An original Thinka practice paper modelled on the structure and difficulty of the June 2025 — Foundation OCR GCSE (9-1) Gateway Science - Chemistry A - J248 paper. Not affiliated with or reproduced from OCR.
J248/01 Section A
Answer all 15 single-best-answer items, selecting one of four options and recording the letter in the answer box. The paper advises a maximum of 30 minutes for this section.
15 Question · 15 marks
Question 1 · multiple choice
1 marks
A sulfide ion is represented as \(^{32}_{16}\text{S}^{2-}\). Which row gives the correct numbers of protons, neutrons and electrons in this ion?
Protons / Neutrons / Electrons
A.A: 16 / 16 / 14
B.B: 16 / 16 / 18
C.C: 16 / 18 / 16
D.D: 18 / 16 / 18
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Worked solution
The atomic number is 16, so there are 16 protons. Neutrons = 32 - 16 = 16. The 2- charge means two extra electrons: 16 + 2 = 18 electrons. The answer is B.
Marking scheme
B (1 mark). Distractors: A subtracts the charge instead of adding it; C gives the neutral atom's electron count with a wrong neutron number; D confuses protons with electrons.
Question 2 · multiple choice
1 marks
In a paper chromatography experiment, the solvent front moves 9.0 cm from the baseline. One spot in a food dye moves 3.6 cm from the baseline. What is the Rf value of this spot?
A.A: 0.25
B.B: 0.40
C.C: 2.5
D.D: 5.4
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Worked solution
Rf = distance moved by spot / distance moved by solvent front = 3.6 / 9.0 = 0.40. The answer is B.
Marking scheme
B (1 mark). Distractor C is the inverted ratio (9.0 / 3.6); A and D come from incorrect processing of the two distances. Rf has no units.
Question 3 · multiple choice
1 marks
Graphite is used to make electrodes. Which statement explains why graphite conducts electricity?
A.A: Each carbon atom forms four strong covalent bonds.
B.B: It contains ions that are free to move.
C.C: It contains delocalised electrons between the layers.
D.D: It has weak forces between its layers.
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Worked solution
In graphite each carbon atom bonds to only three others, leaving one delocalised electron per atom. These electrons can move through the structure and carry charge. The answer is C.
Marking scheme
C (1 mark). A describes diamond, which does not conduct. B is wrong because graphite has no ions. D explains why graphite is soft, not why it conducts.
Question 4 · multiple choice
1 marks
What is the relative formula mass of calcium hydroxide, \(\text{Ca(OH)}_2\)? Relative atomic masses: H = 1, O = 16, Ca = 40.
A.A: 57
B.B: 58
C.C: 74
D.D: 91
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Worked solution
Ca(OH)2 contains one Ca and two OH groups: 40 + 2 × (16 + 1) = 40 + 34 = 74. The answer is C.
Marking scheme
C (1 mark). A counts only one OH group (40 + 17); B gives 40 + 16 + 2, ignoring the bracket; D comes from 40 + 3 × 17.
Question 5 · multiple_choice
1 marks
A potassium ion is written as \(^{39}_{19}\text{K}^{+}\).
Which row shows the numbers of protons, neutrons and electrons in this ion?
A.19 protons, 20 neutrons, 18 electrons
B.19 protons, 20 neutrons, 20 electrons
C.19 protons, 39 neutrons, 18 electrons
D.20 protons, 19 neutrons, 19 electrons
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Worked solution
Protons = atomic number = 19. Neutrons = mass number - atomic number = 39 - 19 = 20. The ion has a 1+ charge, so it has lost one electron: 19 - 1 = 18 electrons. The answer is A.
Marking scheme
A (1 mark). Guidance: B has not removed an electron correctly; C uses the mass number as the number of neutrons; D swaps protons and neutrons. No half marks.
Question 6 · multiple_choice
1 marks
A student runs a paper chromatography experiment. The solvent front moves 7.5 cm. One spot moves 3.0 cm from the baseline.
What is the Rf value of this spot?
A.0.25
B.0.40
C.2.5
D.4.5
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Worked solution
\(R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \dfrac{3.0}{7.5} = 0.40\). The answer is B.
Marking scheme
B (1 mark). Guidance: C is the formula inverted (7.5 / 3.0); A and D are incorrect calculations. An Rf value cannot be greater than 1.
Question 7 · multiple_choice
1 marks
A technician carries out a flame test on a solid compound. The flame turns lilac.
Which metal ion is present in the compound?
A.Calcium
B.Copper
C.Lithium
D.Potassium
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Worked solution
A lilac flame is the flame-test colour for potassium ions. Calcium gives an orange-red flame, copper a blue-green flame and lithium a crimson (red) flame. The answer is D.
Marking scheme
D (1 mark). Guidance: A, B and C give different flame colours (orange-red, blue-green, crimson).
Question 8 · multiple_choice
1 marks
Graphite conducts electricity but diamond does not.
Which statement explains why graphite conducts electricity?
A.Each carbon atom forms four covalent bonds
B.There are delocalised electrons between the layers
C.The layers are held together by strong covalent bonds
D.The structure contains positive and negative ions
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Worked solution
In graphite each carbon atom forms three covalent bonds, leaving one delocalised electron per atom. These electrons can move through the structure and carry charge. In diamond all four outer electrons are used in bonding, so there are no free electrons. The answer is B.
Marking scheme
B (1 mark). Guidance: A describes diamond, which does not conduct; C is incorrect because the layers are held by weak intermolecular forces; D is incorrect because graphite is not ionic.
Question 9 · multiple_choice
1 marks
An aluminium ion has the symbol \(^{27}_{13}\mathrm{Al}^{3+}\). Which row gives the numbers of protons, neutrons and electrons in this ion?
A.13 protons, 13 neutrons, 10 electrons
B.13 protons, 14 neutrons, 10 electrons
C.13 protons, 14 neutrons, 16 electrons
D.14 protons, 13 neutrons, 10 electrons
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Worked solution
Atomic number 13 gives 13 protons. Neutrons = 27 - 13 = 14. A 3+ ion has lost three electrons: 13 - 3 = 10 electrons. The answer is B.
Marking scheme
B (1 mark). Distractors: A uses the wrong neutron number; C adds electrons instead of removing them; D swaps protons and neutrons.
Question 10 · multiple_choice
1 marks
In a paper chromatography experiment, a spot of dye travels 3.0 cm. The solvent front travels 7.5 cm. What is the \(R_f\) value of the dye?
A.0.25
B.2.5
C.0.40
D.4.5
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Worked solution
\(R_f\) = distance moved by spot / distance moved by solvent front = 3.0 / 7.5 = 0.40. The answer is C.
Marking scheme
C (1 mark). A and B result from inverting or mis-dividing the distances; D is the sum-type error (7.5 - 3.0). An Rf value has no units.
Question 11 · multiple_choice
1 marks
A student adds dilute nitric acid and then silver nitrate solution to a solution of an unknown salt. A cream precipitate forms. Which ion is present in the salt?
A.Bromide
B.Chloride
C.Iodide
D.Sulfate
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Worked solution
With acidified silver nitrate, chloride gives a white precipitate, bromide a cream precipitate and iodide a yellow precipitate. Sulfate is tested with acidified barium chloride or nitrate. The answer is A.
Marking scheme
A (1 mark).
Question 12 · multiple_choice
1 marks
Graphite conducts electricity. Which statement explains why?
A.It has strong covalent bonds between its layers
B.It contains ions that are free to move in the solid
C.Each carbon atom is bonded to four other carbon atoms
D.It has delocalised electrons that can move along its layers
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Worked solution
In graphite each carbon atom forms three covalent bonds, leaving one electron per atom delocalised. These electrons can move through the structure and carry charge. The answer is D.
Marking scheme
D (1 mark). A is incorrect because the layers are held by weak forces; B is incorrect because graphite has no ions; C describes diamond, which does not conduct.
Question 13 · multiple_choice
1 marks
A calcium ion is written as \( {}^{40}_{20}\text{Ca}^{2+} \).
Which row shows the numbers of protons, neutrons and electrons in this ion?
A.20 protons, 20 neutrons, 18 electrons
B.20 protons, 20 neutrons, 22 electrons
C.18 protons, 20 neutrons, 20 electrons
D.20 protons, 22 neutrons, 18 electrons
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Worked solution
The atomic number is 20, so there are 20 protons. Neutrons = 40 - 20 = 20. A 2+ ion has lost two electrons, so electrons = 20 - 2 = 18. The answer is A.
Marking scheme
A (1 mark). Accuracy mark only. B: electrons wrongly increased for a positive ion. C: protons and electrons confused. D: neutrons miscalculated using the charge.
