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Thinka Jun 2025 (V3) Cambridge International A Level-Style Mock — Science - Combined (0653)

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An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge International A Level Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.

Paper 23 (Multiple Choice)

Answer all forty questions on the multiple choice answer sheet.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · Multiple Choice
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Four potato cylinders of equal initial length are placed in four different test-tubes, each containing a different concentration of sucrose solution. After two hours, the change in length of each cylinder is measured. Which concentration of sucrose solution is closest to the concentration of the cell sap inside the potato cells?
  1. A.the solution where the potato cylinder increased in length by 2.0 mm
  2. B.the solution where the potato cylinder decreased in length by 1.5 mm
  3. C.the solution where the potato cylinder showed no change in length
  4. D.the solution where the potato cylinder increased in length by 0.5 mm
PastPaper.showAnswers

PastPaper.workedSolution

At the concentration of sucrose that is equal to (isotonic with) the cell sap of the potato cells, there is no net movement of water into or out of the cells by osmosis. As a result, the cells neither gain nor lose water, and the length of the potato cylinder remains unchanged.

PastPaper.markingScheme

1 mark: Identify that no change in length represents zero net osmosis, meaning the concentrations are equal.
PastPaper.question 2 · Multiple Choice
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Which statement about alkanes and alkenes is correct?
  1. A.Alkanes decolourise aqueous bromine, whereas alkenes do not.
  2. B.Alkanes undergo addition reactions with steam to form alcohols.
  3. C.Alkenes are saturated hydrocarbons containing only single bonds.
  4. D.Alkenes are unsaturated hydrocarbons and can be formed by cracking longer chain alkanes.
PastPaper.showAnswers

PastPaper.workedSolution

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond, and they are produced alongside alkanes or hydrogen during the catalytic cracking of long-chain alkanes. Alkenes decolourise bromine water, while alkanes do not react with bromine water in the absence of UV light.

PastPaper.markingScheme

1 mark: Correctly identify that alkenes are unsaturated and are formed by the cracking of longer-chain alkanes.
PastPaper.question 3 · Multiple Choice
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Which row correctly describes the differences between evaporation and boiling of liquid water?
  1. A.Evaporation occurs only at a constant temperature of 100 °C; Boiling occurs at any temperature.
  2. B.Evaporation occurs only at the surface; Boiling occurs throughout the liquid.
  3. C.Evaporation requires a source of heat; Boiling does not require a source of heat.
  4. D.Evaporation forms bubbles of water vapour within the liquid; Boiling does not form bubbles.
PastPaper.showAnswers

PastPaper.workedSolution

Evaporation is a slow process that occurs only at the surface of a liquid at any temperature below the boiling point. Boiling is a rapid process that occurs throughout the liquid at a specific temperature (the boiling point) where bubbles of vapour form within the liquid.

PastPaper.markingScheme

1 mark: Correctly identify the site of evaporation (surface) versus boiling (throughout the liquid).
PastPaper.question 4 · Multiple Choice
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A car travels along a straight road. It accelerates uniformly from rest to a speed of 15 m/s in 5 s, then travels at a constant speed of 15 m/s for 10 s, and finally decelerates uniformly to rest in a further 5 s. What is the total distance travelled by the car?
  1. A.150 m
  2. B.225 m
  3. C.300 m
  4. D.375 m
PastPaper.showAnswers

PastPaper.workedSolution

The motion can be represented by a speed-time graph. The area under the graph is equal to the distance travelled.
- Phase 1 (0 to 5 s): Area = \(0.5 \times 5 \text{ s} \times 15 \text{ m/s} = 37.5 \text{ m}\)
- Phase 2 (5 to 15 s): Area = \(10 \text{ s} \times 15 \text{ m/s} = 150 \text{ m}\)
- Phase 3 (15 to 20 s): Area = \(0.5 \times 5 \text{ s} \times 15 \text{ m/s} = 37.5 \text{ m}\)
Total distance = \(37.5 + 150 + 37.5 = 225 \text{ m}\). (Alternatively, area of trapezium = \(0.5 \times (10 + 20) \times 15 = 225 \text{ m}\).)

PastPaper.markingScheme

1 mark: Calculate the correct distance using the area under a speed-time graph.
PastPaper.question 5 · Multiple Choice
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Which row correctly describes the structural features of arteries compared to veins?
  1. A.Arteries have a wider lumen, thinner wall, and contain valves.
  2. B.Arteries have a narrower lumen, thicker muscular wall, and do not contain valves.
  3. C.Arteries have a narrower lumen, thinner wall, and contain valves.
  4. D.Arteries have a wider lumen, thicker muscular wall, and do not contain valves.
PastPaper.showAnswers

PastPaper.workedSolution

Arteries carry blood under high pressure away from the heart, so they have thick walls containing muscle and elastic fibres, and a relatively narrow lumen. They do not have valves (except for the semi-lunar valves at the exit of the heart). Veins have thinner walls, a wider lumen, and contain valves to prevent the backflow of blood under low pressure.

PastPaper.markingScheme

1 mark: Correctly identify the structural characteristics of arteries (narrower lumen, thicker wall, no valves).
PastPaper.question 6 · Multiple Choice
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A plant is grown in a culture solution deficient in magnesium ions. Which symptom will this plant show, and what is the reason for this?
  1. A.poor root growth because magnesium is needed for cell division
  2. B.yellow leaves because magnesium is required to make chlorophyll
  3. C.purple leaves because magnesium is needed for respiration
  4. D.stunted growth with very dark green leaves because magnesium prevents glucose transport
PastPaper.showAnswers

PastPaper.workedSolution

Magnesium ions are required by plants to synthesise chlorophyll, which is the green pigment that absorbs light for photosynthesis. Without magnesium, chlorophyll cannot be made, leading to yellowing of the leaves (chlorosis).

PastPaper.markingScheme

1 mark: Correctly identify both the symptom (yellow leaves) and the biochemical cause (magnesium is needed for chlorophyll synthesis).
PastPaper.question 7 · Multiple Choice
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Which statement about the electromagnetic spectrum is correct?
  1. A.Infrared waves have a shorter wavelength than ultraviolet waves.
  2. B.All electromagnetic waves travel at the same speed in a vacuum.
  3. C.Gamma rays travel faster than radio waves in a vacuum.
  4. D.Microwaves have a higher frequency than X-rays.
PastPaper.showAnswers

PastPaper.workedSolution

All electromagnetic waves travel at the same constant speed of approximately \(3 \times 10^8 \text{ m/s}\) in a vacuum. Ultraviolet waves have shorter wavelengths than infrared waves, and X-rays have a much higher frequency than microwaves.

PastPaper.markingScheme

1 mark: Correctly identify that all electromagnetic waves travel at the same speed in a vacuum.
PastPaper.question 8 · Multiple Choice
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A student performs a paper chromatography experiment to identify a dye. The solvent front moves 8.0 cm from the baseline. The spot of the dye moves 6.0 cm from the baseline. What is the \(R_f\) value of this dye?
  1. A.0.48
  2. B.0.75
  3. C.1.33
  4. D.14.0
PastPaper.showAnswers

PastPaper.workedSolution

The retention factor (\(R_f\)) is calculated using the formula: \(R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent front}}\). In this case: \(R_f = \frac{6.0 \text{ cm}}{8.0 \text{ cm}} = 0.75\).

PastPaper.markingScheme

1 mark: Apply the correct \(R_f\) formula to obtain 0.75.
PastPaper.question 9 · multiple_choice
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A plant cell is placed in a concentrated salt solution. Which row correctly describes what happens to the vacuole and the cell membrane?
  1. A.The vacuole increases in size and the cell membrane is pushed against the cell wall.
  2. B.The vacuole decreases in size and the cell membrane pulls away from the cell wall.
  3. C.The vacuole increases in size and the cell membrane pulls away from the cell wall.
  4. D.The vacuole decreases in size and the cell membrane is pushed against the cell wall.
PastPaper.showAnswers

PastPaper.workedSolution

When plant cells are placed in a concentrated salt solution (which has a lower water potential than the cell sap), water moves out of the vacuole and cytoplasm by osmosis. This causes the vacuole to shrink (decrease in size) and the cytoplasm/cell membrane to pull away from the cell wall, a process known as plasmolysis.

