An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Biology Section
Answer all questions. Show working for any calculations and use appropriate biological terminology.
12 Question · 40.5 marks
Question 1 · short_answer
2.5 marks
A student cuts three identical cylinders from a fresh potato and places them into a concentrated sucrose solution for 2 hours.
(a) State the change in mass of the potato cylinders after 2 hours. [0.5]
(b) Explain this change in mass in terms of water potential. [2]
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Worked solution
(a) The mass of the potato cylinders decreases (or decreases in mass / becomes lighter). (b) The water potential inside the potato cells is higher than the water potential of the concentrated sucrose solution (or solution has lower water potential). Therefore, water leaves the cells by osmosis across the partially permeable cell membrane down a water potential gradient (from high to low water potential).
Marking scheme
(a) (mass) decreases / mass is reduced / gets lighter [0.5]; (b) (water potential) higher inside the cells / lower in the sucrose solution [1]; water moves out (of potato / cells) by osmosis [0.5]; down a water potential gradient / across a partially permeable membrane [0.5];
Question 2 · short_answer
2.5 marks
The gas exchange system in humans is adapted for efficient diffusion of gases.
(a) Name the specific structures in the lungs where gas exchange occurs. [0.5]
(b) Describe two structural adaptations of these structures that speed up the rate of gas diffusion. [2]
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Worked solution
(a) Gas exchange occurs across the walls of the alveoli (air sacs). (b) Any two adaptations: 1. Large surface area (due to folding / large number of alveoli) allows more gas molecules to diffuse simultaneously. 2. Thin surface / walls are one cell thick, providing a short diffusion distance. 3. Good blood supply / surrounded by a dense network of capillaries to maintain a steep concentration gradient. 4. Moist surface allows gases to dissolve before diffusing.
Marking scheme
(a) alveoli / alveolus / air sacs [0.5]; (b) Any two adaptations with matching explanations for [1 mark] each (max 2): - thin wall / one cell thick wall [0.5] leading to short diffusion distance / short pathway [0.5]; - large surface area [0.5] allowing greater area for diffusion / more gas exchange [0.5]; - rich capillary network / good blood supply [0.5] maintaining a steep concentration gradient [0.5]; - moist lining [0.5] allowing gases to dissolve [0.5];
Question 3 · short_answer
2.5 marks
Vaccination is used to protect individuals against pathogenic diseases.
(a) Define the term pathogen. [0.5]
(b) Explain how the introduction of a harmless form of a pathogen during vaccination leads to long-term immunity. [2]
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Worked solution
(a) A pathogen is a disease-causing organism (such as a bacterium or virus). (b) The harmless pathogen in the vaccine carries specific antigens. These antigens stimulate specific white blood cells (lymphocytes) to produce complementary antibodies and memory cells. If the body is later exposed to the active pathogen, the memory cells recognize the antigens and produce antibodies much faster and in larger quantities, destroying the pathogen before symptoms develop.
Marking scheme
(a) disease-causing organism / organism causing disease [0.5]; (b) antigens (on pathogen) stimulate lymphocytes / white blood cells to produce specific antibodies [1]; memory cells are formed / produced [0.5]; rapid / greater antibody production upon reinfection [0.5];
(a) Identify the trophic level occupied by the lizard in this food chain. [0.5]
(b) Explain why the hawk receives only a very small fraction of the energy originally captured by the grass. [2]
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Worked solution
(a) The lizard occupies the 3rd trophic level (it is the secondary consumer). (b) Energy is lost at each trophic level between grass and hawk because not all biomass is eaten, some parts are indigestible and lost in faeces (egestion), some energy is lost in excretory products (urea), and a large portion is released as heat during cellular respiration for movement and metabolic maintenance.
Marking scheme
(a) third (trophic level) / secondary consumer [0.5]; (b) Any two valid points of energy loss for [1 mark] each (max 2): - energy lost as heat (to surroundings) from respiration [1]; - energy lost through excretion / urine / egestion / faeces / undigested material [1]; - not all of the organism is eaten / consumed [1];
Question 5 · Structured Explanation
3.5 marks
A student investigates water movement in plant tissue using potato cylinders placed in different concentrations of sucrose solution.
(a) Define the term osmosis. [2]
(b) Explain why a potato cylinder placed in a highly concentrated sucrose solution decreases in mass. [1.5]
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Worked solution
(a) Osmosis is formally defined as the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution) through a partially permeable membrane.
(b) The concentrated sucrose solution outside the potato cells has a lower water potential than inside the potato cells (or the cytoplasm has a higher water potential). As a result, water moves out of the potato cells by osmosis down the water potential gradient, leading to a net loss of water and therefore a decrease in total mass.
Marking scheme
(a) 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; [max 2]
(b) (external) solution has lower water potential than inside cells / cells have higher water potential; water leaves / moves out of (cells / potato) by osmosis; [1.5]
Question 6 · Structured Explanation
3.5 marks
Gas exchange in humans takes place across the walls of the alveoli in the lungs.
(a) State two features of the alveoli that allow efficient gas exchange. [1]
(b) Explain the mechanism of ventilation that causes air to be drawn into the lungs during inspiration. [2.5]
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Worked solution
(a) Alveoli have several adaptations: a very large total surface area, thin walls (only one cell thick for a short diffusion distance), a good blood supply / dense capillary network, and a moist surface layer.
