2025 Edexcel IGCSE Chemistry Practice Paper with Answers

The heat energy change is calculated using the formula \(Q = m \times c \times \Delta T\). Here, the mass of the solution \(m = 55.0\text{ g}\), the specific heat capacity \(c = 4.2\text{ J/g/}^{\circ}\text{C}\), and the temperature change \(\Delta T = 6.5\text{ }^{\circ}\text{C}\). Substituting these values into the formula: \(Q = 55.0 \times 4.2 \times 6.5 = 1501.5\text{ J}\).

1 mark for the correct calculation leading to 1501.5 J.

First, find the amount in moles of \(\text{Fe}_2\text{O}_3\): \(\text{moles} = \frac{80\text{ g}}{160\text{ g/mol}} = 0.5\text{ mol}\). Each mole of \(\text{Fe}_2\text{O}_3\) contains 2 moles of Fe atoms. Therefore, the moles of Fe produced is \(0.5 \times 2 = 1.0\text{ mol}\). Finally, find the mass of Fe: \(\text{mass} = 1.0\text{ mol} \times 56\text{ g/mol} = 56\text{ g}\).

1 mark for the correct calculation leading to 56 g.

Bitumen is at the bottom of the fractionating column, consisting of very large hydrocarbons, making it highly viscous (thick and sticky). Kerosene is higher up the column and is much runnier (less viscous).

1 mark for selecting B.

Ammonia (\(M_r = 17\)) has a lower relative molecular mass than hydrogen chloride (\(M_r = 36.5\)). Lighter gas molecules diffuse faster. Therefore, ammonia molecules diffuse more quickly along the tube, traveling a greater distance before meeting the slower-diffusing hydrogen chloride molecules, which is why the white ring forms closer to the hydrochloric acid end.

1 mark for selecting C.

Increasing the concentration of the acid increases the number of acid particles (H\(^{+}\) ions) in a given volume. This increases the collision frequency between the reacting particles, which increases the rate of reaction.

1 mark for selecting A.

Metal X is the most reactive because it reacts with cold water. Metal Z is moderately reactive because it does not react with water but reacts slowly with acid. Metal Y is the least reactive because it does not react with acid but can be reduced by carbon. Thus, the correct order from most to least reactive is X, Z, Y.

1 mark for selecting A.

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

1 mark for selecting B.

Chlorine is more reactive than iodine and displaces it from potassium iodide solution. The colorless solution of potassium iodide turns brown due to the formation of aqueous iodine (\(\text{I}_2\)). Since chlorine is reduced and iodide is oxidized, this displacement reaction is a redox reaction.

1 mark for selecting A.

The rate of diffusion is determined by:
1. Temperature: higher temperature increases the kinetic energy of particles, causing them to move and diffuse faster.
2. Relative molecular mass (or particle mass): lighter gas particles travel and diffuse faster than heavier gas particles at the same temperature.

M1: Temperature (1)
M2: Relative molecular mass / mass / size of the gas particles (1)
Accept: Concentration gradient.

The empirical formula is defined as the simplest whole-number ratio of the atoms of each element present in a compound. To find the empirical formula of hexane (\(\text{C}_6\text{H}_{14}\)), we divide both subscripts by their highest common factor, which is 2, giving \(\text{C}_3\text{H}_7\).

M1: Simplest whole-number ratio of atoms of each element (in a compound) (1)
M2: \(\text{C}_3\text{H}_7\) (1)

Clean, dry air is composed of approximately 78% nitrogen and 21% oxygen by volume. The remaining 1% is made up of noble gases (such as argon, which is about 0.9%) and carbon dioxide (about 0.04%).

M1: 78% (accept 78% to 79%) (1)
M2: Argon / Carbon dioxide / Helium / Neon / Krypton / Xenon (1)
Reject: Water vapour (as the question states 'dry' air).

As you descend Group 7, the halogens become darker in colour and their melting and boiling points increase due to stronger intermolecular forces. Since iodine is a grey/black solid, astatine (which is below iodine) is expected to be a black solid.

M1: Black / dark grey (1)
M2: Solid (1)

As carbon chain length increases, the size of the molecules increases, leading to stronger intermolecular forces that require more thermal energy to break, thus increasing the boiling point. Longer molecules also tangle more easily, which increases the viscosity (making the liquid thicker and less runny).

M1: Boiling point increases (1)
M2: Viscosity increases / becomes thicker / becomes less runny (1)

Addition polymers are non-biodegradable because they are inert due to strong carbon-carbon single bonds, which means they do not decay in landfill sites. If disposed of by burning (combustion), they release toxic/greenhouse gases into the atmosphere.

M1: Non-biodegradable / inert / do not decay (1)
M2: Combustion/burning releases toxic gases / greenhouse gases (1)
Do not accept: 'causes pollution' without qualification.

