An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 Multiple Choice (Extended)
Answer all 40 multiple-choice questions on the separate answer sheet using soft pencil. Each question carries 1 mark.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
A constant resultant force of \(15\text{ N}\) acts on a stationary object of mass \(3.0\text{ kg}\) on a frictionless horizontal surface. What is the kinetic energy of the object after \(4.0\text{ s}\)?
A.\(60\text{ J}\)
B.\(300\text{ J}\)
C.\(600\text{ J}\)
D.\(1200\text{ J}\)
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Worked solution
First, calculate the acceleration using Newton's second law: \(a = \frac{F}{m} = \frac{15\text{ N}}{3.0\text{ kg}} = 5.0\text{ m/s}^2\). Next, calculate the final velocity after \(4.0\text{ s}\): \(v = u + at = 0 + (5.0 \times 4.0) = 20\text{ m/s}\). Finally, calculate the kinetic energy: \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 3.0\text{ kg} \times (20\text{ m/s})^2 = 1.5 \times 400 = 600\text{ J}\).
Marking scheme
C is correct [1]. A uses \(F \times t\) incorrectly as energy. B calculates momentum or omits the squared term. D omits the factor of \(\frac{1}{2}\) in the kinetic energy formula.
Question 2 · Multiple Choice
1 marks
Which row correctly identifies the type of reaction that occurs between ethene and aqueous bromine, and the colour change observed in the reaction mixture?
A.type of reaction: addition ; colour change: orange to colourless
B.type of reaction: addition ; colour change: colourless to orange
C.type of reaction: substitution ; colour change: orange to colourless
D.type of reaction: substitution ; colour change: colourless to orange
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Worked solution
Ethene is an unsaturated hydrocarbon containing a carbon-carbon double bond (\(\text{C}=\text{C}\)). It undergoes an addition reaction with bromine water, forming 1,2-dibromoethane. This causes the aqueous bromine to decolourise, changing from orange (or reddish-brown) to colourless.
Marking scheme
A is correct [1]. Ethene undergoes addition across the double bond causing the orange bromine solution to become colourless. B gives the reversed colour change. C and D incorrectly identify the reaction as substitution (which is typical of alkanes in the presence of UV light).
Question 3 · Multiple Choice
1 marks
A plant is exposed to increasing light intensity while the carbon dioxide concentration is kept constant at a low level and temperature is maintained at \(20\text{ }^\circ\text{C}\). Which statement describes what happens to the rate of photosynthesis as light intensity continues to rise?
A.It decreases continuously as light intensity increases.
B.It increases continuously at a constant rate.
C.It increases initially and then remains constant.
D.It remains constant at all light intensities.
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Worked solution
At low light intensities, light is the limiting factor, so the rate of photosynthesis increases as light intensity increases. However, because carbon dioxide concentration is fixed at a low level, carbon dioxide becomes the limiting factor at higher light intensities, causing the rate to plateau and remain constant.
Marking scheme
C is correct [1]. Rate increases when light is limiting, then plateaus when carbon dioxide concentration becomes the limiting factor. A is incorrect because rate does not drop. B is incorrect because rate cannot increase indefinitely without sufficient \(\text{CO}_2\). D is incorrect because rate is not constant at low light levels.
Question 4 · Multiple Choice
1 marks
A \(4.0\,\Omega\) resistor and a \(6.0\,\Omega\) resistor are connected in parallel across a \(12\text{ V}\) power supply of negligible internal resistance. What is the total current supplied by the power supply?
A.\(1.2\text{ A}\)
B.\(2.0\text{ A}\)
C.\(5.0\text{ A}\)
D.\(10\text{ A}\)
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Worked solution
The potential difference across each parallel branch is \(12\text{ V}\). Current through the \(4.0\,\Omega\) resistor: \(I_1 = \frac{V}{R_1} = \frac{12\text{ V}}{4.0\,\Omega} = 3.0\text{ A}\). Current through the \(6.0\,\Omega\) resistor: \(I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{6.0\,\Omega} = 2.0\text{ A}\). Total current \(I = I_1 + I_2 = 3.0\text{ A} + 2.0\text{ A} = 5.0\text{ A}\).
Marking scheme
C is correct [1]. Total current \(I = \frac{12}{4.0} + \frac{12}{6.0} = 3.0 + 2.0 = 5.0\text{ A}\). A incorrectly treats the resistors as in series (\(12 / 10 = 1.2\text{ A}\)). B is the current through the \(6.0\,\Omega\) branch only. D adds the resistance values.
Question 5 · Multiple Choice
1 marks
Which statement correctly describes the energy changes involved in bond breaking and bond making during a chemical reaction?
A.Bond breaking is endothermic and bond making is endothermic.
B.Bond breaking is endothermic and bond making is exothermic.
C.Bond breaking is exothermic and bond making is endothermic.
D.Bond breaking is exothermic and bond making is exothermic.
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Worked solution
Bond breaking requires an input of energy to overcome the attractive forces between atoms, so it is an endothermic process. Bond making releases energy as stable bonds are formed, so it is an exothermic process.
Marking scheme
B is correct [1]. Bond breaking is endothermic (requires energy) and bond making is exothermic (releases energy). A, C, and D contain incorrect classifications of one or both processes.
Question 6 · Multiple Choice
1 marks
A block of mass \(2.5\text{ kg}\) is pushed along a rough horizontal floor by a constant horizontal forward force of \(15\text{ N}\). A constant frictional force of \(5.0\text{ N}\) opposes the motion. What is the acceleration of the block?
A.\(2.0\text{ m/s}^2\)
B.\(4.0\text{ m/s}^2\)
C.\(6.0\text{ m/s}^2\)
D.\(8.0\text{ m/s}^2\)
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Worked solution
First, calculate the resultant force \(F\) acting on the block: \(F = 15\text{ N} - 5.0\text{ N} = 10\text{ N}\). Then, apply Newton's second law \(F = ma\): \(a = \frac{F}{m} = \frac{10\text{ N}}{2.5\text{ kg}} = 4.0\text{ m/s}^2\).
Marking scheme
B [1]
Question 7 · Multiple Choice
1 marks
A \(12\text{ V}\) power supply is connected across two resistors in parallel. The resistances of the two resistors are \(6.0\ \Omega\) and \(12\ \Omega\). What is the total current drawn from the power supply?
A.\(0.67\text{ A}\)
B.\(1.5\text{ A}\)
C.\(3.0\text{ A}\)
D.\(4.0\text{ A}\)
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Worked solution
In a parallel circuit, each branch experiences the full potential difference of \(12\text{ V}\). Current through the \(6.0\ \Omega\) resistor: \(I_1 = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). Current through the \(12\ \Omega\) resistor: \(I_2 = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}\). Total current \(I = I_1 + I_2 = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A}\).
Marking scheme
C [1]
Question 8 · Multiple Choice
1 marks
Which method and set of reagents is most suitable for preparing pure, dry crystals of the soluble salt zinc sulfate?
A.Titrating aqueous sodium hydroxide against dilute sulfuric acid, then adding zinc metal
B.Adding excess zinc carbonate to dilute sulfuric acid, filtering, and crystallising the filtrate
C.Mixing aqueous zinc chloride with dilute sulfuric acid and collecting the precipitate by filtration
D.Adding excess copper(II) sulfate solution to zinc powder, filtering, and evaporating to dryness
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Worked solution
Zinc sulfate is a soluble salt made from an insoluble base/carbonate and an acid. An excess of insoluble zinc carbonate is reacted with dilute sulfuric acid until effervescence ceases. The excess unreacted zinc carbonate is removed by filtration, leaving an aqueous zinc sulfate solution that is gently heated to the point of crystallisation and allowed to cool to form crystals.
Marking scheme
B [1]
Question 9 · Multiple Choice
1 marks
Which row correctly identifies the leaf tissue containing the highest density of chloroplasts and the process by which carbon dioxide reaches these cells from the air spaces?
A.palisade mesophyll | diffusion
B.palisade mesophyll | active transport
C.spongy mesophyll | diffusion
D.spongy mesophyll | active transport
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Worked solution
Palisade mesophyll cells are situated near the upper surface of the leaf and are densely packed with chloroplasts to maximize light absorption for photosynthesis. Carbon dioxide moves from the intercellular air spaces into the mesophyll cells down a concentration gradient via diffusion.