PastPaper.markingScheme

1 mark for identifying that the vacuole decreases in size and the cell membrane pulls away from the cell wall.
PastPaper.question 10 · multiple_choice
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Which row correctly describes the relative thickness of the muscular wall of the right ventricle compared to the left ventricle, and the relative oxygen concentration of the deoxygenated blood in the pulmonary artery?
  1. A.Right ventricle: thicker than left ventricle | Pulmonary artery: high oxygen concentration
  2. B.Right ventricle: thicker than left ventricle | Pulmonary artery: low oxygen concentration
  3. C.Right ventricle: thinner than left ventricle | Pulmonary artery: low oxygen concentration
  4. D.Right ventricle: thinner than left ventricle | Pulmonary artery: high oxygen concentration
PastPaper.showAnswers

PastPaper.workedSolution

The muscular wall of the right ventricle is thinner than that of the left ventricle because the right ventricle only pumps blood to the lungs (short distance, lower pressure required), whereas the left ventricle pumps blood to the rest of the body at high pressure. The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs, meaning its oxygen concentration is low.

PastPaper.markingScheme

1 mark for identifying the right ventricle wall is thinner and the pulmonary artery has low oxygen concentration.
PastPaper.question 11 · multiple_choice
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An aqueous solution of an unknown salt is tested. Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. Adding dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of the salt?
  1. A.chromium(III) sulfate
  2. B.iron(II) sulfate
  3. C.iron(III) sulfate
  4. D.iron(II) chloride
PastPaper.showAnswers

PastPaper.workedSolution

The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(Fe^{2+}\). The formation of a white precipitate with dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions, \(SO_4^{2-}\). Therefore, the salt is iron(II) sulfate.

PastPaper.markingScheme

1 mark for identifying the cation as iron(II) and the anion as sulfate to determine the salt is iron(II) sulfate.
PastPaper.question 12 · multiple_choice
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Which statement correctly describes a property or reaction of hydrocarbons?
  1. A.Alkanes are unsaturated hydrocarbons that readily decolourise aqueous bromine.
  2. B.Alkanes undergo addition reactions with hydrogen to form alkenes.
  3. C.Alkenes are saturated hydrocarbons that burn in oxygen to form only carbon monoxide and hydrogen.
  4. D.Alkenes react with bromine water in an addition reaction, changing the colour of the mixture from orange to colourless.
PastPaper.showAnswers

PastPaper.workedSolution

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond, allowing them to undergo addition reactions. When mixed with bromine water (which is orange), they react to form a colourless dibromoalkane, decolourising the solution. Alkanes are saturated and do not readily react with bromine water without UV light.

PastPaper.markingScheme

1 mark for identifying that alkenes react with bromine water in an addition reaction, turning the solution from orange to colourless.
PastPaper.question 13 · multiple_choice
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A student sits near the side of a hot metal canister filled with boiling water. The student feels thermal energy transferring to their hand, even though they are not touching the canister. Which processes are primarily responsible for transferring thermal energy through the metal wall of the canister, and then through the air to the student's hand at the side?
  1. A.Through metal wall: conduction | Through air to hand: convection
  2. B.Through metal wall: conduction | Through air to hand: radiation
  3. C.Through metal wall: convection | Through air to hand: conduction
  4. D.Through metal wall: radiation | Through air to hand: radiation
PastPaper.showAnswers

PastPaper.workedSolution

Thermal energy is conducted through the solid metal wall of the canister because metals are excellent thermal conductors. Since the student's hand is to the side of the canister (and convection currents carry hot air vertically upwards), the primary mode of heat transfer through the air to the hand at the side is thermal radiation (infrared radiation).

PastPaper.markingScheme

1 mark for selecting conduction for the metal wall and radiation for the transfer through the air to the side.
PastPaper.question 14 · multiple_choice
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A wooden box of mass \(4.0\text{ kg}\) is pushed along a horizontal wooden floor by a constant force of \(18\text{ N}\). A frictional force of \(6.0\text{ N}\) acts in the opposite direction to the motion. What is the acceleration of the box?
  1. A.\(1.5\text{ m/s}^2\)
  2. B.\(3.0\text{ m/s}^2\)
  3. C.\(4.5\text{ m/s}^2\)
  4. D.\(6.0\text{ m/s}^2\)
PastPaper.showAnswers

PastPaper.workedSolution

First, find the resultant force acting on the box: \(F = 18\text{ N} - 6.0\text{ N} = 12\text{ N}\). Next, use Newton's second law, \(F = ma\), to find the acceleration: \(a = F / m = 12\text{ N} / 4.0\text{ kg} = 3.0\text{ m/s}^2\).

PastPaper.markingScheme

1 mark for calculating the correct acceleration of \(3.0\text{ m/s}^2\) using the resultant force and mass.
PastPaper.question 15 · multiple_choice
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Chlorine, bromine, and iodine are elements in Group VII of the Periodic Table. Which statement about the trends or properties of these elements is correct?
  1. A.Chemical reactivity increases down the group, making chlorine the least reactive halogen.
  2. B.The elements become darker in colour down the group, so iodine is darker than chlorine.
  3. C.The melting point decreases down the group, so chlorine has the highest melting point.
  4. D.Chlorine can be displaced from an aqueous solution of sodium chloride by adding liquid bromine.
PastPaper.showAnswers

PastPaper.workedSolution

In Group VII, the halogens become darker in colour down the group (chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid). Reactivity decreases down the group, and melting/boiling points increase down the group.

PastPaper.markingScheme

1 mark for identifying that the elements become darker in colour down the group, so iodine is darker than chlorine.
PastPaper.question 16 · multiple_choice
1 PastPaper.marks
Which statement about the properties of electromagnetic waves is correct?
  1. A.Microwaves have a higher frequency than ultraviolet waves and travel slower in a vacuum.
  2. B.Radio waves have a longer wavelength than infrared waves and travel at the same speed in a vacuum.
  3. C.Gamma rays have a lower frequency than X-rays and travel faster in a vacuum.
  4. D.Red light has a shorter wavelength than blue light and travels at the same speed in a vacuum.
PastPaper.showAnswers

PastPaper.workedSolution

All electromagnetic waves travel at the same speed in a vacuum (approximately \(3 \times 10^8\text{ m/s}\)). In the electromagnetic spectrum, radio waves have a longer wavelength than infrared waves (the order from longest to shortest wavelength is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays).

PastPaper.markingScheme

1 mark for identifying that radio waves have a longer wavelength than infrared waves and travel at the same speed in a vacuum.
PastPaper.question 17 · multiple_choice
1 PastPaper.marks
Identical cylinders of potato tissue are placed into four different test-tubes, each containing a different concentration of sucrose solution. In which test-tube will the potato cylinder show the greatest percentage increase in mass?
  1. A.distilled water
  2. B.0.2 mol/dm3 sucrose solution
  3. C.0.5 mol/dm3 sucrose solution
  4. D.1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 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solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 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sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 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solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 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sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution1.0 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.5 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2 mol/dm3 sucrose solution0.2/dm3 sucrose solution
PastPaper.showAnswers

PastPaper.workedSolution

Distilled water has the highest water potential. Water enters the potato cells by osmosis from a region of higher water potential (distilled water) to a region of lower water potential (inside cells), causing them to swell and gain the most mass. In sucrose solutions, the water potential gradient is lower or reversed, leading to less mass gain or mass loss.

PastPaper.markingScheme

1 mark for identifying that distilled water causes the highest net movement of water into the cells by osmosis.
PastPaper.question 18 · multiple_choice
1 PastPaper.marks
Which row correctly describes the relative wall thickness, lumen diameter, and the presence of valves in a typical vein compared to a typical artery?
  1. A.vein has thicker walls, narrower lumen, and valves are present
  2. B.vein has thicker walls, wider lumen, and valves are absent
  3. C.vein has thinner walls, wider lumen, and valves are present
  4. D.vein has thinner walls, narrower lumen, and valves are absent
PastPaper.showAnswers

PastPaper.workedSolution

Veins carry blood under low pressure back to the heart, so they have thinner walls, wider lumens to reduce resistance to blood flow, and valves to prevent the backflow of blood.

