▶️ Answer/Explanation
(a) C (a neutron)
The only correct answer is C because the nucleus contains protons and neutrons. Protons are identified as the white dots.
A is not correct because electrons occur in the shells.
B is not correct because a molecule is not a particle found in the nucleus.
D is not correct because the nucleus contains protons and neutrons.
(b) A (an electron)
The only correct answer is A because electrons have a relative mass of \( \frac{1}{1836} \) compared to a proton or a neutron.
B is not correct because a neutron has a relative mass of 1.
C is not correct because the nucleus contains 4 protons and 5 neutrons.
D is not correct because a proton has a relative mass of 1.
(c) C (9)
The only correct answer is C because the mass number is the sum of the protons and neutrons (\(4 + 5 = 9\)).
A is not correct because the atomic number is 4.
B is not correct because 5 is the number of neutrons.
D is not correct because 13 is the total number of protons, neutrons and electrons.
(d) A (4)
The only correct answer is A because the atomic number is equal to the number of protons which is 4.
B is not correct because 5 is the number of neutrons.
C is not correct because 9 is the total number of particles in the nucleus.
D is not correct because 13 is the total number of protons, neutrons and electrons.
(e)(i) beryllium / Be
(e)(ii) (positive) ion
ALLOW ecf from the element given in (e)(i).
ACCEPT any positive beryllium ion (or other ecf ion).
REJECT any negative ion.
▶️ Answer/Explanation
(a)(i) The diagram for the liquid state should show particles that are close together, filling from the bottom of the box, with some particles touching. The arrangement should be irregular, not in a regular lattice.
(a)(ii) Gas
A is not the answer as particles in a solid have the least energy.
B is not the answer as particles in a liquid have more energy than a solid but less than a gas.
(b)
• Water evaporates: Before = \( (l) \), After = \( (g) \)
• Crystals of iodine sublime: Before = \( (s) \), After = \( (g) \)
• Ice melts: Before = \( (s) \), After = \( (l) \)
(c)
• Particles/molecules in hot water have more (kinetic) energy (1).
• Therefore, they can more easily overcome/break the (intermolecular) forces between water molecules and escape from the liquid surface (1).
Allow: “move faster”.
▶️ Answer/Explanation
(a)
most reactive: \( Z_2 \)
least reactive: \( X_2 \)
Order: \( Z_2 > Y_2 > X_2 \)
Explanation: A more reactive halogen can displace a less reactive halogen from its halide solution. \( Z_2 \) reacts with NaX and NaY, so is more reactive than \( X_2 \) and \( Y_2 \). \( Y_2 \) reacts with NaX but not NaZ, so \( Y_2 \) is more reactive than \( X_2 \) but less reactive than \( Z_2 \).
(b) bromine / \( Br_2 \)
Explanation: Bromine solution is orange and is decolourised when it reacts with an alkene in an addition reaction (test for unsaturation).
(c)(i)
Fluorine state at room temperature: gas / vapour
Chlorine boiling point: between -150°C and 10°C (e.g., -34°C)
Astatine colour: dark grey / black
(c)(ii) C
Explanation: Halogens are in Group 7 and have similar chemical properties because they all have 7 electrons in their outer shell.
(d)(i) Chlorine is toxic / poisonous.
Explanation: The reaction should be done in a fume cupboard to prevent inhalation of chlorine gas.
(d)(ii)
\( 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \)
Explanation: The balanced equation shows the formation of iron(III) chloride from its elements.
▶️ Answer/Explanation
(a) \( \text{NH}_4^+ \) or \( \text{NH}_4 \) (ALLOW both)
(b) M1: Add sodium hydroxide solution (and warm).
M2: Test the gas with damp red litmus paper or expose the gas to concentrated hydrochloric acid.
M3: (If using litmus) turns blue or (if using HCl) white smoke (ammonium chloride) is produced.
(If universal indicator is used, allow blue/purple for M3. M3 is dependent on correct method in M2. If sodium hydroxide is not added, max = 1 mark)
(c) The reaction is reversible. (ACCEPT: reaction goes both ways / both forwards and backwards reactions occur. IGNORE references to equilibrium.)
