An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Paper 23
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
Potato cylinders of equal initial length are placed into four separate sucrose solutions: W, X, Y and Z. The table below shows the percentage change in the length of the cylinders after a period of one hour: Solution W: +4.5% change; Solution X: +1.2% change; Solution Y: -2.0% change; Solution Z: -5.5% change. Which solution has the lowest water potential, and what is the state of the potato cells in this solution?
A.solution W, turgid cells
B.solution W, plasmolysed cells
C.solution Z, turgid cells
D.solution Z, plasmolysed cells
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Worked solution
Solution Z shows the greatest decrease in cylinder length (-5.5%), indicating the highest net movement of water out of the potato cells by osmosis. This means Solution Z has the lowest water potential (highest solute concentration) relative to the cell cytoplasm. When water leaves plant cells by osmosis, the vacuole and cytoplasm shrink, and the cells become plasmolysed.
Marking scheme
1 mark for the correct option D.
Question 2 · multiple-choice
1 marks
The table shows the time taken for starch to be completely broken down by amylase at different pH values: at pH 4 it takes 12 minutes; at pH 5 it takes 6 minutes; at pH 6 it takes 2 minutes; at pH 7 it takes 1 minute; at pH 8 it takes 4 minutes; at pH 9 the starch is not broken down after 30 minutes. Which statement is supported by these results?
A.Amylase is denatured at pH 7.
B.The optimum pH for this amylase is between pH 6 and pH 8.
C.Amylase is more active in acidic conditions than in neutral conditions.
D.At pH 9, the kinetic energy of amylase is at its maximum.
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Worked solution
At pH 7, the starch is broken down in the shortest amount of time (1 minute), which indicates that the rate of enzyme activity is highest. Therefore, the optimum pH lies between pH 6 and pH 8. At pH 9, the enzyme is denatured, so it is inactive.
Marking scheme
1 mark for the correct option B.
Question 3 · multiple-choice
1 marks
Which statement correctly compares sexual and asexual reproduction in plants?
A.Genetic variation is present in the offspring of sexual reproduction, but absent in asexual reproduction.
B.Only one parent is required for sexual reproduction, while two are always required for asexual reproduction.
C.A zygote is produced during asexual reproduction, but not during sexual reproduction.
D.Mitosis is the only cell division involved in sexual reproduction, while meiosis is the only cell division in asexual reproduction.
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Worked solution
Sexual reproduction involves the fusion of gametes, which leads to genetic variation in the offspring. Asexual reproduction involves only one parent and produces genetically identical clones (no variation). Azygote is produced in sexual reproduction. Mitosis is used in asexual reproduction, while meiosis is used in the production of gametes for sexual reproduction.
Marking scheme
1 mark for the correct option A.
Question 4 · multiple-choice
1 marks
An ion of element E is represented by \(^{39}_{19}\text{E}^{+}\). How many protons, neutrons, and electrons are there in this ion?
A.19 protons, 20 neutrons, 18 electrons
B.19 protons, 20 neutrons, 19 electrons
C.19 protons, 20 neutrons, 20 electrons
D.20 protons, 19 neutrons, 18 electrons
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Worked solution
The atomic number is 19, so there are 19 protons. The mass number is 39, so there are 39 - 19 = 20 neutrons. The ion has a 1+ charge, meaning it has lost 1 electron, so there are 19 - 1 = 18 electrons.
Marking scheme
1 mark for the correct option A.
Question 5 · multiple-choice
1 marks
During the blast furnace extraction of iron, iron(III) oxide reacts with carbon monoxide according to the equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). Which statement regarding this redox process is correct?
A.The iron(III) oxide undergoes reduction because it gains oxygen atoms.
B.Iron(III) oxide behaves as the reducing agent in this reaction.
C.Carbon monoxide is oxidised because it gains oxygen atoms.
D.Carbon monoxide serves as the oxidising agent.
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Worked solution
Carbon monoxide (CO) gains oxygen to form carbon dioxide (CO2), so it is oxidised. Fe2O3 loses oxygen, so it is reduced (making it the oxidising agent, while CO is the reducing agent).
Marking scheme
1 mark for the correct option C.
Question 6 · multiple-choice
1 marks
How do the chemical reactivity and physical properties of the halogens change as Group VII is descended?
A.Their chemical reactivity increases while their colors become less intense.
B.Their chemical reactivity increases while their melting points rise.
C.Their chemical reactivity decreases while their melting points rise.
D.Their chemical reactivity decreases while their colors become less intense.
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Worked solution
Down Group VII, reactivity decreases (the elements become less reactive) and the melting and boiling points increase (fluorine is a gas, bromine is a liquid, iodine is a solid). The colors also get darker, not lighter.
Marking scheme
1 mark for the correct option C.
Question 7 · multiple-choice
1 marks
A car of mass 1200 kg is accelerated uniformly from rest to a speed of 20 m/s in 5.0 s. It then travels at a constant speed of 20 m/s for another 5.0 s. What is the kinetic energy of the car at 8.0 s?
A.12 kJ
B.24 kJ
C.120 kJ
D.240 kJ
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Worked solution
At 8.0 s, the car is in the constant speed phase, traveling at 20 m/s. Kinetic energy is calculated as: E_k = 0.5 * m * v^2 = 0.5 * 1200 kg * (20 m/s)^2 = 600 * 400 = 240,000 J = 240 kJ.
Marking scheme
1 mark for the correct option D.
Question 8 · multiple-choice
1 marks
A student connects two resistors in parallel: one of resistance 6.0 \(\Omega\) and another of resistance 12 \(\Omega\). This parallel combination is then connected in series with a 4.0 \(\Omega\) resistor and a 12 V battery. What is the total current drawn from the battery?
A.1.0 A
B.1.5 A
C.2.0 A
D.3.0 A
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Worked solution
First, calculate the resistance of the parallel combination: R_p = (6 * 12) / (6 + 12) = 72 / 18 = 4.0 \(\Omega\). Next, add the series resistor to find the total resistance: R_total = 4.0 + 4.0 = 8.0 \(\Omega\). Finally, use Ohm's law to find the total current: I = V / R_total = 12 V / 8.0 \(\Omega\) = 1.5 A.
Marking scheme
1 mark for the correct option B.
Question 9 · multiple_choice
1 marks
Which statement correctly describes active transport?
A.It is the net movement of water molecules from a region of higher water potential to a region of lower water potential across a partially permeable membrane.
B.It is the movement of particles from a region of their lower concentration to a region of their higher concentration using energy from respiration.
C.It is the net movement of particles from a region of their higher concentration to a region of their lower concentration down a concentration gradient.
D.It is the passive movement of ions through carrier proteins without the use of energy.
