \left[ -1182 - 1144 \right] + 2220$$

$$\Delta H_f^\ominus = -2326 + 2220 = -106\text{ kJ mol}^{-1}$$

Hence, the standard enthalpy change of formation of propane is \(-106\text{ kJ mol}^{-1}\).

1 mark for the correct option (A).

To determine the valence shell of element \(Y\), we look for the largest jump in successive ionization energies:

- 1st to 2nd: Increase of \(1239\text{ kJ mol}^{-1}\)
- 2nd to 3rd: Increase of \(928\text{ kJ mol}^{-1}\)
- 3rd to 4th: Increase of \(8833\text{ kJ mol}^{-1}\) (a very large jump)

The huge increase between the 3rd and 4th ionization energies shows that the 4th electron is removed from an inner quantum shell that is closer to the nucleus and experiences less shielding. This means \(Y\) has 3 valence electrons and forms a \(Y^{3+}\) ion.

Since oxide ions are \(O^{2-}\), the formula of the oxide formed by a \(Y^{3+}\) cation is \(Y_2O_3\) (similar to \(Al_2O_3\), as aluminum is the Period 3 element in Group 13).

1 mark for the correct option (D).

Step 1: Calculate the amount of Ca(NO3)2. n = 8.20 g / 164 g mol-1 = 0.0500 mol. Step 2: Determine the molar ratio of gas products to reactant. According to the balanced equation, 2 moles of Ca(NO3)2 produce 5 moles of gas (4 moles of NO2 and 1 mole of O2). Therefore, n(gas) = 0.0500 * (5/2) = 0.125 mol. Step 3: Calculate the volume of gas. V = 0.125 mol * 24.0 dm3 mol-1 = 3.00 dm3.

1 mark: Correctly identifies the final volume of 3.00 dm3 (Option C).

There is a very large increase between the third and fourth ionisation energies (2745 to 11577 kJ mol-1). This indicates that the fourth electron is being removed from an inner shell, meaning there are 3 valence electrons. Therefore, the element is in Group 3 (Group 13) of the Periodic Table. In its most common oxidation state, the metal forms a 3+ ion (X3+). This combines with chloride ions (Cl-) to form XCl3.

1 mark: Correctly identifies XCl3 (Option C) by analyzing the successive ionization energy values.

Butane (CH3CH2CH2CH3) has only weak London forces. Propanal (CH3CH2CHO) has permanent dipole-dipole forces in addition to London forces, which are stronger than London forces alone. Propan-1-ol (CH3CH2CH2OH) can form intermolecular hydrogen bonds, which are the strongest type of intermolecular force. Therefore, the boiling points increase in the order: butane < propanal < propan-1-ol.

1 mark: Correctly identifies butane < propanal < propan-1-ol (Option A) based on the relative strengths of London forces, dipole-dipole interactions, and hydrogen bonding.

As the group is descended, the ionic radius of the M2+ cation increases while keeping the same 2+ charge. Therefore, the charge density of the cation decreases, reducing its power to polarise (distort) the carbonate ion. A lesser degree of polarisation makes the carbon-oxygen bonds in the carbonate ion harder to break, resulting in higher thermal stability.

1 mark: Correctly identifies the trend in cation radius and polarising ability (Option A).

The chemical equation for the formation of propane is: 3C(graphite) + 4H2(g) -> C3H8(g). Using Hess's Law with enthalpy of combustion data: dHf = 3 * dHc[C] + 4 * dHc[H2] - dHc[C3H8]. Substituting the values: dHf = 3(-393.5) + 4(-285.8) - (-2219.2) = -1180.5 - 1143.2 + 2219.2 = -104.5 kJ mol-1.

1 mark: Correctly calculates -104.5 kJ mol-1 (Option A).

The rate of hydrolysis of halogenoalkanes is determined by both the bond enthalpy of the carbon-halogen bond and the mechanism of substitution. The C-I bond is much weaker than the C-Cl bond, so iodoalkanes react faster than chloroalkanes. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster than primary halogenoalkanes because they proceed via the stable tertiary carbocation intermediate in an SN1 pathway. Therefore, 2-iodo-2-methylpropane reacts the fastest.

1 mark: Correctly identifies 2-iodo-2-methylpropane as the fastest reacting halogenoalkane (Option D).

Electrophilic addition of HBr to 2-methylbut-2-ene, (CH3)2C=CHCH3, begins with the addition of H+ to form a carbocation intermediate. The two possible carbocations are: (1) a tertiary carbocation (CH3)2C+-CH2CH3, or (2) a secondary carbocation (CH3)2CH-CH+-CH3. The tertiary carbocation is more stable due to the positive inductive effect of three electron-donating alkyl groups. Attack by Br- on the tertiary carbocation yields the major product, 2-bromo-2-methylbutane.

1 mark: Correctly identifies the major product as 2-bromo-2-methylbutane and attributes it to the formation of a more stable tertiary carbocation (Option A).

At constant temperature, the Maxwell-Boltzmann distribution of molecular kinetic energies is unchanged, so the shape of the curve does not shift. A catalyst works by providing an alternative reaction pathway with a lower activation energy (Ea). Because the activation energy is lowered, a greater fraction of the reactant molecules possess kinetic energies equal to or exceeding the activation energy, leading to an increased rate of successful collisions.

1 mark: Correctly identifies the action of a catalyst on activation energy and reactant molecules (Option C).

To find the empirical formula:

1. Calculate the mass and moles of carbon from \(\text{CO}_2\):
\(\text{Mass of C} = 2.64\text{ g} \times \frac{12.0}{44.0} = 0.72\text{ g}\)
\(\text{Moles of C} = \frac{0.72}{12.0} = 0.06\text{ mol}\)

2. Calculate the mass and moles of hydrogen from \(\text{H}_2\text{O}\):
\(\text{Mass of H} = 1.44\text{ g} \times \frac{2.0}{18.0} = 0.16\text{ g}\)
\(\text{Moles of H} = \frac{0.16}{1.0} = 0.16\text{ mol}\)

3. Calculate the mass and moles of oxygen in the compound:
\(\text{Mass of O} = 1.20\text{ g} - (0.72\text{ g} + 0.16\text{ g}) = 0.32\text{ g}\)
\(\text{Moles of O} = \frac{0.32}{16.0} = 0.02\text{ mol}\)

4. Find the simplest whole-number ratio of moles:
\(\text{C} : \text{H} : \text{O} = 0.06 : 0.16 : 0.02 = 3 : 8 : 1\)

Therefore, the empirical formula is \(\text{C}_3\text{H}_8\text{O}\).

1 mark for the correct option.

There is a very large increase (jump) between the fourth and the fifth ionisation energies (from \(4356\) to \(16091\text{ kJ mol}^{-1}\)). This indicates that the fifth electron is being removed from an inner, fully closed shell which is closer to the nucleus and experiences significantly less shielding. Thus, the element has four valence electrons and must belong to Group 4 (or Group 14).

1 mark for the correct option.

The amide ion, \(\text{NH}_2^-\), has 2 bonding pairs and 2 lone pairs on the nitrogen atom (total of 4 electron pairs, which arrange tetrahedrally). The strong repulsion between the two lone pairs reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) by approximately \(2 \times 2.5^\circ = 5^\circ\) to around \(104.5^\circ\), giving a non-linear shape similar to water.

1 mark for the correct option.

A termination step involves two free radicals combining to form a stable covalent molecule with no new radicals produced. The formation of ethane is explained by the combination of two methyl radicals: \(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\).

1 mark for the correct option.

Addition of \(\text{H}^+\) to the double bond of 2-methylbut-2-ene (\(\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3\)) can yield either a tertiary carbocation (\(\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\)) or a secondary carbocation (\(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}^+-\text{CH}_3\)). The tertiary carbocation is more stable because of the electron-donating inductive effect of three alkyl groups compared to two in the secondary carbocation. Therefore, the major product is formed via the more stable tertiary carbocation intermediate.

1 mark for the correct option.

The chemical equation for the standard enthalpy of formation of propane is:
\(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\)

Using Hess's Law with enthalpies of combustion:
\(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\)
\(\Delta_f H^\ominus = \left[3 \times (-393.5) + 4 \times (-285.8)\right] - (-2220.0)\)
\(\Delta_f H^\ominus = [-1180.5 - 1143.2] + 2220.0\)
\(\Delta_f H^\ominus = -2323.7 + 2220.0 = -103.7\text{ kJ mol}^{-1}\).

1 mark for the correct option.

The ionic equation for the reaction is:
\(\text{Cl}_2\text{(g)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\)

In this reaction, the oxidation state of iodine increases from \(-1\) in \(\text{I}^-\text{(aq)}\) to \(0\) in \(\text{I}_2\text{(aq)}\). Because the iodide ion (\(\text{I}^-\text{(aq)}\)) loses electrons and undergoes oxidation, it acts as the reducing agent.

1 mark for the correct option.

The rate of nucleophilic substitution (hydrolysis) of halogenoalkanes is determined by the strength of the carbon-halogen bond (bond enthalpy), not the bond polarity. Since the bond enthalpy of the \(\text{C}-\text{I}\) bond is much lower than that of the \(\text{C}-\text{Br}\) and \(\text{C}-\text{Cl}\) bonds, the \(\text{C}-\text{I}\) bond breaks most easily. Therefore, 1-iodobutane reacts fastest, and the rate order is 1-iodobutane > 1-bromobutane > 1-chlorobutane.

1 mark for the correct option.