(d)(i)
M1: (Molecules/Particles of) ammonia move/diffuse faster.
M2: Because the ammonium chloride forms nearer to the HCl end or because the ammonia has travelled further in the same time.
(IGNORE references to masses/sizes of particles.)
(d)(ii)
Any two from:
M1: (Gas particles) move in random directions / do not travel in straight lines.
M2: (Gas particles) collide with air/other particles.
M3: (Gas particles) collide with the walls/sides of the tube.
(ALLOW: air/other particles slow them down. IGNORE any references to rate of reaction/collisions.)
▶️ Answer/Explanation
(a) The results show that all the oxygen has reacted because the volume of gas in the tube becomes constant/stops decreasing (at 37.0 cm3 from 5 minutes onwards).
(b) Use a glass tube or scale with smaller divisions (e.g., 0.1 cm3 divisions) to measure the volume more precisely.
(c)
Step 1: Find the volume of oxygen used.
Initial volume = 48.5 cm3
Final constant volume = 37.0 cm3
Volume of oxygen = \( 48.5 – 37.0 = 11.5 \) cm3
Step 2: Calculate the percentage.
Percentage of oxygen = \( \frac{11.5}{48.5} \times 100 \)
Step 3: Compute and round.
\( \frac{11.5}{48.5} \approx 0.2371 \)
\( 0.2371 \times 100 = 23.71\% \)
Rounded to one decimal place: 23.7%
| Question number | Answer | Notes | Marks |
|---|---|---|---|
| (a) | Results are the same at the end | ALLOW there is a constant volume in the tube ALLOW the volume of gas stops decreasing ALLOW no change after 4 / 5 minutes | 1 |
| (b) | (use a glass tube / scale with) smaller divisions. | ALLOW use a glass tube / scale with 0.1cm³ divisions ALLOW use a smaller scale | 1 |
| (c) | M1 Volume of oxygen = 11.5(cm³) M2 (11.5 ÷ 48.5) x 100 M3 23.7% | Correct answer to 1 dp with or without working scores 3 ALLOW ecf from M1 M3 must be to 1dp | 3 |
▶️ Answers/Explanations
Question 6(a)(i)
Magnesium is more reactive than copper / Magnesium is higher than copper in the reactivity series.
This allows magnesium to displace copper from its compound.
Question 6(a)(ii)
magnesium + copper(II) sulfate → magnesium sulfate + copper
Both products required in either order. “Copper” not “copper(II)”.
Question 6(b)(i)
1. Calculate \(\Delta T\): \(\Delta T = 56.3^\circ \text{C} – 20.2^\circ \text{C} = 36.1^\circ \text{C}\)
2. Mass of solution: \(100\ \text{cm}^3 \times 1.00\ \text{g/cm}^3 = 100\ \text{g}\)
3. Use \(Q = mc\Delta T\): \(Q = 100\ \text{g} \times 4.2\ \text{J g}^{-1}\text{°C}^{-1} \times 36.1^\circ \text{C} = 15162\ \text{J}\)
Answer: \(\mathbf{15162\ J}\) (or \(\mathbf{15.2\ kJ}\))
Question 6(b)(ii)
Polystyrene is a thermal insulator / poor conductor of heat.
• This reduces heat loss to the surroundings.
• Therefore, the measured temperature change is more accurate / closer to the true value.
Question (c)(i)
1. Moles of zinc: \(n = \frac{0.500\ \text{g}}{65.4\ \text{g/mol}} = 0.007645\ \text{mol}\) (using \(M_r = 65.4\))
2. Molar enthalpy change: \(\Delta H = \frac{1.67\ \text{kJ}}{0.007645\ \text{mol}} = 218.4\ \text{kJ/mol}\)
3. Sign: Reaction is exothermic, so \(\Delta H\) is negative.
Answer (to 3 s.f.): \(\mathbf{\Delta H = -218\ \text{kJ/mol}}\)
Note: Using \(M_r = 65\) gives \(-217\ \text{kJ/mol}\) as per mark scheme.