Galvanising involves coating iron with zinc. If the coating is scratched, the iron is still protected because zinc is more reactive than iron. The zinc oxidises (loses electrons) preferentially to the iron, protecting the iron sacrificially.

M1: Zinc is more reactive than iron (1)
M2: Zinc reacts/corrodes/oxidises in preference to iron / acts sacrificially / prevents iron from losing electrons (1)

Calcium ions impart a characteristic orange-red (or brick-red) color to the flame. Before doing the test, the wire loop (nichrome or platinum) must be cleaned of impurities by dipping it into concentrated hydrochloric acid and holding it in a hot Bunsen burner flame until the flame shows no colour.

M1: Orange-red / brick-red (1)
M2: Dip the wire in hydrochloric acid AND heat it in a (roaring/blue) Bunsen burner flame (1)

1. The change of state from a solid directly to a gas without passing through the liquid state is called sublimation.
2. In the gas state, the particles move rapidly, randomly, and in all directions.

M1: Sublimation (1 mark)
M2: (Particles move) randomly / rapidly / in all directions (1 mark)
[Reject: vibrate / slide over each other]

As the molecular size of the hydrocarbons increases:
1. The boiling point increases because the intermolecular forces between the larger molecules become stronger, requiring more energy to overcome.
2. The viscosity increases (the liquid becomes thicker and flows less easily) because the longer chains easily become tangled.

M1: Boiling point increases (1 mark)
M2: Viscosity increases / becomes more viscous (1 mark)
[Accept: flows less easily for M2]

1. Dry, unpolluted air contains approximately 78% nitrogen by volume.
2. Argon is the most abundant noble gas in the atmosphere, making up about 0.9% of dry air.

M1: 78% (1 mark) [Accept 78.1% or 78-79%]
M2: Argon (1 mark) [Accept Ar]

At room temperature (approx. 20 °C), bromine is a liquid. Its colour is characteristically red-brown (or brown).

M1: Liquid (1 mark)
M2: Red-brown / brown (1 mark) [Reject: orange, yellow, red]

1. The line of best fit for a rate graph is a smooth curve. It should not be drawn point-to-point.
2. Anomalous points are those that do not follow the overall trend and lie significantly away from the drawn curve.
3. When the reaction stops, no more gas is produced, so the volume remains constant. This is shown on the graph when the curve levels off to become a horizontal line (gradient is zero).

M1: Draw a single smooth curve passing through or close to most points (do not accept joining points with straight lines / dot-to-dot) (1)
M2: Identify the point that lies significantly away from/does not fit the trend of the curve of best fit (1)
M3: Find the time at which the curve becomes completely horizontal / flat / gradient is zero (1)

To find the exact neutralisation point from a temperature-volume graph with rising and falling values:
1. Draw a straight line of best fit through the rising temperature data points.
2. Draw a second straight line of best fit through the falling temperature data points.
3. Extend (extrapolate) both straight lines until they cross each other. The point of intersection represents the maximum temperature rise at complete neutralisation.
4. Read the corresponding volume of sodium hydroxide on the x-axis directly below this intersection point.

M1: Draw two straight lines of best fit (one through the rising points and one through the falling points) (1)
M2: Extend / extrapolate both lines until they cross / intersect (1)
M3: Read the volume (value on the x-axis) at this point of intersection (1)

1. Since the boiling points span from negative to positive values (e.g. methane is \(-162\ ^\circ\text{C}\) and pentane is \(+36\ ^\circ\text{C}\)), the y-axis must have \(0\ ^\circ\text{C}\) positioned above the bottom of the axis so negative numbers can be plotted below it and positive numbers above it.
2. The change in boiling point is non-linear, so the line of best fit must be a single smooth curve passing near all the points (not a straight line or point-to-point straight lines).
3. To estimate the boiling point of hexane (6 carbons), the student should extend/extrapolate the curve smoothly to the vertical line corresponding to 6 carbon atoms on the x-axis, then read across to the y-axis to find the temperature.

M1: Place the zero (\(0\ ^\circ\text{C}\)) of the y-axis in the middle/upper part of the axis so negative values scale downwards and positive values scale upwards (1)
M2: Draw a single smooth curve of best fit (reject straight line or point-to-point lines) (1)
M3: Extrapolate/extend the curve to 6 carbon atoms on the x-axis and read the corresponding temperature on the y-axis (1)

1. Calculate \( M_r \) of \( \text{MgCO}_3 = 24 + 12 + (3 \times 16) = 84 \). 2. Calculate \( M_r \) of \( \text{MgO} = 24 + 16 = 40 \). 3. Calculate the theoretical yield of \( \text{MgO} \): moles of \( \text{MgCO}_3 = \frac{8.40}{84} = 0.100 \text{ mol} \). Since 1 mole of \( \text{MgCO}_3 \) produces 1 mole of \( \text{MgO} \), theoretical moles of \( \text{MgO} = 0.100 \text{ mol} \). Theoretical mass of \( \text{MgO} = 0.100 \times 40 = 4.00 \text{ g} \). 4. Calculate percentage yield: \( \frac{3.40}{4.00} \times 100 = 85.0\% \).

