2025 OCR GCSE Twenty First Century Science - Chemistry B - J258 模擬試題連答案詳解

Rutherford's scattering results showed a small, dense, positive nucleus (true). An electron's relative mass is very small, about 1/2000, not 1 (false). Thomson's 'plum pudding' model matches the description (true). Neutrons are neutral (false).

All four rows correct = 2 marks (✓✓). Three rows correct = 1 mark. Two or fewer correct = 0 marks. Correct answers: row 1 True; row 2 False; row 3 True; row 4 False. AO1.1. Sliding scale: one fewer correct = one mark fewer. Tick both boxes in a row = row not credited.

Reactivity increases down Group 1, so potassium is more reactive than lithium (A) and sodium is more reactive than lithium (B false). Group 1 metals react with water to give hydrogen and a metal hydroxide (C); the hydroxide solution is alkaline (D false). Group 1 metals are soft (E false).

A ✓ (1)
C ✓ (1)
If more than two boxes are ticked, 1 mark is deducted for each extra tick (minimum 0). AO1.1.

Melting requires overcoming strong electrostatic attraction between ions (C). In the solid the ions cannot move so no current flows (B). When molten the ions are free to move (A).

1-C, 2-B, 3-A.
All three correct = 2 marks; two correct = 1 mark; one or none correct = 0 marks. More than one line from a property = not credited. D is a distractor. AO1.1.

A larger surface area exposes more particles so collisions are more frequent (true). Lower concentration means fewer acid particles per unit volume, so fewer collisions (false). Higher temperature gives particles more energy, so a greater proportion of collisions are successful (true). A catalyst changes rate, not the final amount of product (false).

All four rows correct = 2 marks; three correct = 1 mark; two or fewer = 0 marks. Answers: True, False, True, False. AO1.1 / AO2.1.

The temperature fell, so energy was taken in from the surroundings: the reaction is endothermic (B, C). In an endothermic reaction the products have more energy than the reactants, and more energy is needed to break bonds than is released on forming bonds, so D and E are wrong.

B ✓ (1)
C ✓ (1)
If more than two boxes are ticked, deduct 1 mark per extra tick (minimum 0). AO2.1.

Filtration removes insoluble solids, chlorine kills microorganisms, and distillation leaves dissolved salt behind while pure water evaporates and is condensed.

1-C, 2-A, 3-B.
All three correct = 2 marks; two correct = 1 mark; otherwise 0. D is a distractor. AO1.1.

CO2 is a greenhouse gas that absorbs infrared (true). A correlation does not prove causation without a mechanism and control of other factors (false). Fossil fuel combustion releases CO2 (true). Growing trees take in CO2 by photosynthesis so reduce it (false).

All four correct = 2 marks; three correct = 1 mark; two or fewer = 0. Answers: True, False, True, False. AO1.1 / AO3.1a.

Nanoparticles are 1–100 nm across (A). Their surface area to volume ratio is large (B false), so smaller quantities are effective (C). Long-term risks are still being studied (D false) and they are too small to see (E false).

A ✓ (1)
C ✓ (1)
If more than two boxes are ticked, deduct 1 mark per extra tick (minimum 0). AO1.1.

Reactivity of Group 1 increases down the group, so potassium is more reactive than lithium (true). Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide (true). Group 7 atoms gain one electron to form 1- ions (false). Boiling points of Group 7 elements increase down the group because of stronger forces between larger molecules (false).

Rows in order: True, True, False, False.
All four correct = 2 marks (AO1.1).
Three correct = 1 mark.
Two or fewer correct = 0 marks.
If more than one box is ticked in a row, that row scores 0.
Sliding scale: all correct = full marks, one fewer = one mark fewer.

Moles = mass / M\(_r\) = 5.3 / 106 = 0.05 mol.

Correct method: moles = mass / M\(_r\) (5.3 / 106) ✓ (AO2.1)
Answer 0.05 (mol) ✓ (AO2.1)
ALLOW 5 x 10^-2. Full marks for correct answer without working. ALLOW 1 mark for 106 / 5.3 = 20 (inverted method) only if method is clearly shown.

Chlorine is above bromine in Group 7, so it is more reactive. It displaces bromine from the bromide solution. Bromine in solution is orange.

Chlorine is more reactive than bromine / chlorine displaces bromine (from potassium bromide) ✓ (AO2.1)
Bromine / Br\(_2\) is formed (which is orange) ✓ (AO2.1)
ALLOW chlorine is higher up the group. ALLOW equation: Cl\(_2\) + 2KBr → 2KCl + Br\(_2\). IGNORE 'bromide turns orange' and 'bromine is more reactive'.

Powder has a larger surface area than the same mass of chips, so more marble particles are exposed to the acid. This gives more frequent successful collisions per second.

