An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 Multiple Choice
Answer all forty multiple choice questions. Choose the best response (A, B, C or D).
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
A toy car of mass 0.5 kg travels along a straight path. It starts from rest, accelerates uniformly to a velocity of 6 m/s in 4 seconds, travels at this constant velocity for 4 seconds, and then decelerates uniformly to rest in 2 seconds. What is the total distance travelled by the toy car?
A.24 m
B.30 m
C.42 m
D.60 m
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PastPaper.workedSolution
The total distance travelled can be found by calculating the area under the velocity-time graph.
- Phase 1 (0 to 4 s): Uniform acceleration from 0 to 6 m/s. \(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 6 = 12\text{ m}\).
- Phase 3 (8 to 10 s): Uniform deceleration from 6 m/s to 0. \(\text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 6 = 6\text{ m}\).
Total distance = \(12 + 24 + 6 = 42\text{ m}\).
PastPaper.markingScheme
1 mark for the correct calculation of total distance under the v-t graph, leading to 42 m (C).
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
An object of mass \(2.0\text{ kg}\) is dropped from a balcony. When it has fallen through a vertical height of \(5.0\text{ m}\), what is its kinetic energy, assuming there is no air resistance? (Take the acceleration of free fall, \(g\), to be \(10\text{ m/s}^2\).)
A.\(10\text{ J}\)
B.\(20\text{ J}\)
C.\(100\text{ J}\)
D.\(200\text{ J}\)
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PastPaper.workedSolution
By the principle of conservation of energy, the gain in kinetic energy (KE) is equal to the loss in gravitational potential energy (GPE) when air resistance is neglected.
\(\text{Loss in GPE} = mgh\) \(\text{Loss in GPE} = 2.0\text{ kg} \times 10\text{ m/s}^2 \times 5.0\text{ m} = 100\text{ J}\)
Since no energy is lost to the surroundings, the gain in kinetic energy is exactly \(100\text{ J}\).
PastPaper.markingScheme
1 mark for calculating the loss of GPE as 100 J and equating it to the gain in KE, leading to 100 J (C).
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
Which method is most suitable for preparing a pure, dry sample of the insoluble salt, barium sulfate?
A.Add excess barium carbonate to dilute sulfuric acid, filter, evaporate the filtrate to the crystallization point, and dry the crystals.
B.Mix aqueous barium chloride with dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry it.
C.Mix aqueous barium hydroxide with dilute hydrochloric acid, and evaporate the solution to dryness.
D.Heat barium metal with sulfur in a crucible until the reaction is complete.
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PastPaper.workedSolution
Barium sulfate is an insoluble salt. The standard method for preparing insoluble salts is precipitation. This involves mixing two soluble solutions (aqueous barium chloride and dilute sulfuric acid) to form the insoluble salt precipitate. The mixture is then filtered to isolate the precipitate, which is washed with distilled water to remove any remaining soluble ions and then dried.
PastPaper.markingScheme
1 mark for identifying that mixing soluble reactants to precipitate, filter, wash, and dry is the correct method for preparing an insoluble salt like barium sulfate (B).
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
An oxide of element \(X\) dissolves in water to form a solution with a pH of 2. Which statement about element \(X\) and its oxide is correct?
A.Element \(X\) is a metal and its oxide is basic.
B.Element \(X\) is a non-metal and its oxide is basic.
C.Element \(X\) is a metal and its oxide is acidic.
D.Element \(X\) is a non-metal and its oxide is acidic.
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PastPaper.workedSolution
A pH of 2 represents a strongly acidic solution. Acidic oxides are formed by non-metals (such as carbon, sulfur, or nitrogen) when they react with oxygen. In contrast, metal oxides are generally basic or amphoteric. Therefore, element \(X\) must be a non-metal and its oxide is acidic.
PastPaper.markingScheme
1 mark for identifying that a low pH indicates an acidic oxide, which is formed by a non-metal (D).
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
A uniform metal wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0\ \Omega\). A second wire made of the same metal has a length of \(2L\) and a cross-sectional area of \(2A\). What is the resistance of the second wire?
A.\(2.0\ \Omega\)
B.\(4.0\ \Omega\)
C.\(8.0\ \Omega\)
D.\(16.0\ \Omega\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The resistance \(R\) of a wire is given by the formula: \(R = \rho \frac{L}{A}\) where \(\rho\) is the resistivity of the metal.
For the second wire: \(R_2 = \rho \frac{2L}{2A} = \rho \frac{L}{A} = R = 8.0\ \Omega\).
Thus, the resistance remains unchanged at \(8.0\ \Omega\).
PastPaper.markingScheme
1 mark for using the resistance-dimensions relationship to find that doubling both length and area leaves the resistance unchanged at 8.0 ohms (C).
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
Three identical resistors, each of resistance \(R\), are connected in parallel. What is the total resistance of this combination?
A.\(\frac{1}{3}R\)
B.\(\frac{2}{3}R\)
C.\(R\)
D.\(3R\)
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PastPaper.workedSolution
For resistors in parallel, the reciprocal of the equivalent resistance \(R_{total}\) is the sum of the reciprocals of each individual resistance:
Taking the reciprocal of both sides gives: \(R_{total} = \frac{1}{3}R\).
PastPaper.markingScheme
1 mark for using the parallel resistance formula to show that the equivalent resistance is one-third of R (A).
PastPaper.question 7 · multiple-choice
1 PastPaper.marks
A student tests an aqueous solution of an unknown salt. - Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. - Adding dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of the unknown salt?
A.copper(II) chloride
B.iron(II) sulfate
C.iron(III) sulfate
D.nickel(II) chloride
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PastPaper.workedSolution
- The reaction with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess, which is characteristic of iron(II) ions, \(\text{Fe}^{2+}\). - The reaction with dilute nitric acid followed by aqueous barium nitrate produces a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, the unknown salt is iron(II) sulfate.
PastPaper.markingScheme
1 mark for correctly matching the positive cation test to iron(II) and the positive anion test to sulfate, identifying the salt as iron(II) sulfate (B).
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
A student runs a paper chromatography experiment to analyze the pigments in a sample of ink. The solvent front travels \(8.0\text{ cm}\) from the baseline. One of the separated dyes travels \(5.6\text{ cm}\) from the baseline. What is the \(R_f\) value of this dye?
A.\(0.70\)
B.\(1.43\)
C.\(2.40\)
D.\(5.60\)
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PastPaper.workedSolution
The retention factor (\(R_f\)) is calculated using the formula:
\(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}\)
Note that \(R_f\) values are dimensionless ratios and are always less than or equal to 1.00.
PastPaper.markingScheme
1 mark for correctly calculating the retention factor as the ratio of substance distance to solvent distance, resulting in 0.70 (A).
PastPaper.question 9 · MCQ
1 PastPaper.marks
An object of mass \(4.0\text{ kg}\) is released from rest from a height of \(5.0\text{ m}\). Assuming there is no air resistance and the acceleration due to gravity is \(9.8\text{ m/s}^2\), what is the kinetic energy of the object just before it hits the ground?
A.20 J
B.98 J
C.196 J
D.392 J
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PastPaper.workedSolution
The gravitational potential energy lost by the object is converted entirely into kinetic energy just before hitting the ground. Using the formula for gravitational potential energy: \(E_p = mgh = 4.0\text{ kg} \times 9.8\text{ m/s}^2 \times 5.0\text{ m} = 196\text{ J}\). Therefore, the kinetic energy is also \(196\text{ J}\).
PastPaper.markingScheme
Award 1 mark for the correct option C. Deduct 0 marks for incorrect attempts.
PastPaper.question 10 · MCQ
1 PastPaper.marks
A student is given an unknown white solid X. Solid X reacts with dilute hydrochloric acid to produce a gas that turns limewater milky. When aqueous sodium hydroxide is added to a solution of X, a white precipitate forms which is soluble in excess sodium hydroxide to form a colorless solution. What is the identity of X?
A.Calcium carbonate
B.Zinc carbonate
C.Copper(II) carbonate
D.Zinc chloride
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PastPaper.workedSolution
The reaction of X with acid producing a gas that turns limewater milky indicates that X is a carbonate. When aqueous sodium hydroxide is added to a solution containing zinc ions, a white precipitate of zinc hydroxide forms, which dissolves in excess sodium hydroxide to yield a colorless solution. Calcium carbonate forms a white precipitate with sodium hydroxide that is insoluble in excess. Thus, X is zinc carbonate.
PastPaper.markingScheme
Award 1 mark for the correct option B.
PastPaper.question 11 · MCQ
1 PastPaper.marks
Three identical resistors, each of resistance \(R\), are connected in a circuit. Two of the resistors are connected in parallel, and this parallel combination is connected in series with the third resistor. What is the total equivalent resistance of this network?
A.0.33 R
B.0.67 R
C.1.5 R
D.3.0 R
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PastPaper.workedSolution
The equivalent resistance of the two parallel resistors is \(R_p = \frac{R \times R}{R + R} = 0.5R\). This combination is connected in series with the third resistor, so the total equivalent resistance is \(R_{total} = R + 0.5R = 1.5R\).
PastPaper.markingScheme
Award 1 mark for the correct option C.
PastPaper.question 12 · MCQ
1 PastPaper.marks
A chromatogram of a sample of food dye is run using a polar solvent. The solvent front moves a distance of \(8.0\text{ cm}\) from the baseline. A yellow spot travels \(2.4\text{ cm}\) and a blue spot travels \(6.0\text{ cm}\) from the baseline. What are the calculated \(R_f\) values of the yellow and blue dyes?
A.yellow = 0.30; blue = 0.75
B.yellow = 0.33; blue = 1.33
C.yellow = 3.33; blue = 1.33
D.yellow = 0.75; blue = 0.30
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PastPaper.workedSolution
The retention factor \(R_f\) is calculated using the formula: \(R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}}\). For the yellow dye, \(R_f = \frac{2.4\text{ cm}}{8.0\text{ cm}} = 0.30\). For the blue dye, \(R_f = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\).
PastPaper.markingScheme
Award 1 mark for the correct option A.
PastPaper.question 13 · MCQ
1 PastPaper.marks
A car starts from rest and accelerates uniformly to a speed of \(15\text{ m/s}\) in \(5.0\text{ s}\). It then maintains this constant speed for \(10\text{ s}\) before decelerating uniformly to rest in \(3.0\text{ s}\). What is the total distance travelled by the car?
