2024 OCR GCSE Gateway Science - Chemistry A - J248 模擬試題連答案詳解

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond. They undergo addition reactions with bromine, causing the orange bromine water to become colourless. Propene (\(C_3H_6\)) is an alkene, so it decolourises bromine water. Propane is an alkane, propanoic acid is a carboxylic acid, and propanol is an alcohol, none of which rapidly decolourise bromine water.

1 mark for identifying the alkene (propene) as the correct compound.

Magnesium oxide is a giant ionic compound. It has a high melting point due to strong electrostatic forces of attraction between oppositely charged ions. It does not conduct electricity in the solid state because the ions are in fixed positions, but it does conduct electricity when molten because the ions are free to move and carry charge.

1 mark for selecting the correct set of ionic properties: high melting point, non-conductive when solid, conductive when molten.

Chlorine is more reactive than iodine because it is higher up in Group 7. Chlorine displaces iodide ions from potassium iodide, forming aqueous iodine, which is brown in solution: \(Cl_2(aq) + 2KI(aq) \rightarrow 2KCl(aq) + I_2(aq)\).

1 mark for identifying that the solution turns brown because chlorine is more reactive than iodine and displaces the iodide ions.

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

1 mark for calculating the correct overall energy change of \(-185\text{ kJ/mol}\).

In the electrolysis of concentrated aqueous sodium chloride, chloride ions (\(Cl^-\)) are discharged at the anode to produce chlorine gas because they are halide ions in high concentration. Hydrogen ions (\(H^+\)) from water are discharged at the cathode to produce hydrogen gas because hydrogen is less reactive than sodium.

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

Increasing concentration increases the number of particles in a given volume, which increases collision frequency. It does not increase the kinetic energy of the particles, as kinetic energy is only increased by raising the temperature.

1 mark for identifying that increasing concentration increases collision frequency without increasing particle energy.

First, calculate the moles of magnesium: \(n(Mg) = \frac{4.80}{24.3} \approx 0.198\text{ mol}\). According to the balanced equation, the mole ratio of \(Mg\) to \(MgO\) is \(1:1\), so \(0.198\text{ mol}\) of \(MgO\) is formed. Finally, calculate the mass of \(MgO\): \(Mr(MgO) = 24.3 + 16.0 = 40.3\). \(\text{Mass} = 0.198 \times 40.3 \approx 7.96\text{ g}\).

1 mark for the correct calculation yielding \(7.96\text{ g}\).

The forward reaction is exothermic, so decreasing temperature shifts the equilibrium to the right, favoring the production of ammonia. There are more moles of gas on the left (4 moles) than the right (2 moles), so increasing pressure shifts the equilibrium to the side with fewer gas moles (to the right). Thus, low temperature and high pressure maximize the equilibrium yield.

1 mark for identifying that low temperature and high pressure shift the equilibrium to favor ammonia production.

When a carboxylic acid reacts with an alcohol, an ester and water are formed. The ester's name is derived by taking the alkyl group from the alcohol ("ethyl" from ethanol) and the carboxylate group from the carboxylic acid ("propanoate" from propanoic acid). Therefore, ethanol + propanoic acid \(\rightarrow\) ethyl propanoate + water.

[1 mark] A - Ethyl propanoate is the correct product.

Buckminsterfullerene (\(C_{60}\)) has a simple molecular structure composed of spherical cages of carbon atoms. Because it consists of separate molecules held together by relatively weak intermolecular forces, it has a lower melting point compared to giant covalent structures like diamond or graphite.

[1 mark] B - Correctly identifies buckminsterfullerene as a simple molecular structure with a relatively low melting point compared to diamond.

Chlorine is more reactive than iodine and is situated above it in Group 7. Therefore, chlorine will displace the iodide ions from potassium iodide solution to form potassium chloride and molecular iodine: \(\text{Cl}_2(\text{g}) + 2\text{KI}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{I}_2(\text{aq})\). The elemental iodine dissolved in water gives the solution a characteristic brown color.

[1 mark] B - Correctly identifies that a brown solution of iodine is formed due to displacement.

To find the relative atomic mass: \(A_{\text{r}} = \frac{(10 \times 20.0) + (11 \times 80.0)}{100} = \frac{200 + 880}{100} = \frac{1080}{100} = 10.8\).

[1 mark] C - Correctly calculates the relative atomic mass as 10.8.

An energy profile where products have less chemical energy than the reactants indicates that energy was released into the surroundings during the reaction. This means the reaction is exothermic, and the thermal energy transferred to the surroundings causes their temperature to increase.

[1 mark] B - Correctly identifies that the reaction is exothermic and increases the temperature of the surroundings.

At the cathode, \(\text{H}^+\) and \(\text{Na}^+\) ions are present. Since hydrogen is less reactive than sodium, \(\text{H}^+\) is discharged to form hydrogen gas, \(\text{H}_2\). At the anode, \(\text{OH}^-\) and \(\text{SO}_4^{2-}\) ions are present. Since hydroxide ions are oxidized more readily than sulfate ions, oxygen gas, \(\text{O}_2\), is produced at the anode.

[1 mark] B - Correctly identifies hydrogen at the cathode and oxygen at the anode.

The retention factor is given by \(R_{\text{f}} = \frac{\text{Distance traveled by substance}}{\text{Distance traveled by solvent front}} = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\).

[1 mark] A - Correct calculation of the Rf value as 0.65.

Using the same mass of calcium carbonate as a fine powder instead of large lumps increases the surface area exposed to the acid. This leads to a higher frequency of successful collisions, increasing the reaction rate. Because the total mass (and therefore moles) of the limiting reactant (calcium carbonate) remains the same, the final volume of carbon dioxide produced remains unchanged.

[1 mark] D - Correctly identifies that using powder increases rate of reaction while maintaining the total yield of gas.

Oxidation of primary alcohols like propan-1-ol using an oxidising agent such as acidified potassium dichromate(VI) yields a carboxylic acid. Since the starting alcohol has three carbon atoms, the resulting carboxylic acid is propanoic acid, which has the carboxyl functional group \(-\text{COOH}\).

[1 mark] C is correct. Reject A: Propene is formed via dehydration. Reject B: Propanone is a ketone formed by oxidising a secondary alcohol (propan-2-ol). Reject D: Ethyl propanoate is an ester, formed by reacting an alcohol with a carboxylic acid.

To find the monomer of an addition polymer, replace the single bond between the carbon atoms in the main chain of the repeating unit with a double bond and remove the extension bonds. This gives \(\text{CH}_2=\text{CH}-\text{CH}_3\), which is propene.

[1 mark] B is correct. Reject A: Ethene produces poly(ethene), \(-(\text{CH}_2-\text{CH}_2)_n-\). Reject C: But-1-ene produces poly(but-1-ene). Reject D: Propane is an alkane and cannot undergo addition polymerisation because it does not contain a carbon-carbon double bond.

Ionic compounds have high melting points due to strong electrostatic attractions within a giant ionic lattice. They cannot conduct electricity as solids because the ions are fixed in position. However, when molten or dissolved in water, the ions are free to move and carry a charge, allowing electrical conductivity.

[1 mark] C is correct. Reject A: Giant covalent substances (like diamond or silica) do not conduct electricity (except graphite) and do not dissolve to form conducting solutions. Reject B: Simple molecular substances have low melting points. Reject D: Metals conduct electricity in both solid and liquid states.

