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Question 1

(a) Fig. 1.1 is a diagram of a wind-pollinated flower.
(i) State which letter in Fig. 1.1 identifies the part where: fertilisation occurs; pollen is produced.
(ii) Describe two visible pieces of evidence in Fig. 1.1 that show the flower is adapted for wind-pollination.
(b) Fig. 1.2 is a photomicrograph of pollen from an insect-pollinated flower.
Describe two ways the appearance of pollen from a wind-pollinated flower is different from the pollen from an insect-pollinated flower.
(c) Some plants can reproduce asexually and sexually.
(i) State two advantages of sexual reproduction compared to asexual reproduction in plants.
(ii) Suggest a situation where asexual reproduction is more useful to a plant in the wild than sexual reproduction.
(d) Reproduction is one of the characteristics of living organisms. State two other characteristics of living organisms.

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

• Topic B15.3 — Sexual reproduction in plants (Parts (a), (b))
• Topic B15.1/B15.2 — Asexual and sexual reproduction (Part (c))
• Topic B1.1 — Characteristics of living organisms (Part (d))

▶️ Answer/Explanation

(a)(i) Fertilisation: C; Pollen produced: B

C is the carpel/stigma region leading to the ovule where fertilisation occurs.
B is the anther where pollen grains are produced.

(a)(ii) Feathery, exposed stigma (E); anthers hanging outside the flower (B/C).

A feathery stigma increases surface area to catch airborne pollen.
Anthers hanging loosely outside the flower allow pollen to be released easily into the wind.

(b) Wind-pollinated pollen is smoother (not spiky); wind-pollinated pollen is smaller.

Insect-pollinated pollen (Fig. 1.2) is spiky/sticky to attach to insect bodies.
Wind-pollinated pollen lacks these spikes and is lighter/smaller so it can be carried by air currents.

(c)(i) Sexual reproduction increases genetic diversity; allows adaptation to a changing environment.

Genetic variation means a disease is unlikely to wipe out all the plants.
Variation also allows natural selection/evolution to occur, helping the species adapt.

(c)(ii) Asexual reproduction is useful when the plant is isolated from other plants or pollinators are scarce.

\(\text{No mate/pollinator needed for asexual reproduction}\)
This allows rapid colonisation of a favourable habitat without depending on pollination.

(d) Any two of: movement, respiration, sensitivity, growth, excretion, nutrition.

Movement is a change of position caused by the organism.
Respiration releases energy from nutrients in cells.
Excretion removes waste products of metabolism.

Question 2

A student heats three substances X, Y and Z in a water-bath. Table 2.1 shows the state of the three substances before heating, during heating and after cooling.
(a) Draw one line from substance X and one line from substance Y to show the arrangement of the particles before heating.
(b) Describe the difference in the movement of the particles in a solid and in a liquid.
(c) Explain how we know that the change to substance X is a physical change and not a chemical change.
(d) Substance Z is the ionic compound sodium chloride, NaCl. Draw a dot-and-cross diagram to show the ionic bonding in sodium chloride.
(e) Fig. 2.1 shows the electrolysis of concentrated aqueous sodium chloride. Complete the three labels on Fig. 2.1 to show the products made.

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

• Topic C1.1 — States of matter, particle theory (Parts (a), (b))
• Topic C6.1 — Physical and chemical changes (Part (c))
• Topic C2.4 — Ions and ionic bonds (Part (d))
• Topic C4.1 — Electrolysis (Part (e))

▶️ Answer/Explanation

(a) X (solid before heating) connects to the regular, closely-packed particle diagram; Y (liquid before heating) connects to the closely-packed, disordered particle diagram.

X is a solid before heating, so its particles are arranged in a regular pattern.
Y is already a liquid before heating, so its particles are close together but randomly arranged.

(b) Solid: particles vibrate about fixed positions. Liquid: particles move around/slide over each other.

In a solid, particles only vibrate in place and cannot change position.
In a liquid, particles have enough energy to move past one another while staying close together.

(c) It is reversible and no new substance is formed.