Marking scheme
A [1]
Question 10 · Multiple Choice
1 marks
Ethene reacts with steam at high temperature and pressure in the presence of a phosphoric acid catalyst to produce ethanol. What type of chemical reaction is this, and what is the chemical formula of ethanol?
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Worked solution
The reaction of an alkene (ethene, \(\text{C}_2\text{H}_4\)) with steam (\(\text{H}_2\text{O}\)) is an addition reaction because the carbon-carbon double bond breaks and the elements of water add across the double bond to form a single product. The formula of ethanol is \(\text{C}_2\text{H}_5\text{OH}\).
Marking scheme
A [1]
Question 11 · Multiple Choice
1 marks
A student places a piece of pondweed in a beaker of water and records the number of bubbles released per minute as a lamp is positioned at different distances from the beaker.
Which statement correctly describes and explains the effect of increasing the distance between the lamp and the pondweed?
A.The number of bubbles increases because light intensity increases.
B.The number of bubbles decreases because light intensity decreases.
C.The number of bubbles increases because the temperature of the water decreases.
D.The number of bubbles decreases because carbon dioxide becomes more concentrated.
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Worked solution
As the distance between the lamp and the pondweed increases, light intensity decreases. Because light intensity is a limiting factor for photosynthesis under these conditions, the rate of photosynthesis decreases, resulting in fewer oxygen bubbles released per minute.
Marking scheme
B [1 mark]
Question 12 · Multiple Choice
1 marks
The reaction between hydrogen and chlorine gas is exothermic:
Which statement correctly explains why this reaction is exothermic?
A.More bonds are broken than are formed during the reaction.
B.More energy is absorbed to break bonds than is released when new bonds form.
C.More energy is released when new bonds form than is absorbed to break bonds.
D.The total energy of the reactants is less than the total energy of the products.
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Worked solution
A chemical reaction is exothermic if the energy released during bond formation in the products is greater than the energy absorbed to break the chemical bonds in the reactants (\(\Delta H < 0\)).
Marking scheme
C [1 mark]
Question 13 · Multiple Choice
1 marks
A trolley of mass \(4.0\text{ kg}\) is pulled along a horizontal runway by a horizontal forward force of \(14\text{ N}\).
A constant frictional force of \(6.0\text{ N}\) opposes the motion.
What is the acceleration of the trolley?
A.\(0.50\text{ m/s}^2\)
B.\(2.0\text{ m/s}^2\)
C.\(3.5\text{ m/s}^2\)
D.\(5.0\text{ m/s}^2\)
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Worked solution
First calculate the resultant force \(F_{\text{resultant}} = 14\text{ N} - 6.0\text{ N} = 8.0\text{ N}\). Then use Newton's second law: \(a = \frac{F}{m} = \frac{8.0\text{ N}}{4.0\text{ kg}} = 2.0\text{ m/s}^2\).
Marking scheme
B [1 mark]
Question 14 · Multiple Choice
1 marks
Two resistors of resistance \(6.0\,\Omega\) and \(12\,\Omega\) are connected in parallel across a \(12\text{ V}\) power supply.
What is the total current drawn from the supply?
A.\(0.67\text{ A}\)
B.\(1.0\text{ A}\)
C.\(2.0\text{ A}\)
D.\(3.0\text{ A}\)
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Worked solution
The combined resistance \(R_p\) of the two parallel resistors is given by: \[\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4.0\,\Omega} \implies R_p = 4.0\,\Omega\] The total current is calculated using Ohm's law: \[I = \frac{V}{R_p} = \frac{12\text{ V}}{4.0\,\Omega} = 3.0\text{ A}\]
Marking scheme
D [1 mark]
Question 15 · Multiple Choice
1 marks
Two tests are carried out on an aqueous solution of compound \(\text{X}\):
1. When aqueous sodium hydroxide is added, a green precipitate forms which is insoluble in excess. 2. When dilute nitric acid and aqueous barium nitrate are added, a white precipitate forms.
What is compound \(\text{X}\)?
A.copper(II) sulfate
B.iron(II) sulfate
C.iron(III) chloride
D.iron(II) chloride
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Worked solution
The green precipitate with aqueous sodium hydroxide indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). The white precipitate formed with aqueous barium nitrate in the presence of dilute nitric acid indicates the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, compound \(\text{X}\) is iron(II) sulfate.
Marking scheme
B [1 mark]
Question 16 · multiple_choice
1 marks
A motorized toy cart of mass \(0.80\text{ kg}\) accelerates uniformly from rest to a velocity of \(6.0\text{ m/s}\) in a time of \(4.0\text{ s}\).
What is the resultant force acting on the cart?
A.\(0.60\text{ N}\)
B.\(1.2\text{ N}\)
C.\(2.4\text{ N}\)
D.\(4.8\text{ N}\)
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Worked solution
Step 1: Calculate the acceleration \(a\) of the cart. \[a = \frac{v - u}{t} = \frac{6.0\text{ m/s} - 0\text{ m/s}}{4.0\text{ s}} = 1.5\text{ m/s}^2\]
Step 2: Use Newton's second law \(F = ma\) to find the resultant force. \[F = 0.80\text{ kg} \times 1.5\text{ m/s}^2 = 1.2\text{ N}\]
Therefore, the correct option is B.
Marking scheme
B [1 mark]
Question 17 · multiple_choice
1 marks
Ethene undergoes addition reactions with aqueous bromine and with steam under suitable conditions.
Which row correctly identifies what is observed with aqueous bromine and the product formed with steam?
A.Bromine colour turns orange to colourless; steam reaction produces ethanol
B.Bromine colour turns orange to colourless; steam reaction produces ethane
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Worked solution
Ethene is an unsaturated hydrocarbon containing a carbon-carbon double bond (\(\text{C=C}\)). 1. It rapidly decolourises aqueous bromine from orange (or brown-red) to colourless by an addition reaction. 2. When ethene reacts with steam in the presence of an acid catalyst, an addition reaction occurs to form ethanol (\(\text{C}_2\text{H}_5\text{OH}\)).
Thus, row A is correct.
Marking scheme
A [1 mark]
Question 18 · multiple_choice
1 marks
A destarched potted plant has one of its leaves sealed inside a transparent flask containing potassium hydroxide solution, which absorbs carbon dioxide. Another leaf is sealed in a flask containing water. The plant is placed in bright sunlight for 6 hours.
Both leaves are then tested for starch using iodine solution.
What is the expected result and explanation for the green parts of the leaf kept with potassium hydroxide solution?
A.It turns blue-black because photosynthesis has produced starch.
B.It turns blue-black because chlorophyll reacts directly with iodine.
C.It remains brown because carbon dioxide was not available for photosynthesis.
D.It remains brown because the transparent flask absorbed all light energy.
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Worked solution
Potassium hydroxide absorbs carbon dioxide from the air inside the flask. Without carbon dioxide, the leaf cannot carry out photosynthesis to produce glucose and store it as starch. When tested with iodine solution, the leaf lacks starch, so the iodine solution remains yellow-brown (or brown) rather than turning blue-black.
Therefore, option C is correct.
Marking scheme
C [1 mark]
Question 19 · multiple_choice
1 marks
Two resistors with resistances of \(6.0\,\Omega\) and \(12\,\Omega\) are connected in parallel across a \(12\text{ V}\) power supply.
What is the total current supplied by the power supply?
A.\(0.67\text{ A}\)
B.\(1.5\text{ A}\)
C.\(3.0\text{ A}\)
D.\(18\text{ A}\)
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Four different oxides are tested by adding samples to separate test-tubes of dilute hydrochloric acid and aqueous sodium hydroxide.
Which oxide reacts with both dilute hydrochloric acid and aqueous sodium hydroxide?
A.aluminium oxide
B.calcium oxide
C.carbon dioxide
D.copper(II) oxide
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Worked solution
An oxide that reacts with both acids and bases is an amphoteric oxide. Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is an amphoteric oxide.
- Calcium oxide and copper(II) oxide are basic oxides and only react with acids. - Carbon dioxide is an acidic oxide and only reacts with bases.
Hence, the correct answer is A.