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Worked solution
Active transport moves particles from an area of low concentration to an area of high concentration, which is against the concentration gradient. This requires chemical energy released during respiration.
Marking scheme
1 mark for the correct option (B).
Question 10 · multiple_choice
1 marks
An enzyme digests proteins in an acidic environment of pH 2.0. In which part of the human alimentary canal is this enzyme active?
A.duodenum
B.ileum
C.mouth
D.stomach
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Worked solution
The stomach contains hydrochloric acid, creating a highly acidic environment of approximately pH 1.5 to 2.0. This is the optimal pH for the protease pepsin, which digests proteins.
Marking scheme
1 mark for the correct option (D).
Question 11 · multiple_choice
1 marks
An atom of element Y has 11 protons, 12 neutrons and 11 electrons. What are the nucleon number and the Periodic Table group number of element Y?
A.nucleon number = 12, group number = II
B.nucleon number = 23, group number = I
C.nucleon number = 23, group number = II
D.nucleon number = 12, group number = I
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Worked solution
The nucleon number is the sum of protons and neutrons in the nucleus: \(11 + 12 = 23\). With 11 electrons, its electronic configuration is 2, 8, 1. Since it has 1 outer-shell electron, it is in Group I.
Marking scheme
1 mark for the correct option (B).
Question 12 · multiple_choice
1 marks
Which row correctly describes the products formed at the anode and the cathode during the electrolysis of molten lead(II) bromide using inert electrodes?
A.anode: bromine, cathode: lead
B.anode: lead, cathode: bromine
C.anode: oxygen, cathode: hydrogen
D.anode: hydrogen, cathode: oxygen
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Worked solution
During the electrolysis of molten lead(II) bromide, negative bromide ions (\(\text{Br}^-\)) migrate to the positive anode and form bromine. Positive lead ions (\(\text{Pb}^{2+}\)) migrate to the negative cathode and form lead.
Marking scheme
1 mark for the correct option (A).
Question 13 · multiple_choice
1 marks
Which statement about alkenes is correct?
A.They are saturated hydrocarbons.
B.They turn aqueous bromine from orange to colourless.
C.They have the general formula \(\text{C}_n\text{H}_{2n+2}\).
D.They do not contain any carbon-carbon double bonds.
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Worked solution
Alkenes are unsaturated hydrocarbons that contain a carbon-carbon double bond, which allows them to react readily with aqueous bromine, decolourising it from orange to colourless.
Marking scheme
1 mark for the correct option (B).
Question 14 · multiple_choice
1 marks
A crane lifts a load of mass \(500\text{ kg}\) vertically upwards through a height of \(20\text{ m}\) in a time of \(10\text{ s}\). The gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the useful power developed by the crane?
A.1.0 kW
B.10 kW
C.50 kW
D.100 kW
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Worked solution
First calculate the work done: \(W = F \times d = (m \times g) \times h = 500\text{ kg} \times 10\text{ N/kg} \times 20\text{ m} = 100,000\text{ J}\). Next, calculate useful power: \(P = \frac{W}{t} = \frac{100,000\text{ J}}{10\text{ s}} = 10,000\text{ W} = 10\text{ kW}\).
Marking scheme
1 mark for the correct option (B).
Question 15 · multiple_choice
1 marks
A sound wave travels through air. If the frequency of the wave increases and its amplitude decreases, how do the pitch and the loudness of the sound change?
A.The pitch becomes higher and the loudness decreases.
B.The pitch becomes lower and the loudness increases.
C.The pitch becomes higher and the loudness increases.
D.The pitch becomes lower and the loudness decreases.
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Worked solution
The frequency of a wave is directly related to its pitch; a higher frequency produces a higher pitch. The amplitude of a wave is directly related to its loudness; a smaller amplitude produces a quieter sound (decreased loudness).
Marking scheme
1 mark for the correct option (A).
Question 16 · multiple_choice
1 marks
Two resistors, with resistances of \(3.0\ \Omega\) and \(6.0\ \Omega\), are connected in parallel to a \(12\text{ V}\) power supply. What is the total current in the circuit?
A.1.3 A
B.2.0 A
C.4.0 A
D.6.0 A
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Worked solution
Calculate the combined resistance \(R_p\) in parallel: \(\frac{1}{R_p} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{3}{6.0}\), so \(R_p = 2.0\ \Omega\). The total current in the circuit is \(I = \frac{V}{R_p} = \frac{12\text{ V}}{2.0\ \Omega} = 6.0\text{ A}\).
Marking scheme
1 mark for the correct option (D).
Question 17 · multiple-choice
1 marks
Red blood cells are placed in a solution with a higher water potential than the cytoplasm of the cells. What is the effect on the cells?
A.Water enters the cells by osmosis, causing them to swell and burst.
B.Water enters the cells by osmosis, but the cell walls prevent them from bursting.
C.Water leaves the cells by osmosis, causing them to shrink and shrivel.
D.Salt enters the cells by active transport, causing them to burst.
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Worked solution
Water potential is higher outside the cell than inside. Therefore, water moves down the water potential gradient into the cells by osmosis. Since animal cells (such as red blood cells) do not have a cell wall to resist the turgor pressure, they swell and eventually burst (undergo lysis).
Marking scheme
1 mark for the correct option A.
Question 18 · multiple-choice
1 marks
Which statement correctly describes the effect of temperature on enzyme activity?
A.As temperature increases up to the optimum, the kinetic energy of the molecules increases, increasing the rate of successful collisions.
B.At temperatures above the optimum, the active site changes shape, allowing more substrate molecules to bind.
C.At very low temperatures, the enzyme is denatured and cannot function again even if the temperature is warmed up.
D.The rate of reaction always doubles for every 10 °C increase in temperature regardless of the temperature range.
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Worked solution
Increasing the temperature up to the optimum increases the kinetic energy of both the enzyme and substrate molecules. This results in faster movement and a higher frequency of successful collisions, thereby increasing enzyme activity. Above the optimum temperature, enzymes denature (active site changes shape so substrate no longer fits), and low temperatures deactivate enzymes but do not denature them.
Marking scheme
1 mark for the correct option A.
Question 19 · multiple-choice
1 marks
A comparison is made between Group I elements (alkali metals) and transition elements. Which statement is correct?
A.Group I elements have higher densities than transition elements.
B.Group I elements are more reactive with water than transition elements.
C.Transition elements form only white or colourless compounds.
D.Transition elements have lower melting points than Group I elements.
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Worked solution
Group I metals are extremely reactive and react vigorously with water to produce hydrogen gas and an alkaline solution. Transition metals are much less reactive and react very slowly or not at all with cold water. Additionally, transition metals have higher densities, higher melting points, and form colored compounds.
