An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Section A: Biology
Answer all four structured theory questions covering cellular biology, plant transport, enzymology, and nervous/hormonal coordination.
18 Question · 43 marks
Question 1 · matching
3 marks
Table 1.1 lists three biological structures found in plants and animals.
Complete Table 1.1 by writing the correct level of organisation for each biological structure.
Choose your answers from the list. Each term may be used once, more than once, or not at all.
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Worked solution
1. Palisade mesophyll is a group of similar, specialised cells working together to perform photosynthesis, so it is a tissue. 2. The stomach consists of multiple tissues (such as muscular, epithelial, and glandular tissue) working together to digest food, so it is an organ. 3. A ciliated epithelial cell is an individual basic unit of life, so it is a cell.
A student views two different plant cells using a light microscope: a palisade mesophyll cell and a root hair cell. State two structural features that are present in a palisade mesophyll cell but absent from a root hair cell.
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Worked solution
Palisade mesophyll cells are located in the leaves where they absorb sunlight for photosynthesis, so they contain chloroplasts (with chlorophyll). Root hair cells are underground in the soil where there is no light, so they lack chloroplasts. In addition, root hair cells have a long cytoplasmic projection (root hair extension) to increase surface area, whereas palisade cells are rectangular/columnar without this extension.
An image of a plant cell has an observed length of \(48\text{ mm}\). The actual length of the cell is \(80\,\mu\text{m}\). Calculate the magnification of the image. Show your working.
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Worked solution
First, convert the image length from millimetres to micrometres: \(48\text{ mm} = 48 \times 1000 = 48\,000\,\mu\text{m}\). Next, apply the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{48\,000}{80} = 600\). Thus, the magnification is \(\times 600\).
Describe what happens to a plant cell when it is placed into pure water, in terms of water movement and cell state.
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Worked solution
Pure water has a higher water potential than the cytoplasm/cell sap inside the plant cell. Water enters the cell by osmosis across the partially permeable cell membrane. The vacuole expands, pushing the cytoplasm against the cell wall, and the cell becomes turgid. The cell wall prevents it from bursting.
Marking scheme
1. Water enters (the cell) by osmosis / down a water potential gradient; 2. Cell becomes turgid / vacuole swells / cytoplasm pushes against the cell wall (accept: does not burst due to cell wall); [Total: 2]
Question 5 · short_answer
2 marks
Mature mammalian red blood cells are specialised cells adapted for transporting oxygen. State one structural adaptation of a mature red blood cell and explain how it helps the cell perform its function.
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Worked solution
A mature mammalian red blood cell has a biconcave disc shape, which provides a large surface area to volume ratio for the rapid diffusion of oxygen into and out of the cell. Alternatively, it lacks a nucleus, which provides more internal space to contain haemoglobin to carry oxygen.
Marking scheme
1. Adaptation: biconcave disc (shape) / lack of nucleus / contains haemoglobin; 2. Linked explanation: (biconcave disc) gives high surface area to volume ratio for faster diffusion OR (lack of nucleus) allows more space for haemoglobin / more oxygen transport OR (haemoglobin) binds reversibly with oxygen; [Total: 2]
Question 6 · short_answer
2 marks
Prokaryotic organisms such as bacteria have distinct cellular features compared to eukaryotic plant cells. (a) Identify the region or structure where genetic material is located in a bacterium. (b) State one cellular organelle present in plant cells that is always absent in bacteria.
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Worked solution
(a) In a bacterium, genetic material is found free in the cytoplasm as a single circular loop of chromosomal DNA (the nucleoid) and sometimes in small circular DNA molecules called plasmids. (b) Bacteria are prokaryotes and lack membrane-bound organelles such as a true nucleus, mitochondria, chloroplasts, or vacuoles.
Marking scheme
(a) circular DNA / chromosome / loop of DNA / plasmid / nucleoid; [1] (b) nucleus / mitochondrion / chloroplast / vacuole / membrane-bound organelle; [1] [Total: 2]
Question 7 · theory
2.5 marks
A student investigates osmosis using cylinders cut from a fresh potato.
The initial mass of each cylinder is recorded before immersing it into a different concentration of sucrose solution for 60 minutes.
(a) Calculate the percentage change in mass for the potato cylinder placed in \(0.8\text{ mol}/\text{dm}^3\) sucrose solution.
percentage change in mass = ............ % [1]
(b) Explain, in terms of water potential and osmosis, the change in mass of the cylinder placed in \(0.8\text{ mol}/\text{dm}^3\) sucrose solution. [1.5]
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(b) The sucrose solution has a lower water potential (is more concentrated) than the cytoplasm/cell sap of the potato cells. Water moves out of the potato cells by osmosis across the partially permeable cell membrane, down a water potential gradient, causing a decrease in mass.