A.150 m
B.180 m
C.210 m
D.270 m
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PastPaper.workedSolution
The total distance is the total area under the speed-time graph. 1) Acceleration phase (triangle): \(\text{Area}_1 = \frac{1}{2} \times 5.0\text{ s} \times 15\text{ m/s} = 37.5\text{ m}\). 2) Constant speed phase (rectangle): \(\text{Area}_2 = 10\text{ s} \times 15\text{ m/s} = 150\text{ m}\). 3) Deceleration phase (triangle): \(\text{Area}_3 = \frac{1}{2} \times 3.0\text{ s} \times 15\text{ m/s} = 22.5\text{ m}\). Total distance = \(37.5 + 150 + 22.5 = 210\text{ m}\).
PastPaper.markingScheme
Award 1 mark for the correct option C.
PastPaper.question 14 · MCQ
1 PastPaper.marks
Which method is the most suitable for preparing a pure, dry sample of the insoluble salt, barium sulfate?
A.Reacting solid barium carbonate with dilute sulfuric acid, then evaporating the solution.
B.Mixing aqueous barium chloride with aqueous sodium sulfate, filtering the mixture, washing the residue with distilled water, and drying it.
C.Adding excess barium metal to dilute sulfuric acid, filtering, and evaporating the filtrate to dryness.
D.Titrating barium hydroxide solution with dilute sulfuric acid using an indicator, then heating the solution until it crystallizes.
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PastPaper.workedSolution
Insoluble salts like barium sulfate are prepared by precipitation. This involves mixing two soluble salt solutions (e.g., barium chloride and sodium sulfate). The insoluble barium sulfate precipitate is filtered from the mixture, washed with distilled water to remove soluble ions, and dried.
PastPaper.markingScheme
Award 1 mark for the correct option B.
PastPaper.question 15 · MCQ
1 PastPaper.marks
An ideal transformer has \(200\text{ turns}\) on its primary coil and \(600\text{ turns}\) on its secondary coil. An alternating voltage of \(12\text{ V}\) is applied across the primary coil. What is the output voltage across the secondary coil, and what type of transformer is this?
A.4.0 V, step-down
B.4.0 V, step-up
C.36 V, step-down
D.36 V, step-up
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PastPaper.workedSolution
Using the transformer equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\). Substituting the values: \(\frac{V_s}{12\text{ V}} = \frac{600}{200} = 3\), which gives \(V_s = 36\text{ V}\). Because the output voltage is larger than the input voltage, this is a step-up transformer.
PastPaper.markingScheme
Award 1 mark for the correct option D.
PastPaper.question 16 · MCQ
1 PastPaper.marks
A student performs chemical tests on an unknown aqueous salt solution Y. Adding dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. In a separate test, adding aqueous sodium hydroxide and aluminum foil, and then gently warming the mixture, produces a gas that turns damp red litmus paper blue. Which ions are present in salt Y?
A.Chloride and ammonium
B.Bromide and nitrate
C.Iodide and nitrate
D.Bromide and ammonium
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PastPaper.workedSolution
Acidified silver nitrate produces a cream precipitate in the presence of bromide (\(Br^-\)) ions. Warming a nitrate containing solution with aqueous sodium hydroxide and aluminum reduces the nitrate (\(NO_3^-\)) ions to ammonia gas, which is alkaline and turns damp red litmus paper blue. Therefore, salt Y contains bromide and nitrate ions.
PastPaper.markingScheme
Award 1 mark for the correct option B.
PastPaper.question 17 · MCQ
1 PastPaper.marks
A toy car of mass 2.0 kg is moving. A velocity-time graph shows that the car decelerates uniformly from 12 m/s to 0 m/s in 4.0 s. What is the magnitude of the average resultant force acting on the car during this deceleration?
A.1.5 N
B.3.0 N
C.6.0 N
D.24.0 N
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PastPaper.workedSolution
First, calculate the deceleration of the toy car: \(a = \frac{v - u}{t} = \frac{0\text{ m/s} - 12\text{ m/s}}{4.0\text{ s}} = -3.0\text{ m/s}^2\). The magnitude of the deceleration is \(3.0\text{ m/s}^2\). Then, use Newton's second law to find the resultant force: \(F = m \times a = 2.0\text{ kg} \times 3.0\text{ m/s}^2 = 6.0\text{ N}\).
PastPaper.markingScheme
1 mark for the correct calculation of force using \(F = m \times a\), leading to 6.0 N.
PastPaper.question 18 · MCQ
1 PastPaper.marks
An unstretched spring has a length of 12.0 cm. When a load of 4.0 N is suspended from it, the length of the spring becomes 15.0 cm. Assuming the limit of proportionality is not exceeded, what load is required to stretch the spring to a total length of 19.5 cm?
A.6.0 N
B.10.0 N
C.13.0 N
D.26.0 N
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PastPaper.workedSolution
Find the initial extension: \(\Delta x_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). The spring constant is \(k = \frac{F_1}{\Delta x_1} = \frac{4.0\text{ N}}{3.0\text{ cm}} = \frac{4}{3}\text{ N/cm}\). For a total length of 19.5 cm, the required extension is \(\Delta x_2 = 19.5\text{ cm} - 12.0\text{ cm} = 7.5\text{ cm}\). The required load is \(F_2 = k \times \Delta x_2 = \frac{4}{3}\text{ N/cm} \times 7.5\text{ cm} = 10.0\text{ N}\).
PastPaper.markingScheme
1 mark for the correct calculation of load from spring extension, leading to 10.0 N.
PastPaper.question 19 · MCQ
1 PastPaper.marks
Which sequence of experimental steps is correct for preparing pure, dry crystals of copper(II) sulfate from solid copper(II) carbonate and dilute sulfuric acid?
A.Add excess copper(II) carbonate to dilute sulfuric acid, filter to remove unreacted solid, evaporate the filtrate until the crystallization point is reached, leave to cool, filter the crystals, and dry them.
B.Add excess dilute sulfuric acid to copper(II) carbonate, filter to remove unreacted acid, evaporate the filtrate to dryness, and dry the crystals.
C.Add equal amounts of copper(II) carbonate and dilute sulfuric acid, evaporate the mixture to dryness, and wash the crystals with water.
D.Add excess copper(II) carbonate to dilute sulfuric acid, evaporate the mixture to dryness, and filter to obtain the crystals.
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PastPaper.workedSolution
To prepare a soluble salt from an acid and an insoluble base, the base must be added in excess to ensure all acid is neutralised. The excess base is removed by filtration. The filtrate containing the salt is partially evaporated to the crystallization point and left to cool. The crystals are then filtered and dried. Evaporating to dryness would produce anhydrous powder or decompose the salt rather than forming nice crystals.
PastPaper.markingScheme
1 mark for identifying the correct experimental sequence including excess reactant addition, filtration of excess solid, and partial crystallization.
PastPaper.question 20 · MCQ
1 PastPaper.marks
An oxide of element X reacts with dilute hydrochloric acid to form a salt and water. The same oxide also reacts with aqueous sodium hydroxide to form a salt and water. What type of oxide is this, and what is the nature of element X?
A.acidic oxide, non-metal
B.amphoteric oxide, metal
C.basic oxide, metal
D.neutral oxide, non-metal
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PastPaper.workedSolution
An oxide that reacts with both acids and bases to form a salt and water is an amphoteric oxide. Amphoteric oxides (such as zinc oxide or aluminium oxide) are formed by metals.
PastPaper.markingScheme
1 mark for identifying the oxide as amphoteric and the element as a metal.
PastPaper.question 21 · MCQ
1 PastPaper.marks
Three identical resistors, each of resistance \(R\), are connected in a circuit. Two of the resistors are connected in parallel, and this combination is connected in series with the third resistor. What is the total resistance of this combined network?
A.\(0.33R\)
B.\(0.67R\)
C.\(1.5R\)
D.\(3.0R\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The two resistors in parallel have a combined resistance of \(R_p = \frac{R \times R}{R + R} = 0.5R\). This parallel combination is in series with the third resistor of resistance \(R\). Thus, the total resistance is \(R_{\text{total}} = R_p + R = 0.5R + R = 1.5R\).
PastPaper.markingScheme
1 mark for calculating the combined resistance of the network as \(1.5R\).
PastPaper.question 22 · MCQ
1 PastPaper.marks
An electric heater connected to a 230 V mains supply has a power rating of 1.15 kW. What is the current in the heater, and how much charge flows through it in 5.0 minutes?
A.current = 0.20 A, charge = 60 C
B.current = 5.0 A, charge = 25 C
C.current = 5.0 A, charge = 1500 C
D.current = 250 A, charge = 75 000 C
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PastPaper.workedSolution
First, find the current using \(I = \frac{P}{V}\). Since \(P = 1.15\text{ kW} = 1150\text{ W}\) and \(V = 230\text{ V}\), \(I = \frac{1150}{230} = 5.0\text{ A}\). Next, find the charge using \(Q = I \times t\). The time is \(5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\). So, \(Q = 5.0\text{ A} \times 300\text{ s} = 1500\text{ C}\).
PastPaper.markingScheme
1 mark for correctly identifying both the current as 5.0 A and the total charge as 1500 C.
PastPaper.question 23 · MCQ
1 PastPaper.marks
A student carries out a paper chromatography experiment to analyse a sample of green food dye. The solvent front travels 8.0 cm from the baseline. One of the separated blue spots travels 6.4 cm from the baseline. What is the \(R_f\) value of this blue dye, and is chromatography classified as a chemical or a physical process?
A.\(R_f = 0.80\), chemical process
B.\(R_f = 0.80\), physical process
C.\(R_f = 1.25\), chemical process
D.\(R_f = 1.25\), physical process
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PastPaper.workedSolution
The retardation factor \(R_f\) is calculated as: \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} = \frac{6.4\text{ cm}}{8.0\text{ cm}} = 0.80\). Chromatography is a physical process because it separates components of a mixture based on physical properties (solubility and affinity) without forming any new chemical substances.
PastPaper.markingScheme
1 mark for calculating \(R_f = 0.80\) and identifying chromatography as a physical process.
PastPaper.question 24 · MCQ
1 PastPaper.marks
A student performs qualitative tests on an unknown white solid. They dissolve the solid in distilled water to make a solution, and divide it into two portions. To the first portion, they add dilute nitric acid followed by aqueous silver nitrate, and a cream precipitate forms. To the second portion, they add aqueous sodium hydroxide and warm the mixture, which produces a gas that turns damp red litmus paper blue. What is the identity of the white solid?
A.ammonium bromide
B.ammonium chloride
C.sodium bromide
D.calcium bromide
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PastPaper.workedSolution
The cream precipitate with acidified silver nitrate confirms the presence of bromide ions (\(\text{Br}^-\)). Warming with aqueous sodium hydroxide to produce an alkaline gas (ammonia, which turns damp red litmus paper blue) confirms the presence of ammonium ions (\(\text{NH}_4^+\)). Therefore, the unknown white solid is ammonium bromide.