Melting and solidifying are reversible processes, unlike a chemical change.
Since substance X returns to a solid identical to the original on cooling, no new substance has been formed.

(d) \(\text{Na}^{+}\) has no outer electrons shown (transferred away); \(\text{Cl}^{-}\) shows 8 outer electrons (7 of its own plus 1 from Na, shown as a cross).

Sodium loses one electron to form \(\text{Na}^{+}\), leaving an empty outer shell.
Chlorine gains that electron to form \(\text{Cl}^{-}\), giving it a full outer shell of 8 electrons.
Square brackets with charges (+ and −) are drawn around each ion.

(e) Anode (top left): chlorine; Cathode (top right): hydrogen; Bottom outlet: sodium hydroxide.

Chlorine gas is released at the positive electrode (anode).
Hydrogen gas is released at the negative electrode (cathode).
Sodium hydroxide solution remains/forms in the electrolyte and drains from the bottom.

Question 3

Fig. 3.1 shows a sea turtle.
(a)(i) On Fig. 3.1, draw an arrow to show the direction of the weight force acting on the sea turtle.
Label your arrow with the letter W.
(ii) Complete the sentence to describe weight.
Weight is a force caused by the effect of a ………. field on a ………. .
(b) The sea turtle travels a distance of 1200 km in 20 days.
Calculate the average speed of the sea turtle.
Give your answer in km/h.
(c) A team of scientists fits a tracker unit to the sea turtle to monitor its location.
The tracker unit sends a signal using radio waves each time the sea turtle moves to the surface of the water.
(i) Radio waves are part of the electromagnetic spectrum.
Complete the sentences to compare radio waves to visible light.
Radio waves have a ………. frequency and a ………. wavelength than visible light.
Radio waves and visible light both travel at ………. m/s in a vacuum.
(ii) The radio waves emitted by the tracker unit have a frequency of \(1.5 \times 10^{9}\,\text{Hz}\).
Calculate the wavelength of the radio waves.
(iii) The tracker unit uses a battery with an electromotive force (e.m.f.) of 11 V that provides a power output of 22 mW.
The battery can transfer a total charge of 24 000 C before it needs replacing.
Calculate the time for which the battery operates before it needs replacing.

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

• Topic P1.3 — Mass and weight (Part (a))
• Topic P1.2 — Motion, average speed (Part (b))
• Topic P3.3 — Electromagnetic spectrum (Part (c)(i), (ii))
• Topic P4.2.2/P4.2.3 — Electric current, charge and e.m.f. (Part (c)(iii))

▶️ Answer/Explanation

(a)(i) A downward arrow labelled W, drawn from the turtle’s centre.

Weight always acts vertically downwards, toward the centre of the Earth.

(a)(ii) Gravitational field on a mass.

Weight is the force produced when a gravitational field acts on an object’s mass: \(W = mg\).

(b) 2.5 km/h

Convert time: \(20 \text{ days} = 480\) hours.
Average speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{1200}{480}\).
Average speed \(= 2.5\,\text{km/h}\).

(c)(i) Lower frequency and longer wavelength than visible light; both travel at \(3 \times 10^{8}\,\text{m/s}\).

In the electromagnetic spectrum, radio waves sit below visible light, so they have lower frequency and longer wavelength.
All electromagnetic waves travel at the same speed in a vacuum, \(3 \times 10^{8}\,\text{m/s}\).

(c)(ii) 0.20 m

Use \(\lambda = \dfrac{v}{f} = \dfrac{3 \times 10^{8}}{1.5 \times 10^{9}}\).
\(\lambda = 0.20\,\text{m}\).

(c)(iii) \(1.2 \times 10^{7}\,\text{s}\)

Current \(I = \dfrac{P}{V} = \dfrac{0.022}{11} = 0.002\,\text{A}\).
Time \(t = \dfrac{Q}{I} = \dfrac{24000}{0.002}\).
\(t = 1.2 \times 10^{7}\,\text{s}\).