Marking scheme
A [1 mark]
Question 21 · Multiple Choice
1 marks
A plant is kept in an environment with a constant temperature of \( 20\,^\circ\text{C} \) and a fixed, low concentration of carbon dioxide. The light intensity is gradually increased from very low to very high.
What describes the effect on the rate of photosynthesis, and what is the limiting factor at high light intensity?
A.The rate increases and then levels off; the limiting factor is carbon dioxide concentration.
B.The rate increases and then levels off; the limiting factor is light intensity.
C.The rate increases continuously; the limiting factor is carbon dioxide concentration.
D.The rate increases continuously; the limiting factor is light intensity.
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Worked solution
At low light intensities, light intensity is the limiting factor, so increasing light increases the rate of photosynthesis. However, as light intensity continues to increase, the rate reaches a plateau (levels off) because another factor is in short supply. Since the carbon dioxide concentration is fixed at a low level, carbon dioxide becomes the factor that limits any further increase in the rate of photosynthesis.
Marking scheme
A is correct [1]; carbon dioxide is the factor in shortest supply at high light intensity, causing the rate to level off.
Question 22 · Multiple Choice
1 marks
One molecule of decane, \( \text{C}_{10}\text{H}_{22} \), undergoes catalytic cracking to form one molecule of propene (\( \text{C}_3\text{H}_6 \)), one molecule of ethene (\( \text{C}_2\text{H}_4 \)), and one molecule of compound \( \text{X} \).
What is the formula of compound \( \text{X} \), and what is observed when aqueous bromine is added to compound \( \text{X} \)?
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Worked solution
To find compound \( \text{X} \), balance the carbon and hydrogen atoms from the cracking equation: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_2\text{H}_4 + \text{X} \) Number of C atoms in \( \text{X} = 10 - (3 + 2) = 5 \). Number of H atoms in \( \text{X} = 22 - (6 + 4) = 12 \). Thus, \( \text{X} \) is pentane (\( \text{C}_5\text{H}_{12} \)), which is an alkane (saturated hydrocarbon). Saturated hydrocarbons do not readily react with aqueous bromine without UV light, so the aqueous bromine remains orange/brown.
Marking scheme
D is correct [1]; cracking of decane gives pentane (\( \text{C}_5\text{H}_{12} \)), which does not decolourise aqueous bromine.
Question 23 · Multiple Choice
1 marks
A constant resultant force of \( 8.0\,\text{N} \) acts on a trolley of mass \( 4.0\,\text{kg} \) initially at rest on a frictionless horizontal track. The force acts while the trolley moves a distance of \( 9.0\,\text{m} \).
What is the kinetic energy and the final speed of the trolley after moving \( 9.0\,\text{m} \)?
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Worked solution
Work done on the trolley = \( \text{Force} \times \text{distance} = 8.0\,\text{N} \times 9.0\,\text{m} = 72\,\text{J} \). Since the track is frictionless and the trolley starts from rest, all work done is transferred to kinetic energy, so \( E_k = 72\,\text{J} \). Using \( E_k = \frac{1}{2}mv^2 \): \( 72 = \frac{1}{2} \times 4.0 \times v^2 = 2.0 v^2 \) \( v^2 = 36 \implies v = 6.0\,\text{m/s} \).
Marking scheme
B is correct [1]; \( W = F \times d = 72\,\text{J} \) and \( v = \sqrt{2E_k/m} = \sqrt{144/4} = 6.0\,\text{m/s} \).
Question 24 · Multiple Choice
1 marks
A student intends to prepare a pure, dry sample of the insoluble salt barium sulfate.
Which pair of solutions should be mixed together, and which technique is used to separate the solid barium sulfate from the reaction mixture?
A.aqueous barium chloride and aqueous sodium sulfate; filtration
B.aqueous barium chloride and aqueous sodium sulfate; crystallisation
C.solid barium oxide and dilute sulfuric acid; filtration
D.solid barium carbonate and dilute sulfuric acid; crystallisation
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Worked solution
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation, which requires mixing two soluble solutions (e.g., aqueous barium chloride and aqueous sodium sulfate). The insoluble precipitate of barium sulfate is then collected by filtration, washed with distilled water to remove soluble impurities, and dried.
Marking scheme
A is correct [1]; precipitation of an insoluble salt requires two aqueous solutions, and the solid product is separated by filtration.
Question 25 · Multiple Choice
1 marks
A \( 12\,\text{V} \) power supply of negligible internal resistance is connected to a combination of resistors. Two \( 6.0\,\Omega \) resistors are connected in parallel with each other. This parallel combination is connected in series with a \( 3.0\,\Omega \) resistor.
What is the total resistance of the circuit and the total current supplied by the power supply?
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Worked solution
First, calculate the equivalent resistance of the two \( 6.0\,\Omega \) resistors in parallel: \( \frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} = \frac{1}{3.0} \implies R_p = 3.0\,\Omega \). Next, add the series resistor: \( R_{\text{total}} = R_p + 3.0\,\Omega = 3.0 + 3.0 = 6.0\,\Omega \). Then, calculate the current using Ohm's law: \( I = \frac{V}{R_{\text{total}}} = \frac{12\,\text{V}}{6.0\,\Omega} = 2.0\,\text{A} \).
Marking scheme
A is correct [1]; parallel combination \( = 3.0\,\Omega \), total resistance \( = 6.0\,\Omega \), and current \( = 12/6.0 = 2.0\,\text{A} \).
Question 26 · Multiple Choice
1 marks
An object of mass \(4.0\text{ kg}\) accelerates uniformly from rest to a speed of \(6.0\text{ m/s}\) in a time of \(3.0\text{ s}\). What is the resultant force acting on the object during this time?
A.\(0.50\text{ N}\)
B.\(2.0\text{ N}\)
C.\(8.0\text{ N}\)
D.\(24\text{ N}\)
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Worked solution
First, calculate the acceleration using \(a = \frac{v - u}{t} = \frac{6.0\text{ m/s} - 0\text{ m/s}}{3.0\text{ s}} = 2.0\text{ m/s}^2\).
Then, use Newton's second law: \(F = ma = 4.0\text{ kg} \times 2.0\text{ m/s}^2 = 8.0\text{ N}\).
Marking scheme
C [1]
Question 27 · Multiple Choice
1 marks
Decane, \(\text{C}_{10}\text{H}_{22}\), undergoes catalytic cracking to form two molecules of ethene, \(\text{C}_2\text{H}_4\), and one molecule of an unknown hydrocarbon \(\text{Y}\).
What is the molecular formula of hydrocarbon \(\text{Y}\)?
A.\(\text{C}_6\text{H}_{12}\)
B.\(\text{C}_6\text{H}_{14}\)
C.\(\text{C}_8\text{H}_{14}\)
D.\(\text{C}_8\text{H}_{18}\)
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Worked solution
The cracking equation is: \(\text{C}_{10}\text{H}_{22} \rightarrow 2\text{C}_2\text{H}_4 + \text{Y}\).
Therefore, the formula of \(\text{Y}\) is \(\text{C}_6\text{H}_{14}\).
Marking scheme
B [1]
Question 28 · Multiple Choice
1 marks
Which statement correctly describes how the palisade mesophyll layer of a dicotyledonous leaf is adapted for its function?
A.It consists of closely packed vertical cells containing many chloroplasts to maximise light absorption.
B.It contains loosely arranged cells with large air spaces to maximise the rate of gas diffusion.
C.It consists of transparent cells without chloroplasts to reduce water loss from the leaf.
D.It contains hollow, lignified vessels to transport sucrose and amino acids away from the leaf.
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Worked solution
Palisade mesophyll cells are located near the upper surface of the leaf and are packed closely together vertically, containing many chloroplasts to maximise the absorption of light for photosynthesis.
Marking scheme
A [1]
Question 29 · Multiple Choice
1 marks
A \(12\text{ V}\) power supply is connected across two resistors connected in parallel. One resistor has a resistance of \(6.0\ \Omega\) and the other has a resistance of \(12\ \Omega\).
What is the total current supplied by the power source?
A.\(0.67\text{ A}\)
B.\(1.5\text{ A}\)
C.\(3.0\text{ A}\)
D.\(18\text{ A}\)
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Worked solution
Method 1: Find the current through each branch. \(I_1 = \frac{V}{R_1} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\) \(I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}\) \(I_{\text{total}} = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A}\).