PastPaper.markingScheme
1 mark for identifying the cation as ammonium and the anion as bromide, giving ammonium bromide.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
A spring has an unstretched length of 12.0 cm. When a load of 4.0 N is suspended from it, the length becomes 15.0 cm. Assuming the limit of proportionality is not exceeded, what is the length of the spring when a load of 10.0 N is suspended from it?
A.7.5 cm
B.19.5 cm
C.22.5 cm
D.37.5 cm
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PastPaper.workedSolution
First, find the extension caused by the 4.0 N load: \(e_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Using Hooke's Law, extension is directly proportional to load: \(e_2 = \frac{10.0\text{ N}}{4.0\text{ N}} \times 3.0\text{ cm} = 7.5\text{ cm}\). The new length is the unstretched length plus the new extension: \(12.0\text{ cm} + 7.5\text{ cm} = 19.5\text{ cm}\).
PastPaper.markingScheme
1 mark for the correct option B.
PastPaper.question 26 · multiple-choice
1 PastPaper.marks
A toy car of mass 0.5 kg is accelerated from rest to a speed of 4.0 m/s. The work done by the motor is 6.0 J. How much energy is wasted as heat and sound?
A.2.0 J
B.4.0 J
C.6.0 J
D.8.0 J
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PastPaper.workedSolution
Calculate the useful kinetic energy gained by the car: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5\text{ kg} \times (4.0\text{ m/s})^2 = 4.0\text{ J}\). The total work done (energy input) is 6.0 J. Wasted energy is the difference between total work done and useful kinetic energy gained: \(6.0\text{ J} - 4.0\text{ J} = 2.0\text{ J}\).
PastPaper.markingScheme
1 mark for the correct option A.
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
Which method is most suitable to prepare a pure, dry sample of the insoluble salt barium sulfate?
A.Add excess barium carbonate to dilute sulfuric acid, filter, and crystallise the filtrate.
B.Mix aqueous barium chloride and dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid using an indicator, then evaporate the solution to dryness.
D.Heat solid barium metal with dilute sulfuric acid, filter, and evaporate the filtrate.
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PastPaper.workedSolution
Barium sulfate is an insoluble salt, which must be prepared by a precipitation reaction using two soluble solutions. Mixing aqueous barium chloride and dilute sulfuric acid produces insoluble barium sulfate. The solid residue is then separated by filtration, washed with distilled water to remove remaining soluble impurities, and dried.
PastPaper.markingScheme
1 mark for the correct option B.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
Oxides can be classified as acidic, basic, or amphoteric. Which row correctly classifies the oxides of calcium, carbon, and zinc? Row A: Calcium oxide = basic, Carbon dioxide = acidic, Zinc oxide = amphoteric. Row B: Calcium oxide = acidic, Carbon dioxide = basic, Zinc oxide = amphoteric. Row C: Calcium oxide = basic, Carbon dioxide = amphoteric, Zinc oxide = acidic. Row D: Calcium oxide = amphoteric, Carbon dioxide = acidic, Zinc oxide = basic.
A.Row A
B.Row B
C.Row C
D.Row D
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PastPaper.workedSolution
Calcium is a metal, so its oxide is basic. Carbon is a non-metal, so its oxide is acidic. Zinc oxide is amphoteric because it can react with both acids and bases. This corresponds to the classification in Row A.
PastPaper.markingScheme
1 mark for the correct option A.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
A wire of length \(L\) and diameter \(d\) has a resistance of \(R\). What is the resistance of a wire made of the same material that has a length of \(2L\) and a diameter of \(2d\)?
A.\(0.5 R\)
B.\(R\)
C.\(2 R\)
D.\(4 R\)
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PastPaper.workedSolution
Resistance is proportional to length and inversely proportional to the cross-sectional area: \(R = \rho \frac{L}{A}\). The area is proportional to the square of the diameter: \(A \propto d^2\). Doubling the diameter increases the area by a factor of 4. Doubling the length increases resistance by a factor of 2. Combining these, the new resistance is: \(R' = R \times 2 / 4 = 0.5 R\).
PastPaper.markingScheme
1 mark for the correct option A.
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
An electric kettle is rated at \(230\text{ V}\), \(2.3\text{ kW}\). Which fuse is the most suitable for use in the plug of this kettle?
A.3 A
B.5 A
C.10 A
D.13 A
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PastPaper.workedSolution
Calculate the current flowing through the kettle under normal operation using \(I = P / V = 2300\text{ W} / 230\text{ V} = 10\text{ A}\). The fuse rating must be slightly higher than this normal operating current to prevent blowing during normal use, making the standard 13 A fuse the most suitable option.
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PastPaper.question 31 · multiple-choice
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Paper chromatography is used to analyse a mixture of food colorings. The solvent front travels 8.0 cm from the baseline. A spot of a yellow dye travels 6.0 cm from the baseline. What is the \(R_f\) value of the yellow dye?
A.0.75
B.1.33
C.2.00
D.6.00
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PastPaper.workedSolution
The retention factor is given by the formula: \(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}} = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\).
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PastPaper.question 32 · multiple-choice
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An aqueous solution of salt \(X\) is tested as follows: - Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. - Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of salt \(X\)?
A.copper(II) sulfate
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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PastPaper.workedSolution
The green precipitate insoluble in excess sodium hydroxide indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). The white precipitate formed with nitric acid and barium nitrate indicates the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, the salt is iron(II) sulfate.
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PastPaper.question 33 · MCQ
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A cart of mass \(5.0\text{ kg}\) is moving at a constant speed of \(4.0\text{ m/s}\). A constant force of \(10\text{ N}\) is applied to the cart in the direction of motion over a distance of \(6.0\text{ m}\). What is the final kinetic energy of the cart?
A.40 J
B.60 J
C.100 J
D.140 J
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PastPaper.workedSolution
The initial kinetic energy of the cart is \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 5.0\text{ kg} \times (4.0\text{ m/s})^2 = 40\text{ J}\). The work done by the applied force is \(W = F \times d = 10\text{ N} \times 6.0\text{ m} = 60\text{ J}\). The final kinetic energy is the sum of the initial kinetic energy and the work done on the cart: \(E_{f} = 40\text{ J} + 60\text{ J} = 100\text{ J}\).
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PastPaper.question 34 · MCQ
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A spring has an unstretched length of \(12.0\text{ cm}\). When a load of \(4.0\text{ N}\) is suspended from it, its length becomes \(15.0\text{ cm}\). Under a different load, the length of the spring becomes \(19.5\text{ cm}\). Assuming the limit of proportionality is not exceeded, what is the value of this second load?
A.6.0 N
B.8.0 N
C.10.0 N
D.12.0 N
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PastPaper.workedSolution
First, determine the extension for the first load: \(x_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). The spring constant \(k\) is given by \(k = \frac{F_1}{x_1} = \frac{4.0\text{ N}}{3.0\text{ cm}} = \frac{4}{3}\text{ N/cm}\). For the second load, the length is \(19.5\text{ cm}\), so the extension is \(x_2 = 19.5\text{ cm} - 12.0\text{ cm} = 7.5\text{ cm}\). The second load is \(F_2 = k \times x_2 = \frac{4}{3}\text{ N/cm} \times 7.5\text{ cm} = 10.0\text{ N}\).
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PastPaper.question 35 · MCQ
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Which pair of substances, when mixed together as aqueous solutions, will react to form a precipitate of barium sulfate?
A.barium hydroxide and dilute hydrochloric acid
B.barium chloride and sodium sulfate
C.barium carbonate and dilute sulfuric acid
D.barium nitrate and copper(II) carbonate
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PastPaper.workedSolution
To prepare an insoluble salt such as barium sulfate by precipitation, we must mix two soluble salts. Barium chloride is a soluble barium salt, and sodium sulfate is a soluble sulfate salt, so mixing them will yield insoluble barium sulfate and soluble sodium chloride. In (A), barium hydroxide and hydrochloric acid undergo neutralisation without sulfate. In (C), barium carbonate is insoluble. In (D), copper(II) carbonate is insoluble.
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PastPaper.question 36 · MCQ
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An oxide of element X dissolves in water to form a solution with a pH of 12. Which type of element is X and what is the nature of its oxide?
A.X is a metal; the oxide is acidic
B.X is a metal; the oxide is basic
C.X is a non-metal; the oxide is acidic
D.X is a non-metal; the oxide is basic
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PastPaper.workedSolution
A pH of 12 is strongly alkaline. Metal oxides are basic and dissolve in water to form alkaline solutions, whereas non-metal oxides dissolve to form acidic solutions. Therefore, element X must be a metal, and its oxide is basic.
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PastPaper.question 37 · MCQ
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Two resistors, with resistances of \(4.0\ \Omega\) and \(12.0\ \Omega\), are connected in parallel. This combination is connected in series with a \(3.0\ \Omega\) resistor and a \(9.0\text{ V}\) d.c. power supply. What is the current flowing through the \(3.0\ \Omega\) resistor?
A.0.56 A
B.0.75 A
C.1.1 A
D.1.5 A
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PastPaper.workedSolution
First, find the equivalent resistance of the parallel pair: \(1/R_p = 1/4.0 + 1/12.0 = 4/12.0 \implies R_p = 3.0\ \Omega\). Next, the total resistance of the circuit is \(R_{\text{total}} = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). The total current is \(I = V / R_{\text{total}} = 9.0\text{ V} / 6.0\ \Omega = 1.5\text{ A}\). Since the \(3.0\ \Omega\) resistor is in series with the battery and the parallel block, the total current of \(1.5\text{ A}\) flows through it.
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PastPaper.question 38 · MCQ
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A transformer has 200 turns on its primary coil and 800 turns on its secondary coil. An alternating voltage of \(12\text{ V}\) is applied to the primary coil. What is the output voltage from the secondary coil, and what type of transformer is this?
A.3.0 V, step-down
B.3.0 V, step-up
C.48 V, step-down
D.48 V, step-up
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PastPaper.workedSolution
Using the transformer equation: \(V_s / V_p = N_s / N_p\). Substituting values: \(V_s / 12\text{ V} = 800 / 200 = 4 \implies V_s = 48\text{ V}\). Since the secondary voltage is greater than the primary voltage, this is a step-up transformer.
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PastPaper.question 39 · MCQ
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A student performs paper chromatography on a sample of black ink. The solvent front travels \(12.0\text{ cm}\) from the baseline. One of the dye components in the ink travels \(9.0\text{ cm}\) from the baseline. What is the \(R_f\) value of this dye component?