Question 4

(a) Blood glucose concentration is controlled so that it remains within set limits.
State the name given to this type of control.
(b) Fig. 4.1 shows the blood glucose concentration of a person after they have eaten a meal.
Complete the sentences to describe and explain the changes seen in Fig. 4.1.
Carbohydrates such as starch are broken down by the enzyme ………. to form simpler sugars. These simpler sugars are absorbed into the blood. After 30 minutes, the blood glucose concentration reaches a maximum of ………. mmol/dm³. The increase in blood glucose concentration is detected and the hormone insulin is released from the ………. . Insulin causes glucose to be converted to ………. . This is then stored in the ………. reducing the blood glucose concentration to its previous level.
(c) State the names of two hormones that increase blood glucose concentration.
(d) State the name of the component of blood that transports hormones.
(e) Table 4.1 compares nervous and hormonal control.
Complete Table 4.1.

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

• Topic B13.3 — Homeostasis, control of blood glucose (Parts (a), (b))
• Topic B13.2 — Hormones (Part (c))
• Topic B9.4 — Blood, function of plasma (Part (d))
• Topic B13.1/B13.2 — Nervous vs hormonal control (Part (e))

▶️ Answer/Explanation

(a) Homeostasis

Homeostasis is the maintenance of a constant internal environment, keeping factors like blood glucose within set limits.

(b) amylase; 8.0 mmol/dm³; pancreas; glycogen; liver/muscles.

Amylase breaks starch down into simpler sugars in digestion.
Reading the graph at 30 minutes gives a peak of \(8.0\,\text{mmol/dm}^3\).
Insulin, released from the pancreas, converts excess glucose to glycogen, which is stored in the liver/muscles.

(c) Adrenaline; glucagon.

Adrenaline raises blood glucose during ‘fight or flight’ responses.
Glucagon, also from the pancreas, converts stored glycogen back into glucose.

(d) Plasma

Plasma is the liquid component of blood that carries hormones, ions, nutrients and carbon dioxide around the body.

(e) 

Nervous transmission uses electrical impulses along neurones, which act very quickly but briefly.
Hormonal transmission relies on chemicals carried in the blood, which act more slowly but the effect lasts longer.

Question 5

Some students investigate the reaction between marble chips and dilute hydrochloric acid.
They react marble chips of three different sizes, A, B and C, with excess dilute hydrochloric acid, using the same mass of marble chips, the same concentration of acid and the same temperature for each experiment.
They measure the volume of carbon dioxide gas every 30 seconds until the reaction finishes.
Fig. 5.1 shows a graph of their results.
(a)(i) State which marble chips, A, B or C, are the smallest.
(ii) Look at the line for marble chips B.
State when the rate of reaction is the greatest.
Choose your answer from the list:
0–30 s
30–60 s
60–90 s
90–120 s
(b) The students did the experiments at 20°C.
State how the rate of reaction will change if they do the experiments again at 40°C.
Explain your answer using ideas about collisions between particles.
(c) Calculate the volume occupied by 1.1 g of carbon dioxide gas at room temperature and pressure.
The volume of one mole of any gas is 24 dm³ at r.t.p.
[\(A_r\): C, 12; O, 16]
(d) Carbon dioxide is a greenhouse gas.
State two problems caused by increased concentrations of greenhouse gases.

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

• Topic C6.2 — Rate of reaction, collision theory (Parts (a), (b))
• Topic C3.3 — The mole and molar gas volume (Part (c))
• Topic C10.2 — Air quality and climate, greenhouse gases (Part (d))

▶️ Answer/Explanation

(a)(i) A

Smaller chips have a larger surface area to volume ratio, so reaction A (which finishes fastest, with the steepest initial gradient) has the smallest chips.

(a)(ii) 0–30 s

The rate of reaction is greatest where the graph is steepest.
For curve B, the steepest section is the very first interval, 0–30 s.

(b) The rate will be faster.

At a higher temperature, particles have more kinetic energy and move faster.
This increases the frequency of collisions and the proportion of collisions with energy greater than the activation energy.
There are therefore more successful collisions per second, increasing the rate.

