An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Paper 21 Multiple Choice
Answer all forty multiple choice questions. Choose the single best answer A, B, C or D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · MCQ
1 PastPaper.marks
A hydrocarbon molecule \(X\) is cracked to produce one molecule of propene and one molecule of butane. What is the molecular formula of \(X\) and what color change is observed when propene is bubbled into aqueous bromine?
A.\(C_7H_{14}\), orange to colorless
B.\(C_7H_{16}\), orange to colorless
C.\(C_7H_{14}\), colorless to orange
D.\(C_7H_{16}\), colorless to orange
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PastPaper.workedSolution
1. Write down the molecular formulas of the products: Propene is an alkene with 3 carbon atoms: \(C_3H_6\). Butane is an alkane with 4 carbon atoms: \(C_4H_{10}\). 2. Determine the formula of \(X\): Since cracking one molecule of \(X\) produces one molecule of propene and one molecule of butane, we add the atoms together: Number of C atoms = \(3 + 4 = 7\), Number of H atoms = \(6 + 10 = 16\). Therefore, the formula of \(X\) is \(C_7H_{16}\). 3. Determine the chemical test reaction: Propene is an unsaturated hydrocarbon (alkene) containing a double bond. When an alkene is bubbled through orange bromine water, an addition reaction occurs, decoloring the bromine water. The color changes from orange to colorless. Combining these gives option B as the correct answer.
PastPaper.markingScheme
[1 mark] B is correct. Award 1 mark for identifying the correct molecular formula of \(X\) (\(C_7H_{16}\)) and the correct color change (orange to colorless).
PastPaper.question 2 · MCQ
1 PastPaper.marks
Copper(II) sulfate is prepared by reacting excess copper(II) carbonate with dilute sulfuric acid. Which sequence of experimental steps produces a pure, dry sample of copper(II) sulfate crystals from the reaction mixture?
A.filter the mixture \(\rightarrow\) heat filtrate to dryness \(\rightarrow\) wash the residue with cold water
B.filter the mixture \(\rightarrow\) heat filtrate until a saturated solution is formed \(\rightarrow\) leave to crystallize \(\rightarrow\) filter and dry crystals between filter papers
C.evaporate the mixture to dryness \(\rightarrow\) wash the residue with hot water \(\rightarrow\) dry the crystals in an oven
D.leave the mixture to crystallize \(\rightarrow\) filter \(\rightarrow\) wash the crystals with excess warm sulfuric acid
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PastPaper.workedSolution
To prepare a pure, dry sample of a soluble salt from an acid and an insoluble carbonate: 1. React excess insoluble copper(II) carbonate with the acid to ensure all acid is used up. 2. Filter the mixture to remove the unreacted excess copper(II) carbonate solid. 3. Heat the filtrate (the copper(II) sulfate solution) to evaporate some water until a saturated solution is formed (crystallization point). 4. Leave the hot saturated solution to cool and crystallize slowly. 5. Filter the crystals from the remaining solution and dry them between sheets of filter paper.
PastPaper.markingScheme
[1 mark] B is correct. Award 1 mark for the correct sequence of experimental procedures to obtain pure, dry crystals from an insoluble reactant.
PastPaper.question 3 · MCQ
1 PastPaper.marks
Which statement correctly describes the role of the placenta and the umbilical cord during human development?
A.Oxygen and digested nutrients diffuse from the maternal blood system into the fetal blood system across the placenta.
B.Maternal blood flows directly through the umbilical artery into the fetus to supply oxygen.
C.Urea and carbon dioxide pass from the mother's blood into the fetus through the umbilical vein.
D.The placenta protects the fetus from all pathogens and prevents any harmful substances from passing into the fetal blood.
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PastPaper.workedSolution
A is correct: Oxygen and digested nutrients diffuse from the mother's blood (maternal blood) into the fetus's blood across the placenta. B is incorrect: Maternal blood does not mix or flow directly into the fetal circulatory system. C is incorrect: Urea and carbon dioxide are waste products of the fetus, so they pass from the fetal blood to the mother's blood, not the other way around. D is incorrect: While the placenta acts as a selective barrier, it cannot protect the fetus from all pathogens (such as Rubella virus) or all harmful substances (such as nicotine and alcohol).
PastPaper.markingScheme
[1 mark] A is correct. Award 1 mark for identifying the correct movement of nutrients and oxygen by diffusion across the placenta without mixing blood.
PastPaper.question 4 · MCQ
1 PastPaper.marks
An experiment was carried out to investigate the effect of pH on the rate of an enzyme-controlled reaction. At pH 2, the rate of reaction was high. At pH 8, no product was formed. When the enzyme was returned to pH 2 from pH 8, still no product was formed. Which statement explains these observations?
A.At pH 8, the enzyme was denatured because the shape of its active site was permanently altered.
B.At pH 8, the kinetic energy of the substrate and enzyme molecules was too low to cause collisions.
C.The enzyme is an alkaline protease which only works in acidic conditions.
D.The substrate was completely used up at pH 8, leaving no reactants.
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PastPaper.workedSolution
At pH 8, which is far from the enzyme's optimum pH of 2, the enzyme is denatured. Denaturation causes a permanent change in the 3D shape of the active site of the enzyme, so it can no longer bind to the substrate. Since this change is permanent, returning the enzyme to its optimum pH 2 does not restore its activity, and no product is formed.
PastPaper.markingScheme
[1 mark] A is correct. Award 1 mark for identifying that irreversible denaturation explains why no activity was recovered when the pH was returned to optimum.
PastPaper.question 5 · MCQ
1 PastPaper.marks
A circuit contains a \(12\text{ V}\) power supply connected to a network of resistors. The network consists of two \(6.0\ \Omega\) resistors connected in parallel with each other. This parallel combination is connected in series with a \(5.0\ \Omega\) resistor. What is the total current drawn from the power supply?
A.\(1.0\text{ A}\)
B.\(1.5\text{ A}\)
C.\(1.8\text{ A}\)
D.\(4.0\text{ A}\)
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PastPaper.workedSolution
1. Find the equivalent resistance of the two parallel resistors: \(R_{p} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). 2. Find the total resistance of the circuit by adding the series resistor: \(R_{\text{total}} = R_{p} + 5.0 = 3.0 + 5.0 = 8.0\ \Omega\). 3. Use Ohm's law to calculate the total current drawn from the supply: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
PastPaper.markingScheme
[1 mark] B is correct. Award 1 mark for the correct calculation of parallel resistance (\(3.0\ \Omega\)), total resistance (\(8.0\ \Omega\)), and applying Ohm's law to find \(1.5\text{ A}\).
PastPaper.question 6 · MCQ
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The list shows some of the events that occur in a lake after untreated sewage or fertilizer runoff enters the water: 1. Increased growth of producers (algal bloom). 2. Algae block light from reaching plants deeper in the water. 3. Under-surface plants die. 4. [Event X]. 5. Concentration of dissolved oxygen decreases. 6. Aquatic animals die. Which process is represented by [Event X]?
A.Increased decomposition of dead plants by aerobic bacteria
B.Decreased respiration by anaerobic bacteria
C.Increased rate of photosynthesis by deep-water plants
D.Decreased concentration of mineral ions in the water
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PastPaper.workedSolution
During eutrophication, after aquatic plants die due to lack of sunlight, aerobic bacteria (decomposers) multiply rapidly and break down the dead organic matter. During this decomposition, these bacteria respire aerobically, consuming the dissolved oxygen in the water. This rapid consumption leads to a decrease in dissolved oxygen concentration, which causes aquatic animals to die from suffocation. Therefore, Event X represents increased decomposition by aerobic bacteria.
PastPaper.markingScheme
[1 mark] A is correct. Award 1 mark for identifying the key role of aerobic bacteria in decomposing dead plants and consuming oxygen during eutrophication.
PastPaper.question 7 · MCQ
1 PastPaper.marks
Which statement correctly describes the properties of electromagnetic waves?
A.Gamma rays have a longer wavelength than radio waves and travel faster in a vacuum.
B.Microwaves have a higher frequency than ultraviolet waves and are longitudinal waves.
C.Infrared waves have a lower frequency than visible light and travel at the same speed as X-rays in a vacuum.
D.Sound waves and ultraviolet waves are both electromagnetic waves that can travel through a vacuum.
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PastPaper.workedSolution
Let's check the options: A is incorrect because gamma rays have the shortest wavelength in the electromagnetic spectrum, and all electromagnetic waves travel at the same speed in a vacuum. B is incorrect because microwaves have a lower frequency than ultraviolet waves, and all electromagnetic waves are transverse waves, not longitudinal waves. C is correct because infrared waves have a longer wavelength and therefore a lower frequency than visible light. All electromagnetic waves, including infrared waves and X-rays, travel at the same speed (\(3 \times 10^8\text{ m/s}\)) in a vacuum. D is incorrect because sound waves are longitudinal mechanical waves and cannot travel through a vacuum.
PastPaper.markingScheme
[1 mark] C is correct. Award 1 mark for identifying that infrared has a lower frequency than visible light and that all EM waves share the same speed in a vacuum.
PastPaper.question 8 · MCQ
1 PastPaper.marks
Which statement correctly compares aerobic and anaerobic respiration in humans?
Aerobic respiration completely breaks down glucose in the presence of oxygen, releasing a relatively large amount of energy per glucose molecule. Anaerobic respiration only partially breaks down glucose in the absence of oxygen, yielding much less energy per molecule of glucose and producing lactic acid in humans. Therefore, option B is correct.
PastPaper.markingScheme
[1 mark] B is correct. Award 1 mark for identifying that aerobic respiration releases a much larger amount of energy per glucose molecule compared to anaerobic respiration.
PastPaper.question 9 · MCQ
1 PastPaper.marks
An object of mass 4.0 kg is pulled along a friction-free horizontal surface by a constant horizontal force of 12 N. The object starts from rest. What is the kinetic energy of the object after it has travelled a distance of 5.0 m?
A.2.4 J
B.15 J
C.60 J
D.120 J
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PastPaper.workedSolution
Work done on the object is calculated by the formula: \(W = F \times d\). Here, \(F = 12\text{ N}\) and \(d = 5.0\text{ m}\). Therefore, \(W = 12 \times 5.0 = 60\text{ J}\). Since there is no friction and the object starts from rest, all the work done is converted into kinetic energy. Thus, the kinetic energy is 60 J.
PastPaper.markingScheme
1 mark for the correct answer C. [1] for calculating work done as force times distance and relating it directly to the gain in kinetic energy.
PastPaper.question 10 · MCQ
1 PastPaper.marks
Which statement about the reaction of ethene with bromine is correct?
A.It is a substitution reaction that occurs only in the presence of ultraviolet light.
B.It is an addition reaction that causes the bromine water to turn from colourless to brown.
C.It is a substitution reaction that produces bromoethane.
D.It is an addition reaction that causes the bromine water to decolourise.
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PastPaper.workedSolution
Ethene is an unsaturated hydrocarbon containing a carbon-carbon double bond. It undergoes an addition reaction with bromine, which does not require ultraviolet light. During this reaction, reddish-brown bromine water is decolourised, which serves as a test to distinguish unsaturated hydrocarbons from saturated ones.