M1: Correctly calculate \( M_r \) values: \( \text{MgCO}_3 = 84 \) and \( \text{MgO} = 40 \) (1 mark). M2: Calculate amount in moles of \( \text{MgCO}_3 \) (0.100 mol) (1 mark). M3: Calculate theoretical mass of \( \text{MgO} \) (4.00 g) (1 mark). M4: Calculate percentage yield as 85.0% or 85% (1 mark).

1. Calculate the heat energy absorbed by water: \( \Delta T = 45.0 - 20.0 = 25.0^\circ\text{C} \). \( Q = 100.0 \times 4.18 \times 25.0 = 10450\text{ J} = 10.45\text{ kJ} \). 2. Calculate the moles of methanol burned: \( M_r(\text{CH}_3\text{OH}) = 12 + (4 \times 1) + 16 = 32 \). Moles = \( \frac{0.80}{32} = 0.025\text{ mol} \). 3. Calculate molar enthalpy change: \( \Delta H = -\frac{10.45}{0.025} = -418\text{ kJ/mol} \).

M1: Calculate \( Q \) as \( 10450\text{ J} \) or \( 10.45\text{ kJ} \) (1 mark). M2: Calculate moles of methanol as 0.025 mol (1 mark). M3: Divide \( Q \) by moles (1 mark). M4: Give correct final value with negative sign (-418 kJ/mol) (1 mark). Accept -420 kJ/mol if rounded. Reject positive sign.

1. Calculate amount in moles of \( \text{NaOH} \): \( \text{moles} = \frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol} \). 2. Find moles of \( \text{H}_2\text{SO}_4 \) using stoichiometry: ratio is \( 2\text{NaOH} : 1\text{H}_2\text{SO}_4 \), so moles of \( \text{H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol} \). 3. Calculate concentration of \( \text{H}_2\text{SO}_4 \): \( \text{concentration} = \frac{0.00125}{20.0 / 1000} = 0.0625\text{ mol/dm}^3 \).

M1: Calculate moles of \( \text{NaOH} \) (0.00250 mol) (1 mark). M2: Divide moles of \( \text{NaOH} \) by 2 to find moles of \( \text{H}_2\text{SO}_4 \) (0.00125 mol) (1 mark). M3: Divide moles of acid by the volume in dm\(^3\) (1 mark). M4: State the final answer of 0.0625 mol/dm\(^3\) (1 mark).

1. Calculate moles of Mg: \( \text{moles of Mg} = \frac{1.20}{24} = 0.050\text{ mol} \). 2. Determine moles of \( \text{O}_2 \) reacted: from equation, ratio of Mg to \( \text{O}_2 \) is \( 2:1 \), so moles of \( \text{O}_2 \) reacted = \( \frac{0.050}{2} = 0.025\text{ mol} \). 3. Calculate volume of \( \text{O}_2 \) reacted: \( 0.025\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 0.60\text{ dm}^3 \). 4. Calculate volume remaining: \( 1.50\text{ dm}^3 - 0.60\text{ dm}^3 = 0.90\text{ dm}^3 \).

M1: Calculate moles of Mg as 0.050 mol (1 mark). M2: Determine moles of oxygen reacted as 0.025 mol (1 mark). M3: Calculate volume of oxygen reacted as \( 0.60\text{ dm}^3 \) (1 mark). M4: Subtract to find final volume remaining of \( 0.90\text{ dm}^3 \) (1 mark).

Part (a): Loss in mass in first 60 s = \( 150.00\text{ g} - 149.34\text{ g} = 0.66\text{ g} \). Average rate = \( \frac{0.66\text{ g}}{60\text{ s}} = 0.011\text{ g/s} \). Part (b): Total loss in mass at 120 s = \( 150.00\text{ g} - 149.12\text{ g} = 0.88\text{ g} \). \( M_r(\text{CO}_2) = 12 + (16 \times 2) = 44 \). Moles of \( \text{CO}_2 \) escaped = \( \frac{0.88\text{ g}}{44\text{ g/mol}} = 0.020\text{ mol} \).

Part (a): M1: Determine mass loss of 0.66 g (1 mark). M2: Calculate rate of 0.011 (g/s) (1 mark). Part (b): M3: Calculate total mass loss of 0.88 g (1 mark). M4: Divide mass by 44 to obtain 0.020 (mol) (1 mark).

