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Question 1

(a) Fig. 1.1 is a diagram of parts of the eye.
(i) State the letter in Fig. 1.1 which identifies the part that contains:
circular and radial muscles ………………………………
receptor cells ………………………………
neurones. ………………………………
(ii) Draw an X on Fig. 1.1 to identify the position of the blind spot.
(iii) Describe the changes to the parts labelled A, E and F when someone changes their focus from a near object to a distant object.
(iv) State the name of the hormone that causes part D in Fig. 1.1 to widen in response to a ‘fight or flight’ situation.
(b) Accommodation of the eye is an example of sensitivity.
Complete the sentence to describe the term sensitivity.
Sensitivity is the ability to detect or sense ……………………… , which are changes in the internal or external …………………….. and to make appropriate responses.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B13.1 — Coordination and response (Parts (a)–(b))

▶️ Answer/Explanation

(a)(i) B; H; G or H

The iris (B) contains circular and radial muscles that control pupil size.
The retina (H) contains the light receptor cells.
The retina (H) or optic nerve (G) contains neurones that carry impulses to the brain.

(a)(ii) X marked on the retina where the optic nerve leaves the eye

The blind spot is the point where the optic nerve exits the eyeball.
No light receptors are present at this point, so no image can be detected there.

(a)(iii) A – narrows; E – tightens; F – relaxes

Part A, the pupil, narrows/gets less convex when focusing on a distant object.
Part E, the ciliary muscle, tightens (contracts).
Part F, the suspensory ligament, relaxes as the lens becomes thinner.

(a)(iv) Adrenaline

Adrenaline is released during a ‘fight or flight’ response.
It causes the circular muscles of the iris (part D) to relax, widening the pupil to let in more light.

(b) stimuli; environment

Sensitivity is the ability to detect or sense stimuli.
Stimuli are changes in the internal or external environment, allowing the organism to make appropriate responses.

Question 2

Alkanes are saturated hydrocarbons.
(a) State what is meant by the term saturated hydrocarbon.
(b) Alkenes are also hydrocarbons.
Table 2.1 shows information about three alkenes.
Complete Table 2.1.
(c) Propane and propene are both colourless gases.
Describe a test to show which gas is propane and which gas is propene.
(d) The monomer \(\text{C}_4\text{H}_8\) in Fig. 2.1 can be used to make a polymer.
Complete Fig. 2.2 to show the structure of the polymer made.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C11.4 — Alkanes (Part (a))
• Topic C11.5 — Alkenes (Parts (b)–(c))
• Topic C11.7 — Polymers (Part (d))

▶️ Answer/Explanation

(a) Saturated hydrocarbon

A saturated hydrocarbon contains only single (covalent) bonds between carbon atoms.
It is a compound containing only hydrogen and carbon atoms.

(b) Table completed: ethene; propene structure; butene

\(\text{C}_2\text{H}_4\) is named ethene, with the double bond \(\text{H}_2\text{C}=\text{CH}_2\).
Propene, \(\text{C}_3\text{H}_6\), has the structural formula \(\text{CH}_3-\text{CH}=\text{CH}_2\).
\(\text{C}_4\text{H}_8\) with a straight chain and one double bond is named butene.

(c) Test with aqueous bromine

Add aqueous bromine (bromine water) to each gas sample.
With propane (an alkane, saturated), the orange/brown bromine water stays orange/brown, showing no reaction.
With propene (an alkene, unsaturated), the bromine water is decolourised (turns colourless) as it adds across the double bond.

(d) Poly(but-1-ene) repeat unit

During addition polymerisation, the C=C double bond in the monomer opens up.
The monomer units join together in a long chain through single C–C bonds, with the \(\text{CH}_2\text{CH}_3\) side group retained on alternate carbons.

