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Question 1

(a) Fig. 1.1 is a diagram of a cross-section of skin.
(i) State the letter in Fig. 1.1 that identifies a part:
of the peripheral nervous system ……………………
that produces sweat ……………………
that requires energy for contraction. ……………………
(ii) Describe how the blood vessels labelled in Fig. 1.1 try to maintain a constant internal body temperature if internal body temperature increases.
(iii) State the term used to describe the homeostatic mechanism used to control internal body temperature.
(b) The control of glucose concentration in the blood is an example of homeostasis.
(i) State the name of the hormone that reduces the concentration of glucose in the blood.
(ii) State the type of organs that produce hormones.
(c) Stimuli cause the body to make responses.
(i) State the name of the organ that detects the change in temperature of the blood.
(ii) State the name of the characteristic of living organisms that describes the detection and response to stimuli.

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

• Topic B13.1 — Coordination and response (Part (a))
• Topic B13.3 — Homeostasis (Part (a)(ii)–(iii))
• Topic B13.2 — Hormones (Part (b))
• Topic B13.1 — Coordination and response (Part (c))

▶️ Answer/Explanation

(a)(i) peripheral nervous system = A; produces sweat = D; requires energy for contraction = B

A is a sensory nerve ending, part of the peripheral nervous system.
D is the sweat gland, which produces sweat.
B is the erector muscle, which requires energy (ATP) to contract.

(a)(ii) Vasodilation of the arteriole increases blood flow to the capillaries

Arterioles supplying the skin capillaries widen (vasodilate).
This increases blood flow to the capillaries close to the skin surface.
More heat is lost from the blood to the surroundings, cooling the body.

(a)(iii) Negative feedback

A change away from the set point triggers a response that reverses the change.
This return-to-normal mechanism is called negative feedback.

(b)(i) Insulin

Insulin is secreted by the pancreas when blood glucose concentration rises.
It causes glucose uptake by cells and conversion to glycogen, lowering blood glucose.

(b)(ii) Glands

Hormones are produced by (endocrine) glands.
These glands secrete hormones directly into the bloodstream.

(c)(i) Brain

The brain contains thermoreceptors.
These detect changes in the temperature of the blood flowing through it.

(c)(ii) Sensitivity

Sensitivity is the characteristic of living organisms involving detection of, and response to, stimuli.

Question 2

(a) (i) Fig. 2.1 shows the three states of matter.
Complete the labels on Fig. 2.1.
(ii) Describe what happens to the kinetic energy of the particles in a gas when it is heated.
(b) A scientist analyses a food colouring X.
The scientist also analyses four dyes A, B, C, and D.
Fig. 2.2 shows the chromatogram produced.
(i) State why the start line is drawn using pencil instead of ink.
(ii) Identify which of the dyes, A, B, C and D, are in the food colouring X.
(iii) One of the substances in dye D remains on the pencil line. Explain why.
(iv) Use Fig. 2.2 to calculate the \(R_f\) value for dye A.
(c) Table 2.1 shows the melting points of tin, silver and the alloy solder.
Explain how the melting points show that solder is a mixture, but tin and silver are not.

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

• Topic C1.1 — Solids, liquids and gases (Part (a))
• Topic C12.3 — Chromatography (Part (b))
• Topic C12.4 — Separation and purification (Part (c))

▶️ Answer/Explanation

(a)(i) liquid → gas = evaporation/boiling; liquid → solid = freezing/solidification

The forward arrow from liquid to gas represents a change of state called evaporation or boiling.
The reverse arrow from liquid to solid represents freezing (solidification).

(a)(ii) Kinetic energy increases

Heating transfers thermal energy to the gas particles.
This increases their kinetic energy, so they move faster.

(b)(i) Pencil marks are insoluble in the solvent (water)

Pencil (graphite) does not dissolve in the solvent used for chromatography.
Ink would dissolve and run up the paper with the solvent, distorting results.

(b)(ii) Dyes A, B and D

Comparing the spot positions of X with A, B, C and D on the chromatogram.
X matches the heights of spots A, B and D exactly, so these dyes are present in X.