A.0.33
B.0.75
C.1.33
D.3.0
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PastPaper.workedSolution
The retention factor \(R_f\) is calculated as: \(R_f = \frac{\text{distance travelled by component}}{\text{distance travelled by solvent front}} = \frac{9.0\text{ cm}}{12.0\text{ cm}} = 0.75\).
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PastPaper.question 40 · MCQ
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A student is given an unknown salt solution. She adds dilute nitric acid followed by aqueous barium nitrate. A thick white precipitate forms. Which anion is present in the solution?
A.carbonate
B.chloride
C.nitrate
D.sulfate
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PastPaper.workedSolution
Adding dilute nitric acid followed by aqueous barium nitrate is the standard qualitative test for the sulfate ion (\(\text{SO}_4^{2-}\)). The barium ions (\(\text{Ba}^{2+}\)) react with sulfate ions to form an insoluble white precipitate of barium sulfate (\(\text{BaSO}_4\)).
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Paper 4 Theory (Extended)
Answer all questions. Show your working where required and provide appropriate units.
12 PastPaper.question · 120 PastPaper.marks
PastPaper.question 1 · Structured
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A cyclist accelerates from rest to a speed of \(8.0\text{ m/s}\) in a time of \(5.0\text{ s}\). She then travels at a constant speed of \(8.0\text{ m/s}\) for \(15.0\text{ s}\). Finally, she decelerates uniformly to rest in a further \(4.0\text{ s}\). (a) Calculate the acceleration during the first \(5.0\text{ s}\). (b) Calculate the total distance traveled during the entire \(24.0\text{ s}\) journey. (c) The combined mass of the cyclist and her bicycle is \(75\text{ kg}\). Calculate the kinetic energy of the cyclist and bicycle when traveling at \(8.0\text{ m/s}\). (d) Describe the energy transfer that occurs during deceleration, assuming some kinetic energy is transferred to the brakes.
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(a) Acceleration \(a = \frac{\Delta v}{t} = \frac{8.0\text{ m/s} - 0}{5.0\text{ s}} = 1.6\text{ m/s}^2\). (b) Total distance is the area under the speed-time graph: Phase 1: \(d_1 = \frac{1}{2} \times 5.0\text{ s} \times 8.0\text{ m/s} = 20\text{ m}\). Phase 2: \(d_2 = 15.0\text{ s} \times 8.0\text{ m/s} = 120\text{ m}\). Phase 3: \(d_3 = \frac{1}{2} \times 4.0\text{ s} \times 8.0\text{ m/s} = 16\text{ m}\). Total distance \(= 20 + 120 + 16 = 156\text{ m}\). (c) Kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 75\text{ kg} \times (8.0\text{ m/s})^2 = 0.5 \times 75 \times 64 = 2400\text{ J}\). (d) Kinetic energy is transferred to thermal energy (internal energy) in the brakes and surroundings, and also a small amount of sound energy.
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(a) 1 mark for correct formula or substitution: \(8.0 / 5.0\), 1 mark for correct answer with unit: \(1.6\text{ m/s}^2\). (b) 1 mark for calculating distance of first phase (\(20\text{ m}\)), 1 mark for calculating distance of second phase (\(120\text{ m}\)), 1 mark for calculating distance of third phase (\(16\text{ m}\)), 1 mark for final correct sum: \(156\text{ m}\). (c) 1 mark for correct formula or substitution: \(0.5 \times 75 \times 8.0^2\), 1 mark for correct answer with unit: \(2400\text{ J}\) (or \(2.4\text{ kJ}\)). (d) 1 mark for identifying kinetic energy as the input energy, 1 mark for identifying thermal/internal energy as the output energy.
PastPaper.question 2 · Structured
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A crane lifts a container of mass \(1200\text{ kg}\) vertically upwards through a height of \(15\text{ m}\) at a constant speed. The lift takes a time of \(20\text{ s}\). Take the gravitational field strength \(g = 9.8\text{ N/kg}\). (a) Calculate the weight of the container. (b) Calculate the work done in lifting the container. (c) Calculate the useful power output of the crane's motor during this lift. (d) The electrical power input to the crane is \(15\text{ kW}\). Calculate the efficiency of the crane.
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PastPaper.workedSolution
(a) Weight \(W = mg = 1200\text{ kg} \times 9.8\text{ N/kg} = 11760\text{ N}\). (b) Work done \(W_d = F \times d = W \times h = 11760\text{ N} \times 15\text{ m} = 176400\text{ J}\) (or \(176.4\text{ kJ}\)). (c) Power \(P = \frac{W_d}{t} = \frac{176400\text{ J}}{20\text{ s}} = 8820\text{ W}\) (or \(8.82\text{ kW}\)). (d) Efficiency \(= \frac{\text{Useful power output}}{\text{Total power input}} \times 100\% = \frac{8820\text{ W}}{15000\text{ W}} \times 100\% = 58.8\%\) (or \(59\%\)).
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(a) 1 mark for substitution \(1200 \times 9.8\), 1 mark for correct answer: \(11760\text{ N}\) (accept \(12000\text{ N}\) if \(g=10\text{ N/kg}\) is used). (b) 1 mark for formula \(W = F \times d\), 1 mark for correct substitution, 1 mark for correct answer with unit: \(176400\text{ J}\) (or \(176.4\text{ kJ}\); accept \(180000\text{ J}\) if \(g=10\text{ N/kg}\) is used). (c) 1 mark for formula or substitution: \(176400 / 20\), 1 mark for correct answer with unit: \(8820\text{ W}\) (or \(8.82\text{ kW}\); accept \(9000\text{ W}\) if \(g=10\text{ N/kg}\) is used). (d) 1 mark for converting \(15\text{ kW}\) to \(15000\text{ W}\) (or using consistent kW), 1 mark for formula or substitution: \(8820 / 15000\), 1 mark for correct percentage: \(58.8\%\) (accept \(60\%\) if \(g=10\text{ N/kg}\) is used).
PastPaper.question 3 · Structured
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A circuit consists of a \(12\text{ V}\) d.c. power supply connected to three resistors. Resistor \(R_1\) has a resistance of \(4.0\ \Omega\) and is connected in series with a parallel combination of resistor \(R_2 = 6.0\ \Omega\) and resistor \(R_3 = 3.0\ \Omega\). (a) Calculate the combined resistance of the parallel pair of resistors, \(R_2\) and \(R_3\). (b) Calculate the total resistance of the entire circuit. (c) Calculate the total current flowing from the power supply. (d) Calculate the electrical power dissipated in the \(4.0\ \Omega\) resistor.
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PastPaper.workedSolution
(a) For a parallel combination: \(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}\), so \(R_p = 2.0\ \Omega\). (b) Since \(R_1\) is in series with the parallel pair: \(R_{\text{total}} = R_1 + R_p = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega\). (c) Total current \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). (d) Since \(R_1\) is in series, the total current of \(2.0\text{ A}\) flows through it. Power dissipated \(P = I^2 R_1 = (2.0\text{ A})^2 \times 4.0\ \Omega = 4.0 \times 4.0 = 16\text{ W}\).
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(a) 1 mark for correct formula \(1/R_p = 1/R_2 + 1/R_3\), 1 mark for substitution, 1 mark for correct calculation: \(2.0\ \Omega\). (b) 1 mark for formula \(R_{\text{total}} = R_1 + R_p\), 1 mark for correct answer: \(6.0\ \Omega\) (ecf from (a)). (c) 1 mark for formula \(I = V/R\), 1 mark for correct calculation: \(2.0\text{ A}\) (ecf from (b)). (d) 1 mark for identifying current through \(R_1\) as \(2.0\text{ A}\), 1 mark for power formula \(P = I^2 R\) or equivalent, 1 mark for correct answer: \(16\text{ W}\).
PastPaper.question 4 · Structured
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A student prepares a pure, dry sample of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), by reacting dilute sulfuric acid with excess copper(II) oxide. (a) State why copper(II) oxide is added in excess. (b) Describe how the unreacted copper(II) oxide is removed from the reaction mixture. (c) Describe the steps required to obtain pure, dry crystals of hydrated copper(II) sulfate from the filtrate. (d) Write a balanced chemical equation, including state symbols, for the reaction between copper(II) oxide and dilute sulfuric acid.
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(a) Copper(II) oxide is added in excess to ensure that all the sulfuric acid is completely neutralized/reacted, so that no unreacted acid remains in the product solution. (b) The mixture is filtered. The unreacted, insoluble copper(II) oxide remains on the filter paper as the residue, while the copper(II) sulfate solution passes through as the filtrate. (c) The filtrate is heated in an evaporating dish until it is saturated (the crystallization point). The solution is then left to cool, allowing hydrated copper(II) sulfate crystals to form. The crystals are separated from the remaining liquid by filtration, washed with a small amount of cold distilled water to remove impurities, and dried between sheets of filter paper or in a warm oven. (d) The balanced chemical equation with state symbols is: \(\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}\).
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(a) 1 mark for: to ensure all acid is reacted/neutralized. (b) 1 mark for: filtration / filtering, 1 mark for: copper(II) oxide is the residue / remains on filter paper (or copper(II) sulfate is filtrate). (c) 1 mark for: heating filtrate to crystallization point / evaporating some water, 1 mark for: leaving to cool to allow crystals to form, 1 mark for: filtering the crystals (from excess solution), 1 mark for: washing with cold distilled water AND drying with filter paper. (d) 1 mark for correct reactants: \(\text{CuO}\) and \(\text{H}_2\text{SO}_4\), 1 mark for correct products: \(\text{CuSO}_4\) and \(\text{H}_2\text{O}\), 1 mark for all correct state symbols: \(\text{(s)}\), \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(l)}\).
PastPaper.question 5 · Structured
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(a) A solid substance, **X**, is analyzed. (i) A flame test is performed on **X**. The flame turns lilac. Identify the metal ion present in **X**. (ii) An aqueous solution of **X** is reacted with dilute nitric acid followed by aqueous silver nitrate. A cream precipitate is formed. Identify the anion present in **X**. (iii) Write an ionic equation, with state symbols, for the formation of the cream precipitate. (b) Paper chromatography is used to separate a mixture of food dyes. (i) Explain why the start line on a chromatogram is drawn in pencil and not in ink. (ii) In a chromatogram, a dye moves a distance of \(4.5\text{ cm}\) from the start line. The solvent front moves a distance of \(7.5\text{ cm}\) from the start line. Calculate the \(R_f\) value of this dye. (iii) State two factors that affect the \(R_f\) value of a substance.
