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Question 1

(a) Fig. 1.1 is a diagram of the gas exchange system.
(i) Identify the parts labelled A and B in Fig. 1.1.
(ii) Alveoli are the gas exchange surface in humans.
State two features of an efficient gas exchange surface.
(iii) State the name of the cells in the gas exchange system that produce mucus.
(b) A student measures their breathing rate at rest and during exercise.
Table 1.1 shows the results.
(i) Calculate the difference in breathing rate between rest and during exercise in Table 1.1.
(ii) Tick (✓) two boxes to explain the difference in breathing rate during exercise shown in Table 1.1.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B11 — Gas exchange in humans (Parts (a)(i), (a)(ii), (a)(iii))
• Topic B12 — Respiration (Parts (b)(i), (b)(ii))

▶️ Answer/Explanation

(a)(i) A = trachea, B = diaphragm

A is the tube carrying air down to the lungs, identified as the trachea.
B is the muscular sheet at the base of the thorax, identified as the diaphragm.

(a)(ii) Large surface area; thin walls

Any two of: large surface area, thin (single-cell) walls, good blood supply, or good ventilation with air.
These features all maximise the rate of gas diffusion across the alveolar surface.

(a)(iii) Goblet cells

Goblet cells line the airways and secrete mucus.
The mucus traps dust and microorganisms before they reach the alveoli.

(b)(i) 25 breaths per minute

Difference = breathing rate during exercise − breathing rate at rest.
Difference \( = 40 – 15 = 25\) breaths per minute.

(b)(ii) 

During exercise, muscles respire aerobically at a faster rate to release more energy.
This raises the carbon dioxide concentration in the blood, which stimulates an increased breathing rate.
The other three statements are factually incorrect for exercise (oxygen use, water vapour, and water intake all increase or are unrelated, not decrease).

Question 2

Fig. 2.1 shows the displayed formulae of some carbon compounds.
(a) State which compound in Fig. 2.1 is not a hydrocarbon.
Explain your answer.
(b) State which compound in Fig. 2.1 will decolourise aqueous bromine.
(c) Compound D is used as a fuel.
State one other use of compound D.
(d) Compound C is a saturated compound.
State what is meant by saturated.
(e) Compound C, \(\text{C}_3\text{H}_8\), undergoes complete combustion in oxygen.
Construct the balanced symbol equation for this reaction.
(f) Compound A is ethene.
Complete Fig. 2.2 to show the displayed formula of the polymer poly(ethene).
(g) The formation of poly(ethene) from ethene is an example of addition polymerisation.
Explain the differences between addition polymerisation and condensation polymerisation.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C11.1 — Formulas and terminology (Part (a))
• Topic C11.5 — Alkenes (Part (b))
• Topic C11.3 — Fuels (Part (c))
• Topic C11.4 — Alkanes (Part (d))
• Topic C6.2 — Rate of reaction / stoichiometry link (Part (e))
• Topic C11.7 — Polymers (Parts (f), (g))

▶️ Answer/Explanation

(a) Compound D

D contains an −OH group, meaning it is made of carbon, hydrogen, and oxygen.
A hydrocarbon must contain only carbon and hydrogen atoms, so D does not qualify.

(b) Compound A

A has a C=C double bond (it is ethene).
Compounds containing a C=C bond decolourise aqueous bromine via an addition reaction.

(c) Solvent

Compound D is ethanol.
Besides being a fuel, ethanol is commonly used as a solvent.

(d) Contains only C–C single bonds

A saturated compound contains only single covalent bonds between carbon atoms.
This means no further addition reactions are possible across a C–C bond.

(e) \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\)

Balance carbon atoms first: 3 carbons in propane gives \(3\text{CO}_2\).
Balance hydrogen atoms: 8 hydrogens in propane gives \(4\text{H}_2\text{O}\).
Balance oxygen atoms: right side has \( (3\times2) + 4 = 10\) oxygen atoms, so \(5\text{O}_2\) is needed on the left.

(f) 

Ethene’s C=C double bond opens up during addition polymerisation.
Each monomer joins to the next using single C–C bonds, forming a long repeating chain.

(g) Addition vs condensation polymerisation

Addition polymerisation uses monomers containing a C=C bond, joining many identical units with no by-product formed.
Condensation polymerisation uses monomers with two functional groups (one at each end), and a small molecule such as water is released at each linkage.

