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

(a) In the first part of the curve, light intensity is the limiting factor. An increase in light intensity means more light energy is absorbed by the chlorophyll molecules, which increases the rate of chemical reactions in photosynthesis (converting carbon dioxide and water into glucose and oxygen).

(b) When the curve levels off, light intensity is no longer the limiting factor. The limiting factor is now either carbon dioxide concentration or temperature. To increase the rate further, the student could add a source of carbon dioxide (such as sodium hydrogencarbonate) to the water or slightly raise the temperature using a warm water bath.

(c) Palisade mesophyll cells have several adaptations:
1. They contain a very high density of chloroplasts, which contain chlorophyll to absorb maximum light energy.
2. They are tall, column-shaped cells arranged vertically and packed closely together near the upper surface of the leaf, which maximizes the absorption of light as it enters the leaf.

(d) The word equation for photosynthesis is:
carbon dioxide + water -> glucose + oxygen

(a)
- light energy is absorbed by chlorophyll [1]
- higher light intensity provides more energy for the reaction to convert \(\text{CO}_2\) and water to glucose [1]

(b)
- carbon dioxide concentration / temperature [1]
- add sodium hydrogencarbonate / increase temperature / warm the water [1]

(c)
- Feature 1: contain many chloroplasts / high density of chlorophyll [1]
- Explanation 1: to absorb maximum light energy [1]
- Feature 2: tall / column-shaped / tightly packed / located near the upper surface [1]
- Explanation 2: to maximize light interception / absorb more light per unit area [1]

(d)
- carbon dioxide + water -> glucose + oxygen [1]
[Total: 9 marks]

(a) (i) The reaction is exothermic because the temperature of the reaction mixture increased, indicating that thermal energy was released to the surroundings.
(ii) In an exothermic reaction pathway diagram:
- The energy level of the products is lower than the energy level of the reactants.
- The activation energy (\(E_a\)) is represented by an arrow pointing upwards from the reactants' level to the peak of the energy curve.
- The overall energy change (\(\Delta H\)) is represented by an arrow pointing downwards from the reactants' energy level to the products' energy level.

(b) Bond breaking is an endothermic process (requires energy input), and bond making is an exothermic process (releases energy). In this reaction, the energy released when new bonds are formed in the products (\(\text{ZnSO}_4\) and \(\text{Cu}\)) is greater than the energy required to break the bonds in the reactants (\(\text{CuSO}_4\)).

(c) Any one of the following observations:
- The blue color of the copper(II) sulfate solution fades or becomes colorless.
- A red-brown / pink-brown solid (copper metal) is deposited on the zinc.

(a) (i)
- exothermic [1]
- because the temperature increases / heat is released to surroundings [1]
(a) (ii)
- products drawn at a lower energy level than reactants [1]
- activation energy (\(E_a\)) shown as upward arrow from reactants to peak of curve [1]
- energy change (\(\Delta H\)) shown as downward arrow from reactants to products [1]

(b)
- bond breaking absorbs energy AND bond making releases energy [1]
- energy released in bond making is greater than energy absorbed in bond breaking [1]
- resulting in a net release of energy to the surroundings [1]

(c)
- blue color of solution fades / becomes colorless OR reddish-brown/pink solid forms [1]
[Total: 9 marks]

(a) Acceleration \(a\) is calculated using:
\(a = \frac{v - u}{t}\)
Given \(u = 0\text{ m/s}\), \(v = 15\text{ m/s}\), and \(t = 10\text{ s}\):
\(a = \frac{15 - 0}{10} = 1.5\text{ m/s}^2\)

(b) The total distance is the area under the speed-time graph:
- For the first \(10\text{ s}\) (acceleration):
\(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 10 \times 15 = 75\text{ m}\)
- For the next \(20\text{ s}\) (constant speed):
\(\text{Distance}_2 = \text{base} \times \text{height} = 20 \times 15 = 300\text{ m}\)
- Total distance = \(75 + 300 = 375\text{ m}\)

(c) (i) Kinetic energy \(KE\) is calculated using:
\(KE = \frac{1}{2} m v^2\)
\(KE = 0.5 \times 1200 \times 15^2\)
\(KE = 600 \times 225 = 135\ 000\text{ J}\) (or \(135\text{ kJ}\))

(c) (ii) Electrical energy is converted into kinetic energy.

