Question 1
Coronary heart disease is caused by a ………………………………….. of the coronary arteries.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B9.2 — Heart (Part (a))
• Topic B9.3 — Blood vessels (Part (b))
▶️ Answer/Explanation
(a)(i) blockage
Coronary heart disease results from fatty deposits building up inside the coronary arteries.
This narrows the artery lumen, restricting blood flow to the heart muscle.
(a)(ii) Stop smoking; do more exercise
From Fig. 1.1, the person smokes tobacco, which increases CHD risk — so stopping smoking reduces risk.
The person only walks for 10 minutes a week, so increasing exercise also reduces risk.
(a)(iii) Age
Age (68) is a risk factor the person cannot change or control.
Diet, weight, exercise, and smoking are all lifestyle factors that can be modified.
(b)(i) Thick, elastic/muscular wall
Feature \( X \) (the artery wall) is thick to withstand the high pressure of blood pumped from the heart.
The wall is elastic and muscular, allowing it to stretch and recoil.
This recoil helps maintain the high pressure of blood as it flows through the artery.
(b)(ii) Double circulation
Mammals have a double circulatory system.
Blood passes through the heart twice for each complete circuit of the body — once to the lungs (pulmonary circulation) and once to the rest of the body (systemic circulation).
Question 2
Explain why aluminium is resistant to corrosion.
Fig. 2.1 shows the structures of pure aluminium and an alloy of aluminium.
Explain, in terms of their structures and properties, why the alloy is used instead of pure aluminium.
The steel is usually coated with zinc before it is painted.
The zinc prevents rusting, even if the zinc layer is damaged.
Describe how the zinc prevents rusting.
Carbon, a non-metal, is also shown in Table 2.1.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C9.2 — Uses of metals (Part (a), (b))
• Topic C9.3 — Alloys and their properties (Part (c))
• Topic C9.5 — Corrosion of metals (Part (d))
• Topic C9.4/C9.6 — Reactivity series / Extraction of metals (Part (e))
▶️ Answer/Explanation
(a) Low density
Aircraft need to be as light as possible to reduce fuel consumption and allow flight.
Aluminium’s low density makes it much lighter than most other structural metals of similar strength.
(b) Forms a protective oxide layer
Aluminium reacts with oxygen in the air to form a thin layer of aluminium oxide on its surface.
This oxide layer adheres tightly to the metal and prevents further oxygen or water from reaching the aluminium beneath, so corrosion does not continue.
(c) Alloy structure prevents layers sliding
In the alloy, atoms of other elements (different sizes) are mixed in among the aluminium atoms, creating an irregular arrangement.
This distorts the regular layers of atoms found in pure aluminium.
The layers can no longer slide over each other as easily, so more energy/force is needed to deform the metal, making duralumin harder and stronger than pure aluminium.
(d) Sacrificial protection
Zinc is more reactive than iron.
Since zinc is more reactive, it oxidises/corrodes in preference to the iron, even where the zinc coating is scratched or damaged.
This is called sacrificial protection.
(e) Metal: sodium (or calcium or magnesium)
This metal is more reactive than carbon in the reactivity series.
Since carbon cannot displace it from its compound/ore by heating, it cannot be extracted by reduction with carbon.
Instead, it must be extracted using electrolysis, which is used for metals above carbon in reactivity.
Question 3
A small meteorite has a mass of \( 1720 \, \text{g} \) and a volume of \( 200 \, \text{cm}^3 \).
Calculate the density of the meteorite.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.2 — Motion (Part (a))
• Topic P1.4 — Density (Part (b))
• Topic P3.4 — Sound (Part (c))
▶️ Answer/Explanation
(a)(i) \( t = 0.04 \, \text{s} \)
Speed is defined as \( v = \dfrac{d}{t} \), so \( t = \dfrac{d}{v} \).
\( t = \dfrac{1000}{25\,000} = 0.04 \, \text{s} \).
(a)(ii) Constant speed, then decelerates, then stops
From \( 0 \) to \( 3 \, \text{s} \), the meteoroid moves at a constant speed of \( 25\,000 \, \text{m/s} \) (horizontal line on the graph).
From \( 3 \, \text{s} \) onward, it slows down / decelerates, with the deceleration increasing (curve gets steeper).
At \( 7 \, \text{s} \), the speed reaches zero — the meteoroid stops or hits the ground.
(b) \( 8.60 \, \text{g/cm}^3 \)
Density is calculated using \( \rho = \dfrac{m}{V} \).
\( \rho = \dfrac{1720}{200} = 8.60 \, \text{g/cm}^3 \).