Which method is used to prepare a pure, dry sample of barium sulfate?
A.Add excess barium carbonate to dilute sulfuric acid, filter, and crystallise the filtrate.
B.Mix aqueous barium chloride with dilute sulfuric acid, filter off the precipitate, wash, and dry.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid using an indicator, then evaporate the solution.
D.React barium metal with dilute sulfuric acid, filter off the remaining metal, and evaporate the filtrate to dryness.
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Worked solution
Barium sulfate is an insoluble salt, so it must be prepared by precipitation (mixing two soluble salt solutions, e.g. aqueous barium chloride and dilute sulfuric acid/aqueous sodium sulfate). The insoluble precipitate is then filtered off, washed with distilled water to remove soluble impurities, and dried.
Marking scheme
B [1]
Question 31 · Multiple Choice
1 marks
A trolley of mass \( 0.80\text{ kg} \) is pulled across a smooth horizontal surface by a horizontal force of \( 3.6\text{ N} \). A constant frictional force of \( 1.2\text{ N} \) opposes the motion.
What is the acceleration of the trolley?
A.\( 1.5\text{ m/s}^2 \)
B.\( 3.0\text{ m/s}^2 \)
C.\( 4.5\text{ m/s}^2 \)
D.\( 6.0\text{ m/s}^2 \)
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Worked solution
First calculate the resultant horizontal force acting on the trolley: \[ F_{\text{resultant}} = 3.6\text{ N} - 1.2\text{ N} = 2.4\text{ N} \]
Using Newton's second law: \[ F = ma \implies a = \frac{F}{m} = \frac{2.4\text{ N}}{0.80\text{ kg}} = 3.0\text{ m/s}^2 \]
Marking scheme
B [1]
Question 32 · Multiple Choice
1 marks
Which statement about endothermic chemical reactions is correct?
A.Bond breaking releases energy, and bond making takes in energy.
B.More energy is absorbed to break bonds than is released when new bonds form.
C.The temperature of the reaction mixture increases as the reaction proceeds.
D.The total energy of the products is less than the total energy of the reactants.
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Worked solution
In an endothermic reaction, heat energy is taken in from the surroundings. This occurs because the energy absorbed to break existing chemical bonds in the reactants is greater than the energy released when new bonds are formed in the products.
Marking scheme
B [1]
Question 33 · Multiple Choice
1 marks
A variegated plant is destarched and then placed in bright sunlight for six hours. A leaf from the plant is tested for starch using iodine solution.
Which row correctly shows the colour of the white area and the green area of the leaf after testing?
A.White area: yellow-brown; Green area: blue-black
B.White area: blue-black; Green area: yellow-brown
C.White area: yellow-brown; Green area: yellow-brown
D.White area: blue-black; Green area: blue-black
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Worked solution
The green area contains chlorophyll, enabling photosynthesis and the synthesis of starch; thus it turns blue-black when tested with iodine solution. The white area lacks chlorophyll, so photosynthesis cannot take place and no starch is produced; the iodine solution remains yellow-brown.
Marking scheme
A [1]
Question 34 · Multiple Choice
1 marks
Two identical resistors, each of resistance \( R \), are connected in parallel across a \( 12\text{ V} \) power supply. The total current drawn from the supply is \( 3.0\text{ A} \).
What is the resistance \( R \) of each resistor?
A.\( 2.0\ \Omega \)
B.\( 4.0\ \Omega \)
C.\( 8.0\ \Omega \)
D.\( 16\ \Omega \)
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Worked solution
The total combined resistance \( R_{\text{total}} \) of the parallel circuit is: \[ R_{\text{total}} = \frac{V}{I} = \frac{12\text{ V}}{3.0\text{ A}} = 4.0\ \Omega \]
For two identical resistors connected in parallel: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R = 2 \times R_{\text{total}} = 2 \times 4.0\ \Omega = 8.0\ \Omega \]
Marking scheme
C [1]
Question 35 · Multiple Choice
1 marks
An aqueous solution of compound \( \text{X} \) is tested: - Dilute nitric acid and aqueous silver nitrate are added to a sample of \( \text{X} \); a yellow precipitate forms. - Aqueous sodium hydroxide is added to another sample of \( \text{X} \); a green precipitate forms that is insoluble in excess.
What is compound \( \text{X} \)?
A.copper(II) chloride
B.iron(II) iodide
C.iron(III) bromide
D.iron(II) chloride
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Worked solution
A yellow precipitate formed with dilute nitric acid and aqueous silver nitrate confirms the presence of iodide ions (\( \text{I}^- \)). A green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (\( \text{Fe}^{2+} \)). Therefore, compound \( \text{X} \) is iron(II) iodide.
Marking scheme
B [1]
Question 36 · multiple-choice
1 marks
An electric winch lifts a load of mass \(40\text{ kg}\) vertically upwards through a height of \(15\text{ m}\) in \(12\text{ s}\). The gravitational field strength \(g = 10\text{ N/kg}\). The electrical power supplied to the winch is \(600\text{ W}\). What is the overall efficiency of the winch system?
A.12%
B.50%
C.83%
D.120%
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Worked solution
First, determine the useful work done against gravity: \(\text{Work} = \Delta E_p = mgh = 40\text{ kg} \times 10\text{ N/kg} \times 15\text{ m} = 6000\text{ J}\).
Next, calculate the useful power output: \(P_{\text{out}} = \frac{\text{Work}}{\text{time}} = \frac{6000\text{ J}}{12\text{ s}} = 500\text{ W}\).
C [1] (Calculation of useful output power = 500 W, efficiency = (500/600) * 100% = 83%)
Question 37 · multiple-choice
1 marks
Which row correctly identifies the tissue in a dicotyledonous leaf with the greatest concentration of chloroplasts, and the main mechanism by which carbon dioxide reaches these cells from the atmosphere?
A.Palisade mesophyll; diffusion through open stomata and intercellular spaces
B.Palisade mesophyll; active transport across the waxy cuticle
C.Spongy mesophyll; diffusion through the waxy cuticle
D.Spongy mesophyll; active transport through open stomata
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Worked solution
The palisade mesophyll layer contains the highest density of chloroplasts to maximise light absorption near the upper surface of the leaf. Carbon dioxide enters through the stomata and travels through the intercellular air spaces of the spongy mesophyll to the palisade cells by diffusion down a concentration gradient. Active transport and movement through the waxy cuticle are incorrect.
Marking scheme
A [1] (Palisade mesophyll has highest chloroplast density; CO2 moves by diffusion through stomata and air spaces)
Question 38 · multiple-choice
1 marks
Compound X is a liquid hydrocarbon. When shaken with aqueous bromine at room temperature in the absence of light, the mixture rapidly decolourises. What could be the identity of compound X?
A.ethane
B.hex-1-ene
C.hexane
D.poly(ethene)
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Worked solution
Alkenes are unsaturated hydrocarbons containing a \(\text{C=C}\) double bond that react rapidly by addition with aqueous bromine, decolourising it without requiring light. Hex-1-ene is a liquid alkene at room temperature. Ethane and hexane are alkanes, which require ultraviolet light for substitution. Poly(ethene) is a solid polymer that does not contain reactive \(\text{C=C}\) bonds.
Marking scheme
B [1] (Hex-1-ene is a liquid unsaturated hydrocarbon/alkene that undergoes rapid addition with aqueous bromine)
Question 39 · multiple-choice
1 marks
A narrow ray of monochromatic light travels inside a solid glass block towards an interface with air. The critical angle for this glass-air boundary is \(42^\circ\). The angle of incidence of the ray at the boundary is \(50^\circ\). Which statement correctly describes what happens to the ray?
A.Total internal reflection occurs with an angle of reflection of \(50^\circ\).
B.Total internal reflection occurs with an angle of reflection of \(40^\circ\).
C.The ray refracts into the air along the normal.
D.The ray refracts into the air at an angle of refraction greater than \(50^\circ\).
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Worked solution
Total internal reflection occurs when light travels from an optically denser medium to a less dense medium and the angle of incidence exceeds the critical angle. Here, the angle of incidence \(i = 50^\circ\) is greater than the critical angle \(c = 42^\circ\). Therefore, no refraction into the air occurs; the light undergoes total internal reflection back into the glass, obeying the law of reflection (angle of reflection \(r = 50^\circ\)).
