▶️ Answer/Explanation
(a)(i) Six particles should be arranged randomly. (Maximum of two particles should touch.)
Mark scheme note: Allow between 5 and 10 particles.
(a)(ii) Particles are (much) closer together.
Mark scheme note: Allow “the density is (much) greater / less space between particles / packed more tightly”.
(a)(iii) Any one from:
• High pressure
• Risk of explosion
• Oxidising agent
• Risk in contact with combustible materials
Mark scheme note: Allow “risk of fire”. Ignore “flammable”.
(b)(i) Blue flame.
Mark scheme note: Allow “choking / pungent gas produced / sulfur darkens”. Ignore any original colour.
(b)(ii)
M1: (Final colour) red/orange/yellow.
M2: An acidic solution / an acid / sulfurous acid is formed.
Mark scheme note: Allow “sulfuric acid” or “sulfur dioxide is acidic”. Accept “the solution formed contains hydrogen ions”.
▶️ Answer/Explanation
(a) Table completion:
- Solid sodium chloride from aqueous sodium chloride → crystallisation
- Water from aqueous copper(II) sulfate → simple distillation
- Sand from sand and water → filtration
Award 1 mark per correct technique. “Distillation” without “simple” is not accepted.
(b) The box contains two different types of particles / two different substances / more than one type of particle.
Do not accept “compounds”.
(c) Any two of the following:
- Food colouring D contains three food dyes.
- Food colouring D contains dyes A and C.
- Food colouring D does not contain dye B.
- Food colouring D contains another dye which is not A, B or C.
(d)(i) 4 different elements: Ca, H, C, O.
(d)(ii) Total number of atoms = \(1\ \text{Ca} + 2 \times (1\ \text{H} + 1\ \text{C} + 3\ \text{O}) = 1 + 2 \times 5 = 11\) atoms.
▶️ Answer/Explanation
(a)(i) A (electron)
B is not the correct answer because neutrons exist in the nucleus.
C is not the correct answer because particle W is not a nucleus.
D is not the correct answer because protons exist in the nucleus.
(a)(ii) C (25)
A is not the correct answer as the proton number is 12.
B is not the correct answer as the number of neutrons is 13.
D is not the correct answer as the number of protons, neutrons and electrons is not 49.
(a)(iii) (atoms have the) same number of protons and electrons
(a)(iv) B (2+)
A is not the correct answer as this atom will not form an ion with a + charge.
C is not the correct answer as this atom will not form an ion with a – charge.
D is not the correct answer as this atom will not form an ion with a 2- charge.
(b)(i) (both isotopes have) the same number of electrons / the same electron configuration.
IGNORE same number of protons / same number of electrons in the outer shell.
(b)(ii)
M1: \( (7.60 \times 6) + (92.4 \times 7) \)
M2: \( 692.4 \div 100 = 6.924 \)
M3: \( 6.92 \)
Answer of 6.92 scores 3 marks.
ACCEPT \( (6 \times 0.076) + (7 \times 0.924) \).
ALLOW ecf if use of 6 and 7 in calculation.
▶️ Answer/Explanation
(a)
(b)(i) copper(II) oxide / CuO / copper oxide forms
(b)(ii) the volume of a gas changes with temperature / gas expands when hot / gas shrinks when cools (OWTTE)
(b)(iii)
M1: \(138 – 108 = 30 \, \text{cm}^3\) (volume of oxygen reacted)
M2: \(\frac{30}{138} \times 100 = 21.7\%\)
Answer: \(21.7\%\) (accept \(21.74\%\), \(22\%\))
▶️ Answer/Explanation
(a)(i)
M1: water
M2: oxygen
Answers can be in either order. ALLOW moisture / H₂O; air / O₂.
(a)(ii)
oxidation
ALLOW oxidisation / oxidising / redox.
(b)(i)
M1: Paint acts as a barrier / layer.
M2: It prevents air / oxygen / water from getting to / reacting with the iron.
(b)(ii)
galvanising
ALLOW galvanisation. IGNORE sacrificial protection.
(b)(iii)
M1: Zinc is more reactive than iron.
M2: Therefore, zinc is oxidised / reacts with oxygen / loses electrons more readily / in preference to / instead of iron.
ALLOW corrodes instead of iron. REJECT zinc rusts.
(c)(i)
M1: Aluminium is more reactive than iron.
M2: Because aluminium displaces iron from iron(III) oxide.
(c)(ii)
M1: Iron(III) oxide (is the oxidising agent).
