2023 OCR GCSE Gateway Science - Chemistry A - J248 模擬試題連答案詳解

Fermentation of glucose uses enzymes in yeast as a biological catalyst. The optimum temperature for yeast enzymes is around \(37^\circ\text{C}\). The process must be anaerobic (without oxygen) to prevent the ethanol from being oxidised into ethanoic acid.

1 mark: A - Correct set of fermentation conditions. Reject all other options.

The empirical formula mass of \(CH_2O\) is calculated as: \(12.0 + (2 \times 1.0) + 16.0 = 30.0\). The relative molecular mass of the compound is 60.0. To find the molecular formula multiplier, divide the molecular mass by the empirical formula mass: \(60.0 / 30.0 = 2\). Therefore, we multiply the number of atoms in the empirical formula by 2, which gives \(C_2H_4O_2\).

1 mark for the correct option B. Reject all other options.

According to the law of conservation of mass, the total mass of the reactants must equal the total mass of the products. Therefore: \(\text{Mass of iron} + \text{Mass of sulfur} = \text{Mass of iron(II) sulfide}\). Substituting the given values: \(5.60\text{ g} + \text{Mass of sulfur} = 8.80\text{ g}\). This gives: \Ref{Mass of sulfur} = 8.80\text{ g} - 5.60\text{ g} = 3.20\text{ g}\).

1 mark for the correct option A. Reject all other options.

In the electrolysis of aqueous copper(II) sulfate, the ions present are \(Cu^{2+}\), \(H^+\), \(SO_4^{2-}\), and \(OH^-\). At the cathode, copper ions are discharged in preference to hydrogen ions because copper is less reactive than hydrogen, forming copper metal. At the anode, hydroxide ions are discharged in preference to sulfate ions, forming oxygen gas and water. Therefore, the products are oxygen at the anode and copper at the cathode.

1 mark for the correct option B. Reject all other options.

Fractional distillation is the most suitable technique for separating miscible liquids with relatively close boiling points. The fractionating column allows the vapors of the liquid with the lower boiling point (liquid X, \(78^\circ\text{C}\)) to rise and be condensed first, separating it from liquid Y.

1 mark for the correct option C. Reject all other options.

Silicon dioxide (silica) has a giant covalent structure where every silicon atom is covalently bonded to four oxygen atoms, with all valence electrons localized in bonds, meaning there are no free-moving electrons or ions to conduct electricity. Graphite is giant covalent but has delocalised electrons. Sodium chloride is ionic. Copper has metallic bonding.

1 mark for the correct option C. Reject all other options.

The reaction of an acid with a metal carbonate produces carbon dioxide gas, \(CO_2\). The chemical test for carbon dioxide is to bubble it through limewater (calcium hydroxide solution), which turns cloudy/milky due to the formation of a white precipitate of calcium carbonate.

1 mark for the correct option D. Reject all other options.

The atomic number is the number of protons, which is 19. In a neutral atom, the number of electrons is equal to the number of protons, which is 19. The number of neutrons is found by subtracting the atomic number from the mass number: \(39 - 19 = 20\). Therefore, the atom has 20 neutrons, 19 protons, and 19 electrons.

1 mark for the correct option B. Reject all other options.

First, calculate the energy required to break the bonds in the reactants: \(1 \times (\text{H}-\text{H}) + 1 \times (\text{Cl}-\text{Cl}) = 436 + 243 = 679\text{ kJ/mol}\). Second, calculate the energy released when new bonds are formed in the products: \(2 \times (\text{H}-\text{Cl}) = 2 \times 432 = 864\text{ kJ/mol}\). Finally, subtract the energy released from the energy required: \(\text{Overall energy change} = 679 - 864 = -185\text{ kJ/mol}\).

1 mark for the correct option A. Reject all other options.

Ethyl ethanoate is formed by the condensation reaction between ethanol and ethanoic acid. Ethanol provides the ethyl group and ethanoic acid provides the ethanoate group.

1 mark for identifying the correct reactants as ethanol and ethanoic acid.

First, calculate the relative formula mass of \(\text{CaCO}_3\): \(40.1 + 12.0 + (3 \times 16.0) = 100.1\text{ g/mol}\). Calculate the moles of \(\text{CaCO}_3\) used: \(25.0 / 100.1 = 0.250\text{ mol}\). According to the equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CaO}\) is 1:1, so \(0.250\text{ mol}\) of \(\text{CaO}\) is produced. Calculate the relative formula mass of \(\text{CaO}\): \(40.1 + 16.0 = 56.1\text{ g/mol}\). Mass of \(\text{CaO}\) produced: \(0.250\text{ mol} \times 56.1\text{ g/mol} = 14.0\text{ g}\).

1 mark for the correct calculation showing 14.0 g.

At the anode, chloride ions (\(\text{Cl}^-\)) are discharged preferentially over hydroxide ions because the solution is concentrated, forming chlorine gas. At the cathode, hydrogen ions (\(\text{H}^+\)) are discharged preferentially over sodium ions because hydrogen is less reactive than sodium, forming hydrogen gas.

1 mark for identifying chlorine at the anode and hydrogen at the cathode.

The retention factor (\(R_f\)) is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front: \(R_f = 6.0\text{ cm} / 8.0\text{ cm} = 0.75\).

1 mark for the correct Rf calculation of 0.75.

Silicon dioxide is a giant covalent structure. Melting it requires breaking many strong covalent bonds throughout the macromolecule, which takes a massive amount of energy. It does not conduct electricity because all valence electrons are localised in covalent bonds and cannot move freely.

1 mark for identifying the giant covalent structure and lack of mobile charged particles.

The atomic number represents the number of protons, which is 19. In a neutral atom, the number of electrons equals the number of protons, which is 19. The number of neutrons is found by subtracting the atomic number from the mass number: \(39 - 19 = 20\).