Question 14 · multiple_choice
1 marks
A student runs a paper chromatography experiment. A spot of dye travels 3.6 cm from the start line. The solvent front travels 8.0 cm from the start line.
What is the \( R_f \) value of the dye?
A.0.45
B.0.55
C.2.2
D.4.4
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Worked solution
\( R_f = \dfrac{\text{distance moved by dye}}{\text{distance moved by solvent}} = \dfrac{3.6}{8.0} = 0.45 \). The answer is A.
Marking scheme
A (1 mark). Accuracy mark only. B: used (8.0 - 3.6) / 8.0. C: inverted the ratio (8.0 / 3.6). D: subtracted the distances (8.0 - 3.6).
Question 15 · multiple_choice
1 marks
A technician adds dilute nitric acid and then silver nitrate solution to a colourless solution. A cream precipitate forms.
Which ion is present in the solution?
A.chloride
B.bromide
C.iodide
D.sulfate
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Worked solution
Acidified silver nitrate gives a white precipitate with chloride, a cream precipitate with bromide and a yellow precipitate with iodide. Sulfate is tested with barium chloride or nitrate. The answer is B.
Marking scheme
B (1 mark). Accuracy mark only. A: chloride gives a white precipitate. C: iodide gives a yellow precipitate. D: sulfate is not identified with silver nitrate.
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Answer all eight numbered questions, totalling 75 marks. Whole-question marks in order: 6, 11, 6, 9, 14, 11, 9, 9. Use lettered and Roman-numeral subparts as needed, including short responses, completion, calculations, diagrams and practical reasoning. Include one six-mark level-of-response task within this section; it is already included in the whole-question totals. Show calculation working.
8 Question · 75 marks
Question 1 · Structured question
6 marks
A technician tests three solid substances, P, Q and R. Each substance has a different type of structure. The results are shown in Table 1.
Table 1
| Substance | Melting point in \(^\circ\)C | Conducts electricity as a solid? | Conducts electricity when molten? | |---|---|---|---| | P | 801 | no | yes | | Q | 114 | no | no | | R | 1085 | yes | yes |
(a) Substance P, Q or R has a giant ionic structure. Identify this substance and give one reason for your answer. [2]
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Worked solution
(a) P has a high melting point and conducts only when molten, because its ions can move only when the lattice is broken down. (b) Q is made of small molecules. Melting only overcomes the weak forces between molecules, which needs little energy. The strong covalent bonds inside the molecules stay intact. (c) In a metal, outer-shell electrons are delocalised and free to move through the lattice, so they carry charge even when the metal is solid.
Marking scheme
(a) 2 marks - P (1) - reason: conducts when molten but not when solid / high melting point (1) ALLOW ions free to move when molten (AW). DO NOT ALLOW the reason alone without P. Reason mark is dependent on P being chosen.
(b) 2 marks - weak intermolecular forces (between molecules) (1) - little energy needed to overcome them / break them (1) DO NOT ALLOW if covalent bonds are said to break (max 1). IGNORE 'weak covalent bonds'.
(c) 2 marks - delocalised / free / mobile electrons (1) - (electrons) move through the structure and carry charge / current (1) DO NOT ALLOW free ions. IGNORE 'electrons' alone without being free to move.
Question 2 · Structured question
9 marks
A technician tests three solid substances, P, Q and R. The results are shown in Table 1.
Table 1
| Substance | Melting point in °C | Conducts electricity when solid | Conducts electricity when molten | |---|---|---|---| | P | 801 | no | yes | | Q | 3550 | no | no | | R | -7 | no | no |
(a) Each substance has one type of structure and bonding. Complete Table 2 by writing P, Q or R in each blank box.
(b) Explain why substance P conducts electricity when molten but not when solid. ______________________________________________ ______________________________________________ [2]
(c) Explain why substance Q has a very high melting point. ______________________________________________ ______________________________________________ [2]
(d) Graphite is a giant covalent form of carbon. Graphite conducts electricity but diamond does not. Explain this difference in terms of the bonding in each. ______________________________________________ ______________________________________________ [2]
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Worked solution
(a) P conducts only when molten, so it is ionic. Q has a very high melting point and never conducts, so it is a giant covalent structure. R has a very low melting point, so it is made of simple molecules. (b) In the solid the ions are held in fixed positions in the lattice. When molten the ions are free to move and carry charge. (c) Q has a giant structure with a huge number of strong covalent bonds. A large amount of energy is needed to break them. (d) In graphite each carbon atom forms three covalent bonds, leaving one delocalised electron that can move and carry charge. In diamond each carbon atom forms four covalent bonds, so there are no free electrons.
Marking scheme
(a) P = giant ionic; Q = giant covalent; R = simple molecules (3 correct = 3 marks; 2 correct = 2 marks; 1 correct = 1 mark). DO NOT ALLOW more than one letter per box. (b) ions free to move / can move when molten (1); ions fixed in position / cannot move in solid (1). IGNORE 'electrons move' for P (no credit for electrons). (c) many / strong covalent bonds (1); a lot of energy needed to break them (1). DO NOT ALLOW 'intermolecular forces' or 'molecules' for the first mark. (d) graphite: delocalised / free electrons (1); diamond: all outer electrons in bonds / no free electrons (1). ALLOW 'each carbon forms 3 bonds in graphite and 4 in diamond' for supporting reasoning.
Question 3 · Structured question
9 marks
A student uses paper chromatography to investigate the dyes in a green sweet coating. The student sets up the experiment using three known dyes, A, B and C, and the green coating, X. The solvent front travels 8.0 cm. The distances travelled by the spots are shown in Table 1.
Table 1
| Sample | Distance moved by each spot in cm | |---|---| | Dye A | 2.0 | | Dye B | 4.8 | | Dye C | 6.4 | | X | 2.0 and 6.4 |
(a) The student draws the start line on the paper in pencil, not ink. Suggest why. ______________________________________________ [1]
(b) Explain why the start line must be above the level of the solvent in the beaker. ______________________________________________ ______________________________________________ [2]
(c) Calculate the Rf value of dye B. Use the equation: Rf = distance moved by spot / distance moved by solvent front
Rf = ______________ [2]
(d) Identify the dyes in X. Explain your answer. ______________________________________________ ______________________________________________ [2]
(e) The student says that dye C is a pure substance. Describe how the student could use chromatography to give more evidence that dye C is pure. ______________________________________________ ______________________________________________ [2]
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Worked solution
(a) Pencil is insoluble, so the line does not move with the solvent or add coloured spots. (b) If the line were below the solvent level the dyes would dissolve into the solvent in the beaker and would not move up the paper. (c) Rf = 4.8 / 8.0 = 0.60. (d) X has spots at 2.0 cm and 6.4 cm, the same as dye A and dye C. So X contains A and C. (e) Run a chromatogram of dye C in at least one other solvent. If dye C is pure it gives only one spot in every solvent.
Marking scheme
(a) pencil does not dissolve / is insoluble in the solvent / does not run with the solvent (1). IGNORE 'pencil is clearer'. (b) dyes would dissolve in the solvent in the beaker (1); would not travel up the paper / would be lost from the paper (1). (c) correct answer 0.60 / 0.6 = 2 marks. If answer is wrong, 4.8 / 8.0 shown = 1 mark. ALLOW 0.6. DO NOT ALLOW a unit. (d) A and C (1); spots at same distance / same Rf as those of A and C (1). ECF: for the second mark if the dyes named are wrong, no credit. (e) use a different solvent / a second solvent (1); one spot only in each solvent / still a single spot (1). ALLOW 'repeat with a different solvent and compare Rf'.
Question 4 · Structured question
9 marks
A student tests a blue solution, Y, to identify the ions in it.
Test 1: Sodium hydroxide solution is added to Y. A pale blue precipitate forms. Test 2: Dilute hydrochloric acid and then barium chloride solution are added to Y. A white precipitate forms.
(a) (i) Identify the positive ion in Y. ______________________________________________ [1] (ii) Identify the negative ion in Y. ______________________________________________ [1]
(b) Explain why dilute hydrochloric acid is added before the barium chloride solution in Test 2. ______________________________________________ ______________________________________________ [1]
(c)* Three colourless solutions are labelled 1, 2 and 3. One is sodium chloride, one is potassium iodide and one is sodium sulfate. Plan an investigation, using chemical tests, to identify which solution is which. Include the reagents you would use, the results you would expect for each solution and how you would draw your conclusions. ______________________________________________ ______________________________________________ ______________________________________________ ______________________________________________ ______________________________________________ ______________________________________________ [6]
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Worked solution
(a)(i) A pale blue precipitate with sodium hydroxide shows copper(II) ions. (ii) A white precipitate with acidified barium chloride shows sulfate ions. (b) Carbonate ions would also form a white precipitate with barium chloride. The acid reacts with any carbonate first, so the precipitate can only be barium sulfate. (c) Put a sample of each solution in separate clean test tubes. Add dilute nitric acid and then silver nitrate solution. The chloride gives a white precipitate, the iodide gives a yellow precipitate and the sulfate gives no precipitate. To confirm, add dilute hydrochloric acid and barium chloride solution to fresh samples. Only the sulfate gives a white precipitate. Both results agree for each solution.