Question 3

A student investigates an NTC thermistor.
(a) The student connects the thermistor in series with a cell and an ammeter.
The student also connects a voltmeter to measure the potential difference across the thermistor.
Fig. 3.1 shows an incomplete circuit diagram of the circuit used by the student.
(i) Complete Fig. 3.1.
(ii) When the thermistor is at room temperature, the ammeter reads \(3.0\,\text{A}\).
Calculate the charge that flows through the thermistor in \(60\,\text{s}\).
State the unit for your answer.
(b) The student places the thermistor into hot water.
The student records the values shown by the ammeter and the voltmeter as the temperature of the thermistor increases.
Describe how the power output of the thermistor changes as the temperature of the thermistor increases.
(c) Fig. 3.2 shows the current–voltage characteristic for an ohmic resistor.
Explain the shape of the graph shown in Fig. 3.2.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P4.3.1 — Circuit diagrams and circuit components (Part (a)(i))
• Topic P4.2.2 — Electric current (Part (a)(ii))
• Topic P4.2.5 — Electrical energy and electrical power (Part (b))
• Topic P4.2.4 — Resistance (Part (c))

▶️ Answer/Explanation

(a)(i) Thermistor symbol added in series; ammeter and voltmeter correctly placed

The thermistor symbol must be drawn in the main series loop with the cell and ammeter.
The voltmeter must be connected in parallel across the thermistor only.

(a)(ii) \(180\,\text{C}\)

Charge is calculated using \(Q = It\).
\(Q = 3.0 \times 60 = 180\,\text{C}\) (coulombs).

(b) Power increases as temperature increases

As temperature rises, the resistance of the NTC thermistor decreases.
A lower resistance at constant voltage causes the current through the thermistor to increase.
Since \(P = IV\), an increase in current (with p.d. roughly maintained) means the power output increases.

(c) Straight line through the origin

The graph is a straight line passing through the origin, showing current is directly proportional to voltage.
This means the resistance of the component stays constant, which is the definition of an ohmic conductor obeying Ohm’s law.

Question 4

Fig. 4.1 summarises one pathway of water through a plant.
(a) Identify the cells labelled Y and Z in Fig. 4.1.
(b) Describe two ways that cell X in Fig. 4.1 is adapted for photosynthesis.
(c) Explain the effect of an increase in humidity on the process occurring at A in Fig. 4.1.
(d) Complete the sentences to explain how water is moved in part B in Fig. 4.1.
Water is moved up part B by ………… pull.
This creates a ………… gradient, drawing up a column of water molecules that are held together by ………… .

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B8.2 — Water uptake (Part (a))
• Topic B6.2 — Leaf structure (Part (b))
• Topic B8.3 — Transpiration (Part (c))
• Topic B8.1 — Xylem and phloem (Part (d))

▶️ Answer/Explanation

(a) Y = root hair cell, Z = (root) cortex cell

Y sits at the outer surface of the root in contact with soil water, identifying it as a root hair cell.
Z is positioned further along the water pathway inside the root, identifying it as a cortex cell.

(b) Many chloroplasts; positioned for maximum light absorption

Cell X contains numerous chloroplasts, increasing the chlorophyll available to absorb light.
It is column-shaped and positioned near the top of the leaf, maximising light capture for photosynthesis.

(c) Evaporation decreases

Increased humidity raises the water vapour concentration in the air surrounding the leaf.
This reduces the concentration (diffusion) gradient between the leaf’s air spaces and the atmosphere.
As a result, evaporation from the mesophyll cell surfaces at A decreases.

(d) Transpiration; water potential; cohesion

Water is moved up the xylem (part B) by transpiration pull.
This creates a water potential gradient along the xylem.
The column of water molecules is held together by cohesion (hydrogen bonding between water molecules).