Powder has a larger surface area (than the chips) / more surface exposed ✓ (AO1.1)
More frequent collisions (between particles) / more collisions per second ✓ (AO1.1)
ALLOW greater chance of collisions. IGNORE 'more collisions' without reference to frequency or time unless clear. DO NOT ALLOW 'more energy' / 'particles move faster'.

Energy to break bonds = 436 + 243 = 679 kJ/mol. Energy released making bonds = 2 x 432 = 864 kJ/mol. Energy change = 679 - 864 = -185 kJ/mol (exothermic).

Energy in = 679 and energy out = 864 (both correct) ✓ (AO2.1)
Answer -185 (kJ/mol) ✓ (AO2.1)
ALLOW 185 released / exothermic for the final mark if the sign is missing. ECF for one slip in the first mark. ALLOW 1 mark for +185.

The Earth emits infrared radiation. Greenhouse gases absorb it and re-emit some back to Earth, so more thermal energy is retained and the temperature rises.

Greenhouse gases absorb (infrared / long-wave / thermal) radiation (from the Earth's surface) ✓ (AO1.1)
Re-emit / radiate it back towards the Earth so less escapes into space / more energy is retained ✓ (AO1.1)
ALLOW 'trap heat' for 1 mark only. IGNORE 'holes in the ozone layer'. DO NOT ALLOW 'reflects sunlight'.

Positive lead ions move to the cathode and gain electrons to form lead. Negative bromide ions move to the anode and lose electrons to form bromine.

Cathode: lead / Pb ✓ (AO1.1)
Anode: bromine / Br\(_2\) ✓ (AO1.1)
ALLOW bromine gas / vapour. DO NOT ALLOW bromide / Br. DO NOT ALLOW if the products are reversed.

(a) Chlorine is a disinfectant that kills harmful microorganisms. (b) Damp litmus paper is bleached white in chlorine.

(a) To kill bacteria / microorganisms / pathogens / disinfect ✓ (AO1.1)
(b) Damp (blue) litmus paper turns white / is bleached ✓ (AO1.1)
ALLOW 'to sterilise / make safe to drink' for (a). For (b) the result needs 'damp' and 'bleached/white'. IGNORE 'turns red' alone. DO NOT ALLOW 'dry' litmus.

Surface area = 6 x 5 x 5 = 150 nm\(^2\). Volume = 5 x 5 x 5 = 125 nm\(^3\). Ratio = 150 / 125 = 1.2, so 1.2 : 1.

Surface area = 150 and volume = 125 (both correct) ✓ (AO2.1)
1.2 (: 1) ✓ (AO2.1)
ALLOW 150 : 125 or 6 : 5 for 1 mark only. ECF from an incorrect surface area or volume.

Relative formula mass of Na\(_2\)CO\(_3\) = (2 × 23) + 12 + (3 × 16) = 106. Moles = mass ÷ M\(_r\) = 5.3 ÷ 106 = 0.05 mol.

M1: M\(_r\) = 106 (or moles = mass ÷ M\(_r\) with a correct M\(_r\)) ✓
A1: 0.05 (mol) ✓
ALLOW 5 × 10\(^{-2}\).
ECF for one error in M\(_r\) for 1 mark (method).
If the answer is correct, award 2 marks.
AO2.1

Chlorine is higher in Group 7 and so more reactive than bromine. It displaces bromine, which is orange in solution. Chlorine + potassium bromide → potassium chloride + bromine.

Any two from:
chlorine is more reactive than bromine / chlorine displaces bromine (from the bromide) ✓
bromine is formed (which is orange) ✓
correct word equation: chlorine + potassium bromide → potassium chloride + bromine ✓
DO NOT ALLOW 'bromine is displaced' without reference to chlorine being more reactive for the first point.
AO2.1

A higher concentration means more acid particles in each unit volume. Collisions with magnesium atoms are therefore more frequent, so the rate increases.

more particles (of acid) in the same volume / particles closer together ✓
more frequent (successful) collisions (per second) ✓
IGNORE 'more collisions' without reference to frequency/time.
DO NOT ALLOW 'collisions have more energy'.
AO1.1, AO2.1

Energy is transferred to the surroundings because the products have less energy than the reactants, so the reaction is exothermic.

exothermic ✓
products are lower in energy than reactants / energy is given out (to the surroundings) / temperature of surroundings rises ✓
Second mark can only be awarded if the first is correct.
AO1.1, AO2.1

Short-wavelength radiation from the Sun passes through the atmosphere and warms the surface. The surface emits infrared radiation, which greenhouse gases absorb and re-emit, partly back towards the Earth.

absorb (infrared) radiation emitted/radiated by the Earth's surface ✓
(re-)emit/radiate (some of) it back towards the Earth / heat is trapped ✓
ALLOW 'heat' for infrared radiation.
DO NOT ALLOW 'reflect' for the first point.
IGNORE references to the ozone layer.
AO1.1