Marking scheme
(a) -14 / -14.0 (%) ; [1]
(b) Any three from: - water potential of solution is lower than inside cells / potato (ORA) ; - water leaves / moves out of the cells / potato ; - by osmosis / down a water potential gradient ; - through a partially permeable membrane ; [max 1.5]
Question 8 · theory
2.5 marks
A student uses a bubble potometer to investigate the effect of air movement on the rate of water uptake by a leafy shoot.
Table 2.1 shows the distance moved by the air bubble in 10 minutes under two different conditions.
Table 2.1
| condition | distance moved in 10 min / \(\text{mm}\) | rate of water uptake / \(\text{mm}/\text{min}\) | | :--- | :--- | :--- | | still air | 18 | 1.8 | | moving air (fan) | 54 | 5.4 |
(a) State the relationship between air movement and the rate of water uptake shown in Table 2.1. [0.5]
(b) Explain why moving air increases the rate of transpiration from the leaves of the shoot. [2]
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Worked solution
(a) As air movement increases (or when moving air is present), the rate of water uptake increases (from 1.8 mm/min to 5.4 mm/min).
(b) Moving air blows away the layer of humid air / water vapour that accumulates immediately outside the stomata. This maintains a steep diffusion/concentration gradient for water vapour between the air spaces inside the leaf (spongy mesophyll) and the external atmosphere, leading to a faster rate of diffusion/evaporation of water through the stomata.
Marking scheme
(a) as air movement increases, rate of water uptake increases / rate is higher in moving air (ORA) ; [0.5]
(b) (moving air) removes / blows away water vapour / humid air from leaf surface / outside stomata ; maintains / increases the diffusion gradient / concentration gradient of water vapour (between inside and outside of leaf) ; increases rate of diffusion of water vapour out through stomata ; [max 2]
Question 9 · theory
2.5 marks
A student investigates the breakdown of starch by the enzyme amylase at different pH values.
At pH 7, the starch is completely broken down in 2.0 minutes. At pH 2, the starch is not broken down even after 30 minutes.
(a) State the name of the chemical reagent used to test for the presence of starch. [0.5]
(b) Explain why amylase is unable to break down starch at pH 2. [2]
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Worked solution
(a) Iodine solution (or iodine in potassium iodide solution) is used to test for starch; it turns from yellow-brown to blue-black in the presence of starch.
(b) Amylase has an optimum pH near pH 7. At an extreme acidic pH such as pH 2, the bonds holding the enzyme's tertiary structure together are disrupted, causing the active site to change shape (denaturation). As a result, starch molecules can no longer fit into the active site, preventing the formation of enzyme-substrate complexes and catalysis.
Marking scheme
(a) iodine (solution) / \(\text{I}_2\) in \(\text{KI}\) ; [0.5]
(b) enzyme / amylase is denatured ; shape of active site changes / is altered ; substrate / starch is no longer complementary (to active site) / cannot fit / cannot bind ; no enzyme-substrate complexes formed ; [max 2]
Question 10 · theory
2.5 marks
A student investigates reaction time using the ruler drop test.
(a) Complete the sequence showing the pathway of an electrical impulse in a voluntary nervous response: [1]
(b) In an experiment, the student tests five participants before and after consuming a sugary caffeinated beverage. The mean drop distance decreases from \(18.4\text{ cm}\) to \(12.1\text{ cm}\).
Explain what this change indicates about the effect of the beverage on nervous coordination. [1.5]
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Worked solution
(a) The impulse travels from the receptor via the sensory neurone to the central nervous system (relay neurone), and then via the motor neurone to the effector (muscle).
(b) A smaller drop distance before catching the ruler means less time elapsed before responding; therefore, reaction time decreased (became faster). Caffeine acts as a nervous system stimulant, speeding up the transmission of impulses across synapses.
Marking scheme
(a) sensory neurone (in first gap) ; motor neurone (in second gap) ; [1]
(b) shorter drop distance corresponds to a shorter / faster reaction time (AW) ; beverage / caffeine acts as a stimulant / increases speed of impulse transmission / neurotransmitter release ; [1.5]
Question 11 · open-ended
2.5 marks
A student investigates the movement of glucose through a model cell membrane using dialysis tubing submerged in pure water at two temperatures: \(20\,^\circ\text{C}\) and \(40\,^\circ\text{C}\).
Explain, using kinetic theory and ideas about particles, why glucose diffuses through the membrane at a higher rate at \(40\,^\circ\text{C}\) than at \(20\,^\circ\text{C}\).
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Worked solution
1. At \(40\,^\circ\text{C}\), glucose molecules absorb thermal energy and gain more kinetic energy. 2. The molecules move faster / have a higher average velocity. 3. This leads to more frequent random movement and collisions with the microscopic pores of the dialysis membrane, resulting in a faster net movement from the region of higher concentration to lower concentration.
Marking scheme
• (glucose) molecules / particles gain kinetic energy [1] • molecules move faster / have higher velocity / greater random motion [0.5] • more frequent collisions with membrane / faster net movement down the concentration gradient [1]
Question 12 · open-ended
2.5 marks
Catalase is an intracellular enzyme that catalyses the breakdown of hydrogen peroxide to water and oxygen.