1 mark for identifying 19 protons, 20 neutrons, and 19 electrons.

Energy required to break reactant bonds: \(436\text{ (H-H)} + 243\text{ (Cl-Cl)} = 679\text{ kJ/mol}\). Energy released making product bonds: \(2 \times 432\text{ (H-Cl)} = 864\text{ kJ/mol}\). Overall energy change = Energy in - Energy out = \(679 - 864 = -185\text{ kJ/mol}\).

1 mark for the correct calculation showing -185 kJ/mol.

Chlorine is more reactive than iodine because it has a smaller atomic radius and can attract incoming electrons more easily. Therefore, chlorine gas displaces iodide ions from solution to form molecular iodine (which turns the solution brown) and chloride ions.

1 mark for the correct observation of turning brown and the explanation of halogen displacement.

Crude oil is preheated to about 350-400 °C to vaporise most of the hydrocarbons before they enter the bottom of the fractionating column. As the vapour rises through the column, which has a temperature gradient (hotter at the bottom and cooler at the top), different fractions condense at their respective boiling points. Larger hydrocarbons with higher boiling points condense near the bottom where it is hottest, while smaller hydrocarbons with lower boiling points rise further up and condense near the cooler top.

[1 mark] C - Crude oil is heated to vaporise it before it enters the fractionating column.
Reject other options:
- A is incorrect because larger molecules condense near the bottom where it is hot.
- B is incorrect because the temperature is highest at the bottom and lowest at the top.
- D is incorrect because heavier fractions are less flammable and more viscous.

To find the empirical formula, calculate the number of moles of each element:
- Moles of carbon: \(\frac{1.20\text{ g}}{12.0\text{ g/mol}} = 0.10\text{ mol}\)
- Moles of hydrogen: \(\frac{0.30\text{ g}}{1.0\text{ g/mol}} = 0.30\text{ mol}\)
Next, find the simplest whole number ratio by dividing both values by the smallest mole value (0.10):
- Carbon: \(\frac{0.10}{0.10} = 1\)
- Hydrogen: \(\frac{0.30}{0.10} = 3\)
So, the empirical formula is \(CH_3\).

[1 mark] C - \(CH_3\)
Award 1 mark for the correct calculation and identification of the empirical formula.

In the electrolysis of aqueous sodium sulfate:
- At the cathode (negative electrode), both \(Na^+\) and \(H^+\) (from water) are attracted. Since sodium is more reactive than hydrogen, \(H^+\) is preferentially reduced, producing hydrogen gas (\(H_2\)).
- At the anode (positive electrode), both \(SO_4^{2-}\) and \(OH^-\) (from water) are attracted. Hydroxide ions are preferentially oxidised compared to stable sulfate ions, producing oxygen gas (\(O_2\)) and water.
Therefore, the products are hydrogen at the cathode and oxygen at the anode.

[1 mark] B - Cathode: Hydrogen; Anode: Oxygen.
Reject other options because they incorrectly identify either sodium or sulfur dioxide being formed.

The retention factor (\(R_f\)) is calculated using the formula:
\(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}\)
Substituting the given values:
\(R_f = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\).

[1 mark] C - 0.75
Award 1 mark for the correct division resulting in 0.75. Reject D which is the inverse calculation (8.0 / 6.0).

Silicon dioxide exists as a giant covalent lattice where each silicon atom is strongly bonded to four oxygen atoms. To melt silicon dioxide, a vast number of these strong covalent bonds must be broken, requiring a large amount of thermal energy. In contrast, carbon dioxide is a simple molecular structure. While the covalent bonds within the \(CO_2\) molecules are strong, the intermolecular forces between the molecules are very weak and require very little thermal energy to overcome.

[1 mark] A - Silicon dioxide has a giant covalent structure with many strong covalent bonds that require a large amount of energy to break, whereas carbon dioxide consists of simple molecules with weak intermolecular forces between them.
Reject B because silicon-oxygen bonds are covalent, not ionic. Reject C and D because they misrepresent the structures and forces.

When the concentration of a reactant is increased, there are more reactant particles present per unit volume. This reduces the average space between the particles, meaning they collide with each other more frequently. Since the frequency of collisions increases, the rate of successful collisions also increases, which results in an increased rate of reaction.

[1 mark] C - It increases the number of reacting particles per unit volume, which increases the frequency of successful collisions.
Reject A (activation energy is unchanged by concentration). Reject B (kinetic energy is only changed by temperature). Reject D (catalysts change the reaction pathway, not concentration).

To find the empirical formula: 1. Divide the mass percentage of each element by its relative atomic mass (Ar). For Carbon: 85.7 / 12 = 7.14. For Hydrogen: 14.3 / 1 = 14.3. 2. Divide each number of moles by the smallest value obtained (7.14). For Carbon: 7.14 / 7.14 = 1. For Hydrogen: 14.3 / 7.14 = 2. This gives the ratio 1 C : 2 H, so the empirical formula is CH2.

1 mark for calculating correct mole ratio (C = 7.14 and H = 14.3). 1.42 marks for arriving at the correct simplest whole-number ratio and final empirical formula CH2.

Percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. Percentage yield = (9.38 / 12.50) * 100 = 75.04%. Rounding to 3 significant figures gives 75.0%.

1 mark for substituting the values correctly into the percentage yield equation: (9.38 / 12.50) * 100. 1.42 marks for the final correct answer of 75.0% with 3 significant figures.

At the cathode, hydrogen ions are reduced by gaining electrons. Two hydrogen ions gain two electrons to form one hydrogen molecule: 2H+ + 2e- -> H2.

1 mark for showing hydrogen ions and electrons on the left hand side and hydrogen molecules on the right hand side. 1.42 marks for the fully balanced equation.

