Question 1
Table 1.1 shows the results.
During exercise, the pulse rate increases because the heart is pumping blood ……………
The pumping action is caused by contraction of the …………… wall of the heart.
To provide the body with more energy, the process of …………… increases.
This process requires increased blood flow to the cells to deliver more …………… and ……………
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B9.2 — Heart (Parts (a)(i)–(ii))
• Topic B7.1 — Diet (Part (b)(i))
• Topic B16.1 — Chromosomes and genes (Part (b)(ii))
• Topic B16.2 — Cell division (Part (c))
▶️ Answer/Explanation
(a)(i) 58 beats per minute
Difference \( = 122 – 64 \).
\( = 58 \) beats per minute.
(a)(ii)
… pumping blood faster.
… contraction of the muscular wall of the heart.
… process of (aerobic) respiration increases.
… deliver more glucose and oxygen.
(b)(i) Any two of:
Don’t consume too much fat.
Don’t consume too much salt / carbohydrate / calories.
Don’t consume excessive alcohol.
Consume a balanced diet / plenty of fibre.
(b)(ii) XY
Males have one X chromosome and one Y chromosome.
Females have two X chromosomes (XX).
(c) Any two of:
Cells produced by meiosis are genetically different from each other.
Cells produced by meiosis contain half the number of chromosomes (are haploid), unlike mitosis which produces genetically identical diploid cells.
Question 2
Nitrogen monoxide, NO, is made.
Explain why. Use ideas about collisions between particles.
Explain why. Use ideas about collisions between particles.
Nitrogen monoxide reacts with carbon monoxide.
Nitrogen and carbon dioxide are made.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C10.2 — Air quality and climate (Part (a))
• Topic C6.2 — Rate of reaction (Parts (b)(i)–(iii))
• Topic C5.1 — Exothermic and endothermic reactions (Parts (c)(i)–(ii))
▶️ Answer/Explanation
(a) nitrogen gas = 78%, oxygen gas = 21%
Clean, dry air is approximately 78% nitrogen and 21% oxygen.
The remaining ~1% is made up of argon, carbon dioxide, and other trace gases.
(b)(i) \( N_2 + O_2 \rightarrow 2NO \)
One nitrogen molecule reacts with one oxygen molecule to form two molecules of NO.
Atoms are balanced on both sides of the equation.
(b)(ii)
Particles move faster and have more kinetic energy at higher temperature.
More particles have energy greater than or equal to the activation energy.
This increases the rate of successful collisions.
(b)(iii)
Higher concentration means more particles per unit volume.
There is less space between particles, so collisions are more frequent.
This increases the rate of successful collisions.
(c)(i) Circle the \( N{=}O \) bonds and \( C{\equiv}O \) bonds on the reactant side.
These are the bonds present in the reactants, \( NO \) and \( CO \).
These bonds must be broken before new bonds can form in the products.
(c)(ii)
Bond breaking is endothermic and absorbs energy.
Bond making is exothermic and releases energy.
More energy is released in bond making than is absorbed in bond breaking, so the reaction is overall exothermic.
Question 3
Point X is 18 m vertically above point Y.
The gravitational field strength, g, is 10 N/kg.
The speed of the child at point Y is 14 m/s.
Calculate the kinetic energy of the child at point Y.
The zipline’s steel cable heats up as the child slides from point X to point Y.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.6.1 — Energy (Parts (a)(i)–(ii))
• Topic P1.4 — Density (Part (c))
• Topic P1.5.1 — Effects of forces (Parts (b)(i)–(iii), (d)(i)–(ii))
▶️ Answer/Explanation
(a)(i) GPE = 6300 J
\( GPE = mgh = 35 \times 10 \times 18 \).
\( = 6300 \, \text{J} \).
(a)(ii) KE = 3430 J
\( KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 35 \times 14^2 \).
\( = 3430 \, \text{J} \).
(b)(i) friction
Friction between the cable and the moving pulley/harness converts kinetic energy into heat energy.
(b)(ii) conduction
Steel is a solid metal, so thermal energy transfers through it by conduction.
(b)(iii)
Vibrations/oscillations pass from particle to particle along the metal.
Free electrons gain kinetic energy and transfer it through collisions, carrying energy through the metal quickly.
(c) density = 8400 kg/m³
\( \rho = \dfrac{m}{V} = \dfrac{4.2}{5.0 \times 10^{-4}} \).
\( = 8400 \, \text{kg/m}^3 \).
(d)(i) P at (100, 0.5)
The limit of proportionality is the point where the graph stops being a straight line.
This occurs at approximately 100 kN, 0.5 mm on Fig. 3.3.