Describe and explain the effect on the rate of reaction when the temperature is increased from \(10\,^\circ\text{C}\) to the optimum temperature of \(37\,^\circ\text{C}\), in terms of particle movement and collisions.
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Worked solution
1. State trend: The rate of reaction increases up to the optimum temperature. 2. Kinetic energy: Raising temperature increases the kinetic energy of both catalase and hydrogen peroxide molecules, causing them to move faster. 3. Collisions: There are more frequent collisions per unit time between the hydrogen peroxide substrate and the active site of the enzyme, forming more enzyme-substrate complexes.
Marking scheme
• rate of reaction increases [0.5] • molecules (enzyme and substrate) gain kinetic energy / move faster [1] • higher frequency of successful collisions (between substrate and active site) / more enzyme-substrate complexes formed per unit time [1]
Question 13 · open-ended
2.5 marks
Transpiration involves the loss of water vapour from plant leaves.
Explain how water evaporates from the mesophyll cell walls and diffuses out of the leaf, and suggest why moving air (wind) increases the rate of this diffusion process.
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Worked solution
1. Water on the wet surfaces of spongy mesophyll cell walls evaporates into the intercellular air spaces as water vapour. 2. Water vapour diffuses out of the leaf through open stomata down a concentration gradient (from higher water potential inside to lower outside). 3. Wind sweeps away transpired water vapour from around the stomata, preventing a humid boundary layer and maintaining a steep diffusion gradient.
Marking scheme
• water evaporates from the surface of mesophyll cells into intercellular air spaces [0.5] • water vapour diffuses out of the leaf through stomata down a concentration gradient [1] • wind / moving air removes water vapour from the leaf surface / prevents accumulation outside stomata, maintaining a steep concentration / water potential gradient [1]
Question 14 · open-ended
2.5 marks
Nerve impulses are transmitted across a synapse between two neurones.
Explain how neurotransmitters travel across the synaptic cleft and explain why synaptic transmission ensures that nerve impulses travel in only one direction.
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Worked solution
1. Mechanism: Vesicles fuse with the presynaptic membrane, releasing neurotransmitter chemical molecules into the synaptic gap. The molecules diffuse across the cleft down a concentration gradient. 2. Binding: Neurotransmitter molecules bind to complementary protein receptors on the postsynaptic membrane, triggering a new impulse. 3. Directionality: Unidirectional transmission is ensured because vesicles containing the chemical transmitter are only found in the presynaptic knob, and specific receptors are only found on the postsynaptic membrane.
Marking scheme
• (neurotransmitter) diffuses across the synaptic cleft / down a concentration gradient [1] • binds to (specific) receptors on the postsynaptic membrane [0.5] • unidirectional because vesicles / neurotransmitter are only in the presynaptic neurone AND receptors are only on the postsynaptic neurone [1]
Question 15 · open-ended
2.5 marks
Gas exchange in humans takes place in the alveoli.
State two structural adaptations of the alveoli that facilitate rapid diffusion of oxygen into blood capillaries, and explain how continuous blood flow helps maintain a high rate of diffusion.
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Worked solution
1. Adaptations (any two): Thin walls / one cell thick providing a short diffusion pathway; large surface area (millions of alveoli); thin layer of moisture lined with surfactant. 2. Blood flow role: Continuous circulation carries newly absorbed oxygen away from the alveoli and delivers deoxygenated blood, keeping the oxygen concentration in the blood lower than in the alveolar air, thus maintaining a steep concentration gradient for rapid diffusion.
Marking scheme
• thin wall / one cell thick / short diffusion pathway [0.5] • large surface area (due to large number of alveoli) [0.5] • continuous blood flow transports absorbed \(\text{O}_2\) away / brings deoxygenated blood [0.5] • maintains a steep concentration gradient between alveolar air and blood [1]
Question 16 · structured
2.5 marks
A student investigates the factors affecting the movement of gases in a plant leaf during the daytime.
(a) Name the process by which carbon dioxide moves into the leaf through the stomata. [0.5]
(b) Explain, using ideas about particles and kinetic energy, why the rate of diffusion of carbon dioxide increases as the temperature increases from \(15\ ^\circ\text{C}\) to \(30\ ^\circ\text{C}\). [2]
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Worked solution
(a) Carbon dioxide enters the leaf from the atmosphere by diffusion.
(b) At higher temperatures, carbon dioxide molecules absorb thermal energy which is converted to kinetic energy. The molecules move faster with higher random speeds, resulting in more rapid movement from the higher concentration outside the leaf to the lower concentration inside the spongy mesophyll spaces.
Marking scheme
(a) diffusion [0.5];
(b) any two from: * (carbon dioxide) particles / molecules gain (more) kinetic energy [1]; * particles / molecules move faster / have higher speed [1]; * (result in) faster / greater rate of random movement / faster net movement (down a concentration gradient) [1].
Question 17 · Flowchart / Diagram Labelling
2.5 marks
Fig. 1.1 represents a flowchart showing the pathway of water through the tissues of a leaf during transpiration.