PastPaper.markingScheme

1 mark for the correct combination of thin walls, wide lumen, and valves present in veins.
PastPaper.question 19 · multiple_choice
1 PastPaper.marks
Which row correctly identifies the function of the palisade mesophyll cells and the guard cells in a leaf?
  1. A.palisade mesophyll: main site of photosynthesis; guard cells: control the opening and closing of stomata
  2. B.palisade mesophyll: main site of photosynthesis; guard cells: prevent water loss from the upper leaf surface
  3. C.palisade mesophyll: transport of water and mineral ions; guard cells: control the opening and closing of stomata
  4. D.palisade mesophyll: transport of water and mineral ions; guard cells: prevent water loss from the upper leaf surface
PastPaper.showAnswers

PastPaper.workedSolution

Palisade mesophyll cells are packed with chloroplasts and are located near the upper surface of the leaf, making them the main site of photosynthesis. Guard cells surround the stomata and control their opening and closing to regulate gas exchange and transpiration.

PastPaper.markingScheme

1 mark for identifying palisade mesophyll as the main site of photosynthesis and guard cells as controlling stomatal aperture.
PastPaper.question 20 · multiple_choice
1 PastPaper.marks
A student tests an unknown salt solution. The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. Dilute nitric acid followed by aqueous barium nitrate is then added, which produces a white precipitate. What is the identity of the salt?
  1. A.iron(II) chloride
  2. B.iron(II) sulfate
  3. C.iron(III) sulfate
  4. D.chromium(III) sulfate
PastPaper.showAnswers

PastPaper.workedSolution

A green precipitate with sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, Fe2+. The white precipitate formed with barium nitrate under acidic conditions confirms the presence of sulfate ions, SO42-. Thus, the salt is iron(II) sulfate.

PastPaper.markingScheme

1 mark for correctly identifying Fe2+ from the NaOH test and SO42- from the barium nitrate test to name iron(II) sulfate.
PastPaper.question 21 · multiple_choice
1 PastPaper.marks
Which statement about the trends as we go down Group VII of the Periodic Table is correct?
  1. A.The reactivity of the elements increases.
  2. B.The color of the elements becomes lighter.
  3. C.The melting point of the elements increases.
  4. D.The elements form positive ions in their reactions.
PastPaper.showAnswers

PastPaper.workedSolution

As you go down Group VII, the melting points and boiling points of the halogens increase. Reactivity decreases down the group, colors become darker (from pale yellow fluorine to grey-black iodine), and they form negative halide ions (1- charge) when reacting with metals.

PastPaper.markingScheme

1 mark for identifying that melting point increases down Group VII.
PastPaper.question 22 · multiple_choice
1 PastPaper.marks
A sample of a gaseous hydrocarbon is bubbled through aqueous bromine. The orange-brown bromine solution immediately becomes colorless. Which statement about the hydrocarbon is correct?
  1. A.It is an alkane and it is saturated.
  2. B.It is an alkane and it is unsaturated.
  3. C.It is an alkene and it is saturated.
  4. D.It is an alkene and it is unsaturated.
PastPaper.showAnswers

PastPaper.workedSolution

The immediate decolorization of bromine water is a test for unsaturation (presence of double bonds). Therefore, the hydrocarbon is an alkene, which is unsaturated.

PastPaper.markingScheme

1 mark for identifying the hydrocarbon as an unsaturated alkene.
PastPaper.question 23 · multiple_choice
1 PastPaper.marks
A hot metal block is suspended inside a vacuum chamber. By which method or methods of thermal energy transfer can energy leave the block and reach the walls of the chamber?
  1. A.conduction, convection and radiation
  2. B.radiation only
  3. C.conduction and convection only
  4. D.radiation and convection only
PastPaper.showAnswers

PastPaper.workedSolution

Conduction and convection both require a physical medium containing particles to transfer thermal energy. Since the chamber is a vacuum, thermal energy can only travel through it via electromagnetic radiation (infrared radiation).

PastPaper.markingScheme

1 mark for recognizing that radiation is the only method that can transfer thermal energy through a vacuum.
PastPaper.question 24 · multiple_choice
1 PastPaper.marks
Which row correctly describes the speed of all electromagnetic waves in a vacuum, and how the wavelength of these waves changes from radio waves to gamma rays?
  1. A.speed in vacuum: 3.0 x 10^8 m/s; wavelength from radio waves to gamma rays: decreases
  2. B.speed in vacuum: 3.0 x 10^8 m/s; wavelength from radio waves to gamma rays: increases
  3. C.speed in vacuum: varies depending on wave type; wavelength from radio waves to gamma rays: decreases
  4. D.speed in vacuum: varies depending on wave type; wavelength from radio waves to gamma rays: increases
PastPaper.showAnswers

PastPaper.workedSolution

All electromagnetic waves travel at the same constant speed of approximately \(3.0 \times 10^8\text{ m/s}\) in a vacuum. The electromagnetic spectrum ordered from longest wavelength to shortest wavelength starts with radio waves and ends with gamma rays, meaning wavelength decreases from radio waves to gamma rays.

PastPaper.markingScheme

1 mark for identifying that all EM waves travel at the same speed of 3.0 x 10^8 m/s in a vacuum and that wavelength decreases from radio waves to gamma rays.
PastPaper.question 25 · multiple_choice
1 PastPaper.marks
In a paper chromatography experiment, a student measures the distance from the baseline to the solvent front as 12.0 cm. A blue spot of food colouring travels 4.8 cm from the baseline. What is the Rf value of the blue food colouring?
  1. A.0.40
  2. B.0.60
  3. C.2.50
  4. D.7.20 Gold standard is 0.40 since 4.8 / 12 = 0.40, other options represent incorrect divisions or differences of numbers (e.g. 12.0 - 4.8 = 7.2).
PastPaper.showAnswers

PastPaper.workedSolution

The Rf value is calculated using the formula: Rf = (distance moved by substance) / (distance moved by solvent). Substituting the given values: Rf = 4.8 cm / 12.0 cm = 0.40.

PastPaper.markingScheme

1 mark for the correct calculation of Rf as 0.40.
PastPaper.question 26 · multiple_choice
1 PastPaper.marks
A student has four identical flasks containing equal amounts of hot water at 80 degrees Celsius. Each flask has a different surface coating: Flask 1 is dull black, Flask 2 is shiny silver, Flask 3 is dull white, and Flask 4 is shiny white. The flasks are left to cool in a laboratory. Which flask will cool down the fastest, and which will cool down the slowest?
  1. A.Fastest: Flask 1; Slowest: Flask 2
  2. B.Fastest: Flask 1; Slowest: Flask 3
  3. C.Fastest: Flask 2; Slowest: Flask 1
  4. D.Fastest: Flask 4; Slowest: Flask 3
PastPaper.showAnswers

PastPaper.workedSolution

Dull black surfaces are the best emitters of thermal radiation, so Flask 1 will lose heat the fastest. Shiny silver surfaces are the worst emitters of thermal radiation, so Flask 2 will lose heat the slowest.

PastPaper.markingScheme

1 mark for correctly identifying Flask 1 as the fastest cooler and Flask 2 as the slowest cooler.
PastPaper.question 27 · multiple_choice
1 PastPaper.marks
A toy car of mass 2.0 kg accelerates from rest to a speed of 6.0 m/s in a time of 3.0 s. What is the average useful power required to produce this acceleration?
  1. A.4.0 W
  2. B.6.0 W
  3. C.12 W
  4. D.36 W
PastPaper.showAnswers

PastPaper.workedSolution

The work done to accelerate the car is equal to its gain in kinetic energy: E_k = 0.5 * m * v^2 = 0.5 * 2.0 kg * (6.0 m/s)^2 = 36 J. The average useful power is the energy transferred per unit time: P = E / t = 36 J / 3.0 s = 12 W.

PastPaper.markingScheme

1 mark for calculating the work done as 36 J and successfully dividing by the time to obtain 12 W.
PastPaper.question 28 · multiple_choice
1 PastPaper.marks
Which option correctly describes the bonding in ethene and the change observed when it is shaken with aqueous bromine?
  1. A.Ethene has a single covalent C-C bond, and it turns aqueous bromine from orange to colourless.
  2. B.Ethene has a double covalent C-C bond, and it turns aqueous bromine from orange to colourless.
  3. C.Ethene has a single covalent C-C bond, and aqueous bromine remains orange.
  4. D.Ethene has a double covalent C-C bond, and aqueous bromine remains orange.
PastPaper.showAnswers

PastPaper.workedSolution

Ethene is an unsaturated hydrocarbon containing a double covalent carbon-carbon bond. Because it is unsaturated, it undergoes an addition reaction with aqueous bromine, decolourising it from orange to colourless.