Chlorine is more reactive than iodine because it is higher up in Group 7. Therefore, chlorine will displace iodine from potassium iodide: \(\text{Cl}_2(\text{g}) + 2\text{KI}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{I}_2(\text{aq})\). The formation of aqueous iodine causes the solution to turn brown.

[1 mark] C is correct. Reject A: Chlorine is more reactive than iodine, so a reaction does occur. Reject B: The solution starts colorless and turns brown, not the other way around. Reject D: Iodine is formed in solution (aqueous) and remains dissolved, rather than evolving immediately as a purple gas.

The balanced equation is \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\). One mole of carbon reacts to produce one mole of carbon dioxide. Number of moles of \(\text{C} = 12.0\text{ g} / 12.0\text{ g/mol} = 1.0\text{ mol}\). Thus, \(1.0\text{ mol}\) of \(\text{CO}_2\) is produced. Mass of \(\text{CO}_2 = 1.0\text{ mol} \times (12.0 + 2 \times 16.0)\text{ g/mol} = 44.0\text{ g}\).

[1 mark] D is correct. Reject A: This is the mass of the carbon reactant. Reject B: This is the molecular mass of carbon monoxide (CO). Reject C: This is the molecular mass of oxygen gas used.

Increasing pressure forces gas particles closer together in a given volume. This increases the concentration of reactant particles, resulting in a higher frequency of collisions per unit time, which increases the rate of reaction.

[1 mark] C is correct. Reject A: Rate increases, not decreases. Reject B: Increasing temperature (not pressure) increases the kinetic energy of particles. Reject D: Pressure does affect the rate of reaction, in addition to potentially shifting the equilibrium position.

The relative atomic mass (\(A_r\)) is calculated as the weighted average: \(A_r = \frac{(63 \times 69.0) + (65 \times 31.0)}{100} = \frac{4347 + 2015}{100} = \frac{6362}{100} = 63.62\). Rounding to 1 decimal place gives \(63.6\).

[1 mark] B is correct. Reject A: This is the mass of the first isotope. Reject C: This is the simple average of 63 and 65, which does not account for the abundances. Reject D: This is incorrect calculation.

Concentrated aqueous sodium chloride contains \(\text{Na}^+\), \(\text{H}^+\), \(\text{Cl}^-\), and \(\text{OH}^-\). At the anode (positive electrode), negative ions are attracted. Since the solution is concentrated, halide ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)), producing chlorine gas (\(\text{Cl}_2\)).

[1 mark] D is correct. Reject A: Hydrogen gas is produced at the cathode. Reject B: Sodium ions remain in solution; hydrogen is discharged instead because hydrogen is less reactive than sodium. Reject C: Oxygen is produced at the anode only when the solution is dilute or does not contain halide ions.

The mass number of the chlorine ion is 37 and its atomic number is 17. The number of neutrons is calculated as: \(\text{Neutrons} = 37 - 17 = 20\). Since the ion has a single negative charge (\(-\)), it has gained one electron, meaning the number of electrons is \(17 + 1 = 18\).

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

Propanoic acid belongs to the homologous series of carboxylic acids with 3 carbon atoms. Its structural formula is \(\text{CH}_3\text{CH}_2\text{COOH}\). Adding up all the constituent atoms gives the molecular formula of \(\text{C}_3\text{H}_6\text{O}_2\).

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

Silicon dioxide has a giant covalent macromolecular structure where every silicon atom is bonded to four oxygen atoms by strong covalent bonds, requiring high temperatures to break. Carbon dioxide consists of simple molecules held together by weak intermolecular forces, which require very little energy to overcome.

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

The balanced equation for the reaction is \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\). First, calculate the moles of \(\text{Mg}\): \(\text{moles} = 6.0\text{ g} / 24\text{ g/mol} = 0.25\text{ mol}\). Because the molar ratio of \(\text{Mg}\) to \(\text{MgO}\) is 1:1, \(0.25\text{ mol}\) of \(\text{MgO}\) is produced. The relative formula mass of \(\text{MgO} = 24 + 16 = 40\text{ g/mol}\) is used to find the mass: \(\text{Mass} = 0.25\text{ mol} \times 40\text{ g/mol} = 10.0\text{ g}\).

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

During the electrolysis of aqueous sodium chloride, chloride ions (\(\text{Cl}^-\) ions) are attracted to the positive anode and oxidised to form chlorine gas. Hydrogen ions (\(\text{H}^+\) ions) from water are attracted to the negative cathode and reduced to form hydrogen gas, because sodium is more reactive than hydrogen and remains in the solution as ions.

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

In an exothermic reaction, chemical energy is released as heat energy to the surroundings. Consequently, the chemical system loses energy, meaning that the products have lower energy than the reactants, resulting in a negative enthalpy change (\(\Delta H < 0\)).

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

1. Calculate the number of moles of magnesium used: \( \text{Moles} = \frac{\text{mass}}{\text{A}_r} = \frac{4.80}{24.0} = 0.200 \text{ mol} \). 2. Determine the theoretical moles of copper produced: Since the molar ratio is 1:1, theoretical moles of \( \text{Cu} = 0.200 \text{ mol} \). 3. Calculate the theoretical mass of copper: \( \text{Mass} = 0.200 \text{ mol} \times 63.5 = 12.7 \text{ g} \). 4. Calculate the percentage yield: \( \text{Percentage yield} = \frac{9.12}{12.7} \times 100 = 71.81\% \), which rounds to 71.8\%.

M1: Moles of Mg = 0.20 mol (1 mark). M2: Theoretical mass of Cu = 12.7 g (1 mark). M3: Correct calculation setup: \( \frac{9.12}{12.7} \times 100 \) (1 mark). M4: Correct final answer to 3 sig figs: 71.8\% (1 mark).

1. Average rate of reaction = \( \frac{\text{Change in volume}}{\text{Change in time}} = \frac{39 - 15}{60 - 20} = \frac{24}{40} = 0.60 \text{ cm}^3/\text{s} \). 2. As the reaction progresses, reactant particles are used up, meaning their concentration decreases. 3. This leads to a lower frequency of collisions between reactant particles, causing the reaction rate to slow down.

M1: Correct rate calculation of 0.60 (1 mark). M2: Correct unit of \( \text{cm}^3/\text{s} \) or \( \text{cm}^3\text{ s}^{-1} \) (1 mark). M3: Explanation that reactant concentration decreases as they are used up (1 mark). M4: Explanation that this leads to a lower frequency of successful collisions (1 mark).

1. Energy required to break bonds: \( 4 \times (\text{C}-\text{H}) + 2 \times (\text{O}=\text{O}) = 4(413) + 2(498) = 1652 + 996 = 2648 \text{ kJ/mol} \). 2. Energy released in making bonds: \( 2 \times (\text{C}=\text{O}) + 4 \times (\text{O}-\text{H}) = 2(805) + 4(463) = 1610 + 1852 = 3462 \text{ kJ/mol} \). 3. Overall energy change = \( 2648 - 3462 = -814 \text{ kJ/mol} \). 4. Since the value is negative, the reaction is exothermic (energy released making bonds is greater than energy needed to break them).