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Worked solution
Water travels out of the xylem vessels in the leaf veins into the surrounding mesophyll cells (A) by osmosis. From the wet cell walls of the mesophyll cells, water evaporates into the intercellular air spaces (B) inside the leaf, forming water vapour. This water vapour then diffuses out of the leaf through the stomata (C) into the drier surrounding atmosphere down a concentration gradient.
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Worked solution
(a) The substrate has a complementary shape to the active site of the enzyme and binds directly to it. (b) When the substrate binds to the active site of the enzyme, an intermediate called the enzyme-substrate complex (X) is formed. (c) After the reaction takes place, the substrate is converted into product(s) (Y), which leave the active site, leaving the enzyme molecule unchanged and ready to catalyse further reactions.
Marking scheme
(a) active site [1]; (b) enzyme-substrate complex / ESC [1]; (c) product(s) [0.5]; [Total: 2.5]
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Answer all four structured theory questions covering organic chemistry, atomic structure and bonding, metallurgy/stoichiometry, and reaction kinetics.
16 Question · 33.010000000000005 marks
Question 1 · Chemical Nomenclature & Formula Deduction
1.5 marks
A straight-chain carboxylic acid contains four carbon atoms in each molecule.
(a) State the systematic chemical name of this carboxylic acid. [1]
(b) Deduce the molecular formula of this compound. [0.5]
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Worked solution
(a) A carboxylic acid containing a chain of four carbon atoms with no branches is named using the stem 'butan-' and the suffix '-oic acid', giving butanoic acid.
(b) The general formula for an aliphatic carboxylic acid is \(\text{C}_n\text{H}_{2n}\text{O}_2\). For \(n = 4\), the molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).
Marking scheme
(a) butanoic acid [1];
(b) \(\text{C}_4\text{H}_8\text{O}_2\) [0.5] (allow \(\text{C}_3\text{H}_7\text{COOH}\) / numbers must be correctly subscripted or clear).
Question 2 · Chemical Nomenclature & Formula Deduction
1.5 marks
Aluminium reacts with sulfuric acid to produce the ionic salt aluminium sulfate.
(a) Deduce the chemical formula of aluminium sulfate, given that aluminium forms \(\text{Al}^{3+}\) ions and sulfate is \(\text{SO}_4^{2-}\). [1]
(b) State the total number of oxygen atoms present in one formula unit of aluminium sulfate. [0.5]
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Worked solution
(a) To achieve electrical neutrality between \(\text{Al}^{3+}\) and \(\text{SO}_4^{2-}\): \(2 \times (+3) = +6\) \(3 \times (-2) = -6\) Thus, the formula unit requires two aluminium ions and three sulfate ions: \(\text{Al}_2(\text{SO}_4)_3\).
(b) In \(\text{Al}_2(\text{SO}_4)_3\), there are 3 sulfate groups each containing 4 oxygen atoms: \(3 \times 4 = 12\) oxygen atoms.
Marking scheme
(a) \(\text{Al}_2(\text{SO}_4)_3\) [1] (reject \(\text{Al}_2\text{SO}_{12}\) / brackets around sulfate are essential);
(b) 12 [0.5] (allow ECF from incorrect formula in (a) if calculated correctly).
Question 3 · Chemical Nomenclature & Formula Deduction
1.5 marks
Transition metals can form compounds where the metal ion has different oxidation states.
(a) State the systematic chemical name of the iron oxide with the formula \(\text{Fe}_2\text{O}_3\). [1]
(b) Deduce the chemical formula of iron(II) nitrate. [0.5]
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Worked solution
(a) In \(\text{Fe}_2\text{O}_3\), three oxide ions (\(\text{O}^{2-}\)) contribute a total charge of \(-6\). The two iron ions must contribute \(+6\), so each iron ion is \(\text{Fe}^{3+}\). The systematic name includes Roman numeral (III): iron(III) oxide.
(b) Iron(II) has the ion \(\text{Fe}^{2+}\) and nitrate is \(\text{NO}_3^-\). Balancing the charges gives \(\text{Fe}(\text{NO}_3)_2\).
Marking scheme
(a) iron(III) oxide [1] (allow iron (III) oxide / ferric oxide; reject iron oxide without oxidation state);
Question 4 · Chemical Nomenclature & Formula Deduction
1.5 marks
Solid calcium carbonate reacts with dilute hydrochloric acid to produce a salt, a colourless gas, and water.
(a) State the systematic chemical name of the salt produced in this reaction. [0.5]
(b) Deduce the chemical formula of this salt. [1]
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Worked solution
(a) The reaction between calcium carbonate and hydrochloric acid produces calcium chloride as the salt: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). The name of the salt is calcium chloride.
(b) Calcium forms \(\text{Ca}^{2+}\) ions (Group II) and chloride forms \(\text{Cl}^-\). Balancing the charges requires two chloride ions for one calcium ion: \(\text{CaCl}_2\).