The Rf value is calculated by dividing the distance moved by the substance by the distance moved by the solvent front: Rf = 5.1 / 8.5 = 0.6.

1 mark for correct division formula set up (5.1 / 8.5). 1.42 marks for correct final decimal answer of 0.6 (or 0.60).

In CO2, the carbon atom forms two double covalent bonds, one with each oxygen atom (O=C=O). Each double bond consists of two shared pairs of electrons (4 electrons). Therefore, the total number of shared electrons in the molecule is 2 * 4 = 8 electrons.

1 mark for identifying that carbon dioxide contains two double bonds or four shared pairs of electrons. 1.42 marks for stating the correct total number of shared electrons as 8.

The reactant ions in solution are barium ions Ba2+(aq) and sulfate ions SO42-(aq). They react to form the solid precipitate barium sulfate: Ba2+(aq) + SO42-(aq) -> BaSO4(s).

1 mark for the correct reactant and product ions without state symbols. 1.42 marks for the fully correct balanced equation including correct state symbols (aq) and (s).

Relative atomic mass is calculated by taking the weighted average of the isotopic masses: Ar = ((90.0 * 20) + (10.0 * 22)) / 100 = (1800 + 220) / 100 = 2020 / 100 = 20.2.

1 mark for the correct calculation setup: ((90 * 20) + (10 * 22)) / 100. 1.42 marks for the correct final relative atomic mass of 20.2.

The mean rate of reaction is calculated by dividing the volume of gas produced by the time taken: Mean rate = Volume / Time = 45 cm3 / 30 s = 1.5 cm3/s.

1 mark for substituting the correct numbers into the rate equation (45 / 30). 1.42 marks for the correct numerical rate of 1.5.

1. Find the moles of water produced:
\(\text{moles of H}_2\text{O} = \frac{\text{mass}}{\text{relative formula mass}} = \frac{9.0\text{ g}}{18.0\text{ g/mol}} = 0.5\text{ mol}\).

2. Find the total moles of hydrogen atoms in the water:
\(0.5\text{ mol} \times 2 = 1.0\text{ mol}\) of hydrogen atoms.

3. Determine the number of hydrogen atoms per alkene molecule:
Since all hydrogen comes from the alkene:
\(\frac{1.0\text{ mol of H atoms}}{0.10\text{ mol of alkene}} = 10\).

4. Write the formula using the general formula \(\text{C}_n\text{H}_{2n}\):
If \(2n = 10\), then \(n = 5\).
Thus, the molecular formula is \(\text{C}_5\text{H}_{10}\).

- 1 mark for calculating the amount of water as 0.5 mol (or 1.0 mol of H atoms).
- 1 mark for setting up the molar ratio and finding that 1 molecule of alkene contains 10 H atoms.
- 0.42 marks for the correct molecular formula \(\text{C}_5\text{H}_{10}\).

1. Find the moles of magnesium reacting:
\(\text{moles of Mg} = \frac{4.80\text{ g}}{24.3\text{ g/mol}} \approx 0.1975\text{ mol}\).

2. Determine the moles of hydrogen produced:
According to the stoichiometry of the equation, 1 mole of Mg produces 1 mole of \(\text{H}_2\).
Therefore, \(\text{moles of H}_2 = 0.1975\text{ mol}\).

3. Calculate the volume of hydrogen gas:
\(\text{Volume} = \text{moles} \times 24.0\text{ dm}^3/\text{mol}\)
\(\text{Volume} = 0.1975\text{ mol} \times 24.0\text{ dm}^3/\text{mol} \approx 4.74\text{ dm}^3\) (to 3 significant figures).

- 1 mark for calculating the moles of magnesium as approx. 0.198 mol (accept 0.20 mol if rounded prematurely).
- 1 mark for multiplying the calculated moles of hydrogen by 24.0.
- 0.42 marks for the correct volume of 4.74 (accept 4.8 if 24 was used for Mg, or 4.7).

1. Identify the relationship between charge, current, and time:
\(Q = I \times t\)

2. Convert the time from minutes to seconds:
\(t = 40.0\text{ minutes} \times 60\text{ s/min} = 2400\text{ s}\).

3. Calculate the charge:
\(Q = 1.50\text{ A} \times 2400\text{ s} = 3600\text{ C}\).

- 1 mark for converting time to seconds (2400 s).
- 1 mark for recalling and substituting values into \(Q = I \times t\).
- 0.42 marks for the correct final charge value (3600).

1. Nitrogen is in Group 5 of the Periodic Table and has 5 outer-shell electrons. Hydrogen has 1 outer-shell electron.
2. In \(\text{NH}_3\), nitrogen shares three of its outer-shell electrons with three hydrogen atoms to form three covalent bonds. This results in 3 shared pairs of electrons.
3. This leaves 2 unshared outer-shell electrons on the nitrogen atom, which form 1 non-bonding lone pair.

- 1 mark for stating that there are 3 shared pairs of electrons (or 3 single covalent bonds).
- 1 mark for stating that there is 1 lone pair of electrons (or 2 non-bonding electrons) on the nitrogen atom.
- 0.42 marks for a clear, complete correct response.

1. Calculate the energy required to break reactants' bonds:
- Energy to break 1 mol of \(\text{H-H}\) = \(436\text{ kJ}\)
- Energy to break 1 mol of \(\text{Cl-Cl}\) = \(243\text{ kJ}\)
- Total energy input = \(436 + 243 = 679\text{ kJ/mol}\).

2. Calculate the energy released when making products' bonds:
- Energy released forming 2 mol of \(\text{H-Cl}\) = \(2 \times 432 = 864\text{ kJ/mol}\).

3. Calculate the overall energy change:
\(\text{Energy change} = \text{Energy input} - \text{Energy output}\)
\(\text{Energy change} = 679 - 864 = -185\text{ kJ/mol}\).