Question 5

A student investigates the reaction between calcium carbonate and dilute hydrochloric acid.
Fig. 5.1 shows the apparatus the student uses.
The student measures the volume of gas in the gas syringe every 20 seconds. Fig. 5.2 shows a graph of the student’s results.
(a) State the name of the gas made in this reaction.
(b) Complete the sentence.
The reaction is fastest between ………… seconds and ………… seconds.
(c) The total volume of gas made in the experiment is \(50\,\text{cm}^3\).
Calculate the total number of moles in \(50\,\text{cm}^3\) of the gas measured at r.t.p.
The volume of one mole of any gas is \(24\,\text{dm}^3\) at r.t.p.
Show your working.
(d) The student repeats the experiment using dilute hydrochloric acid at a higher temperature.
Explain why the reaction is faster.
Use ideas about collisions between particles.
(e) The reaction between calcium carbonate and dilute hydrochloric acid is exothermic. Complete Fig. 5.3 to show an energy level diagram for an exothermic reaction. Label the activation energy and the energy change on your diagram.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C7.1 — Characteristic properties of acids and bases (Part (a))
• Topic C6.2 — Rate of reaction (Parts (b), (d))
• Topic C3.3 — The mole and the Avogadro constant (Part (c))
• Topic C5.1 — Exothermic and endothermic reactions (Part (e))

▶️ Answer/Explanation

(a) Carbon dioxide

Calcium carbonate reacting with hydrochloric acid produces calcium chloride, water, and carbon dioxide gas.
It is this carbon dioxide that collects in the gas syringe.

(b) Between 0 seconds and 16 seconds

The graph is steepest (gradient is greatest) in the first 16 seconds of the reaction.
A steeper gradient on a volume–time graph indicates a faster rate of gas production.

(c) \(0.0021\) mol (to 2 s.f.)

Convert volume to \(\text{dm}^3\): \(50\,\text{cm}^3 = 0.050\,\text{dm}^3\).
Apply moles \( = \dfrac{\text{volume}}{24}\): moles \( = \dfrac{0.050}{24} = 0.0021\) mol.

(d) Faster due to more frequent successful collisions

At a higher temperature, particles have greater average kinetic energy and move faster.
A larger proportion of particles now possess energy equal to or greater than the activation energy.
This increases the frequency of collisions, and specifically the frequency of successful collisions, speeding up the reaction.

(e)

For an exothermic reaction, the products’ energy level must be drawn lower than the reactants’ energy level.
Activation energy is the energy difference between the reactants and the peak of the curve.
The energy change (ΔH) is the energy difference between the reactants and the products, shown as a downward arrow.

Question 6

Fig. 6.1 shows a car suspension system.
The suspension system uses four identical coil springs.
(a) The weight of the car causes compression in the springs. The length of each spring is reduced from its original length.
Hooke’s Law can be used for compression as well as extension:
\(F = kx\)
where \(F\) = load, \(k\) = spring constant and \(x\) = compression.
The weight of the car is \(17000\,\text{N}\).
Each spring has a spring constant of \(2500\,\text{N/cm}\).
Each spring is reduced to a length of \(24\,\text{cm}\).
Calculate the original length of each spring.
(b) Ultrasound waves are used to check for cracks in the springs of the car.
Ultrasound waves are high-frequency sound waves.
(i) The frequency of the ultrasound waves is above the audible range of a healthy human ear.
Suggest a frequency for ultrasound waves.
(ii) Ultrasound waves are longitudinal waves.
Complete the sentences about longitudinal waves.
Longitudinal waves are produced by vibrations occur ………… to the direction of energy transfer
Longitudinal waves travel through air in compressions and ………… .
(iii) A transmitted ultrasound wave travels through the metal of the spring and is reflected by a crack as shown in Fig. 6.2.
The reflected wave is detected after the transmitted wave is sent, as shown in Fig. 6.3.
The ultrasound wave travels at \(5200\,\text{m/s}\) in the metal of the spring.
Use Fig. 6.3 to determine the distance to the crack.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.5.1 — Effects of forces (Part (a))
• Topic P3.4 — Sound (Parts (b)(i), (b)(iii))
• Topic P3.1 — General properties of waves (Part (b)(ii))

▶️ Answer/Explanation

(a) \(25.7\,\text{cm}\)

Find the force per spring: \(F = \dfrac{17000}{4} = 4250\,\text{N}\).
Find the compression using \(F = kx\): \(x = \dfrac{4250}{2500} = 1.7\,\text{cm}\).
Original length = compressed length + compression \( = 24 + 1.7 = 25.7\,\text{cm}\).

(b)(i) \(20\,000\,\text{Hz}\) (or any value above \(20\,000\,\text{Hz}\))

The upper limit of human hearing is approximately \(20\,000\,\text{Hz}\).
Ultrasound, by definition, has a frequency above this audible threshold.