Positive lead ions move to the cathode and gain electrons to form lead. Negative bromide ions move to the anode and lose electrons to form bromine.

cathode: lead ✓
anode: bromine ✓
ALLOW Pb and Br\(_2\).
DO NOT ALLOW bromide or lead ions.
AO1.1

Filtration removes insoluble solids such as sand and debris. Chlorine kills harmful microorganisms.

filtration: removes (insoluble) solids / suspended particles / debris ✓
chlorination: kills microorganisms / bacteria / pathogens ✓
ALLOW 'germs' for microorganisms.
IGNORE 'cleans the water'.
AO1.1

Surface area = 6 × 5 × 5 = 150 nm\(^2\). Volume = 5 × 5 × 5 = 125 nm\(^3\). Ratio = 150 ÷ 125 = 1.2 nm\(^{-1}\), i.e. 1.2 : 1.

M1: surface area 150 nm\(^2\) and volume 125 nm\(^3\) (or 150/125) ✓
A1: 1.2 (per nm / nm\(^{-1}\) / 1.2 : 1) ✓
If the answer is correct, award 2 marks.
ALLOW ECF for one arithmetic slip.
AO2.1 (M5)

In solid sodium chloride the ions are held in fixed positions in the lattice. When molten the lattice breaks down and the ions can move, carrying the electric current.

in the solid, ions are fixed in position / cannot move ✓
when molten, ions are free to move (and carry charge) ✓
DO NOT ALLOW 'electrons move' or 'atoms move'.
AO1.1, AO2.1

Powdering the marble increases the surface area exposed to the acid. More particles of acid can reach the marble at once, so the frequency of successful collisions increases and the rate increases.

Powder has a larger surface area (than chips) (AO1.1) [1]
more frequent collisions / more collisions per second between acid particles and marble particles (AO2.1) [1]
ALLOW greater surface area exposed to acid.
DO NOT ALLOW 'more collisions' with no reference to time or frequency AND 'particles have more energy'.
IGNORE 'more particles' (the mass is the same).

moles = mass / \(M_r\) = \(5.3 \div 106\) = 0.05 mol

moles = mass / \(M_r\) or \(5.3 \div 106\) (AO2.1) [1]
0.05 / 5.0 \(\times\) 10\(^{-2}\) (mol) (AO2.1) [1]
ALLOW 1 mark for \(106 \div 5.3\) = 20 (inverted).
Correct answer scores 2 marks.

Chlorine is higher in Group 7 than bromine, so it is more reactive and displaces bromine from its salt. Orange colour is bromine. Potassium chloride is the other product.

(a) chlorine is more reactive than bromine / chlorine displaces bromine (AO1.1) [1]
ALLOW bromine is less reactive than chlorine.
DO NOT ALLOW 'chlorine is more reactive' with no comparison.
(b) chlorine + potassium bromide → potassium chloride + bromine (AO2.1) [1]
ALLOW correct balanced symbol equation Cl\(_2\) + 2KBr → 2KCl + Br\(_2\).

Surface area = \(6 \times 5^2\) = 150 nm\(^2\). Volume = \(5^3\) = 125 nm\(^3\). Ratio = \(150 \div 125\) = 1.2, so 1.2 : 1.

surface area = 150 AND volume = 125 (AO2.1) [1]
1.2 (: 1) (AO2.1) [1]
ALLOW ECF for one correct calculation used in the ratio.
ALLOW 6 : 5 for 1 mark only if not in the form x : 1.
Correct answer scores 2 marks.

Moles of Ca = 10.0 / 40 = 0.25 mol.
From the equation 2 mol Ca gives 2 mol CaO, so moles of CaO = 0.25 mol.
Mr of CaO = 40 + 16 = 56.
Mass of CaO = 0.25 x 56 = 14.0 g.

Moles of Ca = 10.0 / 40 = 0.25 (mol) [1] (AO2.1)
Moles CaO = moles Ca (1:1 ratio) AND Mr of CaO = 56 [1] (AO2.1)
Mass = 0.25 x 56 = 14.0 (g) [1] (AO2.1)
ALLOW 14 g. ECF from an incorrect number of moles or Mr. Correct answer scores 3 marks without working.

Moles of HCl = 0.100 x 20.0 / 1000 = 0.00200 mol.
The ratio NaOH : HCl is 1 : 1, so moles NaOH = 0.00200 mol in 25.0 cm\(^3\).
Concentration = 0.00200 / 0.0250 = 0.0800 mol/dm\(^3\).
In g/dm\(^3\): 0.0800 x 40 = 3.2 g/dm\(^3\).

Moles HCl = 0.100 x 0.0200 = 0.00200 mol (and equals moles NaOH) [1] (AO2.1)
Concentration in mol/dm3 = 0.00200 / 0.0250 = 0.0800 [1] (AO2.1)
x 40 = 3.2 g/dm3 [1] (AO2.1)
ALLOW 3.20. ECF between steps. Correct answer scores 3 marks without working.

