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

First, convert the orbital period from hours to seconds: T = 2.5 hours * 3600 seconds/hour = 9000 seconds. Next, use the formula for orbital speed: v = (2 * pi * r) / T. Substituting the given values: v = (2 * 3.14159 * 8.2 * 10^3) / 9000 = 51522 / 9000 = 5.72 km/s. This rounds to 5.7 km/s.

[1 mark] Correct conversion of time to seconds (9000 s). [1 mark] Correct formula for orbital speed (v = 2 * pi * r / T). [1 mark] Correct calculation of speed as 5.7 km/s (or 5.72 km/s).

Antibodies are proteins produced by lymphocytes that have a complementary shape to specific antigens on the surface of pathogens. They bind to these antigens, which can cause the pathogens to clump together (agglutination). This prevents the pathogens from spreading and makes it easier for phagocytes to engulf and destroy them.

[1 mark] Antibodies bind to specific antigens on the surface of pathogens. [1 mark] They cause pathogens to clump together / agglutinate. [1 mark] This marks them for destruction / makes it easier for phagocytes to engulf them.

The manufacture of ethanol by the hydration of ethene is a catalytic addition reaction. Ethene reacts with steam under the following specific industrial conditions: 1. A phosphoric acid catalyst (H3PO4). 2. A temperature of 300 degrees Celsius. 3. A pressure of 60 atmospheres (or 6000 kPa).

[1 mark] Phosphoric acid (H3PO4) catalyst. [1 mark] Temperature of 300 degrees Celsius (accept 250 to 350 degrees Celsius). [1 mark] Pressure of 60 atmospheres (accept 50 to 70 atmospheres / 6000 kPa).

First, calculate the work done (or increase in gravitational potential energy): E_p = m * g * h = 120 kg * 9.8 N/kg * 8.0 m = 9408 J. Next, calculate the power output using the formula: P = Work Done / time = 9408 J / 5.0 s = 1881.6 W. Rounding to 2 significant figures gives 1900 W (or 1880 W to 3 significant figures).

[1 mark] Correct formula for work done (m*g*h) or power (W/t). [1 mark] Correct calculation of energy / work done as 9408 J. [1 mark] Correct calculation of power as 1900 W (accept 1880 W, or 1920 W / 1900 W if g = 10 N/kg is used).

Palisade mesophyll cells are highly adapted for photosynthesis: 1. They are positioned directly under the upper epidermis to receive maximum sunlight. 2. They contain a large number of chloroplasts containing chlorophyll to absorb light energy. 3. Their column-like, vertically elongated shape allows light to penetrate through the cell and lets many cells be packed closely together.

[1 mark] Positioned near the upper surface of the leaf to maximize light absorption. [1 mark] Contain a very high density / large number of chloroplasts. [1 mark] Elongated / column-shaped / tightly packed to maximize light exposure and efficiency.

Bond breaking is an endothermic process because it requires the absorption of energy to overcome attractive forces. Bond making is an exothermic process because energy is released when new bonds are established. If the total energy absorbed during bond breaking is greater than the total energy released during bond making, the overall reaction is endothermic.

[1 mark] States that bond breaking requires energy / is endothermic. [1 mark] States that bond making releases energy / is exothermic. [1 mark] Explains that the energy absorbed to break bonds is greater than the energy released when bonds are formed.

During the electrolysis of molten lead(II) bromide, Pb2+ ions migrate to the negative electrode (cathode) where they gain electrons to form lead metal, observed as a silvery-grey liquid. Br- ions migrate to the positive electrode (anode) where they lose electrons to form bromine gas, observed as red-brown fumes. The product at the anode is therefore bromine.

[1 mark] Silvery-grey liquid / shiny metal beads at the cathode. [1 mark] Red-brown fumes / brown gas at the anode. [1 mark] Product at the anode is bromine (reject bromide).

Proteins are large molecules that must be chemically digested into smaller soluble molecules. This process is catalyzed by protease enzymes (such as pepsin). The chemical digestion of proteins begins in the stomach (where acid conditions suit pepsin) and continues in the small intestine. The final products of protein digestion are amino acids, which can then be absorbed into the bloodstream.

[1 mark] Enzymes: proteases (accept pepsin or trypsin). [1 mark] Site of origin: stomach (accept small intestine / duodenum). [1 mark] Final product: amino acids.

After the stable main sequence stage, a massive star expands and cools to become a red supergiant. This red supergiant eventually explodes in a supernova. The remnants of the supernova collapse to form either a neutron star or a black hole, depending on its remaining mass.

Award 1 mark for each correct stage in the sequence: 1 mark for stating it forms a red supergiant; 1 mark for stating it explodes in a supernova; 1 mark for stating the final remnant is a neutron star or a black hole.