M1: Calculate energy in = 2648 kJ/mol (1 mark). M2: Calculate energy out = 3462 kJ/mol (1 mark). M3: Calculate overall energy change = -814 kJ/mol (1 mark). M4: State that the reaction is exothermic because the overall energy change is negative or energy released is greater than energy taken in (1 mark).

1. Find the total mass of the desired product: \( 2 \times \text{M}_r(\text{C}_2\text{H}_5\text{OH}) = 2 \times 46.0 = 92.0 \). 2. Find the total mass of reactants: \( 180 \). 3. Atom economy = \( \frac{92.0}{180} \times 100 = 51.11\% \), which rounds to 51.1\%. 4. Fermentation is sustainable because glucose is a renewable resource obtained from plants, or because it runs at low temperatures compared to chemical synthesis, saving energy.

M1: Determine mass of desired product = 92.0 (1 mark). M2: Formula setup: \( \frac{92.0}{180} \times 100 \) (1 mark). M3: Correct calculation of atom economy: 51.1\% (1 mark). M4: Correct sustainability reason (e.g., uses renewable feedstock, operates at lower temperatures/pressures) (1 mark).

1. Increasing pressure shifts the equilibrium to the side with fewer moles of gas. There are 4 moles of gas on the left and 2 moles on the right, so the yield of ammonia increases. 2. Since the forward reaction is exothermic, increasing temperature shifts the equilibrium to the left (endothermic direction) to absorb heat, which decreases the yield of ammonia.

M1: Pressure increase increases the yield of ammonia (1 mark). M2: Because there are fewer moles of gas on the right-hand product side (2 vs 4) (1 mark). M3: Temperature increase decreases the yield of ammonia (1 mark). M4: Because the forward reaction is exothermic, so the position of equilibrium shifts in the endothermic direction to oppose the temperature rise (1 mark).

1. At the cathode, hydrogen ions (\( \text{H}^+ \)) are discharged rather than sodium ions because hydrogen is less reactive. Product is hydrogen gas: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \). 2. At the anode, chloride ions (\( \text{Cl}^- \)) are discharged because they are in high concentration. Product is chlorine gas: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \).

M1: Identify hydrogen gas as the product at the cathode (1 mark). M2: Correct cathode half-equation: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \) (1 mark). M3: Identify chlorine gas as the product at the anode (1 mark). M4: Correct anode half-equation: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \) or \( 2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2 \) (1 mark).

1. Calculate \( R_f \) value: \( R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent}} = \frac{3.5}{8.4} = 0.41666... \) 2. Rounding to 2 decimal places gives 0.42. 3. Chromatography distinguishes pure vs mixture: a pure substance will only produce one spot on the paper, whereas a mixture will separate into two or more spots.

M1: Formula setup: \( \frac{3.5}{8.4} \) (1 mark). M2: Correct calculation of \( R_f \) to 2 decimal places: 0.42 (1 mark). M3: State that a pure substance produces only one spot (1 mark). M4: State that a mixture produces multiple spots (1 mark).

1. Set up the weighted average calculation: \( A_r = \frac{(69 \times 60.1) + (71 \times 39.9)}{100} \). 2. Compute the numerator: \( 4146.9 + 2832.9 = 6979.8 \). 3. Divide by 100: \( 69.798 \). 4. Round to 2 decimal places: 69.80. 5. Both isotopes have identical chemical properties because they have the same electronic configuration (same number of outer-shell electrons).

M1: Correct calculation setup: \( \frac{(69 \times 60.1) + (71 \times 39.9)}{100} \) (1 mark). M2: Correct calculation of relative atomic mass: 69.80 (must be 2 d.p.) (1 mark). M3: State that they have the same number of electrons / electronic configuration (1 mark). M4: Explain that chemical reactions depend on outer-shell electron configurations (1 mark).

1. Find the moles of each element in \(100\text{ g}\) 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. Determine the simplest ratio by dividing by the smallest number of moles:
- Ratio of \(\text{C} : \text{H} = \frac{7.14}{7.14} : \frac{14.3}{7.14} = 1 : 2\)
- Thus, the empirical formula is \(\text{CH}_2\).

3. Calculate the empirical formula mass:
- \(M_r(\text{CH}_2) = 12.0 + (2 \times 1.0) = 14.0\)

4. Find the multiplier for the molecular formula:
- Ratio = \(\frac{56}{14.0} = 4\)
- Thus, the molecular formula is \(4 \times \text{CH}_2 = \text{C}_4\text{H}_8\).

Award 1 mark for calculating the moles of carbon (7.14) and hydrogen (14.3).
Award 1 mark for finding the empirical formula as \(\text{CH}_2\).
Award 1 mark for calculating the empirical formula mass of 14.0 and dividing the molecular mass of 56 by 14.0 to get 4.
Award 1 mark for concluding the molecular formula is \(\text{C}_4\text{H}_8\).

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

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

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

4. Determine the thermodynamic nature:
- Since the value of \(\Delta H\) is negative (more heat is released making bonds than is absorbed breaking bonds), the reaction is exothermic.

Award 1 mark for calculating total bond-breaking energy = \(679\text{ kJ/mol}\).
Award 1 mark for calculating total bond-making energy = \(864\text{ kJ/mol}\).
Award 1 mark for calculating overall energy change of \(-185\text{ kJ/mol}\) (accept \(185\text{ kJ/mol}\) released).
Award 1 mark for correctly stating the reaction is exothermic and linking it to more energy being released during bond formation than absorbed during bond breaking.

1. Calculate the mass of water of crystallisation lost:
- Mass of \(\text{H}_2\text{O} = 5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}\)

2. Find the molar masses:
- \(M_r(\text{CuSO}_4) = 63.5 + 32.1 + (4 \times 16.0) = 159.6\text{ g/mol}\)
- \(M_r(\text{H}_2\text{O}) = (2 \times 1.0) + 16.0 = 18.0\text{ g/mol}\)

3. Calculate the amount in moles of each substance:
- Moles of \(\text{CuSO}_4 = \frac{3.20}{159.6} = 0.02005\text{ mol}\)
- Moles of \(\text{H}_2\text{O} = \frac{1.80}{18.0} = 0.100\text{ mol}\)

4. Find the ratio of moles:
- Ratio = \(\frac{0.100}{0.02005} \approx 5\)
- Therefore, \(x = 5\).

Award 1 mark for finding the mass of water lost is \(1.80\text{ g}\).
Award 1 mark for calculating the moles of anhydrous \(\text{CuSO}_4\) as \(0.020\text{ mol}\) (allow ecf from incorrect \(M_r\)).
Award 1 mark for calculating the moles of water as \(0.100\text{ mol}\).
Award 1 mark for dividing moles of water by moles of anhydrous copper(II) sulfate to find \(x = 5\).

1. Let the percentage abundance of \({}^{69}\text{Ga}\) be \(y\%\).
- Then, the percentage abundance of \({}^{71}\text{Ga}\) is \((100 - y)\%\).

2. Express the relative atomic mass equation:
- \(A_r = \frac{69y + 71(100 - y)}{100} = 69.72\)

3. Expand and solve the equation:
- \(69y + 7100 - 71y = 6972\)
- \(-2y = 6972 - 7100\)
- \(-2y = -128\)
- \(y = 64\)

4. Deduce the abundances:
- Percentage abundance of \({}^{69}\text{Ga} = 64\%\)
- Percentage abundance of \({}^{71}\text{Ga} = 100 - 64 = 36\%\).