(b)(iii) It is insoluble in the solvent

A substance that stays on the pencil line does not dissolve in the solvent.
Since it cannot move with the solvent front, it remains at the origin.

(b)(iv) \(R_f = 0.65\)

\( R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent}} \)
\( R_f = \dfrac{2.6}{4.0} = 0.65 \)

(c) Solder melts over a range (220–229°C); tin and silver melt at a fixed point

Pure substances (tin, silver) have one specific, sharp melting point.
Mixtures (solder) melt over a range of temperatures because the different substances interfere with each other’s regular lattice arrangement.

Question 3

An Olympic triathlon event consists of a 1500 m swim, a 40 km cycle ride and a 10 km run.
(a) Fig. 3.1 shows an athlete swimming at a constant speed.
(i) Describe how the size of force A compares with the size of force B.
(ii) The athlete has a weight of 750 N and moves with a kinetic energy of 13.5 J.
Calculate the speed of the athlete.
The gravitational field strength, \(g\), is 10 N/kg.
(b) Fig. 3.2 shows a speed–time graph for the start of the cycle ride.
(i) Show that the maximum speed of the athlete during the first 40 seconds of the cycle ride is 12.5 m/s.
(ii) Calculate the acceleration of the athlete during the first 25 seconds of the cycle ride. Give your answer in \(\text{m/s}^2\).
(iii) Calculate the distance covered by the athlete during the first 35 seconds of the cycle ride.
(iv) Fig. 3.3 shows the pedal of the bicycle as the athlete pedals.
The moment of the force applied by the athlete is 35.7 N m.
Use Fig. 3.3 to calculate the force exerted by the athlete on the pedal.
(c) During the run, the athlete starts to sweat. Explain, in terms of the motion and energy of water molecules, how sweating cools the athlete’s skin.

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

• Topic P1.5.1 — Effects of forces (Part (a)(i))
• Topic P1.6.1 — Energy (Part (a)(ii))
• Topic P1.2 — Motion (Part (b)(i)–(iii))
• Topic P1.5.2 — Turning effect of forces (Part (b)(iv))
• Topic P2.2.2 — Melting, boiling and evaporation (Part (c))

▶️ Answer/Explanation

(a)(i) Force A equals force B

Since the athlete swims at a constant speed, there is no resultant force.
So the forward force (B) and backward drag force (A) must be equal and opposite.

(a)(ii) speed = 0.6 m/s

Mass \(= \dfrac{750}{10} = 75\ \text{kg}\)
\(E_k = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{\dfrac{2 \times 13.5}{75}} = 0.6\ \text{m/s}\)

(b)(i) 45 000 m ÷ 3600 s = 12.5 m/s

45 km/h is converted to m/s by dividing by 3600 and multiplying by 1000.
\( \dfrac{45\,000}{3600} = 12.5\ \text{m/s} \), confirming the given value.

(b)(ii) acceleration = 0.5 m/s²

\( a = \dfrac{\Delta v}{t} = \dfrac{12.5}{25} = 0.5\ \text{m/s}^2 \)
This is the gradient of the speed–time graph over the first 25 s.

(b)(iii) distance = 281.25 m

Area under the graph = triangle (0–25 s) + rectangle (25–35 s):
\( (\tfrac{1}{2} \times 25 \times 12.5) + (12.5 \times 10) = 156.25 + 125 = 281.25\ \text{m} \)

(b)(iv) force = 210 N

\( \text{Moment} = \text{force} \times \text{distance} \)
\( F = \dfrac{35.7}{0.17} = 210\ \text{N} \)

(c) Evaporation of the most energetic water molecules cools the skin

Thermal energy is transferred from the skin to the water molecules on its surface.
The most energetic molecules gain enough energy to escape as vapour (evaporate).
This lowers the average kinetic energy (temperature) of the remaining water and skin.

