2025 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

At 60 °C, the high temperature has denatured the catalase enzyme. Denaturation changes the shape of the active site, meaning the substrate (hydrogen peroxide) can no longer fit or bind to it, and no reaction takes place. At 35 °C, the enzyme is highly active as it is near the optimum temperature, and at 20 °C, it is active but slower due to lower kinetic energy.

1 mark for identifying that denaturation at 60 °C changes the active site shape, preventing substrate binding.

The relative formula mass is calculated by adding the relative atomic masses of all atoms present in the formula: Mr = (14 + (4 * 1)) * 2 + 32 + (4 * 16) = (18 * 2) + 32 + 64 = 36 + 32 + 64 = 132.

1 mark for the correct calculation showing 132.

The distance can be found using the area under a speed-time graph. First part (acceleration): area of a triangle = 0.5 * base * height = 0.5 * 3.0 s * 4.0 m/s = 6.0 m. Second part (constant speed): area of a rectangle = base * height = 5.0 s * 4.0 m/s = 20 m. Total distance = 6.0 m + 20 m = 26 m.

1 mark for calculating the two components of distance (6.0 m and 20 m) and adding them to get 26 m.

A concentrated salt solution has a lower water potential than the cytoplasm/cell sap of a plant cell. Therefore, water moves out of the cell down the water potential gradient by osmosis. As water leaves, the vacuole shrinks and the cell membrane pulls away from the cell wall, making the cell plasmolysed.

1 mark for identifying that water leaves the cell and the cell becomes plasmolysed.

The evolution of a gas that turns limewater cloudy indicates the presence of carbon dioxide, which means solid X is a carbonate. The formation of a blue solution indicates the presence of copper(II) ions in solution. Therefore, solid X must be copper(II) carbonate.

1 mark for identifying copper(II) carbonate based on gas and solution tests.

First, find the equivalent resistance of the two 4.0 ohm resistors in parallel: 1/Rp = 1/4 + 1/4 = 2/4 = 1/2, so Rp = 2.0 ohms. Next, calculate the total resistance of the series circuit: Rtotal = 2.0 ohms (series resistor) + 2.0 ohms (parallel combination) = 4.0 ohms. Finally, use Ohm's law to find the total current: I = V / Rtotal = 6.0 V / 4.0 ohms = 1.5 A.

1 mark for correctly calculating the total equivalent resistance (4.0 ohms) and using Ohm's law to get 1.5 A.

Ethene is an alkene containing a carbon-carbon double bond, which allows it to undergo addition polymerisation to form poly(ethene). Ethane is an alkane with only single bonds (saturated) and cannot undergo addition polymerisation. Options A, B, and D are incorrect because they reverse the properties of alkanes and alkenes.

1 mark for identifying that ethene can undergo addition polymerisation while ethane cannot.

All electromagnetic waves are transverse waves that travel at the speed of light (3.0 * 10^8 m/s) in a vacuum. As frequency increases, wavelength decreases according to the wave equation v = f * lambda.

1 mark for selecting the correct constant speed of electromagnetic waves in a vacuum.

At \(10\text{ }^\circ\text{C}\), the enzyme and substrate molecules have low kinetic energy, meaning fewer successful collisions, resulting in a slow reaction rate (takes a relatively long time, e.g., 15 minutes). At \(35\text{ }^\circ\text{C}\), the temperature is near the optimum for amylase, so the rate of reaction is high and digestion is fast (takes a short time, e.g., 3 minutes). At \(80\text{ }^\circ\text{C}\), the high temperature denatures the active site of the amylase enzyme. As a result, the enzyme can no longer bind to starch, and starch is not digested at all.

1 mark for the correct option (A). Distractors test common misconceptions: Option B incorrectly suggests enzymes work fastest at low temperatures; Option C incorrectly suggests enzymes are completely inactive/denatured at low temperatures; Option D suggests the enzyme is denatured at the optimum temperature.

Arteries carry blood away from the heart under high pressure, so they have thick, muscular, and elastic walls with a narrow lumen. Veins carry blood towards the heart under low pressure, so they have thinner walls, a wider lumen, and valves to prevent backflow. Capillaries are sites of substance exchange, having walls only one cell thick to minimize diffusion distance.

1 mark for the correct option (A). Options B, C, and D confuse the structural specifications (wall thickness and presence of valves) or the directions of blood flow between arteries and veins.

In humans, sperm cells are produced in the testes (testis). Fertilization (the fusion of the nuclei of the male and female gametes) occurs in the oviduct. The zygote then divides to form an embryo, which implants and develops as a fetus in the uterus.

1 mark for the correct option (A). Option B incorrectly lists the urethra as the site of production and the oviduct as the site of development. Option C incorrectly lists the vagina as the site of development. Option D incorrectly identifies the scrotum as the site of sperm production and the ovary as the site of development.

The atomic number is 7, so there are 7 protons. The mass number is 14, so the number of neutrons is \(14 - 7 = 7\). The \(\text{N}^{3-}\) ion has a charge of \(3-\), which means it has gained 3 electrons compared to a neutral nitrogen atom. Therefore, the number of electrons is \(7 + 3 = 10\).

1 mark for the correct option (A). Option B incorrectly subtracts electrons instead of adding them. Option C confuses protons and electrons. Option D incorrectly uses the mass number (14) as the neutron number.