(a) A uniform deceleration is represented by a straight line with a negative gradient. It starts at point \((8.0, 4.0)\) and ends at \((10.0, 0)\).
(b) Total distance = Area under the speed-time graph.
The area can be split into three parts:
- From \(t = 0\) to \(t = 4.0\text{ s}\) (triangle): \(\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} × 4.0\text{ m/s} = 8.0\text{ m}\).
- From \(t = 4.0\) to \(t = 8.0\text{ s}\) (rectangle): \(\text{Area}_2 = \text{base} \times \text{height} = 4.0\text{ s} × 4.0\text{ m/s} = 16.0\text{ m}\).
- From \(t = 8.0\) to \(t = 10.0\text{ s}\) (triangle): \(\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} × 4.0\text{ m/s} = 4.0\text{ m}\).
Total distance = \(8.0\text{ m} + 16.0\text{ m} + 4.0\text{ m} = 28.0\text{ m}\).

Alternatively, using the area of a trapezium:
\(\text{Distance} = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} × (4.0\text{ s} + 10.0\text{ s}) × 4.0\text{ m/s} = 28\text{ m}\).

(a) [1 mark]:
- Correctly states that the line is straight and slopes downwards (negative gradient) (0.5 marks), starting at \(4.0\text{ m/s}\) and ending at \(0\text{ m/s}\) (or at \(t = 10.0\text{ s}\)) (0.5 marks).
(b) [1.5 marks]:
- Shows a correct method for calculating the area under the graph (either by sum of individual shapes or trapezium formula) (0.5 marks).
- Obtains correct final distance of \(28\text{ m}\) (1.0 mark). [Award full 1.5 marks for correct numerical answer with no working shown].

(a) The palisade mesophyll layer is located near the upper surface of the leaf to receive maximum sunlight. It contains a high concentration of chloroplasts which are clearly visible in diagrams as small green organelles.
(b) Carbon dioxide gas diffuses into the leaf down its concentration gradient. On a diagram, this pathway is represented by an arrow starting in the external atmosphere, entering the leaf through a stoma in the lower epidermis, passing through the air spaces of the spongy mesophyll layer, and terminating at the cell wall/membrane of a palisade mesophyll cell.

(a) [1 mark]:
- Identifies the palisade mesophyll (or palisade layer) (0.5 marks).
- States a correct visible adaptation: closely packed together, vertically elongated, or contains numerous chloroplasts (0.5 marks).
(b) [1.5 marks]:
- Starts outside the leaf / in the atmosphere and enters through a stoma (0.5 marks).
- Passes through the air spaces of the spongy mesophyll (0.5 marks).
- Ends at a palisade cell / spongy mesophyll cell (0.5 marks).

(a) The reaction is exothermic because the product energy level (\(40\text{ kJ/mol}\)) is lower than the reactant energy level (\(120\text{ kJ/mol}\)). This difference means energy is lost to the surroundings.
(b) Activation energy is the energy required to go from the reactants' level to the top of the energy barrier:
\(E_a = 180\text{ kJ/mol} - 120\text{ kJ/mol} = +60\text{ kJ/mol}\).
The enthalpy change (\(\Delta H\)) is the difference between product energy and reactant energy:
\(\Delta H = 40\text{ kJ/mol} - 120\text{ kJ/mol} = -80\text{ kJ/mol}\).

(a) [1 mark]:
- Identifies the reaction as exothermic (0.5 marks).
- Explains that the products have lower energy than the reactants (or energy is released / \(\Delta H\) is negative) (0.5 marks).
(b) [1.5 marks]:
- Calculates \(E_a = +60\text{ kJ/mol}\) (or \(60\text{ kJ/mol}\)) (0.75 marks).
- Calculates \(\Delta H = -80\text{ kJ/mol}\) (0.75 marks). Deduct 0.25 marks if the negative sign is missing for the enthalpy change.

(a) With current on the y-axis and potential difference on the x-axis, the graph is a non-linear curve. Initially, it is straight, but as current increases, the temperature of the filament rises. This increases the vibrations of metal ions, which increases the resistance. Consequently, the gradient of the graph (which represents \(1/R\)) decreases.
(b) The standard symbol for a filament lamp is a circle with a cross (X) inside. In order to collect the necessary values, an ammeter must be placed in series with the lamp to measure current, and a voltmeter must be connected in parallel across the lamp to measure potential difference.

(a) [1 mark]:
- Describes the curve as having a decreasing gradient (curve gets less steep as potential difference increases) (0.5 marks).
- Explains that resistance increases as the temperature of the filament increases (0.5 marks).
(b) [1.5 marks]:
- Correctly describes the symbol of a filament lamp (circle with an X inside) (0.5 marks).
- States that the ammeter must be connected in series with the lamp (0.5 marks).
- States that the voltmeter must be connected in parallel across the lamp (0.5 marks).