Question 3

Fig. 3.1 shows an insect called a pond skater.
Pond skaters spread their weight over their 6 legs so that they can move over the surface of water.
(a) The pond skater has a mass of 0.25 g and is stationary on the surface of the water.
(i) Use the values in the list to complete the sentences about the pond skater.
The gravitational field strength, g, is 10 N / kg.
You can use each value once, more than once or not at all.
0 N   0.0025 kg   0.0025 N   0.25 g   0.25 kg   2.5 N
The weight of the pond skater is ………………………………. .
The force acting upwards on the pond skater by the water is ………………………………. .
The resultant force acting on the pond skater is ………………………………. .
(ii) The pond skater stands on all 6 legs, with the foot of each leg making contact with the surface of the water.
The area of each foot is \(1.2 \times 10^{-7} \, \text{m}^2\).
Calculate the pressure exerted by each foot on the surface of the water.
(b) The pond skater moves across the surface of a pond.
Fig. 3.2 shows a speed–time graph for part of the pond skater’s journey.
(i) Place an X on Fig. 3.2 to show a time at which the pond skater is travelling at a constant speed.
(ii) Use Fig. 3.2 to calculate the maximum acceleration of the pond skater.
(c) The movement of the pond skater on the surface of the water produces waves.
Fig. 3.3 shows a diagram of a wave produced by the pond skater.
(i) Use Fig. 3.3 to determine the wavelength of the water wave in m.
(ii) An observer sees 10 full waves pass a point in 5 seconds.
Use your answer to (c)(i) to calculate the speed of the wave.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.5 — Effects of forces (Part (a)(i))
• Topic P1.7 — Pressure (Part (a)(ii))
• Topic P1.2 — Motion (Part (b))
• Topic P3.1 — General properties of waves (Part (c))

▶️ Answer/Explanation

(a)(i) weight = 0.0025 N; upward force = 0.0025 N; resultant = 0 N

Weight = mass × g = \(0.00025 \, \text{kg} \times 10 \, \text{N/kg} = 0.0025 \, \text{N}\).
Since the pond skater is stationary and floating, the upward force from the water equals its weight, 0.0025 N.
As these forces are balanced, the resultant force is 0 N.

(a)(ii) 3500 Pa

Total contact area = \(6 \times 1.2 \times 10^{-7} = 7.2 \times 10^{-7} \, \text{m}^2\).
Pressure = \(\dfrac{F}{A} = \dfrac{0.0025}{7.2\times10^{-7}} \approx 3500 \, \text{Pa}\).

(b)(i) X placed between 4.0 s and 8.0 s

Between 4.0 s and 8.0 s the graph is a horizontal line, showing speed does not change.
A constant, unchanging speed corresponds to zero acceleration.

(b)(ii) 0.0075 m/s²

Maximum acceleration occurs on the steepest part of the graph, from 0 s to 4.0 s.
Acceleration = \(\dfrac{\Delta v}{\Delta t} = \dfrac{0.03}{4.0} = 0.0075 \, \text{m/s}^2\).

(c)(i) 0.016 m

Reading the distance for one full wave cycle from Fig. 3.3 gives 1.6 cm.
Converting to metres: \(\dfrac{1.6}{100} = 0.016 \, \text{m}\).

(c)(ii) 0.032 m/s

Frequency = \(\dfrac{10 \text{ waves}}{5 \text{ s}} = 2 \, \text{Hz}\).
Speed = \(f \lambda = 2 \times 0.016 = 0.032 \, \text{m/s}\).

Question 4

(a) Photosynthesis is an enzyme-controlled reaction.
Fig. 4.1 is a graph showing the effect of temperature on the rate of photosynthesis.
(i) Explain the results between 0–30°C in Fig. 4.1.
(ii) Explain the result at 40°C in Fig. 4.1.
(b) State the name of the substance that converts light energy to chemical energy during photosynthesis.
(c) State the names of the two raw materials used in photosynthesis.
(d) Describe two ways the cells in the palisade mesophyll layer are adapted for photosynthesis.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B6.1 — Photosynthesis (Parts (a)–(c))
• Topic B6.2 — Leaf structure (Part (d))

▶️ Answer/Explanation

(a)(i) Rate increases as kinetic energy of particles increases

As temperature rises, particles gain more kinetic energy and move faster.
This increases the frequency of effective collisions between substrate and the enzyme’s active site, speeding up the rate of the enzyme-controlled reaction.

(a)(ii) Enzyme denatured at 40°C

At this high temperature the enzyme becomes denatured.
The active site changes shape/is deformed.
The substrate is no longer complementary to the active site, so no reaction (photosynthesis) occurs.

(b) Chlorophyll

Chlorophyll, found in chloroplasts, absorbs light energy and converts it into chemical energy for photosynthesis.

(c) Carbon dioxide; water

The two raw materials required for photosynthesis are carbon dioxide, taken in through stomata, and water, absorbed by the roots.