1. Multiply each isotopic mass by its relative abundance: \( (85 \times 72.17) = 6134.45 \) and \( (87 \times 27.83) = 2421.21 \). 2. Add these values together: \( 6134.45 + 2421.21 = 8555.66 \). 3. Divide by 100: \( \frac{8555.66}{100} = 85.5566 \). 4. Round to 2 decimal places: \( 85.56 \).

M1: Set up numerator calculation: \( (85 \times 72.17) + (87 \times 27.83) \) (1 mark). M2: Divide by 100 (1 mark). M3: Obtain correct evaluation of 85.5566 (1 mark). M4: Give final answer to 2 decimal places as 85.56 (1 mark).

(a) The gaseous ammonia reacts with gaseous hydrogen chloride to produce solid ammonium chloride:
\(\text{NH}_3(\text{g}) + \text{HCl}(\text{g}) \rightarrow \text{NH}_4\text{Cl}(\text{s})\)

(b) Relative molecular mass (\(M_r\)) of ammonia (\(\text{NH}_3\)) is 17. Relative molecular mass (\(M_r\)) of hydrogen chloride (\(\text{HCl}\)) is 36.5. Because ammonia molecules are lighter/have a lower mass, they diffuse faster and cover a greater distance than the heavier hydrogen chloride molecules in the same amount of time. Therefore, they meet and react closer to the hydrogen chloride end (B).

(c) If the room is warmer (higher temperature), the gas particles gain kinetic energy. As a result, they move and diffuse faster, so the white ring of ammonium chloride forms in less time.

(a)
* M1: Correct reactants and product: \(\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}\) (1 mark)
* M2: Correct state symbols: \((\text{g})\) for both reactants and \((\text{s})\) for the product (1 mark)

(b)
* M1: Ammonia has a lower relative molecular mass (\(M_r\)) than hydrogen chloride / ammonia molecules are lighter / HCl molecules are heavier (1 mark)
* M2: Ammonia molecules diffuse faster / move faster / travel further in the same time (1 mark)

(c)
* M1: The ring forms in less time / faster (1 mark)
* M2: Particles have more kinetic energy and therefore move/diffuse faster (1 mark) (reject "react faster" on its own without mention of kinetic energy or speed of movement)

(a)
* From the first observation, X reacts with Y-nitrate, meaning X is more reactive than Y because it displaces Y.
* From the second observation, Y does not react with Z-nitrate, meaning Y cannot displace Z, so Z is more reactive than Y.
* From the third observation, Z does not react with X-nitrate, meaning Z cannot displace X, so X is more reactive than Z.
* Combining these facts, X is the most reactive, followed by Z, and Y is the least reactive. Therefore, the order is X > Z > Y.

(b) Metal X is oxidized to \(X^{2+}\) ions, while the \(Y^{2+}\) ions are reduced to metal Y:
\(X(\text{s}) + Y^{2+}(\text{aq}) \rightarrow X^{2+}(\text{aq}) + Y(\text{s})\)

(a)
* M1: Deduce the correct order: X, Z, Y (or X > Z > Y) (1 mark)
* M2: State that X is more reactive than Y because X displaces Y / a reaction occurs (1 mark)
* M3: State that Z is more reactive than Y because Y cannot displace Z / no reaction occurs (1 mark)
* M4: State that X is more reactive than Z because Z cannot displace X / no reaction occurs (1 mark)

(b)
* M1: Correct ionic equation: \(X + Y^{2+} \rightarrow X^{2+} + Y\) (1 mark)
* M2: Correct state symbols: \((\text{s})\) for metals and \((\text{aq})\) for ions (1 mark)

(a)
1. Calculate the mass of anhydrous magnesium sulfate (\(\text{MgSO}_4\)):
\(27.80\text{ g} - 25.40\text{ g} = 2.40\text{ g}\)

2. Calculate the mass of water of crystallization lost:
\(30.32\text{ g} - 27.80\text{ g} = 2.52\text{ g}\)

3. Calculate the number of moles of \(\text{MgSO}_4\):
\(\text{Moles} = \frac{2.40\text{ g}}{120\text{ g/mol}} = 0.020\text{ mol}\)

4. Calculate the number of moles of \(\text{H}_2\text{O}\):
\(\text{Moles} = \frac{2.52\text{ g}}{18\text{ g/mol}} = 0.140\text{ mol}\)

5. Determine the simplest whole-number molar ratio:
\(x = \frac{0.140}{0.020} = 7\)

Therefore, \(x = 7\).

(b) Heating to "constant mass" (which involves heating, cooling, weighing, and repeating until the mass remains identical) ensures that all water of crystallization has evaporated from the sample. If the heating is insufficient, some water remains, which would make the calculated mass of water lost too low, resulting in an underestimated value of \(x\).

