2023 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

Work done is calculated using the formula: \(\text{Work done} = \text{force} \times \text{distance} = \text{weight} \times \text{height}\). \(\text{Weight} = m \times g = 120\text{ kg} \times 10\text{ N/kg} = 1200\text{ N}\). \(\text{Work done} = 1200\text{ N} \times 2.5\text{ m} = 3000\text{ J}\).

1 mark for correct calculation: \(120 \times 10 \times 2.5 = 3000\). 1 mark for correct unit: J or Joules (accept 3 kJ).

Acceleration is calculated as: \(a = \frac{v - u}{t}\), where \(v = 8.0\text{ m/s}\), \(u = 0\text{ m/s}\), and \(t = 5.0\text{ s}\). \(a = \frac{8.0 - 0}{5.0} = 1.6\text{ m/s}^2\).

1 mark for correct numerical value of 1.6. 1 mark for correct unit: \(\text{m/s}^2\) or \(\text{m s}^{-2}\).

Complete combustion of a hydrocarbon produces carbon dioxide and water. The unbalanced equation is: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). Balancing carbon atoms: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}\). Balancing hydrogen atoms: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\). Finally, balancing oxygen atoms on the left side: \(3 \times 2 + 4 = 10\) oxygen atoms on the right, so we need \(5\text{O}_2\) on the left. The balanced equation is: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\).

1 mark for correct reactants and products: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). 1 mark for correct balancing: \(5\text{O}_2\), \(3\text{CO}_2\), and \(4\text{H}_2\text{O}\).

To test for unsaturation, aqueous bromine (or bromine water) is added to the hydrocarbon. An unsaturated hydrocarbon (containing a carbon-carbon double bond) undergoes an addition reaction and decolourises the bromine water, changing its colour from orange-brown to colourless. A saturated hydrocarbon does not react and the mixture remains orange-brown.

1 mark for the test: add bromine water / aqueous bromine. 1 mark for the positive result with an unsaturated hydrocarbon: turns from orange/brown/yellow to colourless / decolourises (reject 'clear').

When the light intensity is low, light is the limiting factor, so increasing the light intensity increases the rate of photosynthesis proportionally. When the light intensity is very high, light is no longer the limiting factor. The rate of photosynthesis levels off and remains constant because another factor (such as temperature or carbon dioxide concentration) is limiting.

1 mark: at low light intensity, rate of photosynthesis increases (as light intensity increases). 1 mark: at very high light intensity, rate of photosynthesis levels off / remains constant (because another factor becomes limiting).

Stomata are pores in the epidermis of a leaf that allow gases (carbon dioxide and oxygen) and water vapor to diffuse into and out of the leaf. Guard cells surround the stomata and control their size. When guard cells absorb water by osmosis, they become turgid, swell, and curve outward, which opens the stomatal pore.

1 mark for stating the function of stomata: allows gas exchange / diffusion of \(\text{CO}_2\) / \(\text{O}_2\) / water vapour. 1 mark for describing guard cell action: guard cells absorb water / become turgid to open the stomata.

Increasing the temperature increases the kinetic energy of the reacting particles, making them move faster. This results in more frequent collisions between particles. Additionally, a much higher proportion of the colliding particles possess energy greater than or equal to the activation energy, leading to a higher frequency of successful collisions.

1 mark: particles gain kinetic energy / move faster, leading to more frequent collisions. 1 mark: a higher proportion of collisions have energy greater than or equal to the activation energy / more successful collisions per unit time.

In this reaction, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses its oxygen atoms to form metallic iron (\(\text{Fe}\)). Loss of oxygen is defined as reduction, so iron(III) oxide is the substance that is reduced.

1 mark for stating that iron(III) oxide / \(\text{Fe}_2\text{O}_3\) is reduced. 1 mark for explaining that it loses oxygen (to form iron).

