2025 OCR GCSE Gateway Science - Chemistry A - J248 Practice Paper with Answers

The atomic number 13 gives 13 protons. Neutrons = 27 - 13 = 14. The 3+ charge means three electrons have been lost, so electrons = 13 - 3 = 10.

C (1 mark, AO2). Two or contradictory answers score 0. Distractors: A ignores the charge; B uses the mass number as neutrons; D swaps protons and electrons.

\(M_r = 24 + 2\times(16+1) = 58\). Moles = mass / \(M_r\) = 2.9 / 58 = 0.05 mol.

A (1 mark, AO2). Distractors: C inverts the ratio (58/2.9); D counts only one H per OH group incorrectly / uses \(M_r\) = 29.

Sodium chloride is an ionic compound. In the solid the ions are held in a lattice and cannot move. When molten the ions are free to move and carry charge.

D (1 mark, AO1). Distractors refer to the wrong particle (electrons, molecules) or the wrong bonding (covalent).

Percentage yield = actual yield / theoretical yield x 100 = 6.0 / 8.0 x 100 = 75%.

B (1 mark, AO2). Distractors: A is the percentage lost; C inverts the fraction; D multiplies the two masses.

Higher temperature gives particles more kinetic energy, so they collide more frequently and more of the collisions have energy equal to or greater than the activation energy.

C (1 mark, AO1). A describes a catalyst; B is incorrect; D describes an increase in concentration.

Potassium manganate(VII) is an oxidising agent that oxidises an alcohol to the carboxylic acid with the same number of carbon atoms: propan-1-ol gives propanoic acid.

A (1 mark, AO1). Propene and propane would need dehydration or reduction, not oxidation.

Barium ions react with sulfate ions to form insoluble white barium sulfate. Carbonates would fizz with acid and halides are tested using silver nitrate.

D (1 mark, AO1).

Copper is less reactive than hydrogen, so copper ions are discharged at the negative cathode. At the positive anode hydroxide ions from water are discharged instead of sulfate ions, forming oxygen.

A (1 mark, AO2). Distractors reverse the electrodes or give products that do not form.

The bracket means there are two OH groups. \(M_r = 40 + 2 \times (16 + 1) = 40 + 34 = 74\). Answer B.

B (74) = 1 mark (AO2). Distractor A (57) comes from counting only one OH group; C (58) from adding 2 x H only to one O. Two answers or contradictory answers score 0.

Protons = atomic number = 13. Neutrons = 27 - 13 = 14. The atom has 13 electrons; the 3+ ion has lost three, so 10 electrons. Answer A.

A (13, 14, 10) = 1 mark (AO2). B adds electrons instead of removing them; C uses the mass number as neutrons; D reverses protons and electrons. Two answers score 0.

\(M_r\) of NaOH = 40. Moles = 4.0 / 40 = 0.10 mol. Volume = 500 cm\(^3\) = 0.5 dm\(^3\). Concentration = 0.10 / 0.5 = 0.20 mol/dm\(^3\). Answer C.

C (0.20) = 1 mark (AO2). B is the number of moles, not divided by 0.5 dm3; A is from multiplying by 0.5; D is from dividing the mass by 5 without using Mr. Two answers score 0.

Sodium chloride is an ionic compound. In the solid the ions are held in a lattice and cannot move. When molten the ions are free to move and carry charge. Answer A.

A = 1 mark (AO1). B, C and D refer to delocalised electrons, covalent bonds or a change in electron number, which are incorrect for an ionic compound. Two answers score 0.

Positive lead ions move to the negative cathode and gain electrons to form lead. Negative bromide ions move to the positive anode and lose electrons to form bromine. The compound is molten, so there is no water and no hydrogen or oxygen. Answer A.

A = 1 mark (AO1/AO2). B reverses the electrodes; C and D include products from water, which is not present. Two answers score 0.

Higher temperature gives particles more kinetic energy, so they collide more often and a greater proportion of collisions have at least the activation energy. The activation energy itself does not change and the number of particles in the volume stays the same. Answer B.

B = 1 mark (AO1). A and C wrongly change the activation energy (that is the effect of a catalyst); D describes a concentration or pressure change. Two answers score 0.