(d) Any two adaptations

Palisade cells contain many chloroplasts/chlorophyll to absorb maximum light.
They are positioned near the upper surface of the leaf and are column-shaped, tightly packed to maximise light absorption.

Question 5

Ammonia is made from the reaction of nitrogen and hydrogen in the Haber process.
\(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\)
(a) State the sources of the nitrogen and hydrogen used in the Haber process.
(b) Fig. 5.1 shows the energy level diagram for the reaction between nitrogen and hydrogen.
(i) Draw an arrow on Fig. 5.1 to show the energy change in the reaction. Label your arrow A.
(ii) Draw an arrow on Fig. 5.1 to show the activation energy of the reaction. Label your arrow B.
(iii) Explain if Fig. 5.1 represents an exothermic or endothermic reaction. Use ideas about bond breaking and bond making in your answer.
(c) 58.8 kg of nitrogen gas reacts with hydrogen gas to make 142.8 kg of ammonia gas.
\(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\)
Calculate the volume occupied by 142.8 kg of ammonia gas.
The volume of one mole of any gas is \(24 \, \text{dm}^3\) at room temperature and pressure (r.t.p.).

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C5.1 — Exothermic and endothermic reactions (Parts (a)–(b))
• Topic C3.3 — The mole and the Avogadro constant (Part (c))

▶️ Answer/Explanation

(a) Nitrogen from the air; hydrogen from methane/natural gas and steam

Nitrogen gas is obtained from the air (fractional distillation of liquid air).
Hydrogen is obtained from the reaction of methane/natural gas with steam.

(b)(i)–(ii) Arrows A and B

Arrow A is drawn vertically between the reactants’ energy level and the products’ energy level, showing the overall energy released.
Arrow B is drawn vertically from the reactants’ level up to the peak of the curve, showing the activation energy.

(b)(iii) Exothermic

The energy of the products is lower than the energy of the reactants, so the energy change is negative and energy is given out.
Bond breaking (in \(\text{N}_2\) and \(\text{H}_2\)) is endothermic, while bond making (in \(\text{NH}_3\)) is exothermic.
Since more energy is released in bond making than is absorbed in bond breaking, the reaction is exothermic overall.

(c) 201 600 dm³

\(M_r\) of \(\text{NH}_3 = 17\).
Moles of \(\text{NH}_3 = \dfrac{142\,800}{17} = 8400\) mol.
Volume = \(8400 \times 24 = 201\,600 \, \text{dm}^3\).

Question 6

(a) Fig. 6.1 shows the average power output over a summer’s day from a solar panel made from solar cells.
(i) State the time at which the power output of the solar panel is at its maximum.
(ii) Suggest one reason why the power output of the solar panel is at a maximum at this time.
(b) Table 6.1 gives some data about different types of power stations.
Use Table 6.1 to complete each sentence. Each letter, P, Q or R, can be used once, more than once or not at all.
The power station that produces the least carbon dioxide is power station ………. .
The power station that releases the most energy per second is power station ………. .
The power station with the generator that produces the largest current is power station ………. .
(c) Power stations use transformers to increase the output voltage.
Fig. 6.2 shows a simple transformer.
(i) State the name of the part of the transformer labelled X.
(ii) Describe how the output voltage across the secondary coil is produced.
(iii) Fig. 6.2 shows a step-up transformer containing 8 turns in the primary coil and 17 turns in the secondary coil.
The input voltage across the primary coil is 22 kV.
Calculate the output voltage across the secondary coil.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.6.3 — Energy resources (Parts (a)–(b))
• Topic P4.5.6 — The transformer (Part (c))

▶️ Answer/Explanation

(a)(i) 12:00

The graph in Fig. 6.1 peaks at around 12:00 noon.

(a)(ii) Sun highest in the sky

At noon the light from the sun is incident at almost 90° to the panel.
This corresponds to the sun being at its highest point in the sky, giving maximum light intensity on the panel.

(b) R; P; P

Power station R (uranium/nuclear) produces the least \(\text{CO}_2\) since it does not burn fossil fuel.
Power station P (coal) has the largest output power (1500 MW), so it releases the most energy per second.
Using \(P = VI\), for the same output power, station P (lowest voltage, 22 kV) requires the largest current.

(c)(i) Soft-iron core

Part X is the soft-iron core, which links the primary and secondary coils magnetically.

(c)(ii) Electromagnetic induction

An alternating current in the primary coil produces a changing magnetic field in the soft-iron core.
This changing magnetic field induces an alternating voltage/emf in the secondary coil.