Award 1 mark for setting up a correct algebraic equation, e.g., \(\frac{69y + 71(100 - y)}{100} = 69.72\).
Award 1 mark for correctly expanding and simplifying the algebraic expression, e.g., \(-2y = -128\) or equivalent.
Award 1 mark for solving to find \(y = 64\%\) (abundance of \({}^{69}\text{Ga}\)).
Award 1 mark for calculating the remaining abundance of \({}^{71}\text{Ga} = 36\%\).

1. Find the relative formula mass (\(M_r\)) of the desired product, ethanol (\(\text{C}_2\text{H}_5\text{OH}\)):
- \(M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\text{ g/mol}\)
- Total mass of desired product = \(2 \times 46.0 = 92.0\text{ g/mol}\)

2. Find the relative formula mass (\(M_r\)) of the reactant, glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)):
- \(M_r(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 180.0\text{ g/mol}\)

3. Calculate the percentage atom economy:
- \(\text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired products}}{\text{Total } M_r \text{ of all reactants}} \times 100\)
- \(\text{Atom Economy} = \frac{92.0}{180.0} \times 100 = 51.111\dots \%\)

4. Round to 3 significant figures:
- \(51.1\%\)

Award 1 mark for calculating the total mass of the desired product, ethanol, as \(2 \times 46.0 = 92.0\).
Award 1 mark for calculating the total mass of reactants (glucose) as \(180.0\) (or total mass of products: \(92.0 + (2 \times 44.0) = 180.0\)).
Award 1 mark for setting up the correct calculation: \(\frac{92.0}{180.0} \times 100\).
Award 1 mark for the final answer of \(51.1\%\) (must be rounded to 3 significant figures; do not accept 51 or 51.11).

1. Calculate the change in gas volume over the specified time interval:
- Volume at \(40\text{ s} = 48\text{ cm}^3\)
- Volume at \(10\text{ s} = 18\text{ cm}^3\)
- \(\Delta \text{Volume} = 48 - 18 = 30\text{ cm}^3\)

2. Calculate the length of the time interval:
- \(\Delta \text{Time} = 40\text{ s} - 10\text{ s} = 30\text{ s}\)

3. Calculate the mean rate of reaction:
- \(\text{Mean rate} = \frac{\Delta \text{Volume}}{\Delta \text{Time}} = \frac{30\text{ cm}^3}{30\text{ s}} = 1\text{ cm}^3/\text{s}\)

4. Format to 2 decimal places:
- \(1.00\text{ cm}^3/\text{s}\)

Award 1 mark for calculating the change in volume as \(30\text{ cm}^3\).
Award 1 mark for calculating the time interval as \(30\text{ s}\).
Award 1 mark for dividing volume by time (e.g., \(\frac{30}{30}\)).
Award 1 mark for the final rate of \(1.00\) (must state to 2 decimal places as requested).

1. Calculate the relative formula mass (\(M_r\)) of \(\text{Fe}_2\text{O}_3\):
- \(M_r(\text{Fe}_2\text{O}_3) = (2 \times 55.8) + (3 \times 16.0) = 111.6 + 48.0 = 159.6\)

2. Determine the moles of \(\text{Fe}_2\text{O}_3\) used:
- \(\text{Moles} = \frac{24.0 \times 10^6\text{ g}}{159.6} = 150,376\text{ mol}\) (or keeping units in proportional tonnes: \(\frac{24.0}{159.6} = 0.1504\text{ million moles}\))

3. Use the stoichiometric ratio from the balanced equation (\(1\text{ mol } \text{Fe}_2\text{O}_3 : 2\text{ mol } \text{Fe}\)):
- Moles of \(\text{Fe} = 2 \times 150,376 = 300,752\text{ mol}\)

4. Calculate the mass of iron produced:
- \(\text{Mass of Fe} = 300,752\text{ mol} \times 55.8\text{ g/mol} = 1.678 \times 10^7\text{ g}\)
- Convert to tonnes: \(1.678 \times 10^7\text{ g} = 16.78\text{ tonnes}\) (which rounds to \(16.8\text{ tonnes}\)).

Award 1 mark for calculating the formula mass of \(\text{Fe}_2\text{O}_3\) as \(159.6\).
Award 1 mark for identifying the mole ratio of reactant to product is \(1 : 2\).
Award 1 mark for a correct expression linking mass, formula mass, and mole ratio (e.g., \(\frac{24.0}{159.6} \times 2 \times 55.8\)).
Award 1 mark for the final answer of \(16.8\text{ tonnes}\) (accept values in the range of \(16.7\) to \(16.8\)).

a) Convert percentage to ppm:
- \(1\% = 10,000\text{ ppm}\)
- \(0.0850\% = 0.0850 \times 10,000 = 850\text{ ppm}\)

b) Convert volumes to the same unit and find the ratio:
- Volume of greenhouse = \(250\text{ m}^3 \times 1000 = 250,000\text{ dm}^3\)
- Concentration increase by volume ratio = \(\frac{110\text{ dm}^3}{250,000\text{ dm}^3} = 0.00044\)
- Convert this fraction to ppm: \(0.00044 \times 1,000,000 = 440\text{ ppm}\)

For part a):
Award 1 mark for showing the conversion factor of multiplying by 10,000.
Award 1 mark for the final answer of \(850\text{ ppm}\).

For part b):
Award 1 mark for converting the greenhouse volume to the same units as the gas released (\(250,000\text{ dm}^3\)).
Award 1 mark for dividing the added volume by the total volume and scaling to find \(440\text{ ppm}\).

Step 1: Calculate the amount in moles of magnesium reacted. \(\text{Moles of Mg} = \frac{\text{mass}}{A_r} = \frac{4.86}{24.3} = 0.20\text{ mol}\). Step 2: Use the stoichiometry of the reaction. According to the balanced equation, \(1\text{ mol}\) of \(\text{Mg}\) produces \(1\text{ mol}\) of \(\text{H}_2\). Therefore, \(0.20\text{ mol}\) of \(\text{Mg}\) produces \(0.20\text{ mol}\) of \(\text{H}_2\). Step 3: Calculate the volume of hydrogen gas. \(\text{Volume} = \text{moles} \times 24.0\text{ dm}^3\text{/mol} = 0.20 \times 24.0 = 4.80\text{ dm}^3\).

Award 1 mark for calculating the moles of magnesium: \(4.86 / 24.3 = 0.20\text{ mol}\). Award 1 mark for identifying the 1:1 molar ratio, hence moles of \(\text{H}_2 = 0.20\text{ mol}\). Award 1 mark for setting up the volume calculation: \(0.20 \times 24.0\). Award 1 mark for the correct final volume: \(4.80\text{ dm}^3\) (allow \(4.8\text{ dm}^3\)). Note: Award full marks for correct final answer with working shown.