Marking scheme
A [1] (Angle of incidence > critical angle so total internal reflection occurs at angle of reflection = 50°)
Question 40 · multiple-choice
1 marks
A student prepares a pure, dry sample of hydrated copper(II) sulfate crystals using solid copper(II) oxide and dilute sulfuric acid. Which sequence gives the correct experimental procedure?
A.Add excess acid to copper(II) oxide \(\rightarrow\) filter \(\rightarrow\) evaporate all water to complete dryness
B.Mix stoichiometric amounts using an indicator \(\rightarrow\) boil to dryness \(\rightarrow\) dry in an oven
C.Filter cold sulfuric acid \(\rightarrow\) add copper(II) oxide until pH is 1 \(\rightarrow\) dry by heating strongly
D.Add excess copper(II) oxide to warm acid \(\rightarrow\) filter unreacted solid \(\rightarrow\) heat filtrate to crystallisation point \(\rightarrow\) cool, filter, and dry crystals
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Worked solution
Copper(II) oxide is an insoluble base. To ensure all acid is reacted, excess copper(II) oxide is added to warm dilute sulfuric acid. The excess unreacted oxide is removed by filtration. The filtrate (aqueous copper(II) sulfate) is heated until the crystallisation point is reached, then left to cool to form crystals. The crystals are then separated by filtration and dried between sheets of filter paper.
Marking scheme
D [1] (Correct sequence for preparation of soluble salt from insoluble base: excess base -> filter -> crystallise -> dry)
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Answer all questions in the spaces provided. Show all working and state appropriate units for numerical answers.
9 Question · 80.45 marks
Question 1 · structured
8.89 marks
A motorized drone of mass \(1.5\text{ kg}\) accelerates uniformly from rest vertically upwards.
(a) State the formula that relates acceleration, change in velocity, and time taken. [1]
(b) The drone reaches a vertical speed of \(12.0\text{ m/s}\) in a time of \(4.0\text{ s}\). (i) Calculate the acceleration of the drone. State the unit. [2] (ii) Calculate the kinetic energy of the drone when travelling at \(12.0\text{ m/s}\). [2]
(c) The gravitational field strength \(g\) is \(9.8\text{ N/kg}\). (i) Calculate the weight of the drone. [1] (ii) Calculate the upward thrust force exerted by the drone's motors during this acceleration. [3]
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Worked solution
(a) Acceleration is given by \(a = \frac{\Delta v}{t} = \frac{v - u}{t}\).
Hydrocarbons from petroleum are separated and converted into useful chemical products.
(a) State the definition of a hydrocarbon. [1]
(b) Decane, \(\text{C}_{10}\text{H}_{22}\), is cracked at high temperature in the presence of a catalyst to form ethene, \(\text{C}_2\text{H}_4\), and another hydrocarbon \(\text{X}\). \(\text{C}_{10}\text{H}_{22} \rightarrow 2\text{C}_2\text{H}_4 + \text{X}\) (i) Deduce the molecular formula of hydrocarbon \(\text{X}\). [1] (ii) State two essential conditions required for catalytic cracking. [2]
(c) Ethene reacts with aqueous bromine in an addition reaction. (i) State the colour change observed when ethene is bubbled through aqueous bromine. [1] (ii) Draw the displayed formula of the organic product formed in this reaction and state its name. [2]
(d) Write a balanced chemical equation for the complete combustion of propane, \(\text{C}_3\text{H}_8\). [2]
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Worked solution
(a) A hydrocarbon is a compound consisting solely of hydrogen and carbon atoms.
(b)(i) Subtracting \(2 \times \text{C}_2\text{H}_4 = \text{C}_4\text{H}_8\) from \(\text{C}_{10}\text{H}_{22}\): Number of C = \(10 - 4 = 6\) Number of H = \(22 - 8 = 14\) Formula of \(\text{X}\) is \(\text{C}_6\text{H}_{14}\).
(b)(ii) Conditions: High temperature (approx. \(450\text{--}750\,^\circ\text{C}\)) and an aluminium oxide / silicon dioxide / zeolite catalyst.
(c)(i) Orange/brown to colourless (bromine water is decolourised).
(c)(ii) 1,2-dibromoethane: each carbon is bonded to one bromine atom and two hydrogen atoms via single covalent bonds.
(c)(ii) Correct displayed formula showing all single bonds: \(\text{H}-\text{C}(\text{H})(\text{Br})-\text{C}(\text{H})(\text{Br})-\text{H}\); [1] 1,2-dibromoethane / dibromoethane; [1]
Green plants produce carbohydrates by photosynthesis and transport them throughout their structure.
(a) Complete the balanced chemical equation for photosynthesis. \(\dots\dots \text{CO}_2 + \dots\dots \text{H}_2\text{O} \xrightarrow{\text{light, chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + \dots\dots \text{O}_2\) [2]
(b) Describe two structural adaptations of the palisade mesophyll layer that maximize the rate of photosynthesis. [2]
(c) Plants take up mineral ions from the soil through root hair cells. (i) Name the mineral ion required by plants to synthesize chlorophyll. [1] (ii) Root hair cells absorb nitrate ions from low concentrations in soil water by active transport. Explain why root hair cells need a constant supply of oxygen to absorb these ions. [3]
(d) State the name of the specialized plant tissue that transports sucrose and amino acids. [1]
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(b) Palisade mesophyll adaptations: 1. Positioned near the upper surface of the leaf to receive maximum sunlight. 2. Cells are columnar and tightly packed with high numbers of chloroplasts per cell.
(c)(i) Magnesium ions (\(\text{Mg}^{2+}\)) are needed to form the chlorophyll molecule.
(c)(ii) Active transport moves mineral ions against their concentration gradient, which requires energy in the form of ATP. ATP is generated through aerobic cellular respiration, a metabolic process that depends directly on the availability of oxygen.
(d) Phloem tissue conducts sucrose and amino acids (translocation).
(b) Any two from: - Packed closely/tightly together vertically; [1] - Near upper surface / receives maximum light; [1] - Contain large numbers of / densely packed chloroplasts; [1] (max [2])
(c)(i) Magnesium (ion) / \(\text{Mg}^{2+}\); [1]
(c)(ii) Active transport moves ions against a concentration gradient; [1] Requires energy / ATP; [1] Energy/ATP is provided by aerobic respiration which requires oxygen; [1]
(d) Phloem; [1]
Question 4 · structured
8.89 marks
A circuit contains a \(12.0\text{ V}\) battery of negligible internal resistance connected to a resistor network. The network consists of a \(16.0\ \Omega\) resistor connected in series with a parallel combination of a \(20.0\ \Omega\) resistor and a \(30.0\ \Omega\) resistor.
(a) State Ohm's law in words or as an equation. [1]
(b) Calculate: (i) the combined resistance of the two resistors in parallel [2] (ii) the total resistance of the complete circuit [1] (iii) the total current supplied by the battery [2]
(c) Calculate the electrical power dissipated in the \(16.0\ \Omega\) series resistor. State the unit. [3]
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Worked solution
(a) \(V = IR\), or potential difference is directly proportional to current at constant temperature.
(c) Formula \(P = I^2 R\) or \(P = V_{\text{resistor}} \times I\); [1] \((0.429)^2 \times 16.0 = 2.94\) (allow ecf from (b)(iii)); [1] W / Watts / \(\text{J/s}\); [1]
Question 5 · structured
8.89 marks
A student prepares a pure, dry sample of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\), using insoluble copper(II) carbonate powder and dilute sulfuric acid.
(a) Write a balanced chemical equation, including state symbols, for the reaction between solid copper(II) carbonate and aqueous sulfuric acid. [2]
(b) Describe the practical method to obtain pure, dry crystals of copper(II) sulfate from the reaction mixture after adding copper(II) carbonate to warm dilute sulfuric acid. [4]
(c) The student conducts chemical tests on an aqueous sample of the prepared copper(II) sulfate. (i) Describe the observations when aqueous ammonia is added dropwise until in excess to a portion of the copper(II) sulfate solution. [2] (ii) Describe the chemical test to confirm the presence of sulfate ions in the solution, including the reagent and the expected observation. [1]
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Worked solution
(a) Solid copper(II) carbonate reacts with dilute sulfuric acid to give aqueous copper(II) sulfate, carbon dioxide gas, and liquid water: \(\text{CuCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\).