The equation for the cracking reaction is:
\[\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_3\text{H}_6 + \text{X}\]
To balance the carbon atoms: \(10 - 2 - 3 = 5\).
To balance the hydrogen atoms: \(22 - 4 - 6 = 12\).
Therefore, the molecular formula of X is \(\text{C}_5\text{H}_{12}\). Since it fits the general formula \(\text{C}_n\text{H}_{2n+2}\), it belongs to the alkane homologous series.

1 mark for the correct option (A). Option B incorrectly suggests \(\text{C}_5\text{H}_{10}\) is formed. Option C has the correct formula but incorrectly classifies it as an alkene. Option D has an incorrect formula and incorrectly classifies it as an alkane.

The useful work done is equal to the gain in gravitational potential energy (\(\Delta E_p\)) of the package, which depends only on the vertical height raised:
\[\Delta E_p = mgh = 40.0\text{ kg} \times 10.0\text{ m/s}^2 \times 3.00\text{ m} = 1200\text{ J}\]
The average useful power (\(P\)) is the work done divided by the time taken:
\[P = \frac{\text{Work Done}}{\text{time}} = \frac{1200\text{ J}}{5.00\text{ s}} = 240\text{ W}\]

1 mark for the correct option (A). Distractors test common errors: Option B incorrectly uses the slope distance instead of the vertical height; Option C forgets to multiply by \(g\); Option D is the total work done (1200 J) without dividing by time.

First, calculate the equivalent resistance of the two parallel resistors (\(R_p\)):
\[\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0}\text{ }\Omega^{-1}\]
So, \(R_p = 3.0\ \Omega\).
Next, find the total resistance of the circuit (\(R_{\text{total}}\)) by adding the series resistor:
\[R_{\text{total}} = R_p + 5.0\ \Omega = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\]
Finally, calculate the total current (\(I\)) using Ohm's law:
\[I = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{8.0\text{ }\Omega} = 1.5\text{ A}\]

1 mark for the correct option (A). Option B is based on simply adding all resistors in series (21.0 \(\Omega\)). Option C is based on omitting the series resistor (using only 3.0 \(\Omega\)). Option D is based on incorrectly treating the parallel network as having a combined resistance of 16.0 \(\Omega\) to get a total of 21.0 \(\Omega\) then dividing wrongly.

The frequency of a wave is determined by its source and does not change when the wave transitions from one medium to another. Thus, the frequency in water remains \(4.0\text{ kHz} = 4000\text{ Hz}\).
The wavelength (\(\lambda\)) in water can be calculated using the wave equation \(v = f \lambda\):
\[\lambda = \frac{v}{f} = \frac{1500\text{ m/s}}{4000\text{ Hz}} = 0.375\text{ m}\]

1 mark for the correct option (A). Option B incorrectly uses the wavelength in air (0.085 m). Option C incorrectly assumes the wavelength stays constant and calculates an increased frequency. Option D incorrectly scales the frequency down and calculates an incorrect wavelength.

At temperatures above the optimum (around 40 °C), the high thermal energy causes the bonds holding the three-dimensional structure of the enzyme to break. This alters the shape of the active site (denaturing) so that the substrate can no longer fit. Thus, the rate of reaction drops rapidly.

[1 mark] - Correctly identifies that high temperatures denature enzymes by changing the shape of the active site.

Arteries have thick muscular and elastic walls to withstand high blood pressure, a relatively narrow lumen, and carry blood away from the heart. Veins have thinner walls, a wider lumen, and carry blood towards the heart.

[1 mark] - Correctly identifies an artery's thick muscular wall, narrow lumen, and direction of flow away from the heart.

Sperm cells are much smaller than egg cells, have a flagellum that makes them mobile, and are produced in millions. Egg cells are large (containing energy stores), immobile, and are typically released one at a time during the menstrual cycle.

[1 mark] - Correct comparison of size, mobility, and relative number of male and female gametes.

Both X and Y have 17 protons. The number of neutrons in X is 37 - 17 = 20. The number of neutrons in Y is 35 - 17 = 18. Therefore, X has two more neutrons than Y. Since they have the same proton number, they are the same element, have the same number of electrons (17), and identical chemical properties.

[1 mark] - Correctly calculates the number of neutrons for both isotopes and identifies that X has two more neutrons than Y.

Ethene is an alkene (unsaturated, contains a C=C double bond) and reacts rapidly with aqueous bromine to decolourise it. Ethane is an alkane (saturated, single bonds only) and does not react with aqueous bromine under normal conditions. Ethene, not ethane, can undergo addition polymerisation to form poly(ethene).

[1 mark] - Identifies that ethene decolourises aqueous bromine because it is unsaturated, unlike ethane.

Adding an excess of the insoluble reactant (copper(II) oxide) ensures that all of the acid is fully neutralised. The unreacted copper(II) oxide can then be easily filtered off, leaving a pure solution of copper(II) sulfate.

[1 mark] - Identifies that adding an excess of insoluble reactant ensures the acid is completely reacted.

Work done is calculated using the formula: \(W = F \times d = 12\text{ N} \times 6.0\text{ m} = 72\text{ J}\). Since the floor is frictionless and there are no other energy transfers, all of the work done is transferred into kinetic energy. Therefore, the final kinetic energy of the box is also 72 J.