Question 4

(a) Oxygen is one of the products of photosynthesis. Complete the balanced symbol equation for photosynthesis.
……………………… + ……………………… \(\xrightarrow[\text{chlorophyll}]{\text{light}}\) …………………….. + \(6O_2\)
(b) Complete the energy transfer that takes place using chlorophyll.
………………………………… energy → ………………………………… energy
(c) State the name of the cells in a leaf that contain the highest concentration of chlorophyll.
(d) A student investigates the effect of light intensity on the rate of photosynthesis.
Fig. 4.1 shows the apparatus she uses.
The student:
  • places the lamp at 10 cm from the aquatic plant
  • counts the number of oxygen bubbles released in 2 minutes
  • repeats this two more times and calculates a mean
  • repeats the process with the lamp at different distances from the aquatic plant.
Table 4.1 shows the results. The number of oxygen bubbles released indicates the rate of photosynthesis.
(i) Calculate the mean number of oxygen bubbles released when the lamp is 30 cm from the aquatic plant.
Give your answer to the nearest whole number.
Write your answer in Table 4.1.
(ii) Describe the effect of light intensity on the rate of photosynthesis using the data in Table 4.1.
(iii) The aquatic plant releases less oxygen into the water than it produces during photosynthesis.
Suggest one reason for this difference.

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

• Topic B6.1 — Photosynthesis (Part (a)–(b))
• Topic B6.2 — Leaf structure (Part (c))
• Topic B6.1 — Photosynthesis, limiting factors (Part (d))

▶️ Answer/Explanation

(a) \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\)

Carbon dioxide and water are the reactants of photosynthesis.
Glucose (\(C_6H_{12}O_6\)) is the organic product, alongside the given oxygen.

(b) Light energy → chemical energy

Chlorophyll absorbs light energy.
This is transferred into chemical energy stored in glucose.

(c) Palisade (mesophyll) cells

Palisade mesophyll cells are packed with chloroplasts.
Their position near the leaf’s upper surface maximises light absorption.

(d)(i) mean = 15

Mean \( = \dfrac{14+15+17}{3} = \dfrac{46}{3} = 15.33 \)
Rounded to the nearest whole number, mean = 15.

(d)(ii) As light intensity decreases, the rate of photosynthesis decreases, until it becomes constant

As the lamp distance increases (light intensity decreases), the mean number of bubbles/rate of photosynthesis decreases.
Beyond 40 cm, the rate becomes constant, showing light is no longer the limiting factor.

(d)(iii) Some of the oxygen produced is used in respiration

The plant respires continuously, consuming some of the oxygen it produces.
Only the surplus oxygen is released as bubbles, so less is measured than is actually produced.

Question 5

(a) Fig. 5.1 shows a diagram of a lithium atom.
(i) Complete the labels on Fig. 5.1.
(ii) State the electronic structure of a lithium atom.
(b) (i) A lithium atom bonds with a chlorine atom by ionic bonding.
Fig. 5.2 shows the formation of a lithium ion, \(Li^+\), from a lithium atom.
Draw a similar diagram to show the formation of a chloride ion, \(Cl^-\), from a chlorine atom.
(ii) Ionic compounds, such as lithium chloride, have a lattice structure.
Describe the lattice structure of ionic compounds.
You may include a labelled diagram if you wish.
(c) (i) Carbon has three naturally occurring isotopes: carbon-12, carbon-13 and carbon-14.
Complete Table 5.1 to show the numbers of protons, neutrons and electrons in an atom of each isotope.
(ii) Explain, in terms of particles, why these isotopes have the same chemical properties.

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

• Topic C2.1 — Atomic structure (Part (a))
• Topic C2.4 — Ions and ionic bonds (Part (b))
• Topic C2.3 — Isotopes (Part (c))

▶️ Answer/Explanation

(a)(i) Nucleus contains protons and neutrons; outer particle is an electron

The central nucleus label should read “protons and neutrons.”
The particle orbiting in the outer shell is labelled “electron.”

(a)(ii) 2,1

Lithium has 3 electrons total.
These are arranged with 2 in the first shell and 1 in the second shell (2,1).

(b)(i) Chlorine atom gains one electron to form \(Cl^-\)

A chlorine atom (2,8,7) gains one electron from the lithium atom.
This forms a chloride ion with electronic structure (2,8,8) and a single negative charge, shown in square brackets with a “–” superscript.

