Question 1
Define the term sensitivity.
contains receptor cells ………………………….
controls the amount of light entering the eye ………………………….
refracts light. ………………………….
Describe the changes that occur to the parts labelled A, E and F in Fig. 1.1 when focusing on a distant object.
Name the two parts of the central nervous system.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B1.1 — Characteristics of living organisms (Part (a))
• Topic B13.1 — Coordination and response (Parts (b)(i)–(iii), (c))
▶️ Answer/Explanation
(a) Sensitivity is the ability to detect and respond to changes in the environment
Living organisms sense stimuli in their surroundings.
They then respond appropriately to these changes.
This ability is essential for survival and adaptation.
(b)(i) X drawn on the blind spot (where the optic nerve leaves the eye)
The blind spot is where the optic nerve exits the retina.
No photoreceptor cells are present at this point.
Light falling here therefore cannot be detected.
(b)(ii) H — receptor cells; B — controls light entering; C — refracts light
H is the retina, which contains the light-sensitive receptor cells.
B is the iris, which controls pupil size and hence light entering the eye.
C is the cornea, which does most of the refracting of light entering the eye.
(b)(iii) A becomes thinner; E tightens/stretches; F relaxes
A (the lens) becomes thinner to reduce its refractive power for distant objects.
E (the ciliary muscle) tightens/contracts, pulling the lens flatter.
F (the suspensory ligaments) become taut, increasing tension on the lens.
(c) Brain; spinal cord
The central nervous system (CNS) is made up of the brain and spinal cord.
It processes information received from receptors via the peripheral nervous system.
The eye, as a receptor organ, is part of the peripheral nervous system, not the CNS.
Question 2
- a potassium atom
- a potassium ion.
Sodium reacts with water.
Sodium hydroxide, NaOH, and hydrogen are made.
Construct the balanced symbol equation for this reaction.
The flame colour turns from blue to yellow.
- bromide
- chloride
- iodide
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.3 — Isotopes (Part (a)(i))
• Topic C2.2 — Atomic structure and the Periodic Table (Part (a)(ii))
• Topic C8.2 — Group I properties (Part (b))
• Topic C2.4 — Ions and ionic bonds (Part (c))
• Topic C12.5 — Identification of ions and gases (Parts (d)(i)–(ii))
▶️ Answer/Explanation
(a)(i) Atoms with the same proton number but different nucleon number
Isotopes are atoms of the same element with identical proton numbers.
They have different nucleon (mass) numbers.
This means they differ in the number of neutrons only.
(a)(ii) Atom: 19 protons, 19 electrons, 20 neutrons; Ion (K⁺): 19 protons, 18 electrons, 20 neutrons
Number of neutrons = nucleon number − proton number = \(39 – 19 = 20\).
A potassium atom has 19 protons and 19 electrons (neutral).
A K⁺ ion has lost one electron, giving 18 electrons but still 19 protons and 20 neutrons.
(b) \(2Na + 2H_2O \rightarrow 2NaOH + H_2\)
Sodium reacts with water to give sodium hydroxide and hydrogen gas.
Balancing sodium, oxygen, and hydrogen atoms gives the coefficients above.
Two moles of sodium react with two moles of water to give two moles of NaOH and one mole of H₂.
(c) Na⁺ has 10 electrons (2,8); F⁻ has 10 electrons (2,8), with charges shown
Sodium (2,8,1) loses its outer electron to form Na⁺ (2,8).
Fluorine (2,7) gains that electron to form F⁻ (2,8).
Both ions must be shown in square brackets with the correct charge.
(d)(i) Sodium ion
A yellow flame colour is characteristic of sodium compounds.
This flame test is used to identify sodium ions (Na⁺) in a sample.
(d)(ii) Chloride
A white precipitate with acidified silver nitrate indicates a halide ion.
Silver chloride (AgCl) forms a white precipitate.
Silver bromide is cream and silver iodide is yellow, so the ion must be chloride.
Question 3
During one journey, the train travels 180 km in 1 hour.
Calculate the speed of sound in air.