(b) During inspiration: external intercostal muscles contract while internal intercostal muscles relax, pulling the ribs upwards and outwards. At the same time, the diaphragm contracts and flattens. This increases the volume of the thorax / thoracic cavity, which decreases the air pressure inside the lungs to below atmospheric pressure. Air is consequently drawn into the lungs down the pressure gradient.
Marking scheme
(a) Any two from: large surface area; thin walls / one cell thick / short diffusion pathway; moist surface; dense capillary network / rich blood supply; [1]
Vaccination is used worldwide to protect populations from infectious diseases.
(a) Define the term pathogen. [1]
(b) Explain how the administration of a vaccine provides long-term active immunity against a specific disease. [2.5]
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Worked solution
(a) A pathogen is a disease-causing organism (such as a bacterium, virus, or fungus).
(b) A vaccine contains weakened, dead, or inactive forms of a pathogen, or isolated antigens. When injected, lymphocytes recognise the antigens as foreign and produce complementary/specific antibodies. The lymphocytes also divide to produce memory cells. If the individual is exposed to the live pathogen in the future, these memory cells recognise the antigen quickly and produce specific antibodies much faster and in higher concentrations, destroying the pathogen before illness occurs.
Marking scheme
(a) disease-causing organism / agent; [1]
(b) vaccine contains weakened / dead / harmless pathogen / antigens; stimulates lymphocytes to produce antibodies; memory cells are produced; on subsequent infection, antibodies are produced more rapidly / in larger quantities; [max 2.5]
Question 8 · Structured Explanation
3.5 marks
A woodland ecosystem contains food chains involving oak trees, caterpillars, blue tits, and sparrowhawks.
(a) State the principal source of energy for this ecosystem. [1]
(b) Explain why the transfer of energy between trophic levels is inefficient, and why food chains rarely exceed four or five trophic levels. [2.5]
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Worked solution
(a) The principal source of energy for almost all biological ecosystems is sunlight / the Sun.
(b) Only about 10% of energy is transferred from one trophic level to the next. Energy is lost as heat generated during cellular respiration, through metabolic activities (e.g., movement), via excretion/egestion (urine and faeces), and because not all parts of the organism are eaten or digested. Because of these cumulative losses at each stage, there is not enough energy remaining after 4 or 5 trophic levels to sustain another viable population of top predators.
Marking scheme
(a) the Sun / sunlight / light energy; [1]
(b) energy is lost as heat / through respiration; energy lost in excretion / faeces / uneaten parts; only a small percentage / ~10% of energy is transferred to next level; not enough energy remains to sustain a viable population at higher trophic levels; [max 2.5]
Question 9 · Structured Explanation
3.5 marks
Enzyme activity is strongly influenced by physical factors such as temperature.
(a) State what is meant by the term optimum temperature for an enzyme. [1]
(b) Explain why the rate of an enzyme-catalysed reaction decreases rapidly at temperatures well above the optimum. [2.5]
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Worked solution
(a) The optimum temperature is the specific temperature at which an enzyme catalyzes a reaction at its maximum / fastest rate.
(b) At temperatures significantly above the optimum, excessive kinetic and thermal energy causes the enzyme molecules to vibrate violently. This disrupts and breaks the chemical bonds (such as hydrogen and ionic bonds) that maintain the specific three-dimensional shape of the protein. As a result, the shape of the active site is irreversibly altered (denaturation). The substrate is no longer complementary and cannot bind to form enzyme-substrate complexes, causing the reaction rate to fall sharply to zero.
Marking scheme
(a) the temperature at which the enzyme activity / reaction rate is highest / maximum; [1]
(b) high temperature causes bonds (holding protein shape) to break; the active site changes shape / loses its complementary shape; enzyme becomes denatured; substrate can no longer bind / fit (into the active site) / no enzyme-substrate complexes form; [max 2.5]
Question 10 · structured
4 marks
Gas exchange in humans takes place in the alveoli located in the lungs.
Explain how the structure of the alveoli is adapted to maximize the rate of gas exchange between the air and the blood.
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Worked solution
The alveoli are adapted for efficient gas exchange in several key ways: 1. Thin walls: The alveolar wall consists of a single layer of epithelial cells (one cell thick), which minimizes the diffusion distance for oxygen and carbon dioxide. 2. Large surface area: The millions of microscopic alveoli collectively create a very large surface area-to-volume ratio, allowing a higher volume of gas to diffuse at any given moment. 3. Rich capillary network: Each alveolus is surrounded by a dense network of capillaries with continuous blood flow, which rapidly removes oxygen and brings carbon dioxide, maintaining a steep concentration gradient. 4. Moist inner surface: A thin layer of moisture lines the alveoli, allowing oxygen to dissolve before diffusing across the cell membranes into the blood.
Marking scheme
Any four from: • (alveolar wall is) one cell thick / very thin giving a short diffusion distance / pathway; • large surface area (to volume ratio) to increase diffusion rate / allow more gas exchange; • dense / extensive capillary network / rich blood supply maintaining a steep concentration gradient; • ventilation / breathing maintains a steep concentration gradient; • moist lining / layer of moisture allowing gases / oxygen to dissolve (before diffusing);
Question 11 · Data Interpretation / Calculation
4.5 marks
A student investigated osmosis by placing identical potato cylinders into sucrose solutions of different concentrations for 2 hours.