(c) 0.60 dm³

\(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\).
Moles of \(\text{CO}_2 = \dfrac{1.1}{44} = 0.025\,\text{mol}\).
Volume \(= 0.025 \times 24 = 0.60\,\text{dm}^3\).

(d) Any two of: enhanced greenhouse effect; climate change; rising sea levels; flooding; extreme weather/melting ice caps.

Increased greenhouse gases trap more thermal energy, enhancing the greenhouse effect.
This leads to global warming, which causes climate change, melting ice caps and rising sea levels.

Question 6

Fig. 6.1 shows an electric pressure-washer being used to wash a car.
(a) The pressure-washer pumps water at a high pressure through a small nozzle.
The cross-sectional area of the nozzle is \(5.0 \times 10^{-6}\,\text{m}^2\).
The water leaves the nozzle with a pressure of \(9.0 \times 10^{6}\,\text{Pa}\).
Calculate the force exerted by the water as it leaves the nozzle.
(b) The pressure-washer uses a d.c. motor to pump the water out of the nozzle. Fig. 6.2 shows a diagram of a simple d.c. motor.
(i) The arrows on Fig. 6.2 show the direction of the current.
Draw an arrow to show the direction of the force acting on the coil at the point labelled X.
(ii) Describe the function of the split-ring commutator in a simple d.c. motor.
(c) After the car has been washed, droplets of cold water remain on the roof of the car.
After a few minutes, the droplets of water have disappeared.
(i) State the name of the process which causes the droplets of water to disappear.
(ii) Describe the process which causes the droplets of water to disappear in terms of molecules.

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

• Topic P1.7 — Pressure (Part (a))
• Topic P4.5.5 — The d.c. motor (Part (b))
• Topic P2.2.2 — Melting, boiling and evaporation (Part (c))

▶️ Answer/Explanation

(a) 45 N

Use \(F = P \times A = (9.0 \times 10^{6}) \times (5.0 \times 10^{-6})\).
\(F = 45\,\text{N}\).

(b)(i) Arrow drawn upwards at X.

Using the left-hand (motor) rule with current flowing into the page at X between the N pole and S pole, the force on the coil at X acts upwards.

(b)(ii) It reverses the current direction in the coil every half-turn.

The split-ring commutator swaps the contacts every half-rotation, reversing the current in the coil.
This keeps the force on each side of the coil acting in a consistent rotational direction, so the coil continues turning the same way.

(c)(i) Evaporation

The change of liquid water to water vapour at a temperature below boiling point is called evaporation.

(c)(ii) The most energetic molecules escape from the surface of the liquid.

Within the liquid, molecules have a range of kinetic energies.
The most energetic molecules near the surface have enough energy to overcome intermolecular forces and escape as vapour, leaving the droplet to gradually disappear.

Question 7

(a) A student investigates antibiotic resistance in one strain of bacteria.
They use five different antibiotics on paper discs.
The antibiotic discs are placed in a Petri dish with the bacteria and left for three days.
Fig. 7.1 shows the results.
Identify the antibiotic in Fig. 7.1 that is most effective against this strain of bacteria.
Give one reason for your answer.
(b) The differences in antibiotic resistance in bacteria are caused by random mutation.
(i) State the structure in a cell where mutation occurs.
(ii) State the type of radiation that increases the rate of mutation.
(c) Explain why the development of antibiotic resistance in bacteria is an example of evolution.

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

• Topic B14.1 — Drugs, antibiotic resistance (Part (a))
• Topic B17.1 — Variation, mutation (Part (b)(i)); Topic P5.2.5 — Effects of ionising radiation (Part (b)(ii))
• Topic B17.2 — Selection, natural selection (Part (c))

▶️ Answer/Explanation

(a) Antibiotic B

B has the largest clear area of no bacterial growth around the disc.
A larger clear zone means that antibiotic diffused further while still killing bacteria, showing it is the most effective.

(b)(i) Gene/chromosome/nucleus/DNA.