(c)(iii) 46.75 kV (≈ 47 kV)

Using \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\): \(V_s = 22 \times \dfrac{17}{8} \approx 47 \, \text{kV}\).

Question 7

(a) Fig. 7.1 shows a food web.
(i) Use the information in Fig. 7.1 to state the name of one:
herbivore
quaternary consumer
organism that occupies the first trophic level.
(ii) State three ways that energy is lost between trophic levels.
(iii) State the principal source of energy for all biological systems.
(b) Fertilisers are sometimes used to improve the growth of crop plants.
Excessive use of fertilisers pollutes rivers.
    (i) Explain why this pollution increases the growth of producers in rivers.
    (ii) Explain how death of producers in rivers reduces the oxygen concentration in the water.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B18.2 — Food chains and food webs (Part (a))
• Topic B19.1 — Habitat destruction (Part (b))

▶️ Answer/Explanation

(a)(i) grasshopper/fruit fly; eagle/snake; mango/corn

A herbivore is an organism that eats only plants, e.g. grasshopper or fruit fly.
A quaternary consumer is a 4th-level consumer, e.g. eagle or snake.
The first trophic level is occupied by producers, e.g. mango or corn.

(a)(ii) Any three from: respiration/heat loss; incomplete consumption; egestion/excretion

Energy is lost as heat through respiration and other metabolic processes.
Not all of an organism is eaten by the next consumer.
Not all that is eaten is digested, so energy is lost in faeces/urine, and energy is also used for movement.

(a)(iii) The Sun

The Sun is the principal source of energy for almost all biological systems, providing light energy for photosynthesis.

(b)(i) Fertilisers provide nitrate ions

Fertilisers contain nitrate ions, which run off into rivers.
These nitrate ions are needed by producers for protein synthesis/amino acid production, boosting their growth.

(b)(ii) Decomposition uses up oxygen

When producers die, decomposers break down the dead material.
Decomposers carry out aerobic respiration, using up dissolved oxygen from the water, reducing the oxygen concentration.

Question 8

(a) Copper has two isotopes, copper-63 and copper-65.
Table 8.1 shows information about copper-63 and copper-65.
Complete Table 8.1.
(b) Complete the sentence to describe what is meant by isotopes. Choose words from the list.
compound
electron
element
group
nucleon
nucleus
proton
Isotopes are atoms of the same ………………….. which have the same ………………….. number but a different ………………….. number.
(c) An element is in Group II and Period 3 of the Periodic Table.
Predict the electronic structure of this element.
Tick (✓) one box.
(d) Complete Fig. 8.1 to show the covalent bonding in a molecule of nitrogen, \(\text{N}_2\).
(e) The element nitrogen is covalently bonded.
The melting point of nitrogen is –210°C.
Calcium nitride, \(\text{Ca}_3\text{N}_2\), is an ionic compound.
The melting point of calcium nitride is 1195°C.
Explain the difference in melting points in terms of attractive forces.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C2.3 — Isotopes (Parts (a)–(b))
• Topic C2.4 — Ions and ionic bonds (Part (e))
• Topic C2.5 — Simple molecules and covalent bonds (Parts (c)–(d), (e))

▶️ Answer/Explanation

(a) Copper-63: 29 electrons; Copper-65: 29 protons, 36 neutrons

In a neutral atom, the number of electrons equals the number of protons, so copper-63 has 29 electrons.
Copper-65 has 29 protons (same as copper-63); neutrons = nucleon number − proton number = \(65-29=36\).

(b) element; proton; nucleon

Isotopes are atoms of the same element which have the same proton number but a different nucleon number.

(c) 2.8.2

Group II means 2 outer-shell electrons; Period 3 means 3 occupied shells.
The electronic structure filling 2, then 8, then 2 in the outer shell is 2.8.2.

(d) Shared pair of electrons forming a triple bond

Each nitrogen atom shares three pairs of electrons with the other nitrogen atom to form a triple covalent bond.
The diagram shows all outer-shell electrons, including 3 shared pairs between the two N atoms and one lone pair on each.

(e) Weak intermolecular forces vs strong ionic bonds

Nitrogen consists of covalent molecules held together by weak intermolecular forces, needing little energy to overcome, giving a low melting point.
Calcium nitride is an ionic compound with strong electrostatic forces of attraction between oppositely charged ions.
More energy is required to overcome these stronger ionic forces, giving calcium nitride a much higher melting point.