Table 1.1 shows their results.
Table 1.1
| Concentration of sucrose solution / (mol/dm³) | Initial mass / g | Final mass / g | Change in mass / g | Percentage change in mass / % | | :--- | :--- | :--- | :--- | :--- | | 0.0 | 3.20 | 3.68 | +0.48 | +15.0 | | 0.2 | 3.25 | 3.48 | +0.23 | +7.1 | | 0.4 | 3.15 | 3.03 | -0.12 | .................... | | 0.6 | 3.30 | 2.97 | -0.33 | -10.0 | | 0.8 | 3.20 | 2.72 | -0.48 | -15.0 |
(a) Calculate the percentage change in mass of the potato cylinder placed in the \(0.4\text{ mol/dm}^3\) sucrose solution. Give your answer to one decimal place.
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Worked solution
(a) Percentage change in mass = \(\frac{\text{Change in mass}}{\text{Initial mass}} \times 100\) \(\text{Percentage change} = \frac{-0.12}{3.15} \times 100 = -3.8095\% \approx -3.8\%\)
(b) The internal concentration is the sucrose concentration where there is no net movement of water (0% change in mass). Looking at the data, the percentage change is positive at \(0.2\text{ mol/dm}^3\) (+7.1%) and negative at \(0.4\text{ mol/dm}^3\) (-3.8%). The point of zero change lies approximately halfway between, around \(0.30\text{ mol/dm}^3\) (acceptable range: \(0.28\text{ to }0.32\text{ mol/dm}^3\)).
(c) Pure water (\(0.0\text{ mol/dm}^3\)) has a higher water potential than the potato cell cytoplasm/sap. Water moves into the cells down a water potential gradient across the partially permeable cell membrane by osmosis.
Marking scheme
(a) - Correct working: \(\frac{-0.12}{3.15} \times 100\) [1]; - Final answer: -3.8(%) (minus sign required, allow 1 mark if 3.8 without sign or rounded to -3.81) [1];
(b) - Value between 0.28 and 0.32 (\(\text{mol/dm}^3\)) [0.5]; - (Because) this is the concentration where there is zero / no change in mass / no net movement of water [1];
(c) - Higher water potential outside (in solution) than inside cells / water potential gradient AND water enters cells by osmosis [1].
Question 12 · Data Interpretation / Calculation
4.5 marks
A scientist investigated changes in breathing parameters of an athlete at rest and while running on a treadmill.
(c) Adaptations of alveoli include: large surface area, thin walls (one cell thick) giving a short diffusion distance, surrounded by a dense network of blood capillaries to maintain a steep concentration gradient, or a moist inner surface.
(b) - Subtraction to find increase: \(72 - 7.0 = 65\) (allow ecf from part (a)) [1]; - Calculation of percentage: \(\frac{65}{7.0} \times 100 = 929(\%)\) / 928.6(%) [1];
(c) - Any one valid adaptation: large surface area / thin wall / short diffusion pathway / well supplied with capillaries (good blood supply) / moist surface [1].
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Answer all questions. You may use the Periodic Table provided on the back page.
13 Question · 34 marks
Question 1 · structured
2.5 marks
Phosphorus is in Group V of the Periodic Table and reacts with hydrogen to form phosphine, \(\text{PH}_3\).
(a) State the number of shared pairs of electrons (covalent bonds) and the number of lone pairs on the phosphorus atom in one molecule of \(\text{PH}_3\).
(b) Describe the bonding in a molecule of phosphine in terms of outer-shell electrons.
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Worked solution
(a) A phosphorus atom (Group V) has 5 outer-shell electrons. It shares 1 electron with each of the 3 hydrogen atoms, forming 3 shared pairs (covalent bonds) and leaving 2 non-bonding electrons as 1 lone pair on phosphorus.
(b) Phosphorus contributes 3 valence electrons to form three single P–H covalent bonds with three hydrogen atoms (each contributing 1 electron). The remaining 2 valence electrons of phosphorus form one lone pair, giving phosphorus a stable outer octet of 8 electrons and each hydrogen a complete shell of 2 electrons.
Marking scheme
(a) 3 shared / bonding pairs AND 1 lone pair [1.0] (b) Three single covalent bonds formed by sharing one electron from P and one from each H [1.0]; phosphorus has one unshared pair / lone pair of electrons (or completes outer shell of 8 / hydrogen completes shell of 2) [0.5]
Question 2 · structured
2.5 marks
Magnesium reacts with nitrogen gas at elevated temperatures to form the ionic compound magnesium nitride.
(a) Deduce the chemical formula of magnesium nitride by stating the charges on the magnesium ion and the nitride ion.
(b) Describe, in terms of electron transfer, how magnesium ions and nitride ions are formed from their neutral atoms.
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Worked solution
(a) Magnesium is in Group II and forms \(\text{Mg}^{2+}\) cations. Nitrogen is in Group V and forms \(\text{N}^{3-}\) anions. To balance the charges: \(3 \times (+2) + 2 \times (-3) = 0\), giving the formula \(\text{Mg}_3\text{N}_2\).
(b) Magnesium atoms lose two outer electrons each (oxidation/cation formation), while nitrogen atoms gain three electrons each (reduction/anion formation) to achieve stable noble gas electronic configurations.