(a)
- formula: acceleration = change in speed / time [1]
- calculation: 15 / 10 = 1.5 [1]
- unit: \(\text{m/s}^2\) [1]

(b)
- distance during acceleration = 0.5 * 10 * 15 = 75 (m) [1]
- distance during constant speed = 20 * 15 = 300 (m) [1]
- total distance = 75 + 300 = 375 m [1]

(c) (i)
- formula: \(KE = \frac{1}{2} m v^2\) [1]
- calculation: \(0.5 \times 1200 \times 15^2 = 135\ 000\text{ J}\) (or \(135\text{ kJ}\)) [1]

(c) (ii)
- electrical energy -> kinetic energy [1]
[Total: 9 marks]

(a) The balanced equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) Palisade mesophyll cells are adapted for photosynthesis in several ways:
1. They contain a very high density of chloroplasts, which contain chlorophyll to absorb maximum light energy.
2. The cells are columnar (vertically elongated) and closely packed together in the upper region of the leaf, allowing maximum exposure and capture of incoming sunlight.
3. They have large central vacuoles that push chloroplasts to the edges of the cell, minimizing the diffusion distance for carbon dioxide entering from intercellular spaces.

(c) Photosynthesis is a series of reactions controlled by enzymes. At 60 °C, the temperature exceeds the optimum range for these enzymes. The high thermal energy causes the enzymes' molecules to vibrate excessively, breaking the hydrogen and ionic bonds holding their tertiary structures. This changes the shape of their active sites (denaturation). Consequently, the substrate molecules can no longer fit into the active sites, and the rate of photosynthesis decreases significantly or stops.

(a) [3 marks total]
- Correct reactants (\(6\text{CO}_2 + 6\text{H}_2\text{O}\)) [1]
- Correct products (\(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)) [1]
- Fully balanced equation [1]

(b) [3 marks total]
- Identifying a valid feature (e.g., many chloroplasts) [1]
- Explaining its role (to absorb maximum light) [1]
- Identifying a second valid feature with its explanation (e.g., closely packed cells near the upper epidermis to maximize light capture, or large vacuole to decrease CO2 diffusion distance) [1]

(c) [3 marks total]
- Mentioning enzymes are denatured at high temperature (60 °C) [1]
- Explaining that denaturation involves a change in the shape of the active site [1]
- Explaining that the substrate can no longer fit / bind to the active site [1]

(a) The combustion of methane is an exothermic reaction because heat energy is transferred from the chemical system to the surroundings, causing the temperature of the surroundings to increase.

(b) An energy level diagram for an exothermic reaction features:
1. A horizontal line for reactants at a higher energy level than a horizontal line for products.
2. A curve rising from the reactant level to a peak and then descending to the product level.
3. The activation energy (\(E_a\)) is represented by an upward arrow from the reactant energy level to the peak of the curve.
4. The overall enthalpy change (\(\Delta H\)) is represented by a downward arrow pointing from the reactant energy level to the product energy level.