First, calculate the work done (change in gravitational potential energy, \(\Delta E_p\)):
\(\Delta E_p = mgh = 2.0 \text{ kg} \times 10 \text{ N/kg} \times 18 \text{ m} = 360 \text{ J}\)

Next, calculate the power output:
\(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{360 \text{ J}}{6.0 \text{ s}} = 60 \text{ W}\)

1 mark for correct calculation of work done or potential energy change (360 J) OR correct formula shown: \(P = \frac{mgh}{t}\)
1 mark for correct final answer of 60 with correct unit (W or J/s)

First, calculate the extension (\(x\)) in metres:
\(x = 27.0 \text{ cm} - 15.0 \text{ cm} = 12.0 \text{ cm} = 0.12 \text{ m}\)

Using Hooke's Law (\(F = kx\)), calculate the spring constant (\(k\)):
\(k = \frac{F}{x} = \frac{6.0 \text{ N}}{0.12 \text{ m}} = 50 \text{ N/m}\)

1 mark for calculating correct extension in metres (0.12 m) OR showing the correct formula \(k = \frac{F}{x}\)
1 mark for the correct final value of 50 (N/m)

Since the bromine water does not change colour (remains orange-brown), no addition reaction has occurred. This indicates that the hydrocarbon is saturated, meaning it contains only single carbon-carbon (C-C) bonds and no double bonds (C=C).

1 mark for stating that it is saturated / contains only single carbon-carbon bonds.
1 mark for explaining that there are no double carbon-carbon bonds / it does not react with bromine water (without UV light).

Propene belongs to the alkene homologous series with the general formula \(C_nH_{2n}\). For 3 carbon atoms, its molecular formula is \(C_3H_6\). Because it is unsaturated, it reacts with aqueous bromine, decolourising it from orange-brown to colourless.

1 mark for the correct molecular formula: \(C_3H_6\)
1 mark for describing the colour change correctly: decolourises / turns colourless (reject 'clear')

Chlorophyll is a green pigment that traps or absorbs light energy from the Sun and transfers it into chemical energy during photosynthesis. It is located inside the chloroplasts of plant cells.

1 mark for stating that it absorbs / traps light energy (and transfers it to chemical energy).
1 mark for stating that it is located in the chloroplasts.

When stomata are closed, carbon dioxide gas from the atmosphere cannot diffuse into the leaves of the plant. Since carbon dioxide is a necessary reactant for photosynthesis, its absence acts as a limiting factor, causing the rate of photosynthesis to decrease significantly.

1 mark for stating that carbon dioxide cannot enter / diffuse into the leaf.
1 mark for explaining that carbon dioxide is a reactant / starting material needed for photosynthesis.

Decreasing the size of the solid pieces increases the total surface area exposed to the acid. This means there are more reactant particles available on the surface to collide, which increases the frequency of collisions (more collisions per second) between the reactant particles, thus increasing the rate of reaction.

1 mark for stating that smaller pieces have a larger total surface area (to volume ratio).
1 mark for explaining that this results in more frequent collisions / higher rate of collisions / more collisions per second.

During the reaction, copper(II) oxide (CuO) reacts with carbon (C) to form copper (Cu) and carbon dioxide (\(CO_2\)). The copper(II) oxide loses its oxygen atoms to the carbon, meaning it undergoes reduction.

1 mark for identifying copper(II) oxide / CuO as the reactant that is reduced.
1 mark for explaining that it loses oxygen during the reaction.

1. Convert the extension from centimetres to metres:
\(x = 4.0\text{ cm} = 0.04\text{ m}\)

2. Use Hooke's Law formula \(F = kx\) rearranged to solve for \(k\):
\(k = \frac{F}{x}\)

3. Substitute the values:
\(k = \frac{12\text{ N}}{0.04\text{ m}} = 300\text{ N/m}\)

1 mark: Correct conversion of \(4.0\text{ cm}\) to \(0.04\text{ m}\) (or correct formula \(k = F/x\))
1 mark: Correct calculation of the spring constant to give \(300\text{ (N/m)}\)

During addition polymerisation, the double bonds between the carbon atoms (\(\text{C=C}\)) in the ethene monomers open up (break) to form single covalent bonds (\(\text{C-C}\)) in the polymer chain, linking the monomer units together.

1 mark: for mentioning that the carbon-carbon double bond (\(\text{C=C}\)) breaks / opens up.
1 mark: for mentioning that carbon-carbon single bonds (\(\text{C-C}\)) are formed (to link the monomers).

The palisade mesophyll layer is located near the upper surface of the leaf where light intensity is highest. The cells are column-shaped and tightly packed together to maximize the area that can intercept sunlight. Furthermore, they contain a very high concentration of chloroplasts, allowing for maximum absorption of light energy.