(d)(ii) spring constant = 200,000,000 N/m
\( k = \dfrac{F}{x} = \dfrac{100\,000}{0.0005} \) (using values from the linear region, converting kN to N and mm to m).
\( = 2.0 \times 10^{8} \, \text{N/m} \).
Question 4
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B18.3 — Carbon cycle (Parts (a)(i)–(iv))
• Topic B7.1 — Diet (Part (b)(i))
• Topic B4 — Biological molecules (Part (b)(ii))
▶️ Answer/Explanation
(a)(i) fossilisation
Process E shows carbon from dead organisms being converted into fossil fuels.
This long-term burial and transformation process is called fossilisation.
(a)(ii) Arrow drawn from “carbon in plants” to “carbon in animals”.
Feeding represents animals eating plants (or other animals), transferring carbon compounds from the food source to the consumer.
(a)(iii) \( 6CO_2 + 6H_2O \xrightarrow{\text{light, chlorophyll}} C_6H_{12}O_6 + 6O_2 \)
Process A is photosynthesis, converting atmospheric carbon dioxide into glucose in plants.
Light energy and chlorophyll are required for this reaction.
(a)(iv) Name: combustion (of fossil fuels)
Burning fossil fuels releases stored carbon back into the atmosphere as carbon dioxide.
This increases the concentration of carbon dioxide in the atmosphere.
(b)(i) kwashiorkor / marasmus
Both are diseases resulting from insufficient protein and/or energy intake, common where diets lack adequate nutrition.
(b)(ii) amino acids
Proteins are polymers made up of chains of amino acid monomers joined by peptide bonds.
Question 5
Describe how a sodium atom forms a sodium ion.
Use \( e^- \) to represent an electron.
Draw a dot-and-cross diagram to show the bonding in sodium chloride.
Include the charges on the ions.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.2 — Atomic structure and the Periodic Table (Part (a))
• Topic C2.4 — Ions and ionic bonds (Parts (b)(i)–(ii), (d)(i)–(ii))
• Topic C8.3 — Group VII properties (Part (c))
▶️ Answer/Explanation
(a)
Electrons have negligible mass compared to protons/neutrons and carry a charge of −1.
Protons and neutrons each have a relative mass of 1; protons are +1, neutrons are neutral.
(b)(i) Sodium atom loses one electron.
Sodium has one electron in its outer shell.
Losing this electron gives it a stable, full outer shell and an overall +1 charge, forming \( Na^+ \).
(b)(ii) \( Na \rightarrow Na^+ + e^- \)
This half-equation shows sodium losing one electron to become a positively charged sodium ion.
(c)
Fluorine’s 9 electrons fill as 2 in the first shell and 7 in the second (2.7).
Bromine’s atomic number of 35 matches its total electron count of 2+8+18+7.
(d)(i) \( 2Na + Cl_2 \rightarrow 2NaCl \)
Two sodium atoms react with one chlorine molecule to form two units of sodium chloride.
The equation is balanced with equal Na and Cl atoms on both sides.
(d)(ii)
Sodium transfers its one outer electron to chlorine.
This forms \( Na^+ \) (no outer electrons shown) and \( Cl^- \) (8 outer electrons, gained one), held together by ionic bonding.
Question 6
Calculate the maximum distance that can be covered by an elephant in 120 seconds.
A switch controls the a.c. power supply to the lamps.
Include the switch in your diagram.
The a.c. power supply has been drawn for you.
Draw a circle around the correct current through each lamp.
Calculate the power output of each lamp.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.3 — Mass and weight (Part (a))
• Topic P1.2 — Motion (Part (b))
• Topic P4.3.2 — Series and parallel circuits (Parts (c)(i)–(iii))
▶️ Answer/Explanation
(a) 480 kg is the mass, not the weight; weight should be measured in Newtons.
Mass (kg) measures the amount of matter in an object.
Weight (N) is the force of gravity acting on that mass, so the unit should be Newtons, not kilograms.
(b) distance = 1320 m
\( d = v \times t = 11 \times 120 \).
\( = 1320 \, \text{m} \).
(c)(i) Four lamps drawn in parallel, each on its own branch, with the switch placed in the main circuit (in series with the power supply) so it controls all lamps together.
A parallel arrangement means each lamp has its own separate loop back to the power supply.
Placing the switch in the main line ensures it can turn all lamps on or off together.
(c)(ii) 4 A
In a parallel circuit with four identical lamps, the total current splits equally between them.
\( \dfrac{16}{4} = 4 \, \text{A} \) through each lamp.
(c)(iii) power = 960 W
\( P = IV = 4 \times 240 \).
\( = 960 \, \text{W} \).
Question 7
- labelling the x-axis
- drawing a line to predict the expected results.