(c)(i) Compressions and rarefactions
Sound waves are longitudinal waves.
They are transmitted through air as a series of compressions (regions of higher pressure) and rarefactions (regions of lower pressure) as air particles vibrate back and forth.
(c)(ii) 
Sound travels fastest through solids, since particles are closely packed and transmit vibrations most efficiently.
It travels faster in liquids than in gases, since liquid particles are more closely packed than gas particles.
So: air (least dense, particles far apart) = 340 m/s; rock (solid) = 1500 m/s; water (liquid) = 4200 m/s — matching the given values exactly as per the mark scheme.
Question 4
Fertilisers entering the water contain ……………. ions.
An increase in the availability of these ions enables an increase in the ……………… of producers on the surface of the water.
Producers underneath the water are unable to ……………. due to a lack of light and they die.
The population of bacteria increases as they …………… the dead material.
The concentration of dissolved oxygen decreases because the bacteria need oxygen for the process of ……………. ……………. .
This entire process is called ………….. .
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B19.1 — Habitat destruction / freshwater pollution (Part (a))
• Topic B16.2 — Cell division; B15.1/B15.2 — Asexual/Sexual reproduction (Part (b))
▶️ Answer/Explanation
(a) nitrate; growth; photosynthesise; decompose; aerobic respiration; eutrophication
Fertilisers add nitrate ions, which increase the growth of producers (algae) on the water’s surface.
Producers below the surface are shaded from light and cannot photosynthesise, so they die.
Bacteria increase in number as they decompose this dead plant material, using up oxygen for aerobic respiration.
This whole process — from nutrient enrichment to oxygen depletion — is called eutrophication.
(b)(i) Mitosis
Asexual reproduction produces genetically identical offspring through mitosis, a type of nuclear division.
(b)(ii) Any two: identical offspring / one parent / no gametes
Asexual reproduction creates genetically identical offspring (clones), whereas sexual reproduction creates genetically varied offspring.
Asexual reproduction only requires one parent, while sexual reproduction requires two parents.
Asexual reproduction does not involve gametes or fertilisation, unlike sexual reproduction.
(b)(iii) Gamete (egg or sperm)
Gametes (the egg cell and sperm cell) are the specialised sex cells adapted for sexual reproduction in humans.
Question 5
[\(A_r\) : H, 1; N, 14]
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.1 — Elements, compounds and mixtures (Part (a))
• Topic C1.2 — Diffusion (Part (b)(i))
• Topic C3.3 — The mole and the Avogadro constant (Part (b)(ii))
▶️ Answer/Explanation
(a)
(b)(i) Ammonia molecules are lighter and diffuse faster
The molecular mass of ammonia (\( M_r = 17 \)) is less than that of hydrogen chloride (\( M_r = 36.5 \)).
Lighter molecules move faster and diffuse a greater distance in the same time, so ammonia travels further along the tube before meeting the hydrogen chloride, forming the cloud nearer the HCl end.
(b)(ii) \( 7.20 \, \text{dm}^3 \)
\( M_r \) of \( \text{NH}_3 = 14 + (3 \times 1) = 17 \).
moles of \( \text{NH}_3 = \dfrac{5.1}{17} = 0.3 \, \text{mol} \).
volume \( = 0.3 \times 24 = 7.2 \, \text{dm}^3 \).
Question 6
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P2.3.2 — Convection (Part (a))
• Topic P2.1.3 — Pressure changes (Part (b))
• Topic P4.5.5 — The d.c. motor (Part (c))
▶️ Answer/Explanation
(a)(i) Convection
Most thermal energy is transferred from the food to the cooling unit by convection.
(a)(ii) Cooled air sinks, warm air rises
Air at the top is cooled by the cooling unit, becoming denser.
This cooled, denser air sinks to the bottom of the refrigerator.
Warmer air, which has a lower density, rises to take its place near the cooling unit, setting up a convection current that cools all the air inside.
(b) 
At constant volume, increasing the temperature increases the pressure of a fixed mass of gas.
At constant temperature, increasing the volume decreases the pressure — so completing the table: temperature increases → pressure increases; volume increases (temperature constant) → pressure decreases.
(c) Any two: decrease current; fewer turns; weaker magnet
Decreasing the current flowing through the coil reduces the turning effect, slowing the motor.
Using fewer turns on the coil also reduces the turning effect.