[1 mark] - Calculates work done as 72 J and identifies that the final kinetic energy is also 72 J.

First, calculate the current through each parallel branch: \(I_1 = \frac{V}{R_1} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\) and \(I_2 = \frac{V}{R_2} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\). The total current drawn is the sum of the currents in each branch: \(I_{\text{total}} = I_1 + I_2 = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A}\). Alternatively, find the combined resistance of the parallel pair first: \(\frac{1}{R_p} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{3}{6.0}\) which gives \(R_p = 2.0\ \Omega\). Then, \(I_{\text{total}} = \frac{V}{R_p} = \frac{6.0\text{ V}}{2.0\ \Omega} = 3.0\text{ A}\).

[1 mark] - Correctly calculates individual currents or combined resistance to find a total current of 3.0 A.

Palisade mesophyll cells are located in leaves and carry out photosynthesis, so they contain chloroplasts. Root hair cells are underground plant cells that absorb water and mineral ions; they do not receive light, so they do not contain chloroplasts. Both types of plant cells contain a cell wall, a large permanent vacuole, and a nucleus.

[1 mark] A: correct cellular structure (chloroplast) identified as present in palisade cells but absent in root hair cells.

During melting (a change of state from solid to liquid), the temperature remains constant because the thermal energy supplied is used to overcome the attractive forces holding the particles in their fixed positions, rather than increasing their average kinetic energy. The particles change from a highly ordered, regular lattice in the solid to a randomly arranged, fluid state in the liquid.

[1 mark] B: correctly states that temperature is constant during a phase change and the regular lattice breaks down into a random arrangement.

First, find the work done (which equals the gain in gravitational potential energy): \(W = m \times g \times h = 200\text{ kg} \times 10\text{ N/kg} \times 15\text{ m} = 30000\text{ J}\). Next, find the useful power output: \(P = \frac{W}{t} = \frac{30000\text{ J}}{6.0\text{ s}} = 5000\text{ W}\).

[1 mark] C: calculated correctly using \(P = \frac{mgh}{t}\).

High temperatures cause the weak bonds holding the three-dimensional structure of the enzyme to break. This alters the shape of the active site permanently (denaturation), preventing the substrate from binding.

[1 mark] B: correctly identifies that denaturation involves a permanent change in the shape of the active site, preventing substrate binding.

For the representation \({}_{Z}^{A}\text{X}\), \(Z\) is the proton number (atomic number) and \(A\) is the nucleon number (mass number). Here, the number of protons in the nucleus is 17. The number of neutrons in the nucleus is \(A - Z = 37 - 17 = 20\). The nucleus contains only protons and neutrons; electrons are located in shells orbiting the nucleus.

[1 mark] A: correct number of protons and neutrons identified as the only constituents of the nucleus.

In the electromagnetic spectrum, moving from gamma rays to radio waves corresponds to decreasing frequency and increasing wavelength. Therefore, gamma rays (shortest wavelength) < X-rays < ultraviolet < visible light (longest wavelength of the four).

[1 mark] A: Correctly orders the electromagnetic waves from shortest wavelength to longest wavelength.

The reaction between dilute hydrochloric acid (an acid) and calcium carbonate (a metal carbonate) produces calcium chloride, water, and carbon dioxide gas. The production of carbon dioxide gas causes effervescence (bubbles of gas seen) in the flask. Carbon dioxide turns limewater cloudy (milky) when bubbled through it.

[1 mark] A: identifies both the effervescence in the flask and the cloudiness of limewater as positive observations.

Current is calculated using the formula: \(I = \frac{Q}{t}\). The time must be converted from minutes to seconds: \(3.0\text{ minutes} = 3.0 \times 60 = 180\text{ s}\). Substituting the values: \(I = \frac{180\text{ C}}{180\text{ s}} = 1.0\text{ A}\).

[1 mark] A: Correct calculation including conversion of minutes to seconds.

Palisade mesophyll cells (plant) and liver cells (animal) both contain a cell membrane, cytoplasm, and a nucleus. Animal cells lack cell walls, chloroplasts, and large permanent vacuoles.

1 mark for identifying the correct combination of structures common to both plant and animal cells.

Denaturation of an enzyme involves a change in the shape of its active site, preventing the substrate from fitting and reacting. Enzymes are proteins, not carbohydrates, so they do not break down into glucose. Kinetic energy increases with temperature rather than decreasing to zero.

1 mark for identifying that denaturation involves the permanent change of the active site shape.

First, react the excess insoluble copper(II) oxide with dilute sulfuric acid. Filter the mixture to remove the unreacted copper(II) oxide. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until the crystallisation point is reached. Allow it to cool so crystals form, then filter and dry the crystals.

1 mark for the correct, logical chemical preparation sequence starting with reaction, followed by filtration, crystallisation, and drying.

Unsaturated hydrocarbons (alkenes) contain double bonds and react with aqueous bromine, causing it to decolourise. Saturated hydrocarbons (alkanes) do not react with bromine water under normal conditions.

1 mark for correctly matching decolourisation with alkenes (unsaturated) and no reaction with alkanes (saturated).