Marking scheme
(a) calcium chloride [0.5];
(b) \(\text{CaCl}_2\) [1] (reject \(\text{CaCl}\) or \(\text{CaCl}^2\)).
Question 5 · theory
2 marks
Ethene, \(\text{C}_2\text{H}_4\), undergoes addition polymerisation to form the synthetic polymer poly(ethene).
Draw the displayed formula of the repeat unit of poly(ethene).
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Worked solution
To draw the repeat unit of poly(ethene): 1. Break the double bond of the ethene monomer into a single \(\text{C}-\text{C}\) bond. 2. Show two single-bonded hydrogen atoms on each carbon atom (four \(\text{C}-\text{H}\) single bonds in total). 3. Draw horizontal extension (continuation) bonds extending through brackets on both sides to show that the chain continues, labelled with an 'n' (optional subscript).
Marking scheme
Single \(\text{C}-\text{C}\) bond with two continuation bonds extending beyond the carbon atoms / brackets; [1] Correct four \(\text{H}\) atoms attached by single bonds to the two carbon atoms; [1]
Question 6 · theory
2 marks
Propene, \(\text{C}_3\text{H}_6\), is an unsaturated hydrocarbon used as the monomer in the production of poly(propene).
Draw the displayed formula of propene, showing all atoms and all covalent bonds.
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Worked solution
Propene contains three carbon atoms and six hydrogen atoms. - It has one carbon-carbon double bond (\(\text{C}=\text{C}\)) and one carbon-carbon single bond (\(\text{C}-\text{C}\)). - The first carbon has 2 \(\text{C}-\text{H}\) single bonds. - The central carbon has 1 \(\text{C}-\text{H}\) single bond. - The third carbon has 3 \(\text{C}-\text{H}\) single bonds. All 8 covalent bonds (including both lines of the double bond) must be clearly shown.
Marking scheme
One \(\text{C}=\text{C}\) double bond between two carbon atoms and one \(\text{C}-\text{C}\) single bond to the third carbon; [1] All 6 \(\text{H}\) atoms correctly placed with all individual single covalent bonds displayed; [1]
Question 7 · theory
2 marks
Synthetic polymers can be formed by addition polymerisation or by condensation polymerisation.
State two differences between addition polymerisation and condensation polymerisation.
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Worked solution
Key differences between addition and condensation polymerisation: 1. Number of products: In addition polymerisation, monomer units add together without forming any other product (100% atom economy for the polymer). In condensation polymerisation, a small molecule (such as \(\text{H}_2\text{O}\) or \(\text{HCl}\)) is eliminated as a by-product. 2. Nature of monomers: Addition polymerisation requires unsaturated monomers (containing \(\text{C}=\text{C}\) double bonds), while condensation polymerisation involves monomers with reactive functional groups (e.g. \(-\text{COOH}\), \(-\text{OH}\), \(-\text{NH}_2\)).
Marking scheme
Addition forms only a single product / polymer only, whereas condensation also produces a small molecule / by-product / water / \(\text{HCl}\); [1] Addition requires monomers with a \(\text{C}=\text{C}\) double bond / unsaturated monomers, whereas condensation requires monomers with two (different) functional groups; [1]
Question 8 · theory
2 marks
Hydrogen sulfide, \(\text{H}_2\text{S}\), is a simple covalent compound.
Draw a dot-and-cross diagram to show the arrangement of outer-shell electrons in a molecule of hydrogen sulfide.
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Worked solution
1. Sulfur is in Group VI and has 6 outer-shell electrons. Hydrogen is in Group I and has 1 outer-shell electron. 2. Sulfur forms a single covalent bond with each hydrogen atom, sharing two pairs of electrons in total (one dot and one cross per S–H bond). 3. Sulfur retains 4 unshared electrons (2 lone pairs) in its outer shell to complete its octet (8 electrons). 4. Each hydrogen atom has 2 electrons in its outer shell (a complete outer shell).
Marking scheme
one shared pair of electrons (one dot and one cross) between sulfur and each of the two hydrogen atoms [1]; two lone pairs of electrons (four non-bonding electrons) on the sulfur atom and no extra electrons on hydrogen [1]
Question 9 · theory
2 marks
Magnesium fluoride, \(\text{MgF}_2\), is an ionic compound.
Draw a dot-and-cross diagram to show the electronic configurations and charges of the ions present in magnesium fluoride. Show outer-shell electrons only.
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Worked solution
1. Magnesium transfers its 2 valence electrons, forming a \(\text{Mg}^{2+}\) ion with an empty outer shell (or full shell from the previous level). 2. Each of the two fluorine atoms gains 1 electron to complete its outer octet of 8 electrons, forming two \(\text{F}^-\)(fluoride) ions. 3. The diagram must show the correct charges (\(2+\) for \(\text{Mg}\), \(1-\) for each \(\text{F}\)) and correct numbers of outer-shell electrons in brackets.