Particle A: surface area = 6 x 16 = 96 nm\(^2\); volume = 64 nm\(^3\); ratio = 96 / 64 = 1.5.
Particle B: surface area = 6 x 1600 = 9600 nm\(^2\); volume = 64000 nm\(^3\); ratio = 9600 / 64000 = 0.15.
Ratio A / ratio B = 1.5 / 0.15 = 10.

Ratio for A = 96 / 64 = 1.5 (surface area 96 AND volume 64 seen) [1] (AO2.1)
Ratio for B = 9600 / 64000 = 0.15 [1] (AO2.1)
1.5 / 0.15 = 10 (times) [1] (AO2.1)
ALLOW 1.5:1 and 0.15:1 forms. ECF. Correct answer scores 3 marks without working.

Energy to break bonds = 436 + 243 = 679 kJ/mol.
Energy released making bonds = 2 x 432 = 864 kJ/mol.
Energy change = 679 - 864 = -185 kJ/mol (exothermic).

Bonds broken = 436 + 243 = 679 [1] (AO2.1)
Bonds made = 2 x 432 = 864 [1] (AO2.1)
Energy change = 679 - 864 = -185 (kJ/mol) [1] (AO2.1)
DO NOT ALLOW +185 for the final mark, but ALLOW 185 released / exothermic. ECF from one arithmetic slip. Correct answer scores 3 marks without working.

Al2O3 contains 2 Al and 3 O. Two Al2O3 need 4 Al and 6 O atoms, so 3 O2 molecules are needed. Equation: \( 4\mathrm{Al} + 3\mathrm{O_2} \rightarrow 2\mathrm{Al_2O_3} \).

4 (Al) and 3 (O2) and 2 (Al2O3) all correct = 2 marks (AO2.1)
Two of the three numbers correct = 1 mark
ALLOW multiples e.g. 8, 6, 4 for 1 mark only if all three correct (scores 2). DO NOT ALLOW changes to the formulae.

Mg2+ has lost 2 electrons: 12 − 2 = 10. Oxygen has 8 protons (it has 10 electrons after gaining 2, so 8 protons). Neutrons = 16 − 8 = 8.

All three correct (10, 8, 8) = 2 marks
Two correct = 1 mark
One or none correct = 0 marks (AO2.1)

Bonds formed: 2 × 432 = 864 kJ/mol. Overall change = energy in − energy out = 679 − 864 = −185 kJ/mol (exothermic).

864 ✓
−185 ✓ (ECF from an incorrect first answer: 679 − their value)
ALLOW 185 kJ/mol released / exothermic. IGNORE a missing minus sign if 'released' or 'exothermic' is stated (AO2.1)

Pb2+ has a 2+ charge so gains 2 electrons to become neutral. Two Br− carry a total charge of 2−, so 2 electrons are lost.

2 in the first box ✓
2 in the second box ✓
DO NOT ALLOW 1 or 4 (AO2.1)

X: 2.0 ÷ 8.0 = 0.25. Y: 6.0 ÷ 8.0 = 0.75.

0.25 for X ✓
0.75 for Y ✓
ALLOW 0.3 for X and 0.8 for Y (rounded). ECF not applicable. (AO2.1)

Moles = 5.0 ÷ 100 = 0.050 mol. Mass = moles × Mr = 0.20 × 100 = 20 g.

0.05 / 0.050 ✓
20 / 20.0 ✓
ALLOW 5.0 × 10^-2 for the first answer. (AO2.1)

Curve B is steeper, so the reaction is faster: this is Experiment 2. Zinc powder has a larger surface area than granules, so more zinc particles are exposed to acid and collisions are more frequent. At 40 °C the particles move faster and have more energy, so collisions are more frequent and a greater proportion of collisions have energy greater than the activation energy, so more collisions are successful per second. Both curves level off at 400 cm\(^3\) because the same mass of zinc (the limiting reactant) was used, so the same amount of hydrogen is made; the acid is in excess so the temperature and surface area do not change the total amount of product. To check repeatability, repeat each experiment (at least three times) under the same conditions and compare the volumes/curves, or calculate a mean.

Level 3 (5-6 marks): Correctly identifies curve B as Experiment 2 AND gives a clear collision-theory explanation covering both surface area and temperature (frequency of collisions and energy / activation energy), AND explains the equal final volume by the same limiting reactant/mass of zinc, AND describes a valid repeatability check. Well-structured, logical, correct scientific terms. (AO2.1, AO3.2b)
Level 2 (3-4 marks): Identifies curve B and explains the rate difference using collisions for at least one factor, AND makes a correct attempt at one of the other two bullets. Some structure.
Level 1 (1-2 marks): Some relevant points, e.g. curve B is faster, or powder has larger surface area, or same amount of zinc. Limited or no use of collision theory. Disorganised.
Level 0 (0 marks): No relevant content.