PastPaper.markingScheme
1 mark for correct option D. [1] for identifying the reaction as addition and recognizing the correct colour change of bromine water.
PastPaper.question 11 · MCQ
1 PastPaper.marks
An enzyme-catalysed reaction is investigated at different temperatures. Which statement correctly describes the effect of temperature on the enzyme and its activity?
A.As temperature increases from \(10\ ^\circ\text{C}\) to \(35\ ^\circ\text{C}\), the kinetic energy of the substrate and enzyme molecules decreases.
B.At temperatures above \(60\ ^\circ\text{C}\), the active site of the enzyme changes shape permanently, preventing substrate binding.
C.At \(0\ ^\circ\text{C}\), the enzyme is completely denatured and cannot function even if warmed.
D.The rate of reaction continues to double for every \(10\ ^\circ\text{C}\) rise in temperature up to \(100\ ^\circ\text{C}\).
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PastPaper.workedSolution
Enzymes are protein molecules that can be denatured at high temperatures. Above their optimum temperature (typically above \(50-60\ ^\circ\text{C}\)), the active site of the enzyme changes shape permanently. As a result, the substrate can no longer fit, and the reaction stops. Low temperatures do not denature enzymes; they only reduce their kinetic energy.
PastPaper.markingScheme
1 mark for option B. [1] for identifying that high temperatures cause permanent denaturation of the active site.
PastPaper.question 12 · MCQ
1 PastPaper.marks
A student wants to prepare a pure, dry sample of the insoluble salt, barium sulfate. Which pair of aqueous solutions should be mixed together to obtain barium sulfate as a precipitate?
A.Barium carbonate and dilute sulfuric acid
B.Barium chloride and sodium sulfate
C.Barium nitrate and copper(II) carbonate
D.Barium hydroxide and dilute hydrochloric acid
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PastPaper.workedSolution
To prepare an insoluble salt by precipitation, two soluble starting materials must be used. Barium chloride and sodium sulfate are both highly soluble salts in water. When mixed, they react to form the precipitate barium sulfate and soluble sodium chloride.
PastPaper.markingScheme
1 mark for correct option B. [1] for identifying that both starting reactants must be soluble aqueous solutions to produce an insoluble salt precipitate.
PastPaper.question 13 · MCQ
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A resistor of resistance \(R\) is connected in series with a parallel combination of two identical resistors, each of resistance \(R\). The total resistance of this combined circuit is \(12\ \Omega\). What is the value of \(R\)?
A.\(4\ \Omega\)
B.\(6\ \Omega\)
C.\(8\ \Omega\)
D.\(24\ \Omega\)
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PastPaper.workedSolution
The equivalent resistance of two identical resistors of resistance \(R\) connected in parallel is \(R_p = \frac{R \times R}{R + R} = \frac{R}{2} = 0.5R\). This parallel combination is in series with another resistor of resistance \(R\), so the total resistance is \(R_{\text{total}} = R + 0.5R = 1.5R\). Given that \(1.5R = 12\ \Omega\), solving for \(R\) gives \(R = \frac{12}{1.5} = 8\ \Omega\).
PastPaper.markingScheme
1 mark for option C. [1] for correctly determining the parallel equivalent resistance as 0.5R and using the series resistance formula to solve for R.
PastPaper.question 14 · MCQ
1 PastPaper.marks
Which row correctly identifies the products of anaerobic respiration in yeast and in human muscle cells?
A.Yeast: carbon dioxide and water | Human muscle: lactic acid
B.Yeast: ethanol and carbon dioxide | Human muscle: lactic acid
C.Yeast: lactic acid | Human muscle: ethanol and carbon dioxide
D.Yeast: ethanol and carbon dioxide | Human muscle: carbon dioxide and water
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PastPaper.workedSolution
In yeast cells, anaerobic respiration (fermentation) produces ethanol and carbon dioxide as waste products. In human muscle cells during vigorous exercise, anaerobic respiration produces lactic acid only; no carbon dioxide is released during this specific pathway.
PastPaper.markingScheme
1 mark for option B. [1] for correctly identifying products of anaerobic respiration in both yeast and human muscles.
PastPaper.question 15 · MCQ
1 PastPaper.marks
During the electrolysis of concentrated aqueous sodium chloride (brine) using inert electrodes, what are the primary products formed at the positive electrode (anode) and the negative electrode (cathode)?
A.Anode: oxygen; Cathode: hydrogen
B.Anode: chlorine; Cathode: sodium
C.Anode: chlorine; Cathode: hydrogen
D.Anode: oxygen; Cathode: sodium
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PastPaper.workedSolution
Concentrated aqueous sodium chloride contains \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). At the anode (positive electrode), chloride ions are in high concentration, so they are discharged preferentially to form chlorine gas. At the cathode (negative electrode), hydrogen ions are discharged preferentially because hydrogen is less reactive than sodium, forming hydrogen gas.
PastPaper.markingScheme
1 mark for option C. [1] for correctly identifying the gaseous products discharged at both electrodes under concentrated conditions.
PastPaper.question 16 · MCQ
1 PastPaper.marks
A water wave travels from deep water into shallow water. What happens to its speed, frequency, and wavelength as it enters the shallow water?
A.Speed decreases, frequency decreases, wavelength remains unchanged
B.Speed decreases, frequency remains unchanged, wavelength decreases
C.Speed increases, frequency remains unchanged, wavelength increases
D.Speed remains unchanged, frequency increases, wavelength decreases
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PastPaper.workedSolution
When water waves travel from deep to shallow water, their speed decreases due to friction with the seabed. The frequency of a wave is determined only by its source, so it remains unchanged. Using the wave equation \(v = f \lambda\), since speed \(v\) decreases while frequency \(f\) remains constant, the wavelength \(\lambda\) must also decrease.
PastPaper.markingScheme
1 mark for option B. [1] for identifying that speed decreases, frequency remains constant, and wavelength decreases.
PastPaper.question 17 · MCQ
1 PastPaper.marks
An experiment was carried out to investigate the effect of temperature on the rate of an enzyme-controlled reaction. The rate of reaction increased from \(10^\circ\text{C}\) to \(40^\circ\text{C}\), but then fell rapidly above \(45^\circ\text{C}\). Which statement explains why the rate of reaction decreased above \(45^\circ\text{C}\)?
A.The active site of the enzyme changed shape, so substrate molecules could no longer fit.
B.The substrate molecules gained too much kinetic energy and bounced off the enzyme.
C.The enzyme molecules were completely used up by the reaction at higher temperatures.
D.The activation energy of the reaction increased, making it harder for the reaction to occur ; occur if temperature is high but the actual reason is denaturation of the enzyme active site shape change.
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PastPaper.workedSolution
At temperatures above \(45^\circ\text{C}\), the thermal energy breaks the bonds that maintain the specific three-dimensional shape of the enzyme. This leads to denaturation of the enzyme, meaning that the shape of its active site is altered. Consequently, the substrate molecule is no longer complementary in shape and cannot bind to form an enzyme-substrate complex. This reduces the rate of reaction.
PastPaper.markingScheme
[1 mark] Selects option A. Accept only A. Award 1 mark for the correct explanation that thermal denaturation of the enzyme's active site prevents substrate binding.
PastPaper.question 18 · MCQ
1 PastPaper.marks
Hydrocarbon X is cracked to produce propene and hydrocarbon Y. One molecule of X produces exactly one molecule of propene and one molecule of Y. If the relative molecular mass (\(M_r\)) of X is 114, and Y is an alkane, which formula represents Y? (Ar: C = 12, H = 1)
A.\(\text{C}_5\text{H}_{12}\)
B.\(\text{C}_5\text{H}_{10}\)
C.\(\text{C}_8\text{H}_{18}\)
D.\(\text{C}_3\text{H}_8\)
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PastPaper.workedSolution
1. Assume X is an alkane with formula \(\text{C}_n\text{H}_{2n+2}\). 2. Calculate the number of carbons in X: \(12n + (2n + 2) = 114\) which simplifies to \(14n = 112\), so \(n = 8\). Thus, X is octane, \(\text{C}_8\text{H}_{18}\). 3. Cracking equation: \(\text{C}_8\text{H}_{18} \rightarrow \text{C}_3\text{H}_6\text{ (propene)} + \text{Y}\). 4. Deduce the formula of Y: Number of C atoms = \(8 - 3 = 5\). Number of H atoms = \(18 - 6 = 12\). Hence, Y is \(\text{C}_5\text{H}_{12}\).
PastPaper.markingScheme
[1 mark] Selects option A. Accept only A. Award 1 mark for calculating the molecular formula of octane (X) and successfully subtracting the atoms of propene to find pentane (Y).
PastPaper.question 19 · MCQ
1 PastPaper.marks
A student wants to prepare a pure, dry sample of the insoluble salt, silver chloride (\(\text{AgCl}\)). Which pair of aqueous solutions should be mixed together to prepare this salt, and which method should be used to obtain the dry solid from the mixture?
A.Reactants: silver nitrate and sodium chloride; Method: filtration, washing the residue with distilled water, and drying.
B.Reactants: silver metal and hydrochloric acid; Method: evaporation of the water to leave the crystals.
C.Reactants: silver oxide and hydrochloric acid; Method: filtration, then crystallization of the filtrate.
D.Reactants: silver nitrate and sodium chloride; Method: crystallization of the filtrate, followed by drying.
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PastPaper.workedSolution
Silver chloride is an insoluble salt. The standard preparation method for insoluble salts is precipitation, where two soluble salt solutions are mixed. Silver nitrate and sodium chloride are both highly soluble. When mixed, they react to form a precipitate of silver chloride: \(\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)\). The insoluble precipitate is obtained by filtration as the residue, washed with distilled water to remove soluble impurities, and dried.
PastPaper.markingScheme
[1 mark] Selects option A. Accept only A. Award 1 mark for recognizing that precipitation from soluble reactants followed by filtration, washing, and drying is the correct procedure for an insoluble salt.
PastPaper.question 20 · MCQ
1 PastPaper.marks
Which statement correctly compares asexual and sexual reproduction?
A.In asexual reproduction, gametes are produced; in sexual reproduction, gametes are not produced.
B.In asexual reproduction, a zygote is formed; in sexual reproduction, a zygote is not formed.
C.In asexual reproduction, offspring are genetically identical to the parent; in sexual reproduction, offspring show genetic variation.
D.Asexual reproduction requires two parents; sexual reproduction requires only one parent.
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PastPaper.workedSolution
Asexual reproduction involves only one parent and produces offspring that are genetically identical (clones) to the parent, as it relies on mitosis. Sexual reproduction involves the fusion of male and female gametes from two parents, introducing genetic variation in the offspring due to meiosis and fertilization.
PastPaper.markingScheme
[1 mark] Selects option C. Accept only C. Award 1 mark for identifying the correct biological differences regarding genetic variation and parent-offspring identity.