Using a weaker magnet (weaker magnetic field) similarly reduces the force on the coil, slowing its rotation.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B6.2 — Leaf structure (Part (a))
• Topic B6.1 — Photosynthesis (Part (b) — use and storage of carbohydrates)
▶️ Answer/Explanation
(a)(i) A – (upper) epidermis; B – palisade mesophyll
Layer A is the upper epidermis, a protective outer layer.
Layer B is the palisade mesophyll, containing tightly packed cells rich in chloroplasts for photosynthesis.
(a)(ii) Large air spaces between loosely packed cells
The spongy mesophyll (C) has large air spaces between loosely packed cells.
This increases the surface area available for gas diffusion, allowing carbon dioxide and oxygen to exchange efficiently.
(a)(iii) Guard cell
A guard cell, found in pairs surrounding a stoma (pore), controls the opening and closing of the stoma, regulating gas entry into the leaf.
(b) Any three: glucose (respiration); starch (storage); cellulose (cell walls); sucrose (transport)
Glucose is used in respiration to release energy for the plant.
Starch is used as an insoluble storage carbohydrate.
Cellulose is used to build strong cell walls.
Sucrose is used to transport sugars around the plant in the phloem.
Question 8
Each word may be used once, more than once, or not at all.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C4.1 — Electrolysis (all parts)
▶️ Answer/Explanation
(a) gain; anions; lose
Positive ions (cations) move to the negative electrode (cathode) and gain electrons to form atoms.
Negative ions (anions) move to the positive electrode (anode) and lose electrons to form atoms.
(b) Anode – oxygen; Cathode – copper
With inert carbon electrodes, oxygen gas is produced at the anode.
Copper metal is deposited at the cathode, since copper is less reactive than hydrogen.
(c) \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \)
Aluminium ions each carry a charge of \( 3+ \).
To form neutral aluminium atoms, each ion must gain 3 electrons, giving the balanced half-equation \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \).
Question 9
The half-life of tellurium-109 is \( 4.63 \, \text{s} \).
Calculate the time taken for a sample of pure tellurium-109 to contain 87.5% tin.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P5.2.3/P5.2.4 — Radioactive decay / Half-life (Part (a))
• Topic P1.6.3/P1.6.4 — Energy resources / Power (Part (b))
▶️ Answer/Explanation
(a)(i) Z reduces by 2; neutrons reduce by 2; A reduces by 4
An alpha particle consists of 2 protons and 2 neutrons (like a helium nucleus).
Emitting one therefore reduces the proton number (\( Z \)) by 2, the number of neutrons by 2, and the nucleon number (\( A \)) by 4.
(a)(ii) \( 13.9 \, \text{s} \)
87.5% tin means 12.5% tellurium-109 remains.
\( 100\% \rightarrow 50\% \rightarrow 25\% \rightarrow 12.5\% \) is 3 half-lives.
Time \( = 3 \times 4.63 = 13.9 \, \text{s} \).
(b)(i) Advantage: no CO₂ emissions; Disadvantage: doesn’t work at night
Solar cells generate electricity without producing carbon dioxide or contributing to climate change.
However, they don’t generate electricity at night, and require a large surface area to produce significant power.
(b)(ii) Black absorbs light/radiation well
A black surface is a good absorber of light and thermal radiation, meaning more of the incoming solar energy is absorbed rather than reflected, improving efficiency.
(b)(iii) \( 168 \, \text{W} \)
Area of the panel \( = 1.5 \times 0.5 = 0.75 \, \text{m}^2 \).
Power input \( = 1400 \times 0.75 = 1050 \, \text{W} \).
Power output \( = 1050 \times 0.16 = 168 \, \text{W} \).
Question 10
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B13.2 — Hormones (Parts (a), (b))
• Topic B7.2 — Digestive system (Parts (c), (d))
▶️ Answer/Explanation
(a)(i) 81%
Blood glucose concentration at 0 minutes \( \approx 1050 \, \text{g/m}^3 \); at 10 minutes \( \approx 1900 \, \text{g/m}^3 \).
Percentage increase \( = \dfrac{1900 – 1050}{1050} \times 100 \approx 81\% \).
(a)(ii) Insulin released, glucose converted to glycogen
The rise in blood glucose is detected by the pancreas.
Insulin is released, which stimulates the conversion of glucose to glycogen.
Glycogen is stored in the liver, causing blood glucose concentration to fall back towards normal.
(b) Glucagon
Glucagon, also secreted by the pancreas, raises blood glucose concentration by stimulating the breakdown of glycogen to glucose.
(c) mechanical digestion; assimilation; absorption
Mechanical digestion breaks food into smaller pieces without chemical change.