(b)(ii) Parallel; rarefactions

In a longitudinal wave, particle vibrations occur parallel to the direction of energy transfer.
The wave travels through air as a series of compressions and rarefactions.

(b)(iii) \(0.0104\,\text{m}\) (or \(1.04\,\text{cm}\))

From Fig. 6.3, the time delay between transmitted and reflected pulses is \(t = 4.0 \times 10^{-6}\,\text{s}\) (peaks at \(1.0\) and \(5.0 \times 10^{-6}\,\text{s}\)), so the one-way travel time is \(2.0 \times 10^{-6}\,\text{s}\).
Using \(d = v \times t\): \(d = 5200 \times 2.0 \times 10^{-6} = 0.0104\,\text{m}\).

Question 7

(a) Fig. 7.1 shows the structure of a villus.
(i) State the name and function of the part labelled A in Fig. 7.1.
(ii) Explain how the structure of the part labelled B in Fig. 7.1 is adapted for its function.
(b) Coeliac disease results in damage to the small intestine when gluten is eaten.
Fig. 7.2 shows villi from a person without coeliac disease and Fig. 7.3 shows villi from a person with coeliac disease.
(i) Describe one way the shape of the villi in Fig. 7.3 are different from the villi in Fig. 7.2.
(ii) Explain the effect of this difference on villi function in a person with coeliac disease.
(c) Gluten is a type of protein.
(i) State the name of one disease caused by protein-energy malnutrition.
(ii) State the chemical test for protein.
(d) Tick (✓) the boxes to show the correct features of mechanical and chemical digestion.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B7.2 — Digestive system (Parts (a), (b))
• Topic B7.1 — Diet (Part (c)(i))
• Topic B4 — Biological molecules (Part (c)(ii))
• Topic B7.3 — Digestion (Part (d))

▶️ Answer/Explanation

(a)(i) Lacteal — absorption/transport of fat

The structure labelled A at the centre of the villus is the lacteal.
Its function is the absorption and transport of digested fats away from the small intestine.

(a)(ii) Thin walls give a short diffusion distance

Part B, the villus surface/wall, is only one cell thick.
This gives a short diffusion distance for the transfer of digested nutrients into the bloodstream.

(b)(i) Shorter / flatter / smaller villi

In Fig. 7.3, the villi appear shorter, flatter, and smaller compared to the tall, finger-like villi in Fig. 7.2.

(b)(ii) Less surface area for absorption of nutrients

Flattened villi have a reduced surface area compared to healthy villi.
This decreases the rate of absorption of digested nutrients into the blood, contributing to malnutrition in coeliac disease.

(c)(i) Marasmus / kwashiorkor

Both marasmus and kwashiorkor are diseases caused by protein-energy malnutrition.

(c)(ii) Biuret solution (biuret test)

The biuret test is used to detect the presence of protein in a sample, giving a purple/lilac colour change when protein is present.

(d) 

Mechanical digestion (e.g. chewing) physically breaks food into smaller pieces and occurs in the mouth, but does not involve enzymes or directly produce soluble molecules.
Chemical digestion involves enzymes which break down large insoluble molecules into smaller, soluble molecules.

Question 8

(a) Table 8.1 shows some information about the structure of atoms.
Complete Table 8.1.
(b) Fig. 8.1 shows two forms of the element carbon, diamond and graphite.
Put a tick (✓) next to the correct description for the structure of diamond and graphite.
(c) Graphite is used to make electrodes for electrolysis because it is a good conductor of electricity.
Explain why graphite is a good conductor of electricity.
Use ideas about structure and bonding.
(d) There are different isotopes of the element carbon.
Two of the isotopes are called carbon-12 and carbon-14.
(i) Table 8.2 shows some information about one atom of each of these isotopes of carbon. Complete Table 8.2.
(ii) The different isotopes of carbon have the same chemical properties. Explain why.
(e) Elements are organised in the Periodic Table in groups.
Carbon is in Group IV of the Periodic Table. Another element, tellurium, is in Group VI of the Periodic Table.
A common compound of tellurium is sodium telluride, \(\text{Na}_2\text{Te}\).
Put a tick (✓) next to the formula of the tellurium ion in sodium telluride, \(\text{Na}_2\text{Te}\).