PastPaper.question 21 · MCQ
1 PastPaper.marks
Two identical resistors of resistance \(R\) are connected in series with a cell of electromotive force (e.m.f.) \(V\) and negligible internal resistance. The current in the circuit is \(I_s\). The two resistors are then reconnected in parallel across the same cell, and the current from the cell is now \(I_p\). What is the ratio \(\frac{I_p}{I_s}\)?
A.\(0.25\)
B.\(0.5\)
C.\(2.0\)
D.\(4.0\)
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PastPaper.workedSolution
1. For the series circuit, the total equivalent resistance is \(R_s = R + R = 2R\). According to Ohm's Law, the current is \(I_s = \frac{V}{2R}\). 2. For the parallel circuit, the total equivalent resistance \(R_p\) is given by \(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}\), which yields \(R_p = \frac{R}{2}\). The total current from the cell is \(I_p = \frac{V}{R_p} = \frac{V}{R/2} = \frac{2V}{R}\). 3. Find the ratio: \(\frac{I_p}{I_s} = \frac{2V/R}{V/2R} = 4\).
PastPaper.markingScheme
[1 mark] Selects option D. Accept only D. Award 1 mark for deriving the current equations for both configurations and obtaining the correct ratio of 4.
PastPaper.question 22 · MCQ
1 PastPaper.marks
Which sequence of events shows the correct order during the eutrophication of a body of water after fertilizer runoff?
A.increased growth of producers \(\rightarrow\) decreased light for submerged plants \(\rightarrow\) death of submerged plants \(\rightarrow\) decomposition by aerobic bacteria \(\rightarrow\) depletion of dissolved oxygen
B.depletion of dissolved oxygen \(\rightarrow\) decomposition by aerobic bacteria \(\rightarrow\) increased growth of producers \(\rightarrow\) death of submerged plants
C.decomposition by aerobic bacteria \(\rightarrow\) increased growth of producers \(\rightarrow\) depletion of dissolved oxygen \(\rightarrow\) death of submerged plants
D.death of submerged plants \(\rightarrow\) depletion of dissolved oxygen \(\rightarrow\) increased growth of producers \(\rightarrow\) decomposition by aerobic bacteria
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PastPaper.workedSolution
During eutrophication: 1. Fertilizer runoff causes nutrient enrichment, prompting increased growth of producers (algae/plants on the surface). 2. The algal bloom blocks sunlight, causing decreased light for submerged plants, leading to their death. 3. Aerobic bacteria decompose the dead organic matter, multiplying rapidly. 4. The bacteria consume large amounts of oxygen during respiration, causing depletion of dissolved oxygen in the water.
PastPaper.markingScheme
[1 mark] Selects option A. Accept only A. Award 1 mark for the correct chronological sequence of ecological events leading to eutrophication.
PastPaper.question 23 · MCQ
1 PastPaper.marks
Which statement about electromagnetic waves is correct?
A.Infrared waves have a shorter wavelength than ultraviolet waves, and both travel at the same speed in a vacuum.
B.Radio waves have a lower frequency than microwaves, and both travel at the same speed in a vacuum.
C.Gamma rays have a longer wavelength than X-rays, and both travel at different speeds in a vacuum.
D.Ultraviolet waves have a lower frequency than visible light, and both travel at different speeds in a vacuum.
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PastPaper.workedSolution
In the electromagnetic spectrum, ordered by increasing frequency: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Therefore, radio waves have a lower frequency than microwaves. Furthermore, all electromagnetic waves travel at the same speed (approximately \(3 \times 10^8\text{ m/s}\)) in a vacuum.
PastPaper.markingScheme
[1 mark] Selects option B. Accept only B. Award 1 mark for identifying that radio waves have a lower frequency than microwaves and that all EM waves travel at the same speed in a vacuum.
PastPaper.question 24 · MCQ
1 PastPaper.marks
A scientist compares the processes of aerobic and anaerobic respiration in yeast cells. Which statement correctly describes a difference between these two processes?
A.Aerobic respiration produces carbon dioxide, whereas anaerobic respiration in yeast does not.
C.Aerobic respiration releases much more energy per glucose molecule than anaerobic respiration.
D.Aerobic respiration occurs only in the cytoplasm, whereas anaerobic respiration occurs in mitochondria.
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PastPaper.workedSolution
Aerobic respiration involves the complete breakdown of glucose in the presence of oxygen, releasing a large amount of energy (around 30-32 molecules of ATP per molecule of glucose). Anaerobic respiration in yeast (fermentation) involves the incomplete breakdown of glucose, producing ethanol and carbon dioxide, and releasing much less energy (only 2 molecules of ATP per glucose molecule).
PastPaper.markingScheme
[1 mark] Selects option C. Accept only C. Award 1 mark for identifying the correct comparative energetic output between aerobic and anaerobic respiration.
PastPaper.question 25 · MCQ
1 PastPaper.marks
An amylase enzyme catalyzes the breakdown of starch into maltose. A student measures the rate of this reaction at different temperatures. Which statement correctly explains why the rate of reaction decreases above the optimum temperature of \(40^\circ\text{C}\)?
A.The starch molecules are denatured, changing their shape so they cannot bind to amylase.
B.The amylase molecules have less kinetic energy, so there are fewer successful collisions.
C.The active site of the amylase enzyme changes shape permanently, so starch can no longer fit.
D.The activation energy of the reaction is increased, preventing starch from breaking down.
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PastPaper.workedSolution
Above its optimum temperature, the active site of the amylase enzyme changes shape permanently (denatures). This means the substrate starch can no longer bind to the active site, decreasing the rate of reaction. Option A describes what happens below the optimum temperature. Option B is incorrect because starch does not have an active site. Option D is incorrect because denaturation does not increase activation energy.
PastPaper.markingScheme
1 mark for identifying the correct explanation of denaturation and its effect on the active site.
PastPaper.question 26 · MCQ
1 PastPaper.marks
A crane lifts a load of mass \(400\text{ kg}\) vertically upwards through a height of \(15\text{ m}\) in a time of \(30\text{ s}\). The acceleration of free fall, \(g\), is \(10\text{ m/s}^2\). What is the useful average power developed by the crane?
A.200 W
B.2000 W
C.6000 W
D.60 000 W
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PastPaper.workedSolution
First, calculate the useful work done (which is equal to the increase in gravitational potential energy): \(W = mgh = 400\text{ kg} \times 10\text{ m/s}^2 \times 15\text{ m} = 60\,000\text{ J}\). Next, calculate the useful average power: \(P = \frac{W}{t} = \frac{60\,000\text{ J}}{30\text{ s}} = 2000\text{ W}\).
PastPaper.markingScheme
1 mark for calculating correct useful average power of 2000 W.
PastPaper.question 27 · MCQ
1 PastPaper.marks
Hydrocarbon X reacts with aqueous bromine, turning it from orange to colorless. Hydrocarbon X also undergoes polymerization to form a long-chain molecule. Which statement about hydrocarbon X and its polymer is correct?
A.Hydrocarbon X is an alkane and the polymer is formed by addition polymerization.
B.Hydrocarbon X is an alkene and the polymer is formed by addition polymerization.
C.Hydrocarbon X is an alkane and the polymer is formed by condensation polymerization.
D.Hydrocarbon X is an alkene and the polymer is formed by condensation polymerization.
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PastPaper.workedSolution
Decolorizing aqueous bromine is the test for unsaturation (a carbon-carbon double bond), which means hydrocarbon X is an alkene. Alkenes undergo addition polymerization to form addition polymers (such as poly(ethene)). Therefore, hydrocarbon X is an alkene and the polymer is formed by addition polymerization.
PastPaper.markingScheme
1 mark for identifying the hydrocarbon as an alkene and the polymerization type as addition.
PastPaper.question 28 · MCQ
1 PastPaper.marks
A student wants to prepare a pure, dry sample of the insoluble salt, barium sulfate (\(\text{BaSO}_4\)). Which pair of aqueous solutions should be mixed together to obtain a precipitate of barium sulfate?
A.barium carbonate and dilute sulfuric acid
B.barium chloride and sodium sulfate
C.barium hydroxide and dilute hydrochloric acid
D.barium metal and dilute sulfuric acid
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PastPaper.workedSolution
To prepare an insoluble salt, we mix two soluble salts (or a soluble salt and a soluble acid) in a precipitation reaction. Barium chloride is soluble, and sodium sulfate is soluble. Mixing their aqueous solutions produces insoluble barium sulfate as a precipitate and soluble sodium chloride. Barium carbonate is insoluble in water, so option A is incorrect. Dilute hydrochloric acid does not contain sulfate ions, so option C is incorrect. Barium metal is a solid, not an aqueous solution, so option D is incorrect.
PastPaper.markingScheme
1 mark for identifying the correct pair of soluble aqueous solutions to form the precipitate.
PastPaper.question 29 · MCQ
1 PastPaper.marks
A cell of electromotive force (e.m.f.) \(6.0\text{ V}\) is connected to a resistor of resistance \(12\,\Omega\). A charge of \(30\text{ C}\) passes through the resistor. How much energy is transferred to the resistor?
A.2.5 J
B.15 J
C.60 J
D.180 J
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PastPaper.workedSolution
The potential difference across the resistor is \(V = 6.0\text{ V}\). The charge passing through is \(Q = 30\text{ C}\). The energy transferred \(E\) is given by \(E = V \times Q\). Substituting the values: \(E = 6.0\text{ V} \times 30\text{ C} = 180\text{ J}\). Note that the resistance \(12\,\Omega\) is not needed for this calculation.
PastPaper.markingScheme
1 mark for calculating correct energy transferred of 180 J.
PastPaper.question 30 · MCQ
1 PastPaper.marks
A species of plant produces flowers with large, feathery stigmas that hang outside the flower, small green petals, and large quantities of light, smooth pollen grains. Which method of pollination is this flower adapted for, and what is the function of its feathery stigmas?
A.insect-pollination; to attract insects by mimicry
B.insect-pollination; to brush pollen off the bodies of visiting insects
C.wind-pollination; to catch pollen grains carried in the air
D.wind-pollination; to release pollen grains into the air
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PastPaper.workedSolution
The features described (small green petals, feathery stigmas hanging outside, light smooth pollen) are characteristics of wind-pollinated flowers. The feathery stigmas provide a large surface area to catch pollen grains carried in the air.
PastPaper.markingScheme
1 mark for identifying wind-pollination and the correct function of feathery stigmas.
PastPaper.question 31 · MCQ
1 PastPaper.marks
Molten lead(II) bromide is electrolyzed using inert carbon electrodes. Which statement describes the correct observations at the electrodes?
A.A grey metal forms at the anode and a brown gas is released at the cathode.
B.A brown gas is released at the anode and a grey metal forms at the cathode.
C.A grey metal forms at both the anode and the cathode.
D.Bubbles of a colorless gas are released at the anode and a grey metal forms at the cathode.