First, calculate the moles of anhydrous \(\text{CaCl}_2\): \(M_{\text{r}}(\text{CaCl}_2) = 40.1 + (2 \times 35.5) = 111.1\text{ g mol}^{-1}\). Moles of \(\text{CaCl}_2 = \frac{2.775\text{ g}}{111.1\text{ g mol}^{-1}} = 0.0250\text{ mol}\). Next, find the mass and moles of water lost: Mass of water = \(3.675\text{ g} - 2.775\text{ g} = 0.900\text{ g}\). Moles of \(\text{H}_2\text{O} = \frac{0.900\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\). Finally, find the mole ratio: \(x = \frac{0.0500\text{ mol}}{0.0250\text{ mol}} = 2\).

1 mark for the correct option B.

\(\text{CO}_2\) is a linear molecule with a bond angle of \(180^\circ\). \(\text{BF}_3\) is trigonal planar with a bond angle of \(120^\circ\). \(\text{NH}_3\) is trigonal pyramidal with a bond angle of \(107^\circ\) due to the repulsion from its single lone pair. \(\text{H}_2\text{O}\) is bent/non-linear with a bond angle of \(104.5^\circ\) because it contains two lone pairs which repel more strongly than the bonding pairs. Therefore, \(\text{H}_2\text{O}\) has the smallest bond angle.

1 mark for the correct option D.

The combustion equation for ethane is: \(\text{C}_2\text{H}_6\text{(g)} + 3.5\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\). Applying Hess's Law: \(\Delta H_{\text{c}}^\ominus = 2\Delta H_{\text{f}}^\ominus(\text{CO}_2) + 3\Delta H_{\text{f}}^\ominus(\text{H}_2\text{O}) - \Delta H_{\text{f}}^\ominus(\text{C}_2\text{H}_6) = 2(-393.5) + 3(-285.8) - (-84.7) = -787.0 - 857.4 + 84.7 = -1559.7\text{ kJ mol}^{-1}\).

1 mark for the correct option B.

Down Group 2, the ionic radius of the cation increases while the charge remains constant at \(2+\). This leads to a decrease in the charge density of the cation. Consequently, the cation polarizes (distorts) the electron cloud of the carbonate ion less easily, which makes the carbonate compound more stable to thermal decomposition.

1 mark for the correct option A.

The carbon-halogen bond strength decreases down Group 7, meaning the \(\text{C-Cl}\) bond is stronger than the \(\text{C-Br}\) bond and requires more energy to break. Therefore, 1-bromobutane undergoes nucleophilic substitution (hydrolysis) faster than 1-chlorobutane. Since bromide ions react with silver ions to form a cream precipitate of \(\text{AgBr}\) and chloride ions form a white precipitate of \(\text{AgCl}\), 1-bromobutane will form a cream precipitate more rapidly than 1-chlorobutane.

1 mark for the correct option B.

On carbon-2, the two attached groups are \(\text{-CH}_3\) (higher priority) and \(\text{-H}\) (lower priority). On carbon-3, the two attached groups are \(\text{-CH}_2\text{CH}_2\text{CH}_3\) (higher priority) and \(\text{-CH}_3\) (lower priority). Since the high-priority groups (the methyl group on C2 and the propyl group on C3) are on opposite sides of the double bond, it is the (E)-isomer. Therefore, the name is (E)-3-methylhex-2-ene.

1 mark for the correct option A.

Decreasing the volume of the reaction vessel increases the pressure. This increases the concentration of gaseous particles, causing more frequent successful collisions and increasing the reaction rate. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium to the side with fewer moles of gas (the right-hand side has 2 moles of gas, whereas the left-hand side has 3 moles), thereby increasing the yield of \(\text{SO}_3\).

1 mark for the correct option B.

There is a very large jump between the 3rd and 4th ionization energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)), indicating that the fourth electron is being removed from an inner electron shell. This means element X has three outer (valence) electrons, placing it in Group 13 (Group 3).

1 mark for the correct option C.

1. Find the percentage of oxygen by mass: \( 100\% - 30.4\% = 69.6\% \).
2. Calculate the number of moles of each element in a 100 g sample:
- Moles of \( \text{N} = \frac{30.4}{14.0} = 2.17\text{ mol} \)
- Moles of \( \text{O} = \frac{69.6}{16.0} = 4.35\text{ mol} \)
3. Determine the simplest whole-number ratio:
- N: \( \frac{2.17}{2.17} = 1 \)
- O: \( \frac{4.35}{2.17} \approx 2 \)

Therefore, the empirical formula is \( \text{NO}_2 \).

1 mark for selecting the correct option (B).

Sodium (\( \text{Na} \)) has the electronic configuration \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^1 \). Its first ionization energy involves removing the single valence electron from the \( 3\text{s} \) orbital. The second ionization energy involves removing an electron from the stable, full \( 2\text{p} \) subshell of the \( \text{Na}^+ \) ion. This electron experiences a much stronger electrostatic attraction to the nucleus due to less shielding and being closer to the nucleus, resulting in an extremely high second ionization energy compared to Mg, Al, or Si (where the second electron is still removed from the outer third shell).

1 mark for selecting the correct element (A).

Phosphorus is in Group 15 and has 5 valence electrons. In \( \text{PCl}_3 \), phosphorus forms 3 single covalent bonds with chlorine atoms and has 1 lone pair of electrons remaining. This gives a total of 4 electron pairs, which arrange themselves tetrahedrally to minimize repulsion. The presence of the lone pair repels the bonding pairs more strongly, reducing the bond angle from the tetrahedral angle of 109.5° to approximately 107°, making the molecular shape trigonal pyramidal.

1 mark for the correct shape and angle (B).

The addition of HBr to an unsymmetrical alkene like 2-methylbut-2-ene, \( \text{(CH}_3)_2\text{C=CHCH}_3 \), follows Markovnikov's rule. The electrophile (\( \text{H}^+ \)) adds to the double-bonded carbon with the greater number of hydrogen atoms (C3) to form the more stable tertiary carbocation, \( \text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3 \), rather than a secondary carbocation. The bromide ion (\( \text{Br}^- \)) then attacks this tertiary carbocation, forming 2-bromo-2-methylbutane as the major product.

1 mark for identifying the correct IUPAC name of the major product (A).

The chemical equation for the standard enthalpy change of formation of propane is:
\( 3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)} \)

According to Hess's Law using combustion data:
\( \Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products}) \)
\( \Delta H^\theta_f = [3 \times \Delta H^\theta_c(\text{C(s)})] + [4 \times \Delta H^\theta_c(\text{H}_2\text{(g)})] - [\Delta H^\theta_c(\text{C}_3\text{H}_8\text{(g)})] \)
\( \Delta H^\theta_f = [3 \times (-393.5)] + [4 \times (-285.8)] - (-2219.2) \)
\( \Delta H^\theta_f = -1180.5 - 1143.2 + 2219.2 = -104.5\text{ kJ mol}^{-1} \)

1 mark for the correct calculation and sign (A).

Water (\( \text{H}_2\text{O} \)) has the highest boiling temperature because each water molecule can participate in an average of two hydrogen bonds, forming a highly extensive 3D hydrogen-bonding network. Although HF has stronger individual hydrogen bonds due to fluorine's higher electronegativity, each HF molecule can only form one hydrogen bond on average because of the 1:3 ratio of hydrogen atoms to lone pairs. Ammonia (\( \text{NH}_3 \)) is limited by having only one lone pair per molecule, and methane (\( \text{CH}_4 \)) only experiences weaker London forces.

1 mark for selecting the correct hydride (D).

As you go down Group 2 from magnesium to barium:
- The solubility of Group 2 hydroxides increases (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble).
- The solubility of Group 2 sulfates decreases (magnesium sulfate is highly soluble, whereas barium sulfate is virtually insoluble).

1 mark for identifying the correct trends (C).

At a higher temperature, the average kinetic energy of the molecules increases, meaning a greater proportion of molecules have higher energy. Consequently, the peak of the curve (the most probable molecular energy) shifts to a higher energy (to the right). To maintain the same total area under the curve (representing the constant number of gas particles), the peak must also become lower.

1 mark for the correct changes to the peak's position and height (D).

Ammonia, \(\text{NH}_3\), has 4 electron pairs around the nitrogen atom: 3 bonding pairs and 1 lone pair. This gives a trigonal pyramidal shape with a bond angle of \(107^\circ\) (or \(107.5^\circ\)). Water, \(\text{H}_2\text{O}\), also has 4 electron pairs around the oxygen atom: 2 bonding pairs and 2 lone pairs, resulting in a non-linear (bent) shape with a bond angle of \(104.5^\circ\). Since lone pair-lone pair repulsion is greater than lone pair-bonding pair repulsion, which in turn is greater than bonding pair-bonding pair repulsion, the two lone pairs in water push the bonding pairs closer together than the single lone pair in ammonia, reducing the bond angle.

1 mark: Correctly state both bond angles (107°/107.5° for ammonia, 104.5° for water). 1 mark: Identify the electron pair distribution (ammonia: 3 bonding, 1 lone; water: 2 bonding, 2 lone). 0.5 mark: Explain that lone pair repulsion is greater than bonding pair repulsion.

The bond formed is a dative covalent (or coordinate) bond. Nitrogen has a lone pair of electrons in its outer shell which it donates into the vacant orbital of the boron atom. Before the reaction, boron has 3 bonding pairs and 0 lone pairs, giving a trigonal planar shape (bond angle \(120^\circ\)). After the reaction, boron has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape (bond angle \(109.5^\circ\)).

1 mark: Identify the bond as a dative covalent / coordinate bond and explain that the lone pair on nitrogen is donated to the empty orbital of boron. 1 mark: Describe the shape change around boron from trigonal planar to tetrahedral. 0.5 mark: State the bond angles correctly (from 120° to 109.5°).