The distance is the area under the speed-time graph. Phase 1 (acceleration): Area of triangle = 0.5 * base * height = 0.5 * 10 s * 20 m/s = 100 m. Phase 2 (constant speed): Area of rectangle = base * height = 15 s * 20 m/s = 300 m. Total distance = 100 m + 300 m = 400 m.

1 mark for using the area under a speed-time graph to correctly calculate total distance as 400 m.

First, find the combined resistance of the two parallel resistors: 1/Rp = 1/12 + 1/12 = 2/12, which gives Rp = 6 ohms. Next, apply Ohm's law to find the current: I = V / Rp = 6.0 V / 6 ohms = 1.0 A.

1 mark for calculating parallel resistance (6 ohms) and applying V = IR to find the current (1.0 A).

A concentrated salt solution has a lower water potential than the cytoplasm of the plant cells. Water therefore moves out of the cells by osmosis, down a water potential gradient, through the partially permeable cell membrane. This causes the cytoplasm to shrink and the cell membrane to pull away from the cell wall (plasmolysis).

1 mark for identifying both correct net water direction (out of the cells) and the resulting state (plasmolysis).

As you go down Group VII: 1. The colour of the elements gets darker (chlorine is a pale green gas, bromine is a red-brown liquid, iodine is a grey-black solid). 2. Reactivity decreases because the atomic radius increases, making it harder for the nucleus to attract and gain an incoming electron.

1 mark for identifying that colour becomes darker and reactivity decreases down Group VII.

(a) Iodine test: Iodine solution (yellow/brown) is added. If starch is present, it turns blue-black.
(b) Raising the temperature increases the kinetic energy of both amylase and starch molecules. They move faster, leading to a higher frequency of collisions, and a greater proportion of collisions have sufficient energy to react, resulting in a higher rate of reaction.
(c) Beyond the optimum temperature (around \(40^\circ\text{C}\) to \(50^\circ\text{C}\)), the enzyme denatures. The thermal energy disrupts the weak bonds that maintain the specific shape of the enzyme, altering the shape of the active site permanently. Consequently, starch can no longer bind to the active site, stopping the catalytic action.

(a) [2 marks total]
- 1 mark: Add iodine solution
- 1 mark: Correct positive result (turns blue-black)

(b) [3 marks total]
- 1 mark: Molecules gain kinetic energy / move faster
- 1 mark: More frequent collisions
- 1 mark: More successful/effective collisions (between substrate and active site)

(c) [3 marks total]
- 1 mark: Enzyme is denatured (do not accept "killed")
- 1 mark: Shape of the active site is altered / changed
- 1 mark: Substrate (starch) can no longer fit/bind to the active site

(a) This is a neutralisation reaction where a basic metal oxide reacts with an acid to produce a salt and water: \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\).
(b) Excess copper(II) oxide is used to guarantee that no unreacted sulfuric acid remains in the solution, which would otherwise contaminate the crystals during evaporation. Since copper(II) oxide is insoluble in water, the excess solid can easily be removed by filtration, leaving a pure copper(II) sulfate solution as the filtrate.
(c) To obtain crystals, the solution is heated to evaporate water until it is saturated (crystallization point). It is then cooled slowly to allow crystals to grow. The crystals are collected by filtration, rinsed with a tiny amount of cold distilled water to remove surface impurities without dissolving the crystals, and dried carefully using filter paper.

(a) [1 mark total]
- 1 mark: Neutralisation / acid-base reaction

(b) [3 marks total]
- 1 mark: Excess ensures all the acid reacts / is neutralized
- 1 mark: Unreacted copper(II) oxide is insoluble
- 1 mark: Removed by filtration

(c) [4 marks total]
- 1 mark: Heat/evaporate filtrate to point of crystallization (or heat to evaporate some water)
- 1 mark: Cool the solution to allow crystals to form
- 1 mark: Filter to isolate/collect the crystals
- 1 mark: Wash crystals with a small amount of cold distilled water AND dry with filter paper

(a) Formula: \(a = \frac{\Delta v}{t}\)
Working: \(a = \frac{8.0\text{ m/s} - 0\text{ m/s}}{4.0\text{ s}} = 2.0\text{ m/s}^2\).

(b) Distance is the area under the speed-time graph.
Area of triangle (0 to 4 s): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 8.0\text{ m/s} = 16.0\text{ m}\).
Area of rectangle (4 to 10 s): \(\text{width} \times \text{height} = (10.0\text{ s} - 4.0\text{ s}) \times 8.0\text{ m/s} = 6.0\text{ s} \times 8.0\text{ m/s} = 48.0\text{ m}\).
Total distance = \(16.0\text{ m} + 48.0\text{ m} = 64.0\text{ m}\).

(c) Formula: \(F = m \times a\)
Working: \(F = 0.50\text{ kg} \times 2.0\text{ m/s}^2 = 1.0\text{ N}\).