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PastPaper.workedSolution
During the electrolysis of molten lead(II) bromide: at the cathode (negative electrode), lead ions (\(\text{Pb}^{2+}\)) gain electrons to form molten lead metal, which is grey: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\). At the anode (positive electrode), bromide ions (\(\text{Br}^-\)) lose electrons to form bromine gas, which is brown: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\). Thus, a brown gas is released at the anode and a grey metal forms at the cathode.
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1 mark for correct observation at both anode (brown gas) and cathode (grey metal).
PastPaper.question 32 · MCQ
1 PastPaper.marks
Which statement correctly compares radio waves and gamma rays traveling in a vacuum?
A.Radio waves have a higher frequency and travel faster than gamma rays.
B.Radio waves have a longer wavelength and travel faster than gamma rays.
C.Radio waves have a lower frequency and travel at the same speed as gamma rays.
D.Radio waves have a shorter wavelength and travel at the same speed as gamma rays.
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PastPaper.workedSolution
All electromagnetic waves travel at the same speed in a vacuum (the speed of light, \(c \approx 3 \times 10^8\text{ m/s}\)). In the electromagnetic spectrum, radio waves have a longer wavelength and a lower frequency, while gamma rays have a shorter wavelength and a higher frequency. Therefore, radio waves have a lower frequency and travel at the same speed in a vacuum as gamma rays.
PastPaper.markingScheme
1 mark for identifying that they have the same speed and radio waves have a lower frequency.
PastPaper.question 33 · MCQ
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A student investigates the effect of temperature on the rate of an amylase-catalyzed reaction. The time taken for starch to be completely broken down is measured at various temperatures: at \(20^\circ\text{C}\) it takes 10 minutes, at \(30^\circ\text{C}\) it takes 5 minutes, at \(40^\circ\text{C}\) it takes 2 minutes, at \(50^\circ\text{C}\) it takes 8 minutes, and at \(60^\circ\text{C}\) starch is still present after 30 minutes. Which statement best explains the result at \(60^\circ\text{C}\)?
A.The enzyme molecules have gained kinetic energy and collide more frequently with substrate molecules.
B.The starch molecules have been denatured and can no longer fit into the active site of the amylase.
C.The active site of the amylase has changed shape, preventing starch from binding.
D.The activation energy of the reaction has increased, preventing the reaction from occurring.
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PastPaper.workedSolution
At \(60^\circ\text{C}\), which is well above the optimum temperature for amylase, the enzyme is denatured. Denaturation occurs because high thermal energy disrupts the bonds holding the enzyme's three-dimensional structure together, causing the active site to change shape irreversibly. Consequently, the substrate (starch) can no longer fit into the active site, and the reaction cannot take place. Starch is a carbohydrate and does not undergo denaturation, ruling out option B. Kinetic energy increases collisions but does not explain the complete loss of reaction, ruling out option A. Enzymes lower activation energy, but denaturation refers specifically to structural changes in the active site, ruling out option D.
PastPaper.markingScheme
1 mark for the correct option C. Correct explanation of denaturation in terms of active site shape changes. Reject options A, B, and D.
PastPaper.question 34 · MCQ
1 PastPaper.marks
Which row correctly describes the pollen grain and where fertilisation occurs in a wind-pollinated flower?
A.pollen grain: light and smooth; location of fertilisation: ovule
B.pollen grain: sticky and spiky; location of fertilisation: stigma
C.pollen grain: light and smooth; location of fertilisation: stigma
D.pollen grain: sticky and spiky; location of fertilisation: ovule
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PastPaper.workedSolution
Wind-pollinated flowers produce pollen grains that are light and smooth so they can easily be suspended in and carried by the wind. In contrast, insect-pollinated flowers have sticky or spiky pollen to adhere to insects. In all flowering plants, fertilisation (the fusion of male and female gametes) occurs in the ovule inside the ovary, not on the stigma. The stigma is the site where pollination (deposition of pollen) occurs. Therefore, the correct combination is light/smooth pollen and fertilisation occurring in the ovule.
PastPaper.markingScheme
1 mark for the correct option A. Candidate correctly identifies characteristics of wind-pollinated pollen (light and smooth) and the location of fertilisation (ovule).
PastPaper.question 35 · MCQ
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An organic compound \(X\) reacts with steam in the presence of an acid catalyst at high temperature and pressure to form ethanol. What is compound \(X\), and what type of reaction is this?
A.compound \(X\): ethane; type of reaction: substitution
B.compound \(X\): ethene; type of reaction: addition
C.compound \(X\): ethane; type of reaction: addition
D.compound \(X\): ethene; type of reaction: substitution
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PastPaper.workedSolution
Industrial production of ethanol by hydration involves reacting ethene (compound \(X\)) with steam (water) in the presence of an acid catalyst (typically phosphoric acid) at high temperature and pressure. This is an addition reaction because water adds across the double bond of the unsaturated ethene molecule to yield a single product, ethanol: \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\). Ethane is a saturated alkane and does not undergo addition reactions, ruling out options A and C. Substitution reactions produce more than one product, ruling out option D.
PastPaper.markingScheme
1 mark for the correct option B. Candidate identifies the starting material as ethene and the reaction type as addition.
PastPaper.question 36 · MCQ
1 PastPaper.marks
To prepare a pure, dry sample of soluble copper(II) sulfate crystals, a student adds excess copper(II) oxide to warm dilute sulfuric acid. Which sequence of steps should the student perform next to obtain the crystals?
A.Filter the mixture, evaporate the filtrate to dryness, and wash the residue with cold water.
B.Filter the mixture, heat the filtrate to its crystallisation point, leave it to cool, then filter and dry the crystals.
C.Evaporate the mixture to dryness, recrystallise from hot water, then filter and dry the crystals.
D.Heat the mixture to its crystallisation point, filter while hot, then leave the residue to dry.
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PastPaper.workedSolution
To prepare a pure, dry sample of a soluble salt from an acid and an insoluble base: 1. Filter the mixture to remove the excess unreacted copper(II) oxide (insoluble solid residue). The filtrate is a copper(II) sulfate solution. 2. Heat the filtrate to evaporate some of the water until the crystallisation point (saturated solution) is reached. 3. Allow the hot saturated solution to cool down slowly to allow copper(II) sulfate crystals to form. 4. Filter the crystals from the remaining solution and dry them between filter papers. Evaporating to dryness (Option A) would yield anhydrous copper(II) sulfate powder instead of hydrated crystals. Skipping the initial filtration (Options C and D) would leave unreacted copper(II) oxide contaminating the product.
PastPaper.markingScheme
1 mark for the correct option B. Candidate correctly sequences the steps: filtration of excess reactant, concentration of filtrate, cooling for crystallisation, followed by separation and drying of the crystals.
PastPaper.question 37 · MCQ
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Three identical \(6.0\ \Omega\) resistors are connected in a circuit. Two of the resistors are connected in parallel with each other, and this combination is connected in series with the third resistor and a \(9.0\text{ V}\) d.c. power supply of negligible internal resistance. What is the total current drawn from the power supply?
A.\(0.50\text{ A}\)
B.\(1.0\text{ A}\)
C.\(1.5\text{ A}\)
D.\(3.0\text{ A}\)
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PastPaper.workedSolution
First, calculate the equivalent resistance of the two identical \(6.0\ \Omega\) resistors in parallel: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). Next, add the resistance of the third \(6.0\ \Omega\) resistor which is in series with this parallel combination: \(R_{\text{total}} = R_p + 6.0\ \Omega = 3.0\ \Omega + 6.0\ \Omega = 9.0\ \Omega\). Finally, use Ohm's law to find the total current drawn from the \(9.0\text{ V}\) power supply: \(I = \frac{V}{R_{\text{total}}} = \frac{9.0\text{ V}}{9.0\ \Omega} = 1.0\text{ A}\).
PastPaper.markingScheme
1 mark for the correct option B. Correct calculation of parallel resistance (3.0 ohms), total resistance (9.0 ohms), and application of Ohm's law to find current (1.0 A).
PastPaper.question 38 · MCQ
1 PastPaper.marks
An electric motor is used to lift a load of \(40\text{ kg}\) vertically through a height of \(15\text{ m}\) in a time of \(6.0\text{ s}\). The acceleration of free fall, \(g\), is \(10\text{ m/s}^2\). Assuming the motor is \(80\%\) efficient, what is the electrical power input to the motor?
A.\(800\text{ W}\)
B.\(1000\text{ W}\)
C.\(1250\text{ W}\)
D.\(6000\text{ W}\)
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PastPaper.workedSolution
First, calculate the useful work done in lifting the load, which is equal to the gain in gravitational potential energy (GPE): \(E = mgh = 40\text{ kg} \times 10\text{ m/s}^2 \times 15\text{ m} = 6000\text{ J}\). Next, calculate the useful power output of the motor: \(P_{\text{out}} = \frac{E}{t} = \frac{6000\text{ J}}{6.0\text{ s}} = 1000\text{ W}\). Since the motor is only \(80\%\) efficient (efficiency = \(0.80\)), the electrical power input \(P_{\text{in}}\) is greater than the useful power output: \(P_{\text{in}} = \frac{P_{\text{out}}}{\text{efficiency}} = \frac{1000\text{ W}}{0.80} = 1250\text{ W}\).
PastPaper.markingScheme
1 mark for the correct option C. Correct steps include finding work (6000 J), finding output power (1000 W), and dividing by efficiency (0.80) to find input power (1250 W).
PastPaper.question 39 · MCQ
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Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at the anode and the cathode, and the change in pH of the electrolyte?
A.anode product: chlorine gas; cathode product: hydrogen gas; change in pH of electrolyte: increases
B.anode product: oxygen gas; cathode product: sodium metal; change in pH of electrolyte: decreases
C.anode product: chlorine gas; cathode product: sodium metal; change in pH of electrolyte: no change
D.anode product: oxygen gas; cathode product: hydrogen gas; change in pH of electrolyte: increases
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PastPaper.workedSolution
During the electrolysis of concentrated aqueous sodium chloride (brine): 1. At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions due to their high concentration, forming chlorine gas (\(\text{Cl}_2\)). 2. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is less reactive than sodium, forming hydrogen gas (\(\text{H}_2\)). 3. The removal of \(\text{H}^+\) and \(\text{Cl}^-\体积\) leaves behind sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) in the solution, forming sodium hydroxide (\(\text{NaOH}\)). Since sodium hydroxide is alkaline, the pH of the electrolyte increases. This makes option A the correct choice.
PastPaper.markingScheme
1 mark for the correct option A. Candidate correctly identifies chlorine at the anode, hydrogen at the cathode, and an increase in pH due to the formation of sodium hydroxide.
PastPaper.question 40 · MCQ
1 PastPaper.marks
Which statement about electromagnetic waves is correct?
A.Infrared waves have a higher frequency than ultraviolet waves and both travel at the same speed in a vacuum.
B.Radio waves have a longer wavelength than microwaves and both travel at different speeds in a vacuum.
C.Gamma rays have a higher frequency than X-rays and both travel at the same speed in a vacuum.
D.Visible light has a shorter wavelength than ultraviolet waves and both travel at different speeds in a vacuum.