Magnesium oxide consists of \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) ions, which carry charges of \(+2\) and \(-2\) respectively. Sodium chloride consists of \(\text{Na}^+\) and \(\text{Cl}^-\) ions, with charges of \(+1\) and \(-1\). Additionally, \(\text{Mg}^{2+}\) is smaller than \(\text{Na}^+\), and \(\text{O}^{2-}\) is smaller than \(\text{Cl}^-\). The combination of higher charges and smaller ionic radii results in a much higher charge density in \(\text{MgO}\). Consequently, the electrostatic forces of attraction between the oppositely charged ions in the giant ionic lattice of \(\text{MgO}\) are significantly stronger than in \(\text{NaCl}\), requiring much more thermal energy to overcome.

1 mark: State that the charges on the ions in MgO (+2 and -2) are higher than in NaCl (+1 and -1). 1 mark: State that Mg2+ and/or O2- are smaller than Na+ and/or Cl-, leading to a higher charge density. 0.5 mark: Conclude that the electrostatic attraction between ions is stronger in MgO, requiring more energy to overcome.

Graphite has a giant covalent structure where each carbon atom is covalently bonded to three others in planar layers, leaving one delocalised electron per carbon atom. These delocalised electrons are free to move throughout the layers, allowing graphite to conduct electricity. The strong covalent bonds throughout the giant lattice require a very large amount of energy to break, resulting in a high sublimation temperature. In contrast, iodine (\(\text{I}_2\)) has a simple molecular structure. The molecules are held together only by weak intermolecular London forces, which require little energy to overcome, resulting in a low sublimation temperature. Since all electrons in iodine are localised within the atoms or covalent bonds, there are no mobile charge carriers to conduct electricity.

1 mark: Describe graphite as giant covalent with delocalised electrons, and iodine as simple molecular with weak London forces. 1 mark: Explain conductivity (delocalised electrons in graphite can move, while iodine has no mobile charge carriers). 0.5 mark: Explain sublimation temperature (breaking strong covalent bonds in graphite requires lots of energy, while overcoming weak London forces in iodine requires very little).

Fluorine is highly electronegative, which allows \(\text{HF}\) to form strong intermolecular hydrogen bonds. In contrast, chlorine is less electronegative, so \(\text{HCl}\) molecules are held together only by weaker permanent dipole-dipole attractions and London forces, resulting in a much lower boiling temperature for \(\text{HCl}\). Water (\(\text{H}_2\text{O}\)) has a higher boiling temperature than \(\text{HF}\) because each water molecule has two hydrogen atoms and two lone pairs on the oxygen atom, allowing it to form, on average, four hydrogen bonds per molecule. Each \(\text{HF}\) molecule, despite having stronger individual hydrogen bonds due to the greater electronegativity of fluorine, can only form an average of two hydrogen bonds per molecule because of a shortage of hydrogen atoms relative to lone pairs. This results in a more extensive network of hydrogen bonds in water, requiring more energy to break.

1 mark: Explain that HF forms hydrogen bonds while HCl only forms weaker dipole-dipole/London forces. 1 mark: Explain that H2O can form up to 4 hydrogen bonds per molecule on average, whereas HF can only form 2. 0.5 mark: Link the number of hydrogen bonds to the higher energy required to boil water compared to HF.

Metallic bonding consists of a giant lattice of positive metal ions (cations) surrounded by a 'sea' of delocalised valence electrons. Electrical conductivity is explained by the fact that these delocalised electrons are mobile and can move through the lattice when a potential difference is applied. Malleability is explained by the ability of the layers of positive ions to slide past one another when a force is applied, without shattering the structure. This is because the delocalised electrons are free to move with the layers, maintaining the electrostatic attraction (the metallic bond) between the cations and the electron cloud at all times.

1 mark: Describe metallic bonding as a lattice of positive metal ions surrounded by a sea of delocalised electrons. 1 mark: Explain malleability by stating that layers of cations can slide past one another without breaking the bond due to the mobile delocalised electrons. 0.5 mark: Explain conductivity by stating that the delocalised electrons are mobile and can carry charge.

The beryllium ion (\(\text{Be}^{2+}\)) has the same charge (\(+2\)) as the calcium ion (\(\text{Ca}^{2+}\)) but has a much smaller ionic radius. Consequently, the \(\text{Be}^{2+}\) ion has a much higher charge density and thus a much greater polarizing power than the \(\text{Ca}^{2+}\) ion. This highly polarizing \(\text{Be}^{2+}\) ion strongly distorts the electron cloud of the neighboring chloride ions (\(\text{Cl}^-\)), pulling the electron density into the space between the nuclei. This significant distortion results in a high degree of electron sharing, which gives \(\text{BeCl}_2\) substantial covalent character compared to the more purely ionic \(\text{CaCl}_2\).

1 mark: Explain that Be2+ is smaller than Ca2+ and therefore has a higher charge density. 1 mark: State that Be2+ has greater polarizing power and polarizes/distorts the electron cloud of the chloride ion (Cl-). 0.5 mark: Link this polarization to the sharing of electron density, which causes covalent character.

The nitrogen atom in the ammonium ion (\(\text{NH}_4^+\)) has 5 valence electrons, minus 1 for the positive charge, leaving 4 valence electrons. These are shared with 4 hydrogen atoms to form 4 bonding pairs and 0 lone pairs in the outer shell of nitrogen. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs around a central atom repel each other to be as far apart as possible to minimize repulsion. Since there are 4 bonding pairs and no lone pairs, they repel each other equally, resulting in a symmetrical tetrahedral geometry with bond angles of \(109.5^\circ\).

1 mark: Identify that there are 4 bonding pairs and 0 lone pairs around the central nitrogen atom. 1 mark: State that the shape is tetrahedral with a bond angle of 109.5°. 0.5 mark: Explain that electron pairs repel to minimize repulsion/maximize separation.

Silicon dioxide (\(\text{SiO}_2\)) exists as a giant covalent lattice. To melt it, strong covalent bonds between silicon and oxygen atoms must be broken, which requires a very large amount of thermal energy. In contrast, silicon tetrachloride (\(\text{SiCl}_4\)) is a simple molecular substance. When it melts, only the weak intermolecular London forces between the individual molecules must be overcome, which requires relatively little energy.

1 mark: Identify that \(\text{SiO}_2\) has a giant covalent structure and \(\text{SiCl}_4\) has a simple molecular structure. 1 mark: Explain that melting \(\text{SiO}_2\) requires breaking strong covalent bonds, while melting \(\text{SiCl}_4\) requires overcoming weak intermolecular/London forces. 0.5 marks: Conclude that significantly more energy is required to break the covalent bonds in \(\text{SiO}_2\) than to overcome the intermolecular forces in \(\text{SiCl}_4\).

The ammonium ion (\(\text{NH}_4^+\)) has 4 bonding pairs and 0 lone pairs of electrons around the central nitrogen atom. These 4 pairs repel each other equally to positions of maximum separation, forming a regular tetrahedral shape with a bond angle of \(109.5^{\circ}\). In contrast, ammonia (\(\text{NH}_3\)) has 3 bonding pairs and 1 lone pair of electrons. Because lone pairs repel more strongly than bonding pairs, the lone pair compresses the bonding pairs together, reducing the \(\text{H}-\text{N}-\text{H}\) bond angle to approximately \(107^{\circ}\).

0.5 marks: Predict the bond angle in \(\text{NH}_4^+\) as \(109.5^{\circ}\) (accept \(109^{\circ}\) to \(110^{\circ}\)). 1 mark: State that \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, whereas \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs. 1 mark: Explain that lone pairs repel more strongly than bonding pairs, which reduces the bond angle in \(\text{NH}_3\) relative to the equal repulsion between the bonding pairs in \(\text{NH}_4^+\).

Both metals have metallic bonding, consisting of a lattice of metal cations surrounded by delocalised valence electrons. An aluminium atom loses three electrons to form an \(\text{Al}^{3+}\) ion, contributing three electrons to the delocalised pool. A sodium atom loses only one electron to form a \(\text{Na}^+\) ion, contributing one electron. Consequently, aluminium has a higher density of mobile, delocalised electrons to carry electrical charge through the lattice when a potential difference is applied.

1 mark: Identify that aluminium contributes more delocalised electrons per atom/ion (three from \(\text{Al}^{3+}\)) than sodium (one from \(\text{Na}^+\)). 1 mark: Explain that this results in a higher density or concentration of mobile/delocalised electrons in aluminium. 0.5 marks: Link the delocalised electrons to their role as mobile charge carriers that conduct electricity.

Oxygen is more electronegative than carbon, which pulls electron density towards the oxygen atoms and makes the \(\text{C}=\text{O}\) bonds polar. However, carbon dioxide (\(\text{CO}_2\)) is a linear molecule, which means it is highly symmetrical. The two bond dipoles are equal in magnitude but point in exactly opposite directions. Therefore, the individual dipoles cancel each other out completely, leaving the molecule with no net dipole moment.

1 mark: Explain that oxygen is more electronegative than carbon, creating polar \(\text{C}=\text{O}\) bonds (or individual bond dipoles). 1 mark: State that the carbon dioxide molecule is linear and symmetrical. 0.5 marks: Explain that this symmetry causes the individual dipoles to pull in opposite directions and cancel each other out.