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(a)(i) A lilac flame test indicates the presence of potassium ions, \(\text{K}^+\). (a)(ii) A cream precipitate with aqueous silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\). (a)(iii) The ionic equation is \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\). (b)(i) Pencil consists of graphite, which is insoluble in chromatography solvents, so it will not dissolve or run. If ink were used, it would dissolve and separate, interfering with the results of the chromatogram. (b)(ii) The \(R_f\) value is calculated as: \(R_f = \frac{\text{distance moved by dye}}{\text{distance moved by solvent front}} = \frac{4.5\text{ cm}}{7.5\text{ cm}} = 0.60\). (b)(iii) The two factors are: 1. The choice of solvent (mobile phase), and 2. The type of chromatography paper (stationary phase).
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(a)(i) 1 mark for potassium (ion) / \(\text{K}^+\). (a)(ii) 1 mark for bromide (ion) / \(\text{Br}^-\). (a)(iii) 1 mark for correct ionic reactants and product, 1 mark for all correct state symbols. (b)(i) 1 mark for: pencil / graphite is insoluble (does not dissolve in solvent), 1 mark for: ink is soluble / would run / would separate and interfere. (b)(ii) 1 mark for correct formula or substitution \(4.5 / 7.5\), 1 mark for correct answer: \(0.60\) (accept \(0.6\)). (b)(iii) 1 mark for solvent (or solubility of solute), 1 mark for stationary phase / paper type.
PastPaper.question 6 · Structured
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(a) Describe the pathway of a reflex arc when a person accidentally touches a hot object. (b) Synapses are gaps between neurones. (i) Describe how an impulse is transmitted across a synapse. (ii) Explain why neurotransmitters ensure that impulses travel in one direction only. (c) Adrenaline is a hormone secreted in response to fight-or-flight situations. State two physiological effects of adrenaline on the body.
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(a) 1. A stimulus (heat) is detected by temperature/pain receptors in the skin. 2. An electrical impulse is generated and transmitted along a sensory neurone to the central nervous system (spinal cord). 3. The impulse is passed across a synapse to a relay neurone in the spinal cord. 4. The impulse passes across another synapse to a motor neurone, which carries the signal to the effector (biceps muscle), causing it to contract and pull the hand away. (b)(i) When an electrical impulse reaches the end of the pre-synaptic neurone, it stimulates the vesicles to release neurotransmitters. These chemical neurotransmitters diffuse across the synaptic cleft/gap. They bind to complementary receptor proteins on the membrane of the post-synaptic neurone, triggering a new electrical impulse. (b)(ii) Neurotransmitters are only stored in vesicles on the pre-synaptic side, and receptor proteins are only located on the post-synaptic membrane. (c) Two physiological effects of adrenaline are: 1. Increased heart rate (to pump more blood/oxygen to muscles), and 2. Increased breathing rate / dilation of airways (to absorb more oxygen).
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(a) 1 mark for: receptor detects heat/stimulus and generates impulse, 1 mark for: impulse travels along sensory neurone to spinal cord, 1 mark for: impulse passes through relay neurone, 1 mark for: impulse travels along motor neurone to effector / muscle causing contraction. (b)(i) 1 mark for: impulse triggers release of neurotransmitter from vesicles, 1 mark for: neurotransmitters diffuse across synaptic gap, 1 mark for: neurotransmitters bind to receptors on post-synaptic neurone. (b)(ii) 1 mark for: vesicles are only present on pre-synaptic side / receptors only on post-synaptic side. (c) 1 mark each for any two of: increased heart rate, increased breathing rate / depth, dilation of pupils, increased blood glucose concentration, redirection of blood flow to skeletal muscles (max 2 marks).
PastPaper.question 7 · Structured
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Calcium carbonate, \(\text{CaCO}_3\), reacts with dilute hydrochloric acid according to the equation: \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\). A student reacts \(5.00\text{ g}\) of calcium carbonate with \(50.0\text{ cm}^3\) of \(1.50\text{ mol/dm}^3\) hydrochloric acid. [Relative atomic masses, \(A_r\): \(\text{H} = 1.0\), \(\text{C} = 12.0\), \(\text{O} = 16.0\), \(\text{Cl} = 35.5\), \(\text{Ca} = 40.0\). Molar volume of gas at r.t.p. is \(24.0\text{ dm}^3/\text{mol}\).] (a) Show by calculation that hydrochloric acid is the limiting reactant. (b) Calculate the mass of calcium chloride, \(\text{CaCl}_2\), produced in this reaction. (c) Calculate the volume of carbon dioxide gas, in \(\text{cm}^3\), produced at room temperature and pressure (r.t.p.).
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PastPaper.workedSolution
(a) First, calculate the moles of calcium carbonate: \(M_r(\text{CaCO}_3) = 40.0 + 12.0 + (3 \times 16.0) = 100.0\). \(\text{Moles of CaCO}_3 = \frac{5.00\text{ g}}{100.0\text{ g/mol}} = 0.0500\text{ mol}\). Next, calculate the moles of hydrochloric acid: \(\text{Moles of HCl} = \text{volume} \times \text{concentration} = \frac{50.0\text{ cm}^3}{1000\text{ cm}^3/\text{dm}^3} \times 1.50\text{ mol/dm}^3 = 0.0750\text{ mol}\). From the stoichiometry of the equation, \(1\text{ mol}\) of \(\text{CaCO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\). Therefore, \(0.0500\text{ mol}\) of \(\text{CaCO}_3\) would require: \(0.0500 \times 2 = 0.100\text{ mol}\) of \(\text{HCl}\). Since only \(0.0750\text{ mol}\) of \(\text{HCl}\) is available, there is insufficient \(\text{HCl}\), making hydrochloric acid the limiting reactant. (b) Since \(\text{HCl}\) is the limiting reactant, the calculations must be based on \(\text{HCl}\). From the equation, \(2\text{ mol}\) of \(\text{HCl}\) produce \(1\text{ mol}\) of \(\text{CaCl}_2\). \(\text{Moles of CaCl}_2 = \frac{0.0750\text{ mol}}{2} = 0.0375\text{ mol}\). \(M_r(\text{CaCl}_2) = 40.0 + (2 \times 35.5) = 111.0\). \(\text{Mass of CaCl}_2 = 0.0375\text{ mol} \times 111.0\text{ g/mol} = 4.16\text{ g}\) (to 3 sig figs). (c) From the equation, \(2\text{ mol}\) of \(\text{HCl}\) produce \(1\text{ mol}\) of \(\text{CO}_2\). \(\text{Moles of CO}_2 = 0.0375\text{ mol}\). \(\text{Volume of CO}_2 = 0.0375\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 0.900\text{ dm}^3\). Since \(1\text{ dm}^3 = 1000\text{ cm}^3\), \(\text{Volume in cm}^3 = 0.900 \times 1000 = 900\text{ cm}^3\).
PastPaper.markingScheme
(a) 1 mark for: \(\text{moles of CaCO}_3 = 0.0500\text{ mol}\), 1 mark for: \(\text{moles of HCl} = 0.0750\text{ mol}\), 1 mark for: stating that \(0.0500\text{ mol}\) of \(\text{CaCO}_3\) requires \(0.100\text{ mol}\) of \(\text{HCl}\) (or showing that \(0.0750\text{ mol}\) of \(\text{HCl}\) can only react with \(0.0375\text{ mol}\) of \(\text{CaCO}_3\)), 1 mark for: concluding \(\text{HCl}\) is limiting based on the comparison. (b) 1 mark for: \(\text{moles of CaCl}_2 = 0.0375\text{ mol}\) (ecf from (a)), 1 mark for: \(M_r(\text{CaCl}_2) = 111\), 1 mark for: mass \(= 4.16\text{ g}\) (accept \(4.1625\text{ g}\) or \(4.2\text{ g}\)). (c) 1 mark for: \(\text{moles of CO}_2 = 0.0375\text{ mol}\) (ecf), 1 mark for: volume \(= 0.900\text{ dm}^3\), 1 mark for: multiplying by 1000 to get \(900\text{ cm}^3\).
PastPaper.question 8 · Structured
10 PastPaper.marks
(a) State the balanced chemical equation for photosynthesis. (b) A student investigates the effect of light intensity on the rate of photosynthesis of an aquatic plant. (i) Describe how the rate of photosynthesis can be measured in this experiment. (ii) Describe the graph showing how the rate of photosynthesis changes as light intensity increases. Explain the shape of the graph. (c) Before testing a leaf for starch, the leaf must be treated. Explain the purpose of the following steps: (i) Placing the leaf in boiling water for 30 seconds. (ii) Boiling the leaf in ethanol.
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(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)(i) To measure the rate of photosynthesis, one can count the number of oxygen gas bubbles released from the cut stem of the aquatic plant per unit time (e.g., bubbles per minute), or collect the gas in a gas syringe over a fixed time to measure its volume. (b)(ii) Graph description: The rate of photosynthesis increases steadily (linearly) as light intensity increases, and then levels off (plateaus) at higher light intensities. Explanation: At low light intensities, light is the limiting factor. As light intensity increases, the rate of reaction increases. At the plateau, light is no longer the limiting factor; some other factor, such as carbon dioxide concentration or temperature, is limiting the rate. (c)(i) Boiling water kills the leaf, denatures enzymes, and breaks down the cell membranes, making them permeable. (c)(ii) Boiling in ethanol extracts/dissolves chlorophyll, which decolorizes the green leaf so that the blue-black color change with iodine (indicating starch) can be clearly observed without green interference.
PastPaper.markingScheme
(a) 1 mark for correct formulas of reactants and products, 1 mark for correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\). (b)(i) 1 mark for: counting bubbles of gas (or measuring volume of gas), 1 mark for: reference to a measured time interval (e.g., per minute). (b)(ii) 1 mark for: rate increases then levels off / reaches plateau, 1 mark for: light is limiting at the start (linear part), 1 mark for: another factor (such as carbon dioxide concentration or temperature) is limiting at the plateau. (c)(i) 1 mark for: to kill the leaf / denature enzymes / break cell membranes. (c)(ii) 1 mark for: to remove/dissolve chlorophyll, 1 mark for: so that color change with iodine can be seen clearly.
PastPaper.question 9 · Structured
10 PastPaper.marks
A small drone of mass \(1.5\text{ kg}\) rises vertically from rest. It accelerates uniformly upwards at \(2.0\text{ m/s}^2\) for the first \(4.0\text{ s}\).
Assume the gravitational field strength \(g = 9.8\text{ N/kg}\).
(a) (i) Calculate the velocity of the drone at \(t = 4.0\text{ s}\). [2] (ii) Calculate the height the drone reaches during these first \(4.0\text{ s}\). [2]
(b) Calculate the upward thrust force acting on the drone during this acceleration. Show your working. [3]
(c) After \(4.0\text{ s}\), the drone hovers at a constant height. State the net force acting on the drone when it is hovering, and compare the magnitude and direction of the thrust force to the weight of the drone. [3]
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(a) (i) Using the equation of motion \(v = u + at\): \(v = 0 + (2.0\text{ m/s}^2 \times 4.0\text{ s}) = 8.0\text{ m/s}\).