(a) [3 marks total]
- 1 mark: State formula: \(a = \frac{v-u}{t}\) (or change in speed over time)
- 1 mark: Correct substitution: \(8.0 / 4.0\)
- 1 mark: Correct calculation with unit: \(2.0\text{ m/s}^2\)

(b) [3 marks total]
- 1 mark: Recognition that distance is the area under the graph
- 1 mark: Calculation of one correct component (triangle = 16 m OR rectangle = 48 m)
- 1 mark: Correct total sum: \(64\text{ m}\) (with unit)

(c) [2 marks total]
- 1 mark: State formula: \(F = ma\) or correct substitution: \(0.50 \times 2.0\)
- 1 mark: Correct answer: \(1.0\text{ N}\) (accept with correct unit)

(a) In a double circulatory system, the blood travels in two loops: pulmonary circulation (between the heart and lungs to pick up oxygen) and systemic circulation (between the heart and the rest of the body to deliver oxygen). Therefore, blood passes through the heart twice during one complete circuit of the body.
(b) Arteries carry blood away from the heart under high pressure; thus, they require thick walls containing muscle fibers and elastic tissue to stretch and recoil. Their narrow lumen maintains this high pressure. Veins return blood to the heart at much lower pressures, so their walls are thinner, and they have a wider lumen to reduce resistance to flow. Due to low pressure, veins also contain valves to prevent the backflow of blood.
(c) The pulmonary artery is the only artery carrying deoxygenated blood, traveling from the right ventricle of the heart to the lungs.
(d) The coronary artery branches off the aorta to supply the cardiac muscle cells with oxygenated blood and nutrients required for respiration.

(a) [2 marks total]
- 2 marks: Statement that blood passes through the heart twice for one complete circuit of the body (accept 1 mark for mentioning pulmonary and systemic circuits separately if double pass is implied).

(b) [4 marks total]
- 1 mark: Artery has thick/muscular/elastic wall AND vein has thin/less muscular wall.
- 1 mark: Artery has narrow lumen AND vein has wide lumen.
- 1 mark: Vein has valves (to prevent backflow) AND artery does not have valves.
- 1 mark: Relates thickness to resisting high pressure OR relates valves to low pressure / preventing backflow.

(c) [1 mark total]
- 1 mark: Pulmonary artery.

(d) [1 mark total]
- 1 mark: Coronary artery.

(a) The balanced equation is: \(\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)\).
(b) Limestone is added to remove sandy impurities (silicon dioxide, \(\text{SiO}_2\)) which would otherwise clog the furnace. Under high temperatures, limestone thermally decomposes:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).
Calcium oxide (basic) reacts with silicon dioxide (acidic) in a neutralisation reaction to produce molten slag (calcium silicate):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).
(c) The reactivity series determines extraction methods. Carbon is positioned above iron in the reactivity series, allowing it to displace/reduce iron from its oxide. In contrast, aluminium is more reactive than carbon, meaning carbon is not strong enough to reduce aluminium oxide. Therefore, a more energetic method, electrolysis, is required to break the strong chemical bonds in aluminium oxide.

(a) [3 marks total]
- 1 mark: Correct reactants: \(\text{Fe}_2\text{O}_3\) and \(\text{CO}\), and correct products: \(\text{Fe}\) and \(\text{CO}_2\)
- 2 marks: Correct balancing: \(1, 3 \rightarrow 2, 3\) (deduct 1 mark for incorrect balancing but correct formulae)

(b) [3 marks total]
- 1 mark: Limestone decomposes to form calcium oxide (\(\text{CaO}\)) / equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
- 1 mark: Calcium oxide reacts with silicon dioxide/sand (acidic impurity)
- 1 mark: Form slag / calcium silicate / equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)

(c) [2 marks total]
- 1 mark: Carbon is more reactive than iron (so can reduce iron oxide)
- 1 mark: Aluminium is more reactive than carbon (so carbon cannot reduce aluminium oxide / requires electrolysis)

(a) For a parallel circuit, the combined resistance \(R_p\) is given by:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0} = \frac{1}{4.0\ \Omega}\).
Therefore, \(R_p = 4.0\ \Omega\).

(b) The parallel pair is in series with \(R_3\). For a series connection:
\(R_{\text{total}} = R_p + R_3 = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).

(c) Using Ohm's Law, \(I = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).

(d) The total current of \(1.5\text{ A}\) flows through both the parallel combination and the series resistor \(R_3\).
The potential difference across the parallel pair is:
\(V_p = I \times R_p = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\).

(a) [2 marks total]
- 1 mark: Correct formula for parallel resistance (e.g., \(1/R_p = 1/R_1 + 1/R_2\) or \(R_p = (R_1 \times R_2)/(R_1 + R_2)\))
- 1 mark: Correct calculation: \(4.0\ \Omega\) (must include units)

(b) [2 marks total]
- 1 mark: State that series resistances add together: \(R_{\text{total}} = R_p + R_3\)
- 1 mark: Complete substitution and calculation showing \(4.0 + 4.0 = 8.0\ \Omega\)

(c) [2 marks total]
- 1 mark: State formula \(I = V/R\) or correct substitution: \(12.0 / 8.0\)
- 1 mark: Correct calculation: \(1.5\text{ A}\) (must include unit)

(d) [2 marks total]
- 1 mark: Correct approach (e.g., \(V_p = I \times R_p\) or \(V_p = V_{\text{total}} - I \times R_3\))
- 1 mark: Correct calculation: \(6.0\text{ V}\) (must include unit)

(a) The small intestine is the main site where digestion is completed and absorption of digested nutrients takes place. Its inner surface is highly folded into finger-like projections called villi (which also possess microvilli), vastly increasing the surface area for absorption. The epithelium of each villus is only one cell thick, which minimizes the diffusion distance. Furthermore, each villus contains a dense capillary network and a lacteal, maintaining a steep concentration gradient by rapidly carrying absorbed nutrients away.
(b) Bile is produced in the liver and stored in the gall bladder. It contains bile salts that emulsify fats. Emulsification is a mechanical process where large fat globules are broken down into small fat droplets. This increases the surface area to volume ratio, enabling the water-soluble enzyme lipase to chemically digest the lipids into fatty acids and glycerol at a much faster rate.
(c) While carbohydrates consist only of carbon, hydrogen, and oxygen, proteins always contain nitrogen (and sometimes sulfur) in addition to carbon, hydrogen, and oxygen.