Step 1: Convert the side length of the cubic crystal from millimeters to centimeters.
\(1.20 \text{ mm} = 0.120 \text{ cm}\)

Step 2: Calculate the volume of the cubic crystal in \(\text{cm}^3\).
\(V = (0.120 \text{ cm})^3 = 1.728 \times 10^{-3} \text{ cm}^3\)

Step 3: Calculate the mass of the crystal using the density.
\(m = \text{density} \times V = 2.17 \text{ g cm}^{-3} \times 1.728 \times 10^{-3} \text{ cm}^3 = 3.74976 \times 10^{-3} \text{ g}\)

Step 4: Calculate the moles of sodium chloride in the crystal.
\(n(\text{NaCl}) = \frac{3.74976 \times 10^{-3} \text{ g}}{58.5 \text{ g mol}^{-1}} = 6.40985 \times 10^{-5} \text{ mol}\)

Step 5: Determine the total moles of ions.
Each formula unit of \(\text{NaCl}\) contains 1 \(\text{Na}^+\)\ ion and 1 \(\text{Cl}^-\)\ ion, giving a total of 2 moles of ions per mole of \(\text{NaCl}\).
\(n(\text{ions}) = 2 \times 6.40985 \times 10^{-5} \text{ mol} = 1.28197 \times 10^{-4} \text{ mol}\)

Step 6: Calculate the total number of ions in the crystal.
\(\text{Total number of ions} = 1.28197 \times 10^{-4} \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 7.71746 \times 10^{19}\)

Rounding to 3 significant figures gives \(7.72 \times 10^{19}\).

- **M1**: Calculation of crystal volume in \(\text{cm}^3\) (\(1.728 \times 10^{-3} \text{ cm}^3\)) [1 Mark]
- **M2**: Calculation of mass of the crystal (\(3.75 \times 10^{-3} \text{ g}\) or \(3.7498 \times 10^{-3} \text{ g}\)) [1 Mark]
- **M3**: Calculation of amount of \(\text{NaCl}\) in moles (\(6.41 \times 10^{-5} \text{ mol}\)) [1 Mark]
- **M4**: Calculation of amount of total ions in moles (\(1.282 \times 10^{-4} \text{ mol}\)) [1 Mark]
- **M5**: Correct final calculation of total number of ions to 3 significant figures (\(7.72 \times 10^{19}\)) [1 Mark]

*(Allow full marks for a correct final answer with no intermediate steps shown. Allow consequential error (TE) throughout intermediate steps.)*

Step 1: Calculate the molar mass of silicon dioxide, \(\text{SiO}_2\).
\(M(\text{SiO}_2) = 28.1 + (2 \times 16.0) = 60.1 \text{ g mol}^{-1}\)

Step 2: Calculate the amount of \(\text{SiO}_2\) in moles.
\(n(\text{SiO}_2) = \frac{1.50 \text{ g}}{60.1 \text{ g mol}^{-1}} = 0.024958 \text{ mol}\)

Step 3: Determine the ratio of covalent bonds to \(\text{SiO}_2\) formula units.
In the giant molecular structure of silica, each silicon atom is bonded to 4 oxygen atoms by single covalent bonds, and each oxygen atom is bonded to 2 silicon atoms. Thus, there are exactly 4 single covalent bonds associated with each silicon dioxide formula unit.

Step 4: Calculate the total amount of covalent \(\text{Si–O}\) bonds in moles.
\(n(\text{bonds}) = 4 \times 0.024958 \text{ mol} = 0.099832 \text{ mol}\)

Step 5: Calculate the total number of covalent \(\text{Si–O}\) bonds.
\(\text{Total number of bonds} = 0.099832 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 6.0099 \times 10^{22}\)

Rounding to 3 significant figures gives \(6.01 \times 10^{22}\).

- **M1**: Calculation of molar mass of \(\text{SiO}_2\) as \(60.1 \text{ g mol}^{-1}\) [1 Mark]
- **M2**: Calculation of the amount of \(\text{SiO}_2\) as \(0.02496 \text{ mol}\) [1 Mark]
- **M3**: Deduction that there are 4 covalent \(\text{Si–O}\) bonds per formula unit of \(\text{SiO}_2\) [1 Mark]
- **M4**: Calculation of the total moles of bonds as \(0.0998 \text{ mol}\) [1 Mark]
- **M5**: Correct evaluation of the total number of covalent bonds to 3 significant figures (\(6.01 \times 10^{22}\)) [1 Mark]

*(Accept range: \(6.0 \times 10^{22}\) to \(6.02 \times 10^{22}\) depending on intermediate rounding. Allow TE throughout.)*

Step 1: Calculate the molar mass of \(\text{C}_{60}\).
\(M(\text{C}_{60}) = 60 \times 12.0 \text{ g mol}^{-1} = 720 \text{ g mol}^{-1}\)

Step 2: Calculate the moles of \(\text{C}_{60}\) in the sample.
\(n(\text{C}_{60}) = \frac{0.450 \text{ g}}{720 \text{ g mol}^{-1}} = 6.25 \times 10^{-4} \text{ mol}\)

Step 3: Determine the number of shared electron pairs per molecule.
Each single bond contains 1 shared pair of electrons, and each double bond contains 2 shared pairs of electrons.
\(\text{Shared pairs per molecule} = 60 + (30 \times 2) = 120\) shared pairs of electrons (which corresponds to all 4 valence electrons of the 60 carbon atoms being shared: \(60 \times 4 / 2 = 120\)).

Step 4: Calculate the total moles of shared electron pairs in the sample.
\(n(\text{shared pairs}) = 6.25 \times 10^{-4} \text{ mol} \times 120 = 0.0750 \text{ mol}\)

Step 5: Calculate the total number of shared pairs of electrons.
\(\text{Total number} = 0.0750 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 4.515 \times 10^{22}\)

Rounding to 3 significant figures gives \(4.52 \times 10^{22}\).

- **M1**: Calculation of molar mass of \(\text{C}_{60}\) as \(720 \text{ g mol}^{-1}\) [1 Mark]
- **M2**: Calculation of the amount of \(\text{C}_{60}\) as \(6.25 \times 10^{-4} \text{ mol}\) [1 Mark]
- **M3**: Deduction that there are 120 shared pairs of electrons (covalent bonds) per molecule [1 Mark]
- **M4**: Calculation of the total moles of shared electron pairs as \(0.0750 \text{ mol}\) [1 Mark]
- **M5**: Final answer of \(4.52 \times 10^{22}\) (accept \(4.5 \times 10^{22}\) or \(4.515 \times 10^{22}\)) [1 Mark]

*(Allow TE throughout.)*

The dot-and-cross diagram of \(\text{H}_3\text{O}^+\) shows: oxygen as the central atom with three single covalent bonds to three hydrogen atoms (each bond consisting of one dot and one cross, or two of the same symbol if representing the dative covalent bond) and one lone pair of electrons remaining on oxygen. The entire structure is enclosed in square brackets with a \(+\) sign outside. Shape: The ion has a trigonal pyramidal shape. Bond angle: The bond angle is approximately \(107^\circ\) (accept \(106^\circ\) to \(108^\circ\)). Explanation: There are four areas of electron density (three bonding pairs and one lone pair) around the central oxygen atom. These electron pairs repel each other to a position of minimum repulsion (tetrahedral arrangement). Since lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, the bond angle is compressed from the tetrahedral angle of \(109.5^\circ\) to approximately \(107^\circ\).

M1: Correct dot-and-cross diagram of \(\text{H}_3\text{O}^+\) showing 3 bonding pairs, 1 lone pair on oxygen, and overall \(+\) charge. [1 mark]
M2: Shape stated as trigonal pyramidal. [1 mark]
M3: Bond angle stated as \(107^\circ\) (accept \(106^\circ\) to \(108^\circ\)). [1 mark]
M4: Explanation of shape: Four areas of electron density / electron pairs around the central oxygen atom repel to a position of maximum separation / minimum repulsion. [1 mark]
M5: Explanation of bond angle: Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion. [1 mark]

The diagram must show a regular, closely packed arrangement of magnesium cations, each labelled with a \(2+\) charge (or \(\text{Mg}^{2+}\)). Surrounding these cations, there should be a 'sea' of delocalised electrons, represented as dots or \(e^-\). To explain the difference in melting temperatures: Magnesium ions have a higher charge (\(2+\) vs \(1+\)) and a smaller ionic radius than sodium ions, giving them a much higher charge density. Furthermore, each magnesium atom contributes two delocalised electrons to the sea, compared to only one from sodium. This results in a significantly stronger electrostatic force of attraction between the magnesium cations and the delocalised electrons, requiring more thermal energy to overcome and break the metallic lattice.

M1: Diagram showing a regular, closely packed arrangement of metal cations with \(2+\) or \(\text{Mg}^{2+}\) labels. [1 mark]
M2: Diagram showing randomly distributed delocalised electrons (labelled \(e^-\)) in the spaces between the cations. [1 mark]
M3: Magnesium has \(2+\) ions which have a greater charge and smaller ionic radius (higher charge density) than sodium's \(1+\) ions. [1 mark]
M4: Magnesium contributes more delocalised electrons per atom (two vs one for sodium). [1 mark]
M5: Stronger electrostatic attraction between cations and delocalised electrons in magnesium (requires more energy to break). [1 mark]

The diagram must show a water molecule and an ethanol molecule. In both molecules, the highly electronegative oxygen atoms are labelled with a partial negative charge (\(\delta-\)) and the hydrogen atoms of the \(\text{O}-\text{H}\) bonds are labelled with a partial positive charge (\(\delta+\)). At least one lone pair of electrons must be drawn on the oxygen atom of the receiving molecule (either water or ethanol). A dashed or dotted line represents the hydrogen bond, which must extend directly from this lone pair to the \(\delta+\) hydrogen atom of the other molecule. The covalent \(\text{O}-\text{H}\) bond, the hydrogen bond, and the oxygen atom of the second molecule must be shown in a linear arrangement (\(\text{O}-\text{H}\cdots\text{O}\)). The bond angle around the hydrogen atom involved in the hydrogen bond is labelled as \(180^\circ\).