- 1 mark for calculating the total energy required to break bonds as +679 kJ/mol.
- 1 mark for calculating the total energy released when forming bonds as 864 kJ/mol.
- 0.42 marks for the correct overall energy change of -185 kJ/mol (must include minus sign for accuracy).

1. Effect of temperature: The forward reaction is endothermic (indicated by the positive value of \(\Delta H\)). Increasing temperature will favor the endothermic process to absorb the heat, shifting the position of equilibrium to the right.
2. Effect of pressure: There is 1 mole of gaseous reactant on the left and 2 moles of gaseous product on the right. Increasing pressure will favor the side with fewer moles of gas to reduce the pressure, shifting the equilibrium to the left.

- 1 mark for explaining that increasing temperature shifts the equilibrium to the right because the forward reaction is endothermic.
- 1 mark for explaining that increasing pressure shifts the equilibrium to the left because there are fewer moles of gas on the left-hand side.
- 0.42 marks for correct terminology ("dynamic equilibrium", "shifts left/right").

1. Recall the retention factor formula:
\(R_f = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}\)

2. Substitute the measurements into the formula:
\(R_f = \frac{6.2\text{ cm}}{8.0\text{ cm}} = 0.775\)

3. Round the value to 2 decimal places:
\(R_f \approx 0.78\).

- 1 mark for using the correct formula to set up the fraction (6.2 / 8.0).
- 1 mark for obtaining the decimal value 0.775.
- 0.42 marks for rounding correctly to two decimal places (0.78).

1. Determine the mass of carbon dioxide gas lost from the flask:
\(\text{Mass loss} = 150.00\text{ g} - 148.24\text{ g} = 1.76\text{ g}\).

2. Calculate the rate of mass loss:
\(\text{Rate} = \frac{\text{change in mass}}{\text{time taken}} = \frac{1.76\text{ g}}{120\text{ s}} \approx 0.01467\text{ g/s}\).

3. Convert to standard form to 2 significant figures:
\(0.01467\text{ g/s} \approx 1.5 \times 10^{-2}\text{ g/s}\).

- 1 mark for calculating the mass loss as 1.76 g.
- 1 mark for dividing the mass loss by the time (1.76 / 120) to get 0.0147 g/s.
- 0.42 marks for stating the final value in correct standard form to 2 significant figures (\(1.5 \times 10^{-2}\)).

First, calculate the moles of each element in 100 g of the compound: Moles of C = 60.0 / 12.0 = 5.0 mol. Moles of H = 13.3 / 1.0 = 13.3 mol. Moles of O = 26.7 / 16.0 = 1.67 mol. Next, divide each value by the smallest number of moles (1.67): Ratio of C = 5.0 / 1.67 = 3. Ratio of H = 13.3 / 1.67 = 8. Ratio of O = 1.67 / 1.67 = 1. Therefore, the empirical formula is C3H8O.

1 mark for calculating correct mole ratios (C = 5, H = 13.3, O = 1.67). 1 mark for dividing by the smallest value to find the simplest whole number ratio (3 : 8 : 1) and writing the final correct empirical formula C3H8O.

First, calculate the moles of iron: Moles of Fe = 11.2 / 56.0 = 0.2 mol. From the balanced equation, 4 moles of Fe produce 2 moles of \(Fe_2O_3\), so the molar ratio of Fe to \(Fe_2O_3\) is 2:1. Moles of \(Fe_2O_3\) = 0.2 / 2 = 0.1 mol. Calculate the relative formula mass (Mr) of \(Fe_2O_3\): (2 * 56.0) + (3 * 16.0) = 160.0. Calculate the mass of \(Fe_2O_3\): 0.1 mol * 160.0 g/mol = 16.0 g.

1 mark for calculating correct moles of Fe (0.2 mol) and identifying the correct mole ratio to find moles of \(Fe_2O_3\) (0.1 mol). 1 mark for calculating Mr of \(Fe_2O_3\) as 160.0 and finding the final mass of 16.0 g.

At the cathode, positive copper(II) ions (\(Cu^{2+}\)) gain electrons (reduction) to form copper metal. The balanced ionic half-equation with state symbols is \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\).

1 mark for the correct reactants and products with charges (\(Cu^{2+}\) and \(e^-\) on the left, \(Cu\) on the right). 1 mark for correct balancing and inclusion of correct state symbols (aq) and (s).

The formula for the Rf value is: Rf = distance moved by substance / distance moved by solvent front. Rf = 5.2 cm / 8.0 cm = 0.65.

1 mark for showing correct substitution into the Rf equation (5.2 / 8.0). 1 mark for the correct final answer of 0.65 (must be to 2 significant figures).

Each oxygen atom shares two pairs of electrons with the carbon atom to form a double covalent bond (C=O). Since there are two double bonds in a \(CO_2\) molecule, there are 4 covalent bonds in total. Each bond consists of 2 shared electrons. Therefore, the total number of shared electrons is 4 * 2 = 8 electrons.

1 mark for identifying that carbon dioxide contains two double covalent bonds. 1 mark for calculating the total number of shared electrons as 8.

Energy required to break reactant bonds: H-H bond + Cl-Cl bond = 436 + 243 = 679 kJ/mol. Energy released when making product bonds: 2 * H-Cl bonds = 2 * 432 = 864 kJ/mol. Overall energy change = Energy in - Energy out = 679 - 864 = -185 kJ/mol.

1 mark for calculating the total energy required to break bonds (679 kJ/mol) and total energy released when making bonds (864 kJ/mol). 1 mark for calculating the correct overall energy change of -185 kJ/mol (must include negative sign for exothermic reaction).

Rate of reaction = Volume of gas produced / Time taken. Rate = 45 \(cm^3\) / 30 s = 1.5 \(cm^3/s\).