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PastPaper.workedSolution
Let's review the electromagnetic spectrum in order of increasing frequency (and decreasing wavelength): radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. All electromagnetic waves travel at the same speed (the speed of light, \(c \approx 3 \times 10^8\text{ m/s}\)) in a vacuum. Gamma rays have higher frequency than X-rays and both travel at the same speed in a vacuum, making option C correct. Option A is incorrect because infrared has lower frequency than ultraviolet. Option B is incorrect because they travel at the same speed in a vacuum. Option D is incorrect because visible light has longer wavelength than ultraviolet, and they travel at the same speed in a vacuum.
PastPaper.markingScheme
1 mark for the correct option C. Candidate correctly compares electromagnetic wave frequencies/wavelengths and recognises that all EM waves travel at the same speed in a vacuum.
Paper 41 Extended Theory
Answer all twelve structured theory questions. Write answers in the spaces provided.
12 PastPaper.question · 120 PastPaper.marks
PastPaper.question 1 · Structured
10 PastPaper.marks
A toy car of mass \(0.5\text{ kg}\) is released from rest at the top of a smooth slope, which is at a vertical height of \(1.2\text{ m}\).
(a) Calculate the gravitational potential energy of the car at the top of the slope. (Take \(g = 9.8\text{ m/s}^2\)) [2]
(b) The car reaches the bottom of the slope. If \(15\%\) of the initial potential energy is lost as thermal energy due to friction, calculate the kinetic energy of the car at the bottom of the slope. [2]
(c) Calculate the speed of the car at the bottom of the slope. [3]
(d) Describe the difference between speed and velocity. [3]
(b) Since \(15\%\) of the energy is lost, the remaining kinetic energy (KE) is \(85\%\) of the initial GPE: \(KE = 5.88\text{ J} \times 0.85 = 4.998\text{ J} \approx 5.0\text{ J}\).
(c) Kinetic energy is given by: \(KE = \frac{1}{2}mv^2\) \(4.998 = \frac{1}{2} \times 0.5 \times v^2\) \(v^2 = \frac{4.998}{0.25} = 19.992\) \(v = \sqrt{19.992} \approx 4.47\text{ m/s}\) (or \(4.5\text{ m/s}\)).
(d) Speed is a scalar quantity, meaning it has magnitude only. Velocity is a vector quantity, meaning it has both magnitude and direction.
PastPaper.markingScheme
(a) - Formula: \(GPE = mgh\) [1] - Correct calculation: \(5.88\text{ J}\) (accept \(5.9\text{ J}\) or \(6.0\text{ J}\) if \(g = 10\text{ m/s}^2\) is used) [1]
(b) - Recognises that \(85\%\) of the energy is kept: \(0.85 \times 5.88\) [1] - Correct calculation: \(5.0\text{ J}\) (accept \(5.1\text{ J}\) if \(g = 10\text{ m/s}^2\) is used) [1]
(d) - Speed is a scalar/has only magnitude [1] - Velocity is a vector/has magnitude and direction [1] - Explicit mention of direction being the difference [1]
PastPaper.question 2 · Structured
10 PastPaper.marks
Ethane, \(C_2H_6\), and ethene, \(C_2H_4\), are hydrocarbons.
(a) (i) Explain why ethene is described as an unsaturated hydrocarbon. [2] (ii) Describe a chemical test to distinguish between ethane and ethene. Give the results for each compound. [3]
(b) Ethene can be polymerized to form poly(ethene). (i) Draw the structure of poly(ethene) showing two repeat units. [2] (ii) State the type of polymerization that occurs when poly(ethene) is formed. [1]
(c) Large alkane molecules can be cracked to produce smaller, more useful molecules. Write a balanced chemical equation for the cracking of decane, \(C_{10}H_{22}\), to produce oct-1-ene, \(C_8H_{16}\), and one other product. [2]
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PastPaper.workedSolution
(a) (i) Ethene is unsaturated because it contains a carbon-carbon double bond (\(C=C\)). It is a hydrocarbon because it is composed of hydrogen and carbon atoms only. (ii) Add bromine water (aqueous bromine) to both compounds. Ethane will show no color change (remains orange/yellow/brown), while ethene will decolorize the bromine water (turns from orange to colorless).
(b) (i) Poly(ethene) with two repeat units is drawn with four carbon atoms in a single-bonded chain, each carbon with two single-bonded hydrogen atoms, and single open-ended bonds at both ends: \(-CH_2-CH_2-CH_2-CH_2-\). (ii) The process is addition polymerization.
(c) Cracking decane, \(C_{10}H_{22}\), produces oct-1-ene, \(C_8H_{16}\), and ethane, \(C_2H_6\). The balanced chemical equation is: \(C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_6\).
PastPaper.markingScheme
(a) (i) - Unsaturated because it contains a double bond (\(C=C\)) [1] - Hydrocarbon because it contains carbon and hydrogen only [1] (ii) - Test: Add bromine water / aqueous bromine [1] - Result with ethane: remains orange / yellow / brown [1] - Result with ethene: turns colorless / decolorizes [1]
(b) (i) - Two correct repeating monomer units connected by single carbon-carbon bonds (\(-C-C-C-C-\)) [1] - Correct hydrogens on carbons and open single bonds shown at both ends [1] (ii) - Addition (polymerization) [1]
(c) - Correct formula of product \(C_2H_6\) (ethane) [1] - Fully correct balanced equation: \(C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_6\) [1]
PastPaper.question 3 · Structured
10 PastPaper.marks
Enzymes are biological catalysts that speed up metabolic reactions.
(a) Explain how enzymes function in terms of the active site, substrate, and complementary shape (the lock and key model). [3]
(b) An experiment was carried out to study the rate of reaction of amylase at different temperatures. (i) Describe and explain the effect of increasing the temperature from \(20^\circ\text{C}\) to \(40^\circ\text{C}\) on the rate of reaction. [3] (ii) Describe and explain the effect of increasing the temperature above \(60^\circ\text{C}\) on the rate of reaction. [4]
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PastPaper.workedSolution
(a) Enzymes have a specifically shaped region called the active site. The substrate has a shape complementary to this active site. According to the lock and key model, the substrate fits precisely into the active site to form an enzyme-substrate complex, enabling the reaction to occur.
(b) (i) As temperature increases from \(20^\circ\text{C}\) to \(40^\circ\text{C}\), the rate of reaction increases. This is because the kinetic energy of both enzyme and substrate molecules increases, causing them to move faster. This results in a higher frequency of successful collisions between the active sites and substrate molecules.
(ii) As temperature increases above \(60^\circ\text{C}\), the rate of reaction drops rapidly to zero. At high temperatures, the thermal energy causes bonds within the enzyme molecule to break. This changes the shape of the active site, a process called denaturation. The substrate can no longer fit into the active site, so no reaction takes place.
PastPaper.markingScheme
(a) - Active site has a specific 3D shape [1] - Substrate shape is complementary to active site [1] - Substrate binds to active site to form an enzyme-substrate complex / lock and key description [1]
(b) (i) - Rate of reaction increases [1] - Kinetic energy of molecules increases / they move faster [1] - Frequency of successful collisions between enzyme and substrate increases [1]
(b) (ii) - Rate of reaction decreases / drops to zero [1] - Bonds holding the enzyme structure break [1] - Active site changes shape / enzyme denatures [1] - Substrate can no longer fit/bind to the active site [1]
PastPaper.question 4 · Structured
10 PastPaper.marks
A student prepares a sample of copper(II) sulfate crystals by reacting dilute sulfuric acid with copper(II) oxide.
(a) (i) Write a balanced chemical equation for this reaction. [2] (ii) State why copper(II) oxide is added in excess. [1] (iii) Describe how the excess copper(II) oxide is removed from the mixture. [1] (iv) Describe how dry crystals of copper(II) sulfate are obtained from the resulting solution. [3]
(b) Explain, in terms of ions, the difference between a strong acid and a weak acid. [3]
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PastPaper.workedSolution
(a) (i) Dilute sulfuric acid reacts with solid copper(II) oxide to produce copper(II) sulfate solution and water: \(H_2SO_4 + CuO \rightarrow CuSO_4 + H_2O\). (ii) Copper(II) oxide is added in excess to ensure that all of the sulfuric acid in the mixture reacts completely. (iii) The excess solid copper(II) oxide can be removed by filtration. (iv) To get dry crystals, heat the copper(II) sulfate solution in an evaporating basin to evaporate some water until a saturated solution is formed (crystallization point). Allow the solution to cool slowly so that crystals form. Finally, filter the crystals from the remaining liquid and dry them using filter paper.
(b) A strong acid completely ionizes (dissociates) in aqueous solution, releasing a high concentration of hydrogen ions (\(H^+\)). In contrast, a weak acid only partially ionizes in aqueous solution, resulting in a much lower concentration of hydrogen ions (\(H^+\)) in solution at the same concentration.
PastPaper.markingScheme
(a) (i) - Reactants and products correct: \(H_2SO_4 + CuO \rightarrow CuSO_4 + H_2O\) [1] - Correct balancing of the chemical equation [1] (ii) - To ensure all the acid is fully neutralized / reacted [1] (iii) - Filtration / filter the mixture [1] (iv) - Heat/evaporate the filtrate to saturation / crystallization point [1] - Cool to allow crystals to grow [1] - Filter crystals and dry using filter paper [1]
(b) - Strong acid: fully/completely dissociates/ionizes to release \(H^+\) ions [1] - Weak acid: partially/incomplete dissociation/ionization of molecules [1] - Relates this to a higher concentration of \(H^+\) ions in a strong acid than a weak acid of equal concentration [1]
PastPaper.question 5 · Structured
10 PastPaper.marks
An electrical circuit consists of a \(12\text{ V}\) d.c. power supply, a \(6.0\ \Omega\) resistor, and a \(12.0\ \Omega\) resistor connected in parallel.
(a) (i) Calculate the combined resistance of the two resistors in parallel. [2] (ii) Calculate the total current drawn from the power supply. [2] (iii) Calculate the electrical power dissipated in the \(6.0\ \Omega\) resistor. [3]
(b) Explain how the resistance of a thermistor changes as temperature increases, and describe how this property can be utilized in an automatic heating controller. [3]
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PastPaper.workedSolution
(a) (i) The combined resistance \(R_p\) of resistors in parallel is given by: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0}\) \(R_p = 4.0\ \Omega\).
(ii) Total current \(I\) is calculated using Ohm's Law: \(I = \frac{V}{R_p} = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\).
(iii) In a parallel circuit, the voltage across each branch is equal to the supply voltage, so the potential difference across the \(6.0\ \Omega\) resistor is \(12\text{ V}\). Power \(P = \frac{V^2}{R} = \frac{12^2}{6.0} = \frac{144}{6.0} = 24\text{ W}\).
(b) As the temperature of a thermistor increases, its electrical resistance decreases. In an automatic heating controller, the thermistor is connected in a potential divider circuit. A decrease in temperature causes its resistance to increase, which increases the voltage across it. This change in voltage can be used as an input signal to trigger a switch (or relay) to turn on the heater when the temperature drops below a certain level.