M1: Correct structural representations of both \(\text{H}_2\text{O}\) and \(\text{CH}_3\text{CH}_2\text{OH}\) with \(\text{O}-\text{H}\) bonds shown. [1 mark]
M2: Correct dipoles shown on both participating molecules (\(\delta-\)\(\text{O}\) and \(\delta+\)\(\text{H}\)). [1 mark]
M3: Dashed/dotted line representing the hydrogen bond from a lone pair on oxygen to the \(\delta+\)\(\text{H}\) atom of the other molecule. [1 mark]
M4: Lone pair shown on the oxygen atom participating in the hydrogen bond. [1 mark]
M5: Stating/labelling the bond angle around the hydrogen atom involved in the hydrogen bond as \(180^\circ\) (allow \(170^\circ - 180^\circ\)). [1 mark]

First, calculate the heat energy transferred to the water: \(q = m c \Delta T = 100.0\ \text{g} \times 4.18\ \text{J g}^{-1}\ \text{K}^{-1} \times 35.5\ \text{K} = 14839\ \text{J} = 14.839\ \text{kJ}\). Next, calculate the moles of methanol burned: \(n = \frac{\text{mass}}{M_r} = \frac{0.80\ \text{g}}{32.0\ \text{g mol}^{-1}} = 0.0250\ \text{mol}\). Finally, calculate the enthalpy change of combustion per mole of methanol: \(\Delta_c H = -\frac{q}{n} = -\frac{14.839\ \text{kJ}}{0.0250\ \text{mol}} = -593.56\ \text{kJ mol}^{-1}\). Rounding to 3 significant figures gives \(-594\ \text{kJ mol}^{-1}\).

M1: Correct calculation of heat energy, \(q = 14.8\ \text{kJ}\) (or \(14839\ \text{J}\)). M2: Correct calculation of moles of methanol, \(n = 0.025\ \text{mol}\). M3: Correct calculation of enthalpy change of combustion with negative sign, to 3 significant figures: \(-594\ \text{kJ mol}^{-1}\). (Allow ECF from transcription errors).

To obtain propanal, the reaction mixture must be heated under distillation rather than reflux. Acidified potassium dichromate(VI) should be added dropwise to the warm alcohol, or the alcohol should be kept in excess. Because propanal has a lower boiling point than propan-1-ol (due to the lack of hydrogen bonding), it vaporizes and distils out of the reaction flask immediately upon formation, preventing further oxidation to propanoic acid.

M1: State the use of distillation apparatus (or immediate distillation / heating under distillation). M2: State that the oxidizing agent (acidified potassium dichromate(VI)) should be added dropwise OR alcohol should be in excess. M3: Explain that propanal has a lower boiling point than propan-1-ol/propanoic acid, so it distils off as soon as it is formed, which prevents further oxidation.

Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the equilibrium position to the right (products side) to oppose the increase in temperature by absorbing heat. This increases the yield of \(\text{PCl}_3\text{(g)}\). Because the equilibrium shifts to the right, the concentrations of the products increase relative to the reactants, meaning the equilibrium constant, \(K_c\), increases.

M1: State that the yield of \(\text{PCl}_3\) increases. M2: Explain that the forward reaction is endothermic, so increasing the temperature shifts the equilibrium to the right (to oppose the temperature rise). M3: State that the value of \(K_c\) increases.

1-bromobutane is hydrolyzed faster than 1-chlorobutane. This is because the rate of hydrolysis depends on the bond enthalpy of the carbon-halogen bond, not its polarity. The C-Br bond is longer and weaker (has a lower bond enthalpy) than the C-Cl bond because bromine has a larger atomic radius and more shielding, leading to a weaker electrostatic attraction between the bonding pair of electrons and the nuclei. Therefore, the C-Br bond is broken more easily.

M1: Identify that 1-bromobutane reacts faster / hydrolyzes at a faster rate than 1-chlorobutane. M2: State that the C-Br bond is weaker than the C-Cl bond / has a lower bond enthalpy. M3: Explain that bond enthalpy (or bond strength) is the dominant factor determining the rate of reaction, rather than bond polarity.

The equation for the standard enthalpy change of formation of propene is: \(3\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_6\text{(g)}\). Using Hess's cycle based on combustion data: \(\Delta_f H^\ominus = \sum \Delta_c H^\ominus[\text{reactants}] - \sum \Delta_c H^\ominus[\text{products}]\). \(\Delta_f H^\ominus = 3\Delta_c H^\ominus[\text{C(s)}] + 3\Delta_c H^\ominus[\text{H}_2\text{(g)}] - \Delta_c H^\ominus[\text{C}_3\text{H}_6\text{(g)}]\). \(\Delta_f H^\ominus = 3(-394) + 3(-286) - (-2058) = -1182 - 858 + 2058 = +18\ \text{kJ mol}^{-1}\).

M1: Write a correct expression or construct a valid Hess's cycle for the calculation: \(3 \times (-394) + 3 \times (-286) - (-2058)\). M2: Calculate the sum of reactant combustion enthalpies as \(-2040\ \text{kJ mol}^{-1}\). M3: State final value of \(+18\ \text{kJ mol}^{-1}\) (must include '+' sign and units).

Stereoisomerism (specifically geometric or cis-trans isomerism) in alkenes occurs due to restricted rotation about the carbon-carbon double bond (\(\text{C=C}\)) caused by the presence of a \(\pi\) bond. For geometric isomerism to exist, each carbon atom of the \(\text{C=C}\) bond must be attached to two different groups. In but-2-ene, both carbon atoms are attached to a hydrogen atom (\(-\text{H}\)) and a methyl group (\(-\text{CH}_3\)). In but-1-ene, one of the double-bonded carbons is attached to two identical hydrogen atoms (\(-\text{H}\)), making geometric isomerism impossible.

M1: State that there is restricted rotation about the \(\text{C=C}\) double bond (due to the presence of a \(\pi\) bond). M2: Explain that in but-2-ene, each carbon of the double bond is attached to two different groups (\(-\text{H}\) and \(-\text{CH}_3\)). M3: Explain that in but-1-ene, one of the carbon atoms of the double bond is attached to two identical groups / two hydrogen atoms.

The strong, sharp absorption band at \(1715\ \text{cm}^{-1}\) indicates the presence of a carbonyl group, \(\text{C=O}\). The lack of a broad absorption band in the \(3200\text{--}3750\ \text{cm}^{-1}\) range indicates the absence of an \(\text{O-H}\) bond (meaning \(\text{X}\) is not an alcohol). With three carbon atoms and the formula \(\text{C}_3\text{H}_6\text{O}\), a carbonyl-containing compound must be either propanone (a ketone) or propanal (an aldehyde). Both answers are acceptable as both contain the carbonyl group.

M1: Identify the functional group as a carbonyl group (or ketone/aldehyde) due to the peak at \(1715\ \text{cm}^{-1}\) (corresponding to the \(\text{C=O}\) stretch). M2: Explain that the absence of a broad peak at \(3200\text{--}3750\ \text{cm}^{-1}\) indicates that there is no alcohol / no \(\text{O-H}\) bond present. M3: Correctly identify \(\text{X}\) as propanone (or propanal).

A catalyst provides an alternative pathway for the reaction with a lower activation energy (\(E_a\)). On a Maxwell-Boltzmann distribution curve, the activation energy line shifts to the left to a lower energy value. As a result, a larger fraction (greater proportion) of molecules have kinetic energy greater than or equal to this new activation energy. This leads to a higher frequency of successful collisions, thereby increasing the rate of the reaction.

M1: State that a catalyst provides an alternative pathway with a lower activation energy. M2: Explain (with reference to the Maxwell-Boltzmann distribution) that a greater fraction of molecules now have kinetic energy greater than or equal to the lower activation energy. M3: State that this leads to more frequent successful collisions (per unit time).

1. Calculate the heat energy released (q):
\(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)
\(\Delta T = 26.9\ ^\circ\text{C} - 20.2\ ^\circ\text{C} = 6.7\ ^\circ\text{C}\)
\(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 6.7\ ^\circ\text{C} = 2800.6\text{ J} = 2.8006\text{ kJ}\)

2. Calculate the moles of water formed:
\(n(\text{H}_2\text{O}) = n(\text{H}^+) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)

3. Calculate \(\Delta H_{\text{neut}}\):
\(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.012\text{ kJ mol}^{-1}\)

To 3 significant figures, this is \(-56.0\text{ kJ mol}^{-1}\).

**[1 Mark]** Calculates the heat energy released: \(q = 2.80\text{ kJ}\) (or \(2800\text{ J}\)).
**[1 Mark]** Calculates moles of reactant/water formed (\(0.0500\text{ mol}\)) and divides heat energy by this value.
**[1 Mark]** Correct final answer of \(-56.0\text{ kJ mol}^{-1}\) (must have negative sign, 3 significant figures, and correct units. Accept \(-56\text{ kJ mol}^{-1}\) if matching previous rounding, but penalise positive sign or missing units).

1. **Y** is a tertiary alcohol because it has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) and does not undergo oxidation with acidified potassium dichromate(VI).
2. The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol.
3. Tertiary alcohols are resistant to oxidation because the carbon atom holding the hydroxyl group (\(\text{-OH}\)) is bonded to three other carbon atoms and has no C-H bonds to be broken during oxidation. Oxidation would require the cleavage of stable C-C bonds, which does not occur under these mild oxidising conditions.

**[1 Mark]** Identifies **Y** as 2-methylpropan-2-ol (accept formula if unambiguous).
**[1 Mark]** Identifies **Y** as a tertiary (3°) alcohol.
**[1 Mark]** Explains that the carbon atom bonded to the \(\text{-OH}\) group has no hydrogen atom attached to it / requires breaking a C-C bond.