Marking scheme
1 mark for the correct option B.
Question 20 · multiple-choice
1 marks
The equations show four chemical reactions. In which reaction is the underlined substance reduced?
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Worked solution
Reduction is defined as the loss of oxygen. In reaction A, copper(II) oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)), meaning it is reduced. In the other options, the underlined substances are either oxidized or involved in displacement/redox where they do not lose oxygen.
Marking scheme
1 mark for the correct option A.
Question 21 · multiple-choice
1 marks
Which description of propene is correct?
A.It is a saturated hydrocarbon that decolourises aqueous bromine.
B.It is an unsaturated hydrocarbon that decolourises aqueous bromine.
C.It is a saturated hydrocarbon that does not react with aqueous bromine.
D.It is an unsaturated hydrocarbon that does not react with aqueous bromine.
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Worked solution
Propene is an alkene with the formula \(\text{C}_3\text{H}_6\). Because it contains a carbon-carbon double bond (\(\text{C=C}\)), it is an unsaturated hydrocarbon. Unsaturated hydrocarbons undergo an addition reaction with bromine, decolourising aqueous bromine from orange/brown to colourless.
Marking scheme
1 mark for the correct option B.
Question 22 · multiple-choice
1 marks
A car of mass 1200 kg accelerates from rest to a speed of 20 m/s. What is the work done on the car to reach this speed, assuming no energy is lost as heat?
A.12 kJ
B.24 kJ
C.240 kJ
D.480 kJ
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Worked solution
The work done on the car is equal to its gain in kinetic energy. \(\text{Work done} = \Delta E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (20\text{ m/s})^2 = 600 \times 400 = 240,000\text{ J} = 240\text{ kJ}\).
Marking scheme
1 mark for the correct option C.
Question 23 · multiple-choice
1 marks
Which statement about electromagnetic waves is correct?
A.They are longitudinal waves and travel at the same speed in a vacuum.
B.They are transverse waves and travel at the same speed in a vacuum.
C.They are longitudinal waves and travel at different speeds in a vacuum.
D.They are transverse waves and travel at different speeds in a vacuum.
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Worked solution
All electromagnetic waves (radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays) are transverse waves and they all travel at the same high speed of approximately \(3.0 \times 10^8\text{ m/s}\) in a vacuum.
Marking scheme
1 mark for the correct option B.
Question 24 · multiple-choice
1 marks
Two resistors, of resistance \(4.0\ \Omega\) and \(12\ \Omega\), are connected in parallel. This parallel combination is then connected in series with a \(5.0\ \Omega\) resistor. What is the total combined resistance of this circuit?
A.\(3.0\ \Omega\)
B.\(8.0\ \Omega\)
C.\(16\ \Omega\)
D.\(21\ \Omega\)
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Worked solution
First, calculate the resistance of the parallel combination (\(R_p\)): \(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12}\), so \(R_p = \frac{12}{4} = 3.0\ \Omega\). Next, add this in series to the \(5.0\ \Omega\) resistor: \(R_{\text{total}} = R_p + R_{\text{series}} = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\).
Marking scheme
1 mark for the correct option B.
Question 25 · multiple_choice
1 marks
Which structures are present in a palisade mesophyll cell of a plant leaf but absent from a human cheek cell?
A.cell wall, chloroplast and large permanent vacuole
B.cell wall, cell membrane and cytoplasm
C.cell membrane, cytoplasm and nucleus
D.chloroplast, mitochondria and cytoplasm
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Worked solution
Plant cells, such as palisade mesophyll cells, possess a cellulose cell wall, chloroplasts (for photosynthesis), and a large permanent vacuole. Human cheek cells are animal cells and do not have any of these three structures.
Marking scheme
1 mark: correct option selected.
Question 26 · multiple_choice
1 marks
A food sample is mixed with different chemical reagents. A portion of the sample turns orange-brown when iodine solution is added, another portion turns brick-red when heated with Benedict's solution, and a third portion remains blue when Biuret reagent is added. Which biological molecules are present in this food sample?
A.reducing sugar and starch
B.reducing sugar only
C.protein and starch
D.reducing sugar and protein
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Worked solution
The orange-brown color with iodine solution indicates that starch is absent. The brick-red color with Benedict's solution indicates that reducing sugar is present. The blue color with Biuret reagent indicates that protein is absent. Therefore, only reducing sugar is present in the sample.
Marking scheme
1 mark: correct option selected.
Question 27 · multiple_choice
1 marks
An enzyme-catalysed reaction is carried out at different temperatures. At \(20^\circ\text{C}\), the reaction rate is very slow. At \(40^\circ\text{C}\), the reaction rate is at its maximum. At \(60^\circ\text{C}\), the reaction stops completely. Which row correctly explains the findings at \(20^\circ\text{C}\) and \(60^\circ\text{C}\)?
A.At \(20^\circ\text{C}\): molecules have low kinetic energy. At \(60^\circ\text{C}\): enzymes are denatured.
B.At \(20^\circ\text{C}\): enzymes are denatured. At \(60^\circ\text{C}\): molecules have low kinetic energy.
C.At \(20^\circ\text{C}\): enzymes are denatured. At \(60^\circ\text{C}\): enzymes are denatured.
D.At \(20^\circ\text{C}\): molecules have low kinetic energy. At \(60^\circ\text{C}\): molecules have low kinetic energy.
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Worked solution
At a lower temperature of \(20^\circ\text{C}\), the molecules have low kinetic energy, resulting in fewer successful collisions per unit time. At a high temperature of \(60^\circ\text{C}\), the enzymes have denatured because the high temperature has permanently altered the shape of their active sites, preventing them from binding to the substrate.
Marking scheme
1 mark: correct option selected.
Question 28 · multiple_choice
1 marks
Which plant tissue transports sucrose and amino acids, and in which direction(s) can this transport occur?
A.tissue: phloem, direction(s): upwards and downwards
B.tissue: xylem, direction(s): upwards only
C.tissue: phloem, direction(s): downwards only
D.tissue: xylem, direction(s): upwards and downwards
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Worked solution
Sucrose and amino acids are transported by phloem tissue in a process called translocation. This transport occurs from sources to sinks, which can be both upwards and downwards in the plant. In contrast, xylem tissue transports water and mineral ions upwards only.
Marking scheme
1 mark: correct option selected.
Question 29 · multiple_choice
1 marks
An ion of magnesium has the symbol \(^{24}_{12}\text{Mg}^{2+}\). How many protons, neutrons and electrons are present in this ion?