Explain the effect of increasing humidity on the rate of transpiration.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B8.3 — Transpiration (Parts (a)(i)–(ii))
• Topic B8.1 — Xylem and phloem (Parts (b)(i)–(ii))
• Topic B6.2 — Leaf structure (Part (c))
• Topic B8.4 — Translocation (Part (d))
▶️ Answer/Explanation
(a)(i) x-axis labelled “temperature”; line increasing (and levelling off at higher temperatures).
As temperature rises, water molecules gain more kinetic energy and evaporate faster, so transpiration rate rises.
The line typically curves and levels off since stomata may partially close, or the rate becomes limited by water supply.
(a)(ii) Any three of:
Transpiration rate decreases.
There is an increase in water vapour in the air/atmosphere.
This decreases the water potential (concentration) gradient between the leaf and the air, so less evaporation/diffusion of water vapour occurs out of the stomata.
(b)(i)
Loss of water vapour from leaves creates a “transpiration pull” (tension) at the top of the xylem.
This creates a water potential gradient/difference, drawing water up through the xylem to replace what is lost.
(b)(ii) cohesion
Water molecules stick together through hydrogen bonding, allowing them to be pulled up as a continuous column.
(c) spongy mesophyll cells; guard cells
Spongy mesophyll cells have air spaces between them, increasing surface area for gas exchange.
Guard cells control the opening and closing of stomata, regulating gas exchange.
(d) Translocation:
occurs in the phloem.
involves the movement of amino acids.
transports substances to regions of storage in a plant.
Question 8
These elements are found in fertilisers.
A student makes some potassium sulfate.
He reacts potassium carbonate, \( K_2CO_3 \), with sulfuric acid.
Look at the equation for this reaction.
She uses a flame test.
Describe how she will know if the fertiliser contains potassium.
Ammonia is made from nitrogen and hydrogen.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C10.1 — Water (Part (a))
• Topic C3.3 — The mole and the Avogadro constant (Part (b))
• Topic C12.5 — Identification of ions and gases (Part (c))
• Topic C6.2 — Rate of reaction (Parts (d)(i)–(ii), (e)(i)–(ii))
▶️ Answer/Explanation
(a)
Nitrogen, phosphorus and potassium are essential nutrients that plants deplete from soil as they grow.
Fertilisers replace these nutrients to improve crop quality and increase yield, preventing deficiency symptoms such as discoloured leaves.
(b) mass of \( K_2SO_4 \) = 3.48 g
\( M_r(K_2CO_3) = (2\times39)+12+(3\times16) = 138 \); \( M_r(K_2SO_4) = (2\times39)+32+(4\times16) = 174 \).
Using the 1:1 mole ratio: \( \dfrac{174}{138}\times 2.76 = 3.48 \, \text{g} \).
(c) The flame test gives a lilac/purple flame.
Potassium compounds produce a characteristic lilac flame colour when heated, allowing identification.
(d)(i)
A catalyst increases the rate of reaction without being used up.
This allows ammonia to be made faster, sometimes at a lower temperature, reducing energy costs.
(d)(ii)
A reversible reaction can proceed in both the forward and backward directions.
Under the same conditions, reactants can form products, and products can re-form reactants.
(e)(i) The percentage of ammonia made decreases as temperature increases.
The forward reaction (making ammonia) is exothermic.
Increasing temperature favours the reverse (endothermic) reaction, reducing ammonia yield.
(e)(ii) Any one of: 350°C and 125 atm; 400°C and 210 atm; 450°C and 325 atm.
Reading across from 40% on the y-axis to each curve gives the corresponding pressure for that temperature.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P2.1.2 — Particle model (Parts (a)(i)–(ii))
• Topic P3.2.3 — Thin converging lens (Parts (b)(i)–(ii))
• Topic P3.3 — Electromagnetic spectrum (Parts (c)(i)–(ii))
▶️ Answer/Explanation
(a)(i) Brownian (motion)
This random, jittery motion of small particles suspended in a fluid is called Brownian motion.
(a)(ii)
The pollen grains are constantly hit/collided with by light, fast-moving water particles (molecules).
These random collisions from all directions cause the pollen grains to move in short, random paths.
(b)(i)
Two rays are drawn from the tip of the object: one parallel to the axis refracting through the far focal point, and one through the centre of the lens undeviated.
Where these rays cross beyond the lens marks the tip of the real, inverted image, which should be labelled “image”.
(b)(ii)
A real image can be projected onto a screen, since light rays actually meet there.
A virtual image cannot be projected onto a screen, as the rays only appear to diverge from it.
(c)(i) \( 3 \times 10^{8} \, \text{m/s} \)
This is the speed of all electromagnetic waves travelling through a vacuum.