(b) Method: 1. Add excess \(\text{CuCO}_3\) to ensure all acid is neutralized. 2. Filter off the excess unreacted solid using filter paper and funnel. 3. Heat the filtrate gently in an evaporating basin to the crystallisation point (until saturated). 4. Allow the saturated solution to cool and crystallise slowly. 5. Filter the crystals and dry them carefully between sheets of filter paper (or in a low-temperature desiccator/oven).
(c)(i) With aqueous ammonia, a light blue precipitate of \(\text{Cu(OH)}_2\) forms initially. Upon adding excess ammonia, the precipitate dissolves to produce a deep blue solution.
(c)(ii) Acidify with dilute nitric acid (or dilute hydrochloric acid) and add aqueous barium nitrate (or barium chloride). A white precipitate of barium sulfate confirms the sulfate ion.
(b) Add excess \(\text{CuCO}_3\) (until no more dissolves / fizzes); [1] Filter to remove unreacted / excess solid; [1] Heat / evaporate filtrate to crystallisation point / saturation; [1] Leave to cool/crystallise AND dry crystals with filter paper / in warm oven; [1]
(c)(i) Light blue precipitate; [1] Dissolves in excess (ammonia) to give a deep blue solution; [1]
(c)(ii) Add dilute nitric acid / hydrochloric acid and barium nitrate / chloride, giving a white precipitate; [1]
Question 6 · short-answer
9 marks
A student investigates photosynthesis in an aquatic plant, Elodea.
(a) Write the balanced chemical equation for photosynthesis. [2]
(b) Describe two structural features of a leaf that maximize light absorption and gas exchange. [2]
(c) The student places the aquatic plant at various distances from a lamp and counts the number of gas bubbles released per minute. Table 1.1 shows the results.
Table 1.1
| distance from lamp / cm | rate of bubbling / bubbles per minute | | :--- | :--- | | 10 | 64 | | 20 | 32 | | 30 | 16 | | 40 | 8 |
(i) State the relationship between the distance from the lamp and the rate of bubbling shown in Table 1.1. [1]
(ii) Explain the results in Table 1.1 using ideas about light intensity and limiting factors. [2]
(iii) Explain why the student placed a glass beaker of water between the lamp and the aquatic plant during the experiment. [2]
[Total: 9]
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Worked solution
(a) The balanced chemical equation is: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
(b) 1. Palisade mesophyll layer contains a high density of chloroplasts and is positioned near the top of the leaf to absorb maximum sunlight. 2. Spongy mesophyll layer contains large intercellular air spaces / stomata allow diffusion of carbon dioxide into the leaf and oxygen out.
(c)(i) As the distance from the lamp increases, the rate of bubbling decreases (an inverse relationship).
(c)(ii) As distance increases, light intensity decreases. At these distances, light intensity is the limiting factor, so the rate of photosynthesis decreases as light intensity decreases.
(c)(iii) The glass beaker of water acts as a heat shield/filter to absorb infrared/thermal energy from the lamp. This ensures temperature remains constant, preventing temperature from becoming an uncontrolled variable that affects the rate.
Marking scheme
(a) correct formulae for all reactants and products \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\) [1]; correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) [1]
(b) any two from: - palisade cells / chloroplasts packed tightly near upper surface (for maximum light absorption) [1]; - broad / large surface area / thin leaf (for light absorption / short diffusion pathway) [1]; - stomata / guard cells / spongy mesophyll air spaces (for gas exchange / diffusion of \(\text{CO}_2\) / \(\text{O}_2\)) [1]
(c)(i) as distance increases, the rate of bubbling / photosynthesis decreases / inversely proportional [1]
(c)(ii) as distance increases, light intensity decreases [1]; light intensity is the limiting factor (for the rate of photosynthesis) [1]
(c)(iii) absorbs heat / thermal energy / infrared radiation (from the lamp) [1]; to keep temperature constant / prevent temperature from affecting enzyme activity / rate [1]
Question 7 · short-answer
9 marks
Zinc metal reacts with dilute sulfuric acid according to the following equation:
(a) State the type of chemical reaction occurring when zinc displaces hydrogen from sulfuric acid. [1]
(b) The temperature of the reaction mixture was measured before and after the reaction. The temperature increased by \(4.5\,^{\circ}\text{C}\).
(i) State whether this reaction is exothermic or endothermic. Give a reason for your answer. [1]
(ii) In terms of bond breaking and bond making, explain why this reaction produces an increase in temperature. [2]
(c) A student investigates the rate of this reaction by measuring the volume of hydrogen gas collected over time using two different concentrations of sulfuric acid: \(0.50\text{ mol/dm}^3\) and \(1.0\text{ mol/dm}^3\).
All other variables are kept constant.
(i) State which concentration of sulfuric acid produces a faster initial rate of reaction. [1]
(ii) Explain your answer to (c)(i) using ideas about particles and collision theory. [4]
[Total: 9]
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Worked solution
(a) The reaction is a displacement (or redox) reaction.
(b)(i) The reaction is exothermic because thermal energy is transferred to the surroundings, causing the temperature of the reaction mixture to rise.
(b)(ii) Bond breaking is endothermic (requires energy) and bond making is exothermic (releases energy). The energy released in forming bonds in the products (\(\text{ZnSO}_4\) and \(\text{H}_2\)) is greater than the energy absorbed to break bonds in the reactants (\(\text{H}_2\text{SO}_4\)).
(c)(i) \(1.0\text{ mol/dm}^3\) sulfuric acid gives a faster initial rate.
(c)(ii) A higher concentration contains more acid particles / \(\text{H}^+\) ions in a given volume. Therefore, particles are closer together, leading to a higher collision frequency (more frequent collisions between reactant particles). Consequently, there is a greater frequency of successful collisions (collisions with energy greater than or equal to the activation energy), resulting in a faster rate of reaction.
Marking scheme
(a) displacement / redox [1]
(b)(i) exothermic because thermal energy is released / temperature increases [1]
(b)(ii) bond breaking absorbs / requires energy AND bond making releases energy [1]; energy released in forming bonds is greater than energy needed to break bonds [1]
(c)(i) \(1.0\text{ mol/dm}^3\) [1]
(c)(ii) more particles / \(\text{H}^+\) ions per unit volume / per \(\text{cm}^3\) / per \(\text{dm}^3\) [1]; particles collide more frequently / collision frequency increases [1]; more particles possess energy \(\ge\) activation energy [1]; rate of successful / effective collisions increases [1]
Question 8 · short-answer
9 marks
A robotic rover of mass \(6.0\text{ kg}\) moves along a horizontal surface.
Fig. 3.1 shows the speed-time graph for the rover over a \(30\text{ s}\) period.
* From \(t = 0\text{ s}\) to \(t = 6.0\text{ s}\), speed increases steadily from \(0\text{ m/s}\) to \(3.0\text{ m/s}\). * From \(t = 6.0\text{ s}\) to \(t = 24.0\text{ s}\), speed remains constant at \(3.0\text{ m/s}\). * From \(t = 24.0\text{ s}\) to \(t = 30.0\text{ s}\), speed decreases steadily from \(3.0\text{ m/s}\) to \(0\text{ m/s}\).
(a) (i) Calculate the acceleration of the rover during the first \(6.0\text{ s}\). [2]
(ii) Calculate the resultant force acting on the rover during the first \(6.0\text{ s}\). [2]
(b) Calculate the total distance travelled by the rover in the \(30\text{ s}\). [3]
(c) Calculate the kinetic energy of the rover when it is travelling at its maximum speed. State the unit of your answer. [2]
[Total: 9]
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(a)(ii) \(F = ma\) OR \(6.0 \times 0.50\) [1]; \(3.0\text{ N}\) (allow ecf from (a)(i)) [1]
(b) evidence of determining area under the graph [1]; correct substitution: \(\frac{1}{2}(30 + 18) \times 3.0\) OR \(9.0 + 54.0 + 9.0\) [1]; \(72\text{ m}\) (or \(72.0\text{ m}\)) [1]
(c) \(E_k = \frac{1}{2}mv^2\) OR \(\frac{1}{2} \times 6.0 \times (3.0)^2\) [1]; \(27\) AND \(\text{J}\) / joules [1]
Question 9 · short-answer
9 marks
An electric monitoring circuit contains a \(12.0\text{ V}\) power supply connected in series with a fixed resistor \(R_1 = 40.0\,\Omega\) and a parallel combination. The parallel combination consists of a fixed resistor \(R_2 = 60.0\,\Omega\) and a light-dependent resistor (LDR).