A.12 protons, 12 neutrons, 10 electrons
B.12 protons, 12 neutrons, 12 electrons
C.12 protons, 24 neutrons, 10 electrons
D.10 protons, 12 neutrons, 12 electrons
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Worked solution
The atomic number is 12, which represents the number of protons. The mass number is 24, so the number of neutrons is mass number minus atomic number: \(24 - 12 = 12\). The charge is \(2+\), which means it has lost two electrons compared to the neutral atom: \(12 - 2 = 10\) electrons.
Marking scheme
1 mark: correct option selected.
Question 30 · multiple_choice
1 marks
Copper(II) sulfate is a soluble salt prepared by reacting insoluble copper(II) oxide with dilute sulfuric acid. Which step is NOT part of this salt preparation?
A.evaporating the solution to dryness immediately after filtration
B.adding excess copper(II) oxide to warm dilute sulfuric acid
C.filtering the mixture to remove unreacted copper(II) oxide
D.heating the filtrate to obtain a saturated solution
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Worked solution
To obtain large, well-defined hydrated copper(II) sulfate crystals, the filtrate must be heated to obtain a saturated solution and then allowed to cool slowly. Evaporating the solution to dryness immediately after filtration would leave a fine powder or ruin the structure of hydrated crystals.
Marking scheme
1 mark: correct option selected.
Question 31 · multiple_choice
1 marks
A wooden box of mass \(5.0\text{ kg}\) is pulled along a frictionless horizontal surface by a constant horizontal force of \(20\text{ N}\) over a distance of \(10\text{ m}\). What is the work done on the box by this force?
A.50 J
B.100 J
C.200 J
D.1000 J
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Worked solution
The formula for work done is \(W = F \times d\), where \(F\) is the force and \(d\) is the distance moved in the direction of the force. Here, \(W = 20\text{ N} \times 10\text{ m} = 200\text{ J}\). The mass of the box is not needed to calculate the work done.
Marking scheme
1 mark: correct option selected.
Question 32 · multiple_choice
1 marks
Two identical resistors, each having a resistance of \(6.0\ \Omega\), are connected in parallel across a \(12\text{ V}\) power supply. What is the total current drawn from the power supply?
A.1.0 A
B.2.0 A
C.4.0 A
D.12 A
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Worked solution
The combined resistance \(R_p\) of two identical resistors in parallel is \(R_p = \frac{R}{2} = \frac{6.0}{2} = 3.0\ \Omega\). According to Ohm's law, the total current drawn from the power supply is \(I = \frac{V}{R_p} = \frac{12\text{ V}}{3.0\ \Omega} = 4.0\text{ A}\).
Marking scheme
1 mark: correct option selected.
Question 33 · multiple_choice
1 marks
An amylase-controlled reaction is carried out at 60 °C. It is observed that starch is no longer broken down. What is the explanation for this observation?
A.The active site of the amylase has permanently changed shape, so starch can no longer fit.
B.The starch molecules have denatured and can no longer bind to the active site.
C.The activation energy of the reaction has increased, preventing any collisions.
D.The rate of diffusion of starch molecules has decreased to zero.
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Worked solution
At high temperatures, the thermal energy causes the enzyme (amylase) molecules to vibrate excessively, breaking the bonds that maintain its three-dimensional structure. This denatures the enzyme, permanently changing the shape of its active site so that starch molecules can no longer fit.
Marking scheme
1 mark for the correct option.
Question 34 · multiple_choice
1 marks
Which row correctly identifies the raw materials needed for photosynthesis and the form in which carbohydrates are stored in the leaf?
A.raw materials: carbon dioxide and oxygen | stored carbohydrate: glucose
B.raw materials: carbon dioxide and water | stored carbohydrate: starch
C.raw materials: glucose and oxygen | stored carbohydrate: cellulose
D.raw materials: water and oxygen | stored carbohydrate: sucrose
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Worked solution
Photosynthesis requires carbon dioxide and water as raw materials to produce glucose and oxygen in the presence of light and chlorophyll. The glucose produced is converted into starch for storage in the leaf because starch is insoluble and does not affect the osmotic potential of the cells.
Marking scheme
1 mark for the correct option.
Question 35 · multiple_choice
1 marks
An atom of phosphorus is represented by the symbol _{15}^{31}P. Which row shows the correct number of protons, neutrons and electrons in this neutral atom?
A.protons: 15, neutrons: 16, electrons: 15
B.protons: 15, neutrons: 31, electrons: 15
C.protons: 16, neutrons: 15, electrons: 16
D.protons: 31, neutrons: 15, electrons: 15
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Worked solution
The atomic number (bottom number) represents the number of protons, which is 15. In a neutral atom, the number of electrons is equal to the number of protons, which is 15. The mass number (top number) is the sum of protons and neutrons (31). Therefore, the number of neutrons is 31 - 15 = 16.
Marking scheme
1 mark for the correct option.
Question 36 · multiple_choice
1 marks
Which statement about the trends as we go down Group VII of the Periodic Table is correct?
A.The elements become more reactive and their color becomes darker.
B.The elements become more reactive and their melting point decreases.
C.The elements become less reactive and their color becomes darker.
D.The elements become less reactive and their boiling point decreases.
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Worked solution
As we go down Group VII (the halogens), the elements become less reactive because the outer shell is further from the nucleus, making it harder to attract an electron. Their colors also become darker (chlorine is pale green, bromine is red-brown, iodine is grey-black).
Marking scheme
1 mark for the correct option.
Question 37 · multiple_choice
1 marks
A hydrocarbon gas is bubbled through aqueous bromine. The orange-brown solution rapidly becomes colourless. Which statement about the hydrocarbon gas is correct?
A.It is an alkane and contains only single C-C bonds.
B.It is an alkane and contains a double C=C bond.
C.It is an alkene and contains only single C-C bonds.
D.It is an alkene and contains a double C=C bond.
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Worked solution
The rapid decolourisation of bromine water is a test for unsaturation, indicating the presence of a carbon-carbon double bond (C=C). This means the hydrocarbon is an alkene.
Marking scheme
1 mark for the correct option.
Question 38 · multiple_choice
1 marks
A rectangular block of metal has dimensions 2.0 cm x 3.0 cm x 5.0 cm. The mass of the block is 270 g. What is the density of the metal?
A.0.11 g/cm^3
B.9.0 g/cm^3
C.27 g/cm^3
D.8100 g/cm^3
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Worked solution
First, calculate the volume of the block: Volume = 2.0 cm * 3.0 cm * 5.0 cm = 30 cm^3. Next, calculate the density using the formula: Density = mass / volume = 270 g / 30 cm^3 = 9.0 g/cm^3.
Marking scheme
1 mark for the correct option.