Marking scheme
(a)(i) copper(II) / Cu2+ / copper ions (1). DO NOT ALLOW 'copper' alone if charge is wrong, e.g. Cu+. (a)(ii) sulfate / SO4 2- (1). (b) acid removes / reacts with carbonate ions (1). ALLOW 'to prevent false positive from carbonate'. (c) Level 3 (5-6 marks): a complete plan with a correct reagent (acidified silver nitrate and/or acidified barium chloride), correct results for all three solutions (white / yellow / no precipitate; white precipitate for sulfate with barium) and a clear logical conclusion linking results to each solution. Level 2 (3-4 marks): correct reagent(s) and some correct results, but one solution not fully identified or the conclusion is not clearly linked. Level 1 (1-2 marks): a relevant test or result named, with little detail or logic. Level 0 (0 marks): no creditworthy response. Indicative content: separate samples in test tubes; add dilute nitric acid then silver nitrate; chloride gives white precipitate; iodide gives yellow precipitate; sulfate gives no precipitate (or a possible faint one, IGNORE); confirm sulfate with dilute hydrochloric acid then barium chloride giving white precipitate; the other two give no precipitate; conclude by matching results. ALLOW cream/pale yellow for iodide. Flame tests are not creditworthy because cations do not distinguish chloride from iodide.
Question 5 · Structured question
11 marks
A student tests four solid substances, W, X, Y and Z. The results are shown in Table 1.1.
Table 1.1
| Substance | Melting point in °C | Conducts electricity as a solid? | Conducts electricity when molten? | |---|---|---|---| | W | 770 | no | yes | | X | 1085 | yes | yes | | Y | 63 | no | no | | Z | sublimes at 3650 | yes | does not melt |
(a) Substances W, X, Y and Z are an ionic compound, a metal, a simple molecular substance and a giant covalent substance. Use Table 1.1 to identify each substance. Write the letter W, X, Y or Z next to each type of substance.
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Worked solution
(a) W has a high melting point and conducts only when molten, so it is ionic. X conducts as a solid and when molten with a high melting point, so it is a metal. Y has a low melting point and never conducts, so it is simple molecular. Z conducts as a solid and does not melt, so it is giant covalent. (b) Y has weak intermolecular forces between its molecules. Only a small amount of energy is needed to overcome these forces. The strong covalent bonds within the molecules are not broken. (c) Each carbon atom uses three of its four outer electrons for bonding. The fourth electron is delocalised and free to move between layers, so it can carry charge. (d)(i) Surface area = 6 × 20² = 6 × 400 = 2400 nm². Volume = 20³ = 8000 nm³. Ratio = 2400 ÷ 8000 = 0.3 per nm. (ii) For sides of 200 nm the ratio is 6 × 40000 ÷ 8000000 = 0.03 per nm, so it decreases by a factor of ten.
Marking scheme
(a) W = ionic ✓; X = metal ✓; Y = simple molecular ✓; Z = giant covalent ✓. [4] All four correct for 4; one mark per correct identification. DO NOT ALLOW a letter used twice.
(b) weak intermolecular forces (between molecules) ✓; little energy needed to overcome them / covalent bonds are not broken ✓. [2] ALLOW AW for weak forces between molecules. DO NOT ALLOW 'covalent bonds are weak' or 'bonds between atoms break' (max 1 for second point lost).
(c) delocalised / free electrons ✓; (able to) move through the structure / between layers and carry charge ✓. [2] DO NOT ALLOW 'free ions'. IGNORE 'sea of electrons' without movement for second mark only.
(d)(i) correct answer 0.3 (per nm) ✓✓. [2] If answer incorrect: surface area = 2400 (nm²) and volume = 8000 (nm³) seen ✓ (one mark for either correct value). ALLOW 3/10 or 0.30. ECF for one arithmetic slip.
(d)(ii) it decreases / goes down (by a factor of ten / to one tenth) ✓. [1] DO NOT ALLOW 'it gets smaller' with no reference to the ratio being changed incorrectly. ALLOW 'divided by 10'.
Total 11 marks.
Question 6 · Structured question
11 marks
A student adds 0.60 g of calcium carbonate lumps to an excess of dilute hydrochloric acid in a conical flask. The carbon dioxide gas given off is collected and its volume measured every 20 seconds. The equation for the reaction is:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
The results are shown in Table 2.1.
Table 2.1
| Time in s | 0 | 20 | 40 | 60 | 80 | 100 | 120 | |---|---|---|---|---|---|---|---| | Volume of CO₂ in cm³ | 0 | 30 | 50 | 62 | 68 | 70 | 70 |
(a) State the name of a piece of apparatus that can be used to measure the volume of the gas.
(d) The student repeats the experiment using 0.60 g of powdered calcium carbonate. All other conditions are the same. Describe how each of the following changes, compared with the first experiment.
(e) Calculate the mass of calcium carbonate that reacts to produce 70 cm³ of carbon dioxide. The molar gas volume at room temperature and pressure is 24 dm³. Relative formula mass (Mr) of CaCO₃ = 100.
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Worked solution
(a) A gas syringe measures the volume of gas. (b)(i) Mean rate = 30 ÷ 20 = 1.5 cm³/s. (b)(ii) 100 s, because the volume reaches 70 cm³ and does not increase afterwards. (c) As acid is used up, its concentration decreases. There are fewer acid particles in a given volume, so fewer collisions per second with calcium carbonate and a lower frequency of successful collisions. The rate falls. (d)(i) Powder has a larger surface area, so the reaction is faster and finishes in less time. (d)(ii) The same mass of calcium carbonate is used, and it is the limiting reactant, so the same total volume of gas is made (70 cm³). (e) Volume = 70 ÷ 1000 = 0.070 dm³. Moles of CO₂ = 0.070 ÷ 24 = 0.002917 mol. Moles of CaCO₃ = 0.002917 mol (1:1 ratio). Mass = 0.002917 × 100 = 0.29 g (2 significant figures).
Marking scheme
(a) gas syringe ✓. [1] ALLOW (inverted) measuring cylinder / burette over water. DO NOT ALLOW 'syringe' alone if it is not for gas; DO NOT ALLOW beaker or balance.
(b)(i) correct answer 1.5 (cm³/s) ✓✓. [2] If answer incorrect: 30 ÷ 20 seen ✓. DO NOT ALLOW 0.67 (20 ÷ 30).
(b)(ii) 100 s ✓ AND volume stops increasing / stays at 70 cm³ ✓ (both needed for the mark). [1] ALLOW any time up to 100 s only if reasoning stated as 'first time volume reaches 70'; DO NOT ALLOW 120 s.
(c) concentration of acid decreases / acid is used up ✓; fewer collisions per second / lower frequency of collisions between particles ✓. [2] ALLOW AW for 'fewer successful collisions in a given time'. DO NOT ALLOW 'less energy' or 'particles move more slowly'. IGNORE 'fewer particles' without link to acid being used up.
(d)(i) shorter / less / faster (reaction finishes sooner) ✓. (d)(ii) same / unchanged / 70 cm³ ✓. [2] One mark each. DO NOT ALLOW 'more gas' or 'less gas'.
(e) correct answer 0.29 (g) ✓✓✓. [3] If answer incorrect: moles CO₂ = 0.070 ÷ 24 (= 0.0029) ✓ (conversion and division); moles CaCO₃ = moles CO₂ (1:1) ✓; mass = moles × 100 ✓. ALLOW 0.292 / 0.3 g (0.3 g loses no mark only if working shown). ECF for incorrect unit conversion, e.g. 70 ÷ 24 = 2.9 mol gives 290 g – max 2.
Total 11 marks.
Question 7 · Structured question
14 marks
5 A student heats magnesium ribbon in a crucible with a lid. The magnesium burns with a bright white flame and forms magnesium oxide, MgO.
(a) Magnesium oxide contains magnesium ions, \( {}^{24}_{12}\text{Mg}^{2+} \). Complete the table to show the numbers of subatomic particles in one magnesium ion.