Step 1: Calculate the energy required to break reactants' bonds. Energy in = \(4 \times (\text{C}-\text{H}) + 1 \times (\text{Cl}-\text{Cl}) = (4 \times 413) + 243 = 1652 + 243 = 1895\text{ kJ/mol}\). (Alternatively, considering only bonds broken: \(1 \times (\text{C}-\text{H}) + 1 \times (\text{Cl}-\text{Cl}) = 413 + 243 = 656\text{ kJ/mol}\)). Step 2: Calculate the energy released when products' bonds are formed. Energy out = \(3 \times (\text{C}-\text{H}) + 1 \times (\text{C}-\text{Cl}) + 1 \times (\text{H}-\text{Cl}) = (3 \times 413) + 339 + 432 = 1239 + 339 + 432 = 2010\text{ kJ/mol}\). (Alternatively, considering only bonds formed: \(1 \times (\text{C}-\text{Cl}) + 1 \times (\text{H}-\text{Cl}) = 339 + 432 = 771\text{ kJ/mol}\)). Step 3: Calculate the overall energy change. \(\text{Energy change} = \text{Energy in} - \text{Energy out} = 1895 - 2010 = -115\text{ kJ/mol}\) (or \(656 - 771 = -115\text{ kJ/mol}\)). Step 4: Determine the type of reaction. Since the overall energy change is negative, the reaction is exothermic.

Award 1 mark for calculating the energy to break reactant bonds: \(1895\text{ kJ/mol}\) (or \(656\text{ kJ/mol}\)). Award 1 mark for calculating the energy released on making product bonds: \(2010\text{ kJ/mol}\) (or \(771\text{ kJ/mol}\)). Award 1 mark for calculating the overall energy change: \(-115\text{ kJ/mol}\) (allow \(115\text{ kJ/mol}\) of energy released). Award 1 mark for stating the reaction is exothermic with a correct justification (e.g., negative sign / more energy is released making bonds than taken in breaking them).

1. Monomers: Addition polymerisation uses unsaturated monomers that contain a carbon-carbon double bond \(C=C\). Condensation polymerisation uses monomers that have two different functional groups (or two different monomers, each with two functional groups). 2. Products: Addition polymerisation produces only the polymer molecule as the sole product. Condensation polymerisation produces the polymer and also a small molecule, such as water (\(H_2O\)) or hydrogen chloride (\(HCl\)), as a side product.

[1 mark] - Addition monomers contain a \(C=C\) double bond / are unsaturated, whereas condensation monomers have two functional groups. [1 mark] - Addition polymerisation produces only one product (the polymer). [1 mark] - Condensation polymerisation produces a small molecule (such as water or hydrogen chloride) in addition to the polymer.

Chlorine is more reactive than iodine, so it displaces the iodide ions from potassium iodide. The chlorine molecules act as an oxidising agent, oxidising the colourless iodide ions to iodine. The presence of dissolved molecular iodine causes the solution to change to a brown colour. The balanced ionic equation with state symbols is: \(Cl_2(aq) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq)\).

[1 mark] - Colour change: from colourless to brown (accept orange or yellow-brown). [1 mark] - Correct ionic equation: \(Cl_2 + 2I^- \rightarrow 2Cl^- + I_2\). [1 mark] - Correct state symbols: \((aq)\) for all reactants and products.

Silicon dioxide exists as a giant covalent lattice structure. To melt it, many strong covalent bonds between silicon and oxygen atoms must be broken, which requires a massive amount of thermal energy. In contrast, carbon dioxide exists as a simple molecular structure. Although the covalent bonds within each individual \(CO_2\) molecule are strong, the attraction forces between the molecules (intermolecular forces) are very weak and require very little energy to overcome, making it a gas at room temperature.

[1 mark] - Silicon dioxide has a giant covalent structure with strong covalent bonds that require a lot of energy to break. [1 mark] - Carbon dioxide has a simple molecular structure with weak intermolecular forces between molecules. [1 mark] - Less energy is required to overcome the weak intermolecular forces in carbon dioxide than to break the strong covalent bonds in silicon dioxide.

Using the conservation of mass: Mass of reactants = Mass of products. Therefore, Mass of \(CaCO_3\) = Mass of \(CaO\) + Mass of \(CO_2\). Substituting the known values: 12.5 g = 7.0 g + Mass of \(CO_2\). Mass of \(CO_2\) = 12.5 g - 7.0 g = 5.5 g. The law of conservation of mass is fully obeyed because no atoms are created or lost; the apparently lost mass is entirely accounted for by the 5.5 g of carbon dioxide gas that escaped from the open crucible into the surroundings.

[1 mark] - Correct calculation of the mass of carbon dioxide: 5.5 g. [1 mark] - Explanation that total mass of products (solid residue + escaped gas) is equal to the initial mass of the reactant. [1 mark] - Explanation that the decrease in mass in the crucible occurs only because carbon dioxide is a gas and has escaped into the atmosphere.

A catalyst increases the rate of reaction by offering an alternative reaction pathway that has a lower activation energy. This allows a greater proportion of colliding reactant particles to possess energy equal to or greater than the activation energy, leading to more frequent successful collisions. On a reaction profile diagram, the reactant and product energy levels remain unchanged, but the pathway representing the catalysed reaction is shown as a lower curved line with a lower peak than the uncatalysed reaction pathway.

[1 mark] - Catalyst provides an alternative reaction pathway. [1 mark] - This alternative pathway has a lower activation energy. [1 mark] - On a reaction profile diagram, this is shown as a lower curve or peak (lower activation energy barrier) between reactants and products.

Isotopes are defined as atoms of the same element that contain the same number of protons (and thus the same atomic number) but have a different number of neutrons (giving them different mass numbers). Carbon always has an atomic number of 6. A neutral atom of Carbon-13 has 6 protons and 6 electrons. The number of neutrons in Carbon-13 is calculated by subtracting the atomic number from the mass number: 13 - 6 = 7 neutrons.

[1 mark] - Definition: atoms of the same element with the same number of protons but different numbers of neutrons. [1 mark] - Carbon-13 has 6 protons and 6 electrons. [1 mark] - Carbon-13 has 7 neutrons.

1. Energy absorbed to break reactant bonds: \(1 \times (H-H) + 1 \times (Cl-Cl) = 436 + 242 = 678\text{ kJ/mol}\). 2. Energy released in making product bonds: \(2 \times (H-Cl) = 2 \times 431 = 862\text{ kJ/mol}\). 3. Overall energy change = Energy in - Energy out = \(678 - 862 = -184\text{ kJ/mol}\). Since the overall energy change is negative (or since more energy is released when making bonds than is absorbed when breaking bonds), the reaction is exothermic.

[1 mark] - Correct calculation of energy required to break bonds (678 kJ/mol) and energy released when making bonds (862 kJ/mol). [1 mark] - Correct calculation of the overall energy change: -184 kJ/mol (or 184 kJ/mol released). [1 mark] - Correctly identifying the reaction as exothermic with a correct reason (e.g., negative energy value or more energy released in bond-making than absorbed in bond-breaking).

In concentrated aqueous sodium chloride, both sodium ions (\(Na^+\)) and hydrogen ions (\(H^+\), from the ionisation of water) migrate to the negative electrode (cathode). Because hydrogen is less reactive than sodium in the reactivity series, hydrogen ions are preferentially reduced (discharge by gaining electrons) at the cathode rather than sodium ions. The half equation representing this reduction process is: \(2H^+ + 2e^- \rightarrow H_2\).