M2: Iron(III) oxide donates oxygen to aluminium / takes electrons from aluminium / causes aluminium to be oxidised.
ALLOW iron oxide / iron ions / Fe³⁺ ions.
▶️ Answer/Explanation
(a) The gas also contains air (displaced from the conical flask).
(b)
- M1: A catalyst provides an alternative (reaction) pathway / route.
- M2: Of lower activation energy.
(c)
M1: Add hydrogen peroxide solution (to the conical flask) and add one of the catalysts.
M2: Record the time taken to collect a fixed volume of gas OR record the volume of gas collected in a fixed time.
AND any 2 from:
M3: Repeat with the same volume / same concentration of hydrogen peroxide solution.
M4: (Repeat at) same temperature.
M5: Use same mass / same surface area of each catalyst.
AND
M6: The most effective catalyst produces the greatest volume of gas per unit time OR takes the least time to produce a fixed volume of oxygen.
Allowable alternatives: “record the time when no more gas produced”, “same amount”, “the least time taken to complete the reaction is the most effective catalyst”.
(d)
M1: Steeper curve starting at the origin.
M2: Same final volume of oxygen produced.
The curve at 40°C should rise more steeply than the 20°C curve but plateau at the same maximum volume.
▶️ Answer/Explanation
(a)
M1: limewater turns cloudy / milky
M2: because carbon dioxide / CO₂ is produced / one of the products
M3: which reacts with limewater forming calcium carbonate
ALLOW: chalky / white precipitate; carbon dioxide is present; forming an insoluble product
(b)(i)
M1: amount of water = \(2.16 \div 18 = 0.12\) mol
M2: number of hydrogen atoms = \((0.12 \div 0.01) \times 2 = 24\)
M3: formula of alkene = \(C_{12}H_{24}\)
ALLOW: \(12H_2O\); Correct answer of \(C_{12}H_{24}\) scores 3; \(C_6H_{12}\) scores 2
(b)(ii)
Some steam / water vapour is lost (to the atmosphere) / does not condense.
IGNORE references to incomplete combustion
(b)(iii)
M1: Heat (the water) / measure the boiling point
M2: (If it) boils at \(100^\circ C\) (it is pure water) / boiling point is \(100^\circ C\)
ALLOW: find the freezing point / melting point; freezes / melts at \(0^\circ C\)
REJECT: evaporate
(c)
M1: (moles of heptane) = \(30 \div 100 = 0.30\) mol
M2: (moles of oxygen) = \(0.30 \times 11 = 3.3\) mol
M3: (mass of oxygen) = \(3.3 \times 32 = 106\) g
ALLOW: \(105.6\) g; answer of \(106\) g scores 3; answer of \(9.6\) g scores 2
▶️ Answer/Explanation
(a)
M1: Graphite has delocalised electrons.
M2: These electrons are able to flow/move throughout the structure.
Note: Any mention of ions scores 0.
(b)
M1 (Diamond): Diamond is hard because it has a 3D/tetrahedral/rigid lattice where each carbon atom is bonded to four others by strong covalent bonds.
M2 (Graphite): Graphite is soft because it has a layered structure. The layers are held together by weak forces and can slide over one another.
Note: Reject references to intermolecular forces in diamond.
(c)
Any one from:
• The \( C_{60} \) molecule is inert/unreactive (with blood/medicine).
• The \( C_{60} \) molecule is non-toxic.
• The medicine can be enclosed/fit inside the \( C_{60} \) molecule’s cage-like structure.
▶️ Answer/Explanation
(a)(i)
Calcium ion: \(\text{Ca}^{2+}\)
Nitrate ion: \(\text{NO}_3^-\)
(a)(ii)
• Calcium nitrate has a giant ionic lattice / structure.
• There are strong electrostatic forces / attractions.
• Between oppositely charged ions / cations and anions.
• A lot of energy is required to overcome these forces / break the bonds.
(a)(iii)
\(2\text{Ca(NO}_3)_2 \rightarrow 2\text{CaO} + 4\text{NO}_2 + \text{O}_2\)
(b)
• Perform a flame test on each solution.
– Sodium chloride and sodium sulfate give a yellow flame (sodium ions).
– Calcium chloride and calcium bromide give an orange-red / brick-red flame (calcium ions).
• To distinguish between the sodium salts and the calcium salts:
– Add acidified silver nitrate (or nitric acid followed by silver nitrate) to each solution.
– Calcium bromide gives a cream precipitate (bromide ions).
– Calcium chloride and sodium chloride give a white precipitate (chloride ions).