Communication statement: the better-structured and more accurately worded answer in a level gets the higher mark of the band.

Indicative content:
- Curve B = Experiment 2 (steeper / reaches maximum sooner)
- powder has larger surface area / more particles exposed ORA
- higher temperature: particles move faster / have more energy
- more frequent collisions
- greater proportion of collisions have energy above the activation energy / more successful collisions per second
- same mass/amount of zinc, zinc is the limiting reactant, so same amount/volume of hydrogen
- acid in excess
- repeat the experiment (at least 3 times) with the same conditions and calculate a mean / compare the results

ALLOW AW throughout. IGNORE 'particles collide harder' alone. DO NOT ALLOW 'particles expand' or 'activation energy is lowered' for temperature/surface area.

River water: pass through filter beds to remove suspended solids (the cloudiness), then add chlorine to kill bacteria. The low dissolved salt content (0.4 g/dm\(^3\)) means no desalination is needed. Sea water: contains 35 g/dm\(^3\) dissolved salt, about 87 times more than the river water (35 / 0.4 = 87.5), so it must be distilled: heated until the water evaporates, then the vapour is condensed and collected, leaving the salts behind. Heating a large volume of water to boiling and providing the latent heat to evaporate it needs a large amount of energy, so sea water is more costly to treat. Filtration and chlorination need much less energy.

Level 3 (5-6 marks): Gives correct stages for river water (filtration AND chlorination) with purposes, correctly describes distillation for sea water, AND makes a justified judgement that sea water uses more energy, using data from Table 2.1 and energy ideas. Clear, logical, correct terminology. (AO1.1, AO2.1, AO3.2a)
Level 2 (3-4 marks): Describes at least two of the correct treatment stages with some purposes, and attempts a comparison with some use of data or energy. Mostly clear.
Level 1 (1-2 marks): Some relevant statements, e.g. filter removes solids or chlorine kills bacteria or distillation removes salt. No comparison or judgement. Limited structure.
Level 0 (0 marks): No relevant content.

Communication statement: the better-structured and more accurately worded answer in a level gets the higher mark of the band.

Indicative content:
- river water: filtration (through sand/gravel beds) removes suspended solids
- chlorination kills bacteria/microbes
- (ALLOW sedimentation/aeration if correctly described)
- little dissolved salt in river water (0.4 g/dm3) so no distillation needed
- sea water: distillation, heat to evaporate/boil, condense vapour, salt left behind
- sea water has high salt (35 g/dm3), far more than river water, so filtration/chlorination alone is not enough
- distillation needs large amount of (heat) energy / fuel to boil water
- so sea water is more costly/energy-demanding to treat

ALLOW AW. IGNORE 'filters out bacteria' unless membrane filtration is specified. DO NOT ALLOW 'chlorine removes salt' or 'distillation kills bacteria' as the only purpose.

Mr of ZnO = 65 + 16 = 81. Mr of C = 12. Total mass of reactants = 81 + 12 = 93. Mass of desired product (Zn) = 65.
Atom economy = \( \frac{65}{93} \times 100 = 69.9\% \).

First check the answer on the answer line: 69.9 (%) = 2 marks.
If the answer is incorrect, award 1 mark for a correct method: (65 / 93) x 100 / correct sum of reactant Mr values (93) seen.
ALLOW 69.89 / 70 (%) for full marks. ALLOW ECF from an incorrect Mr for one mark.
DO NOT ALLOW 65/81 (= 80.2%) for the final mark.
AO2.1 x1, AO2.2 x1

Percentage yield = \( \frac{\text{actual mass}}{\text{maximum (theoretical) mass}} \times 100 = \frac{6.4}{8.0} \times 100 = 80\% \).

First check the answer on the answer line: 80 (%) = 2 marks.
If the answer is incorrect, award 1 mark for (6.4 / 8.0) x 100 / 0.8 seen.
DO NOT ALLOW 125 (inverted fraction).
DO NOT ALLOW 6.4/4.8 x 100.
AO2.1 x2

Mr of CaCO3 = 40 + 12 + 48 = 100. Moles of CaCO3 = 25.0 / 100 = 0.250 mol. Ratio CaCO3 : CO2 is 1 : 1, so 0.250 mol of CO2. Mr of CO2 = 44. Mass = 0.250 x 44 = 11.0 g.

First check the answer on the answer line: 11 / 11.0 (g) = 2 marks.
If the answer is incorrect, award 1 mark for moles of CaCO3 = 0.25 (mol) OR for 25.0 / 100 x 44 seen with one error in the final step.
ALLOW ECF from an incorrect Mr of CaCO3 or CO2.
AO2.1 x2

Mr of NaOH = 23 + 16 + 1 = 40. Moles = 4.0 / 40 = 0.10 mol. Volume = 250 / 1000 = 0.25 dm3. Concentration = 0.10 / 0.25 = 0.40 mol/dm3.