Marking scheme
correct electronic configuration and \(2+\) charge on \(\text{Mg}^{2+}\) (outer shell shown empty or with 8 electrons) [1]; correct electronic configuration showing 8 electrons (7 of one symbol, 1 of the other) and \(1-\) charge on two \(\text{F}^-\)/fluoride ions (or a coefficient of 2 in front of one \([\text{F}]^-\)) [1]
Question 10 · open-ended
2.67 marks
Magnesium carbonate decomposes on heating according to the equation: \(\text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g})\). A \(0.42\text{ g}\) sample of pure magnesium carbonate is heated until decomposition is complete. Calculate the volume of carbon dioxide gas, in \(\text{cm}^3\), formed at room temperature and pressure (r.t.p.). [\(M_r\): \(\text{MgCO}_3 = 84\); molar gas volume at r.t.p. \(= 24.0\text{ dm}^3\)]
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Worked solution
Step 1: Calculate the amount in moles of \(\text{MgCO}_3\) reacted: \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{0.42\text{ g}}{84\text{ g/mol}} = 0.0050\text{ mol}\). Step 2: Determine moles of \(\text{CO}_2\) produced from the 1:1 molar ratio: \(\text{moles of }\text{CO}_2 = 0.0050\text{ mol}\). Step 3: Calculate the volume of \(\text{CO}_2\) at r.t.p.: \(\text{volume} = 0.0050\text{ mol} \times 24.0\text{ dm}^3 = 0.12\text{ dm}^3\). Step 4: Convert volume to \(\text{cm}^3\): \(0.12 \times 1000 = 120\text{ cm}^3\).
Marking scheme
Moles of \(\text{MgCO}_3 = \frac{0.42}{84} = 0.005(0)\text{ (mol)}\) [1]; Multiplication of moles by \(24.0\text{ dm}^3\) or \(24000\text{ cm}^3\) [1]; Correct final volume: \(120\text{ (cm}^3\text{)}\) [1]
Question 11 · open-ended
2.67 marks
Zinc reacts with dilute hydrochloric acid according to the equation: \(\text{Zn}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{ZnCl}_2(\text{aq}) + \text{H}_2(\text{g})\). A student adds excess dilute hydrochloric acid to \(0.13\text{ g}\) of zinc powder. Calculate the volume of hydrogen gas, in \(\text{cm}^3\), collected at room temperature and pressure (r.t.p.). [\(A_r\): \(\text{Zn} = 65\); molar gas volume at r.t.p. \(= 24.0\text{ dm}^3\)]
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Worked solution
Step 1: Calculate moles of zinc used: \(\text{moles of Zn} = \frac{0.13\text{ g}}{65\text{ g/mol}} = 0.0020\text{ mol}\). Step 2: Use the stoichiometric ratio from the balanced equation (\(1\text{ mol Zn} : 1\text{ mol H}_2\)) to find moles of hydrogen: \(\text{moles of H}_2 = 0.0020\text{ mol}\). Step 3: Calculate the gas volume in \(\text{cm}^3\): \(\text{Volume} = 0.0020\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 48\text{ cm}^3\).
Marking scheme
Moles of \(\text{Zn} = \frac{0.13}{65} = 0.002(0)\text{ (mol)}\) [1]; Moles of \(\text{H}_2 = 0.0020\text{ mol}\) AND multiplication by \(24000\) (or \(24.0\)) [1]; Correct final answer: \(48\text{ (cm}^3\text{)}\) [1]
Question 12 · open-ended
2.67 marks
Propane burns completely in oxygen according to the equation: \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\). A sample containing \(72\text{ cm}^3\) of propane is burned in excess oxygen at room temperature and pressure (r.t.p.). Calculate the mass, in grams, of carbon dioxide produced. [\(M_r\): \(\text{CO}_2 = 44\); molar gas volume at r.t.p. \(= 24.0\text{ dm}^3\)]
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Moles of \(\text{C}_3\text{H}_8 = 0.003(0)\text{ mol}\) OR volume of \(\text{CO}_2 = 216\text{ cm}^3\) [1]; Moles of \(\text{CO}_2 = 0.009(0)\text{ mol}\) [1]; Correct final mass: \(0.396\text{ (g)}\) [1]
Question 13 · structured
2 marks
A student investigates the reaction between magnesium ribbon and excess dilute hydrochloric acid.
Explain, in terms of collision theory, why increasing the temperature of the hydrochloric acid increases the rate of reaction.
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Worked solution
When temperature is increased, the reacting particles gain kinetic energy and move faster. This leads to two effects: 1. Particles collide more frequently (higher collision frequency). 2. A higher proportion of the colliding particles have energy equal to or greater than the activation energy (≥ \(E_a\)), resulting in a higher rate of successful collisions.
Marking scheme
particles gain kinetic energy / move faster [1]; more frequent collisions / higher frequency of successful collisions OR more particles have energy greater than or equal to the activation energy [1]
Question 14 · structured
2 marks
Zinc metal reacts with dilute sulfuric acid to produce aqueous zinc sulfate and hydrogen gas.