(b)(ii) Regular arrangement of alternating positive and negative ions

Ionic lattices consist of oppositely charged ions arranged in a fixed, repeating (regular) 3D pattern.
Strong electrostatic forces of attraction hold the alternating positive and negative ions together.

(c)(i)

Carbon-13: 6 protons, 7 neutrons, 6 electrons.
Carbon-14: 6 protons, 8 neutrons, 6 electrons.
(Protons = atomic number = 6 for all isotopes; neutrons = mass number − protons.)

(c)(ii) Isotopes have the same number of electrons in the outer shell

Chemical properties depend on the arrangement of electrons, especially the outer shell.
Since all isotopes of carbon have the same number of electrons (and same outer shell arrangement), they react in the same way.

Question 6

Fig. 6.1 shows a boiler that uses combustion of natural gas to heat water.
(a) Natural gas is a non-renewable energy source.
Describe one environmental impact of using natural gas in this way.
(b) The boiler has an efficiency of 90%.
The combustion of natural gas provides an input energy of 1.50 kJ.
Calculate the useful energy output from the boiler.
(c) Thermal energy is transferred through the water in the boiler by convection.
Describe the process of convection in terms of density changes.
(d) Light from the gas flame has a wavelength of \(4.6 \times 10^{-7}\ \text{m}\).
(i) Calculate the frequency of the light from the flame.
(ii) The light from the flame is a transverse wave. Complete the sentences to describe the differences between a transverse wave and a longitudinal wave.
Transverse waves are produced by vibrations acting ………………………………………………. to the direction of energy transfer.
Longitudinal waves are produced by vibrations acting ………………………………………………. to the direction of energy transfer.
An example of a longitudinal wave is a ………………………………………………. wave.

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

• Topic C10.2 — Air quality and climate (Part (a))
• Topic P1.6.4 — Power (Part (b))
• Topic P2.3.2 — Convection (Part (c))
• Topic P3.1 — General properties of waves (Part (d))

▶️ Answer/Explanation

(a) Contributes to global warming / enhanced greenhouse effect

Burning natural gas releases carbon dioxide.
This is a greenhouse gas that contributes to global warming/climate change.

(b) useful energy output = 1.35 kJ

\( \text{Output} = \text{efficiency} \times \text{input} \)
\( = 0.90 \times 1.50 = 1.35\ \text{kJ} \)

(c) Heated water becomes less dense and rises

Water near the flame is heated and expands, becoming less dense.
This less dense (hotter) water rises, while cooler, denser water sinks to take its place, setting up a convection current.

(d)(i) frequency \( \approx 6.5 \times 10^{14}\ \text{Hz}\)

Speed of light, \(c = 3 \times 10^8\ \text{m/s}\)
\( f = \dfrac{c}{\lambda} = \dfrac{3\times10^{8}}{4.6\times10^{-7}} \approx 6.5\times10^{14}\ \text{Hz} \)

(d)(ii) Transverse: perpendicular; Longitudinal: parallel; example: sound

In transverse waves, vibrations act perpendicular to the direction of energy transfer.
In longitudinal waves, vibrations act parallel to the direction of energy transfer.
Sound is a common example of a longitudinal wave.

Question 7

(a) Fig. 7.1 is a diagram showing the development of a strain of antibiotic resistant bacteria.
(i) Describe what happens to the bacteria during stage 1 in Fig. 7.1.
(ii) Describe what happens to the bacteria during stage 2 in Fig. 7.1.
(iii) State the name of the process that results in antibiotic resistance shown in Fig. 7.1.
(b) Antibiotic resistance initially occurs due to a mutation. Some chemicals can cause mutation.
(i) Define the term mutation.
(ii) State the type of radiation that increases the rate of mutation.
(c) Components of blood are responsible for protecting the body from disease-causing organisms including some strains of bacteria.
State the name of the component of blood responsible for:
antibody production ……………………………………………………………………………………………………
blood clotting. ……………………………………………………………………………………………………………

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

• Topic B17.2 — Selection (Part (a))
• Topic B17.1 — Variation and mutation (Part (b))
• Topic B9.4 — Blood (Part (c))

▶️ Answer/Explanation

(a)(i) Non-resistant bacteria are killed by the antibiotic; resistant bacteria survive

The antibiotic kills the bacteria that do not have resistance.
The bacteria with the resistance allele/gene survive the treatment.