1. Palisade cells are closely packed / tightly packed / column-shaped. [1 mark]
2. Arranged vertically / near the upper surface of the leaf to receive maximum light. [1 mark]
3. Contain a high density / large number of chloroplasts. [1 mark]

When the marble chips are smaller, they have a larger total surface area exposed to the acid. This means more calcium carbonate particles are available to react at any one time, resulting in a higher frequency of collisions (more collisions per second) between the acid particles and the marble chips, thus increasing the rate of reaction.

1. Smaller chips have a larger surface area (for the same total mass). [1 mark]
2. There are more frequent collisions / more collisions per unit time. [1 mark]
3. Between reacting particles (calcium carbonate particles and hydrogen ions in acid). [1 mark]

During the run, the athlete's chemical energy store in muscles decreases as it is transferred. During acceleration, this chemical energy is transferred into a kinetic energy store, which increases. At constant maximum speed, kinetic energy remains constant while chemical energy continues to be depleted. During deceleration, the kinetic energy decreases to zero, and the energy is transferred to the thermal energy store of the surroundings and the track via friction and air resistance.

1. Chemical (potential) energy store (in the athlete's muscles) decreases. [1 mark]
2. Kinetic energy store increases during acceleration and then decreases to zero during deceleration. [1 mark]
3. Energy is transferred to the thermal energy store of the surroundings (due to friction, air resistance, and metabolic heat). [1 mark]

Since the cargo ship is traveling at a constant velocity, its acceleration is zero. According to Newton's First Law, the resultant force acting on the ship must be zero. Therefore, the forward driving force provided by the propellers is exactly equal in magnitude and opposite in direction to the total resistive force (friction and drag from the water and air), resulting in balanced forces.

1. The driving force is equal in magnitude to the total resistive force. [1 mark]
2. The forces act in opposite directions / the forces are balanced. [1 mark]
3. Since velocity is constant, acceleration is zero, meaning the resultant force is zero (Newton's First Law). [1 mark]

Ethene is a monomer that contains a carbon-carbon double bond (\(\text{C}=\text{C}\)). During polymerization, these double bonds open up to link the monomers together, so the resulting polymer, poly(ethene), consists of a long chain of carbon atoms joined only by single covalent bonds (\(\text{C}-\text{C}\)). Because poly(ethene) has only single bonds between its carbon atoms, it is classified as a saturated compound.

1. Ethene contains a carbon-carbon double bond (\(\text{C}=\text{C}\)). [1 mark]
2. Poly(ethene) contains only carbon-carbon single bonds (\(\text{C}-\text{C}\)). [1 mark]
3. Classified as saturated because all carbon-carbon bonds are single bonds / there are no double bonds. [1 mark]

To distinguish between the two, add a few drops of orange/brown bromine water (aqueous bromine) to each sample and shake. The hexene (alkene) will react rapidly via an addition reaction, decolorizing the bromine water (turning it colorless). The hexane (alkane) will not react, and the mixture will remain orange/brown.

1. Add bromine water / aqueous bromine to both samples. [1 mark]
2. Hexane: stays orange / brown / no color change. [1 mark]
3. Hexene: decolorizes the bromine water / turns from orange to colorless. [1 mark]

Three key structural features of alveoli are: (1) their walls are extremely thin, being only one cell thick, (2) they have a very large total surface area, and (3) they have a rich supply of blood capillaries. The one-cell thick wall increases the rate of diffusion because it provides an extremely short distance for oxygen and carbon dioxide molecules to travel between the air and the blood.

1. States three structural features of alveoli: very thin wall / one-cell thick; large total surface area; surrounded by a dense network of capillaries / good blood supply. [2 marks for 3 features, 1 mark for 2 features]
2. Explanation of any one of the stated features: [1 mark]
- Thin wall: provides a short diffusion distance.
- Large surface area: allows more gas molecules to diffuse simultaneously.
- Capillaries / good blood supply: carries oxygen away / brings carbon dioxide to maintain a steep concentration gradient.

The highly acidic conditions (low pH) in the stomach denature the salivary amylase enzyme. This change in pH disrupts the bonds holding the 3D structure of the protein together, causing the active site to change its shape permanently. As a result, the starch substrate can no longer fit into the active site, preventing any catalytic activity.