Question (c)(ii)
This is a redox reaction because:
• Oxidation occurs: Zinc atoms lose electrons to form Zn²⁺ ions (\(\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-\)).
• Reduction occurs: Copper(II) ions gain electrons to form copper atoms (\(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)).
• Both oxidation and reduction happen simultaneously.
▶️ Answer/Explanation
(a)(i) Measuring cylinder / burette / (volumetric) pipette
(a)(ii) Neutralisation
(b)(i) 12.4
(b)(ii)
• 15 cm3 — red/orange
• 30 cm3 — blue/purple
(b)(iii) OH– / hydroxide (ion)
(c)
M1: The reaction is exothermic (therefore the temperature rises).
M2: (After 25 cm3 of sodium hydroxide) the reaction is complete / neutralisation happens.
M3: So adding more sodium hydroxide / liquid / solution cools the mixture down / no more heat energy is given out.
▶️ Answer/Explanation
(a)
| Answer | Notes | Marks |
|---|---|---|
| M1 calcium loses electrons M2 chlorine gains electrons M3 two atoms of chlorine each gain one electron OR M3 calcium loses 2 electrons and chlorine gains 1 electron | IGNORE references to redox Allow 1 mark from M1 and M2 for electron transfer from chlorine to calcium If chlorine molecules are gaining electrons do not award M3 Any reference to sharing electrons or covalent or metallic bonding scores 0 | 3 |
(b)
| Answer | Notes | Marks |
|---|---|---|
| (test for \( \text{Ca}^{2+} \) ions) M1 flame test (allow description of a flame test) M2 orange-red flame colour (test for \( \text{Cl}^- \) ions) M3 add silver nitrate M4 white precipitate | ALLOW brick-red ALLOW M1 add sodium hydroxide | 4 |
(c)(i) 
| Answer | Notes | Marks |
|---|---|---|
| M1 and M2 all points correct ± half a square M3 2 straight lines of best fit ignoring the anomalous point | One plotting error scores M1 | 3 |
(c)(ii) the conductivity is (directly) proportional (to the number of spatulas of calcium chloride added)
OR
the conductivity increases (as the number of spatulas of calcium chloride increases) (1 mark)
(c)(iii) Any one from:
• The student took the reading before adding the calcium chloride
• The student forgot to stir the mixture OR did not stir the mixture properly
IGNORE any references to human error (1 mark)
(d)
M1 Heat (the calcium chloride)
M2 until molten / melts
IGNORE references to electrons / ions (2 marks)
▶️ Answer/Explanation
(a)(i)
| Molecular formula | \( \text{C}_2\text{H}_4 \) |
|---|---|
| Empirical formula | \( \text{CH}_2 \) |
| General formula | \( \text{C}_n\text{H}_{2n} \) |
(a)(ii)
Any two from:
1. Each member differs from the next by a \( \text{CH}_2 \) group.
2. They have the same functional group.
3. They have similar chemical properties / react in similar ways.
4. There is a trend in their physical properties (e.g., boiling point).
(b)(i) Addition (polymerisation)
(b)(ii) 
$$ n \text{H}_2\text{C}=\text{CH}_2 \rightarrow \text{—[CH}_2\text{—CH}_2\text{—]}_n— $$ or the displayed structure with single bonds between carbons and hydrogens, and trailing bonds through brackets and ‘n’ to the right.
(b)(iii)
Advantages of poly(ethene):
• Cheaper per tonne (£1500 vs £3700).
• Stronger (relative strength 100 vs 50).
• Does not use land that could be used for food crops.
• Comes from crude oil (a well-established industrial process).
Disadvantages of poly(ethene):
• Non-biodegradable / takes much longer to decompose (450 years vs 3–6 months).
• Causes problems with litter and landfill disposal.
• Burning it can release toxic fumes/greenhouse gases.
• Made from a non-renewable, finite resource (crude oil).
(c) 
The monomer is a dicarboxylic acid and a diol. The displayed formula of the monomer pair that forms this polyester is:
HOOC—COOH (Ethanedioic acid) and HO—CH\(_2\)—CH\(_2\)—OH (Ethanediol)
Or, if drawn as a single monomer for the repeat unit, it is not a single alkene but a combination of these two molecules. The displayed structure should show:
H—O—C(=O)—C(=O)—O—H and H—O—CH\(_2\)—CH\(_2\)—O—H
Note: The polymer is formed by condensation polymerisation between these two monomers.