PastPaper.markingScheme

1 mark for identifying the carbon-carbon double bond and the corresponding orange to colourless colour change.
PastPaper.question 29 · multiple_choice
1 PastPaper.marks
The characteristics of three blood vessels are described: Vessel X has a large lumen, a thin wall, and carries blood towards the heart. Vessel Y has a very small lumen (one cell wide), a wall one cell thick, and links arterioles to venules. Vessel Z has a small lumen, a thick muscular wall, and carries blood away from the heart. Which option correctly identifies vessels X, Y, and Z?
  1. A.X is an artery, Y is a capillary, Z is a vein
  2. B.X is a vein, Y is a capillary, Z is an artery
  3. C.X is a vein, Y is an artery, Z is a capillary
  4. D.X is a capillary, Y is a vein, Z is an artery
PastPaper.showAnswers

PastPaper.workedSolution

Vessel X is a vein because it has a large lumen, thin walls, and carries blood towards the heart. Vessel Y is a capillary because its wall is one cell thick to allow rapid diffusion, and it links arterioles to venules. Vessel Z is an artery because it has a thick muscular wall to withstand high pressure and carries blood away from the heart.

PastPaper.markingScheme

1 mark for identifying X as a vein, Y as a capillary, and Z as an artery.
PastPaper.question 30 · multiple_choice
1 PastPaper.marks
An experiment is set up to study osmosis in plant cells. Identical cylinders of potato are placed in salt solutions of different concentrations. In which solution will the plant cells become turgid, and what is the reason?
  1. A.In a concentrated salt solution, because water molecules enter the potato cells by osmosis.
  2. B.In distilled water, because water molecules enter the potato cells by osmosis.
  3. C.In a concentrated salt solution, because salt molecules enter the potato cells by diffusion.
  4. D.In distilled water, because water molecules leave the potato cells by osmosis.
PastPaper.showAnswers

PastPaper.workedSolution

Distilled water has a higher water potential than the cytoplasm of the potato cells. Water molecules enter the potato cells down a water potential gradient by osmosis, causing the cells to swell and become turgid. In concentrated salt solution, water leaves the cells, making them flaccid.

PastPaper.markingScheme

1 mark for identifying distilled water as the solution and the entry of water molecules by osmosis as the correct mechanism.
PastPaper.question 31 · multiple_choice
1 PastPaper.marks
Which statement about the trends in the Group VII elements (the halogens) from chlorine to iodine is correct?
  1. A.Their reactivity increases, and their melting point increases.
  2. B.Their reactivity decreases, and their melting point increases.
  3. C.Their reactivity increases, and their melting point decreases.
  4. D.Their reactivity decreases, and their melting point decreases.
PastPaper.showAnswers

PastPaper.workedSolution

As you go down Group VII from chlorine to iodine: reactivity decreases because the outer shell is further from the positive nucleus, making it harder to attract an electron; melting point increases because the molecules become larger, resulting in stronger intermolecular forces that require more thermal energy to overcome.

PastPaper.markingScheme

1 mark for correctly stating that reactivity decreases and melting point increases.
PastPaper.question 32 · multiple_choice
1 PastPaper.marks
A radio wave travels at a speed of 3.0 x 10^8 m/s in a vacuum and has a frequency of 1.5 x 10^8 Hz. What is its wavelength, and how does its wavelength compare to that of visible light?
  1. A.The wavelength is 0.5 m, which is shorter than visible light.
  2. B.The wavelength is 0.5 m, which is longer than visible light.
  3. C.The wavelength is 2.0 m, which is shorter than visible light.
  4. D.The wavelength is 2.0 m, which is longer than visible light.
PastPaper.showAnswers

PastPaper.workedSolution

Using the wave equation v = f * lambda, we find lambda = v / f = (3.0 x 10^8 m/s) / (1.5 x 10^8 Hz) = 2.0 m. Since visible light has wavelengths in the range of nanometres (approx. 4 x 10^-7 m to 7 x 10^-7 m), a wavelength of 2.0 m is much longer than that of visible light.

PastPaper.markingScheme

1 mark for the correct calculation of wavelength (2.0 m) and identifying that it is longer than visible light.
PastPaper.question 33 · multiple_choice
1 PastPaper.marks
Four potato cylinders of equal initial mass are placed into four different sucrose solutions: P, Q, R, and S. After two hours, the percentage change in mass of each cylinder is measured: Cylinder in P = \(+8\%\), Cylinder in Q = \(-5\%\), Cylinder in R = \(0\%\), and Cylinder in S = \(+12\%\). Which solution has the highest water potential?
  1. A.P
  2. B.Q
  3. C.R
  4. D.S
PastPaper.showAnswers

PastPaper.workedSolution

Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane. A gain in mass indicates water has entered the potato cells. The cylinder in solution S gained the most mass (\(+12\%\)), which means solution S has the highest water potential relative to the potato cells among all the solutions tested.

PastPaper.markingScheme

1 mark for selecting correct option D. [1]
PastPaper.question 34 · multiple_choice
1 PastPaper.marks
Which row correctly identifies the chamber of the heart that pumps blood to the lungs, and the blood vessel that carries oxygenated blood back to the heart?
  1. A.Chamber: Left ventricle; Vessel: Pulmonary artery
  2. B.Chamber: Left ventricle; Vessel: Pulmonary vein
  3. C.Chamber: Right ventricle; Vessel: Pulmonary artery
  4. D.Chamber: Right ventricle; Vessel: Pulmonary vein
PastPaper.showAnswers

PastPaper.workedSolution

The right ventricle pumps deoxygenated blood to the lungs through the pulmonary artery. Oxygenated blood from the lungs returns to the left atrium of the heart via the pulmonary vein. Therefore, the correct row is D.

PastPaper.markingScheme

1 mark for selecting correct option D. [1]
PastPaper.question 35 · multiple_choice
1 PastPaper.marks
A student performs a paper chromatography experiment to analyze a dye. The solvent front travels \(8.0\text{ cm}\) from the baseline. The spot of the dye travels \(6.0\text{ cm}\) from the baseline. What is the \(R_f\) value of the dye?
  1. A.\(0.75\)
  2. B.\(1.33\)
  3. C.\(6.0\text{ cm}\)
  4. D.\(14.0\text{ cm}\)
PastPaper.showAnswers

PastPaper.workedSolution

The retention factor (\(R_f\)) is calculated using the formula: \(R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent}} = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\). It is a dimensionless ratio, so it has no units.

PastPaper.markingScheme

1 mark for selecting correct option A. [1]
PastPaper.question 36 · multiple_choice
1 PastPaper.marks
An organic compound X is a liquid that decolorizes aqueous bromine. When X undergoes complete combustion, it forms carbon dioxide and water. Which statement about X is correct?
  1. A.X is a saturated hydrocarbon.
  2. B.X belongs to the homologous series of alkanes.
  3. C.X contains at least one carbon-carbon double bond.
  4. D.The general formula for X is \(C_nH_{2n+2}\).
PastPaper.showAnswers

PastPaper.workedSolution

Decolorizing aqueous bromine is the standard test for unsaturation (presence of a carbon-carbon double bond, \(C=C\)). Alkanes (saturated hydrocarbons with the general formula \(C_nH_{2n+2}\)) do not react with bromine water under normal conditions. Therefore, X must contain at least one carbon-carbon double bond, which makes it an alkene.

PastPaper.markingScheme

1 mark for selecting correct option C. [1]
PastPaper.question 37 · multiple_choice
1 PastPaper.marks
A student investigates the rate of evaporation of water. She places equal volumes of water into four identical open containers under different environmental conditions. Which set of conditions will produce the fastest rate of evaporation?
  1. A.High temperature, low humidity, strong wind
  2. B.High temperature, high humidity, strong wind
  3. C.Low temperature, low humidity, no wind
  4. D.Low temperature, high humidity, strong wind
PastPaper.showAnswers

PastPaper.workedSolution

The rate of evaporation increases with: 1. Higher temperature, because the molecules have greater average kinetic energy to escape the liquid surface. 2. Lower humidity, because this maintains a steep concentration gradient between the liquid surface and the surrounding air. 3. Strong wind, because it sweeps away water vapor from above the container surface, preventing the air from becoming saturated.