Question 39 · multiple_choice
1 marks
A sound wave travels from a loudspeaker through the air. The frequency of the sound wave is increased while its amplitude is kept constant. How do the pitch and loudness of the sound heard change?
A.The pitch increases and the loudness increases.
B.The pitch increases and the loudness remains the same.
C.The pitch remains the same and the loudness increases.
D.The pitch decreases and the loudness remains the same.
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Worked solution
The pitch of a sound is determined by its frequency; an increase in frequency results in a higher pitch. The loudness of a sound is determined by its amplitude; since the amplitude is kept constant, the loudness remains the same.
Marking scheme
1 mark for the correct option.
Question 40 · multiple_choice
1 marks
A lamp is connected to a 12.0 V battery. The current in the lamp is measured as 0.50 A. What is the resistance of the lamp?
A.0.042 ohms
B.6.0 ohms
C.24 ohms
D.48 ohms
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Worked solution
Using Ohm's law, resistance is given by the formula R = V / I. Substituting the given values: R = 12.0 V / 0.50 A = 24 ohms.
Marking scheme
1 mark for the correct option.
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9 Question · 81 marks
Question 1 · Structured
9 marks
1 (a) Describe the role of cartilage in the walls of the trachea. [1]
(b) During inspiration, the lungs are ventilated with fresh air. Describe how the actions of the diaphragm and the intercostal muscles work to decrease the pressure inside the thorax. [3]
(c) Table 1.1 shows the volume of air inhaled per breath (tidal volume) and the breathing rate for an athlete before and after completing a running race.
Table 1.1 $$\begin{array}{|l|c|c|} \hline \text{condition} & \text{tidal volume / } \text{dm}^3 & \text{breathing rate / breaths per minute} \\ \hline \text{before race (at rest)} & 0.5 & 12 \\ \hline \text{after race} & 1.6 & 25 \\ \hline \end{array}$$ (i) Calculate the total volume of air inhaled per minute (ventilation rate) immediately after the race. Show your working. [2]
(ii) Explain, in terms of aerobic respiration and carbon dioxide levels in the blood, why the athlete’s ventilation rate increases during and after the race. [3]
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Worked solution
(a) Cartilage keeps the trachea open / prevents it from collapsing during breathing when the pressure inside falls. (b) The diaphragm contracts and moves downwards (flattens). The external intercostal muscles contract, pulling the ribcage upwards and outwards. This increases the volume of the thorax, which decreases the pressure inside below atmospheric pressure. (c) (i) Ventilation rate = tidal volume $\times$ breathing rate $\text{Ventilation rate} = 1.6\text{ dm}^3 \times 25\text{ breaths/min} = 40\text{ dm}^3\text{/min}$ (ii) During exercise, muscle cells increase their rate of aerobic respiration to meet the higher demand for energy. This produces more carbon dioxide as a waste product. The increased carbon dioxide levels in the blood are detected by the brain, which sends signals to increase both breathing rate and tidal volume to remove carbon dioxide more rapidly and supply more oxygen.
Marking scheme
(a) • prevents trachea from collapsing / keeps airway open [1]
(b) • diaphragm contracts and moves down / flattens [1] • external intercostal muscles contract and pull ribcage up / out [1] • increases thorax volume which decreases pressure [1]
(c) (i) • calculation: $1.6 \times 25$ [1] • correct evaluation: $40$ with unit $\text{dm}^3\text{/min}$ (or $\text{dm}^3\text{ min}^{-1}$) [1]
(c) (ii) • (exercise causes) increased rate of respiration to supply energy [1] • increases concentration of carbon dioxide in the blood [1] • faster / deeper breathing removes carbon dioxide / supplies more oxygen [1]
Question 2 · Structured
9 marks
2 (a) Define the term osmosis. [2]
(b) A student investigates the effect of sucrose concentration on plant cells. He places equal-sized cylinders of potato tissue in four test-tubes containing sucrose solutions of concentrations: $0.0\text{ mol/dm}^3$, $0.2\text{ mol/dm}^3$, $0.4\text{ mol/dm}^3$, and $0.6\text{ mol/dm}^3$.
(i) Predict the change in mass of the potato cylinder placed in the $0.0\text{ mol/dm}^3$ sucrose solution and explain this change. [3]
(ii) In the $0.35\text{ mol/dm}^3$ sucrose solution, the potato cylinder shows no change in mass. State what this tells us about the internal concentration of the potato cells. [1]
(c) Red blood cells are animal cells. Explain why red blood cells burst when placed in pure water, whereas potato cells do not. [3]
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Worked solution
(a) Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution) through a partially permeable membrane. (b) (i) The mass of the potato cylinder increases. This is because the water potential outside the potato cells (in pure water) is higher than inside the potato cells. Water moves into the cells by osmosis down the water potential gradient. (ii) It indicates that the sucrose concentration inside the potato cells is equal to $0.35\text{ mol/dm}^3$ (isotonic). (c) Red blood cells do not have a cell wall, so when water enters by osmosis, the cell membrane stretches until it bursts. Potato cells have a strong, rigid cellulose cell wall which prevents them from bursting when they become turgid.
Marking scheme
(a) • net movement of water molecules from higher to lower water potential [1] • across a partially permeable membrane [1]
(b) (i) • mass increases [1] • water potential of pure water is higher than inside potato cells [1] • water enters by osmosis [1]
(b) (ii) • internal cell concentration is equal to $0.35\text{ mol/dm}^3$ (or equivalent wording) [1]
(c) • red blood cells lack a cell wall [1] • plant cells have a rigid cell wall / cellulose wall [1] • cell wall resists turgor pressure / expansion, preventing bursting [1]
Question 3 · Structured
9 marks
3 (a) Molten zinc chloride, $\text{ZnCl}_2$, is electrolysed using inert carbon electrodes.
(i) Explain why inert carbon electrodes are used in this process. [2]
(ii) Identify the product formed at each electrode. • product at negative electrode (cathode): • product at positive electrode (anode): [2]
(iii) Describe, in terms of the movement of electrons, the reaction occurring at the negative electrode. Write an ionic half-equation for this reaction. [3]
(b) Explain why solid zinc chloride does not conduct electricity, whereas molten zinc chloride is a good conductor. [2]
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Worked solution
(a) (i) Carbon conducts electricity and is unreactive (inert), so it does not react with the electrolyte or the products of electrolysis (zinc and chlorine). (ii) Cathode: zinc; Anode: chlorine. (iii) Zinc ions ($\text{Zn}^{2+}$) are attracted to the negative electrode, where each ion gains two electrons to form a neutral zinc atom. This is a reduction reaction. Half-equation: $\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}$ (b) In solid zinc chloride, the ions are held in fixed positions within a giant ionic lattice and cannot move. When molten, the ionic bonds are broken, allowing the ions ($\text{Zn}^{2+}$ and $\text{Cl}^-$) to move freely and carry the electric charge.