(c)(ii)
Question 10
The country has a higher percentage of tobacco smokers across all age groups.
Describe and explain the difference you would expect to see in the results.
Complete Table 10.1.
State the effect of an increased concentration of carbon dioxide in the blood on the gas exchange system.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B11 — Gas exchange in humans (Parts (a)–(e))
▶️ Answer/Explanation
(a) age; sex
The graph shows COPD prevalence rising sharply with increasing age.
It also shows males consistently have a higher percentage than females across most age groups.
(b)
You would expect an increased percentage of COPD across all age groups in the country with more smokers.
This is because smoking is a major cause of COPD, so higher smoking rates lead to greater disease prevalence at every age.
(c)
Nicotine is the addictive chemical in tobacco smoke.
Carbon monoxide binds to haemoglobin, reducing the blood’s oxygen-carrying capacity, while tar contains carcinogens that cause cancer.
(d) Increased rate of breathing.
Higher blood \( CO_2 \) is detected by chemoreceptors, which stimulate an increase in breathing rate to remove the excess \( CO_2 \).
(e) Ciliated epithelial cells.
These cells line the airways and use beating cilia to sweep mucus (and trapped particles/pathogens) up and out of the gas exchange system.
Question 11
Three of these fractions are gasoline, gas oil and refinery gas.
Draw a diagram to show the structure of butane.
Cracking breaks large hydrocarbon molecules into smaller molecules.
State two conditions needed for cracking.
Fig. 11.1 shows the structure of dodecane.
During cracking, dodecane can make octane, \( C_8H_{18} \), and ethene, \( C_2H_4 \).
Ethene is an alkene. Alkenes have the general formula \( C_nH_{2n} \).
Dodecane and octane are alkanes.
State the general formula of the alkanes.
The mass of carbon dioxide gas made is 352 g.
Calculate the volume occupied by 352 g of carbon dioxide gas.
Show your working.
The volume of one mole of any gas is 24 dm³ at room temperature and pressure (r.t.p.).
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.4 — Alkanes (Parts (a), (d))
• Topic C11.3 — Fuels (Parts (b), (c))
• Topic C3.3 — The mole and the Avogadro constant (Part (e))
▶️ Answer/Explanation
(a)
Butane is a straight-chain alkane with 4 carbon atoms, each bonded to the maximum number of hydrogens.
Each carbon forms 4 single covalent bonds, satisfying its tetravalency.
(b) Any two of:
A (named) catalyst is used (e.g. zeolite/aluminium oxide/silicon dioxide).
High temperature and/or high pressure are required to break the covalent bonds.
(c)
Any covalent bond within the dodecane chain can break, at a random point along the molecule.
This random breaking produces different-sized fragments, giving a mixture of products such as octane and ethene.
(d) \( C_nH_{2n+2} \)
Alkanes are saturated hydrocarbons with only single bonds, so they have 2 more hydrogen atoms than twice the number of carbons.
(e) volume = 192 dm³
\( M_r(CO_2) = 12 + (2\times16) = 44 \); moles of \( CO_2 = \dfrac{352}{44} = 8 \).
Volume \( = 8 \times 24 = 192 \, \text{dm}^3 \).
Question 12
Calculate the output voltage when the a.c. power supply has an e.m.f. of 34 000 V.
Assume the transformer has an efficiency of 100%.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.5.6 — The transformer (Parts (a)(i)–(ii), (b))
• Topic P4.5.3 — Magnetic effect of current (Part (b))
• Topic P5.2.2 — The three types of nuclear emission (Parts (c)(i)–(ii))
▶️ Answer/Explanation
(a)(i) The soft-iron core is the central bar linking the two coils; it should be labelled X.
The soft-iron core links the primary and secondary coils magnetically, allowing efficient transfer of the changing magnetic field between them.
(a)(ii) output voltage = 16 000 V
Using \( \dfrac{V_p}{V_s} = \dfrac{N_p}{N_s} \): \( V_s = V_p \times \dfrac{N_s}{N_p} = 34\,000 \times \dfrac{8}{17} \).
\( = 16\,000 \, \text{V} \).
(b)
Field lines run parallel and close together inside the solenoid, pointing from one end to the other (like a bar magnet).
Outside the solenoid, the field lines curve round from the north pole end back to the south pole end.
(c)(i) \( {}^{238}_{92}U \rightarrow {}^{234}_{90}Th + {}^{4}_{2}\alpha \)
Mass number decreases by 4 (238 → 234) and atomic number decreases by 2 (92 → 90), consistent with the emission of an alpha particle \( ({}^{4}_{2}\alpha) \).
(c)(ii)
An alpha particle carries a positive charge.
A moving charged particle experiences a force when passing through a magnetic field, causing it to be deflected.