Each decrease of one pH unit means a ten-fold increase in hydrogen ion concentration. X is 3 pH units lower than Y, so \(10 \times 10 \times 10 = 1000\) times greater. Answer C.

C = 1 mark (AO2). A and B use simple multiplication of the pH difference; D reverses the relationship. Two answers score 0.

Relative formula mass of \(\mathrm{Na_2CO_3}\) = (2 × 23) + 12 + (3 × 16) = 106.
Amount = mass ÷ Mr = 5.3 ÷ 106 = 0.050 mol.

First check the answer on the answer line: if 0.050 (mol) / 0.05 award 2 marks.
Otherwise:
- Mr = 106 ✓ (AO2)
- 5.3 ÷ their Mr ✓ (ECF allowed)
DO NOT ALLOW units other than mol.

Zinc is higher in the reactivity series than iron, so it forms its positive ions more readily and corrodes instead of the steel. This is sacrificial protection.

- Zinc is more reactive than iron / steel ✓ (AO1)
- zinc corrodes / loses electrons / is oxidised instead of (in preference to) the iron ✓ (AO2)
ALLOW zinc is sacrificed.
IGNORE zinc forms a barrier / coating (this is not the explanation).

Percentage yield = actual yield ÷ theoretical yield × 100 = 6.0 ÷ 8.0 × 100 = 75%.

First check the answer on the answer line: if 75 (%) award 2 marks.
Otherwise:
- 6.0 ÷ 8.0 (× 100) ✓
- 75 ✓
ALLOW 1 mark for 0.75 or 133 (ECF for inverted ratio gets 1 mark only).

In graphite each carbon atom forms three covalent bonds, so one outer electron per atom is not used in bonding. These delocalised electrons can move along the layers and carry an electric current.

- each carbon atom forms three covalent bonds / has one electron not used in bonding ✓ (AO1)
- delocalised / free electrons can move (through the layers) and carry charge ✓ (AO1)
IGNORE 'free ions'. DO NOT ALLOW electrons 'move between atoms' without 'free' / 'delocalised' idea.

Ethene has a C=C double bond. Bromine adds across it, forming 1,2-dibromoethane, which is colourless, so the bromine water is decolourised.

(i) orange / brown to colourless ✓ (AO1)
ALLOW decolourised / goes clear. DO NOT ALLOW 'colourless' with a precipitate or any other colour mentioned.
(ii) 1,2-dibromoethane ✓ (AO1)
ALLOW dibromoethane / \(\mathrm{C_2H_4Br_2}\).

Sodium chloride is a giant ionic lattice. Strong electrostatic attractions between oppositely charged ions act in all directions, and much energy is needed to break them.

- strong electrostatic attraction between (oppositely charged) ions / strong ionic bonds ✓ (AO1)
- a lot of energy needed to overcome / break these (forces) ✓ (AO1)
DO NOT ALLOW covalent bonds are broken / intermolecular forces / molecules.
IGNORE 'melting point' restated.

Protons = atomic number = 13. Neutrons = 27 − 13 = 14. The 3+ charge means three electrons lost from 13, leaving 10 electrons.

- 13 protons and 14 neutrons ✓ (AO2)
- 10 electrons ✓ (AO2)
Both numbers needed for the first mark.

Hydrochloric acid is completely ionised in water, so more \(\mathrm{H^+}\) ions are present at the same concentration. Ethanoic acid is only partly ionised, so there are fewer \(\mathrm{H^+}\) ions and a higher pH.

- hydrochloric acid is strong / fully (completely) ionised (in solution) ✓ (AO1)
- ethanoic acid is weak / only partly ionised, so fewer hydrogen ions (ORA) ✓ (AO2)
IGNORE references to 'dilute' or 'concentrated'.

Neutrons = mass number - atomic number = \(56 - 26 = 30\). A neutral atom has 26 electrons; the 2+ ion has lost two, so \(26 - 2 = 24\).

30 neutrons (1) AO2
24 electrons (1) AO2
ALLOW 1 mark if the values are reversed. DO NOT ALLOW 26 electrons.

Magnesium oxide is a giant lattice of Mg2+ and O2- ions. The electrostatic forces between these oppositely charged ions are strong in all directions, so a large amount of energy is needed to separate them.