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C2.2 — Atomic structure and the Periodic Table (Part (a))
• Topic C2.6 — Giant covalent structures (Parts (b), (c))
• Topic C2.3 — Isotopes (Part (d))
• Topic C2.4 — Ions and ionic bonds (Part (e))

▶️ Answer/Explanation

(a) 

An electron carries a charge of \(-1\) and has a negligible relative mass (about \(\frac{1}{1835}\) of a proton’s mass).
A neutron is neutral (charge 0) with a relative mass of 1.
A proton carries a charge of \(+1\) with a relative mass of 1.

(b) Giant covalent

Both diamond and graphite consist of carbon atoms held together by a vast network of covalent bonds, making them giant covalent structures.

(c) Free/delocalised electrons that can move

In graphite, each carbon atom forms only three covalent bonds, leaving one delocalised electron per atom.
These free electrons can move through the structure, allowing graphite to conduct electricity.

(d)(i) 

Carbon-12 has a mass number of 12 and atomic number 6, so neutrons \( = 12 – 6 = 6\).
Carbon-14 has a mass number of 14, so neutrons \( = 14 – 6 = 8\), while protons and electrons remain 6 (same element).

(d)(ii) Same number of electrons in the outer shell

Isotopes of the same element have identical electron arrangements (4 electrons in the outer shell for carbon).
Since chemical properties depend on electron configuration, not the number of neutrons, isotopes share the same chemical properties.

(e) \(\text{Te}^{2-}\)

Tellurium is in Group VI, meaning it has 6 outer electrons and gains 2 electrons to achieve a full outer shell.
This forms a \(2-\) ion, consistent with the formula \(\text{Na}_2\text{Te}\) (two \(\text{Na}^+\) ions balance one \(\text{Te}^{2-}\) ion).

Question 9

(a) Fig. 9.1 shows distance–time graphs for a car journey and a bicycle journey.
The car and bicycle both start from the same point and travel in the same direction along the same road.
(i) Use Fig. 9.1 to describe the car journey.
(ii) State the time at which the bicycle passes the car.
(iii) The bicycle and rider have a combined mass of \(80\,\text{kg}\).
Use Fig. 9.1 to calculate the kinetic energy of the bicycle during this journey.
(b) Bicycles are fitted with reflectors which reflect light from car headlights.
Fig. 9.2 shows a diagram of a reflector.
Explain why refraction does not occur:
(i) at point X.
(ii) at point Y.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.2 — Motion (Parts (a)(i), (a)(ii))
• Topic P1.6.1 — Energy (Part (a)(iii))
• Topic P3.2.2 — Refraction of light (Parts (b)(i), (b)(ii))

▶️ Answer/Explanation

(a)(i) Constant speed for 20 s (250 m), then the car stops

The car travels at a constant speed for the first 20 seconds, covering 250 m.
After this point, the line on the graph becomes horizontal, showing the car is stationary.

(a)(ii) 50 s

The bicycle passes the car at the point where the two distance–time lines intersect, which occurs at \(t = 50\,\text{s}\).

(a)(iii) \(1000\,\text{J}\)

From the graph, the bicycle’s speed is found using \(v = \dfrac{\text{distance}}{\text{time}} = \dfrac{300}{60} = 5.0\,\text{m/s}\).
Kinetic energy is calculated using \(KE = \tfrac{1}{2}mv^2 = 0.5 \times 80 \times 5.0^2 = 1000\,\text{J}\).

(b)(i) The incident ray hits the surface at 90° (along the normal)

At point X, the light ray strikes the surface at a right angle to it, meaning the angle of incidence is \(0°\).
A ray travelling along the normal does not bend, so refraction does not occur.

(b)(ii) Total internal reflection occurs instead

At point Y, the angle of incidence inside the plastic exceeds the critical angle for the plastic-air boundary.
When this happens, all the light is reflected internally (total internal reflection) rather than refracting out of the material.