Table 5.1 | Particle | Number in one \( {}^{24}_{12}\text{Mg}^{2+} \) ion | | protons | | | neutrons | | | electrons | | [3]
(b) Magnesium oxide is an ionic compound. Draw a dot-and-cross diagram to show the bonding in magnesium oxide. Show the outer electrons only, and show the charge on each ion. [Drawing space] [2]
(c) Write a balanced equation, including state symbols, for the reaction of magnesium with oxygen to form magnesium oxide. [2]
The student measures the masses shown in Table 5.2.
Table 5.2 | Measurement | Mass in g | | crucible and lid | 20.40 | | crucible, lid and magnesium ribbon | 20.88 | | crucible, lid and product after heating to constant mass | 21.12 |
Use the Periodic Table to help you answer the questions below.
(d) (i) Calculate the mass of magnesium oxide that should be made from the magnesium ribbon used. Show your working.
Mass of magnesium oxide = ........................ g [3]
(ii) Calculate the percentage yield of magnesium oxide obtained by the student.
Percentage yield = ........................ % [1]
(e) Suggest one reason why the student's yield is less than 100%. [1]
(f) Suggest one change to the student's method that would improve the yield. Then explain why this change would improve the yield.
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Worked solution
(a) Atomic number 12 gives 12 protons. Mass number 24 gives 24 - 12 = 12 neutrons. The 2+ charge means 2 electrons lost, so 12 - 2 = 10 electrons. (b) Magnesium transfers its two outer electrons to oxygen. Oxygen then has 8 outer electrons. Each ion is shown in square brackets with its charge: [Mg]2+ and [O]2- (8 electrons). (c) 2Mg(s) + O2(g) -> 2MgO(s) (d)(i) Mass of Mg = 20.88 - 20.40 = 0.48 g. Moles of Mg = 0.48 / 24 = 0.020 mol. 2 mol Mg gives 2 mol MgO, so moles MgO = 0.020 mol. Mr of MgO = 24 + 16 = 40. Mass of MgO = 0.020 x 40 = 0.80 g. (ii) Actual mass of MgO = 21.12 - 20.40 = 0.72 g. Percentage yield = 0.72 / 0.80 x 100 = 90%. (e) Some magnesium oxide escaped as white smoke when the lid was lifted, or some magnesium did not react. (f) Lift the lid only slightly and briefly so less smoke escapes.
Marking scheme
(a) Three correct = 3 marks; two correct = 2 marks; one correct = 1 mark. [3] protons 12 (1); neutrons 12 (1); electrons 10 (1). ECF none.
(b) Mg2+ shown with correct charge and either no outer electrons or a full shell of 8 (1); O2- shown with 8 outer electrons, with the 2 transferred electrons shown as a different symbol from oxygen's own 6 (1) and both charges shown (brackets not required). ALLOW 1 mark for a correct diagram of transfer without charges. DO NOT ALLOW shared electrons between the ions (covalent bond). [2]
(c) Correct formulae/reactants and product (1); balanced with correct state symbols 2Mg(s) + O2(g) -> 2MgO(s) (1). ALLOW multiples. ALLOW 1 mark for a correctly balanced equation with a state symbol error. [2]
(d)(i) Final answer 0.80 (g) = 3 marks. Otherwise: mass of Mg = 0.48 g (1); moles of Mg = 0.48 / 24 = 0.02 and Mr of MgO = 40 (1); 0.02 x 40 = 0.80 g (1). ALLOW ECF from wrong Mg mass. ALLOW 0.8. [3]
(ii) 90 (%) (1). ALLOW ECF from (d)(i) using 0.72 / (d)(i) x 100. [1]
(e) Any one from: some magnesium oxide / white smoke escaped when lid lifted (1); not all the magnesium reacted / reaction incomplete (1); some magnesium oxide stuck to tongs or crucible lid (1). IGNORE 'human error', 'impure magnesium' on their own. [1]
(f) Change linked to the reason in (e) (1) and linked explanation (1). Any one from: lift the lid only slightly / briefly, so less magnesium oxide smoke escapes; heat for longer and reheat until constant mass, so all the magnesium reacts; sand the magnesium ribbon first to remove the oxide layer, so more of the magnesium can react. Explanation mark depends on a sensible change. DO NOT ALLOW 'repeat the experiment' or 'be more careful' alone. [2]
Question 8 · Structured question containing one six-mark level-of-response task
6 marks
A technician has a dry mixture of three solids: iron filings, sand (silicon dioxide) and copper(II) sulfate crystals.
Iron filings are attracted to a magnet. Sand does not dissolve in water. Copper(II) sulfate dissolves in water to form a blue solution.
*Plan an experiment to obtain a sample of dry copper(II) sulfate crystals, a sample of dry sand and a sample of iron filings from the mixture.
Your answer should include the apparatus you would use, the order of the steps and the reason for each separation technique. [6]
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Worked solution
1. Pass a magnet over the dry mixture (or place it in a bag) to remove the iron filings, because iron is magnetic and the other two solids are not. 2. Add the remaining solids to a beaker of water and stir so the copper(II) sulfate dissolves; sand is insoluble. 3. Filter the mixture using filter paper in a funnel. The sand stays in the filter paper (residue) and the blue copper(II) sulfate solution passes through (filtrate). Wash the residue with distilled water and dry it in a warm oven. 4. Heat the filtrate in an evaporating basin to evaporate some of the water until the solution is saturated (a few crystals form on a glass rod dipped in and cooled). 5. Leave the saturated solution to cool so that crystals form (crystallisation), filter off the crystals and dry them between filter papers.
Marking scheme
Level of response, 6 marks. Assess best fit. Scientific completeness and reasoning set the level; clarity, relevance and logical order set the mark within the level.
Level 3 (5–6 marks): A complete, correctly ordered plan that covers all three solids and uses magnetic separation, dissolving, filtration and crystallisation (evaporation) with a reason for each technique linked to a property (magnetism, solubility/insolubility). Apparatus is named and appropriate. The plan is logical and clear.
Level 2 (3–4 marks): A mostly correct plan that includes at least three techniques, OR all four techniques with limited reasons or missing apparatus/drying steps. Order is mostly sensible. Some reasons given.
Level 1 (1–2 marks): One or two correct techniques named or described (e.g. uses a magnet; adds water and filters) with little or no reasoning. Order may be unclear.
Level 0 (0 marks): No response or no creditworthy content.
Indicative content (no single wording or exact sequence is compulsory, but the magnet step must come before dissolving, and filtration before crystallisation): ✓ Use a magnet to remove the iron filings; iron is magnetic, the others are not. ✓ Add water and stir/warm in a beaker so the copper(II) sulfate dissolves; sand is insoluble. ✓ Filter using funnel and filter paper; sand is the residue, copper(II) sulfate solution is the filtrate. ✓ Wash and dry the sand (e.g. warm oven/between filter papers). ✓ Heat the filtrate in an evaporating basin to evaporate some water / until saturated (crystallisation point). ✓ Leave to cool and crystallise; filter off crystals and dry (e.g. filter paper/warm oven, not strong heating).
ALLOW evaporating all the water to dryness for crystals at Level 2 maximum for that step (poorer crystals); ALLOW 'decant' for separating liquid only if sand is then dried. DO NOT ALLOW simple or fractional distillation as the main method to obtain the crystals. DO NOT ALLOW chromatography. IGNORE safety points unless linked to the method.
J248/02 Section A
Answer all 15 single-best-answer items, selecting one of four options and recording the letter in the answer box. The paper advises a maximum of 30 minutes for this section.
15 Question · 15 marks
Question 1 · multiple_choice
1 marks
A magnesium ion has the symbol \( {}^{24}_{12}\text{Mg}^{2+} \).
Which row shows the numbers of protons, neutrons and electrons in this ion?
Put a tick in the answer box for one letter only.
A.A: 12 protons, 12 neutrons, 14 electrons
B.B: 12 protons, 12 neutrons, 10 electrons
C.C: 12 protons, 24 neutrons, 10 electrons
D.D: 10 protons, 14 neutrons, 12 electrons
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Worked solution
The atomic number is 12, so there are 12 protons. Neutrons = mass number − atomic number = 24 − 12 = 12. A 2+ ion has lost two electrons, so electrons = 12 − 2 = 10. The answer is B.
Marking scheme
B (1 mark) A: gained instead of lost electrons. C: used the mass number as the number of neutrons. D: swapped protons and electrons. DO NOT ALLOW more than one letter.
Question 2 · multiple_choice
1 marks
A student runs a paper chromatography experiment on a green food colouring. One spot travels 3.6 cm from the pencil baseline. The solvent front travels 8.0 cm from the baseline.
What is the \( R_f \) value of the spot?
A.A: 0.36
B.B: 2.2
C.C: 0.45
D.D: 4.4
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Worked solution
\( R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \dfrac{3.6}{8.0} = 0.45 \). The answer is C.