(a)(ii) Resistant bacteria multiply, passing on resistance to offspring

The surviving resistant bacteria reproduce (divide).
They pass on the resistance genes/DNA to their offspring, which are all resistant.

(a)(iii) Natural selection

This process, where organisms best suited to their environment survive and reproduce, is called natural selection.

(b)(i) A mutation is a change in a gene or chromosome

A mutation is a random change in the DNA sequence of a gene or the structure/number of a chromosome.

(b)(ii) Ionising radiation

Ionising radiation (e.g. UV, X-rays, gamma rays) can damage DNA, increasing the rate of mutation.

(c) antibody production = white blood cells; blood clotting = platelets

White blood cells (lymphocytes) produce antibodies as part of the immune response.
Platelets are responsible for triggering blood clotting at a wound.

Question 8

A student investigates the rate of reaction between dilute hydrochloric acid, HCl, and magnesium, as shown in Fig. 8.1.
Magnesium chloride, \(MgCl_2\), and hydrogen gas, \(H_2\), are made.
(a) Construct the balanced symbol equation for this reaction.
(b) The student repeats the experiment with five different temperatures of the dilute hydrochloric acid.
The same volume and concentration of hydrochloric acid and the same mass of magnesium ribbon are used in each experiment.
She measures the time for the magnesium to completely react at each temperature.
Table 8.1 shows her results.
(i) The reaction gets faster as the temperature increases.
Explain how you can tell this from Table 8.1.
(ii) Tick (✓) two reasons in Table 8.2 which explain why reactions get faster as the temperature increases.
(c) The reaction between magnesium and dilute hydrochloric acid is an exothermic reaction.
Use the axes shown in Fig. 8.2 to draw and label the energy level diagram for this reaction.
Label:
  • the energy of the reactants and the products
  • the energy change in the reaction
  • the activation energy of the reaction.
(d) Zinc reacts with sulfuric acid, \(H_2SO_4\), to make zinc sulfate, \(ZnSO_4\), and hydrogen gas.
\( Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2 \)
3.35 g of zinc reacts with excess dilute sulfuric acid to make 0.1 g of hydrogen gas.
Calculate the volume occupied by 0.1 g of hydrogen 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, 2025–2027 syllabus):

• Topic C9.4 — Reactivity series (Part (a))
• Topic C6.2 — Rate of reaction (Part (b))
• Topic C5.1 — Exothermic and endothermic reactions (Part (c))
• Topic C3.3 — The mole and the Avogadro constant (Part (d))

▶️ Answer/Explanation

(a) \( Mg + 2HCl \rightarrow MgCl_2 + H_2 \)

Magnesium displaces hydrogen from hydrochloric acid.
Formulae and coefficients must balance atoms on both sides.

(b)(i) The time decreases as temperature increases

Table 8.1 shows the time taken falls steadily from 119 s (20°C) to 31 s (40°C).
A shorter time for the same amount of reactant to react means a faster rate.

(b)(ii)

Higher temperature gives particles more kinetic energy, so they move faster.
Faster-moving particles collide more frequently and with greater force, increasing the reaction rate.

(c)

For an exothermic reaction, the products’ energy level is lower than the reactants’.
The energy change (ΔH) is the vertical drop from reactants to products.
The activation energy is the “hump” from reactants up to the peak of the curve.