Question 10

A farmer planted some seeds.
(a) Explain why the farmer digs the soil to ensure there are air spaces in the soil before the seeds are planted.
(b) The seeds grow and some of the tips are removed.
Fig. 10.1 shows the shoots after a few days.
(i) State the name of the tropic response shown by shoot A in Fig. 10.1.
(ii) Explain why shoot B does not grow towards the light in Fig. 10.1.
(c) State the name of the type of selection the farmer could use to improve the quality of crop plants.
(d) The shoots eventually flower.
Fig. 10.2 is a drawing of one of the flowers.
(i) Label the part in Fig. 10.2 that protects the developing flower. Use a label line and the correct name.
(ii) Describe one visible feature that shows the flower in Fig. 10.2 is insect-pollinated.
(e) Explain the effect of a magnesium deficiency in the soil on the colour of the plant leaves.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B6.1 — Photosynthesis (Parts (a), (e))
• Topic B13.1 — Coordination and response (Part (b))
• Topic B17.2 — Selection (Part (c))
• Topic B15.3 — Sexual reproduction in plants (Part (d))

▶️ Answer/Explanation

(a) Provides oxygen for germination

Air spaces in the soil allow oxygen to reach the seed.
This oxygen is required for aerobic respiration during germination, releasing the energy needed for growth.

(b)(i) Phototropism

Shoot A bends towards the light source, which is the definition of a positive phototropic response.

(b)(ii) No auxin reaches the growing region

Auxin (the growth hormone) is produced in the tip of the shoot.
Since the tip of shoot B has been removed, no auxin is made, so no unequal cell elongation is stimulated and the shoot does not bend towards the light.

(c) Artificial selection

Artificial selection is the process where a farmer deliberately breeds plants with desirable characteristics to improve crop quality.

(d)(i) Sepal

The outermost part that encloses and protects the flower bud before it opens is the sepal.

(d)(ii) Large, brightly coloured petals (or anther/stigma positioned inside the flower)

Insect-pollinated flowers typically have large, conspicuous petals to attract insects.
The reproductive parts (anther and stigma) are usually located inside the flower so insects brush against them.

(e) Less chlorophyll is made, causing yellow leaves

Magnesium is needed to synthesise chlorophyll.
A deficiency means less (or no) chlorophyll can be made.
This results in the leaves appearing yellow rather than green.

Question 11

Fig. 11.1 shows the reactivity series of some metals. The element carbon is also included in the list.
(a)(i) Iron is extracted from the ore hematite by heating with carbon.
Use Fig. 11.1 to state and explain how magnesium is extracted from magnesium ore.
(ii) Sodium is more reactive than magnesium. Explain why.
(b) Carbon is used to extract an element, X, from its oxide.
The equation for the reaction is shown.
\(\text{XO}_2 + \text{C} \rightarrow \text{X} + \text{CO}_2\)
The sum of the relative formula masses of the reactants \((\text{XO}_2 + \text{C})\) is 163.
Calculate the relative atomic mass of X.
\([A_r: \text{C}, 12; \text{O}, 16]\)
(c) Iron is extracted from iron oxide by reacting the iron oxide with aluminium.
The equation for the reaction is shown.
\(2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3\)
A mixture contains \(162\,\text{g}\) of aluminium and \(800\,\text{g}\) of iron oxide.
Show that aluminium is the limiting reactant.
\([A_r: \text{Al}, 27; \text{Fe}, 56; \text{O}, 16]\)
(d) Magnesium displaces copper from copper chloride solution.
The ionic equation is shown.
\(\text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu}\)
Explain why the reaction between magnesium atoms and copper ions involves both oxidation and reduction.
(e) Complete the following sentences about oxidising agents and reducing agents.
An oxidising agent is a substance which ………………………………….. another substance during a redox reaction.
A reducing agent is a substance which ………………………………….. another substance during a redox reaction.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C9.6 — Extraction of metals (Part (a)(i))
• Topic C9.4 — Reactivity series (Part (a)(ii))
• Topic C3.2 — Relative masses of atoms and molecules (Part (b))
• Topic C3.3 — The mole and the Avogadro constant (Part (c))
• Topic C6.3 — Redox (Parts (d), (e))

▶️ Answer/Explanation

(a)(i) Electrolysis

Magnesium is more reactive than carbon, so carbon cannot displace magnesium from its ore by heating.
Instead, magnesium must be extracted using electrolysis.

(a)(ii) Sodium atoms lose electrons more easily than magnesium atoms

Sodium forms positive ions more readily than magnesium because it loses its single outer electron more easily.
This greater tendency to lose electrons makes sodium more reactive than magnesium.