Explain, in terms of particles and collisions, why increasing the concentration of the dilute sulfuric acid increases the rate of reaction.
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Worked solution
Increasing the concentration of the acid means there are more reactant particles (hydrogen ions) in a given volume (per unit volume). This increases the frequency of collisions between the acid particles and the zinc metal, increasing the rate of reaction.
Marking scheme
more particles / ions per unit volume (or per \(\text{cm}^3\)) [1]; more frequent collisions / higher rate of collisions (between particles) [1]
Question 15 · structured
2 marks
A student investigates the rate of reaction between solid calcium carbonate and dilute nitric acid.
Explain, in terms of collision theory, why powdered calcium carbonate reacts at a faster rate than large lumps of calcium carbonate of the same mass.
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Worked solution
Breaking a solid into a powder increases its total surface area exposed to the acid. As more solid particles are exposed at any one time, collisions between the acid particles and the solid occur more frequently per unit time.
Marking scheme
(powder has a) larger / greater surface area (to volume ratio) [1]; more frequent collisions / higher frequency of collisions (between acid particles and solid) [1]
Question 16 · open-ended
3 marks
A student tests a solution for the presence of sulfate ions.
The student adds dilute hydrochloric acid followed by aqueous barium chloride to the sample. A white precipitate of barium sulfate is formed.
Write the ionic equation, including state symbols, for the formation of barium sulfate.
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Worked solution
To write the ionic equation for the precipitation reaction: 1. Identify the reacting ions: barium ions, \(\text{Ba}^{2+}\), and sulfate ions, \(\text{SO}_4^{2-}\). The spectator ions (chloride ions and the original counter-cation) do not participate in the formation of the precipitate. 2. Write the balanced formula of the product: insoluble barium sulfate, \(\text{BaSO}_4\). 3. Assign state symbols: dissolved ions in solution are denoted by \(\text{(aq)}\), and the solid precipitate is denoted by \(\text{(s)}\).
Final ionic equation: \[\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\]
Marking scheme
\(\text{Ba}^{2+}\) and \(\text{SO}_4^{2-}\) as reactants [1] ; \(\text{BaSO}_4\) as single product and correctly balanced [1] ; correct state symbols: \(\text{(aq)}\) on LHS and \(\text{(s)}\) on RHS [1] (dependent on correct or recognizable formulae) ;
Section C: Physics
Answer all four structured theory questions covering energy resources and efficiency, wave optics and Snell's Law, DC electrical circuits, and nuclear decay.
10 Question · 19.25 marks
Question 1 · theory
2.25 marks
An electric motor is used to lift a container of mass \(45\text{ kg}\) through a vertical height of \(8.0\text{ m}\). The total electrical energy supplied to the motor is \(4800\text{ J}\).
(Gravitational field strength \(g = 9.8\text{ N/kg}\))
(a) Calculate the useful energy transferred (work done) in lifting the container. [1]
(b) Calculate the efficiency of the electric motor. Give your answer as a percentage. [1.25]
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A ray of light travelling in air strikes the flat surface of a glass block with an angle of incidence of \(42^\circ\).
The refractive index of the glass is \(1.52\).
Calculate the angle of refraction, \(r\), inside the glass block. Show your working.
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Worked solution
Use Snell's law: \(n = \frac{\sin i}{\sin r}\) Rearrange to solve for \(\sin r\): \(\sin r = \frac{\sin i}{n} = \frac{\sin 42^\circ}{1.52} = \frac{0.669}{1.52} \approx 0.440\) \(r = \arcsin(0.440) \approx 26.1^\circ\) (or \(26^\circ\))
Marking scheme
(\sin r =) \frac{\sin 42^\circ}{1.52} OR 0.440 [1]; 26(°)$ / $26.1(°)$ [1]
Question 3 · open
2 marks
A transparent liquid has a refractive index of \(1.38\).
Calculate the critical angle \(c\) for light travelling from the liquid into air. Show your working.
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Worked solution
Use the critical angle formula: \(\sin c = \frac{1}{n}\) \(\sin c = \frac{1}{1.38} \approx 0.7246\) \(c = \arcsin(0.7246) \approx 46.4^\circ\) (or \(46^\circ\))
Marking scheme
\sin c = \frac{1}{n} OR \sin c = \frac{1}{1.38} OR 0.725 [1]; 46(°)$ / $46.4(°)$ [1]
Question 4 · open
2 marks
The speed of light in air is \(3.0 \times 10^8\text{ m/s}\). Light travels through a transparent acrylic block at a speed of \(2.0 \times 10^8\text{ m/s}\).
Calculate the refractive index of the acrylic block. Show your working.
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Worked solution
Refractive index \(n = \frac{\text{speed of light in air}}{\text{speed of light in medium}}\) \(n = \frac{3.0 \times 10^8\text{ m/s}}{2.0 \times 10^8\text{ m/s}} = 1.5\)
Marking scheme
n = \frac{v_1}{v_2} OR \frac{3.0 \times 10^8}{2.0 \times 10^8} [1]; 1.5 [1]
Question 5 · open
2 marks
A ray of light in air enters a triangular glass prism. The angle of incidence is \(35^\circ\) and the angle of refraction is \(22^\circ\).