(a) Using the calorimetry equation: \(q = m c \Delta T\), where \(m = 100.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1}\), and \(\Delta T = 50.0^\circ\text{C}\). \(q = 100.0 \times 4.18 \times 50.0 = 20900\text{ J} = 20.9\text{ kJ}\). (b) Using the relationship: \(\text{moles} = \frac{q}{\Delta H_c}\), \(\text{moles} = \frac{20.9\text{ kJ}}{2090\text{ kJ mol}^{-1}} = 0.0100\text{ mol}\). (c) Calculating molar mass: \(M = \frac{\text{mass}}{\text{moles}} = \frac{0.880\text{ g}}{0.0100\text{ mol}} = 88.0\text{ g mol}^{-1}\). For the general formula \(\text{C}_n\text{H}_{2n+1}\text{OH}\): \(14n + 18.0 = 88.0 \implies 14n = 70.0 \implies n = 5\). Therefore, the molecular formula of the alcohol is \(\text{C}_5\text{H}_{11}\text{OH}\).

- (a) [1 mark] for correct calculation of \(q = 20.9\text{ kJ}\) (or \(20900\text{ J}\)).
- (b) [1 mark] for calculating the moles of alcohol: \(\frac{\text{Answer to (a)}}{2090} = 0.0100\text{ mol}\) (allow transfer of error from (a)).
- (c) [1 mark] for calculating molar mass \(M = 88.0\text{ g mol}^{-1}\) (allow transfer of error from (b)); [1 mark] for correctly deducing the molecular formula \(\text{C}_5\text{H}_{11}\text{OH}\) (or \(\text{C}_5\text{H}_{12}\text{O}\)) based on their calculated molar mass.

(a) First, find the moles of butane burned: \(\text{moles} = \frac{220\text{ g}}{58.0\text{ g mol}^{-1}} = 3.793\text{ mol}\). From the stoichiometric ratio, \(1\text{ mol}\) of butane produces \(4\text{ mol}\) of \(\text{CO}_2\), so: \(\text{moles of CO}_2 = 3.793 \times 4 = 15.17\text{ mol}\). Volume at RTP: \(V = 15.17\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 364\text{ dm}^3\) (to 3 s.f.). (b) Total heat energy released: \(\text{Heat} = 3.793\text{ mol} \times 2877\text{ kJ mol}^{-1} = 10912\text{ kJ} = 10900\text{ kJ}\) (to 3 s.f.).

- (a) [1 mark] for calculating the moles of butane (\(3.79\text{ mol}\)) and hence moles of \(\text{CO}_2\) (\(15.2\text{ mol}\)); [1 mark] for correct volume of \(364\text{ dm}^3\) (allow answers in the range \(364 - 364.2\)).
- (b) [1 mark] for multiplying the moles of butane by \(2877\); [1 mark] for the correct final answer of \(10900\text{ kJ}\) (or \(1.09 \times 10^4\text{ kJ}\)) to 3 s.f.

(a) \(\Delta H_r^\theta = \sum \Delta H_f^\theta(\text{products}) - \sum \Delta H_f^\theta(\text{reactants})\). \(\Delta H_r^\theta = [(-90.5) + (-285.8)] - [(-277.6) + (-36.3)] = [-376.3] - [-313.9] = -62.4\text{ kJ mol}^{-1}\). (b)(i) \(\text{moles of ethanol} = \frac{4.60\text{ g}}{46.0\text{ g mol}^{-1}} = 0.100\text{ mol}\). Since the stoichiometry is 1:1, \(\text{moles of HBr} = 0.100\text{ mol}\). \(\text{Volume of HBr} = 0.100\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 2400\text{ cm}^3\). (b)(ii) \(\text{Heat energy change} = \text{moles of ethanol} \times \Delta H_r^\theta = 0.100\text{ mol} \times (-62.4\text{ kJ mol}^{-1}) = -6.24\text{ kJ}\).

- (a) [1 mark] for correct cycle/expression: \([(-90.5) + (-285.8)] - [(-277.6) + (-36.3)]\); [1 mark] for correct evaluation to \(-62.4\text{ kJ mol}^{-1}\).
- (b)(i) [1 mark] for calculating the moles of ethanol as \(0.100\text{ mol}\); [1 mark] for correct calculation of \(\text{HBr}\) gas volume to give \(2400\text{ cm}^3\) (accept \(2.4\text{ dm}^3\) if unit is clearly stated).
- (b)(ii) [1 mark] for correct calculation of heat energy change as \(-6.24\text{ kJ}\) (or stating that \(6.24\text{ kJ}\) of heat energy is released; allow transfer of error from part (a)).

(a) Rearranging the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT}\). First, convert volume to \(\text{m}^3\): \(V = 263\text{ cm}^3 = 2.63 \times 10^{-4}\text{ m}^3\). Then substitute the values into the equation: \(n = \frac{1.01 \times 10^5\text{ Pa} \times 2.63 \times 10^{-4}\text{ m}^3}{8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 320\text{ K}} = \frac{26.563}{2659.2} = 0.009989\text{ mol} = 0.0100\text{ mol}\) (to 3 s.f.). (b) Since the stoichiometry of the reaction is 1:1, \(\text{moles of XCO}_3 = 0.0100\text{ mol}\). Molar mass of \(\text{XCO}_3 = \frac{\text{mass}}{\text{moles}} = \frac{1.97\text{ g}}{0.0100\text{ mol}} = 197\text{ g mol}^{-1}\). Subtracting the molar mass of the carbonate group (\(\text{CO}_3 = 12.0 + (3 \times 16.0) = 60.0\text{ g mol}^{-1}\)): \(M_r(\text{X}) = 197 - 60.0 = 137\text{ g mol}^{-1}\). Consulting the Periodic Table, the Group 2 metal with a relative atomic mass close to 137 is Barium (\(\text{Ba}\), \(A_r = 137.3\)).

- (a) [1 mark] for correct conversion of volume to \(\text{m}^3\) (\(2.63 \times 10^{-4}\)); [1 mark] for correct substitution into the rearranged ideal gas equation; [1 mark] for correct calculation of moles as \(0.0100\text{ mol}\) (accept range \(0.00998 - 0.0100\)).
- (b) [1 mark] for calculating the molar mass of the carbonate as \(197\text{ g mol}^{-1}\) (allow transfer of error from part (a)) and subtracting 60 to obtain \(137\text{ g mol}^{-1}\) for \(\text{X}\); [1 mark] for correctly identifying \(\text{X}\) as Barium (or \(\text{Ba}\)).

1. Experimental Procedure and Measurements:
- Measure a known volume (e.g., \(50\text{ cm}^3\)) of water using a measuring cylinder and pour it into a copper calorimeter or glass beaker.
- Clamp the beaker and record the initial temperature of the water using a thermometer.
- Weigh the spirit burner containing methanol, including its cap, and record the initial mass.
- Place the burner under the beaker, remove the cap, light the burner, and stir the water continuously.
- After a temperature rise of approximately \(20\text{ }^\circ\text{C}\), extinguish the flame using the burner cap.
- Record the maximum temperature reached by the water.
- Reweigh the spirit burner and cap, recording the final mass.

2. Calculation Method:
- Calculate the temperature rise: \(\Delta T = T_{\text{final}} - T_{\text{initial}}\).
- Calculate the mass of methanol burned: \(m_{\text{burned}} = m_{\text{initial}} - m_{\text{final}}\).
- Calculate the heat energy transferred to the water using: \(q = m_{\text{water}} \times c \times \Delta T\) (where \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\) and \(m_{\text{water}}\) is the mass of water, assuming density \(= 1\text{ g cm}^{-3}\)).
- Calculate the moles of methanol burned: \(n = \frac{m_{\text{burned}}}{M_r(\text{methanol})}\).
- Calculate the enthalpy change of combustion: \(\Delta H_c = -\frac{q}{n}\) (converting J to kJ by dividing by 1000, and ensuring a negative sign for exothermic reaction).

3. Reasons for Discrepancy:
- Heat loss to the surroundings (air, beaker, thermometer).
- Incomplete combustion of methanol (forming soot/CO instead of \(CO_2\)).
- Evaporation of methanol from the wick while weighing or after extinguishing.
- Non-standard conditions.

4. Improvements:
- Use draught shields to reduce heat loss to the surrounding air.
- Use a copper calorimeter instead of a glass beaker to ensure more efficient heat transfer.
- Place a lid on the beaker to minimize heat loss from the water surface.

This is a Level-of-Response question. Use the following criteria to award marks:

Level 3 (5-6 marks):
- Fully describes the experimental procedure, including all essential measurements (mass of burner before/after, volume of water, initial/final temperature).
- Shows a complete and correct mathematical pathway to calculate \(\Delta H_c\) from the measurements, including both formulas \(q = mc\Delta T\) and \(\Delta H_c = -q/n\).
- Identifies at least two realistic reasons for the experimental value being less exothermic and suggests at least one valid experimental improvement.
- The answer is structured logically with clear chemical terminology used throughout.

Level 2 (3-4 marks):
- Describes most of the experimental procedure but may omit one key measurement.
- Outlines the calculation pathway but may omit a step (e.g., converting to moles or standard units) or make a sign error.
- Identifies at least one reason for the discrepancy and/or an improvement.
- The answer is generally clear but may lack some detail or logical structure.

Level 1 (1-2 marks):
- Identifies some of the required measurements or mentions the basic formula \(q = mc\Delta T\).
- Offers simple, generic reasons for heat loss without clear links to the calculation or improvements.
- The response is unstructured or has significant gaps in chemistry explanation.

Level 0 (0 marks):
- No rewardable material.