Marking scheme
(a) Correct ion charges \(\text{Mg}^{2+}\) and \(\text{N}^{3-}\) [0.5]; correct balanced formula \(\text{Mg}_3\text{N}_2\) [0.5] (b) Magnesium (atom) loses 2 electrons [0.75]; nitrogen (atom) gains 3 electrons [0.75]
Question 3 · structured
2.5 marks
Carbon dioxide, \(\text{CO}_2\), is a covalent compound.
(a) Deduce the total number of shared electrons and the total number of non-bonding outer-shell electrons in one molecule of \(\text{CO}_2\).
(b) State the type of covalent bond between the central carbon atom and each oxygen atom in a molecule of \(\text{CO}_2\).
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Worked solution
(a) In \(\text{CO}_2\), carbon forms a double covalent bond with each oxygen atom (\(\text{O}=\text{C}=\text{O}\)). There are two double bonds, meaning \(2 \times 4 = 8\) shared electrons (4 bonding pairs). Each oxygen atom retains 4 non-bonding valence electrons (2 lone pairs), giving \(2 \times 4 = 8\) non-bonding electrons in total.
(b) The bond between carbon and each oxygen atom is a double covalent bond.
Marking scheme
(a) 8 shared electrons / 4 shared pairs [1.0]; 8 non-bonding outer electrons / 4 lone pairs total (2 on each O atom) [1.0] (b) Double (covalent) bond / double bond [0.5]
Question 4 · structured
2.5 marks
Potassium reacts with sulfur to produce potassium sulfide.
(a) Write the electronic configuration of a sulfide ion, \(\text{S}^{2-}\), and give the chemical formula of potassium sulfide.
(b) Explain why solid potassium sulfide does not conduct electricity, but molten potassium sulfide is a good electrical conductor.
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Worked solution
(a) A sulfur atom has atomic number 16 with electron configuration 2,8,6. Upon gaining 2 electrons, the sulfide ion (\(\text{S}^{2-}\)) has configuration 2,8,8. Potassium forms \(\text{K}^+\) ions, so two potassium ions balance one sulfide ion, giving \(\text{K}_2\text{S}\).
(b) In solid potassium sulfide, the ions are held in fixed lattice positions by strong electrostatic forces and are not free to move. In molten potassium sulfide, the ionic lattice breaks down, allowing the ions to move freely and carry electrical current.
Marking scheme
(a) 2,8,8 (or \(1s^2 2s^2 2p^6 3s^2 3p^6\)) [0.5]; \(\text{K}_2\text{S}\) [0.5] (b) Solid: ions in fixed positions / cannot move [0.75]; Molten: ions are free to move (and carry charge) [0.75] (Note: reject mention of free/delocalised electrons for electrical conduction in ionic compounds)
Question 5 · short_answer
2 marks
During the electrolysis of concentrated aqueous sodium chloride, chlorine gas, \(\text{Cl}_2\), is produced at the anode.
Write the balanced ionic half-equation for the formation of chlorine gas from chloride ions, \(\text{Cl}^-\).
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Worked solution
At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) lose electrons (oxidation) to form diatomic chlorine molecules (\(\text{Cl}_2\)).
1. Identify reactant and product: \(\text{Cl}^- \to \text{Cl}_2\) 2. Balance chlorine atoms: \(2\text{Cl}^- \to \text{Cl}_2\) 3. Balance charges by adding electrons to the right side: \(2\text{Cl}^- \to \text{Cl}_2 + 2\text{e}^-\)
Marking scheme
\(\text{Cl}_2\) and \(\text{e}^-\) on the correct side / correct species; [1] fully balanced: \(2\text{Cl}^- \to \text{Cl}_2 + 2\text{e}^-\) (accept \(2\text{Cl}^- - 2\text{e}^- \to \text{Cl}_2\)); [1]
Question 6 · short_answer
2 marks
When aqueous sodium hydroxide is added to an aqueous solution containing copper(II) ions, a pale blue precipitate of copper(II) hydroxide is formed.
Write the ionic equation for this precipitation reaction. Include state symbols.
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Worked solution
In precipitation reactions, aqueous ions combine to form an insoluble solid.
1. Reacting aqueous ions: \(\text{Cu}^{2+}\text{(aq)}\) and hydroxide ions \(\text{OH}^-\text{(aq)}\). 2. Solid product: \(\text{Cu(OH)}_2\text{(s)}\). 3. Balanced equation with state symbols: \(\text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \to \text{Cu(OH)}_2\text{(s)}\).
Marking scheme
correct formulae and balancing: \(\text{Cu}^{2+} + 2\text{OH}^- \to \text{Cu(OH)}_2\); [1] correct state symbols: \(\text{(aq)}\), \(\text{(aq)}\) and \(\text{(s)}\) dependent on correct or recognisable formulae; [1]
Question 7 · short_answer
2 marks
In the blast furnace, iron(III) oxide, \(\text{Fe}_2\text{O}_3\), is reduced by carbon monoxide, \(\text{CO}\), to produce molten iron and carbon dioxide gas.
Write the balanced chemical equation for this reaction.