(a)
* M1: Calculate mass of \(\text{MgSO}_4\) (\(2.40\text{ g}\)) and mass of \(\text{H}_2\text{O}\) (\(2.52\text{ g}\)) (1 mark)
* M2: Calculate moles of \(\text{MgSO}_4\) = \(0.020\text{ mol}\) (1 mark)
* M3: Calculate moles of \(\text{H}_2\text{O}\) = \(0.140\text{ mol}\) (1 mark)
* M4: Divide moles of water by moles of salt to show \(x = 7\) (1 mark)

(b)
* M1: To ensure all water (of crystallization) has been evaporated / driven off / reaction is complete (1 mark)
* M2: To ensure that the final mass of anhydrous salt recorded is accurate / to avoid underestimating the amount of water lost (1 mark)

Fractional distillation is used to separate crude oil because the compounds within it (hydrocarbons) have different boiling points depending on their chain length.

1. **Pre-heating**: The crude oil is heated and vaporized before it is pumped into the bottom of the fractionating column.
2. **Temperature Gradient**: The fractionating column is designed with a temperature gradient; it is very hot at the bottom and becomes progressively cooler towards the top.
3. **Vapor Movement**: The gaseous hydrocarbons enter the column and rise up.
4. **Condensation**: As the vapors rise, they cool down. When the temperature of the column falls below a hydrocarbon's boiling point, that hydrocarbon condenses back into a liquid.
5. **High Boiling Point Fractions**: Long-chain hydrocarbons have strong intermolecular forces and high boiling points. They condense quickly near the bottom of the column and are drained off as thick liquids (e.g., bitumen, fuel oil).
6. **Low Boiling Point Fractions**: Short-chain hydrocarbons have weak intermolecular forces and low boiling points. They rise high up the column to the cooler regions before they condense and are collected (e.g., refinery gases, gasoline).

* M1: Crude oil is heated and vaporized before entering the fractionating column (1 mark)
* M2: State that the column has a temperature gradient / is hot at the bottom and cool at the top (1 mark)
* M3: State that the vaporized hydrocarbons rise up the column (1 mark)
* M4: State that different fractions / hydrocarbons have different boiling points (1 mark)
* M5: Explain that longer-chain hydrocarbons (with higher boiling points) condense and are collected near the bottom (1 mark)
* M6: Explain that shorter-chain hydrocarbons (with lower boiling points) condense and are collected near the top / or exit the top as gases (1 mark)

(a) When the concentration of hydrochloric acid is increased, there are more reacting acid particles (hydrogen ions) present in a given volume of solution. This means that the particles are packed closer together, which increases the probability of them colliding. Consequently, there is a higher frequency of successful collisions (more successful collisions per unit time), which increases the reaction rate.

(b) When the marble chips are crushed into a fine powder, the surface area increases significantly for the same mass. This means a much larger number of calcium carbonate particles are exposed on the surface and are available to collide with the acid molecules at any one time. This increases the frequency of successful collisions, resulting in a faster rate of reaction.

(a)
* M1: More reactant/acid particles in a given volume / unit volume (1 mark) (reject "more particles" on its own without reference to volume)
* M2: Particles are closer together (1 mark)
* M3: Greater frequency of successful collisions / more successful collisions per second / per unit time (1 mark) (reject "more collisions" on its own without time reference)

(b)
* M1: Fine powder has a larger surface area / larger surface area to volume ratio (1 mark)
* M2: More reactant/calcium carbonate particles are exposed / available to react at the surface (1 mark)
* M3: Greater frequency of successful collisions / more successful collisions per second / per unit time (1 mark)

1. **Calculate energy needed to break reactant bonds:**
- \(4 \times (\text{C}-\text{H}) = 4 \times 413 = 1652\text{ kJ/mol}\)
- \(2 \times (\text{O}-\text{H})\) in \(\text{H}_2\text{O} = 2 \times 463 = 926\text{ kJ/mol}\)
- Total energy input = \(1652 + 926 = 2578\text{ kJ/mol}\)

2. **Calculate energy released when product bonds form:**
- \(1 \times (\text{C}\equiv\text{O})\) in \(\text{CO} = 1077\text{ kJ/mol}\)
- \(3 \times (\text{H}-\text{H})\) in \(\text{H}_2 = 3 \times 436 = 1308\text{ kJ/mol}\)
- Total energy output = \(1077 + 1308 = 2385\text{ kJ/mol}\)

3. **Calculate enthalpy change (\(\Delta H\)):**
\(\Delta H = \text{energy input} - \text{energy output} = 2578 - 2385 = +193\text{ kJ/mol}\).

Award 1 mark for the correct option (A).