Calculate the volume of each length of rail.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.2 — Motion (Parts (a)(i)–(ii))
• Topic P3.4 — Sound (Parts (b)(i)–(ii))
• Topic P1.4 — Density (Part (c)(i))
• Topic P2.2.1 — Thermal expansion of solids, liquids and gases (Part (c)(ii))
▶️ Answer/Explanation
(a)(i) 50 m/s
Distance travelled = \(180 \, \text{km} = 180\,000 \, \text{m}\).
Time taken = \(1 \, \text{hour} = 3600 \, \text{s}\).
Speed \(= \dfrac{180\,000}{3600} = 50 \, \text{m/s}\).
(a)(ii) \(8.5 \times 10^8\) J
Kinetic energy \( = \dfrac{1}{2}mv^2\).
\( = \dfrac{1}{2} \times 680\,000 \times 50^2\).
\( = 850\,000\,000 \, \text{J}\).
(b)(i) 330 m/s
Wave speed \(v = f\lambda\).
\(v = 250 \times 1.32\).
\(v = 330 \, \text{m/s}\).
(b)(ii) Sound travels as a longitudinal wave via vibrating air particles
Air particles vibrate/oscillate parallel to the direction of travel.
This produces regions of compression, where particles are close together.
It also produces regions of rarefaction, where particles are spread apart.
(c)(i) 0.04 m³
Volume \(= \dfrac{\text{mass}}{\text{density}}\).
Volume \(= \dfrac{324}{8100}\).
Volume \(= 0.04 \, \text{m}^3\).
(c)(ii) Gaps allow for thermal expansion and prevent buckling
When temperature increases, the steel rails expand.
Without gaps, the expanding rails would have nowhere to go.
The small gaps prevent the tracks from buckling or bending under this expansion.
Question 4
- sets up three test-tubes, as shown in Table 4.1
- uses an indicator which turns pink when fatty acid is present
- records how long it takes the indicator to turn pink in each test-tube
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B5 — Enzymes (Parts (a)(i)–(iii))
• Topic B7.2 — Digestive system (Part (b))
• Topic B7.3 — Digestion (Parts (c)(i)–(ii))
▶️ Answer/Explanation
(a)(i) The boiled lipase is denatured and cannot digest the fat
Boiling denatures the lipase enzyme.
The shape of its active site is permanently changed.
It is no longer complementary to the substrate, so fat is not broken down and the indicator never turns pink.
(a)(ii) 182 seconds
Time for test-tube 1 = 378 s.
Time for test-tube 3 = 196 s.
Difference \(= 378 – 196 = 182 \, \text{s}\).
(a)(iii) Bile in test-tube 3 emulsifies the fat, speeding up digestion
Bile emulsifies the fat into smaller droplets.
This increases the surface area of fat exposed to lipase.
Digestion is therefore faster in test-tube 3 than in test-tube 1, which has no bile.
(b) Liver
Bile is produced by the liver.
It is stored in the gallbladder before being released into the small intestine.
(c)(i) Villi provide a large surface area for absorption
Villi are finger-like projections lining the small intestine.
They greatly increase the surface area available for absorption of digested nutrients.
(c)(ii) Small intestine
Villi are found lining the small intestine.
This is where most absorption of digested food takes place.
Question 5
Write down the molecular formula of propene.
State which group of hydrocarbons propene belongs to.
State what is meant by the word unsaturated.
State what you would see when aqueous bromine is added to propene and propane.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.4 — Alkanes (Parts (a)–(b))
• Topic C11.5 — Alkenes (Parts (c), (e))
• Topic C6.3 — Redox (Part (d))
• Topic C11.7 — Polymers (Parts (f)(i)–(ii))
▶️ Answer/Explanation
(a) \(C_3H_6\)
Propene has three carbon atoms and one carbon-carbon double bond.
Counting the hydrogens attached gives the molecular formula \(C_3H_6\).
(b) Alkenes
Propene contains a carbon-carbon double bond.
Hydrocarbons with at least one C=C double bond belong to the alkene homologous series.