(c) Calculation:
1. Energy needed to break reactant bonds:
- \(4 \times \text{C–H}\) bonds = \(4 \times 413 = 1652\text{ kJ}\)
- \(2 \times \text{O=O}\) bonds = \(2 \times 496 = 992\text{ kJ}\)
- Total energy input = \(1652 + 992 = 2644\text{ kJ}\)

2. Energy released in forming product bonds:
- \(2 \times \text{C=O}\) bonds = \(2 \times 743 = 1486\text{ kJ}\)
- \(4 \times \text{O–H}\) bonds = \(4 \times 463 = 1852\text{ kJ}\)
- Total energy output = \(1486 + 1852 = 3338\text{ kJ}\)

3. Enthalpy change (\(\Delta H\)):
\(\Delta H = \text{Energy absorbed} - \text{Energy released}\)
\(\Delta H = 2644 - 3338 = -694\text{ kJ/mol}\)

(a) [2 marks total]
- Stating the reaction is exothermic [1]
- Describing heat transfer from system to surroundings / temperature of surroundings increases [1]

(b) [3 marks total]
- Stating or drawing reactants at a higher energy level than products [1]
- Explaining activation energy (\(E_a\)) as the energy barrier from reactants to peak [1]
- Explaining overall energy change (\(\Delta H\)) as the downward difference from reactants to products [1]

(c) [4 marks total]
- Calculating total energy to break bonds (\(2644\text{ kJ}\)) [1]
- Calculating total energy released in making bonds (\(3338\text{ kJ}\)) [1]
- Showing subtraction of energy released from energy absorbed [1]
- Correct final value with negative sign and correct units (\(-694\text{ kJ/mol}\)) [1] (accept \(-694\text{ kJ}\), reject positive \(694\))

(a) (i) The work done in lifting the crate against gravity is calculated as:
\(W = F \times d = m \times g \times h\)
\(W = 55\text{ kg} \times 10\text{ m/s}^2 \times 8.0\text{ m} = 4400\text{ J}\)

(ii) Useful power output is work done divided by time taken:
\(P = \frac{W}{t} = \frac{4400\text{ J}}{5.0\text{ s}} = 880\text{ W}\) (or \(\text{J/s}\))

(b) The efficiency of the motor is:
\(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}} \times 100\%\)
\(\text{Efficiency} = \frac{880}{1100} \times 100\% = 0.80 \times 100\% = 80\%\) (or \(0.80\))

(c) According to Hooke's Law:
\(F = kx\)
First, convert extension from millimeters to meters:
\(x = 12\text{ mm} = 0.012\text{ m}\)
Now, rearrange the formula to solve for the spring constant (\(k\)):
\(k = \frac{F}{x} = \frac{180\text{ N}}{0.012\text{ m}} = 15000\text{ N/m}\)

(a) (i) [2 marks total]
- Correct formula or substitution (\(55 \times 10 \times 8.0\)) [1]
- Correct calculation of useful work done (\(4400\text{ J}\)) [1]

(ii) [2 marks total]
- Correct calculation of power (\(880\)) [1]
- Correct unit (\(\text{W}\) or \(\text{J/s}\)) [1]

(b) [2 marks total]
- Correct substitution into efficiency formula (\(880 / 1100\)) [1]
- Correct calculated efficiency (\(80\%\) or \(0.80\)) [1]

(c) [3 marks total]
- Correct conversion of extension to meters (\(0.012\text{ m}\)) [1]
- Use of \(F = kx\) or rearrangement \(k = F/x\) [1]
- Correct calculation with unit (\(15000\text{ N/m}\)) [1]

(a) Destarching the plant ensures that any starch detected at the end of the experiment was produced during the experiment, validating the results.
(b) (i) 1. Place the leaf in boiling water for about 30 seconds to kill the cells and denature enzymes. 2. Place the leaf into a tube of ethanol and heat in a water bath (not over an open flame because ethanol is highly flammable) to dissolve and remove the green pigment chlorophyll. 3. Dip the leaf in warm water to soften it. 4. Spread the leaf on a white tile and add a few drops of iodine solution.
(ii) The green parts of the leaf contain chlorophyll, allowing photosynthesis to occur and starch to be produced; thus, they turn blue-black. The white parts lack chlorophyll, so no photosynthesis occurs, and they remain the yellow-brown color of the iodine solution.
(c) The balanced chemical equation is: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(a) [1 mark] to destarch the plant / ensure starch from before the experiment is removed.
(b)(i) [4 marks total]
- boiling the leaf in water [1]
- heating the leaf in ethanol [1]
- safety precaution: using a water bath because ethanol is highly flammable [1]
- adding iodine solution to test for starch [1]
(b)(ii) [2 marks total]
- green parts turn blue-black because they contain chlorophyll and make starch [1]
- white parts remain yellow-brown because they lack chlorophyll and cannot photosynthesise [1]
(c) [2 marks total]
- correct formulas of reactants and products: \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\) [1]
- correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) [1]