Mutations are random changes that occur in the DNA/genetic material found in the nucleus of a cell.

(b)(ii) Ionising radiation

Ionising radiation (e.g. gamma rays, X-rays, ultraviolet) can damage DNA and increase the rate of mutation.

(c) Random mutation produces resistant bacteria, which survive and reproduce, passing on the resistance.

Genetic variation within the bacterial population means some individuals are randomly resistant to an antibiotic.
Resistant bacteria survive exposure to the antibiotic and have a greater chance of reproducing.
Their alleles for resistance are passed to the next generation, so the adaptive feature becomes more common over time — the basis of evolution by natural selection.

Question 8

Fig. 8.1 shows the structures of three carbon compounds: ethene, ethanoic acid and ethanol.
(a) Ethene is an unsaturated hydrocarbon.
Explain how the structure of ethene shows that ethene is an unsaturated hydrocarbon.
(b) Ethene, \(\text{C}_2\text{H}_4\), reacts with hydrogen to make an alkane.
Write the balanced symbol equation for this reaction.
(c) Complete the dot-and-cross diagram in Fig. 8.2 to show the bonding in ethene.
Only show the outer-shell electrons.
(d) Ethanol is made by fermentation.
State one condition for making ethanol by fermentation.
(e) Ethanol can also be made from ethene in an addition reaction.
Complete the symbol equation for this reaction:
(f) A scientist makes a solution of ethanol.
250 cm³ of the solution contains 5.75 g of ethanol.
Calculate the concentration of the ethanol solution in mol/dm³.
[\(A_r\): C, 12; H, 1; O, 16]

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

• Topic C11.1/C11.5 — Formulas/terminology, alkenes (Parts (a), (b), (c))
• Topic C11.6 — Alcohols, ethanol production (Parts (d), (e))
• Topic C3.3 — The mole, concentration calculations (Part (f))

▶️ Answer/Explanation

(a) Unsaturated: contains a carbon-to-carbon double bond. Hydrocarbon: contains only hydrogen and carbon atoms.

The structural diagram shows \(\text{C}=\text{C}\), a double bond between the carbon atoms, which makes it unsaturated.
Ethene contains only carbon and hydrogen atoms, with no other elements, making it a hydrocarbon.

(b) \(\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6\)

Ethene reacts with hydrogen in an addition reaction (with a nickel catalyst) to form ethane.
The equation is balanced, with equal numbers of each atom on both sides.

(c) Each carbon shares 2 electrons (one dot, one cross) with each of two hydrogens, and the carbon atoms share 2 pairs of electrons between them (a double bond).

Each C–H bond is shown as one shared pair of electrons (a dot and a cross) between carbon and hydrogen.
The C=C double bond is shown as two shared pairs of electrons between the two carbon atoms.

(d) Absence of air/oxygen (anaerobic conditions), or warm temperature, or presence of yeast/sugar/water.

Fermentation requires anaerobic conditions (no oxygen) using yeast acting on sugar/glucose solution at a warm temperature.

(e) \(\text{H}_2\text{O}\)

Ethene reacts with steam (\(\text{H}_2\text{O}\)) in the presence of an acid catalyst in an addition reaction to form ethanol.

(f) 0.5 mol/dm³

\(M_r(\text{C}_2\text{H}_5\text{OH}) = 46\); moles \(= \dfrac{5.75}{46} = 0.125\,\text{mol}\).
Volume \(= 250\,\text{cm}^3 = 0.250\,\text{dm}^3\).
Concentration \(= \dfrac{0.125}{0.250} = 0.5\,\text{mol/dm}^3\).

Question 9

The element strontium has many naturally occurring isotopes, some of which are unstable.
(a) Table 9.1 shows the half-lives of four unstable isotopes of strontium.
(i) Fig. 9.1 shows a decay curve for one of the isotopes given in Table 9.1.
Determine which isotope of strontium from Table 9.1 would give the data shown in Fig. 9.1.
(ii) A scientist purchases a sample of a strontium isotope to use as a radioactive source in a series of experiments.
The scientist estimates that the experiments will take three months to complete.
Suggest which of the isotopes in Table 9.1 would be best for the scientist to purchase.
Explain your suggestion.
(b) Place ticks (✓) in Table 9.2 to show the nature of a beta particle.
(c) The density of strontium is 2.6 g/cm³.
A sample of strontium has a mass of 7.8 g.
Calculate the volume of the sample of strontium.