(a) In bright light, the resistance of the LDR is \(30.0\,\Omega\).
Calculate the combined resistance of the parallel combination of \(R_2\) and the LDR. [2]
(b) (i) Calculate the total resistance of the entire circuit in bright light. [1]
(ii) Calculate the current leaving the \(12.0\text{ V}\) power supply in bright light. State the unit of your answer. [2]
(c) The circuit is moved into a dark room.
(i) State what happens to the resistance of the LDR. [1]
(ii) Explain what happens to the potential difference across resistor \(R_1\) when the circuit is in the dark room. [3]
[Total: 9]
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(c)(i) In the dark, the resistance of the LDR increases.
(c)(ii) As the resistance of the LDR increases, the combined resistance of the parallel branch and thus the total resistance of the whole circuit increases. Since the supply voltage is constant, the total current in the circuit decreases. Because \(V = IR\) and \(R_1\) is fixed, the smaller current causes the potential difference across \(R_1\) to decrease.
(b)(i) \(60\,\Omega\) / \(60.0\,\Omega\) (allow ecf from (a)) [1]
(b)(ii) \(I = \frac{V}{R}\) OR \(\frac{12.0}{60.0}\) [1]; \(0.20\) AND \(\text{A}\) / amperes (allow ecf from (b)(i)) [1]
(c)(i) (resistance) increases [1]
(c)(ii) total circuit resistance increases [1]; current (in circuit / through \(R_1\)) decreases [1]; potential difference across \(R_1\) decreases (as \(V = IR\)) [1]
Paper 6 Alternative to Practical
Answer all practical questions including graph plotting, table completion, and a 7-mark experimental plan.
4 Question · 40 marks
Question 1 · Practical Skills & Planning
10 marks
A student investigates the requirement of light for starch production in leaves.
(a) The student uses a plant that has been kept in a dark cupboard for 48 hours to destarch its leaves.
(i) Explain why the plant is destarched before the investigation begins. [1]
(ii) One leaf on the plant is partially covered with black paper as shown in Fig. 1.1. The plant is then placed in bright sunlight for 6 hours.
Describe the steps involved in testing this leaf for the presence of starch. State the reason for each step and include a safety precaution. [4]
(b) The student carries out a second experiment to investigate the effect of light intensity on the rate of photosynthesis in an aquatic plant (pondweed). The number of gas bubbles released by the pondweed in 2 minutes is counted at different distances from a light source. The results are shown in Table 1.1.
Table 1.1 | Distance from lamp / cm | Number of bubbles in trial 1 | Number of bubbles in trial 2 | Number of bubbles in trial 3 | Mean number of bubbles in 2 min | | :--- | :--- | :--- | :--- | :--- | | 10 | 84 | 86 | 85 | 85 | | 20 | 58 | 56 | 57 | 57 | | 30 | 36 | 21 | 38 | .............. | | 40 | 22 | 24 | 23 | 23 | | 50 | 11 | 13 | 12 | 12 |
(i) Identify the anomalous result in Table 1.1. [1]
(ii) Calculate the mean number of bubbles in 2 minutes for a distance of 30 cm, excluding the anomalous result. Write your answer in Table 1.1. [1]
(iii) Calculate the mean rate of bubble production per minute when the distance is 10 cm. [1]
(iv) State two variables that must be kept constant during this experiment. [2]
[Total: 10]
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Worked solution
(a)(i) Destarching ensures that no starch is present in the leaf at the start of the experiment, so any starch found at the end was synthesized during the illumination period. (ii) Step 1: Place leaf in boiling water for 1 minute to break cell membranes / kill cells. Step 2: Place leaf in boiling ethanol using a water bath (safety: ethanol is flammable, so use an electric water bath or turn off the Bunsen burner) to dissolve and extract chlorophyll. Step 3: Dip the leaf into warm water to soften it. Step 4: Spread the leaf on a white tile and add drops of iodine solution. (b)(i) The anomalous result is 21 at 30 cm (trial 2) as it is significantly lower than trials 1 (36) and 3 (38). (ii) \(\text{Mean} = \frac{36 + 38}{2} = \frac{74}{2} = 37\). (iii) \(\text{Rate} = \frac{85}{2} = 42.5\text{ bubbles/min}\). (iv) Controlled variables include: temperature of the water, concentration of dissolved carbon dioxide / sodium hydrogencarbonate, length/mass/species of pondweed, wavelength/type of light bulb.
Marking scheme
(a)(i) to ensure any starch present was made during the investigation / leaf starts with no starch [1] (ii) boil leaf in water (to kill/soften cells / break membranes) [1]; boil in ethanol to remove chlorophyll / decolourise leaf [1]; use a water bath / extinguish Bunsen flame because ethanol is flammable [1]; add iodine solution (and observe colour change / yellow-brown to blue-black) [1] (b)(i) 21 (at 30 cm / trial 2) [1] (ii) 37 [1] (iii) 42.5 (accept 43) [1] (iv) any two from: temperature of water; volume/concentration of sodium hydrogencarbonate solution; species/mass/length of pondweed; colour/type of light [2]
Question 2 · Practical Skills & Planning
10 marks
A student carries out tests to identify an unknown solid salt \(\mathbf{X}\) and investigates the thermal energy change when \(\mathbf{X}\) dissolves in water.
(a) Solid \(\mathbf{X}\) is dissolved in distilled water to make solution \(\mathbf{S}\). Solution \(\mathbf{S}\) is divided into three test-tubes.
(i) To the first test-tube, the student adds aqueous sodium hydroxide drop by drop until in excess. A green precipitate forms which is insoluble in excess. Identify the cation present in solid \(\mathbf{X}\). [1]
(ii) To the second test-tube, the student adds dilute nitric acid followed by aqueous barium nitrate. A white precipitate forms. Identify the anion present in solid \(\mathbf{X}\). [1]
(iii) To the third test-tube, the student adds dilute hydrochloric acid. No effervescence is seen. Gas evolved is tested with damp blue litmus paper and it does not bleach. Explain why this test confirms the absence of carbonate and sulfite ions. [1]
(iv) Give the chemical formula of solid \(\mathbf{X}\). [1]
(b) The student determines the temperature change when different masses of solid \(\mathbf{X}\) are dissolved in \(25.0\text{ cm}^3\) of distilled water.
Table 2.1 shows the results.
Table 2.1 | Mass of solid \(\mathbf{X}\) / g | Initial temperature / °C | Final temperature / °C | Temperature change \(\Delta T\) / °C | | :--- | :--- | :--- | :--- | | 1.0 | 20.5 | 18.5 | -2.0 | | 2.0 | 20.5 | 16.5 | -4.0 | | 3.0 | 20.5 | 14.5 | -6.0 | | 4.0 | 20.5 | 12.5 | -8.0 | | 5.0 | 20.5 | 10.5 | .............. |
(i) Complete Table 2.1 by calculating the temperature change for \(5.0\text{ g}\) of solid \(\mathbf{X}\). [1]
(ii) State whether the dissolving of solid \(\mathbf{X}\) is exothermic or endothermic. Give a reason for your choice. [1]
(iii) State the name of a suitable piece of apparatus to measure the \(25.0\text{ cm}^3\) volume of distilled water accurately. [1]
(iv) State two improvements to the apparatus to reduce heat exchange with the surroundings during this experiment. [2]
(v) Predict the temperature change \(\Delta T\) if \(3.5\text{ g}\) of solid \(\mathbf{X}\) is dissolved in \(25.0\text{ cm}^3\) of distilled water. [1]
[Total: 10]
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Worked solution
(a)(i) Green precipitate insoluble in excess aqueous sodium hydroxide indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (ii) Addition of dilute nitric acid followed by aqueous barium nitrate giving a white precipitate indicates sulfate ions, \(\text{SO}_4^{2-}\). (iii) Carbonates react with acid to produce carbon dioxide (effervescence), and sulfites produce sulfur dioxide gas; neither was observed. (iv) Combining \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\) gives \(\text{FeSO}_4\). (b)(i) \(\Delta T = \text{Final} - \text{Initial} = 10.5 - 20.5 = -10.0\text{ }^\circ\text{C}\). (ii) Endothermic because temperature decreases (energy taken in from solution). (iii) A \(25\text{ cm}^3\) volumetric pipette, burette, or measuring cylinder. (iv) 1. Use an insulated container (e.g. polystyrene/expanded plastic cup). 2. Add a lid/cover to minimize heat gain from the surroundings. (v) Since \(\Delta T = -2.0\text{ }^\circ\text{C}\text{ per gram}\), for \(3.5\text{ g}\): \(\Delta T = 3.5 \times (-2.0) = -7.0\text{ }^\circ\text{C}\).