1. Highly acidic environment / low pH denatures the enzyme. [1 mark]
2. Active site changes shape (permanently). [1 mark]
3. Starch / substrate molecule can no longer fit into the active site / no enzyme-substrate complexes can form. [1 mark]

When the skydiver first jumps, they accelerate downwards because their weight is greater than the air resistance. As they fall faster, the air resistance increases. When the air resistance becomes equal to their weight, the forces are balanced (resultant force is zero), meaning the acceleration is zero, resulting in a constant terminal velocity.

1 mark: State that air resistance (drag) increases as speed increases.
1 mark: State that air resistance eventually becomes equal to the skydiver's weight / gravity.
1 mark: State that the resultant force becomes zero (forces are balanced), so there is no further acceleration (constant speed).

Increasing the concentration increases the number of acid particles per unit volume. As a result, the reactant particles are closer together, leading to more frequent collisions (more collisions per second) and therefore a higher rate of reaction.

1 mark: More (acid) particles per unit volume / more closely packed particles.
1 mark: More frequent collisions / higher frequency of collisions / more collisions per second (reject: 'more collisions' without time reference).
1 mark: Higher rate of successful / fruitful collisions.

First, recall the formula for kinetic energy:

\(E_k = \frac{1}{2} m v^2\)

Rearrange the formula to solve for speed, \(v\):

\(v^2 = \frac{2 E_k}{m}\)

Substitute the given values:

\(v^2 = \frac{2 \times 4000}{80} = 100\)

\(v = \sqrt{100} = 10\text{ m/s}\)

- 1 mark for correct substitution or rearrangement showing \(v^2 = 100\) (or \(\sqrt{\frac{2 \times 4000}{80}}\))
- 1 mark for correct final answer with appropriate unit: 10 m/s

First, convert the time from minutes to seconds:

\(t = 5.0\text{ minutes} \times 60 = 300\text{ s}\)

Next, use the formula relating charge, current, and time:

\(I = \frac{Q}{t}\)

Substitute the values:

\(I = \frac{450\text{ C}}{300\text{ s}} = 1.5\text{ A}\)

- 1 mark for converting time to seconds (300 s) OR for correct substitution into the current formula: \(\frac{450}{300}\)
- 1 mark for correct final answer with unit: 1.5 A (accept 1.5 amperes, reject 1.5 without unit)

1. Calculate the relative molecular mass of carbon dioxide, \(\text{CO}_2\):
\(M_r = 12 + (2 \times 16) = 44\)

2. Calculate the number of moles of \(\text{CO}_2\):
\(\text{moles} = \frac{\text{mass}}{M_r} = \frac{11}{44} = 0.25\text{ mol}\)

3. Calculate the volume occupied by this amount of gas:
\(\text{volume} = 0.25\text{ mol} \times 24\text{ dm}^3\text{/mol} = 6.0\text{ dm}^3\)

- 1 mark for calculating the number of moles of \(\text{CO}_2\) as 0.25 mol (or showing \(\frac{11}{44}\))
- 1 mark for correct final answer with unit: 6.0 dm^3 (accept 6 dm^3, accept 6000 cm^3)

Use the wave equation:

\(v = f \lambda\)

Rearrange to solve for wavelength, \(\lambda\):

\(\lambda = \frac{v}{f}\)

Substitute the given values:

\(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^6\text{ Hz}} = 200\text{ m}\)

- 1 mark for correct rearrangement or substitution: \(\frac{3.0 \times 10^8}{1.5 \times 10^6}\)
- 1 mark for correct final answer with unit: 200 m (accept 2.0 \(\times 10^2\) m)

Recall the formula for pressure:

\(p = \frac{F}{A}\)

Rearrange to solve for force, \(F\):

\(F = p \times A\)

Substitute the given values:

\(F = 8.0 \times 10^4\text{ Pa} \times 0.020\text{ m}^2 = 1600\text{ N}\)

- 1 mark for correct formula or substitution: \(8.0 \times 10^4 \times 0.020\)
- 1 mark for correct final answer with unit: 1600 N (accept 1.6 kN)

1. Calculate the useful work done by the motor:
\(W = F \times d = 40\text{ N} \times 3.0\text{ m} = 120\text{ J}\)

2. Calculate the power output of the motor:
\(P = \frac{W}{t} = \frac{120\text{ J}}{8.0\text{ s}} = 15\text{ W}\)

- 1 mark for calculating the useful work done as 120 J OR showing \(W = 40 \times 3.0\)
- 1 mark for correct final answer with unit: 15 W (accept 15 watts, accept 15 J/s)