▶️ Answer/Explanation
(a)(i)
M1: Four electrons (two shared pairs) between the carbon and each oxygen atom.
M2: Rest of the molecule correct (lone pairs on oxygen atoms correctly shown).
(a)(ii)
M1: Shared pair(s) of electrons.
M2: (The electrons are) attracted to the (two) nuclei.
REJECT: singular “nucleus”. Must be plural for M2.
(b)(i)
M1: Graphite has delocalised electrons.
M2: The delocalised electrons can move or flow (throughout the structure), carrying charge.
IGNORE: sea of electrons, free electrons, references to layers. Any mention of ions scores 0.
(b)(ii)
M1: Diamond has a giant covalent / macromolecular structure.
M2: To melt diamond, strong covalent bonds must be broken.
M3: \( C_{60} \) fullerene has a simple molecular structure.
M4: To melt \( C_{60} \), only weak intermolecular forces (of attraction) are overcome.
M5: More energy is needed to break the strong covalent bonds in diamond than to overcome the weak intermolecular forces in \( C_{60} \).
ACCEPT: Breaking bonds in \( C_{60} \) if intermolecular forces are clearly mentioned.
REJECT: Mention of breaking covalent bonds in \( C_{60} \).
▶️ Answer/Explanation
(a)
\[ 4\text{CuO(s)} + \text{CH}_4\text{(g)} \rightarrow 4\text{Cu(s)} + \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l/g)} \]
M1 correct balancing (4,1,4,1,2). M2 correct state symbols.
(b)(i)
Mass of copper oxide = \( 18.63 – 15.05 = 3.58 \text{ g} \)
Mass of copper produced = \( 18.23 – 15.05 = 3.18 \text{ g} \)
Mass of oxygen in oxide = \( 3.58 – 3.18 = 0.40 \text{ g} \)
Moles of copper = \( \frac{3.18}{63.5} = 0.0501 \text{ mol} \)
Moles of oxygen = \( \frac{0.40}{16.0} = 0.0250 \text{ mol} \)
Ratio Cu : O = \( 0.0501 : 0.0250 = 2 : 1 \)
Therefore, empirical formula = \(\text{Cu}_2\text{O}\).
(b)(ii)
Any one from:
• Use a safety screen / wear heat-proof gloves / tie hair back.
• Position the class at a safe distance.
• Conduct the experiment in a fume cupboard.
• Ignite the excess methane immediately to prevent buildup.
(c)(i)
Iron(III) oxide loses oxygen (to form iron).
(c)(ii)
Carbon monoxide is poisonous / toxic. It reduces the blood’s ability to carry oxygen.
(c)(iii)
\( M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 160 \)
Mass of \(\text{Fe}_2\text{O}_3\) in grams = \( 30.0 \times 10^6 \text{ g} \)
Moles of \(\text{Fe}_2\text{O}_3\) = \( \frac{30.0 \times 10^6}{160} = 187,500 \text{ mol} \)
From equation: moles of Fe = \( 2 \times \text{moles of Fe}_2\text{O}_3 = 2 \times 187,500 = 375,000 \text{ mol} \)
Mass of Fe = \( 375,000 \times 56 = 21,000,000 \text{ g} = 21.0 \text{ tonnes} \)
Maximum mass of iron = 21.0 tonnes.
(c)(iv)
Moles of carbon = \( \frac{840,000}{12} = 70,000 \text{ mol} \)
From equation, 3 mol C react with 1 mol \(\text{Fe}_2\text{O}_3\).
Therefore, moles of \(\text{Fe}_2\text{O}_3\) needed = \( \frac{70,000}{3} \approx 23,333 \text{ mol} \)
Moles of \(\text{Fe}_2\text{O}_3\) available = 25,000 mol.
Since available (25,000 mol) > needed (23,333 mol), iron(III) oxide is in excess.