Marking scheme
(a) (i) • conducts electricity [1] • unreactive / will not react with electrolyte or products [1]
(a) (iii) • zinc ions / $\text{Zn}^{2+}$ gain electrons [1] • each ion gains two electrons / reduction occurs [1] • correct ionic half-equation: $\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}$ [1]
(b) • in solid, ions are fixed / cannot move [1] • in molten, ions are free to move and carry charge [1] (reject: electrons are free to move)
Question 4 · Structured
9 marks
4 (a) A student prepares a sample of nickel(II) sulfate crystals by reacting insoluble nickel(II) carbonate with dilute sulfuric acid.
(i) State two observations that indicate a chemical reaction is taking place when nickel(II) carbonate is added to dilute sulfuric acid. [2]
(ii) Write a word equation for this reaction. [1]
(iii) Describe how the student can obtain pure, dry crystals of nickel(II) sulfate from the reaction mixture. Explain why the nickel(II) carbonate must be added in excess. [4]
(b) (i) Suggest the pH value of the dilute sulfuric acid before the reaction begins and name a suitable indicator to confirm this pH. [2]
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Worked solution
(a) (i) Effervescence (fizzing/bubbles of gas) and the green solid nickel(II) carbonate dissolves to form a green solution. (ii) nickel(II) carbonate + sulfuric acid $\rightarrow$ nickel(II) sulfate + carbon dioxide + water (iii) Nickel(II) carbonate is added in excess to ensure that all of the sulfuric acid reacts, leaving no unreacted acid. The mixture is filtered to remove the unreacted nickel(II) carbonate. The filtrate (nickel(II) sulfate solution) is heated to evaporate some water to the point of crystallization. The solution is allowed to cool so crystals form, which are then filtered and dried with filter paper. (b) (i) The pH is around 1 (or 2) as it is a strong acid. A suitable indicator is Universal Indicator (or a pH meter).
Marking scheme
(a) (i) • bubbles / fizzing / effervescence [1] • solid dissolves / solution changes colour to green [1]
(a) (ii) • nickel(II) carbonate + sulfuric acid $\rightarrow$ nickel(II) sulfate + carbon dioxide + water [1]
(a) (iii) • excess solid added to ensure all acid is neutralised / used up [1] • filter to remove unreacted solid [1] • heat filtrate to saturation point / point of crystallization [1] • allow to cool, filter crystals, and dry with filter paper [1]
(b) (i) • pH: 1 or 2 [1] • indicator: Universal Indicator / pH paper / pH meter [1]
Question 5 · Structured
9 marks
5 (a) Decane, $\text{C}_{10}\text{H}_{22}$, is a saturated hydrocarbon.
(i) State what is meant by the terms saturated and hydrocarbon. [2]
(ii) Write a balanced chemical equation for the complete combustion of decane. [2]
(b) Decane can be cracked to produce pentane, $\text{C}_5\text{H}_{12}$, and an alkene, E.
(i) Deduce the chemical formula of alkene E and state the conditions required for industrial cracking. [2]
(ii) Describe a chemical test to distinguish between pentane and alkene E. Give the name of the reagent used and the observations for both compounds. [3]
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Worked solution
(a) (i) Saturated means the compound contains only single covalent carbon-carbon bonds. Hydrocarbon means the compound contains only carbon and hydrogen atoms. (ii) $2\text{C}_{10}\text{H}_{22} + 31\text{O}_2 \rightarrow 20\text{CO}_2 + 22\text{H}_2\text{O}$ (b) (i) Decane $\rightarrow$ pentane + alkene E. $\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_5\text{H}_{12} + \text{C}_5\text{H}_{10}$. Thus, the formula of alkene E is $\text{C}_5\text{H}_{10}$. Industrial cracking conditions: high temperature ($500^{\circ}\text{C}$ to $700^{\circ}\text{C}$) and a catalyst (such as alumina/silica/zeolite). (ii) Add bromine water to both. Pentane (alkane) does not react, so the mixture remains orange/brown. Alkene E (pentene) reacts rapidly, decolourising the bromine water from orange/brown to colourless.
Marking scheme
(a) (i) • saturated: only contains single C-C bonds [1] • hydrocarbon: contains hydrogen and carbon only [1]
6 (a) A solid block of steel has a mass of $450\text{ kg}$. The density of steel is $7800\text{ kg/m}^3$.
(i) Calculate the volume of the steel block. Show your working and state the unit. [3]
(ii) State the weight of the steel block. (Take the gravitational field strength $g = 10\text{ N/kg}$). [1]
(b) A crane lifts the steel block vertically upwards.
(i) At the start of the lift, the crane accelerates the block upwards at $0.50\text{ m/s}^2$. Calculate the resultant force acting on the block. [2]
(ii) The block is lifted to a height of $15\text{ m}$ at a constant speed in a time of $30\text{ s}$. Calculate the average useful power output of the crane during this lift. [3]
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Worked solution
(a) (i) Density $\rho = \frac{m}{V} \Rightarrow V = \frac{m}{\rho}$ $V = \frac{450\text{ kg}}{7800\text{ kg/m}^3} = 0.0577\text{ m}^3$ (or $0.058\text{ m}^3$) Unit: $\text{m}^3$ (ii) Weight $W = mg = 450\text{ kg} \times 10\text{ N/kg} = 4500\text{ N}$. (b) (i) Resultant force $F = ma = 450\text{ kg} \times 0.50\text{ m/s}^2 = 225\text{ N}$. (ii) Work done against gravity $W = F \times d = 4500\text{ N} \times 15\text{ m} = 67\,500\text{ J}$. Power $P = \frac{W}{t} = \frac{67\,500\text{ J}}{30\text{ s}} = 2250\text{ W}$ (or $2.25\text{ kW}$).
(b) (ii) • work done: $4500 \times 15 = 67\,500\text{ J}$ [1] • formula used: $P = W / t$ [1] • evaluation: $2250\text{ W}$ (or $\text{J/s}$) [1]
Question 7 · Structured
9 marks
7 (a) A marine vessel uses a sonar system to measure the depth of the sea. The system emits ultrasound waves.
(i) Explain what is meant by the term ultrasound, with reference to the typical range of human hearing. [2]
(ii) The ultrasound wave has a frequency of $40\text{ kHz}$. The speed of sound in seawater is $1500\text{ m/s}$. Calculate the wavelength of this wave. [2]
(iii) The sonar receiver detects the reflected echo $0.80\text{ s}$ after the pulse is emitted. Calculate the depth of the sea under the vessel. [2]
(b) The vessel also uses electromagnetic waves to communicate with the shore.