(a) [4 marks total]
- 1 mark: Small intestine (accept ileum / duodenum)
- 1 mark each for any two adaptations described (max 3 marks):
* Villi / microvilli present to increase surface area
* Very thin wall / one-cell thick epithelium for short diffusion distance
* Rich capillary network / lacteal to maintain steep concentration gradient

(b) [3 marks total]
- 1 mark: Bile emulsifies fats / breaks large fat droplets into smaller droplets
- 1 mark: This is physical/mechanical digestion (no chemical bonds broken)
- 1 mark: Increases surface area of lipids for lipase to digest them faster

(c) [1 mark total]
- 1 mark: Nitrogen (accept sulfur)

(a) All electromagnetic waves share several key characteristics: they are transverse waves, they can travel through a vacuum (do not require a medium), and they all travel at the same high speed in a vacuum (approximately \(3.0 \times 10^8\text{ m/s}\)).

(b) Formula: \(v = f \lambda\) where \(v\) is wave speed, \(f\) is frequency, and \(\lambda\) is wavelength.
Rearranging for frequency: \(f = \frac{v}{\lambda}\).
Substitution: \(f = \frac{3.0 \times 10^8\text{ m/s}}{7.0 \times 10^{-7}\text{ m}}\).
Calculation: \(f \approx 4.29 \times 10^{14}\text{ Hz}\) (or \(4.3 \times 10^{14}\text{ Hz}\)).

(c) Transverse waves oscillate at a right angle (perpendicularly) to the direction of energy transfer / wave propagation (e.g., light waves, water waves). Longitudinal waves oscillate back and forth parallel to the direction of energy transfer / wave propagation. A common example of a longitudinal wave is a sound wave (or ultrasound, or a compression wave in a spring).

(a) [2 marks total]
- 1 mark each for any two correct properties (max 2 marks):
* Travel at the speed of light (\(3.0 \times 10^8\text{ m/s}\)) in a vacuum
* Do not require a medium / can travel through a vacuum
* Are transverse waves
* Can be reflected / refracted / diffracted

(b) [3 marks total]
- 1 mark: State formula: \(v = f \lambda\) (or \(f = v / \lambda\))
- 1 mark: Correct substitution: \((3.0 \times 10^8) / (7.0 \times 10^{-7})\)
- 1 mark: Correct evaluation: \(4.3 \times 10^{14}\text{ Hz}\) (accept \(4.29 \times 10^{14}\text{ Hz}\), must include correct unit)

(c) [3 marks total]
- 1 mark: Transverse oscillations are perpendicular to wave travel / energy transfer
- 1 mark: Longitudinal oscillations are parallel to wave travel / energy transfer
- 1 mark: Sound waves (or ultrasound, or seismic P-waves) as a longitudinal example

(a) (i) Use the formula for kinetic energy:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = 0.5 \times 1200\text{ kg} \times (18\text{ m/s})^2\)
\(E_k = 600 \times 324 = 194\ 400\text{ J}\) (or \(194.4\text{ kJ}\))

(ii) Use the formula for power:
\(P = \frac{\text{Energy}}{t}\)
\(P = \frac{194\ 400\text{ J}}{8.0\text{ s}} = 24\ 300\text{ W}\) (or \(24.3\text{ kW}\))

(b) Distance travelled is the area under the speed-time graph:
Distance during acceleration phase:
\(d_1 = \text{average speed} \times t = \frac{0 + 18}{2} \times 8.0 = 72\text{ m}\)
Distance during constant speed phase:
\(d_2 = 18\text{ m/s} \times 12\text{ s} = 216\text{ m}\)
Total distance:
\(d_{\text{total}} = 72\text{ m} + 216\text{ m} = 288\text{ m}\)

(c) In real life, some energy is lost/dissipated as thermal energy (heat) and sound to the surroundings. This is because work must be done against friction in the moving parts and air resistance acting on the car. Furthermore, the engine is not 100% efficient at converting chemical energy in fuel into mechanical work.