[1 mark] - Identifies that both sodium (\(Na^+\)) and hydrogen (\(H^+\)) ions are attracted to the cathode. [1 mark] - Explains that hydrogen ions are discharged/reduced because hydrogen is less reactive than sodium. [1 mark] - Correct half equation: \(2H^+ + 2e^- \rightarrow H_2\) (accept state symbols, or the equivalent water reduction equation: \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\)).

Add bromine water (which is orange/brown) to samples of both substances. For the unsaturated alkene, an addition reaction occurs, causing the solution to decolorize (turn colorless). For the saturated alkane, no reaction occurs, so the solution remains orange/brown.

1 mark: Identify bromine water as the reagent. 1 mark: Correct observation for alkene (turns colorless / decolorizes). 1 mark: Correct observation for alkane (remains orange/brown / no change).

Sodium chloride is an ionic compound. In solid form, its ions are held tightly in a fixed giant lattice structure by strong electrostatic forces, so they cannot move to carry a charge. When molten or dissolved in water, the lattice is broken down, enabling the charged ions to move freely and conduct electricity.

1 mark: Mentions that solid sodium chloride has ions in fixed positions or cannot move. 1 mark: Mentions that ions are free to move or mobile in molten or aqueous states. 1 mark: Explains that moving ions are required to carry the electrical charge.

Copper is more reactive than silver, so it displaces silver from its nitrate solution. This produces silver metal (which appears as grey crystals on the copper wire) and copper(II) nitrate (which turns the solution blue). The balanced equation is: \(Cu(s) + 2AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s)\).

1 mark: Observation of silver/grey solid or solution turning blue. 1 mark: Correct formulas for reactants and products. 1 mark: Fully balanced equation with correct state symbols.

At a higher concentration, there are more reactant particles crowded into the same volume of solution. This increases the frequency of collisions between the acid particles and the calcium carbonate. Consequently, there are more successful collisions per second, increasing the reaction rate.

1 mark: More reactant particles in a given volume / closer together. 1 mark: More frequent collisions / higher rate of collision. 1 mark: More successful collisions per unit time / per second.

A proton has a relative mass of 1 and a relative charge of +1. A neutron has a relative mass of 1 and a relative charge of 0. An electron has a negligible mass (often taken as 1/1840 or 1/2000) and a relative charge of -1.

1 mark: Correct mass and charge for proton. 1 mark: Correct mass and charge for neutron. 1 mark: Correct mass and charge for electron.

In fractional distillation, the mixture is heated in a flask. The liquid with the lower boiling point (ethanol, B.P. 78 degrees Celsius) vaporizes more readily and rises up the fractionating column. Water (B.P. 100 degrees Celsius) also evaporates but condenses back down the column because it is cooler at the top. The ethanol vapor reaches the condenser, where it is cooled and condenses back into liquid ethanol, which is then collected.

1 mark: Heating the mixture so that components vaporize. 1 mark: Explaining the role of the fractionating column (allowing the substance with the higher boiling point to condense back down). 1 mark: Condensing the lower-boiling point vapor in the condenser to collect it as a pure liquid.

Short-wavelength electromagnetic radiation (such as visible light) from the Sun passes through the Earth's atmosphere. The Earth's surface absorbs this radiation and re-emits it as longer-wavelength infrared (thermal) radiation. Greenhouse gases in the atmosphere absorb this outgoing infrared radiation, preventing it from escaping into space, which warms the atmosphere.

1 mark: Short-wavelength radiation passes through the atmosphere to Earth. 1 mark: Earth absorbs this and re-emits it as long-wavelength / infrared radiation. 1 mark: Greenhouse gases absorb the outgoing infrared radiation, trapping heat.

At the anode, hydroxide ions from water are discharged in preference to sulfate ions because hydroxide ions are more easily oxidized. This forms oxygen gas and water. The observation is effervescence (bubbles of colorless gas). The balanced half-equation is: \(4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-\).

1 mark: Observation of bubbles / effervescence / gas produced. 1 mark: Correctly identifies oxygen as the product. 1 mark: Correct half-equation \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\).

When the primary alcohol propan-1-ol is heated with acidified potassium dichromate(VI), it is oxidised to propanoic acid. The orange dichromate(VI) ions are reduced to green chromium(III) ions. The structural formula of the propanoic acid formed is CH3CH2COOH.

1 mark: State colour change from orange to green. 1 mark: Give correct structural formula of propanoic acid as CH3CH2COOH (or CH3CH2CO2H).

Chlorine is more reactive than bromine and displaces bromide ions from solution. The ionic equation is \(\text{Cl}_2\text{(g)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}\). The solution changes from colourless to orange (or yellow-brown) because aqueous bromine is formed.

1 mark: Correct species in the equation: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\). 1 mark: Correct state symbols: \(\text{g}\) for chlorine and \(\text{aq}\) for all other species. 1 mark: Describe colour change as colourless to orange / yellow-brown / brown.

Sodium chloride is a giant ionic compound. It consists of a giant lattice with strong electrostatic forces of attraction between oppositely charged sodium and chloride ions, requiring a large amount of energy to break. Hydrogen chloride is a simple covalent molecular substance. There are weak intermolecular forces between the hydrogen chloride molecules which require very little energy to overcome, so it has a low boiling point.

1 mark: Sodium chloride has a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions. 1 mark: A large amount of energy is needed to break these strong ionic bonds. 1 mark: Hydrogen chloride consists of simple molecules with weak intermolecular forces between them that require very little energy to overcome.

First, calculate the moles of iron: \(\text{moles of Fe} = 2.80 / 56.0 = 0.050\text{ mol}\). From the balanced equation, the mole ratio of \(\text{Fe}\) to \(\text{H}_2\) is 1:1, so 0.050 mol of \(\text{H}_2\) is produced. The molecular mass of \(\text{H}_2\) is \(2 \times 1.0 = 2.0\text{ g/mol}\). Mass of hydrogen gas = \(0.050\text{ mol} \times 2.0\text{ g/mol} = 0.10\text{ g}\).

1 mark: Correct calculation of moles of Fe (0.050 mol). 1 mark: Correct mass of hydrogen gas (0.10 g).

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This means that a higher proportion of colliding reactant particles have kinetic energy greater than or equal to the activation energy, leading to a higher frequency of successful collisions. At the end of the reaction, the catalyst is chemically unchanged and is not used up.

1 mark: Provides an alternative reaction pathway. 1 mark: This pathway has a lower activation energy. 1 mark: The catalyst is chemically unchanged at the end of the reaction / not used up.

To find the relative atomic mass, calculate the weighted average based on isotope abundances: \(A_r = \frac{(69 \times 60.11) + (71 \times 39.89)}{100} = \frac{4147.59 + 2832.19}{100} = \frac{6979.78}{100} = 69.7978\). Rounding to 2 decimal places gives 69.80.

1 mark: Correct mathematical expression of the weighted mean: \(\frac{(69 \times 60.11) + (71 \times 39.89)}{100}\). 1 mark: Correct calculation to 2 decimal places: 69.80 (accept 69.8).

The Rf value is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front: \(R_f = 5.4 / 7.5 = 0.72\). A substance with high solubility in the mobile phase (solvent) has a greater affinity for the mobile phase than the stationary phase (paper). Therefore, it spends a larger proportion of time dissolved in the solvent as it moves, resulting in it travelling a greater distance up the chromatogram.