Strong (electrostatic) forces/bonds between oppositely charged ions / strong ionic bonds (1) AO1
Lots of/high energy needed to overcome/break them (1) AO1
IGNORE references to molecules. DO NOT ALLOW 'covalent bonds break' or 'intermolecular forces' (cancels both marks).

\(M_r = 40 + 2 \times (14 + 3 \times 16) = 40 + 2 \times 62 = 40 + 124 = 164\).

First check the answer on the answer line. If 164 award 2 marks.
Otherwise: \(40 + 2 \times 62\) or 40 + 28 + 96 (1) AO2
164 (1) AO2
ALLOW 1 mark for 102 (bracket ignored, 40+14+48) .

\(R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \dfrac{3.6}{8.0} = 0.45\).

First check the answer on the answer line. If 0.45 award 2 marks.
\(3.6 \div 8.0\) (1) AO2
0.45 (1) AO2
ALLOW 0.450. ALLOW 1 mark for 2.2 (inverted ratio). Rf has no unit; IGNORE a unit.

Positive Pb2+ ions move to the cathode and gain electrons to form lead metal. Negative Br- ions move to the anode and lose electrons to form bromine.

Cathode: lead (1) AO1
Anode: bromine (1) AO1
ALLOW Pb and Br2. DO NOT ALLOW bromide or hydrogen/oxygen.

The left side has 4 moles of gas and the right side has 2 moles. By Le Chatelier's principle, increasing pressure shifts the equilibrium to the side with fewer gas molecules, which is the right, so more ammonia.

Amount of ammonia increases / equilibrium shifts to the right / forward reaction favoured (1) AO2
Because fewer (gas) molecules/moles on the right (2 compared with 4) (1) AO2
ALLOW ECF for the explanation only if the prediction is consistent with it.

Powdered marble has a larger surface area exposed to the acid. More particles are available to collide with acid particles at any moment, so the collision frequency and rate rise.

Powder has larger surface area (to volume ratio) (1) AO1
More frequent/more collisions (per second) between particles (1) AO1
DO NOT ALLOW 'more energetic collisions' or 'more particles' in total. IGNORE 'more collisions' without reference to frequency/time only if the surface area mark is not already credited as a link.

Acidify with dilute hydrochloric acid to remove carbonate ions, then add barium chloride solution. Barium sulfate, a white insoluble solid, forms if sulfate is present.

Add (dilute) hydrochloric acid and barium chloride/barium nitrate (solution) (1) AO1
White precipitate (1) AO1
ALLOW acidified barium chloride. ALLOW nitric acid with barium nitrate. DO NOT ALLOW 'white solid' only if no mention of reagents; DO NOT ALLOW silver nitrate.

Relative formula mass of Mg(OH)\(_2\) = 24 + 2 \(\times\) (16 + 1) = 58.
Amount = 2.9 \(\div\) 58 = 0.050 mol.

First check the answer on the answer line. If answer = 0.05 / 0.050 mol award 2 marks.
Otherwise:
- Relative formula mass = 58 (1) [AO2]
- 2.9 \(\div\) 58 = 0.05 (1) [AO2]
ALLOW ECF for a wrong relative formula mass used correctly in the division (e.g. 41 gives 0.071).
IGNORE units if mol is not given on the answer line but working is correct.

Magnesium oxide is a giant lattice of Mg\(^{2+}\) and O\(^{2-}\) ions. There are strong electrostatic forces of attraction between the oppositely charged ions in all directions, so a large amount of energy is needed to break them.

- Strong electrostatic attraction between (oppositely charged) ions / strong ionic bonds (1) [AO1]
- Idea that a lot of energy / high temperature is needed to overcome / break these (forces) (1) [AO2]
ALLOW giant (ionic) lattice for context.
DO NOT ALLOW any reference to breaking covalent bonds or intermolecular forces (cancels credit).
IGNORE 'strong bonds' on its own.

Powdered marble has a larger surface area exposed to the acid than lumps of the same mass. More marble particles are available to collide with acid particles, so the frequency of successful collisions increases and the rate increases.