- **A** is correct because \(\Delta H = \text{total bonds broken} - \text{total bonds made} = +193\text{ kJ/mol}\).
- **B** is incorrect because it is the negative value (representing an exothermic reaction incorrectly).
- **C** is incorrect because it only breaks one \(\text{O}-\text{H}\) bond in water.
- **D** is incorrect because it fails to multiply the \(\text{H}-\text{H}\) bond energy by 3.

During a condensation polymerisation reaction to form a polyester:
- The diol, \(\text{HO-CH}_2\text{-CH}_2\text{-OH}\), loses the hydrogen atom (\(\text{-H}\)) from each of its two \(\text{-OH}\) groups, leaving the residue \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\).
- The dicarboxylic acid, \(\text{HOOC-CH}_2\text{-COOH}\) (which can be drawn as \(\text{HO-CO-CH}_2\text{-CO-OH}\)), loses the hydroxyl group (\(\text{-OH}\)) from each of its two \(\text{-COOH}\) groups, leaving the residue \(\text{-CO-CH}_2\text{-CO-}\).
- Joining these two residues together with an ester linkage yields the repeating unit: \(\text{[-O-CH}_2\text{-CH}_2\text{-O-CO-CH}_2\text{-CO-]}\).

Award 1 mark for the correct option (A).

- **A** is correct because it correctly represents the ester linkage and the carbon chains of both monomers.
- **B** is incorrect because it is missing one of the ester oxygens.
- **C** is incorrect because the dicarboxylic acid residue has an extra \(\text{CH}_2\) group.
- **D** is incorrect because the diol residue is missing a \(\text{CH}_2\) group.

(i) Particles in a solid are held in fixed positions by strong forces and can only vibrate. (ii) Particles in a liquid are close together but arranged in an irregular, random way. (iii) Particles in a gas have high energy, are far apart, and move rapidly and randomly in all directions.

1 mark for identifying (i) as vibrate about fixed positions (accept vibrate only). 1 mark for identifying (ii) as random/irregularly arranged AND closely packed / touching. 1 mark for identifying (iii) as moving rapidly/randomly/quickly in all directions.

(i) Viscosity increases as the carbon chain length increases from top to bottom. (ii) Bitumen, being at the bottom of the fractionating column, has the highest boiling point. (iii) Refinery gases, having the shortest chain lengths, are the most volatile and therefore the easiest to ignite.

1 mark for stating that viscosity increases (or becomes thicker/more viscous). 1 mark for stating that bitumen has the highest boiling point. 1 mark for stating that refinery gases are the easiest to ignite / most flammable.

(i) Copper(II) ions react with hydroxide ions to form a blue precipitate of copper(II) hydroxide. (ii) Bromide ions react with silver ions to form a cream precipitate of silver bromide. (iii) Sulfate ions are tested using acidified barium chloride (or barium nitrate) solution, which forms a white precipitate of barium sulfate.

1 mark for blue precipitate (reject other colours). 1 mark for bromide ion / Br- (reject bromine). 1 mark for barium chloride / BaCl2 / barium nitrate / Ba(NO3)2 (reject barium alone).

(i) The minimum energy required for particles to react when they collide is the activation energy. (ii) Since the products have less chemical energy than the reactants, energy is released, making the overall enthalpy change negative. (iii) A reaction that releases thermal energy to the surroundings is described as exothermic.

1 mark for activation energy (accept Ea). 1 mark for negative (accept minus / -). 1 mark for exothermic.

(i) Condensation polymerisation produces two products: the polymer itself and a small molecule (such as water or hydrogen chloride). (ii) Addition polymerisation requires monomers with an unsaturated carbon-carbon double bond (C=C). (iii) In the formation of a polyester from a dicarboxylic acid and a diol, water (H2O) is the small molecule eliminated.

1 mark for two products (accept the polymer and a small molecule / water / HCl). 1 mark for carbon-carbon double bond / C=C double bond (accept C=C or unsaturation). 1 mark for water / H2O (accept hydrogen chloride / HCl).

First, calculate the energy required to break bonds in the reactants: \(4 \times (C-H) + 1 \times (F-F) = 4 \times 413 + 158 = 1652 + 158 = 1810\text{ kJ/mol}\). Next, write the expression for the energy released when bonds are formed in the products: \(3 \times (C-H) + 1 \times (C-F) + 1 \times (H-F) = 3 \times 413 + (C-F) + 567 = 1239 + 567 + (C-F) = 1806 + (C-F)\). Use the enthalpy change equation: \(\Delta H = \text{energy in} - \text{energy out}\). \(-468 = 1810 - (1806 + (C-F))\). \(-468 = 4 - (C-F)\). \((C-F) = 4 + 468 = 472\text{ kJ/mol}\).