First check the answer on the answer line: 0.4 / 0.40 (mol/dm3) = 2 marks.
If the answer is incorrect, award 1 mark for moles = 0.10 (mol) seen OR for volume converted to 0.25 dm3 with correct use of the equation.
ALLOW ECF from an incorrect Mr.
DO NOT ALLOW 100 or 0.0004 (failure to convert units) for the final mark.
AO2.1 x2

Moles of NaOH = 0.20 x 25.0 / 1000 = 0.0050 mol. The ratio NaOH : HCl is 1 : 1, so moles of HCl = 0.0050 mol. Concentration of HCl = 0.0050 / (20.0 / 1000) = 0.25 mol/dm3.

First check the answer on the answer line: 0.25 (mol/dm3) = 2 marks.
If the answer is incorrect, award 1 mark for moles of NaOH = 0.005 / 5.0 x 10^-3 (mol) seen.
ALLOW ECF from an incorrect number of moles.
DO NOT ALLOW 0.16 (from 0.2 x 20/25) for the final mark.
AO2.1 x1, AO2.2 x1

Moles Mg = 0.48 / 24 = 0.02 mol. The equation shows 1 mol Mg gives 1 mol MgCl2, so 0.02 mol MgCl2. Mr of MgCl2 = 24 + 2 x 35.5 = 95. Mass = 0.02 x 95 = 1.9 g.

Moles of Mg = 0.48 / 24 = 0.02 [1] (AO2.1)
Mr of MgCl2 = 95 [0.5] (AO1.2)
Mass = 0.02 x 95 = 1.9 g [1] (ECF from moles and Mr)
Correct answer with no working = 2.5 marks. ALLOW 1.90 g. IGNORE missing unit only if g is shown on the answer line. DO NOT ALLOW use of Mr = 59.5 or 24 + 35.5.

(i) 36 / 30 = 1.2 cm3/s. (ii) Between 60 s and 90 s only 12 cm3 forms in 30 s (0.4 cm3/s). The reactants are being used up, so the concentration of acid is lower and the marble surface is smaller; particles collide less often, so fewer successful collisions per second.

(i) 36 / 30 = 1.2 [0.5] cm3/s [0.5] (AO2.1). ALLOW cm3 per second / cm3 s-1.
(ii) Any three from (0.5 each, max 1.5) (AO2.2): reactants / acid being used up / concentration of acid decreases; fewer acid particles in the same volume; fewer collisions per second / lower frequency of collisions; fewer successful collisions per second.
DO NOT ALLOW 'collisions are less energetic' or 'particles slow down'. IGNORE 'less chance of reaction' alone.

(i) Boiling points increase down the group, so astatine has a boiling point above 184 °C and is a solid. (ii) Chlorine is more reactive than iodine because it is higher in the group (its atoms gain an electron more easily), so it displaces iodine from its salt, forming potassium chloride and iodine (solution turns brown).

(i) solid [0.5] (AO2.2). ALLOW solid with any colour. DO NOT ALLOW liquid or gas.
(ii) chlorine + potassium iodide -> potassium chloride + iodine [1] (AO1.2)
chlorine is more reactive than iodine / reactivity decreases down the group [0.5]; so chlorine displaces iodine (from the solution) [0.5] (AO2.1)
ALLOW 'iodine is less reactive'. DO NOT ALLOW 'chlorine is heavier'.

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

Bonds broken 436 + 243 = 679 [0.5]
Bonds made 2 x 432 = 864 [0.5]
Overall change = 679 - 864 = -185 kJ/mol [1] (AO2.1; ECF). ALLOW 185 if exothermic is clearly stated.
Exothermic because more energy is released in making bonds than is needed to break bonds [0.5] (AO1.1).

(i) Pb2+ ions are attracted to the negative electrode and gain electrons to form lead; Br- ions go to the positive electrode and lose electrons to form bromine. (ii) In the solid, ions are held in a rigid lattice by strong electrostatic forces and cannot move. When molten the ions are free to move, so charge can flow.

(i) lead [0.5] at negative; bromine [0.5] at positive (AO1.1). DO NOT ALLOW bromide / lead ions.
(ii) solid: ions fixed in position / cannot move [0.5]; molten: ions are free to move [0.5]; ions carry the charge / current [0.5] (AO1.2). IGNORE 'electrons move in the molten state'.

(i) Surface area = 6 x 10 x 10 = 600 nm2. Volume = 10 x 10 x 10 = 1000 nm3. Ratio = 600 / 1000 = 0.6 nm-1. (ii) Smaller particles have a larger surface area for the same mass, so more of the catalyst is exposed and more reactant particles can collide with its surface each second, speeding up the reaction.