1 mark for correct formula or substitution of values (45 / 30). 1 mark for correct calculation of 1.5 and the correct unit \(cm^3/s\).

Percentage yield is calculated using the formula: (Actual yield / Theoretical yield) * 100. Percentage yield = (3.4 g / 17.0 g) * 100 = 20%.

1 mark for setting up the correct calculation ratio (3.4 / 17.0). 1 mark for multiplying by 100 to obtain the correct percentage yield of 20%.

First, calculate the moles of each element: Moles of Fe = 1.12 / 55.8 = 0.02 mol. Moles of O = 0.48 / 16.0 = 0.03 mol. Next, find the simplest whole-number ratio by dividing by the smallest value: Fe = 0.02 / 0.02 = 1, O = 0.03 / 0.02 = 1.5. Multiplying by 2 to get integers gives a ratio of Fe:O = 2:3. Thus, the empirical formula is \(\text{Fe}_2\text{O}_3\).

1 Mark: Correctly calculates moles of Fe (0.02) and O (0.03). 1.42 Marks: Determines the simplest whole-number ratio of 2:3 and states the empirical formula as \(\text{Fe}_2\text{O}_3\).

Complete combustion of pentane reacts with oxygen to form carbon dioxide and water: \(\text{C}_5\text{H}_{12} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). Balancing carbon gives 5 \(\text{CO}_2\). Balancing hydrogen gives 6 \(\text{H}_2\text{O}\). This requires 16 oxygen atoms on the right side, so we need 8 \(\text{O}_2\) molecules. Adding state symbols at room temperature/combustion conditions: \(\text{C}_5\text{H}_{12}(\text{g}) + 8\text{O}_2(\text{g}) \rightarrow 5\text{CO}_2(\text{g}) + 6\text{H}_2\text{O}(\text{g})\) (or liquid water is acceptable).

1 Mark: Correctly identifies all reactants and products with appropriate state symbols. 1.42 Marks: Fully balanced chemical equation.

At the negative electrode (cathode), copper(II) ions in solution are attracted and reduced by gaining electrons to form solid copper metal. The half-equation is: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\).

1 Mark: Identifies the correct reactants (\(\text{Cu}^{2+}\) and \(\text{e}^-\)) and product (\(\text{Cu}\)). 1.42 Marks: Balances charges and adds correct state symbols: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\).

Energy to break reactant bonds = \(\text{Bond}(\text{H}-\text{H}) + \text{Bond}(\text{Cl}-\text{Cl}) = 436 + 243 = 679\text{ kJ/mol}\). Energy released making product bonds = \(2 \times \text{Bond}(\text{H}-\text{Cl}) = 2 \times 432 = 864\text{ kJ/mol}\). Overall energy change = Energy to break bonds - Energy released making bonds = 679 - 864 = -185 kJ/mol.

1 Mark: Shows working for reactant bond breaking (679 kJ/mol) and product bond making (864 kJ/mol). 1.42 Marks: Correct calculation of the final overall energy change as -185 kJ/mol (must include negative sign).

Mean rate of reaction is calculated by dividing the volume of gas produced by the time taken: Mean rate = 45 cm³ / 90 s = 0.5 cm³/s.

1 Mark: Shows the correct division substitution (45 / 90). 1.42 Marks: Correct final calculation of 0.5 cm³/s with appropriate units.

The retention factor is given by the formula: \(R_f = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}} = \frac{5.2}{8.0} = 0.65\).

1 Mark: Recalls the correct formula or correctly substitutes the values (5.2 / 8.0). 1.42 Marks: Correctly calculates \(R_f\) as 0.65.

Increasing the temperature shifts the equilibrium position to favor the endothermic direction to oppose the increase in temperature. Since the forward reaction is exothermic, the reverse reaction is endothermic. Therefore, the equilibrium shifts to the left, decreasing the yield of ammonia.

1 Mark: States that the yield of ammonia decreases. 1.42 Marks: Explains that the equilibrium shifts to the left / in the endothermic direction to oppose the temperature increase.

Sodium chloride has a giant ionic lattice structure. Within this structure, there are strong electrostatic forces of attraction acting in all directions between the oppositely charged sodium ions (\(\text{Na}^+\)) and chloride ions (\(\text{Cl}^-\)). A substantial amount of thermal energy is required to overcome these strong electrostatic attractions to break the lattice.

1 Mark: Identifies the structure as a giant ionic lattice containing strong electrostatic forces of attraction between oppositely charged ions. 1.42 Marks: Explains that a large amount of energy is required to break these strong ionic bonds/forces.

1. Calculate empirical formula: Moles of C = 85.7 / 12 = 7.14. Moles of H = 14.3 / 1 = 14.3. Dividing both by 7.14 gives a ratio of C:H = 1:2, so the empirical formula is CH2. 2. Compare empirical formula mass to relative molecular mass: Mass of CH2 = 12 + (2 * 1) = 14. Scale factor = 56 / 14 = 4. 3. Multiply empirical formula by scale factor: (CH2) * 4 = C4H8.

1 mark for calculating correct empirical formula of CH2. 1 mark for calculating scale factor of 4. 0.42 marks for final correct molecular formula C4H8.

1. Moles of Fe = 11.2 / 56 = 0.20 mol. 2. From the equation, 4 moles of Fe produce 2 moles of Fe2O3, which is a 2:1 ratio. Therefore, moles of Fe2O3 = 0.20 / 2 = 0.10 mol. 3. Mr of Fe2O3 = (2 * 56) + (3 * 16) = 160. 4. Mass of Fe2O3 = 0.10 mol * 160 g/mol = 16.0 g.

1 mark for calculating moles of Fe as 0.20 mol. 1 mark for calculating moles of Fe2O3 as 0.10 mol. 0.42 marks for calculating final mass of 16.0 g.