(c) Contains a carbon-to-carbon double bond
An unsaturated hydrocarbon has at least one C=C double bond.
This allows it to undergo addition reactions, unlike saturated hydrocarbons.
(d) Addition (reaction) / reduction / hydrogenation
Hydrogen is added across the carbon-carbon double bond of propene.
This converts the unsaturated alkene into the saturated alkane, propane.
(e) Propene: orange to colourless; Propane: stays orange (no change)
Propene reacts with aqueous bromine because of its C=C double bond, decolourising it.
Propane has no double bond, so it does not react and the bromine water stays orange.
(f)(i) Repeating unit \(-CH_2-CH(CH_3)-\) with the double bond opened
In addition polymerisation, the C=C double bond in propene opens.
Monomer units join together with single C–C bonds to form a long repeating chain.
(f)(ii) Addition: one large molecule, no by-product; Condensation: two products, often with water lost
Addition polymerisation joins monomers with double bonds, forming only one large polymer molecule.
Condensation polymerisation joins monomers with the loss of a small molecule, usually water.
Nylon is therefore made with water as a by-product of each linkage formed.
Question 6
State the unit for your answer.
The gravitational field strength \(g\) is 10 N/kg.
Describe how the child becomes positively charged.
A white plastic slide will reflect infrared radiation ……… than a black plastic slide.
On a sunny day, a white plastic slide will heat up ……… than a black plastic slide.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.6.2 — Work (Part (a))
• Topic P1.2 — Motion (Part (b))
• Topic P4.2.1 — Electrical charge (Part (c))
• Topic P2.3.3 — Radiation (Part (d))
▶️ Answer/Explanation
(a) 270 J
Work done \(= mgh\).
\(= 15 \times 10 \times 1.8\).
\(= 270 \, \text{J}\) (unit: joules, J).
(b) The child accelerates, with the acceleration decreasing over time
The speed increases throughout the slide, so the child accelerates.
The gradient of the graph is steeper at first, showing higher acceleration.
The gradient decreases later on, showing the acceleration reduces as friction increasingly opposes motion.
(c) Friction transfers electrons from the child to the slide
As the child rubs against the plastic slide, friction occurs.
Electrons are transferred from the child to the slide.
Losing negative electrons leaves the child with an overall positive charge.
(d) Less; more; less
A white surface absorbs less infrared radiation than a black surface.
A white surface reflects more infrared radiation than a black surface.
Because it absorbs less radiation, a white slide heats up less than a black slide on a sunny day.
Question 7
Put ticks (✓) in Table 7.1 to show all the terms used to describe each organism.
One has been done for you.
Some organisms are better adapted to the ……… than others.
These organisms survive and breed, passing on their ……… .
This process takes many ……… .
List two features of gas exchange surfaces.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B18.2 — Food chains and food webs (Part (a))
• Topic B6.1 — Photosynthesis (Part (b))
• Topic B18.1 — Energy flow (Part (c))
• Topic B17.2 — Selection (Part (d)(i))
• Topic B11 — Gas exchange in humans (Part (d)(ii))
▶️ Answer/Explanation
(a) Arctic cod: carnivore, quaternary consumer; Krill: herbivore; Orca: carnivore, quaternary consumer; Phytoplankton: producer
Phytoplankton is a producer, forming the base of the food web.
Krill feeds directly on phytoplankton, making it a herbivore.
Arctic cod and orca both feed on other consumers, so they are carnivores and, following the food chain to the top, quaternary consumers.
(b) Photosynthesis, using light energy, carbon dioxide and water
Producers absorb light energy using chlorophyll.
This energy converts carbon dioxide and water into glucose during photosynthesis.
Glucose can then be built into other carbohydrates such as starch and cellulose.
(c) Heat lost through respiration; energy lost in excretion/egestion; energy used for movement/metabolism
Some energy is lost as heat during respiration.
Some energy is lost in waste products, such as faeces and urea.
Some energy is used for movement and other metabolic processes, and not all parts of an organism are eaten or digested by the next trophic level.