(ii) - Formula: \(I = \frac{V}{R}\) [1] - Calculation: \(3.0\text{ A}\) (accept error carried forward from (a)(i)) [1]
(iii) - Identifies voltage across the \(6.0\ \Omega\) resistor is \(12\text{ V}\) [1] - Formula: \(P = \frac{V^2}{R}\) or calculates current \(I = 2\text{ A}\) then \(P = I^2 R\) [1] - Calculation: \(24\text{ W}\) [1]
(b) - Resistance of the thermistor decreases as temperature increases [1] - Voltage across thermistor (or series resistor) changes in a potential divider [1] - This voltage change triggers a circuit/relay to turn the heater on or off [1]
PastPaper.question 6 · Structured
10 PastPaper.marks
Plant reproduction can occur sexually through pollination.
(a) Define pollination. [2]
(b) Compare insect-pollinated and wind-pollinated flowers in terms of: (i) the appearance and structure of the petals, [2] (ii) the structure and position of the anthers and stigma. [4]
(c) Describe the process of fertilization in a flowering plant, starting from the pollination of the stigma. [2]
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PastPaper.workedSolution
(a) Pollination is defined as the transfer of pollen grains from the anther of a flower to the stigma of a flower.
(b) (i) Insect-pollinated flowers have large, brightly colored petals (often scented or with nectar) to attract pollinators. Wind-pollinated flowers have small, dull-colored (often green) petals and lack scent and nectar.
(ii) In insect-pollinated flowers, the anthers are small and firmly attached inside the flower to rub against visiting insects, and the stigma is sticky and positioned inside the flower. In wind-pollinated flowers, the anthers are large and hang outside on long, flexible filaments to easily release pollen into the wind, and the stigma is large, feathery, and hangs outside the flower to catch drifting pollen.
(c) After landing on the stigma, a pollen grain germinates and grows a pollen tube down through the style into the ovary. The male gamete nucleus travels down this tube and enters the ovule, where it fuses with the female gamete nucleus to form a zygote.
PastPaper.markingScheme
(a) - Transfer of pollen grains [1] - From anther to stigma [1]
(b) (i) - Insect-pollinated: petals are large / brightly colored / scented / have nectar [1] - Wind-pollinated: petals are small / dull / green / absent [1]
(b) (ii) - Insect-pollinated: anthers and stigma are enclosed inside the flower [1]; stigma is sticky / anthers are small/firm [1] - Wind-pollinated: anthers and stigma hang outside the flower [1]; stigma is feathery / anthers are on long/flexible filaments [1]
(c) - Pollen tube grows down the style (into ovary/ovule) [1] - Male nucleus fuses with the female nucleus / ovule egg cell nucleus [1]
PastPaper.question 7 · Structured
10 PastPaper.marks
Electrolysis is the breakdown of an ionic compound by the passage of electricity.
(a) Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. (i) Identify the product formed at the anode and the product formed at the cathode. [2] (ii) Write the ionic half-equation for the reaction occurring at the anode. [2] (iii) Explain why the solution becomes alkaline as the electrolysis proceeds. [2]
(b) Describe how a copper key can be electroplated with nickel. In your answer, specify: - the electrolyte used - the material used for the anode (+) - the material used for the cathode (-) - the ionic half-equation at the cathode. [4]
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PastPaper.workedSolution
(a) (i) During the electrolysis of concentrated aqueous sodium chloride, chloride ions (\(Cl^-\)) are discharged at the anode to produce chlorine gas, and hydrogen ions (\(H^+\)) are discharged at the cathode to produce hydrogen gas.
(ii) Chloride ions lose electrons at the anode (oxidation): \(2Cl^- \rightarrow Cl_2 + 2e^-\).
(iii) Concentrated sodium chloride contains \(Na^+\), \(Cl^-\), \(H^+\), and \(OH^-\). As \(H^+\) and \(Cl^-\) are removed during the reaction, \(Na^+\) and \(OH^-\) remain in the solution. This forms sodium hydroxide, which is a soluble base (alkali), thus increasing the pH.
(b) To electroplate a copper key with nickel: - The electrolyte should be a soluble nickel salt solution, such as aqueous nickel(II) sulfate (\(NiSO_4\)). - The anode (positive electrode) must be made of pure nickel metal. - The cathode (negative electrode) must be the copper key to be plated. - At the cathode, nickel ions are reduced to nickel metal: \(Ni^{2+} + 2e^- \rightarrow Ni\).
PastPaper.markingScheme
(a) (i) - Anode product: Chlorine gas (\(Cl_2\)) [1] - Cathode product: Hydrogen gas (\(H_2\)) [1] (ii) - Reactant and product correct: \(2Cl^- \rightarrow Cl_2\) [1] - Electrons balanced correctly: \(2Cl^- \rightarrow Cl_2 + 2e^-\) [1] (iii) - \(H^+\) and \(Cl^-\) ions are discharged/removed from solution [1] - \(Na^+\) and \(OH^-\) ions remain in solution, forming alkaline sodium hydroxide [1]
A ray of light in air is incident on the flat surface of a glass block at an angle of incidence of \(40^\circ\). The refractive index of the glass is \(1.5\).
(a) (i) Calculate the angle of refraction in the glass. [3] (ii) State what happens to the speed of the light wave as it enters the glass block. [1]
(b) (i) State two properties that are common to all electromagnetic waves. [2] (ii) Name the component of the electromagnetic spectrum that has the longest wavelength. [1]
(c) Describe one hazard and one safety precaution associated with exposure to ultraviolet (UV) radiation. [3]
(ii) The speed of the light wave decreases (it slows down) as it enters the optically denser glass block.
(b) (i) Properties common to all electromagnetic waves: 1. They all travel at the same speed in a vacuum (\(3.0 \times 10^8\text{ m/s}\)). 2. They are all transverse waves (or can travel through a vacuum, or transfer energy). (ii) Radio waves have the longest wavelength in the electromagnetic spectrum.
(c) A major hazard of UV radiation is that it can cause damage to skin cells, leading to sunburn and increasing the risk of skin cancer (or cataracts/eye damage). A precaution is to apply sunscreen with a high SPF (or wear UV-blocking sunglasses, protective clothing, or limit exposure to mid-day sun) which absorbs or blocks the harmful UV rays before they reach the tissue.
(b) (i) - Any two from: travel at \(3 \times 10^8\text{ m/s}\) in a vacuum / are transverse waves / can travel through a vacuum / transfer energy [2] (ii) - Radio waves [1]
A test car of mass \(800\text{ kg}\) is traveling along a straight horizontal track.
(a) Describe the difference between a scalar quantity, such as speed, and a vector quantity, such as velocity. [2]
(b) The car starts from rest and accelerates uniformly to a speed of \(15\text{ m/s}\) in a time of \(6.0\text{ s}\).
(i) Calculate the acceleration of the car. Show your working and state the unit. [2]
(ii) Calculate the kinetic energy of the car when it is traveling at \(15\text{ m/s}\). [2]
(c) After \(6.0\text{ s}\), the car travels at a constant speed of \(15\text{ m/s}\) for a further \(10\text{ s}\). During this time, the engine exerts a forward force of \(1200\text{ N}\).
(i) Calculate the work done by the engine during this \(10\text{ s}\) interval. [2]
(ii) Explain, in terms of forces, why the car travels at a constant speed even though the engine is exerting a forward force of \(1200\text{ N}\). [2]
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PastPaper.workedSolution
(a) A scalar quantity has magnitude (size) only (e.g., speed), whereas a vector quantity has both magnitude and a specific direction (e.g., velocity).
(b)(i) Use the formula for acceleration: \(a = \frac{v - u}{t}\) \(a = \frac{15\text{ m/s} - 0}{6.0\text{ s}} = 2.5\text{ m/s}^2\) Unit: \(\text{m/s}^2\) (or \(\text{m s}^{-2}\)).
(b)(ii) Use the kinetic energy formula: \(KE = \frac{1}{2} m v^2\) \(KE = 0.5 \times 800\text{ kg} \times (15\text{ m/s})^2\) \(KE = 400 \times 225 = 90\,000\text{ J}\) (or \(90\text{ kJ}\)).
(c)(i) First, calculate the distance traveled during the 10 s interval at constant speed: \(d = v \times t = 15\text{ m/s} \times 10\text{ s} = 150\text{ m}\) Then, calculate the work done: \(W = F \times d = 1200\text{ N} \times 150\text{ m} = 180\,000\text{ J}\) (or \(180\text{ kJ}\)).
(c)(ii) The car travels at a constant speed because the forward engine force of 1200 N is exactly balanced by opposing resistive forces (friction and air resistance) summing to 1200 N. The resultant force on the car is zero, which means there is no acceleration.
PastPaper.markingScheme
(a) - Scalar has magnitude only [1] - Vector has magnitude and direction [1]
(b)(ii) - Correct formula or substitution: \(0.5 \times 800 \times 15^2\) [1] - Correct final value: \(90\,000\text{ J}\) (or \(90\text{ kJ}\)) [1]
(c)(i) - Correct distance calculation (\(150\text{ m}\)) or substitution into work formula [1] - Correct work done: \(180\,000\text{ J}\) (or \(180\text{ kJ}\)) [1]
(c)(ii) - Mention of equal and opposite drag / air resistance / friction of \(1200\text{ N}\) [1] - Resultant (net) force is zero [1]
PastPaper.question 10 · Structured
10 PastPaper.marks
(a) Define the term enzyme. [2]
(b) Describe how an enzyme catalyzes a reaction, with reference to the active site and the 'lock and key' hypothesis. [3]
(c) An experiment was conducted to study the effect of temperature on the rate of an amylase-catalyzed reaction.
(i) Explain why the rate of reaction is very low or zero when the enzyme is incubated at \(60^\circ\text{C}\). Use the term denatured in your answer. [3]
(ii) Explain why the rate of reaction decreases when the temperature is lowered from the optimum temperature of \(37^\circ\text{C}\) to \(10^\circ\text{C}\). [2]
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PastPaper.workedSolution
(a) An enzyme is defined as a protein [1] that acts as a biological catalyst [1] (speeds up metabolic reactions without being permanently altered or consumed).
(b) The enzyme has a specifically shaped region called the active site. According to the 'lock and key' hypothesis, the substrate molecule (the 'key') has a complementary shape to the active site (the 'lock') [1]. The substrate binds to the active site [1] to form an enzyme-substrate complex, where the chemical transformation takes place to release products [1].
(c)(i) At high temperatures such as \(60^\circ\text{C}\), the excessive thermal energy causes the atoms in the enzyme to vibrate rapidly, breaking the weak bonds holding its specific 3D tertiary structure together [1]. This alters the specific shape of the active site [1], making it no longer complementary to the substrate. The enzyme is said to be denatured [1].
(c)(ii) At lower temperatures (from \(37^\circ\text{C}\) to \(10^\circ\text{C}\)), the molecules have less kinetic energy [1]. They move more slowly, which results in a lower frequency of successful collisions [1] between enzyme active sites and substrate molecules, reducing the reaction rate.