(d) volume = 1.2 dm³

\( M_r \) of \(H_2 = 2\)
Moles \( = \dfrac{0.1}{2} = 0.05\ \text{mol}\)
Volume \( = 0.05 \times 24 = 1.2\ \text{dm}^3 \)

Question 9

A student investigates how the resistance of a wire changes with length.
Fig. 9.1 shows the equipment she uses.
(a) The student moves the crocodile clips to change the length of the wire.
She measures this length with the ruler and uses the ammeter reading to calculate the resistance of the wire.
When the wire is made longer, the reading on the ammeter decreases.
Explain why the reading on the ammeter decreases.
(b) Fig. 9.2 shows a length of wire connected in series with another component labelled X.
(i) State the name of the component labelled X in Fig. 9.2.
(ii) The student uses the component labelled X to vary the potential difference across the length of wire.
The student records the potential difference across the wire and the current in the wire.
Fig. 9.3 shows her results.
Use Fig. 9.3 to determine the resistance of the wire.
(c) The student chooses to use a maximum electromotive force (e.m.f.) of 1.5 V.
State the meaning of the term electromotive force (e.m.f.).
(d) On Fig. 9.4, draw the shape and direction of the magnetic field around the current-carrying wire.

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

• Topic P4.2.4 — Resistance (Part (a))
• Topic P4.3.1 — Circuit diagrams and circuit components (Part (b)(i))
• Topic P4.2.4 — Resistance, I–V graphs (Part (b)(ii))
• Topic P4.2.3 — Voltage (electromotive force and potential difference) (Part (c))
• Topic P4.5.3 — Magnetic effect of current (Part (d))

▶️ Answer/Explanation

(a) A longer wire has greater resistance, so current decreases

Resistance of a wire increases as its length increases.
Since current is inversely proportional to resistance (for a fixed e.m.f.), the ammeter reading decreases.

(b)(i) Variable resistor

Component X, drawn with an arrow through a rectangle, is a variable resistor (rheostat), used to change the current/p.d. in the circuit.

(b)(ii) resistance ≈ 1.9 Ω

\( R = \dfrac{V}{I} \)
Using the graph, at \(I = 0.80\ \text{A}\), \(V = 1.50\ \text{V}\).
\( R = \dfrac{1.50}{0.80} \approx 1.9\ \Omega \)

(c) e.m.f. is the energy supplied (work done) per unit charge driven around a complete circuit

Electromotive force is the electrical work done by a source in driving charge around a complete circuit, per unit charge.

(d) Concentric circles around the wire, direction shown

The magnetic field forms concentric circles centred on the wire.
Its direction is found using the right-hand grip rule, based on the direction of current flow.

Question 10

Fig. 10.1 shows two red blood cells after they have been immersed in different solutions for an hour.
Cell A was immersed in a concentrated salt solution.
Cell B was immersed in blood plasma.
(a) Complete the sentences to explain the appearance of cell A in Fig. 10.1.
The concentrated salt solution has a lower ………………………………………………. than cell A.
Water crosses the ………………………………………………. and leaves the cell by osmosis.
Water molecules move from a more ………………………………………………. solution to a more ………………………………………………. solution.
(b) Immersion of cell A in concentrated salt solution changes the shape of the cell.
Suggest how this change in shape affects the function of red blood cells in the body.
(c) Concentration gradients affect the rate of osmosis.
Suggest two other factors that affect the rate of osmosis.
(d) Plant cells have additional cell structures that are not present in animal cells.
(i) State the names of two cell structures present in plant cells but not in animal cells.
(ii) State the name of the type of plant cell that is specialised for absorption of water.

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

• Topic B3.2 — Osmosis (Part (a), (c))
• Topic B9.4 — Blood (Part (b))
• Topic B2.1 — Cell structure (Part (d))

▶️ Answer/Explanation

(a) lower water potential; crosses the membrane; more dilute to more concentrated

The salt solution has a lower water potential than the cell.
Water leaves the cell across the (partially permeable) cell membrane by osmosis.
Water moves from a more dilute solution (inside the cell) to a more concentrated solution (outside).

(b) Reduced surface area/volume means less oxygen can be transported

The shrunken, spiky shape reduces the cell’s surface area available for gas exchange.
This means less oxygen can be carried/transported around the body.

(c) Surface area and temperature

A larger surface area increases the rate of osmosis.
Higher temperature increases the kinetic energy of water molecules, increasing the rate.