(ii) Using the equation of motion \(s = ut + \frac{1}{2}at^2\): \(s = 0 + \frac{1}{2} \times 2.0\text{ m/s}^2 \times (4.0\text{ s})^2 = 16\text{ m}\).
(b) First, calculate the net force required for this acceleration: \(F_{\text{net}} = ma = 1.5\text{ kg} \times 2.0\text{ m/s}^2 = 3.0\text{ N}\) upwards.
Next, calculate the weight of the drone: \(W = mg = 1.5\text{ kg} \times 9.8\text{ N/kg} = 14.7\text{ N}\) downwards.
Using Newton's second law for vertical motion: \(F_{\text{net}} = F_{\text{thrust}} - W\) \(3.0\text{ N} = F_{\text{thrust}} - 14.7\text{ N}\) \(F_{\text{thrust}} = 14.7\text{ N} + 3.0\text{ N} = 17.7\text{ N}\).
(c) When hovering, the drone is stationary at a constant height, so its acceleration is zero: - The net force on the drone is \(0\text{ N}\). - The thrust force is equal in magnitude to the weight of the drone (\(14.7\text{ N}\)). - The thrust force acts in the opposite direction (upwards) to the weight (downwards).
PastPaper.markingScheme
(a) (i) - Formula \(v = u + at\) or substitution \(2.0 \times 4.0\) [1] - Correct final value with unit: \(8.0\text{ m/s}\) [1]
(ii) - Formula \(s = \frac{1}{2}at^2\) or substitution \(0.5 \times 2.0 \times 4.0^2\) [1] - Correct final value with unit: \(16\text{ m}\) [1]
(b) - Calculation of weight: \(W = 1.5 \times 9.8 = 14.7\text{ N}\) (accept \(15\text{ N}\) if \(g = 10\text{ N/kg}\) is used) [1] - Calculation of net force: \(F = 1.5 \times 2.0 = 3.0\text{ N}\) [1] - Calculation of upward thrust force: \(17.7\text{ N}\) (accept \(18\text{ N}\) if \(g = 10\text{ N/kg}\) is used) [1]
(c) - Net force is \(0\text{ N}\) (or zero) [1] - Thrust force is equal in magnitude to the weight [1] - Thrust force is opposite in direction to the weight (or acts upwards) [1]
PastPaper.question 10 · Structured
10 PastPaper.marks
A student prepares a sample of insoluble barium sulfate, \(\text{BaSO}_4\), by reacting aqueous barium chloride, \(\text{BaCl}_2\), with dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
(a) Write a balanced chemical equation, including state symbols, for this reaction. [3]
(b) (i) Describe the method used to obtain a pure, dry sample of barium sulfate from the reaction mixture. [4] (ii) Name the type of reaction that occurs in this preparation. [1]
(c) The student wants to confirm that the filtrate contains chloride ions. Describe a chemical test and its positive result for chloride ions. [2]
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(a) Barium chloride reacts with sulfuric acid to produce insoluble barium sulfate and hydrochloric acid. The balanced equation with state symbols is: \(\text{BaCl}_2\text{(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{HCl(aq)}\)
(b) (i) To get a pure, dry sample of an insoluble salt from a reaction mixture: 1. Filter the mixture to separate the solid barium sulfate residue from the soluble filtrate (hydrochloric acid and excess reactants). 2. Wash the residue on the filter paper with distilled/deionised water to remove any remaining soluble impurities. 3. Repeat washing multiple times. 4. Dry the residue by placing it in a warm oven or pressing it gently between filter papers.
(ii) This reaction is a precipitation reaction, as two solutions react to form an insoluble solid.
(c) To test for chloride ions, add dilute nitric acid to acidify the sample, and then add aqueous silver nitrate. A positive result is the formation of a white precipitate of silver chloride.
PastPaper.markingScheme
(a) - Correct formulas of all reactants and products [1] - Correct balancing: \(2\text{HCl}\) [1] - Correct state symbols: \(\text{BaCl}_2\text{(aq)}\), \(\text{H}_2\text{SO}_4\text{(aq)}\), \(\text{BaSO}_4\text{(s)}\), \(\text{HCl(aq)}\) [1]
(b) (i) - Filter the mixture (to obtain the residue) [1] - Wash the residue with distilled / deionised water [1] - Wash multiple times (to remove soluble impurities) [1] - Dry the residue in a warm oven / between filter papers [1]
(b) (ii) - Precipitation [1]
(c) - Add dilute nitric acid and aqueous silver nitrate [1] - White precipitate [1]
PastPaper.question 11 · Structured
10 PastPaper.marks
A circuit contains a \(12.0\text{ V}\) d.c. power supply, an ammeter, a \(4.0\ \Omega\) resistor, and a filament lamp connected in series.
(a) The current measured by the ammeter is \(1.5\text{ A}\). (i) Calculate the total resistance of the circuit. [2] (ii) Calculate the resistance of the filament lamp. [2] (iii) Calculate the electrical energy transferred by the lamp in \(5.0\text{ minutes}\). Show your working. [3]
(b) The \(4.0\ \Omega\) resistor is made of a uniform length of resistance wire. Explain how the resistance of this wire would change if: (i) the length of the wire is doubled. [1] (ii) the cross-sectional area of the wire is doubled. [1]
(c) Draw or describe the standard circuit symbol for a variable resistor. [1]
(ii) For a series circuit: \(R_{\text{total}} = R_{\text{resistor}} + R_{\text{lamp}}\) \(8.0\ \Omega = 4.0\ \Omega + R_{\text{lamp}} \implies R_{\text{lamp}} = 4.0\ \Omega\).
(iii) Convert time to seconds: \(t = 5.0\text{ minutes} \times 60\text{ s/min} = 300\text{ s}\).
Calculate the electrical power dissipated by the lamp: \(P = I^2 R = (1.5\text{ A})^2 \times 4.0\ \Omega = 2.25 \times 4.0 = 9.0\text{ W}\). (Alternatively, \(V_{\text{lamp}} = I R_{\text{lamp}} = 1.5 \times 4.0 = 6.0\text{ V}\), then \(P = V_{\text{lamp}} I = 6.0\text{ V} \times 1.5\text{ A} = 9.0\text{ W}\)).
Calculate the energy transferred: \(E = P \times t = 9.0\text{ W} \times 300\text{ s} = 2700\text{ J}\) (or \(2.7\text{ kJ}\)).
(b) (i) Since resistance is directly proportional to length (\(R \propto L\)), doubling the length doubles the resistance (to \(8.0\ \Omega\)).
(ii) Since resistance is inversely proportional to cross-sectional area (\(R \propto \frac{1}{A}\)), doubling the cross-sectional area halves the resistance (to \(2.0\ \Omega\)).
(c) The symbol is a rectangle representing a fixed resistor, with a diagonal arrow pointing through it.
PastPaper.markingScheme
(a) (i) - Formula \(R = \frac{V}{I}\) or substitution \(\frac{12}{1.5}\) [1] - Correct final value with unit: \(8.0\ \Omega\) (or ohms) [1]
(ii) - Calculation subtraction: \(8.0 - 4.0\) [1] - Correct final value with unit: \(4.0\ \Omega\) (or ohms) [1]
(iii) - Conversion of time to seconds: \(300\text{ s}\) [1] - Correct power calculation formula and value: \(P = 9.0\text{ W}\) (or \(E = VIt = 6.0 \times 1.5 \times 300\)) [1] - Correct final energy value with unit: \(2700\text{ J}\) (or \(2.7\text{ kJ}\)) [1]
(b) (ii) - Resistance halves / decreases to \(2.0\ \Omega\) [1]
(c) - Correct drawing or clear description of a rectangle with a diagonal arrow through it [1]
PastPaper.question 12 · Structured
10 PastPaper.marks
A student uses paper chromatography to investigate the dyes present in three food colorings, **X**, **Y**, and **Z**, and a banned food additive dye, **W**. The chromatogram obtained shows the following measurements: - The solvent front traveled \(8.0\text{ cm}\) from the baseline. - Dye **W** traveled \(6.0\text{ cm}\) from the baseline. - Food coloring **X** produced two spots, at heights of \(4.0\text{ cm}\) and \(6.0\text{ cm}\). - Food coloring **Y** produced one spot at a height of \(5.2\text{ cm}\). - Food coloring **Z** produced two spots, at heights of \(3.2\text{ cm}\) and \(4.8\text{ cm}\).
(a) Explain what the results tell you about: (i) the purity of food coloring **Y**. [1] (ii) the safety of food coloring **X**. [2]
(b) (i) Define the term \(R_{\text{f}}\) value. [1] (ii) Calculate the \(R_{\text{f}}\) value of the banned dye **W**. Show your working. [2]
(c) The baseline must be drawn in pencil rather than ink. Explain why. [2]
(d) State how a chromatogram can be used to identify colorless substances, such as amino acids, on the paper. [2]
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PastPaper.workedSolution
(a) (i) Food coloring **Y** is a pure substance because it only separates into a single spot on the chromatogram. (ii) Food coloring **X** is unsafe because one of its spots traveled \(6.0\text{ cm}\), which matches the distance traveled by the banned dye **W** exactly. This indicates that **X** contains the banned dye **W**.
(b) (i) The \(R_{\text{f}}\) (retardation factor) value is defined as: \(R_{\text{f}} = \frac{\text{distance traveled by the substance (spot)}}{\text{distance traveled by the solvent front}}\)
(ii) For the banned dye **W**: \(R_{\text{f}} = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\). (Note: \(R_{\text{f}}\) values are dimensionless ratios and have no units).
(c) Pencil lead (graphite) is insoluble in the solvent and will remain on the baseline, whereas ink is soluble and would dissolve, separating into components that travel up the paper and interfere with the results.
(d) A locating agent (such as ninhydrin spray) is applied to the paper. This chemical reacts with the colorless substances to form colored spots, making them visible.
PastPaper.markingScheme
(a) (i) - Pure, because it produces only one spot on the chromatogram [1]
(ii) - Unsafe / contains the banned dye **W** [1] - Because one of its spots has the same travel distance (\(6.0\text{ cm}\)) / same \(R_{\text{f}}\) value as **W** [1]
(b) (i) - Ratio of the distance moved by the solute / substance / spot to the distance moved by the solvent front [1]
(b) (ii) - Calculation \(\frac{6.0}{8.0}\) [1] - Correct value: \(0.75\) (must be decimal, reject if unit is given) [1]
(c) - Pencil (graphite) is insoluble in the solvent (does not run) [1] - Ink is soluble and would dissolve / run / contaminate the chromatogram [1]
(d) - Spray / treat with a locating agent [1] - Which reacts to form colored spots / makes the spots visible [1]
Paper 6 Alternative to Practical
Answer all experimental and planning questions based on laboratory observations.