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Worked solution
1. Identify the reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \to \text{Fe} + \text{CO}_2\). 2. Balance the iron (\(\text{Fe}\)) atoms: 2 on left, so 2 on right (\(2\text{Fe}\)). 3. Balance oxygen and carbon atoms using carbon monoxide: each \(\text{CO}\) takes one oxygen atom from \(\text{Fe}_2\text{O}_3\) to form \(\text{CO}_2\), so \(3\text{CO}\) reacts with \(\text{Fe}_2\text{O}_3\) to form \(3\text{CO}_2\). 4. Fully balanced equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2\).
Marking scheme
correct formulae of reactants and products: \(\text{Fe}_2\text{O}_3\), \(\text{CO}\), \(\text{Fe}\), and \(\text{CO}_2\); [1] correct balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2\); [1]
Question 8 · structured
3 marks
A student investigates the reaction between excess dilute hydrochloric acid and \(5.0\text{ g}\) of calcium carbonate.
The reaction is carried out twice: - Experiment 1: using large lumps of calcium carbonate. - Experiment 2: using powdered calcium carbonate.
All other conditions are kept constant.
Explain, using collision theory, why the rate of reaction is greater in Experiment 2 than in Experiment 1.
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Worked solution
When calcium carbonate is in powdered form instead of large lumps, it has a much larger surface area for the same mass. This means a greater number of solid particles are exposed and available to collide with the hydrogen ions in the dilute hydrochloric acid. Consequently, the frequency of collisions between reactant particles increases (i.e. more collisions occur per second), leading to a higher rate of reaction.
Marking scheme
• (powder has a) larger / greater surface area (exposed) [1] • higher frequency of collisions / more collisions per second / more collisions per unit time [1] • more successful collisions per second / greater rate of effective collisions [1]
Note: Do NOT accept 'more collisions' alone without reference to time/rate/frequency for mark point 2.
Question 9 · structured
3 marks
The reaction between sodium thiosulfate solution and dilute hydrochloric acid produces a precipitate of sulfur, which turns the solution cloudy.
Explain, in terms of particles and collision theory, why increasing the temperature of the reaction mixture increases the rate of reaction.
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Worked solution
Increasing the temperature increases the thermal energy of the reactant particles, which is converted to kinetic energy. As a result: 1. The particles move faster, which causes them to collide more frequently (higher collision frequency). 2. More importantly, a much greater fraction of the colliding particles possess energy equal to or greater than the activation energy (\(E_a\)). This means a higher percentage of collisions are successful, significantly increasing the rate of reaction.
Marking scheme
• particles have more kinetic energy / move faster [1] • higher frequency of collisions / more collisions per unit time / more collisions per second [1] • greater proportion / fraction / percentage of particles have energy equal to or greater than activation energy / more collisions are successful (per second) [1]
Question 10 · structured
3 marks
Magnesium ribbon reacts with dilute hydrochloric acid to produce magnesium chloride solution and hydrogen gas.
Explain, in terms of particles and collisions, why the rate of this reaction increases when the concentration of hydrochloric acid is increased from \(0.5\text{ mol/dm}^3\) to \(2.0\text{ mol/dm}^3\).
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Worked solution
When the concentration of the hydrochloric acid is increased, there are more hydrogen ions (acid particles) present in a given volume of solution. Because the particles are crowded closer together, they collide with the surface of the magnesium ribbon more frequently (more collisions per unit time), which increases the rate of successful collisions and thus the rate of the reaction.
Marking scheme
• more particles / ions per unit volume / per \(\text{cm}^3\) / per \(\text{dm}^3\) [1] • higher frequency of collisions / more collisions per second / more collisions per unit time [1] • more successful / effective collisions per unit time / per second [1]
Note: Reject 'more particles' without reference to volume/space for M1. Reject 'more collisions' without time/frequency for M2.
Question 11 · structured
3 marks
Hydrogen peroxide, \(\text{H}_2\text{O}_2\), decomposes slowly at room temperature according to the equation:
When a small amount of solid manganese(IV) oxide is added, oxygen gas is produced much more rapidly. The manganese(IV) oxide acts as a catalyst.
Explain, in terms of activation energy and collision theory, how the catalyst increases the rate of this decomposition.
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Worked solution
A catalyst increases the rate of a chemical reaction without being consumed. It acts by providing an alternative reaction mechanism/pathway that requires a lower activation energy (\(E_a\)). Because the energy barrier is reduced, a greater fraction of the colliding reactant particles have kinetic energy equal to or exceeding this lower activation energy, resulting in a higher rate/frequency of successful (effective) collisions.