(d)(i) Chloroplast and (permanent) vacuole (or cell wall)

Plant cells contain chloroplasts (for photosynthesis) and a permanent vacuole, structures not found in animal cells.
A cellulose cell wall is also unique to plant cells.

(d)(ii) Root hair cell

Root hair cells have a large surface area, specialised for the absorption of water (and mineral ions) from the soil.

Question 11

Look at the structures of the carbon compounds shown in Fig. 11.1.
(a) (i) State which compound is made by the catalytic addition of steam to compound A. Choose from B, C, D or E.
(ii) State which compound is made when compound A reacts with hydrogen gas, \(H_2\). Choose from B, C, D or E.
(iii) State which compound reacts with bromine to form \(C_2H_4Br_2\). Choose from A, B, C, D or E.
(iv) State which compound forms compound E in an addition polymerisation reaction. Choose from A, B, C or D.
(b) Fig. 11.2 represents the formation of the polymer nylon from two monomers, X and Y.
Nylon is made in a condensation polymerisation reaction.
Describe how monomer X and monomer Y react together to make nylon.
Use the information in Fig. 11.2 in your answer.

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

• Topic C11.5 — Alkenes (Part (a))
• Topic C11.7 — Polymers (Part (b))

▶️ Answer/Explanation

(a)(i) B

Catalytic addition of steam to an alkene (A, ethene) produces an alcohol.
Compound B is ethanol.

(a)(ii) C

Addition of hydrogen to an alkene produces the corresponding alkane.
Compound C is ethane.

(a)(iii) A

Compound A (ethene, \(C_2H_4\)) reacts with bromine by addition across the C=C double bond.
This forms \(C_2H_4Br_2\) (1,2-dibromoethane).

(a)(iv) A

Compound E is a repeating polymer unit derived from a \(C=C\) monomer.
Compound A (ethene) undergoes addition polymerisation to form this polymer (polyethene).

(b) Monomer X and Y join, losing a small molecule (water)

The carbon (–COOH) group of monomer X joins with the nitrogen (–NH) group of monomer Y.
An –OH is lost from monomer X and an –H is lost from monomer Y.
These combine to eliminate a molecule of water, forming the amide (C–N) link.

Question 12

A student investigates the penetrating abilities of ionising radiation.
Fig. 12.1 shows the equipment used by the student.
(a) The student places different shielding materials between the source and the detector and uses the counter to record the number of counts in 1 minute.
Table 12.1 shows the student’s results.
(i) Use Table 12.1 to state and explain which type of ionising radiation is emitted by the source.
(ii) The source used in Fig. 12.1 has a half-life of 29 years.
Calculate the time it will take for the activity of the source to drop to 12.5% of the original value.
(b) The lead used in the student’s investigation is a solid.
The melting point of lead is 327°C. When lead melts, it turns from a solid into a liquid.
Describe the changes in the forces between particles when a solid melts.
(c) The density of liquid lead is \(10.6\ \text{g/cm}^3\).
A sample of liquid lead has a mass of 37.1 g.
Calculate the volume of the sample of liquid lead.

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

• Topic P5.2.2 — The three types of nuclear emission (Part (a)(i))
• Topic P5.2.4 — Half-life (Part (a)(ii))
• Topic P2.2.1 — Thermal expansion of solids, liquids and gases (Part (b))
• Topic P1.4 — Density (Part (c))

▶️ Answer/Explanation

(a)(i) Beta radiation

The count rate stays high through paper but drops sharply once thin aluminium is added, then changes little afterwards.
This shows the radiation can penetrate air and paper, but not thin aluminium — the pattern characteristic of beta particles.

(a)(ii) time = 87 years

12.5% remaining corresponds to \( \left(\dfrac{1}{2}\right)^3 \), i.e. 3 half-lives.
Time \( = 3 \times 29 = 87\) years.

(b) The forces between particles decrease

As a solid melts, the strong forces holding particles in fixed positions weaken.
This allows particles to move more freely past one another, forming a liquid.

(c) volume = 3.5 cm³

\( V = \dfrac{m}{d} \)
\( V = \dfrac{37.1}{10.6} \approx 3.5\ \text{cm}^3 \)

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