1. Experimental Procedure:
- Measure equal volumes (e.g., \(1\text{ cm}^3\)) of 1-chlorobutane, 1-bromobutane, and 1-iodobutane into three separate test tubes.
- Add a fixed volume of ethanol to each test tube. Ethanol acts as a mutual solvent because halogenoalkanes are insoluble in water; it allows the halogenoalkanes and the aqueous silver nitrate reagent to mix in a single phase.
- Place the test tubes in a water bath at around \(50-60\text{ }^\circ\text{C}\) to control the temperature variable.
- Add equal volumes of aqueous silver nitrate solution (which contains water acting as the nucleophile for hydrolysis) to each test tube and start a timer.
- Record the time taken for a precipitate to appear in each test tube.

2. Observations and Relative Rates:
- 1-iodobutane forms a yellow precipitate (AgI) almost immediately (fastest rate of reaction).
- 1-bromobutane forms a cream precipitate (AgBr) after some time (moderate rate of reaction).
- 1-chlorobutane forms a white precipitate (AgCl) very slowly or only after prolonged heating (slowest rate of reaction).
- Order of rate of hydrolysis: 1-iodobutane > 1-bromobutane > 1-chlorobutane.

3. Explanation of the Trend:
- The reaction involves breaking the carbon-halogen bond (C-X).
- Although the C-Cl bond is more polar than the C-I bond due to electronegativity differences (which would theoretically attract nucleophiles faster), bond enthalpy is the dominant factor.
- The bond enthalpy decreases down the group: \(\text{C-Cl} > \text{C-Br} > \text{C-I}\).
- The C-I bond is the weakest and requires the least energy to break, making 1-iodobutane the most reactive and hydrolyzed fastest.
- The C-Cl bond is the strongest and requires the most energy to break, making 1-chlorobutane the least reactive and hydrolyzed slowest.

This is a Level-of-Response question. Use the following criteria to award marks:

Level 3 (5-6 marks):
- Comprehensive description of the experimental setup, including the correct reagents (aqueous silver nitrate and ethanol), control of temperature, and the role of ethanol as a mutual solvent.
- Correctly identifies the observations (white, cream, and yellow precipitates) and correctly links these to the order of reactivity (iodo > bromo > chloro).
- Clear and accurate explanation of the trend based on bond enthalpy decreasing down Group 7 (C-I is weakest, C-Cl is strongest), explicitly explaining why bond enthalpy outweighs bond polarity.
- Structured in a highly logical sequence with precise chemical terminology.

Level 2 (3-4 marks):
- Describes the experimental setup with the correct reagents but may omit the control of variables (such as temperature/volume) or the specific role of ethanol.
- Identifies the precipitates and the order of reactivity but may mix up colors or have slight gaps in linking observations to rates.
- Explains the trend in terms of bond strength or polarity, but might not clearly clarify that bond enthalpy is the deciding factor over polarity.
- The response is mostly clear and structured, with some minor omissions.

Level 1 (1-2 marks):
- Mentions using silver nitrate or ethanol, but lacks detail on the method.
- Identifies at least one precipitate color or states which halogenoalkane reacts fastest.
- Mentions bond energy or polarity but without a clear explanation of the trend.
- Unstructured or fragmented chemistry knowledge.

Level 0 (0 marks):
- No rewardable material.

1. Calculate the percentage uncertainty for the \(25.0\text{ cm}^3\) pipette:
\$$\text{Percentage uncertainty} = \frac{0.06}{25.0} \times 100 = 0.24\%\$$

2. Compare with using a \(10.0\text{ cm}^3\) pipette:
To transfer \(25.0\text{ cm}^3\) using a \(10.0\text{ cm}^3\) pipette, multiple measurements are required (e.g., two \(10.0\text{ cm}^3\) transfers and one \(5.0\text{ cm}^3\) transfer, or using it three times). Each transfer introduces its own absolute uncertainty, which accumulates. Even a single transfer of \(10.0\text{ cm}^3\) has a percentage uncertainty of \(\frac{0.04}{10.0} \times 100 = 0.40\%\), which is already higher than \(0.24\%\). Therefore, the overall percentage uncertainty will increase.

M1: Correct calculation of the percentage uncertainty for the \(25.0\text{ cm}^3\) pipette as 0.24% (1)
M2: Identifies that the percentage uncertainty would increase / be larger (1)

1. Calculate the temperature rise (\(\Delta T\)):
\$$\Delta T = 28.5 - 21.5 = 7.0\text{ }^\circ\text{C}\$$

2. Calculate the absolute uncertainty in the temperature change. Because two readings are taken (initial and maximum), the total absolute uncertainty is:
\$$\text{Uncertainty in } \Delta T = 2 \times 0.5 = 1.0\text{ }^\circ\text{C}\$$

3. Calculate percentage uncertainty:
\$$\text{Percentage uncertainty} = \frac{1.0}{7.0} \times 100 = 14.29\% \approx 14.3\%\$$

4. To decrease percentage uncertainty without changing the thermometer, we must increase the value of \(\Delta T\). This can be achieved by using more concentrated solutions of the acid and alkali, which releases more heat energy per unit volume and results in a larger temperature change.

M1: Correct calculation of percentage uncertainty as 14.3% (or 14%) (1)
[Allow 7.14% if student only used single uncertainty of 0.5 instead of 1.0]
M2: Suggestion to use more concentrated solutions of the acid / alkali / reactants (1)
[Do not accept 'use more volume' as that does not increase the temperature rise if concentrations remain the same]

1. If the hydrated salt is only heated once, the water of crystallisation may not be completely driven off. This means the mass of the anhydrous residue recorded is higher than it should be, and the calculated mass of water lost is lower than it should be. Consequently, the calculated moles of water will be lower, resulting in a value of \(x\) that is too low.
2. To ensure all water is removed, the student must heat, cool, and reweigh the sample repeatedly until consecutive mass measurements are identical (constant mass).

M1: Explains that the calculated value of \(x\) will be lower / too small because some water remains / mass of water lost is calculated as too low (1)
M2: States that the student must heat, cool, and reweigh until a constant mass is obtained (1)

1. The thermometer bulb is meant to measure the temperature of the vapour that is about to enter the condenser. Immersing the bulb directly in the boiling mixture measures the temperature of the liquid phase, which contains reactants (cyclohexanol, which has a much higher boiling point of \(161\text{ }^\circ\text{C}\)), phosphoric acid catalyst, and products. This liquid mixture boils at a higher temperature than the pure vapour of cyclohexene (boiling point \(83\text{ }^\circ\text{C}\)). Therefore, the recorded temperature is higher than the true boiling point of the vapour being distilled.
2. Because the recorded temperature does not represent the temperature of the vapour, the student cannot reliably identify the steady boiling point range corresponding to pure cyclohexene, making it difficult to collect only the pure product fraction.

M1: Identifies that the recorded temperature will be higher than the actual vapour temperature / boiling point of cyclohexene (1)
M2: Explains that the student cannot identify when the pure fraction is distilling / cannot separate cyclohexene from impurities based on boiling point range (1)

1. A stiff plunger increases the friction and pressure inside the apparatus. Some gas may escape through joints/stoppers instead of pushing the plunger, or the gas is compressed. Consequently, the recorded volume of gas at any given time is lower than the actual volume of gas produced.
2. Since the measured volume is lower at each time interval, the slope of the volume-time graph at \(t = 0\) will be less steep, leading to an underestimated (lower) calculated initial rate of reaction.

M1: States that the measured volume of gas is lower than the actual volume (1)
M2: States that the calculated initial rate of reaction is lower / underestimated (1)

1. 'Weighing by difference' is preferred because some solid often remains adhering to the sides of the weighing bottle after transfer. Subtracting the mass of the empty bottle with residual solid from the mass of the full bottle gives the precise mass of solid actually transferred to the beaker.
2. If the student fails to reweigh the empty bottle, they will assume that all of the weighed solid was transferred. Since the actual mass transferred is less than assumed, the actual concentration of the standard solution is lower than calculated. Thus, the calculated concentration is too high (overestimated).

M1: Explains that weighing by difference accounts for solid remaining in the weighing bottle / ensures the exact mass of transferred solid is known (1)
M2: Identifies that the calculated concentration will be higher than the actual concentration (1)

1. Moles of HCl used in titration: \(n(\text{HCl}) = 0.100 \text{ mol dm}^{-3} \times 0.02500 \text{ dm}^3 = 2.50 \times 10^{-3} \text{ mol}\). 2. Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0 \text{ cm}^3\) sample: Since 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\), \(n(\text{Na}_2\text{CO}_3) = \frac{1}{2} \times 2.50 \times 10^{-3} \text{ mol} = 1.25 \times 10^{-3} \text{ mol}\). 3. Moles of \(\text{Na}_2\text{CO}_3\) in original \(250 \text{ cm}^3\) solution: \(1.25 \times 10^{-3} \text{ mol} \times 10 = 1.25 \times 10^{-2} \text{ mol}\). 4. Molar mass (\(M_{\text{r}}\)) of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\): \(M_{\text{r}} = \frac{3.58 \text{ g}}{1.25 \times 10^{-2} \text{ mol}} = 286.4 \text{ g mol}^{-1}\). 5. Calculate value of \(x\): \(M_{\text{r}}(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}\). Mass of water per mole of hydrate = \(286.4 - 106.0 = 180.4 \text{ g mol}^{-1}\). \(x = \frac{180.4}{18.0} = 10.02 \approx 10\).

M1: Calculates moles of HCl as \(2.50 \times 10^{-3} \text{ mol}\) (1) M2: Calculates moles of \(\text{Na}_2\text{CO}_3\) in \(250 \text{ cm}^3\) as \(1.25 \times 10^{-2} \text{ mol}\) (1) M3: Calculates molar mass of the hydrate as \(286.4 \text{ g mol}^{-1}\) (1) M4: Calculates molar mass or mass of water as \(180.4 \text{ g}\) (1) M5: Deduces value of \(x\) as 10 (must be a whole number) (1).