Question 9

A student investigates series and parallel circuits using filament lamps.
(a) Fig. 9.1 shows the first circuit the student makes using three identical filament lamps.
(i) Determine the potential difference shown on the voltmeter.
(ii) Explain why the reading on both ammeters is the same.
(b) Fig. 9.2 shows the second circuit made by the student using the same three identical filament lamps.
One of the ammeters shows a current of 0.9 A as shown in Fig. 9.2.
The voltmeter shows a potential difference of 6.0 V.
Calculate the resistance of one of the filament lamps.
(c) The filament lamps emit energy in the form of infrared radiation and visible light.
Complete the sentences to compare infrared radiation with visible light.
Infrared radiation and visible light are both parts of the …………………….. .
The wavelength of infrared radiation is …………………… than the wavelength of visible light.
The frequency of infrared radiation is …………………… than the frequency of visible light.
The speed of infrared radiation and visible light is ………………….. .

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P4.3.2 — Series and parallel circuits (Parts (a)–(b))
• Topic P3.3 — Electromagnetic spectrum (Part (c))

▶️ Answer/Explanation

(a)(i) 2.0 V

The voltmeter is connected across one of the parallel lamps within the series circuit.
Reading the circuit diagram, the potential difference across that lamp is 2.0 V.

(a)(ii) Series circuit — current is the same everywhere

In a series circuit, there is only one path for charge to flow.
Since there are no junctions for current to split, the current is the same at every point, so both ammeters show the same reading.

(b) 20 Ω

First find the current through the branch: \(I = 0.9 \, \text{A}\).
Using \(R = \dfrac{V}{I} = \dfrac{6.0}{0.3} = 20 \, \Omega\) (current through single lamp branch = 0.3 A, as the 0.9 A splits across the parallel branches).

(c) Electromagnetic spectrum; longer; lower; same speed

Infrared radiation and visible light are both parts of the electromagnetic spectrum.
The wavelength of infrared radiation is longer than that of visible light, and its frequency is lower.
Both types of radiation travel at the same speed, \(3 \times 10^8 \, \text{m/s}\), in a vacuum.

Question 10

(a) Fig. 10.1 is a diagram of a sperm cell in humans.
(i) Cell structure X contains unpaired chromosomes.
State the two words used to describe cell structure X in Fig. 10.1.
(ii) State the number of chromosomes in cell structure X in Fig. 10.1.
(iii) Mitochondria are the site of aerobic respiration.
Suggest why part Y in Fig. 10.1 contains many mitochondria.
(b) Table 10.1 compares some of the features of asexual reproduction and sexual reproduction.
Place ticks (✓) in the boxes in Table 10.1 to show the correct features of asexual reproduction and sexual reproduction.
(c) Describe two disadvantages of sexual reproduction.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B15.4 — Sexual reproduction in humans (Part (a))
• Topic B15.1 — Asexual reproduction (Parts (b)–(c))

▶️ Answer/Explanation

(a)(i) Haploid nucleus

Cell structure X is the sperm’s nucleus.
It contains a single, unpaired set of chromosomes, so it is described as a haploid nucleus.

(a)(ii) 23

A human haploid gamete cell contains 23 chromosomes, half the diploid number of 46.

(a)(iii) Provides energy for movement

Aerobic respiration in mitochondria releases energy.
This energy is needed to move the sperm’s flagellum, allowing the sperm to swim towards the egg.

(b) Table ticks completed

Formation of a zygote applies only to sexual reproduction.
No genetic variation applies only to asexual reproduction.
Production of offspring applies to both asexual and sexual reproduction.

(c) Any two disadvantages

Sexual reproduction requires more energy, as organisms must find a mate and produce gametes.
It is a longer process than asexual reproduction and is not always successful, and undesirable new characteristics can arise in the offspring.