Marking scheme
• provides an alternative reaction pathway / route [1] • with a lower activation energy / lowers \(E_a\) [1] • greater fraction / proportion of particles / collisions have energy \(\ge\) activation energy / more successful collisions per second / higher frequency of effective collisions [1]
Question 12 · open_ended
3 marks
Zinc carbonate decomposes when heated according to the equation:
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Worked solution
Step 1: Calculate the relative formula mass (\(M_r\)) of \(\text{ZnCO}_3\) and \(\text{ZnO}\). \(M_r(\text{ZnCO}_3) = 65 + 12 + (3 \times 16) = 125\) \(M_r(\text{ZnO}) = 65 + 16 = 81\)
Step 2: Calculate the amount of moles of \(\text{ZnCO}_3\) reacting. \(\text{Moles of } \text{ZnCO}_3 = \frac{25.0}{125} = 0.200\text{ mol}\)
Step 3: Determine the mass of \(\text{ZnO}\) produced. From the balanced equation, \(1\text{ mol}\) of \(\text{ZnCO}_3\) produces \(1\text{ mol}\) of \(\text{ZnO}\). \(\text{Moles of } \text{ZnO} = 0.200\text{ mol}\) \(\text{Mass of } \text{ZnO} = 0.200 \times 81 = 16.2\text{ g}\)
Marking scheme
\(M_r\) of \(\text{ZnCO}_3 = 125\) AND \(M_r\) of \(\text{ZnO} = 81\) [1]; \(\text{Moles of } \text{ZnCO}_3 = \frac{25.0}{125} = 0.20\text{ (mol)}\) [1]; \(\text{Mass of } \text{ZnO} = 0.20 \times 81 = 16.2\text{ (g)}\) [1]; (ecf for M3 from incorrect moles or incorrect \(M_r\))
Question 13 · open_ended
3 marks
A sample of \(3.60\text{ g}\) of magnesium is added to an excess of dilute hydrochloric acid. The equation for the reaction is:
Calculate the mass of magnesium chloride, \(\text{MgCl}_2\), produced.
\([A_r\text{: Mg} = 24,\; \text{Cl} = 35.5]\)
Show your working.
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Worked solution
Step 1: Calculate the amount of moles of \(\text{Mg}\) reacted. \(\text{Moles of Mg} = \frac{3.60}{24} = 0.150\text{ mol}\)
Step 2: Calculate the relative formula mass (\(M_r\)) of \(\text{MgCl}_2\). \(M_r(\text{MgCl}_2) = 24 + (2 \times 35.5) = 95\)
Step 3: Calculate the mass of \(\text{MgCl}_2\) formed. From the stoichiometric ratio (1:1), \(\text{Moles of } \text{MgCl}_2 = 0.150\text{ mol}\). \(\text{Mass of } \text{MgCl}_2 = 0.150 \times 95 = 14.25\text{ g}\) (or \(14.3\text{ g}\) to 3 sig figs).
Marking scheme
\(\text{Moles of Mg} = \frac{3.60}{24} = 0.15\text{ (mol)}\) [1]; \(M_r\) of \(\text{MgCl}_2 = 24 + (2 \times 35.5) = 95\) [1]; \(\text{Mass of } \text{MgCl}_2 = 0.15 \times 95 = 14.25\text{ (g)} / 14.3\text{ (g)}\) [1]; (ecf for M3 from incorrect moles in M1 or incorrect \(M_r\) in M2)
Physics Section
Answer all questions. Take the acceleration of free fall g to be 9.8 m/s².
13 Question · 34 marks
Question 1 · Formula Calculation
2.5 marks
A motorized winch lifts a crate of mass \(45\text{ kg}\) vertically upwards through a height of \(8.0\text{ m}\). The total electrical energy supplied to the motor during this lift is \(4800\text{ J}\). Calculate the efficiency of the motor as a percentage.
[Take \(g = 9.8\text{ m/s}^2\)]
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Worked solution
Step 1: Calculate the gravitational potential energy (useful energy output) gained by the crate. \(\Delta E_p = mgh = 45\text{ kg} \times 9.8\text{ m/s}^2 \times 8.0\text{ m} = 3528\text{ J}\)
Step 2: Calculate efficiency using the formula: \(\text{Efficiency} = \left(\frac{\text{useful energy output}}{\text{total energy input}}\right) \times 100\%\) \(\text{Efficiency} = \left(\frac{3528}{4800}\right) \times 100\% = 73.5\%\)
Marking scheme
• Formula / calculation for useful energy output: \(\Delta E_p = 45 \times 9.8 \times 8.0 = 3528\text{ (J)}\) [1] • Correct substitution into efficiency formula: \(\frac{3528}{4800} \times 100\) [1] • Correct final answer: \(73.5\%\) (or \(74\%\)) [0.5]
Question 2 · Formula Calculation
2.5 marks
A step-up transformer at an electrical power station has \(600\text{ turns}\) on its primary coil and \(9600\text{ turns}\) on its secondary coil. The alternating potential difference across the primary coil is \(25\text{ kV}\).
Calculate the output potential difference from the secondary coil in \(\text{kV}\).
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Worked solution
Step 1: State the transformer equation relating voltage and turns ratio: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
Step 2: Rearrange to solve for the secondary potential difference \(V_s\): \(V_s = V_p \times \frac{N_s}{N_p}\)
Step 3: Substitute the known values: \(V_s = 25\text{ kV} \times \frac{9600}{600} = 25 \times 16 = 400\text{ kV}\)
An ultrasound scanner emits a sound wave with a frequency of \(2.5 \times 10^6\text{ Hz}\) into human soft tissue. The speed of ultrasound in this tissue is \(1500\text{ m/s}\).
Calculate the wavelength of the ultrasound wave in millimetres (\(\text{mm}\)).
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Worked solution
Step 1: Use the wave speed equation to find the wavelength in metres: \(v = f \lambda \implies \lambda = \frac{v}{f}\) \(\lambda = \frac{1500\text{ m/s}}{2.5 \times 10^6\text{ Hz}} = 6.0 \times 10^{-4}\text{ m}\)
Step 2: Convert the wavelength from metres to millimetres (\(1\text{ m} = 1000\text{ mm}\)): \(\lambda = 6.0 \times 10^{-4} \times 1000 = 0.60\text{ mm}\)
An electric go-kart of mass \(180\text{ kg}\) accelerates uniformly from rest to a velocity of \(22\text{ m/s}\) along a straight horizontal track. The acceleration takes \(4.4\text{ s}\).