(d)(i) Selection; environment; alleles; generations
Adaptation results from the process of natural selection.
Organisms best suited to their environment are more likely to survive and reproduce.
These organisms pass on their advantageous alleles to offspring, and this process occurs gradually over many generations.
(d)(ii) Large surface area; thin walls (short diffusion distance)
Gas exchange surfaces have a large surface area to maximise the rate of diffusion.
They also have thin walls to keep the diffusion distance short, along with a good blood supply and good ventilation.
Question 8
Iron and zinc sulfate are made.
Construct the balanced symbol equation for the reaction.
Include the state symbols.
Construct the balanced ionic half-equation for this reaction.
Use the symbol e⁻ for an electron.
Magnesium chloride, \(MgCl_2\), and hydrogen gas are made.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C9.4 — Reactivity series (Parts (a)–(c))
• Topic C3.3 — The mole and the Avogadro constant (Part (d)(i))
• Topic C12.5 — Identification of ions and gases (Part (d)(ii))
▶️ Answer/Explanation
(a) Fe/CuSO₄: pink/brown metal; Mg/CuSO₄: silver-coloured metal, colourless solution
Iron is more reactive than copper, so it displaces copper metal (appearing pink/brown) from CuSO₄.
Magnesium is more reactive than copper, so it fully displaces the copper, leaving the metal silver-coloured.
The solution becomes colourless as the blue Cu²⁺ ions are removed and replaced by colourless Mg²⁺ ions.
(b) \(Zn(s) + FeSO_4(aq) \rightarrow Fe(s) + ZnSO_4(aq)\)
Zinc is more reactive than iron, so it displaces iron from iron sulfate.
Zinc forms zinc sulfate in solution, while solid iron is deposited.
The equation is already balanced with one mole of each species.
(c) \(Mg \rightarrow Mg^{2+} + 2e^-\)
Magnesium atoms lose two electrons to become Mg²⁺ ions.
This is an oxidation half-equation, balanced for charge and mass.
(d)(i) 1.9 g
Moles of Mg \(= \dfrac{0.48}{24} = 0.02 \, \text{mol}\).
From the equation, 1 mol Mg produces 1 mol MgCl₂, so moles of MgCl₂ = 0.02 mol.
\(M_r\) of MgCl₂ \(= 24 + (2 \times 35.5) = 95\), so mass \(= 0.02 \times 95 = 1.9 \, \text{g}\).
(d)(ii) Test: lighted splint; Observation: squeaky pop
A lighted splint is held at the mouth of the test tube containing the gas.
Hydrogen gas ignites rapidly, producing a characteristic squeaky pop sound.
Question 9
- measures the unloaded lengths of each spring
- hangs identical masses from each spring, measuring the extended lengths.
Calculate the weight of the mass hanging from spring A.
State and explain which of the three springs has the largest spring constant.
The student sets up a circuit to determine the electrical resistance of one of the springs.
Fig. 9.2 shows the circuit used.
The ammeter reads 0.75 A and the voltmeter reads 7.5 V.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.5.1 — Effects of forces (Parts (a)(i)–(ii))
• Topic P4.5.3 — Magnetic effect of current (Parts (b)(i)–(ii))
▶️ Answer/Explanation
(a)(i) 0.6 N
Extension \(= 3.4 – 2.2 = 1.2 \, \text{cm}\).
Force \(= k \times x = 0.50 \times 1.2\).
Force (weight) \(= 0.6 \, \text{N}\).
(a)(ii) Spring B; it has the smallest extension for the same load
All three springs carry the same mass, so the same force acts on each.
Spring B extends the least (from 4.0 cm to 4.3 cm, an extension of 0.3 cm).
Since extension is inversely proportional to spring constant, the smallest extension means Spring B has the largest spring constant.
(b)(i) 2 Ω
Voltage across the spring \(= 9 – 7.5 = 1.5 \, \text{V}\) (supply voltage minus the reading across the known 10 Ω resistor).
Resistance \(= \dfrac{V}{I} = \dfrac{1.5}{0.75}\).