Chloride ions (Cl^-(aq)) lose electrons at the positive anode to form diatomic chlorine gas (Cl2(g)). The balanced half-equation showing state symbols is: 2Cl^-(aq) -> Cl_2(g) + 2e^- (or 2Cl^-(aq) - 2e^- -> Cl_2(g)).

1 mark for correct species Cl^- and Cl2 with state symbols. 1 mark for correct balancing with electrons. 0.42 marks for overall correct representation of oxidation at anode.

Rf value is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front. Rf = 5.2 cm / 8.0 cm = 0.65.

1 mark for recalling Rf formula (distance by spot / distance by solvent). 1 mark for substituting values (5.2 / 8.0). 0.42 marks for final correct answer of 0.65.

1. Energy required to break reactant bonds: (H-H) + (Cl-Cl) = 436 + 243 = 679 kJ/mol. 2. Energy released by forming product bonds: 2 * (H-Cl) = 2 * 432 = 864 kJ/mol. 3. Overall energy change = Energy in - Energy out = 679 - 864 = -185 kJ/mol.

1 mark for calculating energy to break reactant bonds (679 kJ/mol). 1 mark for calculating energy released by making product bonds (864 kJ/mol). 0.42 marks for calculating overall energy change (-185 kJ/mol, negative sign must be present).

The mean rate of reaction is calculated by dividing the volume of gas produced by the time taken: Mean rate = Volume of gas / Time = 45.0 cm^3 / 30 s = 1.5 cm^3/s.

1 mark for stating or using the rate formula (volume / time). 1 mark for correct substitution (45.0 / 30). 0.42 marks for final correct answer of 1.5 cm^3/s.

To form a neutral ionic compound, the total positive charge must equal the total negative charge. Two aluminium ions (2 * +3 = +6) balance three oxide ions (3 * -2 = -6). Thus, the chemical formula is Al2O3.

1 mark for identifying the ratio of Al to O as 2:3. 1.42 marks for the correct formula Al2O3.

1. Calculate moles of NaOH used: moles = concentration * volume = 0.100 * (20.0 / 1000) = 0.00200 mol. 2. Use the balanced equation (1:2 ratio of H2SO4 to NaOH): moles of H2SO4 = 0.00200 / 2 = 0.00100 mol. 3. Calculate concentration of H2SO4: concentration = moles / volume = 0.00100 / (25.0 / 1000) = 0.0400 mol/dm^3.

1 mark for calculating moles of NaOH (0.00200 mol). 1 mark for applying mole ratio to find moles of H2SO4 (0.00100 mol). 0.42 marks for finding concentration (0.04 or 0.0400 mol/dm^3).

Step 1: Calculate the molar mass (\(M_r\)) of propane: \(M_r(\text{C}_3\text{H}_8) = (3 \times 12) + (8 \times 1) = 44\ \text{g/mol}\). Step 2: Calculate the number of moles of propane: \(\text{moles} = 22.0 / 44 = 0.5\ \text{mol}\). Step 3: Use the stoichiometric ratio from the balanced equation (1 mole of propane produces 3 moles of \(\text{CO}_2\)) to find the moles of \(\text{CO}_2\): \(0.5 \times 3 = 1.5\ \text{mol}\). Step 4: Calculate the molar mass of \(\text{CO}_2\): \(M_r(\text{CO}_2) = 12 + (2 \times 16) = 44\ \text{g/mol}\). Step 5: Calculate the mass of \(\text{CO}_2\): \(\text{mass} = 1.5 \times 44 = 66\ \text{g}\).

[1 mark] For calculating moles of propane as 0.5 mol. [1 mark] For multiplying propane moles by 3 to find moles of carbon dioxide (1.5 mol). [0.42 marks] For calculating correct mass of 66 g.

Step 1: Calculate total formula mass of products (or reactants): \(M_r(\text{CO}) = 12 + 16 = 28\), \(M_r(3\text{H}_2) = 3 \times (2 \times 1) = 6\). Total mass of products = \(28 + 6 = 34\). Step 2: Determine mass of desired product (\(3\text{H}_2\)) = 6. Step 3: Calculate percentage atom economy: \(\text{Atom Economy} = (6 / 34) \times 100 = 17.647...\%\). Rounded to 3 significant figures, this is 17.6%.

[1 mark] For calculating total formula mass of products as 34. [1 mark] For calculating mass of desired product as 6. [0.42 marks] For correct final calculation to 3 s.f. (17.6%).

At the anode (positive electrode), chloride ions are oxidised to produce chlorine gas. The half-equation is \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\). Chlorine gas is identified because it turns damp blue litmus paper red and then bleaches it white.

[1 mark] For correct balanced half-equation including electrons on the product side: 2Cl⁻ → Cl₂ + 2e⁻ (or 2Cl⁻ - 2e⁻ → Cl₂). [1.42 marks] For stating damp blue (or red) litmus paper is bleached/turns white.

Using the formula: \(R_f = \text{distance moved by substance} / \text{distance moved by solvent front}\). Here, \(R_f = 4.5 \text{ cm} / 6.0 \text{ cm} = 0.75\).

[1 mark] For showing the division of 4.5 by 6.0. [1.42 marks] For the correct final decimal value of 0.75 (accept no units as Rf is a ratio).

To find the simplest ratio that balances the charges: \(2 \times (+3) = +6\) (from two aluminium ions) and \(3 \times (-2) = -6\) (from three oxide ions). The charges sum to zero, so the empirical formula is \(\text{Al}_2\text{O}_3\).

[1 mark] For stating the correct chemical formula: Al₂O₃. [1.42 marks] For explaining that two aluminium ions (+6 total) balance three oxide ions (-6 total) to create a net neutral compound.