6 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · Practical/Structured
10 PastPaper.marks
A student investigates the relationship between the load applied to a spring and its extension. The student sets up a spring hanging from a clamp, with a pointer pointing to a vertically mounted metre rule. With no load attached, the pointer is at position \( L_0 \). When a load \( F \) is suspended from the spring, the pointer moves to position \( L \).
(a) Figure 1.1 shows the pointer position on the metre rule for the unloaded spring (\( F = 0\text{ N} \)) and when a load of \( F = 3.0\text{ N} \) is attached. - Unloaded spring (\( F = 0\text{ N} \)): The pointer is exactly at \( 12.5\text{ cm} \). - Loaded spring (\( F = 3.0\text{ N} \)): The pointer is exactly on the line for \( 20.0\text{ cm} \).
(i) Record the pointer reading \( L_0 \) for the unloaded spring. \( L_0 = \) .................... \(\text{cm}\)
(ii) Record the pointer reading \( L \) for the \( 3.0\text{ N} \) load. \( L = \) .................... \(\text{cm}\)
Calculate the extension \( e \) of the spring for each load using the formula \( e = L - L_0 \), where \( L_0 \) is your value from (a)(i). Complete Table 1.1 with your calculated values.
(c) Calculate the spring constant \( k \) of the spring using the formula: \( k = \frac{F}{e} \) Use the values for a load of \( F = 4.0\text{ N} \) and its corresponding extension from Table 1.1. State the unit. \( k = \) .................... unit = ....................
(d) (i) The student removes the known loads and hangs an object of unknown weight, \( W \), on the spring. The new pointer reading is \( L_U = 21.0\text{ cm} \). Calculate the extension \( e_U \) produced by the unknown object, and use your value of \( k \) from (c) to calculate the weight \( W \) of the object. \( e_U = \) .................... \(\text{cm}\) \( W = \) .................... \(\text{N}\)
(ii) State one precaution the student should take when reading the scale of the metre rule to ensure the readings are as accurate as possible.
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(a) (i) The pointer is at 12.5 cm. (ii) The pointer is at 20.0 cm. (b) Using \( e = L - L_0 \): - For 1.0 N: \( 15.0 - 12.5 = 2.5\text{ cm} \) - For 2.0 N: \( 17.5 - 12.5 = 5.0\text{ cm} \) - For 3.0 N: \( 20.0 - 12.5 = 7.5\text{ cm} \) - For 4.0 N: \( 22.5 - 12.5 = 10.0\text{ cm} \) - For 5.0 N: \( 25.0 - 12.5 = 12.5\text{ cm} \) (c) \( k = F / e = 4.0\text{ N} / 10.0\text{ cm} = 0.40\text{ N/cm} \). (d) (i) \( e_U = 21.0 - 12.5 = 8.5\text{ cm} \). Weight \( W = k \times e_U = 0.40 \times 8.5 = 3.4\text{ N} \). (ii) To avoid parallax errors, the student should read the scale with their eyes aligned level with the pointer.
PastPaper.markingScheme
- (a)(i) 12.5 cm (1 mark) - (a)(ii) 20.0 cm (1 mark) - (b) All five extensions calculated correctly (2.5, 5.0, 7.5, 10.0, 12.5) (2 marks; deduct 1 mark for any single error) - (c) Value of k = 0.40 (1 mark); correct unit N/cm (1 mark) - (d)(i) Extension calculated as 8.5 cm (1 mark); weight calculated as 3.4 N (allow error carried forward from (c)) (1 mark) - (d)(ii) Any one valid precaution: Read perpendicular to the scale/eye-level (to avoid parallax) OR use a set square/fiducial mark (1 mark); make sure rule is vertical (1 mark).
PastPaper.question 2 · Practical/Structured
10 PastPaper.marks
A student investigates how the resistance of a wire depends on its length. The student sets up a circuit containing a power supply, an ammeter, a switch, and a voltmeter connected across a length \( l \) of resistance wire using a sliding jockey.
(a) The student measures the potential difference \( V \) and the current \( I \) for five different lengths: \( 20.0, 40.0, 60.0, 80.0, \text{ and } 100.0\text{ cm} \). Figure 2.1 shows the voltmeter and ammeter scales when the length is \( l = 60.0\text{ cm} \). - Voltmeter: pointer is on the eighth subdivision after \( 1\text{ V} \) (scale has 10 divisions between \( 1\text{ V} \) and \( 2\text{ V} \)). - Ammeter: pointer is on the third subdivision after \( 0\text{ A} \) (scale has 10 divisions between \( 0\text{ A} \) and \( 1\text{ A} \)).
(i) Record the potential difference \( V \) for \( l = 60.0\text{ cm} \). \( V = \) .................... \(\text{V}\)
(ii) Record the current \( I \) for \( l = 60.0\text{ cm} \). \( I = \) .................... \(\text{A}\) [2]
(b) Calculate the resistance \( R \) for each length of wire using the formula: \( R = \frac{V}{I} \) Complete Table 2.1 by writing the calculated resistances. The experimental results are: - \( 20.0\text{ cm} \): \( V = 0.60\text{ V} \), \( I = 0.30\text{ A} \) - \( 40.0\text{ cm} \): \( V = 1.20\text{ V} \), \( I = 0.30\text{ A} \) - \( 60.0\text{ cm} \): (your values from (a)) - \( 80.0\text{ cm} \): \( V = 2.40\text{ V} \), \( I = 0.30\text{ A} \) - \( 100.0\text{ cm} \): \( V = 3.00\text{ V} \), \( I = 0.30\text{ A} \) [2]
(c) (i) Plotting these points on a grid would show a straight line passing through the origin. Calculate the gradient \( m \) of the line of best fit through these data points. Use the points at \( l = 20.0\text{ cm} \) and \( l = 100.0\text{ cm} \). Show your working. \( m = \) .................... \(\Omega/\text{cm}\) [2]
(ii) Use your value of the gradient \( m \) to predict the resistance \( R_{50} \) of a \( 50.0\text{ cm} \) length of the same wire. \( R_{50} = \) .................... \(\Omega\) [1]
(d) (i) Describe the relationship between the length of the wire and its resistance shown by these results. [1]
(ii) State why the student should open the switch (turn off the current) between taking readings for different lengths of wire. [2]
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(a) (i) Voltmeter reading = \( 1.80\text{ V} \). (ii) Ammeter reading = \( 0.30\text{ A} \). (b) Using \( R = V / I \): - For 20.0 cm: \( 0.60 / 0.30 = 2.0\ \Omega \) - For 40.0 cm: \( 1.20 / 0.30 = 4.0\ \Omega \) - For 60.0 cm: \( 1.80 / 0.30 = 6.0\ \Omega \) - For 80.0 cm: \( 2.40 / 0.30 = 8.0\ \Omega \) - For 100.0 cm: \( 3.00 / 0.30 = 10.0\ \Omega \) (c) (i) \( m = \frac{10.0 - 2.0}{100.0 - 20.0} = \frac{8.0}{80.0} = 0.10\ \Omega/\text{cm} \). (ii) \( R_{50} = 0.10 \times 50.0 = 5.0\ \Omega \). (d) (i) Resistance is directly proportional to length. (ii) Passing current through a wire generates thermal energy. Turning off the switch prevents heating, which would otherwise alter the wire's resistance and introduce systematic error.
PastPaper.markingScheme
- (a)(i) 1.80 V (1 mark) - (a)(ii) 0.30 A (1 mark) - (b) All resistance values calculated correctly (2.0, 4.0, 6.0, 8.0, 10.0) (2 marks; 1 mark if one error) - (c)(i) Working showing change in R over change in length (1 mark); correct calculation of gradient as 0.10 (1 mark) - (c)(ii) Correct value of 5.0 (allow ecf from c(i)) (1 mark) - (d)(i) Correctly states that resistance is directly proportional to length (or increases linearly) (1 mark) - (d)(ii) Mention of preventing heating/temperature rise in the wire (1 mark); explaining that a change in temperature would change the resistance of the wire (1 mark).
PastPaper.question 3 · Practical/Structured
10 PastPaper.marks
A student carries out a thermometric titration to determine the concentration of a sample of hydrochloric acid. The student adds different volumes of aqueous sodium hydroxide to a series of polystyrene cups, each containing \( 25.0\text{ cm}^3 \) of dilute hydrochloric acid. In each run, the initial temperature of the acid is \( 20.5^\circ\text{C} \).
(a) Figure 3.1 shows the thermometer reading for the maximum temperature reached when \( 25.0\text{ cm}^3 \) of sodium hydroxide is added. - Thermometer: The meniscus is exactly on the first line above \( 30^\circ\text{C} \) (the thermometer scale has divisions every \( 0.5^\circ\text{C} \)).
(i) Record the maximum temperature reached. Maximum temperature = .................... \(^\circ\text{C}\) [1]
(ii) Calculate the temperature rise \( \Delta T \) for this volume. \( \Delta T = \) .................... \(^\circ\text{C}\) [1]
(b) Table 3.1 shows the temperature rise \( \Delta T \) for other volumes of sodium hydroxide added: - \( 5.0\text{ cm}^3 \): \( \Delta T = 2.0^\circ\text{C} \) - \( 10.0\text{ cm}^3 \): \( \Delta T = 4.0^\circ\text{C} \) - \( 15.0\text{ cm}^3 \): \( \Delta T = 6.0^\circ\text{C} \) - \( 20.0\text{ cm}^3 \): \( \Delta T = 8.0^\circ\text{C} \) - \( 25.0\text{ cm}^3 \): (your value from (a)(ii)) - \( 30.0\text{ cm}^3 \): \( \Delta T = 8.5^\circ\text{C} \) - \( 35.0\text{ cm}^3 \): \( \Delta T = 7.0^\circ\text{C} \)
Complete Table 3.1 by entering the value of \( \Delta T \) for \( 25.0\text{ cm}^3 \). [1]
(c) Explain the following observations: (i) Why the temperature of the mixture rises during the addition of sodium hydroxide up to \( 25.0\text{ cm}^3 \). [1] (ii) Why the temperature of the mixture begins to decrease after \( 25.0\text{ cm}^3 \) of sodium hydroxide has been added. [2]
(d) (i) State the volume of sodium hydroxide required to exactly neutralise \( 25.0\text{ cm}^3 \) of hydrochloric acid. Volume = .................... \(\text{cm}^3\) [1]
(ii) Explain how you used the data in Table 3.1 to determine this volume. [1]
(e) Suggest two improvements to the apparatus that would reduce heat loss to the surroundings, ensuring more accurate temperature readings. [2]
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(a) (i) Reading is \( 30.5^\circ\text{C} \). (ii) \( \Delta T = 30.5 - 20.5 = 10.0^\circ\text{C} \). (b) Enter \( 10.0 \) in Table 3.1. (c) (i) The reaction is exothermic (releases heat) because acid-base neutralisation is taking place. (ii) Beyond \( 25.0\text{ cm}^3 \), all the acid has been neutralised so no further exothermic reaction occurs; adding more room-temperature sodium hydroxide dilutes and cools the reaction mixture. (d) (i) \( 25.0\text{ cm}^3 \). (ii) This is where the peak temperature rise occurs, indicating the exact end-point of neutralisation. (e) Use a lid on the polystyrene cup and use a double-walled cup or add bubble wrap insulation.