Resistance \(= 2 \, \Omega\).
(b)(ii) Concentric loops forming a bar-magnet-like field, direction found using the right-hand grip rule
The magnetic field inside the solenoid runs parallel to its axis, like a bar magnet.
Field lines emerge from one end (forming a north pole) and loop back into the other end (south pole).
The direction is determined by the direction of current flow shown in the diagram, using the right-hand grip rule.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B3.2 — Osmosis (Parts (a)(i)–(ii))
• Topic B8.4 — Translocation (Part (b))
▶️ Answer/Explanation
(a)(i) Plasmolysis
Plasmolysis is the shrinking of the cell contents away from the cell wall.
It occurs when plant cells lose water in a hypertonic (concentrated) external solution.
(a)(ii) Water leaves the cells by osmosis down a water potential gradient
The glucose solution outside has a lower (more negative) water potential than the cell contents.
Water moves out of the cell, across the partially permeable membrane, from high to low water potential.
This is osmosis, and the loss of water causes the cell membrane and cytoplasm to pull away from the cell wall.
(b) Sucrose; amino acids
Translocation is the movement of dissolved substances through phloem tissue.
Sucrose and amino acids are transported from sources, such as leaves, to sinks, such as roots and growing parts.
Question 11
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.6 — Giant covalent structures (Parts (a)–(d))
• Topic C2.5 — Simple molecules and covalent bonds (Part (e))
▶️ Answer/Explanation
(a) Hardness
Diamond is an extremely hard substance.
This property makes it suitable for cutting and drilling tools.
(b) Lubricant / pencils / electrodes
Graphite’s soft, slippery layers make it useful as a lubricant.
Its ability to leave marks makes it useful in pencils, and its conductivity makes it useful as electrodes.
(c) Delocalised electrons move through the structure
Each carbon atom in graphite is bonded to only three others, leaving one delocalised electron per atom.
These delocalised electrons are free to move between the layers.
The movement of these charged particles constitutes an electric current.
(d) Covalent
Diamond has a giant covalent structure.
Each carbon atom forms four strong covalent bonds to neighbouring carbon atoms.
(e) Two C=O double bonds, each sharing two pairs of electrons
Carbon dioxide has a linear structure, \(O=C=O\).
Carbon shares two pairs of electrons with each oxygen atom, forming two double covalent bonds.
Each oxygen also has two non-bonding lone pairs shown in the diagram.
Question 12
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P3.3 — Electromagnetic spectrum (Part (a))
• Topic P3.2.2 — Refraction of light (Part (b))
• Topic P5.2.2 — The three types of nuclear emission (Parts (c)(i)–(iii))
▶️ Answer/Explanation
(a) P: microwaves; Q: infrared; R: ultraviolet
The electromagnetic spectrum is ordered by increasing frequency from radio waves to gamma rays.
Between radio waves and visible light lie microwaves (P) and infrared (Q).
Between visible light and X-rays lies ultraviolet (R).
(b) 1.93
Refractive index \(n = \dfrac{\sin i}{\sin r}\).
\(n = \dfrac{\sin 30°}{\sin 15°}\).
\(n \approx 1.93\).
(c)(i) α-particle: helium nucleus, high; β-particle: electron, medium; γ-ray: electromagnetic radiation, low
α-particles are helium nuclei and have the highest ionising effect due to their large mass and charge.
β-particles are fast-moving electrons and have a medium ionising effect.
γ-rays are electromagnetic radiation with no mass or charge, giving the lowest ionising effect.
(c)(ii) \(^{210}_{82}Pb \rightarrow \, ^{210}_{83}Bi + \, ^{0}_{-1}\beta\)
In β-decay, a neutron converts into a proton and an electron.
The proton number increases by 1 (82 → 83), while the nucleon number stays the same (210).
The emitted β-particle is represented as \(^{0}_{-1}\beta\).
(c)(iii) 22 years
Half-life is the time for activity to fall to half its original value.
Starting at about 380 counts/min, half of this is 190 counts/min.
Reading across from the graph, this activity occurs at approximately 22 years.