Marking scheme
(a)(i) iron(II) / \(\text{Fe}^{2+}\) [1] (ii) sulfate / \(\text{SO}_4^{2-}\) [1] (iii) no gas/effervescence produced (so no \(\text{CO}_2\) / no carbonate) AND no \(\text{SO}_2\) produced (no sulfite) [1] (iv) \(\text{FeSO}_4\) [1] (b)(i) -10.0 (or 10.0 decrease) [1] (ii) endothermic AND temperature decreases / heat absorbed [1] (iii) measuring cylinder / (volumetric) pipette / burette [1] (iv) use a polystyrene cup / lagged beaker [1]; use a lid / cover [1] (v) -7.0 (°C) (allow 7.0 decrease) [1]
Question 3 · Practical Skills & Planning
10 marks
A student investigates the resistance of different lengths of a resistance wire \(\mathbf{W}\).
The student sets up a circuit containing a power source, an ammeter, a switch, and a length of wire \(\mathbf{W}\) mounted on a metre rule. A voltmeter is connected across the length \(L\) of the wire using a sliding contact (jockey).
(a) Draw a circuit diagram to show this experimental arrangement. Use standard electrical circuit symbols. [2]
(b) The student closes the switch and records the current \(I\) and potential difference \(V\) across various lengths \(L\) of the wire. Table 3.1 shows the results.
(i) Calculate the resistance \(R\) for \(L = 80.0\text{ cm}\) using the formula \(R = \frac{V}{I}\). Write your answer in Table 3.1. [1]
(ii) State the relationship between the length \(L\) of the wire and its resistance \(R\). Justify your answer using data from Table 3.1. [2]
(iii) Calculate the resistance per centimetre of wire \(\mathbf{W}\). Give the unit of your answer. [2]
(c) (i) Explain why the student should switch off the circuit between taking readings. [1]
(ii) State one precaution the student should take when positioning the jockey to ensure that the length measurement is accurate. [1]
(iii) Suggest one modification to the experiment that would allow the student to test whether the resistance is affected by the diameter of the wire. [1]
[Total: 10]
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Worked solution
(a) A complete circuit diagram should show: power supply (cell/battery), switch, and ammeter all in series with the resistance wire; voltmeter connected in parallel across the measured length \(L\) between 0 cm and the sliding contact. (b)(i) \(R = \frac{V}{I} = \frac{2.56}{0.40} = 6.40\text{ }\Omega\). (ii) The resistance \(R\) is directly proportional to length \(L\). Evidence: when length doubles from \(20.0\text{ cm}\) to \(40.0\text{ cm}\), resistance doubles from \(1.60\text{ }\Omega\) to \(3.20\text{ }\Omega\) (or \(\frac{R}{L} = 0.080\text{ }\Omega/\text{cm} = \text{constant}\)). (iii) \(\text{Resistance per cm} = \frac{R}{L} = \frac{1.60}{20.0} = 0.080\text{ }\Omega/\text{cm}\). (c)(i) Passing current through the wire produces heating (Joule heating). Heating increases resistance, which would introduce systematic error; turning off the switch allows the wire to cool down. (ii) View the scale perpendicularly to avoid parallax error / ensure the jockey is held vertically at the exact mark on the metre rule. (iii) Keep the length and material of the wire constant, but repeat the measurements with wires of different diameters / SWG gauges.
Marking scheme
(a) series circuit with power supply, ammeter, and resistance wire [1]; voltmeter connected in parallel across length \(L\) of the wire [1] (b)(i) 6.40 (or 6.4) [1] (ii) directly proportional [1]; justification with numerical pair showing ratio \(R/L\) is constant / doubling \(L\) doubles \(R\) [1] (iii) 0.080 (or 0.08) [1]; \(\Omega/\text{cm}\) (or \(\Omega\text{ cm}^{-1}\) or \(\Omega/\text{m}\) if converted to 8.0) [1] (c)(i) to prevent heating of the wire / to keep temperature constant [1] (ii) view ruler perpendicularly / avoid parallax error / ensure contact is firm without bending the wire [1] (iii) keep material and length constant, use wires of different diameters / thicknesses [1]
Question 4 · Practical Skills & Planning
10 marks
A student investigates the effect of sucrose concentration on the mass of potato cylinders.
Five potato cylinders of identical diameter are prepared using a cork borer and trimmed to an initial length of \(40\text{ mm}\). Each cylinder is weighed on an electronic balance and placed into a test-tube containing a different concentration of sucrose solution for \(60\text{ minutes}\).
After \(60\text{ minutes}\), the cylinders are removed, gently blotted with a paper towel, and re-weighed.
(a) Complete Table 1.1 by calculating: (i) the percentage change in mass for the cylinder in \(0.6\text{ mol}/\text{dm}^{3}\) sucrose solution. Give your answer to 1 decimal place. [1] (ii) the change in mass for the cylinder in \(0.8\text{ mol}/\text{dm}^{3}\) sucrose solution. [1]
(b) Explain why the student blotted the potato cylinders dry before weighing them. [1]
(c) State two variables that were kept constant in this investigation. [2]
(d) (i) Identify one source of experimental error in cutting or preparing the potato cylinders. [1] (ii) Suggest one improvement to make the results more reliable. [1]
(e) Describe how the student could test the initial potato tissue for the presence of reducing sugars, stating the reagent used and the positive result observed. [3]
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Worked solution
(a) (i) Percentage change \(= \frac{-0.15}{2.45} \times 100 = -6.122\% \approx -6.1\%\). (ii) Change in mass \(= 2.24 - 2.55 = -0.31\text{ g}\).
(b) Blotting removes liquid adhering to the outer surface of the cylinder so that only the internal water uptake or loss is measured.
(c) Constant variables include: temperature of solutions, immersion time (60 min), diameter/volume of cylinders, volume of sucrose solution used.
(d) (i) Cutting cylinders by hand may leave uneven ends or variations in surface area; presence of potato skin on some cylinders. (ii) Repeat trials at each concentration and calculate the mean percentage change to identify and discard anomalies.
(e) Cut/crush a piece of potato and place in a test-tube. Add Benedict's reagent. Heat in a water bath at \(80^{\circ}\text{C}\) or above for \(3\text{--}5\text{ minutes}\). A colour change from blue to green/yellow/orange/brick-red indicates the presence of reducing sugars.
Marking scheme
(a)(i) \(-6.1(\%)\) ; [1] (a)(ii) \(-0.31(\text{ g})\) (minus sign required) ; [1] (b) To remove surface liquid / ensure extra solution is not weighed / liquid on surface would increase mass ; [1] (c) Any two from: initial length / diameter / volume of cylinders ; immersion time / duration (60 min) ; temperature of solutions ; volume of sucrose solution ; [2] (d)(i) Inaccurate cutting / uneven lengths / varying surface area / varying thickness / skin present on cylinder ; [1] (d)(ii) Repeat the experiment (at each concentration) AND calculate a mean / average (exclude anomalies) ; [1] (e) Add Benedict's solution / reagent ; heat in a hot water bath / water bath above \(70^{\circ}\text{C}\) / boil ; colour change from blue to green / yellow / orange / red / brick-red ; [3]
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