(a) (i) [3 marks]
- Correct substitution: \(0.5 \times 1200 \times 18^2\) [1]
- Correct calculated value: \(194\ 400\) (or \(194.4\text{ k}\)) [1]
- Correct unit: \(\text{J}\) (or \(\text{kJ}\) if matching the value) [1]

(a) (ii) [2 marks]
- Correct formula or substitution of value from (a)(i) divided by 8 [1]
- Correct final value: \(24\ 300\text{ W}\) (or \(24.3\text{ kW}\)) [1] [Allow ecf from (a)(i)]

(b) [2 marks]
- Correct determination of distance during acceleration (\(72\text{ m}\)) OR distance during constant speed (\(216\text{ m}\)) [1]
- Correct total distance: \(288\text{ m}\) [1]

(c) [2 marks]
Any two points from:
- Energy is lost/dissipated as thermal energy / heat (or sound) [1]
- Work is done against air resistance / friction [1]
- The engine is not 100% efficient [1]

(a) Procedure:
1. Use a syringe to place a fixed volume (e.g., \(2\text{ cm}^3\)) of starch solution into a test-tube.
2. Use a different syringe to add a fixed volume (e.g., \(2\text{ cm}^3\)) of a specific pH buffer solution (e.g., pH 4) to the same test-tube.
3. Add a drop of iodine solution into each well of a spotting tile.
4. Add a fixed volume (e.g., \(1\text{ cm}^3\)) of amylase solution to the starch-buffer mixture, and immediately start the stopwatch.
5. Every 30 seconds, use a pipette to take a sample of the mixture and add it to a well on the spotting tile.
6. Record the time taken for the iodine to remain orange-brown (no longer turn blue-black).
7. Repeat steps 1–6 using the other pH buffer solutions (pH 5, 6, 7, and 8).

(b) Variables to keep constant:
1. Temperature: Controlled by placing all reactant test-tubes in a thermostatically controlled water bath (e.g., at \(35^\circ\text{C}\)) before mixing.
2. Concentration / Volume of amylase / starch solution: Controlled by using the same stock solution and measuring precise volumes with syringes.

(c) End-point detection:
The iodine solution remains orange-brown (does not change color to blue-black), indicating that all starch has been broken down.

(d) Table design:
| pH of solution | Time taken for starch breakdown / s |
| :--- | :--- |
| 4 | |
| 5 | |
| 6 | |
| 7 | |
| 8 | |

(e) Safety:
- Hazard: Amylase enzyme solution / iodine solution can be skin or eye irritants.
- Precaution: Wear safety goggles / lab coat / wash hands immediately if spilled on skin.

(a) Procedure [Max 4 marks]:
- Method of measuring volumes of starch/amylase/buffer (using syringes/graduated pipettes) [1]
- Adding iodine to a spotting tile and testing samples at regular time intervals (e.g., every 30 s) [1]
- Timing until the end-point is reached [1]
- Repeating the entire procedure for at least three different pH values [1]

(b) Controlled variables [2 marks]:
- Identifies one correct variable (e.g., temperature, concentration/volume of starch/amylase) with a valid control method [1]
- Identifies a second correct variable with a valid control method [1]

(c) End-point [1 mark]:
- Iodine solution stops turning blue-black / stays orange-brown [1]

(d) Table [2 marks]:
- Table with clear column headings: 'pH' (no unit) and 'Time' with unit 'seconds' or 's' (e.g., Time / s) [1]
- Table shows space/rows to record data for the five different pH values (4, 5, 6, 7, 8) [1]

(e) Safety [1 mark]:
- Identifies irritation risk from chemicals (amylase/iodine) and proposes safety goggles or gloves; OR glassware breakage risk and proposes safe handling [1]

(a) Titre volumes:
- Titration 1: \(24.2\text{ cm}^3 - 0.0\text{ cm}^3 = 24.2\text{ cm}^3\)
- Titration 2: \(35.4\text{ cm}^3 - 12.5\text{ cm}^3 = 22.9\text{ cm}^3\)
- Titration 3: \(29.1\text{ cm}^3 - 5.0\text{ cm}^3 = 24.1\text{ cm}^3\)

(b) Anomalous result:
- Titration 2 (\(22.9\text{ cm}^3\)) is anomalous because it is not within \(0.2\text{ cm}^3\) of the other values.
- Average of concordant runs (Titration 1 and 3):
\(\text{Average} = \frac{24.2 + 24.1}{2} = 24.15\text{ cm}^3\) (or \(24.2\text{ cm}^3\) rounded to 3 s.f.).

(c) Color change:
- Methyl orange in alkali (NaOH) is yellow. At the neutral end-point, it changes to orange (or red/orange).

(d) Reason for pipette:
- A volumetric pipette has a much lower percentage uncertainty / is much more accurate and precise for measuring fixed volumes than a measuring cylinder.

(e) Crystallization:
1. Heat/evaporate the neutralized solution in an evaporating basin until the crystallization point is reached (or when crystals start forming on a cold glass rod).
2. Leave the hot saturated solution to cool slowly so crystals can grow.
3. Filter off the crystals from the remaining liquid, then dry them by pressing between sheets of filter paper or in a warm oven.