Question 11

Electrolysis is the breakdown of an ionic compound, when molten or in aqueous solution, by the passage of electricity.
The products of electrolysis of some aqueous solutions, using inert electrodes, are shown in Table 11.1.
(a) Part of the reactivity series, from most reactive to least reactive, is shown.
sodium
magnesium
(hydrogen)
copper
Use this information to complete Table 11.1.
(b) State why hydrogen forms at the cathode, rather than sodium, during the electrolysis of aqueous sodium bromide.
(c) Copper is formed at the cathode in the electrolysis of aqueous copper chloride.
Construct the ionic half-equation for the formation of copper at the cathode.
(d) Aqueous copper(II) sulfate is electrolysed using inert electrodes.
Oxygen gas is formed at the anode. Copper metal is formed at the cathode.
(i) The ionic half-equation for the formation of oxygen gas at the anode is shown.
\(4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-\)
State if this reaction is oxidation or reduction.
Explain your answer using ideas about electrons.
(ii) The experiment is repeated using copper electrodes instead of inert electrodes.
Describe what happens at the anode and the cathode using copper electrodes.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C4.1 — Electrolysis (Parts (a)–(d))

▶️ Answer/Explanation

(a) copper chloride → chlorine at anode; magnesium sulfate → hydrogen at cathode

For copper chloride, chlorine gas forms at the anode since copper is deposited at the cathode.
For magnesium sulfate, since magnesium is more reactive than hydrogen, hydrogen forms at the cathode instead.

(b) Hydrogen is less reactive than sodium

In the reactivity series, hydrogen is below sodium.
The less reactive ion (\(\text{H}^+\)) is preferentially discharged at the cathode, so hydrogen forms rather than sodium.

(c) \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)

Copper(II) ions gain two electrons at the cathode to form copper metal.
This ionic half-equation must be balanced for both charge and atoms.

(d)(i) Oxidation

In this half-equation, hydroxide ions lose electrons (\(4\text{e}^-\) are produced).
Since loss of electrons defines oxidation, this reaction is oxidation.

(d)(ii) Anode dissolves; cathode gains copper

With copper electrodes, the copper anode dissolves, forming copper ions in solution.
At the cathode, copper is deposited/copper forms, as copper ions gain electrons.

Question 12

(a) Table 12.1 shows information about three types of ionising radiation.
(i) Complete Table 12.1.
(ii) Fig. 12.1 shows the path taken by a beta particle as it moves through an electric field.
Complete Fig. 12.1 to show the paths taken by alpha particles and gamma rays as they pass through an electric field.
(b) Beta particles are released when carbon-14 (\(^{14}_{6}\text{C}\)) decays into an isotope of nitrogen.
Use the correct nuclide notation to complete the decay equation for carbon-14.
(c) At room temperature, nitrogen is a gas, water is a liquid and carbon is a solid.
(i) Explain why a gas can be compressed and a solid cannot be compressed.
(ii) Suggest if water can be compressed. Give a reason for your answer.
(iii) A sample of nitrogen gas is held in a container with a fixed volume.
The temperature of the nitrogen is increased.
Explain the effect that increasing the temperature of the gas has on the pressure in the gas.
Use ideas about molecules in your answer.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P5.2.2 — The three types of nuclear emission (Parts (a)–(b))
• Topic P2.1.1 — States of matter (Part (c))

▶️ Answer/Explanation

(a)(i) Alpha: He nucleus, low penetrating ability; Gamma: high penetrating ability

An alpha particle is a helium nucleus (2 protons and 2 neutrons) with low penetrating ability.
Gamma radiation, being electromagnetic radiation with low ionising effect, has high penetrating ability.

(a)(ii) Alpha curves towards negative plate; gamma travels straight

Alpha particles are positively charged, so they curve towards the negative plate in the electric field.
Gamma rays have no charge, so they pass straight through the electric field without deviating.

(b) \(^{14}_{6}\text{C} \rightarrow \, ^{14}_{7}\text{N} + \, ^{0}_{-1}\beta\)

In beta decay, the nucleon (mass) number stays the same (14).
The proton (atomic) number increases by 1 (from 6 to 7) as a neutron converts to a proton, emitting a beta particle of charge −1 and negligible mass.

(c)(i) Particles far apart in a gas, touching in a solid

In a gas, the particles are far apart with large spaces between them, so they can be pushed closer together (compressed).
In a solid, the particles are already touching, so there is no space for them to be pushed closer together.

(c)(ii) Water cannot be compressed

In a liquid such as water, the molecules are touching, with little or no space between them.
Since there are no significant gaps to reduce, water cannot be compressed.

(c)(iii) Increasing temperature increases pressure

As temperature increases, the kinetic energy/speed of the gas molecules increases.
The molecules collide with the walls of the container more often and with greater force.
This increases the force per unit area on the walls, so the pressure increases.

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