• To distinguish between sodium sulfate and sodium chloride (or confirm sulfate):
– Add acidified barium chloride (or barium nitrate) to the solutions.
– Sodium sulfate gives a white precipitate (sulfate ions).
▶️ Answer/Explanation
(a)(i)
M1: Four electrons (two shared pairs) between carbon 2 and carbon 3 (the double bond) and two electrons (one shared pair) between carbon 1 and carbon 2.
M2: Two electrons (one shared pair) between each carbon and its bonded hydrogen atoms.
ACCEPT any combination of dots and crosses. Max 1 mark if any extra non-bonding electrons are added.
(a)(ii)
M1: Shared pair(s) of electrons.
M2: (Which are) attracted to the (two) nuclei.
OR
M1: (Two) nuclei.
M2: Attracted to the shared pair(s) of electrons.
REJECT singular “nucleus”. Must be plural for the mark.
(b)
Any two from:
M1: Cracking.
M2: Heat to a temperature of 600–700°C.
M3: With a catalyst of silica/alumina (or aluminosilicates/zeolites).
ALLOW a correct cracking equation including propene.
(c)(i)
M1: Correct repeat unit:
M2: Extension bonds, brackets, and ‘n’ subscript to the right.
ALLOW correct repeat unit with or without extension bonds.
(c)(ii)
M1: Poly(propylene) remains for a long time in landfill / takes up a lot of space.
M2: Because it is inert / unreactive / non-biodegradable / does not decompose.
M2 is dependent on M1.
(d)(i)
M1: Calculate moles of each element:
\( \text{C: } \frac{60}{12} = 5 \) \( \text{H: } \frac{13.3}{1} = 13.3 \) \( \text{O: } \frac{26.7}{16} = 1.66875 \)
M2: Divide by smallest (1.66875) to get ratio:
\( \text{C: } \frac{5}{1.66875} \approx 3 \) \( \text{H: } \frac{13.3}{1.66875} \approx 8 \) \( \text{O: } \frac{1.66875}{1.66875} = 1 \)
Empirical formula = \( C_3H_8O \).
M3: \( M_r \text{ of } C_3H_8O = (3 \times 12) + (8 \times 1) + 16 = 60 \). This matches the given \( M_r \), so molecular formula is \( C_3H_8O \).
ACCEPT alternative correct methods. No marks for using atomic numbers.
(d)(ii)
M1: The intermolecular forces (between molecules of compound X) are stronger than those between propene molecules.
AND any one from:
M2: So more energy is required to overcome them.
M3: Because compound X has a higher \( M_r \) / larger surface area.
M4: Because compound X has more electrons (leading to stronger London/dispersion forces).
M2, M3, M4 are dependent on M1.
▶️ Answer/Explanation
(a)
(i), (ii), (iii):
- All 9 points plotted correctly (± half a square).
- Anomalous result circled: the point at mass = 3 g (Temperature = 18.0 °C).
- A smooth curve of best fit is drawn, passing close to all points except the anomalous one.
(b)
(i) Polystyrene is a good insulator / poor conductor of heat. It reduces heat transfer (gain or loss) between the reaction mixture and the surroundings, leading to more accurate temperature measurements.
(ii) The student might have forgotten to stir the mixture properly / took the temperature reading too soon before it stabilized.
(iii) The temperature remains constant (at 15.3 °C) after adding the last few portions of sodium hydrogencarbonate (6g, 7g, 8g).
(iv) The temperature of the mixture decreases, showing that thermal energy is taken in from the surroundings (an endothermic process).
(c)
\[ \Delta T = 20.8 – 15.3 = 5.5 \, ^\circ\text{C} \] \[ m = 100 \, \text{g} \quad (\text{since volume} = 100 \, \text{cm}^3 \text{ and density} \approx 1 \, \text{g/cm}^3) \] \[ Q = m \times c \times \Delta T = 100 \times 4.2 \times 5.5 = 2310 \, \text{J} \] Answer: \( Q = 2310 \, \text{J} \)
(d)
\[ \text{Moles of NaHCO}_3 = \frac{7.0}{84} = 0.0833 \, \text{mol} \] \[ \Delta H = \frac{Q}{\text{moles}} = \frac{3200 \, \text{J}}{0.0833 \, \text{mol}} = 38400 \, \text{J/mol} = 38.4 \, \text{kJ/mol} \] Since the reaction is endothermic (temperature decreased), \( \Delta H \) is positive.
Answer: \( \Delta H = +38.4 \, \text{kJ/mol} \)