1. Temperature change: \(\Delta T = 34.2 - 18.5 = 15.7\text{ }^\circ\text{C}\). 2. Heat energy transferred: \(q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 15.7 \text{ }^\circ\text{C} = 3281.3 \text{ J} = 3.2813 \text{ kJ}\). 3. Moles of copper(II) sulfate: \(n(\text{CuSO}_4) = 0.0500 \text{ dm}^3 \times 0.250 \text{ mol dm}^{-3} = 0.0125 \text{ mol}\). 4. Enthalpy change: \(\Delta H = -\frac{3.2813 \text{ kJ}}{0.0125 \text{ mol}} = -262.5 \text{ kJ mol}^{-1}\). To 3 significant figures, \(\Delta H = -263 \text{ kJ mol}^{-1}\).

M1: Calculates temperature rise as \(15.7\text{ }^\circ\text{C}\) (1) M2: Calculates heat energy transferred as \(3.28 \text{ kJ}\) or \(3280 \text{ J}\) (1) M3: Calculates moles of copper(II) sulfate as \(0.0125 \text{ mol}\) (1) M4: Divides heat energy by moles of limiting reagent (1) M5: Gives final answer of \(-263 \text{ kJ mol}^{-1}\) (must have negative sign, 3 significant figures, and correct units) (1).

1. Mass of butan-1-ol used: \(\text{mass} = \text{volume} \times \text{density} = 10.0 \text{ cm}^3 \times 0.810 \text{ g cm}^{-3} = 8.10 \text{ g}\). 2. Moles of butan-1-ol: \(n(\text{butan-1-ol}) = \frac{8.10 \text{ g}}{74.0 \text{ g mol}^{-1}} = 0.10946 \text{ mol}\). 3. Theoretical mass of 1-bromobutane: Since 1 mole of butan-1-ol theoretically produces 1 mole of 1-bromobutane, \(\text{theoretical mass} = 0.10946 \text{ mol} \times 137.0 \text{ g mol}^{-1} = 15.00 \text{ g}\). 4. Percentage yield: \(\text{yield} = \frac{11.2 \text{ g}}{15.00 \text{ g}} \times 100\% = 74.67\% \approx 74.7\%\).

M1: Calculates mass of butan-1-ol as \(8.10 \text{ g}\) (1) M2: Calculates moles of butan-1-ol as \(0.109 \text{ mol}\) (1) M3: Calculates theoretical mass of 1-bromobutane as \(15.0 \text{ g}\) (1) M4: Divides actual yield by theoretical yield (1) M5: Gives final percentage yield as \(74.7\%\) (accept \(75\%\) or \(74.67\%\)) (1).

(a) Percentage uncertainty of measuring cylinder = \(\frac{1.0}{48.0} \times 100\% = 2.08\%\). (b) Percentage uncertainty of gas syringe = \(\frac{0.5}{48.0} \times 100\% = 1.04\%\). (c) Mass loss = \(120.45 - 119.95 = 0.50 \text{ g}\). Total uncertainty of two balance readings = \(2 \times 0.01 \text{ g} = 0.02 \text{ g}\). Percentage uncertainty = \(\frac{0.02}{0.50} \times 100\% = 4.0\%\). (d) To reduce percentage uncertainty without changing the balance, use a larger mass of reactant (or higher concentration/volume of hydrogen peroxide) to increase the overall mass loss.

M1: Calculates percentage uncertainty of measuring cylinder as \(2.08\%\) or \(2.1\%\) (1) M2: Calculates percentage uncertainty of gas syringe as \(1.04\%\) or \(1.0\%\) (1) M3: Calculates mass loss as \(0.50 \text{ g}\) and identifies total uncertainty as \(0.02 \text{ g}\) (1) M4: Calculates percentage uncertainty in mass loss as \(4.0\%\) (accept \(2.0\%\) only if student states they assume only one reading is taken) (1) M5: States that using a larger quantity/volume/concentration of reactant will increase the mass loss and decrease percentage uncertainty (1).

1. The brick-red flame test indicates the presence of calcium ions, \(Ca^{2+}\).
2. The formation of a cream precipitate with silver nitrate that is insoluble in dilute ammonia but soluble in concentrated ammonia indicates the presence of bromide ions, \(Br^-\).
3. Combining these gives the formula of compound A as calcium bromide, \(CaBr_2\).
4. The ionic equation for the precipitation is \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\).
5. Dilute nitric acid is added to react with and remove any interfering anions, such as carbonate (\(CO_3^{2-}\)) or hydroxide (\(OH^-\)) ions, which would form silver carbonate or silver hydroxide precipitates, leading to a false-positive result.

• Identify cation: Calcium / \(Ca^{2+}\) (1 mark)
• Identify anion: Bromide / \(Br^-\) (1 mark)
• Formula of compound A: \(CaBr_2\) (0.5 marks)
• Ionic equation with correct state symbols: \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\) (1 mark) [allow 0.5 marks if equation is correct but state symbols are missing or incorrect]
• Purpose of dilute nitric acid: to react with and remove carbonate/hydroxide ions that would otherwise react with silver ions to form a precipitate (1 mark) [Do not accept 'to acidify' without qualification].

To distinguish between an alkene (hex-1-ene), an alkane (cyclohexane), and an alcohol (ethanol):
1. Test 1 (to identify hex-1-ene): Add bromine water to a sample of each liquid. Hex-1-ene has a \(C=C\) double bond and will undergo electrophilic addition, decolourising the orange bromine water. Cyclohexane and ethanol do not react with bromine water under standard room conditions, so they remain orange/yellow.
2. Test 2 (to distinguish cyclohexane and ethanol): Add a small piece of sodium metal (or solid \(PCl_5\)) to separate samples of the remaining two liquids. Ethanol, containing an \(-OH\) group, reacts with sodium to produce hydrogen gas (effervescence) or with \(PCl_5\) to produce misty white fumes of hydrogen chloride. Cyclohexane does not react.

• Test 1 reagent: Bromine water / bromine in organic solvent (1 mark)
• Observation with P (hex-1-ene): Orange/yellow to colourless / decolourises AND observation with Q and R: remains orange/yellow (1 mark)
• Test 2 reagent: Sodium metal (Na) OR Phosphorus(V) chloride (\(PCl_5\)) (1 mark)
• Observation with R (ethanol): Effervescence / bubbles (for Na) OR misty white fumes (for \(PCl_5\)) (1 mark)
• Mentioning that Q (cyclohexane) has no visible reaction in either test (0.5 marks)
[Alternative Test 2: Add acidified potassium dichromate(VI) (\(H^+/Cr_2O_7^{2-}\)) and heat. R (ethanol) turns from orange to green; Q (cyclohexane) remains orange. Accept this with equivalent mark breakdown.]

(a) The formation of a yellow precipitate with aqueous silver nitrate indicates the presence of iodide ions (\(I^-\)). The precipitate is silver iodide, \(AgI\). The ionic equation is: \(Ag^+(aq) + I^-(aq) \rightarrow AgI(s)\).
(b) Halogenoalkanes are insoluble in water, while silver nitrate is an aqueous solution. Ethanol is used as a mutual solvent because it dissolves both the organic halogenoalkane and the inorganic aqueous silver nitrate, allowing them to come into contact and react.
(c) Ethanol and halogenoalkanes are volatile and highly flammable. Direct heating with a Bunsen burner flame poses a significant fire hazard. A water bath provides a safe, flame-free heat source and allows for controlled temperature.

• (a) Identify halide: Iodide / \(I^-\)\) (1 mark)
• (a) Ionic equation with correct state symbols: \(Ag^+(aq) + I^-(aq) \rightarrow AgI(s)\) (1.5 marks; 1 mark for correct formulas, 0.5 marks for correct state symbols)
• (b) Role of ethanol: mutual solvent / cosolvent to dissolve both the halogenoalkane and aqueous silver nitrate / allow them to mix (1 mark)
• (c) Safety explanation: Halogenoalkanes/ethanol are flammable / risk of catching fire with open Bunsen flame (1 mark)

To successfully test for both carbonate and sulfate ions in a single sample:
1. Carbonate ions must be identified and removed first. Add excess dilute nitric acid (\(HNO_3\)). The carbonate ions react to produce carbon dioxide gas, causing visible effervescence: \(CO_3^{2-}(aq) + 2H^+(aq) \rightarrow CO_2(g) + H_2O(l)\).
2. It is critical to use excess nitric acid and ensure all effervescence has stopped. If any carbonate ions remain, adding barium ions in the next step would produce a white precipitate of barium carbonate (\(BaCO_3\)), giving a false-positive result for sulfate.
3. Once the carbonate has been completely removed, add aqueous barium nitrate, \(Ba(NO_3)_2(aq)\). Barium ions react with the remaining sulfate ions to form an insoluble white precipitate of barium sulfate: \(Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\).

• Step 1: Add dilute nitric acid / \(HNO_3(aq)\) AND observe effervescence / bubbles (1 mark)
• Ionic equation for carbonate reaction: \(CO_3^{2-}(aq) + 2H^+(aq) \rightarrow CO_2(g) + H_2O(l)\) (1 mark)
• Step 2: Add excess acid / ensure effervescence stops before adding the barium reagent (to remove all carbonate) (1 mark)
• Step 3: Add barium nitrate / barium chloride solution AND observe white precipitate (1 mark)
• Ionic equation for sulfate reaction: \(Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\) (0.5 marks)
[Note: Reject use of sulfuric acid, as this introduces sulfate ions.]