PastPaper.markingScheme
(a) - Protein [1] - Biological catalyst / speeds up reactions [1]
(b) - Active site has a complementary shape to the substrate [1] - Substrate fits / binds to the active site [1] - Mention of 'lock and key' fit / formation of enzyme-substrate complex [1]
(c)(i) - High temperature breaks bonds holding the protein structure [1] - Shape of the active site changes / substrate can no longer fit [1] - Mention of enzyme being denatured [1]
(c)(ii) - Molecules have less kinetic energy [1] - Fewer / less frequent collisions between enzyme and substrate [1]
PastPaper.question 11 · Structured
10 PastPaper.marks
A student investigates the electrolysis of concentrated aqueous copper(II) sulfate using inert carbon (graphite) electrodes.
(a) Describe what is observed at:
(i) the anode (positive electrode) [1]
(ii) the cathode (negative electrode) [1]
(iii) the electrolyte solution as the reaction proceeds. [1]
(b) Write ionic half-equations, including state symbols, for the reaction occurring at:
(i) the anode [2]
(ii) the cathode [2]
(c) The carbon electrodes are now replaced with copper electrodes.
(i) Describe the change in mass of the copper anode. [1]
(ii) Explain why the blue color of the electrolyte does not fade during this electrolysis. [2]
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PastPaper.workedSolution
(a)(i) At the anode, hydroxide ions (\(\text{OH}^-\)) from water are discharged, producing bubbles of a colorless gas, which is oxygen: \(4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-\).
(a)(ii) At the cathode, copper(II) ions (\(\text{Cu}^{2+}\)) are discharged because copper is lower than hydrogen in the reactivity series. This forms a pink/red-brown solid deposit of copper metal.
(a)(iii) Since \(\text{Cu}^{2+}\) ions (which give the solution its blue color) are being discharged and removed from the solution, the blue color of the electrolyte gradually fades.
(b)(i) The balanced half-equation at the anode is: \(4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-\) (or: \(2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4\text{e}^-\))
(b)(ii) The balanced half-equation at the cathode is: \(Cu^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)
(c)(i) When copper electrodes are used, the copper anode dissolves to form copper ions. Therefore, the mass of the copper anode decreases.
(c)(ii) The blue color does not fade because for every copper ion (\(\text{Cu}^{2+}\)) discharged at the cathode, one copper atom from the anode oxidizes and dissolves to replace it as a \(\text{Cu}^{2+}\) ion in the solution. Hence, the concentration of blue \(\text{Cu}^{2+}\) ions remains constant.
PastPaper.markingScheme
(a)(i) - Bubbles / effervescence of colorless gas [1] (Accept: oxygen gas released)
(a)(iii) - Blue color fades / turns lighter / turns colorless [1]
(b)(i) - \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (1 mark for correct formulas of species, 1 mark for balancing and state symbols) [2]
(b)(ii) - \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\) (1 mark for species, 1 mark for balancing and state symbols) [2]
(c)(ii) - Copper atoms from anode dissolve/oxidize to form \(\text{Cu}^{2+}\) ions [1] - Rate of dissolution at anode equals rate of deposition at cathode / concentration of \(\text{Cu}^{2+}\) remains constant [1]
PastPaper.question 12 · Structured
10 PastPaper.marks
Hydrocarbons are organic compounds containing only hydrogen and carbon.
(a) Decane, \(\text{C}_{10}\text{H}_{22}\), is an alkane. It can be cracked to produce ethene, \(\text{C}_2\text{H}_4\), and another hydrocarbon, \(X\).
(i) Write a balanced chemical equation for the cracking of decane to produce ethene and hydrocarbon \(X\). State the molecular formula of \(X\). [2]
(ii) Describe a chemical test to distinguish between decane and ethene. Give the observation for each compound. [3]
(b) Ethene undergoes addition polymerization to form poly(ethene).
(i) Draw the structure of the monomer, ethene, showing all atoms and bonds. [1]
(ii) Draw a diagram to represent a section of poly(ethene) showing two repeating units. [2]
(iii) Poly(ethene) is non-biodegradable. State two environmental problems caused by the disposal of non-biodegradable plastics in the environment. [2]
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PastPaper.workedSolution
(a)(i) The cracking reaction involves breaking a larger alkane into smaller alkanes and alkenes. \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\) The molecular formula of hydrocarbon \(X\) is \(\text{C}_8\text{H}_{18}\) (octane).
(a)(ii) To distinguish an alkane (decane) from an alkene (ethene), add aqueous bromine (bromine water) to both. - Ethene (unsaturated) will rapidly decolorize the orange/brown bromine water, turning it colorless. - Decane (saturated) will not react, so the mixture remains orange/brown.
(b)(i) Ethene consists of two carbon atoms connected by a double bond, with each carbon atom bonded to two hydrogen atoms: H H \ / C=C / \ H H
(b)(ii) During addition polymerization, the double bond opens up to form single covalent bonds connecting the monomers. A section with two repeating units shows a sequence of 4 carbon atoms with open-ended bonds at both ends: H H H H | | | | - C - C - C - C - | | | | H H H H
(b)(iii) Non-biodegradable plastics persist in the environment for centuries. - They accumulate in landfill sites, taking up valuable space. - They pose ingestion and entanglement hazards to animals and marine life. - Incineration can release toxic gases into the atmosphere.
PastPaper.markingScheme
(a)(i) - Correct balanced chemical equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\) [1] - Correct formula for \(X\) identified as \(\text{C}_8\text{H}_{18}\) [1]
(a)(ii) - Test: Add bromine water / aqueous bromine [1] - Result with decane: stays orange / brown / yellow / no change [1] - Result with ethene: decolourizes / turns colorless [1]
(b)(i) - Correct structure of ethene showing double bond between carbons and four C-H single bonds [1]
(b)(ii) - Correct backbone of 4 carbon atoms linked by single bonds, with open-ended bonds at each end [1] - Correct number and placement of H atoms (8 hydrogens total) [1]
(b)(iii) - Credit any two environmental issues: e.g., fills up landfills / does not rot [1]; harms wildlife by ingestion / entanglement [1]; releases toxic gases if burned [1] (Max 2)
Paper 61 Alternative to Practical
Answer all experimental design and analysis questions. Include units where necessary.
6 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · Practical
10 PastPaper.marks
A student investigates the effect of pH on the activity of amylase. Amylase breaks down starch. In five separate test-tubes, amylase is mixed with starch solution and a buffer solution of a specific pH. Every 30 seconds, a sample from each mixture is added to a drop of iodine solution on a white tile. The time taken for the starch to be completely broken down is measured. (a) Identify: (i) the independent variable. [1] (ii) the dependent variable. [1] (b) At pH 7, the student records that the blue-black color of iodine no longer appears after 2 minutes and 30 seconds. (i) Convert this time into seconds. [1] (ii) State the final color of the iodine mixture when starch has been completely broken down. [1] (c) Explain why the test-tubes containing the amylase and starch solutions are placed in a water bath at 35 degrees Celsius for 5 minutes before mixing them. [2] (d) Explain why a water bath is used to maintain the temperature instead of directly heating the test-tubes with a Bunsen burner. [2] (e) Describe a control experiment that the student could perform to prove that the breakdown of starch is catalyzed by the active enzyme amylase. [2]
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PastPaper.workedSolution
(a)(i) The independent variable is the pH. (a)(ii) The dependent variable is the time taken for the starch to be completely broken down. (b)(i) 2 minutes and 30 seconds = 150 seconds. (b)(ii) The final color is orange / brown / yellow-brown. (c) Placing the solutions in the water bath for 5 minutes before mixing allows both the amylase and starch solutions to reach the target temperature (35 degrees Celsius) so that the reaction begins exactly at the controlled temperature. (d) A water bath distributes heat more evenly and allows for precise temperature control, whereas direct heating with a Bunsen burner causes localized overheating and can denature the amylase enzyme. (e) Set up an identical test-tube but use boiled/denatured amylase instead of active amylase, keeping all other factors (starch solution, pH, temperature) constant. The starch will not be broken down, and the iodine will remain blue-black.
PastPaper.markingScheme
(a)(i) 1 mark: pH. (a)(ii) 1 mark: time taken for starch to be completely broken down / rate of starch digestion. (b)(i) 1 mark: 150 (s). (b)(ii) 1 mark: orange / brown / yellow-brown (reject yellow, accept red-brown). (c) 2 marks: 1 mark for allowing both solutions to reach the target temperature / equilibrate; 1 mark for stating this ensures temperature is constant from the start of the reaction. (d) 2 marks: 1 mark for maintaining a constant and controlled temperature; 1 mark for avoiding hot spots / direct heat which denatures the enzyme. (e) 2 marks: 1 mark for using boiled/denatured amylase (or distilled water instead of amylase); 1 mark for keeping other factors constant and showing that starch remains present (iodine remains blue-black).
PastPaper.question 2 · Practical
10 PastPaper.marks
A student investigates the temperature change when different masses of sodium hydrogen carbonate are added to 25 cubic centimeters of dilute hydrochloric acid in a polystyrene cup. The initial temperature of the acid is measured. The sodium hydrogen carbonate is added, the mixture is stirred, and the lowest temperature reached is recorded. (a) For one mass, the initial temperature of the acid was 21.5 degrees Celsius. The lowest temperature recorded was 14.5 degrees Celsius. (i) Calculate the temperature change. [1] (ii) State whether this reaction is exothermic or endothermic, and justify your answer. [2] (b) State two reasons why a polystyrene cup is used in this experiment instead of a glass beaker. [2] (c) Identify two variables that must be kept constant in this investigation to ensure a fair test. [2] (d) Describe how the student could measure the 25 cubic centimeters of dilute hydrochloric acid as accurately as possible. [1] (e) Suggest one safety hazard in this experiment and describe a suitable precaution to minimize the risk. [2]
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PastPaper.workedSolution
(a)(i) Temperature change = 21.5 - 14.5 = 7.0 degrees Celsius. (a)(ii) The reaction is endothermic because the temperature decreased, which shows that heat energy was absorbed from the surroundings. (b) Polystyrene is a good thermal insulator which minimizes heat transfer from the surroundings, resulting in a more accurate measurement of temperature change. (c) The concentration of dilute hydrochloric acid and the volume of dilute hydrochloric acid (25 cm3) must be kept constant. (d) Use a volumetric pipette or a burette to measure the acid. (e) Dilute hydrochloric acid is an irritant/corrosive and can cause damage if splashed into eyes or on skin; wear safety goggles to protect the eyes and disposable gloves to protect the skin.
PastPaper.markingScheme
(a)(i) 1 mark: 7.0 degrees Celsius (accept 7, ignore minus sign). (a)(ii) 2 marks: Endothermic (1 mark) because the temperature decreased / energy was absorbed (1 mark). (b) 2 marks: 1 mark for polystyrene being a thermal insulator; 1 mark for minimizing heat energy transfer from/to the surroundings. (c) 2 marks: any two from: volume of acid, concentration of acid, initial temperature of acid, rate of stirring (1 mark each). (d) 1 mark: use a pipette / burette (reject measuring cylinder). (e) 2 marks: 1 mark for identifying the hazard (acid is corrosive/irritant); 1 mark for the appropriate precaution (wear safety goggles / wear gloves).