(i) surface area 6 x 10^2 = 600 and volume 10^3 = 1000 [0.5]; 600/1000 = 0.6 [0.5] (AO2.1)
(ii) larger surface area (to volume ratio / for same mass) [0.5]; more catalyst atoms exposed / available to reactants [0.5]; more frequent collisions / reaction faster [0.5] (AO1.2, AO2.2)
DO NOT ALLOW 'bigger particles' or 'more surface area' with no link to mass or volume.

Moles HCl = 0.100 x 20.0 / 1000 = 0.00200 mol. The ratio NaOH : HCl is 1 : 1, so moles NaOH = 0.00200 mol in 25.0 cm3 (0.0250 dm3). Concentration = 0.00200 / 0.0250 = 0.0800 mol/dm3.

Moles HCl = 0.100 x 0.0200 = 0.00200 [1] (AO2.1)
Moles NaOH = 0.00200 (1:1 ratio) [0.5]
Concentration = 0.00200 / 0.0250 [0.5] = 0.0800 mol/dm3 [0.5]
Correct answer with no working = 2.5 marks. ALLOW 0.08. ECF from earlier steps.

(i) As carbon dioxide concentration rises, so does the temperature change. (ii) Scientists need a plausible mechanism: carbon dioxide absorbs infrared radiation emitted from the Earth and re-emits it, warming the atmosphere. They also need to show that other factors such as solar activity are not responsible, and have the data checked by other scientists (peer review).

(i) positive correlation / both increase together [0.5] (AO3.1a)
(ii) Any two x 1 mark: a mechanism is needed / carbon dioxide absorbs infrared radiation (traps heat) [1]; other variables (e.g. solar activity, volcanoes) shown not to explain the rise [1]; data/conclusions checked by other scientists / peer review / more data from other sources [1] (AO3.2a)
IGNORE 'more data' with no detail. DO NOT ALLOW 'carbon dioxide makes a hole in the ozone layer'.

Rate 0-20 s = 24 / 20 = 1.2 cm\(^3\)/s. Rate 40-60 s = (48 - 40) / 20 = 0.4 cm\(^3\)/s. The rate is lower later because hydrochloric acid is used up, so its concentration falls, so there are fewer particle collisions per second and fewer successful collisions.

0.5 mark: 1.2 cm\(^3\)/s (0-20 s). 0.5 mark: 0.4 cm\(^3\)/s (40-60 s); ECF allowed. 0.5 mark: rate is lower / slower later. 1 mark: acid concentration decreases (acid used up) AND fewer collisions per second / lower frequency of collisions. ALLOW AW. IGNORE 'less magnesium' (magnesium is in excess). AO2.1, AO3.1a.

Mass of oxygen = 0.80 - 0.48 = 0.32 g. Moles Mg = 0.48 / 24 = 0.02 mol. Moles O = 0.32 / 16 = 0.02 mol. Ratio Mg : O = 1 : 1, so the formula is MgO.

0.5 mark: mass of oxygen = 0.32 g. 1 mark: moles Mg = 0.02 AND moles O = 0.02 (0.5 each). 1 mark: ratio 1 : 1 therefore MgO. ECF allowed from the oxygen mass. AO2.1, AO2.2.

Rf = distance moved by dye / distance moved by solvent = 5.6 / 8.0 = 0.70. Q is more soluble in the solvent (less strongly attracted to the paper) than P, so it is carried further. Two separate spots show the ink contains at least two dyes, so it is a mixture.

1 mark: 0.70 (ALLOW 0.7; 0.5 mark for 5.6 / 8.0 shown with wrong answer). 1 mark: Q is more soluble in the solvent / less strongly attracted to the paper (ORA). 0.5 mark: ink is a mixture / contains more than one dye. AO1.2, AO2.1.

The melting points decrease as the group is descended (181 to 39 \(^\circ\)C). Caesium is below rubidium so its melting point should be lower than 39 \(^\circ\)C, for example 30 \(^\circ\)C, as the decrease continues down the group. The differences between values are getting smaller, so a small drop is expected.

1 mark: melting point decreases down the group / as atomic number increases. 0.5 mark: prediction lower than 39 \(^\circ\)C but above 0 \(^\circ\)C (ALLOW 20-35). 1 mark: reason based on the trend continuing / caesium is below rubidium in the group / differences between values are getting smaller. AO3.1a, AO2.1.

Lead ions gain electrons at the negative electrode: Pb\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Pb. Bromide ions lose electrons at the positive electrode: 2Br\(^-\) \(\rightarrow\) Br\(_2\) + 2e\(^-\). In the solid the ions are held in a lattice and cannot move; when molten they are free to move and carry charge to the electrodes.