(a) The reaction is exothermic because there is an increase in temperature, meaning thermal energy is transferred from the chemical system to the surroundings.
(b) In an exothermic energy level diagram:
- The horizontal line for reactants is drawn at a higher energy level than the line for products.
- A curve rises from the reactant level to a peak before descending to the product level.
- The activation energy is represented by an upward-pointing vertical arrow from the reactant level to the peak of the curve.
(c) (i) \(\text{Number of moles} = \frac{\text{mass}}{\text{Ar}} = \frac{0.65\text{ g}}{65} = 0.01\text{ mol}\).
(ii) \(\text{Energy per mole} = \frac{\text{Energy released}}{\text{Number of moles}} = \frac{1.4\text{ kJ}}{0.01\text{ mol}} = 140\text{ kJ/mol}\).

(a) [2 marks total]
- Exothermic [1]
- Energy is transferred from the system to the surroundings (causing the temperature to rise) [1]
(b) [3 marks total]
- Reactant level shown higher than product level [1]
- Activation energy shown as energy barrier/peak rising above reactant level [1]
- Correctly labeling reactants, products and activation energy (E_a) [1]
(c)(i) [1 mark]
- \(0.01\text{ mol}\) [1]
(c)(ii) [3 marks total]
- division of energy by moles: \(\frac{1.4}{0.01}\) [1]
- correct value: 140 [1]
- correct unit: \(\text{kJ/mol}\) (or \(\text{J/mol}\) if energy is converted: \(140000\text{ J/mol}\)) [1]

(a) Acceleration is calculated using: \(a = \frac{v - u}{t}\).
Given \(u = 0\text{ m/s}\), \(v = 4.0\text{ m/s}\), and \(t = 2.0\text{ s}\):
\(a = \frac{4.0 - 0}{2.0} = 2.0\text{ m/s}^2\).
(b) Kinetic energy is calculated using: \(\text{KE} = \frac{1}{2}mv^2\).
Given \(m = 1.5\text{ kg}\) and \(v = 4.0\text{ m/s}\):
\(\text{KE} = 0.5 \times 1.5 \times (4.0)^2 = 0.75 \times 16 = 12\text{ J}\).
(c) (i) Work done is calculated using: \(W = F \times d\).
Given \(F = 3.0\text{ N}\) and \(d = 5.0\text{ m}\):
\(W = 3.0 \times 5.0 = 15\text{ J}\).
(ii) Power is calculated using: \(P = \frac{W}{t}\).
Given \(W = 15\text{ J}\) and \(t = 2.5\text{ s}\):
\(P = \frac{15}{2.5} = 6.0\text{ W}\).

(a) [2 marks total]
- formula or substitution: \(a = \frac{4.0}{2.0}\) [1]
- correct value with unit: \(2.0\text{ m/s}^2\) [1]
(b) [2 marks total]
- formula or substitution: \(\text{KE} = 0.5 \times 1.5 \times 4.0^2\) [1]
- correct value: \(12\text{ J}\) [1]
(c)(i) [2 marks total]
- formula or substitution: \(W = 3.0 \times 5.0\) [1]
- correct value: \(15\text{ J}\) [1]
(c)(ii) [3 marks total]
- formula or substitution: \(P = \frac{15}{2.5}\) [1]
- correct value: \(6.0\) [1]
- correct unit: \(\text{W}\) (or \(\text{J/s}\)) [1]