PastPaper.markingScheme
- (a)(i) 30.5 (1 mark) - (a)(ii) 10.0 (1 mark) - (b) Correct entry of 10.0 (1 mark) - (c)(i) Reaction is exothermic / releases thermal energy (1 mark) - (c)(ii) All acid has reacted / neutralisation complete (1 mark); excess cool liquid cools the mixture down (1 mark) - (d)(i) 25.0 (1 mark) - (d)(ii) That is where the maximum temperature rise / maximum temperature was achieved (1 mark) - (e) Use a polystyrene cup (1 mark); add a lid to the cup (1 mark).
PastPaper.question 4 · Practical/Structured
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A student is provided with a sample of a green solid, Compound X, which contains one transition metal cation and one anion. The student performs a series of chemical tests to identify the ions present in X.
(a) The student dissolves a sample of X in distilled water to make Solution X. To a portion of Solution X, the student adds a few drops of aqueous sodium hydroxide, and then adds an excess of the reagent. - Observation: A dirty green precipitate forms, which is insoluble in excess sodium hydroxide. (i) Identify the cation present in X. Cation = .................... [1] (ii) Explain why a transition metal cation is expected based on the color of Compound X. [1]
(b) To another portion of Solution X, the student adds a few drops of aqueous ammonia, and then adds an excess of the reagent. - Observation: A green precipitate forms, which is insoluble in excess ammonia. On standing, the top layer of the precipitate in the test tube slowly turns brown. Explain why the precipitate turns brown on standing. [2]
(c) To a third portion of Solution X, the student adds dilute nitric acid followed by a few drops of aqueous barium nitrate. - Observation: A thick white precipitate forms. (i) Identify the anion present in X. Anion = .................... [1] (ii) Explain why the student adds dilute nitric acid before adding the barium nitrate. [1]
(d) Describe how the student would carry out a flame test on a solid sample of a different metal compound to identify its cation, describing the preparation of the equipment and how the flame is used. [3]
(e) State one safety precaution the student should take when performing these tests. [1]
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PastPaper.workedSolution
(a) (i) Iron(II) (or \( \text{Fe}^{2+} \)) is identified because of the dirty green precipitate that is insoluble in excess sodium hydroxide. (ii) Transition metals characteristically form coloured compounds, and X is a green solid. (b) The iron(II) hydroxide precipitate reacts with oxygen in the air to undergo oxidation, turning into brown iron(III) hydroxide. (c) (i) Sulfate (or \( \text{SO}_4^{2-} \)) since barium sulfate is a white precipitate. (ii) Nitric acid is added to destroy any carbonate ions that might produce white barium carbonate, which would give a false-positive result. (d) Clean a platinum or nichrome wire by dipping it in concentrated hydrochloric acid and heating it in a non-luminous Bunsen burner flame until there is no color. Then dip the clean wire in the solid compound and hold it in the hot/blue part of the Bunsen burner flame, noting the color produced. (e) Wear safety goggles to protect eyes from corrosive reagents (like acids and alkalis).
PastPaper.markingScheme
- (a)(i) Iron(II) / \( \text{Fe}^{2+} \) (1 mark) - (a)(ii) Transition metals form coloured compounds (1 mark) - (b) Oxidised (1 mark); by oxygen in the air / to Iron(III) (1 mark) - (c)(i) Sulfate / \( \text{SO}_4^{2-} \) (1 mark) - (c)(ii) To react with / remove carbonate ions (which would also produce a white precipitate) (1 mark) - (d) Clean nichrome/platinum wire using acid (1 mark); dip wire into solid (1 mark); place in blue/non-luminous Bunsen flame to observe color (1 mark) - (e) Wear safety goggles / gloves / lab coat (1 mark).
PastPaper.question 5 · Practical/Structured
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A student investigates the effect of temperature on the activity of amylase. Amylase is an enzyme that catalyses the breakdown of starch into maltose. The student mixes amylase solution and starch solution in test tubes held at different temperatures: \( 20^\circ\text{C}, 30^\circ\text{C}, 40^\circ\text{C}, 50^\circ\text{C}, \text{ and } 60^\circ\text{C} \). Every 30 seconds, a drop of each mixture is added to a drop of iodine solution on a spotting tile.
(a) (i) State the colour of iodine solution when starch is present. Colour = .................... [1] (ii) State the colour of iodine solution when all the starch has been broken down. Colour = .................... [1]
(b) Table 5.1 shows the time taken for starch to disappear (iodine solution remains yellow-brown) at each temperature: - \( 20^\circ\text{C} \): \( 240\text{ s} \) - \( 30^\circ\text{C} \): \( 120\text{ s} \) - \( 40^\circ\text{C} \): \( 60\text{ s} \) - \( 50^\circ\text{C} \): \( 180\text{ s} \) - \( 60^\circ\text{C} \): Starch still present after \( 600\text{ s} \)
(i) Explain, in terms of kinetic energy and collisions, why the time taken for starch to disappear decreases as the temperature increases from \( 20^\circ\text{C} \) to \( 40^\circ\text{C} \). [2]
(ii) Explain why the starch does not disappear at \( 60^\circ\text{C} \). [2]
(c) (i) Name one variable that must be kept constant in this investigation to ensure a fair test. [1] (ii) Describe how the student can maintain the required temperature for each test tube during the experiment. [1] (iii) The student concludes that the optimum temperature for amylase is \( 40^\circ\text{C} \). Suggest how the student could modify the experiment to find a more accurate value for the optimum temperature. [2]
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(a) (i) Starch turns iodine solution blue-black. (ii) When starch is gone, iodine solution remains its original yellow-brown / orange color. (b) (i) As temperature increases to \( 40^\circ\text{C} \), the molecules gain kinetic energy, moving faster. This results in more frequent successful collisions between the active sites of amylase and the starch molecules, speeding up the reaction. (ii) At \( 60^\circ\text{C} \), the amylase enzyme is denatured. Its active site changes shape permanently, preventing starch from binding to it. (c) (i) Concentration or volume of amylase, concentration or volume of starch, or pH. (ii) Place the test tubes in a thermostatically controlled water bath. (iii) Test at smaller temperature increments (e.g., every \( 2^\circ\text{C} \)) within the range \( 30^\circ\text{C} \) to \( 50^\circ\text{C} \).
PastPaper.markingScheme
- (a)(i) Blue-black / black (1 mark) - (a)(ii) Yellow-brown / orange / brown (1 mark) - (b)(i) Molecules gain kinetic energy (1 mark); more frequent successful collisions / more enzyme-substrate complexes formed (1 mark) - (b)(ii) Enzyme/amylase is denatured (1 mark); active site changes shape so starch cannot bind (1 mark) - (c)(i) Concentration/volume of starch OR amylase solution OR pH (1 mark) - (c)(ii) Use of a water bath (1 mark) - (c)(iii) Test at smaller temperature intervals (1 mark); in the temperature range 30 °C to 50 °C (1 mark)
PastPaper.question 6 · Practical/Structured
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Yeast is a single-celled fungus that can carry out anaerobic respiration, producing carbon dioxide gas and ethanol. Plan an investigation to determine the effect of glucose concentration on the rate of anaerobic respiration in yeast.
You are provided with: - Active yeast suspension - Glucose solutions of different concentrations (e.g., \( 2\%, 4\%, 6\%, 8\%, 10\% \)) - Standard laboratory glassware and apparatus.
In your plan, you should include: - the apparatus you will use - a description of the method, including how you will set up the apparatus and make your measurements - the variables you will control and how you will do this - how you will use your results to draw a conclusion, including any graphs you will plot - safety precautions.
You may draw a diagram to support your answer if you wish.
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An elegant plan to investigate this includes: 1. **Apparatus**: A reaction vessel (boiling tube), delivery tube, gas syringe (or inverted measuring cylinder over a trough of water) to collect the gas, a water bath, and a timer. 2. **Method**: Add equal volumes (e.g., \( 10\text{ cm}^3 \)) of yeast suspension and \( 2\% \) glucose solution into a boiling tube. Seal the tube with a rubber bung connected to a gas syringe. Place the boiling tube in a water bath at \( 35^\circ\text{C} \). Let it equilibrate for 2 minutes, then measure the volume of carbon dioxide gas produced in 5 minutes. Repeat the procedure using the remaining glucose solutions (\( 4\%, 6\%, 8\%, 10\% \)). 3. **Control Variables**: - Temperature: Use a water bath and monitor with a thermometer. - Volume of yeast suspension and glucose solution: Measure accurately using graduated pipettes/measuring cylinders. - Concentration/batch of yeast: Use the same stock suspension for all tests. 4. **Replication**: Perform three trials for each concentration to calculate the mean volume and identify anomalies. 5. **Analysis**: Plot a line graph showing the volume of carbon dioxide gas produced per minute (y-axis) against glucose concentration (x-axis). 6. **Safety**: Wear safety goggles and take care when handling warm water baths to avoid scalds.
PastPaper.markingScheme
- Apparatus: delivery tube and gas syringe OR measuring cylinder inverted over water (1 mark) - Independent variable: using at least 5 different glucose concentrations (1 mark) - Dependent variable: measuring volume of gas collected (or counting bubbles) (1 mark) - Time limit: specifying a clear time period for collection (e.g., 5 mins) (1 mark) - Control variable 1: keeping temperature constant using a water bath (1 mark) - Control variable 2: keeping yeast volume / concentration constant (1 mark) - Replication: repeating each trial at least twice/three times and averaging (1 mark) - Safety: mention of goggles / lab coat / care with hot water or glassware (1 mark) - Graph: plotting rate of gas production / volume against glucose concentration (1 mark) - Conclusion: explaining that a higher concentration increases rate of respiration up to a plateau (1 mark)