PastPaper.markingScheme

1 mark for selecting correct option A. [1]
PastPaper.question 38 · multiple_choice
1 PastPaper.marks
An object starts from rest and accelerates uniformly to a speed of \(12\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed of \(12\text{ m/s}\) for another \(6.0\text{ s}\). What is the total distance traveled by the object during these \(10.0\text{ s}\)?
  1. A.\(72\text{ m}\)
  2. B.\(96\text{ m}\)
  3. C.\(120\text{ m}\)
  4. D.\(144\text{ m}\)
PastPaper.showAnswers

PastPaper.workedSolution

The total distance traveled is the area under the speed-time graph: 1. During the uniform acceleration stage (first \(4.0\text{ s}\)), the distance is the area of a triangle: \(\text{Distance}_1 = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\). 2. During the constant speed stage (next \(6.0\text{ s}\)), the distance is the area of a rectangle: \(\text{Distance}_2 = 6.0\text{ s} \times 12\text{ m/s} = 72\text{ m}\). Total distance = \(24\text{ m} + 72\text{ m} = 96\text{ m}\).

PastPaper.markingScheme

1 mark for selecting correct option B. [1]
PastPaper.question 39 · multiple_choice
1 PastPaper.marks
Which statement correctly compares radio waves and gamma rays?
  1. A.Radio waves have a higher frequency and a longer wavelength than gamma rays.
  2. B.Radio waves have a higher frequency and a shorter wavelength than gamma rays.
  3. C.Radio waves have a lower frequency and a longer wavelength than gamma rays.
  4. D.Radio waves have a lower frequency and a shorter wavelength than gamma rays.
PastPaper.showAnswers

PastPaper.workedSolution

In the electromagnetic spectrum, radio waves have the lowest frequencies and the longest wavelengths, while gamma rays have the highest frequencies and the shortest wavelengths. Therefore, radio waves have a lower frequency and a longer wavelength than gamma rays.

PastPaper.markingScheme

1 mark for selecting correct option C. [1]
PastPaper.question 40 · multiple_choice
1 PastPaper.marks
A wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0\\ \Omega\). A second wire made of the same metal has a length of \(2L\) and a cross-sectional area of \(2A\). What is the resistance of the second wire?
  1. A.\(2.0\\ \Omega\)
  2. B.\(4.0\\ \Omega\)
  3. C.\(8.0\\ \Omega\)
  4. D.\(16.0\\ \Omega\)
PastPaper.showAnswers

PastPaper.workedSolution

The resistance of a wire is given by the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity of the metal. For the second wire, the resistance is \(R_{\text{new}} = \rho \frac{2L}{2A} = \rho \frac{L}{A} = R = 8.0\\ \Omega\). The resistance remains unchanged because both the length and cross-sectional area are doubled.

PastPaper.markingScheme

1 mark for selecting correct option C. [1]

Paper 43 (Theory)

Answer all questions in the spaces provided.
9 PastPaper.question · 80.04 PastPaper.marks
PastPaper.question 1 · Structured
8.88 PastPaper.marks
A student investigates a green solid, compound X.

(a) A solution of X is tested by adding a few drops of aqueous sodium hydroxide. A pale green precipitate forms which is insoluble in excess. State the name of the transition metal ion present in X. [1]

(b) Describe a chemical test, including the observation for a positive result, to confirm that compound X contains carbonate ions. [2]

(c) The student also performs paper chromatography on a food colouring containing three dyes. Explain why the baseline (start line) on the chromatogram must be drawn in pencil and not in ink. [2]

(d) Explain how the chromatogram can be used to determine if the food colouring is a pure substance or a mixture, and how the \(R_f\) value of a component is calculated. [3.88]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Iron(II) ion (or \(Fe^{2+}\)).

(b) Add dilute hydrochloric acid (or any dilute acid) to the solid compound X. Bubbles/effervescence will be produced. Bubble the gas through limewater; the limewater turns cloudy/milky.

(c) Ink is soluble in the chromatography solvent and would run/separate along with the dyes, interfering with the results. Pencil (graphite) is insoluble in the solvent and will not run.

(d) A mixture will separate into two or more distinct spots on the chromatogram, whereas a pure substance will show only one single spot. The \(R_f\) value is calculated by dividing the distance travelled by the substance (spot) by the distance travelled by the solvent front from the baseline.

PastPaper.markingScheme

(a) 1 mark for iron(II) / \(Fe^{2+}\) (reject iron / iron(III)).
(b) 1 mark for adding dilute acid and testing gas with limewater; 1 mark for limewater turning cloudy/milky.
(c) 1 mark for stating ink is soluble / would dissolve/run; 1 mark for stating pencil is insoluble / does not smudge/run.
(d) 1.88 marks for pure vs mixture (1 mark for pure = 1 spot, 0.88 marks for mixture = multiple spots); 2 marks for \(R_f\) formula: distance moved by substance / distance moved by solvent (1 mark for numerator, 1 mark for denominator).
PastPaper.question 2 · Structured
8.88 PastPaper.marks
A student investigates the rate of evaporation of water from two different containers, A and B.

Container A is a wide, shallow dish with a large surface area of \(120\text{ cm}^2\). Container B is a tall, narrow beaker with a surface area of \(30\text{ cm}^2\). Both contain the same volume of water at \(40\text{ }^{\circ}\text{C}\).

(a) Explain, in terms of molecules, why evaporation causes the remaining water to cool down. [3]

(b) State and explain, in terms of molecules, which container will have a higher rate of evaporation. [2.88]

(c) The student wants to reduce the rate of heat loss from Container B. They decide to wrap the sides of the beaker in shiny aluminium foil.
Explain how wrapping the beaker in shiny aluminium foil reduces heat loss by radiation. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) During evaporation, the water molecules with the highest kinetic energy escape from the surface of the liquid. This leaves behind molecules with a lower average kinetic energy. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the remaining water decreases.

(b) Container A will have a higher rate of evaporation. This is because Container A has a larger surface area than Container B, which means more water molecules are present at the surface. Consequently, there is a greater probability of molecules escaping from the surface per unit time.

(c) Shiny surfaces are poor emitters of infrared radiation (thermal radiation). Wrapping the sides of the beaker in shiny aluminium foil reduces the rate at which thermal energy is emitted to the surroundings, compared to the duller glass surface.

PastPaper.markingScheme

(a) 1 mark for stating that molecules with high kinetic energy escape; 1 mark for stating that escape occurs from the surface; 1 mark for explaining that the average kinetic energy of remaining molecules decreases, leading to cooling.
(b) 1 mark for identifying Container A; 1.88 marks for explanation: larger surface area means more molecules at the surface (1 mark) and higher chance of escaping (0.88 marks).
(c) 1 mark for identifying that shiny/silver surfaces are poor emitters; 1 mark for identifying that the heat transfer mechanism is infrared/thermal radiation; 1 mark for concluding that less heat energy is emitted to the surrounding air.
PastPaper.question 3 · Structured
8.88 PastPaper.marks
A toy car of mass \(0.50\text{ kg}\) is pulled along a flat horizontal floor.

(a) Calculate the kinetic energy of the car when it is travelling at a constant speed of \(4.0\text{ m/s}\). State the formula used. [2.88]

(b) The car is then accelerated uniformly from rest to a speed of \(6.0\text{ m/s}\) in \(3.0\text{ seconds}\).
(i) Calculate the acceleration of the car. Show your working and state the unit. [3]
(ii) Calculate the force required to produce this acceleration. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Formula: \(KE = \frac{1}{2}mv^2\)
Calculation: \(KE = \frac{1}{2} \times 0.50\text{ kg} \times (4.0\text{ m/s})^2 = 0.25 \times 16 = 4.0\text{ J}\).

(b) (i) Formula: \(a = \frac{v - u}{t}\)
Calculation: \(a = \frac{6.0\text{ m/s} - 0\text{ m/s}}{3.0\text{ s}} = 2.0\text{ m/s}^2\).
Unit: \(\text{m/s}^2\).