First, calculate the initial kinetic energy of the car: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.80\text{ kg} \times (5.0\text{ m/s})^2 = 10\text{ J}\). Since the car is brought to rest, the work done by the braking force is equal to the initial kinetic energy: \(W = E_k = 10\text{ J}\). Next, use the work equation: \(W = F \times d\). Therefore, \(10\text{ J} = F \times 2.0\text{ m}\), which gives \(F = \frac{10}{2.0} = 5.0\text{ N}\).

1 mark for calculating the kinetic energy (\(10\text{ J}\)). 1 mark for recalling or using \(W = F \times d\). 1 mark for the correct final force (\(5.0\text{ N}\) or \(5\text{ N}\)).

1. Calculate the relative formula mass of \(\text{CaCO}_3\): \(M_r = 40 + 12 + (3 \times 16) = 100\). 2. Calculate the moles of \(\text{CaCO}_3\) reacted: \(\text{moles} = \frac{5.0\text{ g}}{100} = 0.05\text{ mol}\). 3. Use the stoichiometric ratio from the balanced equation (1 mol \(\text{CaCO}_3\) produces 1 mol \(\text{CO}_2\)) to find the moles of \(\text{CO}_2\): \(\text{moles of CO}_2 = 0.05\text{ mol}\). 4. Calculate the volume of \(\text{CO}_2\) at r.t.p.: \(\text{volume} = 0.05\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.2\text{ dm}^3\).

1 mark for finding \(M_r\) of \(\text{CaCO}_3\) is 100 and calculating moles of \(\text{CaCO}_3\) as \(0.05\text{ mol}\). 1 mark for using the 1:1 mole ratio to find moles of \(\text{CO}_2\). 1 mark for calculating the final volume as \(1.2\text{ dm}^3\).

1. Convert time from minutes to seconds: \(t = 3.0 \times 60 = 180\text{ s}\). 2. Use the electrical power formula: \(P = I^2 R = (0.40\text{ A})^2 \times 15\ \Omega = 2.4\text{ W}\). 3. Calculate the electrical energy transferred: \(E = P t = 2.4\text{ W} \times 180\text{ s} = 432\text{ J}\). Alternatively, use \(E = I^2 R t\) directly: \(E = (0.40\text{ A})^2 \times 15\ \Omega \times 180\text{ s} = 432\text{ J}\) (or \(430\text{ J}\) to two significant figures).

1 mark for converting time to seconds (\(180\text{ s}\)). 1 mark for selecting and using a correct formula (e.g., \(E = I^2 R t\) or calculating power as \(2.4\text{ W}\)). 1 mark for the correct energy value (\(432\text{ J}\) or \(430\text{ J}\)).

In the core of the Sun, hydrogen nuclei undergo nuclear fusion to form helium nuclei. This process occurs under extreme gravitational pressures and temperatures, converting mass into energy. The energy released is emitted from the Sun and travels through space as electromagnetic waves (primarily visible light and infrared radiation). Since electromagnetic waves do not require a medium to propagate, they can travel through the vacuum of space to reach the Earth.

Fusion definition: hydrogen nuclei fuse to form helium nuclei [1]. Conditions: high temperature and high pressure in the core [1]. Transfer mechanism: electromagnetic radiation / light / infrared waves [1]. Vacuum propagation: electromagnetic waves do not require a physical medium / can travel through a vacuum [0.5].

First, calculate the force needed to lift the crate: F = m * g = 120 kg * 9.8 N/kg = 1176 N. Second, calculate the work done: W = F * d = 1176 N * 8.0 m = 9408 J (or 9400 J to two significant figures). Third, calculate the power output: P = W / t = 9408 J / 6.0 s = 1568 W (or 1570 W to two significant figures).

State formula for work done: W = F * d or W = mgh [1]. Calculate work done correctly: 9408 J or 9400 J [1]. State formula for power: P = W / t [1]. Calculate power correctly: 1568 W or 1570 W [0.5].

Active immunity is the defense against a pathogen by antibody production in the body. It is acquired after infection or vaccination. It gives long-term protection because memory cells are made, allowing a rapid response to future infections. Passive immunity is temporary defense achieved by transferring antibodies from another individual (such as mother to fetus across placenta or via breast milk). It is short-term because the transferred antibodies are eventually broken down, and no memory cells are produced by the host.

Active immunity definition/acquisition: body makes its own antibodies / via infection or vaccination [1]. Active immunity longevity: long-term because memory cells are produced [1]. Passive immunity definition/acquisition: receiving pre-made antibodies / via breast milk, placenta, or antibody injection [1]. Passive immunity longevity: short-term because no memory cells are produced [0.5].