Neutralisation involves the reaction of hydrogen ions from the acid with hydroxide ions from the alkali to form water molecules. The simplified ionic equation is \(\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}\).

[1.42 marks] For the correct species in the ionic equation: H⁺ + OH⁻ → H₂O. [1 mark] For correct state symbols: (aq) for reactants and (l) for product.

Step 1: Calculate change in volume of gas: \(45\ \text{cm}^3 - 15\ \text{cm}^3 = 30\ \text{cm}^3\). Step 2: Calculate change in time: \(50\ \text{s} - 20\ \text{s} = 30\ \text{s}\). Step 3: Divide volume change by time change: \(\text{rate} = 30\ \text{cm}^3 / 30\ \text{s} = 1.0\ \text{cm}^3/\text{s}\).

[1 mark] For showing the change in volume (30 cm³) and change in time (30 s). [1.42 marks] For the correct final value of 1.0 (allow 1) with correct unit cm³/s (or cm³s⁻¹).

Step 1: Calculate energy needed to break bonds (reactants): \(1 \times \text{H-H} + 1 \times \text{Cl-Cl} = 436 + 243 = 679\ \text{kJ/mol}\). Step 2: Calculate energy released when making bonds (products): \(2 \times \text{H-Cl} = 2 \times 432 = 864\ \text{kJ/mol}\). Step 3: Calculate energy change: \(\text{Energy broken} - \text{Energy made} = 679 - 864 = -185\ \text{kJ/mol}\).

[1 mark] For calculating energy to break bonds (679 kJ/mol) and energy released making bonds (864 kJ/mol). [1.42 marks] For the correct final answer of -185 kJ/mol (must include minus sign for exothermic reaction).

Step 1: Calculate the relative formula mass (
\(M_r\)
) of butane (
\(\text{C}_4\text{H}_{10}\)
):
\(M_r = (4 \times 12.0) + (10 \times 1.0) = 58.0\)
.

Step 2: Calculate the number of moles of butane reacted:
\(\text{Moles} = \frac{5.8\text{ g}}{58.0\text{ g/mol}} = 0.10\text{ mol}\)
.

Step 3: Deduce the moles of carbon dioxide produced using the mole ratio. Each molecule of butane contains 4 carbon atoms, meaning 1 mole of
\(\text{C}_4\text{H}_{10}\)
produces 4 moles of
\(\text{CO}_2\)
:
\(\text{Moles of } \text{CO}_2 = 0.10 \times 4 = 0.40\text{ mol}\)
.

Step 4: Calculate the relative formula mass (
\(M_r\)
) of
\(\text{CO}_2\)
:
\(M_r = 12.0 + (2 \times 16.0) = 44.0\)
.

Step 5: Calculate the mass of
\(\text{CO}_2\)
produced:
\(\text{Mass} = 0.40\text{ mol} \times 44.0\text{ g/mol} = 17.6\text{ g}\)
.

• 1 mark for calculating 0.10 mol of butane (or finding its correct \(M_r = 58\)).
• 1 mark for realizing the 1:4 molar ratio to get 0.40 mol of \(\text{CO}_2\).
• 0.42 marks for calculating the correct mass of 17.6 g (accept correct answers based on ecf).

Step 1: Calculate the moles of magnesium used:
\(\text{Moles of Mg} = \frac{2.4\text{ g}}{24.0\text{ g/mol}} = 0.10\text{ mol}\)
.

Step 2: Determine the theoretical moles of hydrogen gas produced from the stoichiometry. Since the ratio of
\(\text{Mg} : \text{H}_2\)
is 1:1, the theoretical yield of
\(\text{H}_2\)
is 0.10 mol.

Step 3: Calculate the theoretical mass of
\(\text{H}_2\)
(where
\(M_r \text{ of } \text{H}_2 = 2.0\)
):
\(\text{Theoretical mass} = 0.10\text{ mol} \times 2.0\text{ g/mol} = 0.20\text{ g}\)
.

Step 4: Calculate the percentage yield:
\(\text{Percentage yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100 = \left(\frac{0.18\text{ g}}{0.20\text{ g}}\right) \times 100 = 90\%\)
.

• 1 mark for calculating the theoretical mass of hydrogen as 0.20 g (or 0.10 mol).
• 1 mark for calculating the correct percentage yield of 90%.
• 0.42 marks for showing correct formula/steps for percentage yield.

Aluminium ions are reduced at the negative cathode. The liquid aluminium ions gain three electrons each to form molten aluminium metal:
\(\text{Al}^{3+}\text{(l)} + 3\text{e}^- \rightarrow \text{Al(l)}
. Note that state symbols must reflect that this process takes place in molten electrolyte (liquid phase).

• 1 mark for reactants and products with correct charges (\(\text{Al}^{3+}\) and \(\text{Al}\)).
• 1 mark for correct balance of electrons (\(3\text{e}^-\)) on the left side.
• 0.42 marks for including correct state symbols (l).

The retention factor (
\(R_f\)
) is calculated using the following formula:

\(R_f = \frac{\text{Distance moved by the substance}}{\text{Distance moved by the solvent front}}\)

Substitute the values from the question:
\(R_f = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\)
.

• 1 mark for selecting/stating the correct formula for \(R_f\).
• 1.42 marks for the correct numerical calculation of 0.65 (accept 0.65 with no working shown).

Carbon dioxide (
\(\text{CO}_2\)
) exists as small, simple molecules. There are only weak intermolecular forces (forces of attraction between different molecules) holding them together, which do not require much thermal energy to overcome.

Silicon dioxide (
\(\text{SiO}_2\)
), on the other hand, exists as a giant covalent macromolecular structure. Every silicon atom is strongly bonded to oxygen atoms by multiple strong covalent bonds throughout the entire crystal lattice. A vast amount of energy is needed to break these strong bonds.