(ii) Formula: \(F = ma\)
Calculation: \(F = 0.50\text{ kg} \times 2.0\text{ m/s}^2 = 1.0\text{ N}\).

PastPaper.markingScheme

(a) 1 mark for formula \(KE = \frac{1}{2}mv^2\); 1.88 marks for correct calculation and value \(4.0\text{ J}\).
(b) (i) 1 mark for correct formula \(a = \frac{v - u}{t}\) or correct substitution; 1 mark for calculation yielding \(2.0\); 1 mark for correct unit \(\text{m/s}^2\).
(ii) 1 mark for formula \(F = ma\); 1 mark for correct substitution \(0.50 \times 2.0\); 1 mark for final answer \(1.0\text{ N}\).
PastPaper.question 4 · Structured
8.88 PastPaper.marks
Petroleum is a mixture of hydrocarbons. It can be separated into fractions by fractional distillation.

(a) State the physical property on which fractional distillation depends. [1]

(b) Decane (\(\text{C}_{10}\text{H}_{22}\)) can be cracked to form octane (\(\text{C}_8\text{H}_{18}\)) and ethene (\(\text{C}_2\text{H}_4\)).
(i) Write a balanced chemical equation for this cracking reaction. [2]
(ii) Describe a chemical test to distinguish between octane and ethene. State the reagent used and the observation for each compound. [3.88]

(c) Ethene can react with steam to form ethanol. State the type of reaction that occurs and one essential condition for this process. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Boiling point (accept boiling range).

(b) (i) \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\)
(ii) Reagent: aqueous bromine / bromine water.
With octane (alkane): solution remains orange/brown/yellow (no change).
With ethene (alkene): solution turns from orange/brown to colourless (decolourises).

(c) Type of reaction: addition reaction (or hydration).
Condition: phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst OR temperature of \(300\text{ }^{\circ}\text{C}\) OR pressure of \(60-70\text{ atm}\).

PastPaper.markingScheme

(a) 1 mark for boiling point.
(b) (i) 1 mark for correct reactants and products; 1 mark for correct balancing.
(ii) 1 mark for naming bromine water (accept aqueous bromine); 1.88 marks for correct observations (1 mark for ethene decolourising, 0.88 marks for octane remaining orange/no change).
(c) 1 mark for addition / hydration; 1 mark for specifying a correct condition (temperature \(300\text{ }^{\circ}\text{C}\), pressure \(60-70\text{ atm}\), or phosphoric acid catalyst).
PastPaper.question 5 · Structured
8.88 PastPaper.marks
The human circulatory system is responsible for the transport of blood.

(a) Describe the difference between the structure of an artery and the structure of a vein. Relate these differences to their functions. [3.88]

(b) Humans have a double circulation system.
(i) Explain what is meant by the term 'double circulation'. [2]
(ii) State one advantage of a double circulation system. [1]

(c) State the name of the blood vessel that carries oxygenated blood from the lungs to the left atrium of the heart, and the valve that prevents blood from flowing back into this atrium from the ventricle. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Arteries have thick, muscular walls with elastic tissue to withstand and maintain the high pressure of blood leaving the heart. Veins have thin, less muscular walls with a wider lumen because blood is at low pressure, and they contain valves to prevent the backflow of blood.

(b) (i) Double circulation means that for every complete circuit of the body, blood passes through the heart twice (once through the pulmonary system to the lungs, and once through the systemic system to the rest of the body).
(ii) Allows blood to be pumped to the body tissues at a higher pressure, increasing the speed of delivery of oxygen and glucose.

(c) Blood vessel: pulmonary vein.
Valve: bicuspid valve (accept mitral valve or left atrioventricular valve).

PastPaper.markingScheme

(a) 1 mark for artery description (thick walls/elastic tissue); 1 mark for relating artery structure to high pressure; 1 mark for vein description (thin walls/valves); 0.88 marks for relating vein structure to low pressure/preventing backflow.
(b) (i) 2 marks for stating blood passes through the heart twice for each complete circuit of the body (1 mark for 'twice', 1 mark for 'one circuit/loop').
(ii) 1 mark for maintaining high pressure / faster transport of materials.
(c) 1 mark for pulmonary vein; 1 mark for bicuspid valve / mitral valve / left atrioventricular valve.
PastPaper.question 6 · Structured
8.88 PastPaper.marks
An experiment is carried out using potato cylinders to study osmosis.

Two potato cylinders of equal size and mass are placed in different solutions for two hours:
- Cylinder A is placed in distilled water.
- Cylinder B is placed in a concentrated sucrose solution.

(a) Define the term osmosis. [3]

(b) Explain, in terms of water potential, why cylinder A increases in mass, while cylinder B decreases in mass. [3.88]

(c) Plant cells do not burst when placed in distilled water, whereas red blood cells would. State the part of the plant cell that prevents it from bursting. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, down a water potential gradient, through a partially permeable membrane.

(b) Cylinder A: Distilled water has a higher water potential than the cell sap inside the potato cells. Water enters the cells by osmosis down the gradient, causing them to gain mass.
Cylinder B: The concentrated sucrose solution has a lower water potential than the cell sap. Water leaves the potato cells by osmosis, causing them to lose mass.

(c) The cell wall (which is rigid and exerts turgor pressure to prevent bursting).

PastPaper.markingScheme

(a) 1 mark for net movement of water molecules; 1 mark for from higher water potential to lower water potential (or down gradient); 1 mark for through a partially permeable membrane.
(b) 1 mark for identifying water moves into Cylinder A (increases mass); 1 mark for identifying water moves out of Cylinder B (decreases mass); 1.88 marks for explaining in terms of relative water potentials (e.g., distilled water has higher water potential than cells (1 mark), concentrated solution has lower water potential than cells (0.88 marks)).
(c) 2 marks for cell wall (accept 1 mark for cellulose cell wall, 1 mark for reference to its rigid nature/turgor pressure).
PastPaper.question 7 · Structured
8.88 PastPaper.marks
Group I of the Periodic Table contains the alkali metals.

(a) State the trend in reactivity of the alkali metals as you go down Group I. Explain this trend in terms of the arrangement of electrons. [3.88]

(b) Sodium reacts vigorously with water.
(i) State two observations that would be made during this reaction. [2]
(ii) Write a balanced chemical equation for the reaction of sodium with water, including state symbols. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Reactivity increases down the group.
As you go down Group I, the number of electron shells increases, meaning the outer shell electron is further from the nucleus. This results in a weaker electrostatic attraction between the positively charged nucleus and the outer shell electron, making it easier for the atom to lose its outer electron and react.

(b) (i) Any two from:
- sodium melts into a ball
- sodium floats on the surface of the water
- sodium moves rapidly across the surface
- bubbles/effervescence (hydrogen gas released)
- sodium disappears/dissolves

(ii) \(2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}\)

PastPaper.markingScheme

(a) 1 mark for stating reactivity increases down the group; 1 mark for more electron shells / larger atomic radius down the group; 1 mark for weaker attraction between nucleus and outer electron; 0.88 marks for outer electron being lost more easily.
(b) (i) 2 marks for any two correct observations (1 mark each).
(ii) 1 mark for correct reactants and products; 1 mark for correct balancing; 1 mark for all correct state symbols.
PastPaper.question 8 · Structured
8.88 PastPaper.marks
Photosynthesis is the process by which plants manufacture carbohydrates.

(a) Write the balanced chemical equation for photosynthesis. [3]

(b) The leaf is adapted to carry out photosynthesis efficiently.
(i) Explain how the palisade mesophyll layer is adapted for maximum photosynthesis. [2.88]
(ii) Describe how carbon dioxide enters the leaf and reaches the palisade cells. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) (i) Palisade cells are located near the top of the leaf, allowing them to absorb maximum sunlight. They are packed closely together vertically and contain a high density of chloroplasts (which contain chlorophyll to absorb light).
(ii) Carbon dioxide gas enters the leaf from the atmosphere through the stomata by diffusion. Once inside, it diffuses through the large air spaces in the spongy mesophyll layer to reach the palisade mesophyll layer.