Marking scheme
C (1 mark) B and D result from inverting the ratio (8.0 ÷ 3.6 = 2.2) or using an incorrect ratio. A is a distractor with no valid working. DO NOT ALLOW more than one letter.
Question 3 · multiple_choice
1 marks
A technician adds dilute hydrochloric acid and then barium chloride solution to a colourless solution. A white precipitate forms.
Which ion is present in the solution?
A.A: sulfate
B.B: carbonate
C.C: chloride
D.D: iodide
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Worked solution
Acidified barium chloride gives a white precipitate of barium sulfate if sulfate ions are present. Carbonate would fizz with the acid, and chloride and iodide are tested with acidified silver nitrate. The answer is A.
Marking scheme
A (1 mark) DO NOT ALLOW more than one letter.
Question 4 · multiple_choice
1 marks
Graphite conducts electricity but diamond does not.
Which statement explains why graphite conducts electricity?
A.A: Each carbon atom forms four covalent bonds.
B.B: The layers are held together by strong covalent bonds.
C.C: The structure contains positive and negative ions.
D.D: Each carbon atom has one delocalised electron.
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Worked solution
In graphite each carbon atom forms three covalent bonds, leaving one delocalised electron that can move through the layers and carry charge. Diamond has four bonds per carbon atom and no free electrons. The answer is D.
Marking scheme
D (1 mark) A describes diamond. B is incorrect because the layers are held by weak intermolecular forces. C describes ionic compounds. DO NOT ALLOW more than one letter.
Question 5 · multiple_choice
1 marks
Graphite is used to make the electrodes in some industrial cells. Which statement explains why graphite conducts electricity?
A.A. Each carbon atom forms four covalent bonds.
B.B. Each carbon atom has one delocalised electron.
C.C. Graphite contains positive and negative ions.
D.D. Graphite has weak forces between its layers.
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Worked solution
In graphite each carbon atom forms three covalent bonds, leaving one electron per atom delocalised. These electrons can move through the layers and carry charge. A describes diamond. C is wrong because graphite is not ionic. D explains why graphite is soft, not why it conducts.
Marking scheme
B (1 mark) ✓ A: describes diamond, not graphite. C: graphite is a giant covalent structure, not ionic. D: explains softness/slipperiness, not conduction. Only one answer may be given; no mark if more than one box is chosen.
Question 6 · multiple_choice
1 marks
An aluminium ion is written as \(^{27}_{13}\mathrm{Al}^{3+}\). Which row gives the numbers of protons, neutrons and electrons in this ion?
Row | Protons | Neutrons | Electrons A | 13 | 14 | 13 B | 13 | 14 | 16 C | 13 | 14 | 10 D | 10 | 14 | 13
A.A
B.B
C.C
D.D
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Worked solution
Protons = atomic number = 13. Neutrons = mass number − atomic number = 27 − 13 = 14. A 3+ ion has lost three electrons, so electrons = 13 − 3 = 10. This is row C.
Marking scheme
C (1 mark) ✓ A: electrons not adjusted for the charge (atom, not ion). B: three electrons added instead of removed. D: protons and electrons swapped. Only one answer may be given.
Question 7 · multiple_choice
1 marks
A student runs a paper chromatography experiment on a green food dye. The solvent front moves 9.0 cm from the pencil baseline. One yellow spot moves 3.6 cm from the baseline. What is the \(R_f\) value of the yellow spot?
A.A. 0.40
B.B. 0.60
C.C. 2.5
D.D. 5.4
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Worked solution
\(R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent}} = \dfrac{3.6}{9.0} = 0.40\).
Marking scheme
A (1 mark) ✓ B: 1 − 0.40, or 5.4 ÷ 9.0 (distance from the top). C: inverted ratio 9.0 ÷ 3.6. D: 9.0 − 3.6 (subtraction used). Only one answer may be given.
Question 8 · multiple_choice
1 marks
Potassium reacts more vigorously with water than sodium does. Which statement explains this?
A.A. Potassium atoms have fewer electrons in total.
B.B. Potassium atoms have a smaller atomic radius.
C.C. Potassium atoms have more electrons in the outer shell.
D.D. Potassium atoms lose their outer electron more easily.
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Worked solution
Both metals are in Group 1 with one outer electron. Potassium's outer electron is in a shell further from the nucleus, so it is held less strongly and is lost more easily to form K+. This makes potassium more reactive.
Marking scheme
D (1 mark) ✓ A: potassium has more electrons than sodium. B: potassium atoms are larger, not smaller. C: both have one outer electron. Only one answer may be given.
Question 9 · multiple_choice
1 marks
A student runs a paper chromatogram of a green food dye. One spot is 3.6 cm from the start line. The solvent front is 8.0 cm from the start line. What is the \(R_f\) value of the spot?
A.0.45
B.0.55
C.2.2
D.4.4
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Worked solution
\(R_f = \frac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \frac{3.6}{8.0} = 0.45\). Option C is the inverse of the correct ratio. Option B is 1 - 0.45. Option D is the inverse of the ratio doubled/misused.
Marking scheme
A (1 mark) ✓ B, C, D: 0 marks. GUIDANCE: Rf has no units.
Question 10 · multiple_choice
1 marks
A sulfide ion is written as \(^{32}_{16}\text{S}^{2-}\). How many neutrons and how many electrons are in one sulfide ion?
A.16 neutrons and 16 electrons
B.16 neutrons and 18 electrons
C.18 neutrons and 16 electrons
D.32 neutrons and 18 electrons
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Worked solution
Neutrons = mass number - atomic number = 32 - 16 = 16. The 2- charge means two extra electrons: 16 + 2 = 18 electrons.
Marking scheme
B (1 mark) ✓ A, C, D: 0 marks.
Question 11 · multiple_choice
1 marks
A reaction transfers energy to the surroundings. Which statement about this reaction is correct?
A.It is exothermic and the temperature of the surroundings rises.
B.It is exothermic and the temperature of the surroundings falls.
C.It is endothermic and the temperature of the surroundings rises.
D.It is endothermic and the temperature of the surroundings falls.
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Worked solution
A reaction that transfers energy to the surroundings is exothermic, so the surroundings get warmer.
Marking scheme
A (1 mark) ✓ B, C, D: 0 marks.
Question 12 · multiple_choice
1 marks
Calcium carbonate has the formula \(\text{CaCO}_3\). What amount, in moles, is present in 5.0 g of calcium carbonate? (Relative atomic masses: Ca = 40, C = 12, O = 16)
A.0.050 mol
B.0.20 mol
C.0.50 mol
D.20 mol
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Worked solution
\(M_r\) of \(\text{CaCO}_3\) = 40 + 12 + (3 × 16) = 100. Moles = mass / \(M_r\) = 5.0 / 100 = 0.050 mol. Option D comes from 100 / 5.0 (inverted).
Marking scheme
A (1 mark) ✓ B, C, D: 0 marks. GUIDANCE: D is the inverted calculation.
Question 13 · multiple_choice
1 marks
A potassium ion is written as \(^{39}_{19}\mathrm{K}^{+}\). Which row shows the numbers of protons, neutrons and electrons in this ion?
A.19 protons, 20 neutrons, 18 electrons
B.19 protons, 20 neutrons, 20 electrons
C.19 protons, 39 neutrons, 18 electrons
D.20 protons, 19 neutrons, 18 electrons
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Worked solution
The atomic number is 19, so there are 19 protons. Neutrons = 39 - 19 = 20. The ion has a charge of +1, so it has lost one electron: 19 - 1 = 18 electrons. The answer is A.
Marking scheme
A (1 mark) ✓ B: wrong, electrons not reduced for the + charge. C: wrong, mass number used as number of neutrons. D: wrong, protons and neutrons swapped/incorrect.
Question 14 · multiple_choice
1 marks
In a paper chromatography experiment the solvent front moves 8.0 cm from the start line. One spot moves 2.0 cm from the start line. What is the \(R_f\) value of this spot?
A.0.25
B.0.75
C.4.0
D.6.0
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Worked solution
\(R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent}} = \dfrac{2.0}{8.0} = 0.25\). The answer is A.
Marking scheme
A (1 mark) ✓ B: wrong, 1 - Rf calculated. C: wrong, ratio inverted (8.0 / 2.0). D: wrong, distances subtracted (8.0 - 2.0). IGNORE: no units, Rf has none.
Question 15 · multiple_choice
1 marks
A technician adds dilute hydrochloric acid and then barium chloride solution to a colourless solution. A white precipitate forms. Which ion is present in the solution?
A.Carbonate
B.Chloride
C.Iodide
D.Sulfate
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Worked solution
Acidified barium chloride gives a white precipitate of barium sulfate if sulfate ions are present. Carbonate would fizz with acid, and chloride and iodide are tested with acidified silver nitrate. The answer is D.