To distinguish an alkane from an alkene, add bromine water (aqueous bromine) to both samples. In the tube containing the alkane, the mixture remains orange/brown because alkanes are saturated and do not readily react with bromine without UV light. In the tube containing the alkene, the bromine water is decolourised (turns from orange/brown to colourless) because the alkene is unsaturated and undergoes an addition reaction across the carbon-carbon double bond, consuming the bromine.

Identify correct reagent: bromine water / aqueous bromine [1]. Correct observation with alkane: remains orange/brown / no change [1]. Correct observation with alkene: decolourises / turns colourless [1]. Correct explanation: alkene has a double bond / is unsaturated and undergoes addition reaction [0.5].

In any chemical reaction, existing bonds must be broken and new bonds must be formed. Breaking the reactant bonds (H-H and Cl-Cl) is an endothermic process that requires an input of energy. Forming the product bonds (H-Cl) is an exothermic process that releases energy. In an exothermic reaction, the total energy released when forming the new H-Cl bonds is greater than the total energy absorbed to break the H-H and Cl-Cl bonds, resulting in a net release of heat energy to the surroundings.

Explain that bond breaking is an endothermic process / requires energy [1]. Explain that bond making is an exothermic process / releases energy [1]. State that more energy is released during bond making than is absorbed during bond breaking [1]. Specify the bonds involved (H-H and Cl-Cl broken, H-Cl formed) [0.5].

Fats are hydrophobic and difficult for water-soluble lipase enzymes to digest. To aid in this, the liver produces a substance called bile. This bile is stored in the gallbladder until food enters the duodenum. The gallbladder then releases the bile, which travels through the bile duct into the small intestine (duodenum). Here, bile performs two roles: it neutralises the acidic mixture from the stomach to create optimum alkaline conditions for enzymes, and it emulsifies large fat droplets into tiny droplets. This physical emulsification greatly increases the surface area of the fat, allowing lipase to digest it much more rapidly.

Liver produces bile [1]. Gallbladder stores bile [1]. Bile duct transports bile to the small intestine/duodenum [0.5]. Bile emulsifies fats to increase surface area / neutralises stomach acid [1].

Water in the soil has a higher water potential than the cytoplasm of root hair cells. Therefore, water enters the root hair cells by osmosis across the partially permeable cell membrane. The water then travels through the cells of the root cortex into the xylem vessels. Transpiration, which is the evaporation of water from the surfaces of mesophyll cells in the leaves, creates a low pressure at the top of the xylem. This generates a transpiration pull (or tension) that sucks water upwards. Because water molecules are cohesive (stick together due to hydrogen bonding), they form a continuous, unbroken column of water that is drawn up through the stem to the leaves.

Osmosis: water enters root hair cells down a water potential gradient [1]. Pathway: water moves across the root cortex to the xylem [0.5]. Transpiration pull: evaporation of water at the leaves creates tension / pull [1]. Cohesion: water molecules stick together to form a continuous column [1].

Giant ionic compounds are made of a regular lattice arrangement of oppositely charged ions. These are held together by strong, multi-directional electrostatic forces of attraction. Because these attractions require a large amount of energy to break, ionic compounds have high melting and boiling points. They do not conduct electricity in the solid state because the ions are locked in fixed positions. However, when molten or dissolved in water, the lattice breaks down, and the ions are free to move and carry charge. In contrast, simple covalent substances consist of discrete molecules. The covalent bonds within the molecules are strong, but the intermolecular forces between the molecules are very weak. Consequently, very little thermal energy is needed to separate the molecules, resulting in low melting and boiling points. They do not conduct electricity in any state because the molecules are uncharged and there are no free electrons or ions to carry current.

Ionic structure: regular lattice of oppositely charged ions held by strong electrostatic forces [1]. Ionic conductivity: conducts when molten/aqueous but not solid because ions become free to move [1]. Covalent structure: discrete molecules with weak intermolecular forces [1]. Covalent conductivity: non-conductors because they lack mobile/free ions or electrons [0.5].

During a chemical reaction such as the combustion of methane, energy is required to break the existing chemical bonds in the reactants (methane and oxygen). This process of bond breaking is endothermic. Conversely, when new bonds are formed to create the products (carbon dioxide and water), energy is released to the surroundings. This process of bond making is exothermic. For the combustion of methane, the total energy released when new bonds are made is greater than the total energy absorbed to break the initial bonds. Consequently, there is a net transfer of thermal energy to the surroundings, which results in an increase in temperature, making the overall reaction exothermic.

[1 mark] State that energy is absorbed / taken in to break bonds (or that bond breaking is endothermic); [1 mark] State that energy is released / given out when new bonds are formed (or that bond making is exothermic); [1 mark] Explain that the energy released during bond making is greater than the energy absorbed during bond breaking; [0.5 mark] State that thermal energy is transferred to the surroundings (causing an increase in temperature).