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

• Topic P5.2.4 — Half-life (Part (a))
• Topic P5.2.2 — Nature of beta particles (Part (b))
• Topic P1.4 — Density (Part (c))

▶️ Answer/Explanation

(a)(i) Strontium-82

The graph shows activity falling from 700 to 350 counts/min (one half-life) in about 25 days.
This half-life of approximately 25.4 days matches strontium-82 in Table 9.1.

(a)(ii) Strontium-90

Strontium-90 has the longest half-life (28.9 years), so its activity barely decreases over three months.
The source will not need replacing during the experiments and will give a roughly constant count rate.

(b) Has a negative charge; is affected by electric fields; is affected by magnetic fields.

A beta particle is a fast-moving electron, so it carries a negative charge.
Because it is charged, it experiences a force in, and is deflected by, both electric and magnetic fields.

(c) 3.0 cm³

Use \(V = \dfrac{m}{\rho} = \dfrac{7.8}{2.6}\).
\(V = 3.0\,\text{cm}^3\).

Question 10

(a) Red blood cells are specialised to transport oxygen.
Describe two ways that red blood cells are adapted for their function.
(b) A student investigates the effect of different concentrations of salt solution on red blood cells.
The student immerses the red blood cells in different concentrations of salt solution and observes the cells after immersion.
Table 10.1 shows the results.
(i) Identify the salt solution with the same water potential as red blood cells.
(ii) Explain the observation seen at 10.0 g/dm³ in Table 10.1.
(c) The investigation is repeated with plant cells.
(i) Plant cells do not burst when immersed in 2.0 g/dm³ salt solution. Explain why.
(ii) State two uses of water in plant cells.
(iii) State the name of the type of plant cell specialised for absorption of water.

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

• Topic B9.4 — Blood, red blood cell adaptations (Part (a))
• Topic B3.2 — Osmosis (Parts (b), (c))

▶️ Answer/Explanation

(a) Any two of: no nucleus (more room for haemoglobin); large surface area/biconcave shape; contains haemoglobin.

The lack of a nucleus leaves more space to carry oxygen-binding haemoglobin.
A biconcave shape gives a large surface area for efficient diffusion of oxygen.

(b)(i) 8.0 g/dm³

At this concentration there is no change to the cells, meaning the solution has the same water potential as the cell contents.

(b)(ii) Water leaves the cell by osmosis because the salt solution has a lower water potential than the cell.

The 10.0 g/dm³ solution is more concentrated than the cell contents (lower water potential).
Water moves out of the red blood cell, by osmosis, from a region of higher to lower water potential, causing it to shrink.

(c)(i) The cell wall prevents the cell from bursting.

Water entering by osmosis causes the cell to swell and push against the cellulose cell wall.
The strong, rigid cell wall resists this pressure (the cell becomes turgid instead of bursting).

(c)(ii) Any two of: solvent; photosynthesis (raw material); support/turgidity (prevents wilting); transport of minerals/ions.

Water acts as a solvent for many substances inside the cell.
Water also provides turgor pressure, which keeps plant tissues rigid and supported.

(c)(iii) Root hair cell

Root hair cells have a large surface area that increases the rate of water absorption from the soil by osmosis.