- Powder has a larger surface area (than lumps) / more particles exposed (1) [AO1]
- More frequent (successful) collisions between the particles (per second) (1) [AO2]
ALLOW AW for more collisions in a given time.
DO NOT ALLOW particles have more energy / collisions are harder.
IGNORE 'more collisions' with no time idea only if surface area point is absent; 'more collisions' alone may score the second mark.

Zinc is higher in the reactivity series than iron, so it forms positive ions more readily. Zinc is oxidised (corrodes) in preference to the iron, so it protects the iron even though the coating is damaged. This is sacrificial protection.

- Zinc is more reactive than iron (1) [AO1]
- Zinc reacts / is oxidised / loses electrons / corrodes instead of (in preference to) the iron, AW (1) [AO2]
ALLOW sacrificial protection for the second mark if linked to zinc being used up.
IGNORE 'zinc is a barrier' (the coating is scratched).
DO NOT ALLOW iron is more reactive.

Protons = atomic number = 13. Neutrons = 27 - 13 = 14. The 3+ ion has lost 3 electrons: 13 - 3 = 10.

- Protons = 13 and neutrons = 14 (1) [AO2]
- Electrons = 10 (1) [AO2]
All three correct for 2 marks; two correct including protons scores 1 only if matching the above allocation.
ALLOW ECF for electrons = protons - 3 if protons are wrong.

\( M_r \) of \( CaCO_3 \) = 40 + 12 + 48 = 100.
Moles of \( CaCO_3 \) = 25.0 ÷ 100 = 0.250 mol.
The equation shows 1 mol \( CaCO_3 \) gives 1 mol CaO, so moles CaO = 0.250 mol.
\( M_r \) of CaO = 40 + 16 = 56.
Mass of CaO = 0.250 × 56 = 14.0 g.

First check the answer on the answer line. If answer = 14 / 14.0 g award 3 marks.
Otherwise:
- moles of CaCO3 = 25.0 / 100 = 0.25 (mol) (1) [AO2]
- 1:1 ratio so moles CaO = 0.25 (mol) / M_r of CaO = 56 used (1) [AO2]
- mass = 0.25 x 56 = 14 (g) (1) [AO2]
ALLOW ECF from an incorrect M_r or moles. Alternative method: 25.0 x 56/100 scores 3. IGNORE significant figures. Missing or wrong unit: max 2.

Moles of HCl = 0.100 × (20.0 ÷ 1000) = 0.00200 mol.
The equation shows a 1:1 ratio, so moles NaOH = 0.00200 mol.
Concentration of NaOH = 0.00200 ÷ (25.0 ÷ 1000) = 0.0800 mol/dm³.

First check the answer on the answer line. If answer = 0.08 / 0.080 / 0.0800 mol/dm3 award 3 marks.
Otherwise:
- moles HCl = 0.100 x 0.0200 = 0.00200 (mol) (1) [AO2]
- moles NaOH = 0.00200 (mol) (1:1 ratio) (1) [AO2]
- concentration = 0.00200 / 0.0250 = 0.08 (mol/dm3) (1) [AO2]
ALLOW ECF. Failure to convert cm3 to dm3 gives an answer 1000 times too large or small: max 2. Missing or wrong unit: max 2.

\( M_r \) of \( Fe_2O_3 \) = 112 + 48 = 160. \( M_r \) of CO = 28, so 3CO = 84.
Total \( M_r \) of reactants = 160 + 84 = 244.
Mass of iron wanted = 2 × 56 = 112.
Atom economy = (112 ÷ 244) × 100 = 45.9 % (3 s.f.).

First check the answer on the answer line. If answer = 45.9 (%) award 3 marks.
Otherwise:
- total M_r of reactants = 244 (1) [AO2]
- M_r of desired product = 112 (1) [AO2]
- (112 / 244) x 100 = 45.9 (%) given to 3 s.f. (1) [AO2]
ALLOW 45.90 / 45.9016. ALLOW ECF from an incorrect total M_r. 45.9 earns the rounding mark only if the unrounded value is correct or ECF.

Energy needed to break bonds = 436 + 243 = 679 kJ/mol.
Energy released making bonds = 2 × 432 = 864 kJ/mol.
Overall energy change = 679 − 864 = −185 kJ/mol.
More energy is released than taken in, so the reaction is exothermic.