(a) Titre values [3 marks]:
- Titre 1 = \(24.2\text{ cm}^3\) [1]
- Titre 2 = \(22.9\text{ cm}^3\) [1]
- Titre 3 = \(24.1\text{ cm}^3\) [1]

(b) Average calculation [2 marks]:
- Identifies Titration 2 as anomalous [1]
- Calculates average using Titrations 1 & 3: \(24.15\text{ cm}^3\) (allow \(24.1\) or \(24.2\)) [1]

(c) Color change [1 mark]:
- Yellow to orange / peach / pinky-orange (Reject: yellow to red, yellow to clear) [1]

(d) Pipette comparison [1 mark]:
- Pipette is more accurate / precise / has lower volume tolerance than a measuring cylinder [1]

(e) Recovery of crystals [3 marks]:
- Heat solution to evaporate water to crystallization point / saturate [1]
- Cool to allow crystallization to occur [1]
- Filter crystals and dry them with filter paper / in a warm oven (Accept: do not heat to dryness) [1]

(a) Table of results:

| Load / \(\text{N}\) | Length / \(\text{cm}\) | Extension / \(\text{cm}\) |
| :--- | :--- | :--- |
| 1.0 | 4.2 | 1.7 |
| 2.0 | 5.9 | 3.4 |
| 3.0 | 7.5 | 5.0 |
| 4.0 | 9.3 | 6.8 |
| 5.0 | 11.0 | 8.5 |

(b) Precaution:
Read the ruler scale perpendicularly (at eye level) to avoid parallax error. (Alternatively: Ensure the ruler is parallel to the spring, or use a fiducial marker).

(c) Checking limit of proportionality:
Remove the load and check if the spring returns to its original length \(L_0 = 2.5\text{ cm}\). If it does not return to its original length, it has plastically deformed and exceeded its limit.

(d) Finding spring constant \(k\):
1. Plot a graph of Load (on the y-axis) against Extension (on the x-axis).
2. Draw a line of best fit through the data points.
3. Calculate the gradient of this straight line. Since \(F = k e\), the gradient of a Load-Extension graph equals the spring constant \(k\) (in \(\text{N/cm}\)).

(e) Sketch graph:
- A straight line passing directly through the origin \((0,0)\).
- Y-axis labeled 'Load / \(\text{N}\)' and X-axis labeled 'Extension / \(\text{cm}\)' (or vice-versa, with gradient being \(1/k\)).

(a) Table [3 marks]:
- Clear column headings with units: Load / \(\text{N}\), Length / \(\text{cm}\), Extension / \(\text{cm}\) [1]
- All 5 lengths recorded correctly with matching loads [1]
- All 5 extensions calculated correctly (1.7, 3.4, 5.0, 6.8, 8.5) [1]

(b) Precaution [1 mark]:
- Avoid parallax error by reading at eye level / perpendicular to the scale; OR use a set square / align ruler vertically [1]

(c) Limit of proportionality check [1 mark]:
- Remove the mass/load and measure to see if it returns to original length / \(2.5\text{ cm}\) (no permanent deformation) [1]

(d) Calculating \(k\) from graph [3 marks]:
- Plot a graph of Load against Extension [1]
- Draw a line of best fit (straight line through origin) [1]
- State that \(k\) is equal to the gradient of the line (or \(1/\text{gradient}\) if extension is on y-axis) [1]

(e) Sketch [2 marks]:
- Straight line drawn through origin showing direct proportionality [1]
- Axes correctly labeled with variables and units: Load / \(\text{N}\) and Extension / \(\text{cm}\) [1]

(a) Constant variables:
1. Volume of hot water in each beaker (e.g., \(100\text{ cm}^3\)).
2. Initial temperature of the water at the start of timing.
3. Material/size/shape of the beakers used.
4. Location / ambient room temperature (absence of drafts).

(b) Temperature change calculation:
\(\Delta T_B = T_{\text{initial}} - T_{\text{final}} = 85.0^\circ\text{C} - 71.5^\circ\text{C} = 13.5^\circ\text{C}\).

(c) Analysis:
- Best insulator: Cotton wool (Beaker C).
- Explanation: Beaker C had the smallest temperature drop (\(9.5^\circ\text{C}\)), compared to bubble wrap (\(13.5^\circ\text{C}\)) and no insulation (\(21.0^\circ\text{C}\)). A smaller temperature drop indicates a lower rate of thermal energy transfer.

(d) Modification:
- Place a lid (e.g., cardboard, plastic, or foam cover) on top of each beaker to reduce heat loss by convection/evaporation.

(e) Stirring:
- Stirring ensures that thermal energy is distributed evenly throughout the water, making the temperature uniform so the thermometer reading is representative of the whole volume.

(f) Extension:
- Use the same insulating material (e.g., bubble wrap).
- Repeat the experiment using different numbers of layers of this material (e.g., 1 layer, 2 layers, 3 layers, 4 layers).
- Keep all other variables (initial temp, volume, time) constant and compare the resulting temperature changes.

(a) Constant variables [2 marks]:
- State any two: volume of water, initial temperature of water, same beaker type, same surface/ambient conditions [2]

(b) Temperature change [2 marks]:
- Correct subtraction shown: \(85.0 - 71.5\) [1]
- Correct answer: \(13.5^\circ\text{C}\) [1]

(c) Best insulator selection and explanation [2 marks]:
- Identify Cotton Wool (Beaker C) [1]
- Explanation referencing smallest temperature drop (\(9.5^\circ\text{C}\)) compared to Beaker B (\(13.5^\circ\text{C}\)) and control Beaker A (\(21.0^\circ\text{C}\)) [1]

(d) Modification [1 mark]:
- Add a lid / cover on top of each beaker [1]

(e) Stirring [1 mark]:
- To ensure uniform / even temperature distribution throughout the water [1]

(f) Extension [2 marks]:
- Use varying thicknesses / numbers of layers of a single insulating material [1]
- Measure temperature change over the same duration, keeping initial temperature and volume constant [1]