Question 11

Sulfuric acid is made by the Contact process. Fig. 11.1 shows part of the Contact process.
(a) A catalyst is used in the Contact process.
Complete Fig. 11.1 to show the two other essential conditions used.
(b) In the Contact process, sulfur dioxide, \(\text{SO}_2\), reacts with oxygen, \(\text{O}_2\), to make sulfur trioxide, \(\text{SO}_3\).
\(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\)
(i) Calculate the maximum mass of sulfur trioxide that is made from 1.6 kg of sulfur dioxide.
[\(A_r\): O, 16; S, 32]
(ii) Fig. 11.2 shows the energy level diagram for the reaction to make sulfur trioxide.
Draw and label on Fig. 11.2:
  • the energy change in the reaction
  • the activation energy of the reaction.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C6.2 — Rate of reaction, catalysts (Part (a))
• Topic C3.3 — Reacting masses/moles (Part (b)(i))
• Topic C5.1 — Exothermic/endothermic reactions, energy level diagrams (Part (b)(ii))

▶️ Answer/Explanation

(a) Temperature of 450°C; pressure of 1–2 atmospheres.

The Contact process uses a catalyst together with a temperature of about 450°C.
A pressure of around 1–2 atmospheres is also used as an essential condition.

(b)(i) 2.0 kg

\(M_r(\text{SO}_2) = 64\), \(M_r(\text{SO}_3) = 80\).
Mass of \(\text{SO}_3 = \dfrac{80}{64} \times 1.6\).
Mass \(= 2.0\,\text{kg}\).

(b)(ii) 

The energy change is the vertical distance between the reactants and products energy levels.
The activation energy is the vertical distance from the reactants level up to the peak of the curve, representing the minimum energy needed to react.

Question 12

Electricity can be generated in different types of power stations.
(a) Table 12.1 gives some information about six types of power station.
(i) Use data from Table 12.1 to explain why electricity generation is negatively impacting the environment.
(ii) Nuclear power stations are very expensive to build.
Apart from cost, state one advantage and one disadvantage of generating electricity using wind compared to nuclear.
(iii) Use data from Table 12.1 to calculate the mass of natural gas needed to generate the same electrical energy output as 1 kg of nuclear fuel.
(b) A coal power station generates electricity at a voltage of 25 000 V.
A transformer is used to step the voltage up to 132 000 V for transmission.
(i) The step-up transformer contains 3000 turns on the primary coil.
Calculate the number of turns on the secondary coil.
(ii) Explain why electricity is transmitted at a voltage of 132 000 V and not 25 000 V.

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

• Topic P1.6.3 — Energy resources, advantages/disadvantages, efficiency (Part (a))
• Topic P4.5.6 — The transformer, high-voltage transmission (Part (b))

▶️ Answer/Explanation

(a)(i) Coal and natural gas (fossil fuels) generate the highest percentage of world electricity (61% combined); plus they release carbon dioxide, causing climate change/global warming (or release sulfur dioxide, causing acid rain).

Table 12.1 shows coal (37%) and natural gas (24%) together supply 61% of world electricity, more than any other source.
Burning these fossil fuels releases carbon dioxide, contributing to the enhanced greenhouse effect and climate change.

(a)(ii) Advantage: no radioactive waste/accidents, suitable for small scale, no fuel needed. Disadvantage: only works when wind speed is suitable/less efficient/needs many turbines/noise pollution.

Wind power avoids the radioactive waste and accident risks associated with nuclear power.
However, wind output is unreliable (depends on wind speed) and has lower efficiency (40% vs 93%), so more turbines are needed for the same output.

(a)(iii) 21 000 kg

Useful energy per kg nuclear fuel \(= 5.0\times10^{5} \times 0.93 = 4.65\times10^{5}\,\text{MJ}\).
Useful energy per kg natural gas \(= 45 \times 0.49 = 22.05\,\text{MJ}\).
Mass needed \(= \dfrac{4.65\times10^{5}}{22.05} \approx 21\,000\,\text{kg}\).

(b)(i) 16 000 turns

Use \(\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}\), so \(N_s = 3000 \times \dfrac{132\,000}{25\,000}\).
\(N_s = 16\,000\) turns.

(b)(ii) A higher voltage reduces the current for the same power, reducing energy/power loss as heat in the cables.

For the same power transmitted, a higher voltage means a lower current is needed.
Since power loss in cables depends on \(I^2R\), a lower current greatly reduces heat/energy losses during transmission.

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