(b) Relative atomic mass of X \(= 119\)

The relative formula mass of \(\text{CO}_2 = 12 + (2\times16) = 44\).
Relative atomic mass of \(\text{X} = 163 – \text{M}_r(\text{CO}_2) = 163 – 44 = 119\) (working: \(163 – 32 – 12 = 119\)).

(c) Aluminium is the limiting reactant

Moles of \(\text{Fe}_2\text{O}_3 = \dfrac{800}{160} = 5\,\text{mol}\).
Moles of \(\text{Al} = \dfrac{162}{27} = 6\,\text{mol}\).
The equation requires 2 mol Al per 1 mol \(\text{Fe}_2\text{O}_3\), so 5 mol \(\text{Fe}_2\text{O}_3\) needs 10 mol Al, but only 6 mol Al is available, confirming aluminium is the limiting reactant.

(d) Magnesium atoms are oxidised; copper ions are reduced

Oxidation occurs because magnesium atoms lose electrons to form \(\text{Mg}^{2+}\) ions.
Reduction occurs because copper ions gain electrons to form copper atoms.
Since both processes happen simultaneously, the reaction is a redox reaction.

(e) An oxidising agent oxidises another substance; a reducing agent reduces another substance

An oxidising agent causes another substance to lose electrons (oxidises it), while itself being reduced.
A reducing agent causes another substance to gain electrons (reduces it), while itself being oxidised.

Question 12

Fig. 12.1 shows a diagram of a nuclear power station used to generate electricity.
(a)(i) State the process in the reactor that releases energy.
(ii) Complete the sentence about energy resources.
The Sun is the source of energy for all our energy resources except nuclear, ………… and tidal.
(b)(i) Describe, in terms of molecules, how the steam exerts pressure on the walls of the boiler.
(ii) The steam in the boiler has a constant volume.
State what happens to the pressure of the steam if the temperature of the steam is increased.
(c) The power station uses an alternating current (a.c.) generator to generate electricity.
Fig. 12.2 shows a simple a.c. generator.
(i) Describe how a simple a.c. generator produces an output potential difference (p.d.).
(ii) The generator converts kinetic energy into electrical energy.
The efficiency of the generator is 75%.
Calculate the kinetic energy required to produce \(3600\,\text{J}\) of electrical energy.
(d) The fuel rod contains uranium-235 (\(^{235}_{92}\text{U}\)).
Uranium-235 decays by alpha emission.
Use correct nuclide notation to complete the decay equation for uranium-235.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P5.1 — The nucleus (Part (a)(i))
• Topic P1.6.3 — Energy resources (Part (a)(ii))
• Topic P2.1.3 — Pressure changes (Part (b))
• Topic P4.5.2 — The a.c. generator (Part (c)(i))
• Topic P1.6.1 — Energy (Part (c)(ii))
• Topic P5.2.3 — Radioactive decay (Part (d))

▶️ Answer/Explanation

(a)(i) Nuclear fission

In the reactor, uranium nuclei split apart in a process called nuclear fission, releasing large amounts of energy.

(a)(ii) Geothermal

Geothermal energy comes from heat within the Earth, not from the Sun, just like nuclear and tidal energy.

(b)(i) Molecules collide with the walls, exerting a force

Steam molecules move rapidly and randomly within the boiler.
As these molecules collide with the container walls, each collision exerts a small force, and the combined effect of many collisions produces pressure.

(b)(ii) Pressure increases

At constant volume, increasing the temperature increases the average kinetic energy and speed of the molecules.
Faster, more frequent and more forceful collisions with the walls result in increased pressure.

(c)(i) The coil experiences a changing magnetic flux, inducing a p.d.

As the coil rotates within the magnetic field, the magnetic flux passing through it continuously changes.
This changing flux induces an output potential difference across the coil, by electromagnetic induction.

(c)(ii) \(4800\,\text{J}\)

Efficiency \( = \dfrac{\text{output energy}}{\text{input energy}} \times 100\%\).
Rearranging: input (kinetic) energy \( = \dfrac{3600}{75} \times 100 = 4800\,\text{J}\).

(d) \(^{235}_{92}\text{U} \rightarrow\, ^{231}_{90}\text{Th} + \,^{4}_{2}\alpha\)

In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2.
Mass number: \(235 – 4 = 231\); atomic number: \(92 – 2 = 90\), identifying the daughter nuclide as thorium-231.

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