PastPaper.markingScheme

(a) 1 mark for correct reactants; 1 mark for correct products; 1 mark for correct balancing.
(b) (i) 1 mark for stating they are located at the upper surface of the leaf; 1 mark for mentioning high density of chloroplasts; 0.88 marks for mentioning cells are packed vertically.
(ii) 1 mark for entering via stomata; 1 mark for movement by diffusion; 1 mark for moving through air spaces in the spongy mesophyll layer.
PastPaper.question 9 · structured
9 PastPaper.marks
A student investigates how different surfaces absorb thermal radiation. Both cans are placed at an equal distance from a powerful infrared heater. Can X is painted matt black. Can Y is painted shiny silver. Each contains identical volumes of cold water. (a) State the principal method of thermal energy transfer from the infrared heater to the cans through the air. [1] (b) Explain, in terms of particles (including electrons), how thermal energy is conducted through the copper walls of the cans. [3] (c) (i) State and explain which can, X or Y, will show a greater rise in water temperature after 10 minutes. [2] (ii) Copper has a high thermal conductivity. State how this property affects the rate of heat transfer to the water compared to using a plastic can. [1] (d) A lid is placed on each can. Explain how this lid reduces thermal energy loss by convection. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The transfer of thermal energy through the air without requiring a medium or by electromagnetic waves is radiation.

(b) In copper:
- Heat causes copper ions/atoms to vibrate more.
- These vibrations are passed to neighbouring ions/atoms through collisions.
- Copper contains free (delocalised) electrons which gain kinetic energy, move through the metal, and rapidly transfer energy to cooler regions by colliding with other ions.

(c) (i) Can X will show a greater temperature rise because matt black surfaces are excellent absorbers of infrared radiation, whereas shiny silver surfaces are good reflectors.
(ii) A high thermal conductivity means that thermal energy is transferred through the material at a faster rate.

(d) Air heated by the water expands, becomes less dense, and rises (convection). A lid traps this warm air, preventing it from escaping and stopping convection currents from carrying energy away.

PastPaper.markingScheme

(a) Radiation [1] (reject: convection / conduction)

(b)
- Particles/ions near the heat source gain kinetic energy / vibrate more [1]
- Energy is passed to neighbouring particles by collisions [1]
- Free/delocalised electrons gain energy and move through the metal, transferring thermal energy [1]

(c) (i)
- Can X [1]
- Matt black is a better absorber of radiation / shiny silver is a better reflector [1]

(c) (ii)
- Increases the rate of energy transfer / transfers heat faster [1]

(d)
- Warm air rises [1]
- The lid traps the warm air / prevents the escape of warm air [1]

Paper 63 (Alternative to Practical)

Answer all questions in the spaces provided.
4 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · Practical/Investigation
10 PastPaper.marks
A student is investigating a solid sample, salt **X**, which is a mixture of two salts.

They perform several tests and record their observations.

(a) The student dissolves salt **X** in distilled water to make solution **Y**. They divide this solution into three portions.

(i) To the first portion, they add dilute nitric acid followed by aqueous silver nitrate. A cream-coloured precipitate is formed.
Identify the anion present in salt **X**. [1]

(ii) To the second portion, they add aqueous sodium hydroxide dropwise until in excess. A green precipitate is formed which remains insoluble in excess.
Identify the cation present in salt **X**. [1]

(iii) Write down the chemical name of the compound formed from the anion identified in (a)(i) and the cation identified in (a)(ii). [1]

(b) To the third portion, the student adds dilute hydrochloric acid and warms gently. They test the gas evolved using filter paper soaked in acidified potassium manganate(VII).
The purple filter paper turns colourless.

(i) State the name of the gas evolved. [1]

(ii) Identify the other anion present in salt **X**. [1]

(c) The student wants to obtain a pure, dry sample of the green precipitate formed in (a)(ii) from the mixture.
Describe the experimental steps they should take to obtain this pure, dry solid. [3]

(d) Another student suggests using a flame test to check if sodium ions are also present in salt **X**.

(i) Describe how to perform a flame test. [1]

(ii) State the expected observation if sodium ions are present. [1]
PastPaper.showAnswers

PastPaper.workedSolution

(a)(i) The reaction with nitric acid and silver nitrate giving a cream precipitate is the characteristic test for bromide ions (\(\text{Br}^-\)).
(ii) Aqueous sodium hydroxide reacting to give a green precipitate insoluble in excess indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)).
(iii) Combining iron(II) and bromide gives iron(II) bromide (\(\text{FeBr}_2\)).
(b)(i) Dilute acid reacting with a sulfite produces sulfur dioxide gas, which reduces acidified potassium manganate(VII) from purple to colourless.
(ii) The presence of sulfur dioxide gas from reaction with dilute acid confirms the presence of sulfite (\(\text{SO}_3^{2-}\)) anions.
(c) To obtain a pure, dry sample of a precipitate:
1. Filter the mixture to separate the solid precipitate (residue) from the liquid (filtrate).
2. Wash the residue with distilled water to remove any soluble impurities.
3. Dry the washed precipitate in a warm oven or leave it to dry between pieces of filter paper.
(d)(i) A loop of nichrome or platinum wire is cleaned by dipping in concentrated hydrochloric acid and heating in a Bunsen flame. It is then dipped into the test sample and placed in the hot, blue region of the Bunsen flame.
(ii) Sodium ions impart a characteristic persistent yellow colour to the flame.

PastPaper.markingScheme

**Part (a)**
- (i) Bromide [1] (Accept: \(\text{Br}^-\), Reject: bromine)
- (ii) Iron(II) [1] (Accept: \(\text{Fe}^{2+}\), Reject: iron / Iron(III))
- (iii) Iron(II) bromide [1] (Accept: \(\text{FeBr}_2\))

**Part (b)**
- (i) Sulfur dioxide [1] (Accept: \(\text{SO}_2\))
- (ii) Sulfite [1] (Accept: \(\text{SO}_3^{2-}\), Reject: sulfate)

**Part (c)**
- Filter the mixture / filtration [1]
- Wash the residue / precipitate with distilled water [1]
- Dry the precipitate in a warm oven / between filter papers [1] (Reject: heating with direct flame)

**Part (d)**
- (i) Use a clean wire (nichrome/platinum) to place the sample into a blue/non-luminous Bunsen flame [1]
- (ii) Yellow [1] (Accept: orange-yellow, Reject: orange)
PastPaper.question 2 · Practical/Investigation
10 PastPaper.marks
A student investigates how the material of a beaker affects the rate of cooling of hot water.

They use two beakers of identical size and shape: Beaker A is made of copper, and Beaker B is made of glass.

(a) Describe how the student should set up the apparatus and carry out the experiment. State the measurements they should make and the apparatus used to make them. [4]

(b) Suggest two variables that must be kept constant to ensure a fair comparison. [2]

(c) The table below shows the temperature of water in Beaker B (glass) at 60-second intervals.

| Time / s | Temperature / °C |
|---|---|
| 0 | 85.0 |
| 60 | 76.5 |
| 120 | 69.5 |
| 180 | 63.5 |
| 240 | 58.5 |
| 300 | 54.0 |

(i) Calculate the average rate of cooling over the first 120 seconds. Include the unit in your final answer. [2]

(ii) State how the rate of cooling changes as time progresses, and explain this observation in terms of temperature difference. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Detailed steps for the investigation:
- Measure a specific volume of hot water (e.g., 100 cm³) using a measuring cylinder and pour it into each beaker.
- Place a thermometer in each beaker.
- Record the starting temperature (at \(t = 0\) s).
- Start a stopwatch immediately.
- Record the temperature of the water in both beakers every 60 seconds for 5 minutes.
(b) To make this a fair test, control variables must be maintained:
- The same volume of hot water in each beaker (as different volumes have different thermal capacities).
- The same initial temperature of the water.
- Identical environments (e.g., same room, away from drafts, same surface material beneath the beakers).
(c)(i) Temperature change over first 120 seconds = \(85.0 \text{ °C} - 69.5 \text{ °C} = 15.5 \text{ °C}\).
Time period = \(120 \text{ s}\).
Average rate of cooling = \(\frac{15.5 \text{ °C}}{120 \text{ s}} \approx 0.129 \text{ °C/s}\).
Alternatively, in minutes: \(\frac{15.5 \text{ °C}}{2 \text{ min}} = 7.75 \text{ °C/min}\).
(ii) The rate of cooling decreases. This is because the rate of heat transfer depends on the temperature gradient between the system (water) and the surroundings. As the water cools, this temperature difference decreases, leading to slower heat transfer.