M1: Calculates the total energy needed to break reactant bonds as \(1810\text{ kJ/mol}\) (1) M2: Formulates an expression for the total energy released on forming product bonds as \(1806 + x\) or \(1239 + 567 + x\) (1) M3: Sets up the complete equation linking reactants, products, and \(\Delta H\): \(-468 = 1810 - (1806 + x)\) (1) M4: Rearranges the equation correctly: \(x = 1810 - 1806 + 468\) (1) M5: Correct final answer of \(472\text{ kJ/mol}\) (1). Accept 472. Reject -472.

First, calculate the moles of magnesium used: \(\text{moles of Mg} = \frac{\text{mass}}{A_r} = \frac{0.360}{24} = 0.0150\text{ mol}\). According to the equation, \(1\text{ mol}\) of \(Mg\) produces \(1\text{ mol}\) of \(H_2\), so the theoretical moles of \(H_2 = 0.0150\text{ mol}\). Next, calculate the theoretical volume of \(H_2\) gas produced at rtp: \(\text{theoretical volume} = 0.0150\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 360\text{ cm}^3\). Finally, calculate the percentage yield: \(\text{percentage yield} = \frac{\text{actual volume}}{\text{theoretical volume}} \times 100 = \frac{306}{360} \times 100 = 85.0\%\).

M1: Calculates the moles of Mg used: \(\frac{0.360}{24} = 0.0150\text{ mol}\) (1) M2: Deduces the theoretical moles of \(H_2\) is \(0.0150\text{ mol}\) from the 1:1 ratio (1) M3: Calculates the theoretical volume of \(H_2\): \(0.0150 \times 24000 = 360\text{ cm}^3\) (or \(0.360\text{ dm}^3\)) (1) M4: Correctly structures the percentage yield expression: \(\frac{306}{360} \times 100\) (1) M5: Gives the final answer as \(85\%\) or \(85.0\%\) (1).

First, calculate the heat energy transferred using \(Q = m c \Delta T\): \(Q = 150\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 22.0\ ^\circ\text{C} = 13794\text{ J} = 13.794\text{ kJ}\). Second, calculate the moles of ethanol burned: \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{0.46}{46} = 0.010\text{ mol}\). Third, calculate the molar enthalpy change: \(\Delta H = -\frac{Q}{n} = -\frac{13.794\text{ kJ}}{0.010\text{ mol}} = -1379.4\text{ kJ/mol}\). Rounding to three significant figures gives \(-1380\text{ kJ/mol}\).

M1: Calculates heat energy transferred \(Q = 150 \times 4.18 \times 22.0 = 13794\text{ J}\) (or \(13.794\text{ kJ}\)) (1) M2: Calculates moles of ethanol burned: \(\frac{0.46}{46} = 0.010\text{ mol}\) (1) M3: Shows division of heat energy (in kJ) by the calculated moles of ethanol (1) M4: Shows negative sign to indicate exothermic combustion (1) M5: Gives final answer of \(-1380\) or \(-1379.4\text{ kJ/mol}\) (1).

First, calculate the moles of \(CaCO_3\) used: \(\text{moles} = \frac{2.50}{100} = 0.0250\text{ mol}\). Second, calculate the moles of \(HCl\) used: \(\text{moles} = 1.00 \times \frac{40.0}{1000} = 0.0400\text{ mol}\). Third, determine the limiting reactant: from the equation, \(1\text{ mol}\) of \(CaCO_3\) reacts with \(2\text{ mol}\) of \(HCl\). Thus, \(0.0250\text{ mol}\) of \(CaCO_3\) requires \(0.0500\text{ mol}\) of \(HCl\). Since only \(0.0400\text{ mol}\) of \(HCl\) is present, \(HCl\) is the limiting reactant. Fourth, deduce the moles of \(CO_2\) produced: \(\text{moles of } CO_2 = \frac{\text{moles of } HCl}{2} = \frac{0.0400}{2} = 0.0200\text{ mol}\). Fifth, calculate the volume of \(CO_2\) produced: \(\text{volume} = 0.0200\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 480\text{ cm}^3\).

M1: Calculates moles of \(CaCO_3\): \(\frac{2.50}{100} = 0.0250\text{ mol}\) (1) M2: Calculates moles of \(HCl\): \(1.00 \times \frac{40.0}{1000} = 0.0400\text{ mol}\) (1) M3: Identifies \(HCl\) as the limiting reactant (or shows that \(CaCO_3\) is in excess) (1) M4: Deduces moles of \(CO_2\) produced as \(\frac{0.0400}{2} = 0.0200\text{ mol}\) (1) M5: Calculates the final volume of \(CO_2\): \(0.0200 \times 24000 = 480\text{ cm}^3\) (1). Accept \(0.48\text{ dm}^3\) if unit is clearly stated.