Marking scheme
D (1 mark) ✓ A: wrong, carbonate would give bubbles of carbon dioxide with acid. B and C: wrong, halides are tested with acidified silver nitrate.
J248/02 Section B
Answer all eight numbered questions, totalling 75 marks. Whole-question marks in order: 9, 12, 11, 5, 8, 12, 10, 8. Use lettered and Roman-numeral subparts as needed, including short responses, completion, calculations, diagrams and practical reasoning. Include one six-mark level-of-response task within this section; it is already included in the whole-question totals. Show calculation working.
8 Question · 75 marks
Question 1 · Structured question
5 marks
A technician has a colourless solution, solution X, that contains one metal cation and one anion. The technician carries out two tests.
Test 1: Sodium hydroxide solution is added to a sample of X. A green precipitate forms. The precipitate turns brown at the surface when left standing in air.
Test 2: Dilute hydrochloric acid is added to a second sample of X, followed by barium chloride solution. A white precipitate forms.
(a) Identify the metal cation in solution X. [1]
(b) Identify the anion in solution X. [1]
(c) Write the name of the compound in solution X. [1]
(d) Explain why dilute hydrochloric acid is added before the barium chloride solution in Test 2. [1]
(e) Write an ionic equation, including state symbols, for the reaction that forms the white precipitate in Test 2. [1]
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Worked solution
(a) A green precipitate that turns brown in air is iron(II) hydroxide, so the cation is iron(II), \( \mathrm{Fe^{2+}} \). (b) A white precipitate with acidified barium chloride shows sulfate ions, \( \mathrm{SO_4^{2-}} \). (c) The compound is iron(II) sulfate. (d) Carbonate ions would also form a white precipitate (barium carbonate) with barium chloride. Acid reacts with carbonate ions and removes them, so a white precipitate can only be barium sulfate. (e) \( \mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)} \)
Marking scheme
(a) iron(II) / Fe2+ ✓ [1]. ALLOW iron(2+). DO NOT ALLOW iron(III) or Fe3+. DO NOT ALLOW 'iron' alone. (b) sulfate / SO4 2- ✓ [1]. DO NOT ALLOW sulphate if the formula is wrong; ALLOW sulphate spelling if otherwise correct. IGNORE charge errors if the name is correct. (c) iron(II) sulfate ✓ [1]. ALLOW FeSO4. ECF from (a) and (b) for a correctly named compound of the ions given. (d) removes/reacts with carbonate ions (AW) ✓ [1]. ALLOW 'carbonate would also give a white precipitate' / 'to prevent a false positive'. IGNORE 'to make the test work' or 'to clean the solution'. (e) Ba2+(aq) + SO4 2-(aq) -> BaSO4(s) ✓ [1]. Species and state symbols must all be correct for the mark. DO NOT ALLOW equations including spectator ions (Cl-). ALLOW ECF from (b) only if the anion given is consistent with a white barium precipitate and the equation is balanced. Total 5 marks.
Question 2 · Structured question
8 marks
A student adds small pieces of Group 1 metals to separate troughs of water.
The student records these observations.
Table 1.1 | Metal | Observation | | lithium | steady fizzing, metal floats and slowly gets smaller | | sodium | rapid fizzing, metal melts into a ball and darts across the surface | | potassium | very rapid fizzing, metal melts and the gas burns with a lilac flame |
(a) State the number of electrons in the outer shell of a Group 1 atom. [1]
.......................................
(b) Explain, in terms of electron arrangement, why potassium is more reactive than sodium. [3]
(c) Sodium reacts with water to make sodium hydroxide and hydrogen. Write the balanced symbol equation for this reaction. [2]
.......................................
(d) The student adds a few drops of universal indicator to the water after the sodium has finished reacting. State the colour the indicator turns and explain why. [2]
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Worked solution
(a) Group 1 atoms have one outer electron. (b) Potassium has more electron shells than sodium. Its outer electron is further from the nucleus and more shielded, so the attraction to the nucleus is weaker. The electron is lost more easily, so potassium is more reactive. (c) Reactants: Na and H2O. Products: NaOH and H2. Balanced: 2Na + 2H2O -> 2NaOH + H2. (d) The solution contains hydroxide ions, so it is alkaline and universal indicator turns blue/purple.
Marking scheme
(a) 1 (1) [1]
(b) Any three from: - potassium atom has more (electron) shells / outer electron further from nucleus (1) - more shielding (1) - weaker attraction between nucleus and outer electron (1) - outer electron lost more easily (1) [3] ALLOW AW / ORA for sodium. DO NOT ALLOW 'potassium has more electrons' alone. IGNORE 'larger atom' alone.
(c) Correct formulae NaOH and H2 with Na and H2O (1); balanced 2Na + 2H2O -> 2NaOH + H2 (1) [2] ALLOW multiples. ALLOW state symbols if correct. Second mark only if formulae correct.
(d) blue / purple / dark blue (1); hydroxide ions / OH- ions / alkali formed (1) [2] ALLOW pH 12-14 for the colour mark context. DO NOT ALLOW green or red. Total 8
Question 3 · Structured question
8 marks
A student reacts magnesium ribbon with an excess of dilute hydrochloric acid and collects the hydrogen gas in a gas syringe.
Mg + 2HCl -> MgCl2 + H2
The student uses 0.12 g of magnesium. The relative atomic mass (Ar) of magnesium is 24.
(a) Calculate the amount, in mol, of magnesium used. [2]
(b) One mole of any gas occupies 24 dm3 at room temperature and pressure. Calculate the volume of hydrogen, in cm3, that the student should collect. [2]
(c) The magnesium takes 40 seconds to react completely. Calculate the mean rate of reaction in cm3 per second. [1]
mean rate = ................... cm3/s
(d) The student repeats the experiment with the acid at a higher temperature. All other conditions stay the same. Explain why the reaction is faster. [3]
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Worked solution
(a) moles = mass / Ar = 0.12 / 24 = 0.005 mol. (b) The equation shows 1 mol Mg gives 1 mol H2, so 0.005 mol H2. Volume = 0.005 x 24 = 0.12 dm3 = 0.12 x 1000 = 120 cm3. (c) Mean rate = 120 / 40 = 3.0 cm3/s. (d) At higher temperature particles have more energy and move faster, so collisions are more frequent. More collisions have energy equal to or greater than the activation energy, so more collisions are successful per second.
Marking scheme
(a) Correct answer 0.005 / 5 x 10^-3 (2). If answer wrong: 0.12 / 24 (1) [2]
(b) Correct answer 120 (2). If answer wrong: 0.005 x 24 = 0.12 dm3 (1) ECF from (a) for 1 mark for use of 24 and conversion x1000 [2]
(c) 3.0 / 3 (1). ECF from (b) for 'volume / 40' [1] DO NOT ALLOW missing units if the answer line has units given.
(d) Any three from: - particles have more (kinetic) energy / move faster (1) - collide more frequently / more collisions per second (1) - greater proportion of collisions have energy >= activation energy (1) - more successful collisions per second (1) [3] DO NOT ALLOW 'particles collide with more energy' alone as the only point for the frequency mark. DO NOT ALLOW 'activation energy decreases'. Total 8
Question 4 · Structured question
9 marks
A technician investigates the halogens in Group 7.
Table 1 shows the melting point and boiling point of three halogens.
| Halogen | Melting point in \(^\circ\)C | Boiling point in \(^\circ\)C | |---|---|---| | chlorine | -101 | -34 | | bromine | -7 | 59 | | iodine | 114 | 184 |
Table 1
(a) State the physical state of bromine at 20 \(^\circ\)C. Use the data in Table 1.
(c) Both have seven outer-shell electrons. A chlorine atom is smaller, so its outer shell is closer to the nucleus (fewer shells). The attraction to the nucleus is stronger, so it gains an electron more easily than bromine.
(d) Moles of Cl\(_2\) = 7.1 / 71 = 0.10 mol. The ratio Cl\(_2\) : NaCl is 1 : 2, so moles of NaCl = 0.20 mol. Mr of NaCl = 23 + 35.5 = 58.5. Mass = 0.20 x 58.5 = 11.7 g.
Marking scheme
(a) liquid ✓ [1] ALLOW l. DO NOT ALLOW 'liquid and gas'.
(b)(i) chlorine, bromine, iodine in this order ✓✓ [2] Award 1 mark if only one of chlorine as most reactive OR iodine as least reactive is correct.
(b)(ii) chlorine + potassium bromide → potassium chloride + bromine ✓ [1] ALLOW symbol equation Cl2 + 2KBr → 2KCl + Br2. ALLOW names in the correct order only.