1 mark: Correct calculation of Rf = 0.72. 1 mark: Explanation that high solubility means it spends more time in the mobile phase (or has less affinity for the stationary phase) and thus travels further.

Industrialisation over the past 150 years has led to significant rises in greenhouse gases. Carbon dioxide levels have risen due to the widespread combustion of fossil fuels (such as coal, oil, and gas) for electricity generation and transport, as well as deforestation which decreases the amount of carbon dioxide absorbed by photosynthesis. Methane levels have risen due to agricultural activities, such as intensive cattle farming (which releases methane from animal digestion) and rice paddy farming, as well as the decomposition of waste in landfills.

1 mark: Identifies burning fossil fuels / deforestation as a driver of increased carbon dioxide. 1 mark: Identifies cattle ranching / livestock / rice farming / landfill waste decomposition as a driver of increased methane. 1 mark: Links these human activities to the overall rise in global atmospheric concentrations over the industrial era.

1. The oxidation of ethanol produces ethanoic acid (acetic acid).
2. Ethanoic acid is a weak organic acid, meaning that when universal indicator is added, it will change colour to orange, yellow, or red (indicating an acidic pH between 3 and 6).

Mark 1: Identifies the acid as ethanoic acid (accept acetic acid) [1].
Mark 2: Identifies the colour change of universal indicator as turning orange / yellow / red (accept pH values 3 to 6 or 'acidic colour') [1].

Potassium has an electronic configuration of 2,8,8,1, whereas sodium has 2,8,1. Potassium has more electron shells, meaning its outer-shell electron is further from the positively charged nucleus. This extra distance, combined with increased shielding from the inner electron shells, weakens the electrostatic attraction between the nucleus and the outer electron. As a result, less energy is needed to remove the outer electron from potassium, making it more reactive than sodium.

Mark 1: Identifies that potassium has more electron shells (or its outer electron is further from the nucleus) [1].
Mark 2: Explains that there is more shielding or weaker electrostatic attraction between the nucleus and the outer electron in potassium [1].
Mark 3: States that the outer electron is lost more easily/requires less energy to remove [1].

Silicon dioxide has a giant covalent lattice structure. To melt it, many strong covalent bonds between silicon and oxygen atoms must be broken, which requires a huge amount of thermal energy. Carbon dioxide, however, exists as simple covalent molecules. Although the covalent bonds within each carbon dioxide molecule are strong, the forces between the molecules (intermolecular forces) are very weak. Very little energy is needed to overcome these weak intermolecular forces, which is why carbon dioxide is a gas at room temperature.

Mark 1: Identifies that silicon dioxide has a giant covalent structure with strong covalent bonds that must be broken [1].
Mark 2: Identifies that carbon dioxide consists of simple molecules with weak intermolecular forces between them [1].
Mark 3: Explains that breaking strong covalent bonds in \(\text{SiO}_2\) requires significantly more energy than overcoming the weak intermolecular forces in \(\text{CO}_2\) [1].

1. Calculate the number of moles of each element:
- Moles of \(\text{Fe} = \frac{11.2}{56} = 0.2\text{ mol}\)
- Moles of \(\text{O} = \frac{4.8}{16} = 0.3\text{ mol}\)

2. Determine the simplest ratio by dividing by the smallest number of moles (0.2):
- \(\text{Fe} = \frac{0.2}{0.2} = 1\)
- \(\text{O} = \frac{0.3}{0.2} = 1.5\)

3. Convert to whole numbers by multiplying both values by 2:
- \(\text{Fe} = 2\)
- \(\text{O} = 3\)

Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

Mark 1: Calculates the correct number of moles for iron (0.2) and oxygen (0.3) [1].
Mark 2: Correctly determines the molar ratio as 1 : 1.5 [1].
Mark 3: Converts the ratio to the simplest whole-number ratio (2:3) to state the correct formula \(\text{Fe}_2\text{O}_3\) [1].

As temperature increases, the reacting particles gain kinetic energy and move faster. This causes them to collide with each other more frequently (more collisions per second). Crucially, a larger proportion of the colliding particles now possess energy that is equal to or greater than the activation energy. This results in a higher frequency of successful collisions, thereby increasing the rate of reaction.

Mark 1: Mentions that particles gain kinetic energy / move faster, leading to more frequent collisions (or more collisions per unit time) [1].
Mark 2: Explains that a greater proportion of particles now have energy equal to or greater than the activation energy [1].
Mark 3: Concludes that there are more successful collisions per unit time / greater frequency of successful collisions [1].

To calculate the relative atomic mass (\(A_r\)) of chlorine:
\(A_r = \frac{(75.0 \times 35) + (25.0 \times 37)}{100}\)
\(A_r = \frac{2625 + 925}{100}\)
\(A_r = \frac{3550}{100} = 35.5\)

Mark 1: Shows correct working to calculate the weighted average: \(\frac{(75.0 \times 35) + (25.0 \times 37)}{100}\) (or equivalent fractional calculation) [1].
Mark 2: Provides the correct final answer of 35.5 [1].

1. Calculate the \(R_f\) value:
\(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\)

2. Identification:
Compare this experimental \(R_f\) value to a database of known reference standards. To ensure a correct match, the database values must have been obtained using the exact same stationary phase (paper type) and mobile phase (solvent).

Mark 1: Correctly calculates \(R_f = 0.65\) [1].
Mark 2: States that this value is compared to reference standards / a database of known substance values [1].
Mark 3: Mentions that the chromatography conditions (solvent and stationary phase) must be identical for the comparison to be valid [1].

During a chemical reaction, bonds in the reactants must be broken, which is an endothermic process that absorbs energy. Then, new bonds are formed in the products, which is an exothermic process that releases energy. A reaction is exothermic overall when the total energy released during bond making is greater than the total energy absorbed during bond breaking.

Mark 1: States that breaking bonds absorbs energy / is endothermic and forming bonds releases energy / is exothermic [1].
Mark 2: Explains that more energy is released during the formation of bonds than is taken in to break bonds [1].

1. Compare ionic charges: Magnesium ions have a \(2+\) charge and oxide ions have a \(2-\post\) charge, whereas sodium ions have a \(1+\) charge and chloride ions have a \(1-\post\) charge. This means the ions in \(MgO\) have higher charges than those in \(NaCl\).
2. Contrast the electrostatic forces: Due to the higher charges, the electrostatic forces of attraction between the oppositely charged ions in the giant ionic lattice are much stronger in \(MgO\) than in \(NaCl\).
3. Relate to melting point: Consequently, far more thermal energy is needed to overcome these stronger electrostatic forces and break the lattice in \(MgO\), leading to a much higher melting point.

Award 1 mark for stating that magnesium ions (\(Mg^{2+}\)) and/or oxide ions (\(O^{2-}\)) have a higher charge than sodium ions (\(Na^+\)) and/or chloride ions (\(Cl^-\)).
Award 1 mark for stating that the electrostatic forces of attraction between oppositely charged ions in the giant ionic lattice are stronger in magnesium oxide.
Award 1 mark for explaining that more thermal energy is needed to overcome/break these stronger ionic bonds.