PastPaper.question 3 · Practical
10 PastPaper.marks
A student investigates the relationship between the length of a simple pendulum, l, and its period of oscillation, T. The student sets up a pendulum suspended from a clamp stand. (a) The student measures the length of the pendulum l from the bottom of the support clamp to the center of the pendulum bob. On a diagram, a ruler shows the bottom of the clamp is at 0.0 cm and the center of the bob is at 75.5 cm. (i) State the length of the pendulum in meters. [1] (ii) State one precaution the student must take when using a ruler to measure this length to avoid a parallax error. [1] (b) The student displaces the bob slightly, releases it, and measures the time taken for 20 complete oscillations. (i) The stopwatch shows a time of 34.80 seconds. Calculate the period T of one oscillation. [1] (ii) Calculate the value of T squared. [1] (c) Explain why measuring the time for 20 oscillations and dividing by 20 is more accurate than measuring the time for only 1 oscillation. [2] (d) A student repeats the procedure for several lengths, plots a graph of T squared on the y-axis against length l on the x-axis, and draws a straight line of best fit through the origin. (i) The gradient of the line is calculated to be 4.02 seconds squared per meter. Using the equation \( g = \frac{4\pi^2}{\text{gradient}} \), calculate the value of the acceleration of free fall, g. Show your working. Use \( \pi = 3.142 \). [2] (ii) State one adjustment the student could make to the pendulum setup to ensure the oscillations remain in a single plane and do not wobble. [2]
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PastPaper.workedSolution
(a)(i) 75.5 cm = 0.755 m. (a)(ii) View the scale perpendicular to the ruler at the level of the reading to avoid parallax error. (b)(i) T = 34.80 / 20 = 1.74 s. (b)(ii) T^2 = 1.74^2 = 3.03 s^2. (c) Measuring 20 oscillations reduces the percentage error/uncertainty caused by the human reaction time when starting and stopping the stopwatch. (d)(i) g = 4 * (3.142)^2 / 4.02 = 4 * 9.872 / 4.02 = 39.489 / 4.02 = 9.82 m/s^2. (d)(ii) Use a split cork or two wooden blocks in the clamp jaws to firmly grip the thread at the suspension point, ensuring a single pivot point, and release the bob in a straight, steady motion.
PastPaper.markingScheme
(a)(i) 1 mark: 0.755 (m) (accept 0.76). (a)(ii) 1 mark: view perpendicular to the ruler scale / view at eye-level to the reading. (b)(i) 1 mark: 1.74 (s). (b)(ii) 1 mark: 3.03 (s^2) (accept 3.0 or 3.028). (c) 2 marks: 1 mark for stating that it reduces the effect of human reaction time errors; 1 mark for stating that it reduces percentage uncertainty/error in the period. (d)(i) 2 marks: 1 mark for correct substitution: 4 * 3.142^2 / 4.02; 1 mark for final calculated value of 9.82 (or 9.8) m/s^2. (d)(ii) 2 marks: 1 mark for using a split cork to clamp the thread securely; 1 mark for releasing the bob without a side-to-side force.
PastPaper.question 4 · Practical
10 PastPaper.marks
A student investigates the fermentation of glucose solution by yeast at different temperatures. A mixture of glucose solution and yeast is placed in a test-tube connected by a delivery tube to a gas syringe. (a) After 10 minutes at 35 degrees Celsius, the gas syringe shows that 48.0 cubic centimeters of gas has been collected. (i) State the name of this gas. [1] (ii) Describe a chemical test, including the positive observation, to confirm the identity of this gas. [2] (b) State the observation when the liquid mixture from the test-tube is filtered and tested with: (i) aqueous bromine. [1] (ii) acidified potassium manganate(VII) with gentle heating. [2] (c) To determine the concentration of ethanoic acid formed as a side-product in the mixture, the student titrates a 10.0 cubic centimeter sample of the filtered reaction mixture against 0.10 mol/dm3 sodium hydroxide solution using phenolphthalein indicator. (i) State the color change of the indicator at the end-point. [2] (ii) Suggest why the student must filter the mixture before performing the titration. [2]
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(a)(i) Carbon dioxide. (a)(ii) Bubble the gas through limewater. The positive observation is that the limewater turns cloudy or milky. (b)(i) The solution remains orange-brown / no color change occurs because ethanol is saturated and does not react with bromine water. (b)(ii) The purple solution of acidified potassium manganate(VII) turns colorless (is decolorized) because ethanol is oxidized to ethanoic acid. (c)(i) The color of phenolphthalein changes from colorless to pink at the end-point. (c)(ii) The student must filter the mixture to remove yeast cells and suspended solids, which would make the mixture cloudy and prevent the student from clearly observing the indicator's color change at the end-point.
PastPaper.markingScheme
(a)(i) 1 mark: Carbon dioxide (or CO2). (a)(ii) 2 marks: bubble through limewater (1 mark); observation of turning cloudy / milky (1 mark). (b)(i) 1 mark: remains orange / brown / yellow (accept no change, reject colorless). (b)(ii) 2 marks: purple (1 mark) turns colorless (1 mark). (c)(i) 2 marks: from colorless (1 mark) to pink (1 mark). (c)(ii) 2 marks: 1 mark for removing yeast cells / solid particles; 1 mark for explaining that solids make the mixture cloudy, which obscures the color change of the indicator / endpoint.
PastPaper.question 5 · Practical
10 PastPaper.marks
A student investigates how the resistance of a wire depends on its length. The student sets up a circuit to measure the current in the wire and the potential difference across a length of the wire. (a) Draw a circuit diagram containing a cell, a switch, an ammeter, a voltmeter, and a length of resistance wire. The ammeter must measure the current through the resistance wire, and the voltmeter must measure the potential difference across the resistance wire. [3] (b) For a wire length of 50.0 cm, the student records: - Current, I = 0.45 A - Potential difference, V = 1.80 V (i) Calculate the resistance, R, of this 50.0 cm length of wire. State the unit. [2] (ii) Predict the resistance of a 100.0 cm length of the same wire, assuming resistance is directly proportional to length. [1] (c) State one reason why the switch should be opened between taking readings. [1] (d) The student notices that the wire becomes very hot during the experiment. (i) State one safety hazard associated with this. [1] (ii) Suggest one improvement to the method to prevent the wire from becoming too hot. [2]
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PastPaper.workedSolution
(a) A circuit diagram should be drawn showing a cell, switch, ammeter, and resistance wire in series to form a complete circuit loop. A voltmeter must be connected in parallel directly across the resistance wire. All standard circuit symbols must be drawn correctly. (b)(i) R = V / I = 1.80 / 0.45 = 4.0 ohms (or \(\Omega\)). (b)(ii) Since resistance is directly proportional to length, a 100.0 cm wire will have twice the resistance: 4.0 * 2 = 8.0 ohms (or \(\Omega\)). (c) Opening the switch prevents current from flowing continuously, which stops the wire from heating up. Heating changes the resistance of the wire, introducing temperature as an uncontrolled variable. (d)(i) Skin burns from touching the hot wire / fire hazard. (d)(ii) Reduce the voltage of the power supply (or use a variable resistor to reduce the current), or keep the switch closed only for the brief moment required to take the readings.
PastPaper.markingScheme
(a) 3 marks: 1 mark for cell, switch, ammeter, and resistance wire in a correct series loop; 1 mark for voltmeter in parallel across the resistance wire; 1 mark for using correct symbols for all components. (b)(i) 2 marks: 4.0 (1 mark); ohms or \(\Omega\) (1 mark). (b)(ii) 1 mark: 8.0 (ohms). (c) 1 mark: to prevent the wire heating up (which changes the resistance). (d)(i) 1 mark: hazard of skin burns / fire / melting components. (d)(ii) 2 marks: 1 mark for reducing the current / using a lower voltage power source; 1 mark for keeping the switch closed only long enough to record readings.
PastPaper.question 6 · Practical
10 PastPaper.marks
A student investigates the rate of respiration in germinating mung bean seeds using a simple respirometer. The apparatus consists of a test-tube containing the germinating seeds supported on a wire mesh. Below the mesh, there is a chemical that absorbs carbon dioxide. The test-tube is sealed with a rubber bung containing a capillary tube with a small drop of red dye. (a) State the name of a chemical that can be used to absorb carbon dioxide in this experiment. [1] (b) Explain why the drop of red dye moves along the capillary tube towards the test-tube containing the seeds. [3] (c) The student measures the position of the dye every minute for 5 minutes. The results are: - At 0 minutes: 10 mm - At 1 minute: 18 mm - At 2 minutes: 27 mm - At 3 minutes: 35 mm - At 4 minutes: 44 mm - At 5 minutes: 52 mm (i) Calculate the distance moved by the dye in 5 minutes. [1] (ii) Calculate the average rate of movement of the dye in mm/min. [1] (d) Describe a control experiment that should be carried out to prove that the movement of the dye is due to respiration by the living seeds, and not due to changes in environmental temperature or pressure. [2] (e) State one environmental factor that must be kept constant during this experiment, and explain why it must be kept constant. [2]
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PastPaper.workedSolution
(a) Soda-lime, sodium hydroxide (NaOH), or potassium hydroxide (KOH) can be used. (b) The germinating seeds absorb oxygen gas from the tube during aerobic respiration. Carbon dioxide gas produced by the seeds is absorbed by the soda-lime / sodium hydroxide. Consequently, the volume and pressure of the gas inside the test-tube decrease, causing the atmospheric pressure outside to push the drop of red dye towards the test-tube. (c)(i) Distance moved = 52 mm - 10 mm = 42 mm. (c)(ii) Average rate = 42 mm / 5 min = 8.4 mm/min. (d) Set up an identical apparatus but replace the living germinating seeds with either dead (boiled) seeds of the same mass, or with non-living glass beads of the same volume. (e) Temperature must be kept constant because fluctuations in temperature cause the gas inside the test-tube to expand or contract, which would push or pull the dye drop independently of respiration, or alter the rate of enzyme-controlled respiration in the seeds.
PastPaper.markingScheme
(a) 1 mark: soda-lime / sodium hydroxide / potassium hydroxide / calcium hydroxide. (b) 3 marks: 1 mark for stating that oxygen is absorbed by the seeds during respiration; 1 mark for stating that the carbon dioxide produced is absorbed by the chemical; 1 mark for stating that this causes a reduction in gas volume/pressure inside the tube, drawing the dye in. (c)(i) 1 mark: 42 mm (accept 42). (c)(ii) 1 mark: 8.4 mm/min (accept 8.4). (d) 2 marks: 1 mark for using dead (boiled) seeds / glass beads; 1 mark for keeping all other parameters (volume, mass, temperature) the same. (e) 2 marks: 1 mark for identifying temperature (or external atmospheric pressure); 1 mark for explaining that changes in temperature cause volume/pressure changes of the air inside the tube or affect the seeds' respiration rate.