(c) Any two from: [2] - (both) have seven electrons in the outer shell / same number of outer electrons (ignored if not linked) ✓ - chlorine has fewer shells / outer shell is closer to the nucleus / chlorine atom is smaller ✓ - stronger attraction between nucleus and outer electrons / incoming electron ✓ - chlorine gains an electron more easily ✓ Max 2. IGNORE 'chlorine is lighter'. DO NOT ALLOW references to chlorine molecules or ions being attracted if they imply bonds breaking.
(d) Final answer 11.7 (g) = 3 marks. [3] Otherwise: - moles Cl2 = 7.1 / 71 = 0.10 ✓ (method) - moles NaCl = 0.20 ✓ (using 1:2 ratio) - mass = 0.20 x 58.5 = 11.7 g ✓ (accuracy) ALLOW ECF from an incorrect Mr for Cl2 (e.g. 35.5) if the ratio and Mr of NaCl are used correctly. ALLOW 12 g for 3 marks only if 11.7 is shown. Alternative: 7.1 g Cl2 gives 117 g NaCl per 71 g so 11.7 g.
Question 5 · Structured question
10 marks
A technician reacts a strip of magnesium with an excess of dilute hydrochloric acid. The hydrogen gas produced is collected in a gas syringe. The technician records the total volume of gas every 20 seconds.
Table 1
| Time in s | 0 | 20 | 40 | 60 | 80 | 100 | |---|---|---|---|---|---|---| | Total volume of hydrogen in cm\(^3\) | 0 | 28 | 46 | 58 | 64 | 66 |
(a) Complete the balanced symbol equation for the reaction.
(b) Look at Table 1. Explain why the rate of reaction decreases between 20 s and 80 s. [2]
(c) Calculate the mean rate of reaction during the first 20 s. Give your answer in cm\(^3\)/s.
Mean rate = ____ cm\(^3\)/s [2]
(d) The molar volume of a gas at room temperature and pressure is 24 dm\(^3\). The relative atomic mass (\(A_r\)) of magnesium is 24. Use the final volume of hydrogen in Table 1 to calculate the mass of magnesium that reacted. Show your working.
Mass of magnesium = ____ g [3]
(e) The technician repeats the experiment using the same mass of magnesium powder instead of a strip. Everything else stays the same. State how the rate of reaction changes and explain your answer. [2]
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Worked solution
(a) Mg(s) + 2HCl(aq) \(\rightarrow\) MgCl\(_2\)(aq) + H\(_2\)(g). (b) The acid is being used up, so its concentration falls. There are fewer particles of acid per unit volume, so fewer successful collisions per second with magnesium. (c) Mean rate = 28 / 20 = 1.4 cm\(^3\)/s. (d) Final volume = 66 cm\(^3\) = 0.066 dm\(^3\). Moles H\(_2\) = 0.066 / 24 = 0.00275 mol. The equation shows 1 mol Mg gives 1 mol H\(_2\), so moles Mg = 0.00275 mol. Mass = 0.00275 \(\times\) 24 = 0.066 g. (e) The rate increases. Powder has a larger surface area, so more magnesium atoms are exposed to the acid and collisions are more frequent.
Marking scheme
(a) 2 in the first space AND H\(_2\) in the second ✓ [1]. ALLOW H2. DO NOT ALLOW H or H\(_2\)(aq).
(b) Any two from: acid is used up / concentration of acid decreases ✓; fewer acid particles in the same volume / fewer particles available to collide ✓; fewer (successful) collisions per second / lower frequency of collisions ✓. [2] IGNORE 'less energy' or 'magnesium runs out' (acid is in excess but magnesium is used up: ALLOW magnesium being used up / less surface area of magnesium remaining as a partial argument, max 1).
(c) Correct answer 1.4 ✓✓ [2]. If the answer is wrong, 28 / 20 ✓ [1]. ALLOW 1.40.
(d) Correct answer 0.066 (g) ✓✓✓ [3]. Otherwise: 66 / 1000 = 0.066 dm\(^3\) (or 66 / 24000) ✓ [method]; moles H\(_2\) = 0.00275 and moles Mg = 0.00275 (1:1 ratio) ✓ [processing]; mass = moles \(\times\) 24 ✓ [accuracy]. ALLOW 6.6 \(\times\) 10\(^{-2}\) g. ECF from an incorrect moles value to the final mass. If volume is not converted (66 / 24 = 2.75), allow max 2 for 2.75 \(\times\) 24 = 66 g.
(e) Rate increases / reaction is faster ✓; powder has a larger surface area (to volume ratio) ✓ so more frequent collisions (between particles) ✓. [2 marks max; the first mark is for 'increases' with a correct reason, the second for a linking statement about more frequent collisions]. DO NOT ALLOW 'more collisions' without reference to time / frequency, unless surface area is also given. IGNORE 'more particles' on its own.
Question 6 · Structured question
11 marks
A student investigates the rusting of iron nails. She sets up four test tubes, each containing a clean iron nail, and leaves them for one week.
Figure 1 shows the conditions in each tube.
- Tube A: nail in boiled water, with a layer of oil on top of the water. - Tube B: nail in dry air, with anhydrous calcium chloride (a drying agent) in the bottom of the tube. The tube is sealed with a bung. - Tube C: nail half-covered by tap water, open to the air. - Tube D: nail wrapped tightly in a strip of zinc, half-covered by tap water, open to the air.
Table 1 shows her results after one week.
| Tube | Observation | |---|---| | A | no rust | | B | no rust | | C | brown rust on the nail | | D | no rust on the nail |
(a) Rust is hydrated iron(III) oxide. Use Table 1 to state the two substances that must be present for iron to rust. [2]
(c) Most iron is used as steel, which is an alloy. Explain why steel is harder than pure iron. You may draw a labelled diagram in the space below to support your answer. [2]
(d) Iron reacts with oxygen to form iron(III) oxide, Fe2O3.
4Fe + 3O2 → 2Fe2O3
Calculate the mass of iron(III) oxide formed when 5.6 g of iron reacts completely with oxygen. Relative atomic masses: Fe = 56, O = 16. Show your working. [3]
Mass of iron(III) oxide = .......................... g
(e) The student's results in Table 1 only say whether rust forms. They do not show how much rust forms. Suggest one improvement to the method so that the amount of rusting in tubes C and D can be compared, and explain how it would help. [2]
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Worked solution
(a) Tube A has no oxygen (boiled water, oil layer) and does not rust. Tube B has no water and does not rust. Tube C has both and rusts. So both oxygen/air and water are needed.
(b) Zinc is more reactive than iron. Zinc loses electrons / reacts with oxygen and water first, so it corrodes instead of the iron. This is sacrificial protection.
(c) Steel contains iron plus small amounts of another element such as carbon. The different-sized atoms disrupt the regular arrangement of iron atoms, so layers cannot slide over one another easily, making steel harder.
(d) Moles of Fe = 5.6 / 56 = 0.10 mol. From the equation 4 mol Fe gives 2 mol Fe2O3, so moles of Fe2O3 = 0.10 x 2/4 = 0.050 mol. Mr of Fe2O3 = (2 x 56) + (3 x 16) = 160. Mass = 0.050 x 160 = 8.0 g.
(e) Dry each nail and measure its mass before and after the week (or scrape off and mass the rust). The larger the increase in mass, the more rust has formed, so tubes C and D can be compared numerically.
Marking scheme
(a) Two points, 1 mark each: - oxygen / air ✓ - water ✓ ALLOW moisture / damp. DO NOT ALLOW 'oxygen gas and air' as two separate answers. IGNORE salt.
(b) - zinc is more reactive than iron ✓ - so zinc reacts / corrodes / is oxidised (loses electrons) in preference to iron / zinc is sacrificed ✓ ALLOW 'sacrificial protection' for the second mark. DO NOT ALLOW 'zinc is a barrier' as the only explanation. ORA for iron being less reactive.
(c) - steel contains atoms of a different size / other elements (e.g. carbon) ✓ - which distort / disrupt the regular layers of iron atoms so layers cannot slide (easily) ✓ ALLOW a labelled diagram showing layers with a differently sized atom for both marks. DO NOT ALLOW 'bonds are stronger' or 'atoms are bigger' alone. AW.
(d) Check the final answer first: 8.0 (g) = 3 marks. - moles of Fe = 5.6 / 56 = 0.10 ✓ - moles of Fe2O3 = 0.10 x 2/4 = 0.050 ✓ (ECF from moles of Fe) - mass = 0.050 x 160 = 8.0 (g) ✓ (ECF from moles of Fe2O3, using Mr = 160) ALLOW alternative route: Mr Fe2O3 = 160; 2 x 56 x 2 = 224 g Fe gives 320 g Fe2O3; 5.6 x 320/224 = 8.0 for 3 marks. 8 gets full marks. If Mr is wrong, ECF for the final mark only.