Assimilation is the uptake and use of digested nutrients by body cells.
Absorption is the movement of digested food molecules through the intestinal wall into the blood.
(d)(i) Villus
The finger-like projection shown lining the small intestine is called a villus.
(d)(ii) Lacteal – absorbs fats/lipids
Part \( X \) is the lacteal, a small lymphatic vessel at the centre of the villus.
Its function is to absorb digested fats and lipids from the small intestine.
Question 11
Magnesium chloride and a gas are made.
[\(A_r\) : Mg, 24]
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C7.1 — Characteristic properties of acids and bases (Part (a))
• Topic C6.2 — Rate of reaction (Part (b))
• Topic C3.3 — The mole and the Avogadro constant (Part (c) — limiting reactant)
▶️ Answer/Explanation
(a)(i) Any value between 0 and 6.9
Dilute hydrochloric acid is acidic, so its pH must be below 7 (any value from 0 up to 6.9 is accepted).
(a)(ii) Proton donor
An acid is defined as a substance that donates protons (\( \text{H}^+ \) ions) in a reaction.
(b)(i) magnesium + hydrochloric acid → magnesium chloride + hydrogen
This word equation matches the given symbol equation \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \).
(b)(ii) Using powdered magnesium instead of magnesium ribbon
Powdered magnesium has a much greater surface area than ribbon, increasing the frequency of collisions between reacting particles and speeding up the reaction.
(b)(iii) Steeper line from the origin, levelling off at the same volume
A more concentrated acid reacts faster, so the graph should start at the origin but rise more steeply than the original curve.
Since the same mass of magnesium is used (the limiting reactant), the curve should level off at the same final volume of hydrogen gas.
(b)(iv) Particles have more kinetic energy, collide more often and more successfully
At a higher temperature, particles have greater kinetic energy and move faster.
This increases the frequency of collisions between particles.
More particles also have energy greater than or equal to the activation energy, so a greater proportion of collisions are successful, increasing the rate of reaction.
(c) Magnesium is the limiting reactant
Moles of \( \text{HCl} = 0.5 \times \dfrac{40}{1000} = 0.02 \, \text{mol} \).
Moles of \( \text{Mg} = \dfrac{0.1}{24} = 0.00417 \, \text{mol} \).
From the equation, 1 mol Mg needs 2 mol HCl, so \( 0.00417 \, \text{mol} \) Mg needs only \( 0.00833 \, \text{mol} \) HCl — since \( 0.02 \, \text{mol} \) HCl is available (more than enough), magnesium is used up first and is the limiting reactant.
Question 12
State the unit for your answer.
State and explain the effect this has on the ammeter reading.
Calculate the minimum frequency of light emitted by the desk lamp.
State what other measurement is required for this calculation.
The current in the primary coil of the transformer is \( 10.5 \, \text{A} \) and the current in the secondary coil is \( 4.2 \, \text{A} \).
The primary coil contains 360 turns and the transformer can be assumed to be 100% efficient.
Calculate the number of turns in the secondary coil.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.2.2 — Electric current (Part (a)(i))
• Topic P4.2.4 — Resistance (Parts (a)(ii), (b))
• Topic P3.1 — General properties of waves (Part (a)(iii))
• Topic P4.5.6 — The transformer (Part (c))
▶️ Answer/Explanation
(a)(i) \( 14.4 \, \text{C} \)
Charge is calculated using \( Q = It \).
\( Q = 0.24 \times 60 = 14.4 \, \text{coulombs (C)} \), since one minute \( = 60 \, \text{s} \).
(a)(ii) Reading increases
Shining bright light on the LDR decreases its resistance.
With lower resistance in the circuit, the current — and therefore the ammeter reading — increases.
(a)(iii) \( 4.0 \times 10^{14} \, \text{Hz} \)
Using \( v = f\lambda \), so \( f = \dfrac{v}{\lambda} \), with \( v = 3 \times 10^8 \, \text{m/s} \) (speed of light).
Minimum frequency occurs at the longest wavelength: \( f = \dfrac{3 \times 10^8}{7.5 \times 10^{-7}} = 4.0 \times 10^{14} \, \text{Hz} \).
(b) Potential difference (p.d.) across the LDR
Resistance is calculated using \( R = \dfrac{V}{I} \), so the potential difference across the LDR must also be measured (using a voltmeter).
(c) 900 turns
For a 100% efficient transformer, \( I_p N_p = I_s N_s \).
\( N_s = \dfrac{I_p N_p}{I_s} = \dfrac{10.5 \times 360}{4.2} = 900 \) turns.