Calculate the average useful power output developed by the go-kart motor during this acceleration.
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Worked solution
Step 1: Calculate the kinetic energy gained by the go-kart: \(E_k = \frac{1}{2}mv^2 = 0.5 \times 180\text{ kg} \times (22\text{ m/s})^2 = 90 \times 484 = 43560\text{ J}\)
Step 2: Calculate power developed using \(P = \frac{E}{t}\): \(P = \frac{43560\text{ J}}{4.4\text{ s}} = 9900\text{ W}\) (or \(9.9\text{ kW}\))
Marking scheme
• Correct calculation of kinetic energy: \(E_k = \frac{1}{2}(180)(22)^2 = 43560\text{ (J)}\) [1] • Correct use of \(P = \frac{E}{t}\): \(\frac{43560}{4.4}\) [1] • Correct final answer with unit: \(9900\text{ W}\) / \(9.9\text{ kW}\) [0.5]
Question 5 · Formula Calculation
2.5 marks
A \(100\%\) efficient step-down transformer is connected to a \(230\text{ V}\) mains supply. The current drawn by the primary coil is \(0.80\text{ A}\). The secondary coil delivers electrical energy at an output potential difference of \(12\text{ V}\).
Calculate the current in the secondary coil. Give your answer to 2 significant figures.
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Worked solution
Step 1: For a \(100\%\) efficient transformer, input power equals output power: \(I_p V_p = I_s V_s\)
Step 2: Rearrange to find the secondary current \(I_s\): \(I_s = \frac{I_p V_p}{V_s}\)
Step 3: Substitute the known values: \(I_s = \frac{0.80\text{ A} \times 230\text{ V}}{12\text{ V}} = \frac{184}{12} = 15.33...\text{ A}\)
• Formula \(I_p V_p = I_s V_s\) or \(P = IV\) stated or implied [1] • Correct substitution: \(I_s = \frac{0.80 \times 230}{12}\) [1] • Final value to 2 s.f.: \(15\text{ (A)}\) (allow \(15.3\text{ A}\) for calculation mark if rounding is shown) [0.5]
Question 6 · structured
3 marks
A ray of monochromatic light travels inside a semicircular glass block and strikes the flat glass–air boundary at an angle of incidence of \(38^\circ\).
The refractive index of the glass is \(1.52\).
(a) Calculate the critical angle \(c\) for the glass–air boundary.
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Worked solution
(a) Use the critical angle formula: \(\sin c = \frac{1}{n}\) \(\sin c = \frac{1}{1.52} = 0.6579\) \(c = \arcsin(0.6579) = 41.1^\circ\) (or \(41^\circ\))
(b) Since the angle of incidence \(i = 38^\circ\) is less than the critical angle \(c = 41.1^\circ\), total internal reflection does not occur; the ray is refracted out into the air (bending away from the normal).
Marking scheme
(a) \(\sin c = \frac{1}{n}\) / \(\sin c = \frac{1}{1.52}\) [1]; \(c = 41.1^\circ\) / \(41^\circ\) [1];
(b) (Ray is) refracted / passes into air AND angle of incidence is less than critical angle / \(i < c\) [1]; [Total: 3]
Question 7 · structured
3 marks
A student uses a thin converging lens of focal length \(f = 8.0\text{ cm}\) to view an illuminated object. The object is positioned along the principal axis at a distance of \(12.0\text{ cm}\) from the centre of the lens.
(a) State two characteristics of the image formed when the object is at this position.
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Worked solution
(a) The object distance \(u = 12.0\text{ cm}\) is between \(f\) (\(8.0\text{ cm}\)) and \(2f\) (\(16.0\text{ cm}\)). For an object between \(F\) and \(2F\), the image formed is real, inverted, and magnified (larger than the object).
(b) When \(u = 5.0\text{ cm} < f\) (\(8.0\text{ cm}\)), the object is placed inside the focal length, acting as a magnifying glass. The image formed is virtual.
Marking scheme
(a) Any two from: - real [1]; - inverted / upside down [1]; - magnified / enlarged / larger (than object) [1]; (max 2 marks)
(b) virtual [1]; [Total: 3]
Question 8 · structured
3 marks
A ripple tank is used to demonstrate the refraction of plane water waves. The waves travel from a deep region into a shallow region across a straight boundary at a non-zero angle of incidence. The frequency of the wave generator is kept constant.
(a) State the effect of entering the shallow region on:
(i) the speed of the water waves ................................................................. [1]
(ii) the wavelength of the water waves ........................................................ [1]
(b) Describe how the direction of travel of the wavefronts changes as they cross the boundary into the shallow region.
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Worked solution
(a)(i) In shallow water, waves travel slower, so wave speed decreases. (a)(ii) Since \(v = f\lambda\) and frequency \(f\) remains constant, a decrease in speed causes the wavelength \(\lambda\) to decrease. (b) Because the waves slow down upon entering the shallow region at an angle, the direction of propagation refracts towards the normal (the angle of refraction is less than the angle of incidence).