First check the answer on the answer line. If answer = -185 (kJ/mol) and exothermic award 3 marks.
Otherwise:
- bonds broken = 436 + 243 = 679 and bonds made = 2 x 432 = 864 (1) [AO2]
- 679 - 864 = -185 (kJ/mol) (1) [AO2]
- exothermic (1) [AO1]; ALLOW exothermic by ECF from a negative value.
ALLOW 185 released / 185 kJ/mol exothermic. A positive 185 with 'exothermic' is acceptable only if the text says energy is released.

Mass of chlorine = 3.25 − 1.12 = 2.13 g.
Moles of Fe = 1.12 ÷ 56 = 0.0200 mol.
Moles of Cl = 2.13 ÷ 35.5 = 0.0600 mol.
Simplest ratio Fe : Cl = 0.0200 : 0.0600 = 1 : 3.
Empirical formula = \( FeCl_3 \).

First check the answer on the answer line. If answer = FeCl3 award 3 marks.
Otherwise:
- mass of Cl = 2.13 (g) (1) [AO2]
- moles Fe = 0.02 and moles Cl = 0.06 (1) [AO2]
- ratio 1 : 3 so FeCl3 (1) [AO2]
ALLOW ECF. ALLOW Fe1Cl3. IGNORE charges written on the formula.

Reactants: Al and \(\mathrm{H_2SO_4}\). Products: \(\mathrm{Al_2(SO_4)_3}\) and \(\mathrm{H_2}\). Balance sulfate: 3 \(\mathrm{SO_4}\) on the right so 3 \(\mathrm{H_2SO_4}\) on the left. This gives 6 H on the left so 3 \(\mathrm{H_2}\). 2 Al on the right so 2 Al on the left. Add state symbols: Al solid, acid and salt aqueous, hydrogen gas.

1 mark: correct formulae for all reactants and products (\(\mathrm{Al}\), \(\mathrm{H_2SO_4}\), \(\mathrm{Al_2(SO_4)_3}\), \(\mathrm{H_2}\)) (AO2) ✓
1 mark: equation correctly balanced: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 (AO2) ✓ (ALLOW multiples; balancing only credited if formulae are correct)
1 mark: all four state symbols correct: (s), (aq), (aq), (g) (AO1) ✓ (ALLOW ECF for state symbols on correct species only if formulae are right; DO NOT ALLOW (l) for the acid or salt)

(a) Positive lead ions are attracted to the cathode and gain two electrons (reduction) to form lead atoms: \(\mathrm{Pb^{2+} + 2e^- \rightarrow Pb}\).
(b) Bromide ions are attracted to the anode and lose one electron each (oxidation). Two ions are needed to make a \(\mathrm{Br_2}\) molecule, so: \(\mathrm{2Br^- \rightarrow Br_2 + 2e^-}\).

(a) 1 mark: Pb2+ + 2e- -> Pb (AO2) ✓ (DO NOT ALLOW if electrons are on the wrong side; IGNORE state symbols)
(b) 1 mark: Br2 as the product formed from Br- (AO2) ✓
(b) 1 mark: equation balanced for atoms and charge: 2Br- -> Br2 + 2e- (AO2) ✓ (ALLOW 2Br- - 2e- -> Br2; ALLOW ECF for electrons if the correct species are used; IGNORE state symbols)

Protons = atomic number. Neutrons = mass number − atomic number. For a positive ion, electrons = protons − charge; for a negative ion, electrons = protons + charge.
Mg2+: protons 12, neutrons 24 − 12 = 12, electrons 12 − 2 = 10.
F-: protons 9, neutrons 19 − 9 = 10, electrons 9 + 1 = 10.

1 mark: Mg2+ row fully correct: 12, 12, 10 ✓ (AO2)
1 mark: F- row fully correct: 9, 10, 10 ✓ (AO2)
Award the mark for a row only if all three values are correct.
DO NOT ALLOW 14 electrons for Mg2+ or 8 electrons for F-.

Copper(II) ions give a blue precipitate with sodium hydroxide. A green precipitate shows iron(II) ions. Iron(III) ions give an orange-brown precipitate.