(i) State two features that are common to all electromagnetic waves. [2]
(ii) Identify the region of the electromagnetic spectrum that has wavelengths slightly longer than those of visible light. [1]
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Worked solution
(a) (i) Ultrasound consists of sound waves with frequencies higher than the upper limit of human hearing, which is $20\,000\text{ Hz}$ (or $20\text{ kHz}$). (ii) Wave equation: $v = f\lambda \Rightarrow \lambda = \frac{v}{f}$ $f = 40\text{ kHz} = 40\,000\text{ Hz}$ $\lambda = \frac{1500\text{ m/s}}{40\,000\text{ Hz}} = 0.0375\text{ m}$ (or $3.75\text{ cm}$) (iii) Distance travelled by pulse $d = v \times t = 1500\text{ m/s} \times 0.80\text{ s} = 1200\text{ m}$. Since the sound travels to the seabed and back, the depth is half this distance: $\text{depth} = \frac{1200\text{ m}}{2} = 600\text{ m}$. (b) (i) All EM waves travel at the speed of light in a vacuum ($3.0 \times 10^8\text{ m/s}$), are transverse waves, and can travel through a vacuum. (ii) Infrared radiation.
Marking scheme
(a) (i) • sound waves with frequency too high to be heard by humans [1] • frequency greater than $20\,000\text{ Hz}$ / $20\text{ kHz}$ [1]
(a) (ii) • formula used: $\lambda = v / f$ [1] • evaluation: $0.0375\text{ m}$ (or $3.75\text{ cm}$) [1]
(b) (i) • any two from: travel at speed of light in vacuum, transverse, can travel through vacuum [2]
(b) (ii) • infrared [1]
Question 8 · Structured
9 marks
8 (a) A student connects two heating elements, X and Y, in parallel to a $12\text{ V}$ d.c. power supply.
(i) Element X has a resistance of $3.0\ \Omega$ and element Y has a resistance of $6.0\ \Omega$. Calculate the combined resistance of the two elements in parallel. [2]
(ii) Calculate the total current drawn from the power supply. [2]
(iii) Calculate the electrical power dissipated by element X. [2]
(b) (i) The student adds a fuse to protect the circuit. Explain how the fuse protects the circuit and state where in the circuit it must be placed. [3]
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Worked solution
(a) (i) Parallel resistance formula: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{3}{6} \Rightarrow R_p = 2.0\ \Omega$. (ii) Current $I = \frac{V}{R_p} = \frac{12\text{ V}}{2.0\ \Omega} = 6.0\text{ A}$. (iii) Power dissipated by X: $P = \frac{V^2}{R_1} = \frac{12^2}{3.0} = \frac{144}{3.0} = 48\text{ W}$. (b) (i) The fuse protects the circuit by melting and breaking the circuit if the current becomes too high (due to a short circuit or overload). This prevents overheating and fires. It must be connected in series on the live wire before the parallel branches.
(b) (i) • fuse wire melts and breaks the circuit when current exceeds rating [1] • prevents overheating of cables / potential fire hazard [1] • placed in series on the live wire before parallel components [1]
Question 9 · structured
9 marks
A student investigates the rate of the reaction between solid marble chips (calcium carbonate) and dilute nitric acid.
The equation for the reaction is: $$\text{CaCO}_3(\text{s}) + 2\text{HNO}_3(\text{aq}) \rightarrow \text{Ca(NO}_3)_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$$
**(a) (i)** Name the salt formed during this reaction. [1]
**(ii)** Describe the test used to identify the carbon dioxide gas produced and state the positive result of this test.
**(iii)** State two changes to the reaction conditions, other than changing the temperature or adding a catalyst, that would decrease the rate of this reaction.
**(b) (i)** Explain, in terms of the collision theory, why increasing the temperature of the acid increases the rate of the reaction. [3]
**(ii)** State how the addition of a catalyst changes the activation energy of a chemical reaction. [1]
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Worked solution
**(a) (i)** Calcium nitrate
**(ii)** test: bubble the gas through limewater result: limewater turns cloudy / milky
**(iii)** 1. Decrease the concentration of the nitric acid. 2. Use larger marble chips / decrease the surface area of the calcium carbonate.
**(b) (i)** - Particles gain more kinetic energy / move faster. - There are more frequent collisions (between reactant particles). - A greater proportion of particles have energy equal to or greater than the activation energy (so more collisions are successful).
**(ii)** - It lowers / decreases the activation energy.
**(iii)** Any two from: - decrease concentration of nitric acid / use more dilute nitric acid; [1] - use larger marble chips / use pieces with smaller surface area; [1] (Accept any two valid points, e.g. decrease concentration [1], decrease surface area / use larger pieces [1])
**(b) (i)** - particles have more kinetic energy / move faster; [1] - there are more frequent collisions / collision rate increases; [1] - a greater fraction / proportion of the colliding particles have energy greater than or equal to the activation energy / more collisions are successful; [1]
**(ii)** - decreases / lowers (the activation energy); [1]
Paper 63
Answer all questions. Write your answers in the spaces provided on the question paper.
4 Question · 40 marks
Question 1 · structured
10 marks
A student investigates osmosis in sweet potato.
(a) The student measures the mass of a sweet potato cylinder before and after placing it in a sucrose solution of concentration \(0.6\text{ mol/dm}^3\) for 2 hours.
Fig. 1.1 shows the digital balance readings before and after immersion. - Before immersion: \(3.24\text{ g}\) - After immersion: \(2.85\text{ g}\)
(i) Record the mass of the sweet potato cylinder: - mass before immersion = .................... g - mass after immersion = .................... g [2]
(ii) Calculate the change in mass of this cylinder. (Include a negative sign if the mass decreased.)
change in mass = .................... g [1]
(iii) Calculate the percentage change in mass of this sweet potato cylinder. Show your working and give your answer to one decimal place.
percentage change in mass = .................... % [2]
(b) State two variables, other than the duration of immersion, that the student must keep constant in this investigation to ensure a fair test.
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Worked solution
(a)(i) Mass before immersion is 3.24 g and mass after immersion is 2.85 g.
(a)(ii) Change in mass = 2.85 g - 3.24 g = -0.39 g.
(a)(iii) Percentage change in mass = \(\frac{-0.39}{3.24} \times 100 = -12.0\%\).
(b) Variables to keep constant include: temperature of the solution, volume of the sucrose solution, surface area or dimensions of the sweet potato cylinders, variety or source of the sweet potato.
(c) Patted dry to remove excess surface liquid/water that would increase the measured mass.