1 mark: Pb\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Pb. 1 mark: 2Br\(^-\) \(\rightarrow\) Br\(_2\) + 2e\(^-\). 0.5 mark: ions are free to move / mobile in the molten state (ORA for solid). ALLOW e for e\(^-\); ALLOW Br\(^-\) \(\rightarrow\) 1/2Br\(_2\) + e\(^-\). AO1.2, AO2.1.

A: 6 / 100 = 0.06 nm\(^{-1}\). B: 6 / 10 = 0.6 nm\(^{-1}\). B's ratio is ten times larger, so for the same mass more atoms are exposed on the surface where reactant particles can collide and react, so the reaction is faster.

0.5 mark: A = 0.06. 0.5 mark: B = 0.6 (units nm\(^{-1}\) or 1:... accepted; ALLOW ratio 0.06:1 and 0.6:1). 0.5 mark: B has the larger ratio / ten times larger. 1 mark: more surface exposed / more particles can collide with the catalyst at once, so faster rate. AO2.1, AO2.2.

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

0.5 mark: 679 (bonds broken). 0.5 mark: 864 (bonds made). 1 mark: -185 kJ/mol (ALLOW 185 with exothermic stated; 0.5 mark for +185 by ECF). 0.5 mark: exothermic because energy released in bond making is greater than energy taken in to break bonds / overall energy change is negative. AO2.1, AO2.2.

Mean titre = (19.9 + 20.1 + 20.0) / 3 = 20.0 cm\(^3\). Moles NaOH = 0.100 x 25.0 / 1000 = 0.00250 mol. Ratio 1 : 1 so moles HCl = 0.00250 mol. Concentration = 0.00250 / (20.0 / 1000) = 0.125 mol/dm\(^3\).

0.5 mark: mean titre 20.0 cm\(^3\). 1 mark: moles NaOH = 0.00250 mol. 0.5 mark: moles HCl = 0.00250 (1:1 ratio). 0.5 mark: 0.125 mol/dm\(^3\). ECF allowed throughout. AO2.1, AO2.2.

Na\(^+\) has lost one electron: 11 - 1 = 10 electrons. Fluorine neutrons = mass number - atomic number = 19 - 9 = 10. F\(^-\) has gained one electron: 9 + 1 = 10 electrons.

Na+ electrons = 10 (0.5) AO1.1
F- neutrons = 10 (0.5) AO2.1
F- electrons = 10 (0.5) AO2.1
Sliding scale: one mark fewer for each wrong box (0.5 per box). ALLOW ECF is not needed as boxes are independent.

The temperature rise shows an exothermic reaction. In an exothermic reaction the products have less stored energy than the reactants because energy is released to the surroundings.

Row 1 True (0.5); Row 2 False (0.5); Row 3 True (0.5). AO1.1/AO2.1
Sliding scale: all correct = 1.5; each error loses 0.5. IGNORE rows with both boxes ticked (0).

A more reactive halogen displaces a less reactive halide. Reactivity decreases down Group 7: chlorine > bromine > iodine. Chlorine displaces bromide; bromine cannot displace chloride; iodine cannot displace bromide.

Row 1 reaction (0.5); Row 2 no reaction (0.5); Row 3 no reaction (0.5). AO2.1
Sliding scale: one mark fewer for each error. Two ticks in a row = 0 for that row.

Powder has a larger surface area so more frequent collisions. Higher temperature gives particles more energy. A catalyst lowers the activation energy.

1 more collisions per second (0.5); 2 particles with more energy (0.5); 3 lower activation energy (0.5). AO1.1
ALLOW more frequent collisions for 1. Sliding scale: 0.5 per correct gap.

A: 2.0 / 8.0 = 0.25. B: 4.8 / 8.0 = 0.60. C: 6.0 / 8.0 = 0.75.

A = 0.25 (0.5); B = 0.60 / 0.6 (0.5); C = 0.75 (0.5). AO2.1
ALLOW fractions equivalent, e.g. 1/4, 3/5, 3/4. IGNORE units.

Pb\(^{2+}\) ions are attracted to the negative electrode and form lead. Br\(^-\) ions go to the positive electrode and form bromine. In the solid lattice the ions cannot move, so no charge flows.

lead (0.5); bromine (0.5); fixed in position (0.5). AO1.1/AO2.1
DO NOT ALLOW bromide as the product at the positive electrode.

Chlorine kills microorganisms. Distillation separates water from dissolved salt. Filtration removes insoluble solids.

B (0.5); C (0.5); A (0.5). AO1.1
Sliding scale: 0.5 per correct row.

For 1.0 cm: 6 / 1 = 6. For 0.5 cm: surface area = 6 x 0.5 x 0.5 = 1.5 cm\(^2\); ratio = 1.5 / 0.125 = 12.

6 (0.5); 1.5 (0.5); 12 (0.5). AO2.1 / M5
ALLOW ECF for the ratio if the surface area of the 0.5 cm cube is wrong but divided correctly by 0.125.