(b) (bends / turns) towards the normal / angle to the normal decreases [1]; [Total: 3]
Question 9 · structured
2.5 marks
A small steel sphere is released from rest at the surface of a tall cylinder filled with viscous oil.
Explain, in terms of the forces acting on the sphere, why it accelerates initially and eventually falls at a constant terminal velocity.
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Worked solution
1. At the instant of release, the only significant force is the downward gravitational force (weight), which is greater than the upward drag force / resistive force. This creates an unbalanced resultant force downwards, producing acceleration according to \(F = ma\).
2. As the sphere's velocity increases, the upward resistive drag force of the viscous oil increases.
3. Eventually, the upward drag force balances the downward weight (resultant force becomes zero). With no net force acting on it, the sphere continues to fall at a constant terminal velocity (in accordance with Newton's first law).
Marking scheme
• Weight / gravitational force is greater than upward drag / resistive force initially (giving a downward resultant force / acceleration) [1] • As speed increases, drag / resistive force increases [1] • Eventually drag force equals weight / forces are balanced / resultant force is zero (so acceleration is zero / speed is constant) [0.5]
Question 10 · structured
2.5 marks
A student places a drop of liquid ethanol on the back of their hand. The ethanol evaporates rapidly, and the student notices that their skin feels cold.
Explain, in terms of particles and energy, why the evaporation of ethanol causes the skin to cool down.
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Worked solution
1. Evaporation occurs because the particles in the liquid have a distribution of kinetic energies; the particles with the highest kinetic energy near the surface have enough energy to overcome intermolecular attractive forces and escape as vapour.
2. When these faster, more energetic particles escape, the average kinetic energy of the particles remaining in the liquid decreases.
3. Since temperature is directly proportional to average kinetic energy, the temperature of the liquid decreases, and thermal energy is conducted / transferred from the warmer skin into the liquid, cooling the skin.
Marking scheme
• Particles with higher / greatest kinetic energy / fastest particles escape from the liquid / surface [1] • (Leaving behind particles with) lower average kinetic energy / thermal energy of the liquid decreases [1] • Thermal energy / heat is transferred from the skin to the liquid / skin temperature decreases [0.5]
Question 11 · structured
2.5 marks
Optical fibres are used in medical endoscopes to view internal organs.
Explain why light remains inside the glass core of the optical fibre as it travels along a curved path.
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Worked solution
1. The light is travelling inside the glass core, which is optically denser than the surrounding medium (cladding or air).
2. As the light ray strikes the boundary between the core and cladding along the bend, the angle of incidence \(i\) is measured against the normal and is greater than the critical angle \(c\) for the glass-boundary interface.
3. When \(i > c\) in the more dense medium, no refraction out of the core occurs; instead, 100% of the light is reflected internally (total internal reflection).
Marking scheme
• Light travels from a more dense medium to a less dense medium / strikes the glass boundary [0.5] • Angle of incidence is greater than the critical angle [1] • Total internal reflection occurs (so no light is refracted out / escapes) [1]
Question 12 · structured
2.5 marks
A bar magnet is pushed rapidly, north pole first, into a solenoid connected to a sensitive centre-zero galvanometer. The galvanometer needle deflects momentarily to the right.
(a) Explain why an electric current is generated in the solenoid when the magnet moves into it.
(b) State the two changes observed on the galvanometer when the magnet is pulled out of the solenoid at a higher speed.
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Worked solution
(a) When the bar magnet moves into the solenoid, its magnetic field lines cut across the turns of the coil (or there is a changing magnetic flux linkage). This induces an electromotive force (e.m.f.) across the ends of the wire. Because the solenoid forms a closed circuit with the meter, an induced current flows.
(b) Reversing the direction of motion reverses the direction of the induced e.m.f. and current, deflecting the needle in the opposite direction (to the left). Moving the magnet at a higher speed increases the rate of cutting of magnetic field lines, which increases the magnitude of the induced e.m.f., giving a larger deflection.
Marking scheme
(a) • Magnetic field lines cut (by the coil / wire) / changing magnetic field / flux [1] • Induces an electromotive force / e.m.f. / voltage (which causes current in closed circuit) [0.5]
(b) • Deflects in the opposite direction / to the left [0.5] • Greater / larger deflection / larger current [0.5]
Question 13 · structured
2.5 marks
Astronomers observing the spectra of light emitted from distant galaxies notice that absorption lines are shifted towards the red end of the spectrum (redshift).
Explain what redshift is and explain what this observation indicates about the motion of distant galaxies and the nature of the universe.
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Worked solution
1. Redshift refers to the phenomenon where spectral absorption lines from celestial objects are shifted to longer wavelengths (lower frequencies) towards the red end of the visible spectrum due to the Doppler effect / expansion of space.
2. The observed shift to longer wavelengths indicates that the source (distant galaxy) is moving away from the observer (receding from Earth).
3. Because light from almost all distant galaxies is redshifted (and the further the galaxy, the greater the redshift), this provides fundamental evidence that space is stretching and the universe is expanding.
Marking scheme
• Increase in (observed) wavelength / decrease in (observed) frequency of light (emitted by distant galaxies) [1] • Galaxies are moving away (from Earth / from each other) / receding [0.75] • (Provides evidence that) the universe is expanding [0.75]
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