• 1 mark for stating carbon dioxide has a simple molecular structure with weak intermolecular forces.
• 1 mark for stating silicon dioxide has a giant covalent lattice with strong covalent bonds.
• 0.42 marks for relating the structures directly to the energy required to overcome them (little energy for weak intermolecular forces vs high energy to break strong covalent bonds).

Step 1: Calculate the total energy required to break reactants' bonds:
- 4
\(\text{C-H}\)
bonds =
\(4 \times 413 = 1652\text{ kJ/mol}\)

- 2
\(\text{O=O}\)
bonds =
\(2 \times 498 = 996\text{ kJ/mol}\)

Total Energy In =
\(1652 + 996 = 2648\text{ kJ/mol}\)
.

Step 2: Calculate the total energy released when products' bonds are formed:
- 2
\(\text{C=O}\)
bonds =
\(2 \times 805 = 1610\text{ kJ/mol}\)

- 4
\(\text{O-H}\)
bonds =
\(4 \times 463 = 1852\text{ kJ/mol}\)

Total Energy Out =
\(1610 + 1852 = 3462\text{ kJ/mol}\)
.

Step 3: Calculate overall energy change:
\(\text{Energy change} = \text{Total Energy In} - \text{Total Energy Out} = 2648 - 3462 = -814\text{ kJ/mol}\)
.

• 1 mark for calculating correct energy required to break bonds (2648 kJ/mol).
• 1 mark for calculating correct energy released making bonds (3462 kJ/mol).
• 0.42 marks for correct calculation of the final overall energy change of -814 kJ/mol (must include negative sign).

The mean rate of reaction is calculated using the change in quantity (volume of gas) divided by the change in time:

\(\text{Mean rate} = \frac{\text{Volume at 40 s} - \text{Volume at 0 s}}{\text{Time elapsed}} = \frac{58\text{ cm}^3 - 0\text{ cm}^3}{40\text{ s} - 0\text{ s}} = \frac{58}{40} = 1.45\text{ cm}^3/\text{s}\)
.

• 1 mark for setting up the correct quotient (\(\frac{58}{40}\)).
• 1.42 marks for the correct value of 1.45 (with or without unit).

Step 1: Calculate the moles of sulfuric acid reacted:
\(\text{Moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume (dm}^3) = 0.150\text{ mol/dm}^3 \times \left(\frac{20.0}{1000}\right)\text{ dm}^3 = 0.0030\text{ mol}\)
.

Step 2: Determine the moles of sodium hydroxide using the stoichiometric ratio (2:1):
\(\text{Moles of NaOH} = 2 \times 0.0030\text{ mol} = 0.0060\text{ mol}\)
.

Step 3: Calculate the concentration of the sodium hydroxide solution:
\(\text{Concentration of NaOH} = \frac{\text{moles}}{\text{volume (dm}^3)} = \frac{0.0060\text{ mol}}{0.0250\text{ dm}^3} = 0.24\text{ mol/dm}^3\)
.

• 1 mark for calculating the moles of sulfuric acid as 0.0030 mol.
• 1 mark for using the 2:1 stoichiometric ratio to find 0.0060 moles of NaOH.
• 0.42 marks for calculating the final concentration of 0.24 (accept 0.24 mol/dm³).

1. Find the number of moles of each element in 100 g of the compound: Moles of \(\text{C} = \frac{85.7}{12.0} = 7.14\text{ mol}\). Moles of \(\text{H} = \frac{14.3}{1.0} = 14.3\text{ mol}\). 2. Find the simplest ratio by dividing by the smallest value: \(\text{C} = \frac{7.14}{7.14} = 1\), \(\text{H} = \frac{14.3}{7.14} \approx 2\). Therefore, the empirical formula is \(\text{CH}_2\).

1 mark for dividing the percentages by the correct relative atomic masses to find the molar ratio of 7.14 : 14.3. 1.42 marks for finding the simplest whole number ratio of 1 : 2 and correctly stating the empirical formula as \(\text{CH}_2\).

Use the formula: \(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\). Substituting the given values: \(\text{Percentage Yield} = \frac{9.50}{12.50} \times 100 = 76.0\%\).

1 mark for the correct substitution of values into the percentage yield formula: \(\frac{9.50}{12.50} \times 100\). 1.42 marks for the correct final answer of 76.0% (accept 76%).

At the cathode, hydrogen ions (\(\text{H}^+\)) gain electrons (reduction) to form hydrogen molecules (\(\text{H}_2\)). The balanced half-equation is: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).

1 mark for identifying \(\text{H}^+\) and \(\text{e}^-\) as reactants and \(\text{H}_2\) as the product. 1.42 marks for a fully balanced ionic equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).

The retention factor (\(R_{\text{f}}\)) is calculated as: \(R_{\text{f}} = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent}} = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\).

1 mark for substituting the correct distances into the chromatography formula: \(\frac{5.2}{8.0}\). 1.42 marks for the correct final answer of 0.65.

Each carbon atom in diamond forms 4 strong covalent bonds with neighbouring carbon atoms in a giant tetrahedral lattice. To melt diamond, a massive amount of thermal energy is needed to break these many strong covalent bonds throughout the giant covalent structure.

1 mark for identifying that each carbon atom forms 4 covalent bonds. 1.42 marks for explaining that many strong covalent bonds must be broken, requiring a large amount of energy.

The change in volume of gas is: \(45 - 15 = 30\text{ cm}^3\). The time interval is: \(50 - 20 = 30\text{ s}\). The mean rate of reaction is: \(\frac{30\text{ cm}^3}{30\text{ s}} = 1.0\text{ cm}^3/\text{s}\).

1 mark for calculating the correct change in volume (30) and change in time (30). 1.42 marks for the correct division leading to the final answer of 1.0 (or 1) \(\text{cm}^3/\text{s}\).