(d) Repeating allows the identification of anomalous results and the calculation of a more reliable mean/average.
Marking scheme
(a)(i) 3.24 (g) AND 2.85 (g) [2 marks, 1 mark for each]
(a)(ii) -0.39 (g) [1 mark]
(a)(iii) Correct calculation showing working: \(\frac{-0.39}{3.24} \times 100\) [1 mark] AND correct answer of -12.0% (accept 12.0% decrease) [1 mark]
(b) State any two of: variety/source of sweet potato, temperature of the sucrose solution, volume of sucrose solution, dimensions/surface area of sweet potato cylinders [2 marks, 1 mark each]
(c) To remove excess surface liquid/water (which would increase the measured mass) [1 mark]
(d) Allows identification of anomalies [1 mark] AND allows a more reliable mean/average to be calculated [1 mark]
Question 2 · structured
10 marks
A student investigates the rate of reaction between calcium carbonate (marble chips) and dilute hydrochloric acid by measuring the volume of carbon dioxide gas produced over time.
(a) Fig. 2.1 shows the gas syringe readings at 30 seconds and 60 seconds after the start of the reaction. - At 30 seconds: \(18\text{ cm}^3\) - At 60 seconds: \(32\text{ cm}^3\)
(i) Record the volume of gas produced at 30 seconds:
volume = .................... \(\text{cm}^3\) [1]
(ii) Record the volume of gas produced at 60 seconds:
volume = .................... \(\text{cm}^3\) [1]
(b) Explain why the volume of gas stops increasing at the end of the reaction.
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Worked solution
(a)(i) 18 cm3 (a)(ii) 32 cm3
(b) The reaction has finished because one of the reactants (the limiting reactant, which is hydrochloric acid) has been completely used up.
(c) Average rate of gas production over the first 30 seconds = \(\frac{18\text{ cm}^3}{30\text{ s}} = 0.60\text{ cm}^3/\text{s}\).
(d) A measuring cylinder, a volumetric pipette, or a burette.
(e) The independent variable is the temperature of the acid/reaction mixture. The dependent variable is the rate of reaction (or volume of gas produced per unit time / time taken to collect a certain volume of gas).
(f) Place the reaction flask on a balance and block the neck of the flask with cotton wool (to allow gas to escape while preventing liquid spray from escaping). Measure the decrease in mass over time.
(b) Hydrochloric acid/limiting reactant is completely used up [1 mark] (do not accept 'reactants are used up' without specifying/implying one is limiting/completely used)
(c) \(\frac{18}{30} = 0.60\) (or 0.6) [1 mark] AND correct unit \(\text{cm}^3/\text{s}\) (or \(\text{cm}^3\text{ s}^{-1}\)) [1 mark]
(e) Independent variable: temperature [1 mark] Dependent variable: rate of reaction / volume of gas per unit time / time taken [1 mark]
(f) Place reaction flask on a (digital) balance [1 mark] AND plug flask with cotton wool (to prevent acid spray/mist loss but allow gas to escape) [1 mark]
Question 3 · structured
10 marks
A student investigates how the resistance of a metal wire depends on its length.
(a) Fig. 3.1 shows the pointers on the scales of the ammeter and voltmeter used in the experiment. - Ammeter reading: \(0.36\text{ A}\) - Voltmeter reading: \(1.8\text{ V}\)
(i) Record the current, \(I\):
\(I\) = .................... A [1]
(ii) Record the potential difference, \(V\):
\(V\) = .................... V [1]
(b) Calculate the resistance, \(R\), of this length of wire using the equation:
\(R = \frac{V}{I}\)
Show your working and state the unit.
\(R\) = .................... [2]
(c) Describe how the student can change the length of the wire connected in the circuit.
(d) Draw a circuit diagram of the circuit used in this investigation. The circuit must contain a cell, a switch, an ammeter, the test wire, and a voltmeter connected to measure the potential difference across the test wire.
[2]
(e) Explain why the switch should be opened between taking readings.
(c) The student can slide or move the crocodile clip along the wire to a different position.
(d) The circuit diagram should show a cell, a switch, an ammeter, and the test wire connected in series, with the voltmeter connected in parallel across the test wire.
(e) The switch is opened to prevent the wire from heating up, because an increase in temperature would change the resistance of the wire.
(f) The student should view the pointer scale directly from above (perpendicularly).
(c) Move / slide the crocodile clip (along the wire) [1 mark]
(d) Correct symbols for cell, switch, ammeter, voltmeter, and resistor/wire [1 mark] AND ammeter in series, voltmeter in parallel across the test wire [1 mark]
(e) To prevent the wire from heating up [1 mark] AND because heating/temperature increase changes/increases the resistance of the wire [1 mark]
(f) View scale perpendicularly / straight from above [1 mark]
Question 4 · structured
10 marks
A student investigates the rate of photosynthesis in an aquatic plant (elodea) by counting the number of oxygen bubbles produced per minute.
(a) The student records the number of bubbles counted in three trials at a distance of \(10\text{ cm}\) from the lamp: - Trial 1: \(42\text{ bubbles/minute}\) - Trial 2: \(46\text{ bubbles/minute}\) - Trial 3: \(44\text{ bubbles/minute}\)
(i) Calculate the mean number of bubbles produced per minute.
mean = .................... bubbles/minute [1]
(ii) State why the student repeated the bubble count three times at each distance.
(a)(ii) To identify any anomalous results and to increase the reliability of the average rate calculated.
(b)(i) Light intensity decreases.
(b)(ii) The rate of bubble production decreases because light is a limiting factor for photosynthesis.
(c) The beaker of water acts as a heat shield. It absorbs heat from the lamp so that the temperature of the aquatic plant remains constant (since temperature also affects photosynthesis).
(d) Test: A glowing splint is placed in a tube of the collected gas. It relights in the presence of oxygen.
(e) Limitation: bubbles can be of different sizes / bubbles are produced too fast to count accurately. More accurate method: collect the gas in a gas syringe / measuring cylinder / graduated tube and measure the volume of gas produced.
Marking scheme
(a)(i) 44 [1 mark]
(a)(ii) To identify anomalous results / to increase reliability of the calculated average [1 mark]
(b)(i) decreases [1 mark]
(b)(ii) Rate decreases [1 mark] AND because light is a limiting factor for photosynthesis / light intensity is reduced [1 mark]
(c) To act as a heat shield / absorb heat from the lamp [1 mark] AND to keep temperature constant / prevent temperature affecting the rate [1 mark]
(d) (glowing) splint relights [1 mark]
(e) Limitation: bubbles can be different sizes / difficult to count if fast [1 mark] AND more accurate method: collect gas in a gas syringe/measuring cylinder to measure volume [1 mark]
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