(e) - Improvement: measure the mass of each nail before and after the experiment / collect and mass the rust / measure rust-covered area of the nail with squared paper ✓ - Explanation: gives numerical data so the amount of rusting can be compared (between C and D) ✓ ALLOW repeating the experiment and calculating a mean only if linked to reliability of the comparison (1 mark for each part). IGNORE 'leave for longer' unless linked to measurement. The explanation mark depends on a valid improvement.
Question 7 · Structured question
12 marks
A student investigates the reaction between calcium carbonate and excess dilute hydrochloric acid. The student adds small lumps of calcium carbonate to the acid in a conical flask and collects the gas in a gas syringe. The volume of gas is recorded every 20 seconds.
Table 7.1 shows the results.
| Time in s | 0 | 20 | 40 | 60 | 80 | 100 | 120 | 140 | |---|---|---|---|---|---|---|---|---| | Volume of gas in cm\(^3\) | 0 | 30 | 48 | 60 | 67 | 71 | 72 | 72 |
(a) Write the balanced symbol equation for this reaction. [2]
(e) The molar gas volume at room temperature and pressure is 24 dm\(^3\). Calculate the mass of calcium carbonate that reacted to produce 72 cm\(^3\) of gas. The relative formula mass (\(M_r\)) of calcium carbonate is 100. Show your working. [3]
Mass of calcium carbonate = ........................ g
(f) The student repeats the experiment using the same mass of calcium carbonate as a powder instead of small lumps. All other conditions stay the same.
State how the rate of reaction and the final volume of gas compare with the first experiment. Explain the change in rate. [2]
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Worked solution
(a) Calcium carbonate reacts with hydrochloric acid to form calcium chloride, water and carbon dioxide: \(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\). (b) The volume stops changing at 72 cm\(^3\) from 120 s onwards, so the reaction stopped at 120 s. (c) Rate = 30 / 20 = 1.5 cm\(^3\)/s. (d) Acid particles are used up, so concentration falls. Fewer collisions per second between acid particles and carbonate particles, so the reaction slows. (e) 72 cm\(^3\) = 0.072 dm\(^3\). Moles CO\(_2\) = 0.072 / 24 = 0.0030 mol. The ratio CaCO\(_3\):CO\(_2\) is 1:1, so moles CaCO\(_3\) = 0.0030 mol. Mass = 0.0030 x 100 = 0.30 g. (f) The rate is faster; the final volume is the same (same mass of calcium carbonate, acid in excess). Powder has a larger surface area, so more particles are exposed and collisions are more frequent.
Marking scheme
(a) Correct formulae for all reactants and products CaCO3, HCl, CaCl2, H2O, CO2 ✓; correctly balanced (2HCl) ✓. ALLOW multiples. State symbols not required; IGNORE them if given. Balancing mark only awarded if formulae are correct. [2] (b) 120 (s) ✓. ALLOW any value from 110 to 120 only if clearly justified; DO NOT ALLOW 140. [1] (c) Correct answer 1.5 ✓✓. If the answer is wrong, 30 / 20 or volume divided by time ✓. Unit given in the answer line. [2] (d) Concentration of (hydrochloric) acid decreases / acid is used up ✓; fewer (successful) collisions per second / lower frequency of collisions ✓. DO NOT ALLOW 'fewer collisions' with no reference to time or frequency unless AW. DO NOT ALLOW 'particles slow down'. [2] (e) Correct answer 0.30 (g) ✓✓✓. Otherwise: 72 / 1000 = 0.072 dm3 or conversion of cm3 to dm3 ✓; 0.072 / 24 = 0.003 (mol) ✓ ECF from the conversion; 0.003 x 100 = 0.3 g ✓ ECF from the moles calculated, ALLOW 0.3. If no conversion (72 / 24 = 3 mol) is used, maximum 1 mark for the 1:1 mole ratio and multiplication by 100. [3] (f) Faster / greater rate AND same final volume ✓ (both needed); larger surface area (of powder) so more frequent collisions / more particles exposed ✓. IGNORE 'more surface' without reference to larger area. [2]
Question 8 · Structured question containing one six-mark level-of-response task
12 marks
Question 7
A council looks after a steel footbridge near the coast. The steel slowly corrodes.
(a) State the two substances from the air that are needed for iron to rust. [2]
(b) Steel is an alloy of iron. Pure iron is softer than steel. Explain why steel is harder than pure iron. You may draw a labelled diagram to help your answer. [2]
(c) Rust is mainly hydrated iron(III) oxide. The formula of anhydrous iron(III) oxide is Fe2O3. Calculate the percentage by mass of iron in Fe2O3. Use the Periodic Table. Relative atomic masses: Fe = 56, O = 16. Give your answer to 2 significant figures.
Percentage of iron = ........................ % [2]
(d)* The council can protect the steel in three ways: - Method 1: paint the steel - Method 2: coat the steel with a layer of zinc (galvanising) - Method 3: attach blocks of magnesium to the steel below the water line of the supports
Table 7.1 shows information about each method.
Table 7.1 | Method | Cost per year | Protection lasts | Protection if the surface is scratched | | 1 Paint | Low | 5 years | None at the scratch | | 2 Zinc coating | Medium | 25 years | Continues to protect | | 3 Magnesium blocks | Medium | 10 years, then blocks replaced | Protects all connected steel |
Compare how the three methods protect steel from rusting. Use ideas about reactivity in your answer. Recommend which method is best for the steel supports in sea water. [6]
Total: 12 marks
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Worked solution
(a) Oxygen and water.
(b) In pure iron the atoms are all the same size and are in regular layers that slide over each other easily. In steel, carbon atoms (a different size) disrupt the regular layers, so the layers cannot slide as easily and the metal is harder.
(c) Mr of Fe2O3 = (2 × 56) + (3 × 16) = 112 + 48 = 160. Percentage of iron = 112 ÷ 160 × 100 = 70%. To 2 s.f. = 70%.
(d) Paint keeps oxygen and water away from the steel (physical barrier). Once scratched, the steel is exposed and rusts. Zinc is also a barrier but, because zinc is more reactive than iron, it loses electrons in preference to iron (sacrificial protection), so exposed steel at a scratch is still protected. Magnesium is more reactive than iron, so it corrodes instead of the steel (sacrificial protection) and protects all connected steel, even though it is underwater and cannot be painted; the blocks are replaced every 10 years. Recommendation: magnesium blocks (for the submerged parts) because they work in sea water and keep protecting without needing an intact coating. Zinc coating with justification is also acceptable.
Marking scheme
(a) oxygen ✓ water ✓ [2] ALLOW air for oxygen only if water also given (1 mark max for either). ALLOW water vapour/moisture. DO NOT ALLOW carbon dioxide / salt.
(b) atoms/ions of different size (in alloy / carbon atoms added) ✓ distort/disrupt the regular layers so the layers cannot slide (over each other easily) ✓ [2] ALLOW AW. IGNORE 'stronger' alone.
(c) Mr = 160 ✓ (112 ÷ 160 × 100 =) 70 ✓ [2] Correct answer of 70 (%) with or without working scores 2. ALLOW 1 mark for 112/160 or 0.7 if the final answer is wrong. ECF from an incorrect Mr.
(d)* Level of response [6] Level 3 (5–6 marks): Compares all three methods, correctly explaining both barrier protection and sacrificial protection using relative reactivity (zinc and magnesium more reactive than iron, so they lose electrons / corrode first), links this to the scratch data, and gives a justified recommendation using the table. Well organised, clear and relevant. Level 2 (3–4 marks): Describes barrier protection and at least one sacrificial method with some reference to reactivity, and makes a recommendation with partial justification. Comparison may be incomplete. Level 1 (1–2 marks): Simple statements about one or more methods (e.g. paint covers the steel; zinc/magnesium protects) with little or no use of reactivity or the data. Recommendation may be unsupported. Level 0 (0 marks): No response or no response worth credit.
Indicative content (not compulsory, any order): - paint: physical barrier keeps out oxygen and water; no protection once scratched; shortest lifetime - zinc: barrier; zinc is more reactive than iron so also sacrificial protection, protecting at scratches - magnesium: more reactive than iron; sacrificial; loses electrons/corrodes instead of iron; protects all connected steel; can be used underwater - magnesium blocks need replacing but are suited to submerged parts; zinc lasts longest per application - recommendation: magnesium blocks (or zinc) with a valid reason from Table 7.1 or reactivity. ALLOW a combination of methods. IGNORE 'cheapest' used as the only reason.
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