All three entries correct = 2 marks; any two correct = 1 mark; fewer than two = 0 marks. (AO1)
P: blue (precipitate) ✓
Q: iron(II) / Fe2+ ✓
R: (orange-)brown / rust-coloured (precipitate) ✓
IGNORE 'solid' in place of 'precipitate'. DO NOT ALLOW 'blue solution' for P.

Mg(OH)2: \( 24 + 2 \times (16 + 1) = 24 + 34 = 58 \).
(NH4)2SO4: \( 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 36 + 32 + 64 = 132 \).

First check the answers in the table. Correct values (58 and 132) = 2 marks (1 each) ✓✓ (AO2)
If an answer is wrong, award the mark for a row where the working shows correct use of the bracket multiplier (e.g. 2 × 17 or 2 × 18).
No unit is required; IGNORE a unit if given.

X has a high melting point and conducts only when molten because its ions are free to move, so it is a giant ionic lattice. Y has a very low melting point and never conducts because it has no free charged particles and weak intermolecular forces, so it is simple molecular.

1 mark: X = giant ionic lattice ✓ (AO2)
1 mark: Y = simple molecular ✓ (AO2)
DO NOT ALLOW two answers in one row if one is wrong.

1. Data: yield rises as temperature falls (at 200 atm: 18% at 500, 37% at 400, 63% at 300) and as pressure rises (at 400 \(^\circ\mathrm{C}\): 25%, 37%, 47%). The best yield is 71% at 300 \(^\circ\mathrm{C}\) and 300 atm.
2. Temperature: the forward reaction is exothermic, so lowering the temperature moves the equilibrium position to the right (Le Chatelier). But at low temperature particles have less energy, collide less often and fewer collisions exceed the activation energy, so the rate is too slow to be economic. 450 \(^\circ\mathrm{C}\) is a compromise, and the iron catalyst lowers the activation energy so a good rate is achieved without a lower temperature. A catalyst does not change the equilibrium yield.
3. Pressure: there are 4 moles of gas on the left and 2 on the right, so high pressure moves the equilibrium right and also increases the rate (more collisions per second). However, very high pressures need thick-walled pipes and powerful compressors, which cost more to build and run, and are more dangerous. 200 atm gives a reasonable yield at acceptable cost.
4. Overall: the company balances yield, rate and cost. Unreacted nitrogen and hydrogen are recycled so the low yield per pass is not wasteful.

AO1 (recall) 2 marks, AO2 (application) 2 marks, AO3 (analysis/evaluation) 2 marks. Levels of response: the science content decides the level; quality of communication decides the mark within the level.

Level 3 (5-6 marks): Explains BOTH temperature and pressure as compromises, linking equilibrium position AND rate/cost, with at least two correct uses of data from the table. Reasoning is logical, well structured and uses correct terminology. Mentions catalyst role.
Level 2 (3-4 marks): Explains the effect of temperature OR pressure on both yield and rate/cost, or gives partial explanations of both with limited or no data use. Some structure.
Level 1 (1-2 marks): Simple statements, e.g. high pressure or low temperature gives more ammonia, or the conditions are cheaper, with no or little linking to the data. Limited structure.
Level 0 (0 marks): No relevant response.

Indicative content:
- Data: yield increases as temperature decreases and as pressure increases; max 71% at 300 \(^\circ\mathrm{C}\)/300 atm; yield at 400 \(^\circ\mathrm{C}\)/200 atm is 37%, 500 \(^\circ\mathrm{C}\)/200 atm is 18% (ALLOW any correct comparison of values).
- Forward reaction exothermic, so low temperature moves equilibrium right (ALLOW favours products).
- Low temperature gives slow rate (fewer/less energetic collisions, fewer successful collisions); so a higher temperature (~450 \(^\circ\mathrm{C}\)) is a compromise.
- Iron catalyst speeds up the reaction / lowers activation energy, allowing a reasonable rate; does not change the equilibrium position/yield (ALLOW still reaches equilibrium faster).
- 4 moles of gas to 2 moles of gas, so high pressure moves equilibrium right (fewer gas molecules); also increases rate.
- Very high pressure is expensive (equipment, energy for compression) and a safety hazard, so 200 atm is a compromise.
- Unreacted gases recycled (ALLOW to improve overall yield).
IGNORE references to the cost of the catalyst alone. DO NOT ALLOW 'catalyst increases yield'.