1. Measure a known volume (such as \(25\text{ cm}^3\) or \(50\text{ cm}^3\)) of copper(II) sulfate solution using a measuring cylinder or pipette, and transfer it into a polystyrene cup. 2. Place the polystyrene cup inside a glass beaker for stability, and fit it with a lid containing a hole for a thermometer to minimize heat loss to the surroundings. 3. Place a thermometer in the solution and record the temperature every minute for 3 minutes to establish a baseline. 4. At the 4th minute, add a known mass of zinc powder (ensuring it is in excess to react all the copper(II) ions) to the cup, but do not record the temperature. 5. Stir the mixture continuously to ensure rapid reaction and even heat distribution. 6. Record the temperature of the mixture every minute from the 5th to the 10th minute. 7. Plot a graph of temperature against time, draw lines of best fit for the points before and after the addition, and extrapolate both lines back to the 4th minute (the time of mixing) to determine the theoretical maximum temperature change (\(\Delta T\)).

M1: Place a known volume of copper(II) sulfate solution into a polystyrene cup supported by a beaker with a lid to insulate [1]. M2: Record the temperature of the solution at regular intervals (e.g. every minute) for 3 minutes before adding the zinc [1]. M3: Add a known mass of zinc powder in excess at a specific time (e.g., at the 4th minute) [1]. M4: Stir the mixture continuously during the reaction [1]. M5: Record the temperature at regular intervals (e.g. every minute) for several minutes after mixing [1]. M6: Plot a graph of temperature against time and extrapolate the cooling curve back to the time of mixing to find the maximum temperature rise [1]. Accept: Mention of using a specific pipette/measuring cylinder for the solution. Reject: Use of a copper or glass beaker instead of polystyrene for M1.

To conduct a fair and accurate comparison: 1. Measure a fixed volume (e.g., \(25\text{ cm}^3\)) of a known concentration of hydrogen peroxide solution and place it into a conical flask. 2. Weigh out a fixed mass (e.g., \(0.5\text{ g\)}) of manganese(IV) oxide powder. 3. Connect the flask to a gas syringe using a delivery tube and a rubber bung. 4. Add the manganese(IV) oxide to the flask, immediately seal with the bung, and start a stopwatch. 5. Record the volume of gas collected in the syringe at regular intervals (e.g., every 10 seconds) for 2 minutes or until the reaction finishes. 6. Repeat the entire procedure keeping all conditions identical (same temperature, volume/concentration of peroxide, and mass of catalyst) but using \(0.5\text{ g}\) of copper(II) oxide powder instead. 7. Plot a graph of volume of gas (y-axis) against time (x-axis) for both catalysts. The catalyst that produces the steepest initial curve (or produces the largest volume of gas in the first 30 seconds) is the more effective catalyst.

M1: Control variables: keep the volume and concentration of hydrogen peroxide solution constant (and constant initial temperature) [1]. M2: Control variables: keep the mass and state of subdivision (both powders) of each catalyst constant [1]. M3: Describe a suitable gas collection apparatus: flask connected to a gas syringe or a delivery tube over a graduated cylinder inverted over water [1]. M4: Measure the volume of oxygen gas collected at regular time intervals (e.g. every 10 seconds) [1]. M5: Repeat the experiment using the same conditions with the second catalyst [1]. M6: Conclusion: State that the more effective catalyst is the one that produces the steepest initial gradient on a graph of volume vs time / produces a given volume of gas in the shortest time [1]. Reject: Determining effectiveness solely by measuring the final volume of gas produced.

For the ammonium ion (\(\text{NH}_4^+\)) test: 1. Place a sample of the solid or its solution in a test tube. 2. Add an excess of sodium hydroxide solution and warm the mixture gently over a Bunsen burner. 3. Hold a piece of damp red litmus paper near the mouth of the test tube. 4. The gas evolved (ammonia) will turn the damp red litmus paper blue, confirming the presence of ammonium ions. For the sulfate ion (\(\text{SO}_4^{2-}\)) test: 1. Dissolve a sample of the solid in deionised water to prepare an aqueous solution. 2. Add a few drops of dilute hydrochloric acid (or dilute nitric acid) to remove any carbonate impurities that might give a false positive. 3. Add a few drops of barium chloride solution (or barium nitrate solution). 4. A white precipitate of barium sulfate will form, confirming the presence of sulfate ions.

For the ammonium test: M1: Add sodium hydroxide solution and warm/heat [1]. M2: Test the evolved gas with damp red litmus paper [1]. M3: Red litmus paper turns blue [1]. For the sulfate test: M4: Dissolve the sample in water and add dilute hydrochloric acid (or dilute nitric acid) [1]. M5: Add barium chloride solution (or barium nitrate solution) [1]. M6: White precipitate forms [1]. Note: For M4, the acid must be added before the barium reagent. Reject: sulfuric acid used for acidification in M4.