To find why a reaction is exothermic, we compare the energy involved in bond processes:
1. Bond breaking: Energy is taken in to break the covalent bonds in the reactant molecules, \(H_2\) and \(Cl_2\) (endothermic).
2. Bond making: Energy is given out when new covalent bonds are formed to make the product molecules, \(HCl\) (exothermic).
3. Since the overall reaction is exothermic, the quantity of energy released when forming the two moles of \(H-Cl\) bonds is greater than the total energy absorbed to break one mole of \(H-H\) bonds and one mole of \(Cl-Cl\) bonds.

Award 1 mark for stating that bond breaking requires energy (is endothermic) AND bond making releases energy (is exothermic).
Award 1 mark for explaining that more energy is released in making the \(H-Cl\) bonds than is taken in to break the reactant \(H-H\) and \(Cl-Cl\) bonds.

Sodium chloride has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions (sodium ions and chloride ions). It has a high melting point because a large amount of energy is needed to break these strong ionic bonds. It does not conduct electricity when solid because the ions are fixed in position, but it does conduct when molten or in aqueous solution because the ions are free to move and carry charge.

Diamond has a giant covalent macromolecular structure where each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement. It has a very high melting point because many strong covalent bonds must be broken. It does not conduct electricity because all valence electrons are held in localised covalent bonds, meaning there are no free or delocalised electrons to carry charge.

Carbon dioxide has a simple molecular structure with strong covalent bonds between the carbon and oxygen atoms within the molecule, but very weak intermolecular forces of attraction between the molecules. It has a low melting point/boiling point because very little energy is needed to overcome these weak intermolecular forces. It does not conduct electricity because it has no free electrons or ions to carry charge.

Level 3 (5-6 marks): Consistently and accurately describes and compares the structure, bonding, melting point, and electrical conductivity of all three substances, linking properties to structures correctly with clear, logical reasoning.
Level 2 (3-4 marks): Explains the structure and bonding for at least two substances and links them to their melting points or electrical conductivity. Some minor errors or omissions.
Level 1 (1-2 marks): Describes some aspects of bonding or properties for at least one substance, but lacks depth or contains significant misconceptions.

Indicative content:
- Sodium chloride: giant ionic, high melting point (strong electrostatic forces), conducts only when molten/aqueous (mobile ions).
- Diamond: giant covalent, very high melting point (strong covalent bonds), non-conductor (no free electrons).
- Carbon dioxide: simple molecular, low melting/boiling point (weak intermolecular forces), non-conductor (no free charged particles).

Addition polymerisation involves monomer molecules that are unsaturated (alkenes, containing a C=C double bond). During the reaction, the C=C double bonds open up, and the monomer units join together to form a long-chain polymer. There is only one product formed: the polymer itself.

Condensation polymerisation involves monomers with two functional groups (such as a dicarboxylic acid and a diol). When these monomers react, they join together and eliminate a small molecule, usually water, as a side-product. Therefore, two products are formed: the polyester and water.

The equation for the formation of poly(ethene) is: \(n \text{CH}_2=\text{CH}_2 \rightarrow \text{[-CH}_2-\text{CH}_2\text{-]}_n\). The monomer is ethene (with a double bond) and the repeating unit shows single carbon-carbon bonds with trailing bonds extending through the brackets.

Level 3 (5-6 marks): Detailed comparison of both polymerisation types, clearly identifying monomer requirements (unsaturated vs bifunctional), product differences (polymer only vs polymer + small molecule), and provides a correct chemical equation/representation for poly(ethene).
Level 2 (3-4 marks): Describes both types of polymerisation and identifies some differences in monomers or products. Shows a partially correct or slightly incomplete equation/diagram.
Level 1 (1-2 marks): Simple description of one polymerisation type or basic recognition of polymer formation without clear comparison or equations.

Indicative content:
- Addition: Alkenes, C=C opens, single polymer product.
- Condensation: Two functional groups (diol + dicarboxylic acid), forms polyester + water.
- Poly(ethene) equation: \(n \text{C}_2\text{H}_4 \rightarrow [\text{C}_2\text{H}_4]_n\) with correct structures.

According to Le Chatelier's principle:
1. Temperature: The forward reaction is exothermic, so increasing the temperature shifts the equilibrium position to the left (the endothermic direction) to oppose the change, reducing the yield of ammonia. A lower temperature would increase yield but make the reaction rate too slow. 450 °C is a compromise temperature that gives an acceptable yield at a reasonable rate.
2. Pressure: There are 4 moles of gas on the left-hand side and 2 moles of gas on the right-hand side. Increasing the pressure shifts the equilibrium to the right (the side with fewer gas molecules) to reduce the pressure, increasing the yield of ammonia. High pressure also increases the rate of reaction. However, generating and safely containing extremely high pressures is very expensive and dangerous. 200 atmospheres is a compromise pressure.
3. Catalyst: An iron catalyst is used to increase the rate of both the forward and reverse reactions equally. It does not affect the yield, but it allows the equilibrium to be reached much faster and permits the use of a lower temperature, reducing energy costs.

Level 3 (5-6 marks): Complete and coherent explanation of both temperature and pressure on yield (using Le Chatelier's principle) and rate, fully justifying the compromises made for temperature, pressure, and the role of the iron catalyst.
Level 2 (3-4 marks): Explains the effect of temperature or pressure using Le Chatelier's principle, and discusses compromise conditions with some reference to rate/cost.
Level 1 (1-2 marks): Identifies the effect of temperature or pressure on yield, or states that a catalyst speeds up the reaction, but lacks detailed explanation or application of Le Chatelier's principle.

Indicative content:
- High pressure shifts equilibrium right (fewer moles), increases yield and rate. Compromise due to safety/equipment cost.
- Low temp shifts equilibrium right (exothermic), increases yield but decreases rate. Compromise at 450 °C.
- Catalyst increases rate without affecting yield.

During the electrolysis of aqueous sodium chloride, four ions are present in solution: \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\).

At the cathode (negative electrode), both \(\text{Na}^+\) and \(\text{H}^+\) ions are attracted. Hydrogen ions are discharged preferentially because hydrogen is less reactive than sodium. Therefore, hydrogen gas is produced. The half-equation is: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).

At the anode (positive electrode), both \(\text{Cl}^-\) and \(\text{OH}^-\) ions are attracted. Halide ions are discharged preferentially over hydroxide ions when in high concentration, so chlorine gas is produced. The half-equation is: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\).

The remaining ions in solution are \(\text{Na}^+\) and \(\text{OH}^-\), which combine to form sodium hydroxide solution (\(\text{NaOH}\)), which is alkaline.

Level 3 (5-6 marks): Correctly identifies all three products (hydrogen, chlorine, and sodium hydroxide), writes both half-equations accurately, and explains why hydrogen is discharged instead of sodium using the reactivity series.
Level 2 (3-4 marks): Identifies at least two products, writes at least one correct half-equation, and gives a basic explanation of why sodium is not formed.
Level 1 (1-2 marks): Identifies one or two products, but contains errors in equations or explanations.

Indicative content:
- Products: hydrogen (cathode), chlorine (anode), sodium hydroxide (solution).
- Cathode half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (or equivalent).
- Anode half-equation: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) (or equivalent).
- Reactivity explanation: Sodium is more reactive than hydrogen, so hydrogen ions gain electrons more easily / are reduced preferentially.