Calculate the refractive index of the glass prism. Give your answer to 2 decimal places and show your working.
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n = \frac{\sin 35^\circ}{\sin 22^\circ} OR \frac{0.574}{0.375} [1]; 1.53 [1]
Question 6 · calculation
2 marks
A battery supplies a steady current of \(0.45\text{ A}\) to a small filament lamp for \(3.0\text{ minutes}\).
Calculate the total quantity of electric charge that passes through the lamp during this time. Show your working and state the unit.
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Worked solution
Convert time to seconds: \(t = 3.0 \times 60\text{ s} = 180\text{ s}\)
Calculate charge: \(Q = I \times t = 0.45\text{ A} \times 180\text{ s} = 81\text{ C}\)
Marking scheme
Conversion of time to seconds (\(180\text{ s}\)) OR substitution \(0.45 \times 3.0 \times 60\) ; [1] \(81\text{ C}\) (allow \(81\text{ coulombs}\) / \(\text{A s}\)) ; [1]
Question 7 · structured
1.75 marks
An unstable isotope of polonium, \(\mathrm{^{218}_{84}Po}\), decays into an isotope of lead (\(\mathrm{Pb}\)) by emitting an alpha (\(\alpha\)) particle.
(a) State the number of protons and the number of neutrons in an \(\alpha\)-particle.
(b) Complete the nuclear equation for this decay by determining the nucleon number (\(A\)) and proton number (\(Z\)) of the resulting lead nucleus:
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Worked solution
(a) An \(\alpha\)-particle is a helium nucleus containing \(2\) protons and \(4 - 2 = 2\) neutrons.
(b) In a nuclear reaction, both total nucleon number and total proton number are conserved: - Total nucleon number: \(218 = A + 4 \implies A = 218 - 4 = 214\) - Total proton number: \(84 = Z + 2 \implies Z = 84 - 2 = 82\)
Therefore, the daughter nucleus is \(\mathrm{^{214}_{82}Pb}\).
Phosphorus-32 is a radioactive isotope used in agricultural tracers. It decays into a stable isotope of sulfur (\(\mathrm{S}\)) by emitting a beta-minus (\(\beta^-\)) particle.
(a) Identify the nature of a \(\beta^-\)-particle.
(b) Complete the decay equation by giving the values of \(A\) and \(Z\):
A freshly prepared radioactive sample intended for medical diagnosis has an initial activity of \(960\text{ Bq}\). After a period of \(18.0\text{ hours}\), its activity drops to \(120\text{ Bq}\).
(a) State what is meant by the half-life of a radioactive isotope.
(b) Calculate the half-life of this isotope in hours. Show your working.
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Worked solution
(a) Half-life is defined as the time taken for half of the unstable nuclei in a sample to decay (OR the time taken for the activity/count rate of a source to fall to half its initial value).
(b) Determine the number of half-lives, \(n\): \(960\xrightarrow{1} 480\xrightarrow{2} 240\xrightarrow{3} 120\text{ Bq}\) This corresponds to \(3\) half-lives.
(a) Time taken for half the radioactive nuclei / count rate / activity to decay / halve ; [0.75] (b) Evidence of \(3\) half-lives (e.g. \(960 \rightarrow 480 \rightarrow 240 \rightarrow 120\) OR \(\frac{960}{2^n} = 120\)) ; [0.5] \(6.0\text{ (hours)}\) ; [0.5]
Question 10 · structured
1.75 marks
A student measures the radiation emitted by a sample of isotope-X in a school laboratory. The background radiation produces a constant count rate of \(18\text{ counts/minute}\).
At the start of the experiment (\(t = 0\)), the total count rate measured by the detector is \(210\text{ counts/minute}\). The half-life of isotope-X is \(4.0\text{ hours}\).
Calculate the total count rate recorded by the detector after \(12.0\text{ hours}\). Show your working.
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Worked solution
1. Calculate the initial corrected count rate of isotope-X alone: \(\text{Corrected count rate at } t = 0 = 210 - 18 = 192\text{ counts/minute}\)
2. Determine the number of half-lives elapsed in \(12.0\text{ hours}\): \(n = \frac{12.0\text{ h}}{4.0\text{ h}} = 3\text{ half-lives}\)
4. Calculate the total count rate recorded by the detector: \(\text{Total count rate} = 24 + 18 = 42\text{ counts/minute}\)
Marking scheme
Subtraction of background: \(210 - 18 = 192\text{ (counts/minute)}\) ; [0.5] Division by \(2^3\) (or \(8\)) to give corrected count rate \(= 24\text{ (counts/minute)}\) ; [0.75] Addition of background to obtain total count rate: \(24 + 18 = 42\text{ (counts/minute)}\) ; [0.5]
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