PastPaper.markingScheme

**Part (a)**
- Pour equal volumes of hot water into Beaker A and Beaker B [1]
- Record the initial temperature of the water using a thermometer [1]
- Measure temperature at regular time intervals (e.g., every 60 seconds) using a stopwatch/timer [1]
- Record values in a structured table for comparison [1]

**Part (b)**
- Any two control variables: volume of water, initial temperature, thickness of beakers, presence of lid/insulation, draft/room temperature [2]

**Part (c)**
- (i) Correct calculation: \(15.5 / 120 = 0.129\) (or \(0.13\)) or \(15.5 / 2 = 7.75\) [1]
- Correct unit: \(\text{°C/s}\) or \(\text{°C/min}\) corresponding to the calculated rate [1]
- (ii) Rate of cooling decreases/slows down [1]
- Because the temperature difference between the water and room temperature decreases [1]
PastPaper.question 3 · Practical/Investigation
10 PastPaper.marks
A student investigates the effect of light intensity on the rate of photosynthesis of an aquatic plant, *Elodea* (cabomba/pondweed).

The student sets up the apparatus with a light source at a distance \(d\) from the plant and counts the number of gas bubbles produced per minute.

(a) Draw a labelled diagram of the apparatus that could be used for this investigation. [3]

(b) Suggest why the student adds a small amount of sodium hydrogencarbonate to the water. [1]

(c) The student records the following data:

| Distance \(d\) / cm | Number of bubbles in minute 1 | Number of bubbles in minute 2 | Average number of bubbles per minute |
|---|---|---|---|
| 10 | 48 | 50 | 49 |
| 20 | 35 | 33 | 34 |
| 30 | 21 | 19 | 20 |
| 40 | 11 | 9 | 10 |
| 50 | 5 | 3 | 4 |

(i) State the independent variable and the dependent variable in this experiment. [2]

(ii) Describe the relationship shown by the data between the distance \(d\) and the average rate of bubble production. [1]

(iii) State how the light intensity changes as the distance \(d\) increases. [1]

(d) Explain why counting the number of bubbles may not be an accurate method to measure the volume of gas produced, and suggest one way to improve the accuracy of this measurement. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The drawing should include:
- A test tube or beaker containing water and a piece of pondweed (cut end pointing upwards to release bubbles).
- A lamp/light source positioned at some distance from the container.
- A ruler or metre rule positioned between the lamp and the container to measure the distance \(d\).
(b) Sodium hydrogencarbonate (\(\text{NaHCO}_3\)) releases dissolved carbon dioxide into the water. This ensures that carbon dioxide is abundant and does not act as a limiting factor, meaning any changes in the rate of photosynthesis are due solely to changes in light intensity.
(c)(i) The independent variable is the one manipulated by the experimenter, which is the distance \(d\). The dependent variable is the one being measured, which is the average number of bubbles per minute (rate of photosynthesis).
(ii) As the distance \(d\) increases from 10 cm to 50 cm, the average number of bubbles per minute decreases from 49 to 4. This is an inverse relationship.
(iii) As the distance \(d\) increases, the light intensity decreases (it follows an inverse-square relationship, \(\text{Light Intensity} \propto 1/d^2\)).
(d) Bubbles are of unequal volume, so counting them is subjective and imprecise. A much more accurate method is to collect the gas using a gas syringe or an inverted measuring cylinder/capillary tube, allowing the actual volume of oxygen gas (in \(\text{cm}^3\)) to be recorded.

PastPaper.markingScheme

**Part (a)**
- Pondweed shown submerged in water within a container (beaker/test-tube) [1]
- Light source/lamp directed at the container [1]
- Ruler/metre rule shown measuring distance between lamp and container [1]

**Part (b)**
- To provide a constant/excess supply of carbon dioxide [1] (Accept: prevents CO2 being a limiting factor)

**Part (c)**
- (i) Independent: distance \(d\) / light intensity AND Dependent: number of bubbles per minute / rate of bubble production [2] (1 mark for each)
- (ii) As distance increases, rate of bubble production decreases [1] (Accept: inversely proportional/negative relationship)
- (iii) Light intensity decreases [1]

**Part (d)**
- Explanation: Bubbles are different sizes / some gas dissolves in the water [1]
- Improvement: Collect gas in a gas syringe / inverted measuring cylinder / graduated capillary tube to measure actual volume [1]
PastPaper.question 4 · Practical/Investigation
10 PastPaper.marks
A student investigates osmosis using potato cylinders.

They prepare five potato cylinders of equal length and diameter. They record the initial mass of each cylinder and then place each into a different concentration of sucrose solution: 0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm³.

After 2 hours, they remove the potato cylinders, gently dry them, and measure their final mass.

Their results are shown in the table below:

| Concentration of sucrose / mol/dm³ | Initial mass / g | Final mass / g | Percentage change in mass / % |
|---|---|---|---|
| 0.0 | 4.24 | 4.62 | +8.96 |
| 0.2 | 4.15 | 4.31 | +3.86 |
| 0.4 | 4.30 | 4.13 | -3.95 |
| 0.6 | 4.08 | 3.75 | -8.09 |
| 0.8 | 4.21 | 3.72 | **(i)** |

(a)
(i) Calculate the percentage change in mass for the potato cylinder placed in 0.8 mol/dm³ sucrose solution. Show your working. Give your answer to two decimal places. [2]

(ii) Explain why the potato cylinder in 0.0 mol/dm³ sucrose solution gained mass. [2]

(b) Describe how the student should dry the potato cylinders before weighing them and explain why this step is necessary. [2]

(c) State why percentage change in mass is calculated rather than just using the actual change in mass. [1]

(d) Explain how the concentration of sucrose inside the potato cells can be determined from a graph of percentage change in mass against sucrose concentration. [2]

(e) Suggest one safety precaution the student should take when preparing the potato cylinders. [1]
PastPaper.showAnswers

PastPaper.workedSolution

(a)(i) Mass change = \(3.72\text{ g} - 4.21\text{ g} = -0.49\text{ g}\).
Percentage change = \(\frac{-0.49}{4.21} \times 100 = -11.6389...\% \approx -11.64\%\).
(ii) Pure water (0.0 mol/dm³) has a higher water potential than the sap inside the potato cells. Water moves down the water potential gradient, through the partially permeable cell membranes, into the cells by osmosis, causing an increase in mass.
(b) The cylinders should be rolled gently on a paper towel. Doing this removes any liquid adhering to the outside surface of the potato, which would otherwise be weighed and artificially inflate the final mass.
(c) Because the starting masses of the potato cylinders were not exactly identical (ranging from 4.08 g to 4.30 g), comparing the absolute change in mass would be unfair. Percentage change standardizes the results.
(d) If a graph is plotted with sucrose concentration on the x-axis and percentage change in mass on the y-axis, the point where the curve or line of best fit crosses the x-axis (y = 0) represents the isotonic concentration. At this point, there is no net movement of water, meaning the water potential inside the cells is equal to that of the external solution.
(e) Use a cutting board/tile when using the cork borer or scalpel, and always cut away from your hands/fingers to prevent cuts.

PastPaper.markingScheme

**Part (a)**
- (i) Correct working shown: \(\frac{3.72 - 4.21}{4.21} \times 100\) [1]
- Correct calculation to 2 d.p. with negative sign: \(-11.64\%\) [1] (Accept: \(-11.6\%\))
- (ii) Water moves into the potato by osmosis [1]
- Down a water potential gradient / from higher water potential outside to lower water potential inside [1]

**Part (b)**
- Gently dab/blot the cylinders with a paper towel [1]
- To remove excess surface water/solution which would add to the mass [1]

**Part (c)**
- The initial masses of the cylinders were not identical / allows comparison of cylinders of different starting masses [1]

**Part (d)**
- Plot a graph and find where the line of best fit crosses the x-axis / where percentage change in mass is zero [1]
- At this point, there is no net water movement / concentration is isotonic [1]

**Part (e)**
- Cut on a tile/cutting board OR cut away from fingers/hands [1] (Accept: wear safety goggles when handling solutions)

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