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As temperature increases from \(20\ ^\circ\text{C}\) towards the optimum temperature (typically around \(37\ ^\circ\text{C}\) to \(40\ ^\circ\text{C}\)), the kinetic energy of the enzyme and substrate molecules increases, resulting in more frequent successful collisions and thus increasing activity. Above the optimum temperature, the thermal energy disrupts the bonds holding the enzyme's active site in shape, causing denaturation, which rapidly decreases the rate of reaction to zero by \(60\ ^\circ\text{C}\).

1 mark for selecting correct option B. Award 1 mark for identifying that activity increases to an optimum and then decreases due to denaturation.

In molten \(\text{PbBr}_2\), the ions present are \(\text{Pb}^{2+}\) and \(\text{Br}^-\). At the positive electrode (anode), negative bromide ions lose electrons (oxidation) to form bromine gas: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\). At the negative electrode (cathode), positive lead ions gain electrons (reduction) to form lead metal: \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\).

1 mark for selecting B. 1 mark for correct identification of oxidation of bromide ions to bromine gas at the positive anode.

The total distance is the area under the speed-time graph. The motion has three phases: 1) acceleration phase (triangle): \(0.5 \times 5.0\text{ s} \times 15\text{ m/s} = 37.5\text{ m}\); 2) constant speed phase (rectangle): \(10.0\text{ s} \times 15\text{ m/s} = 150.0\text{ m}\); 3) deceleration phase (triangle): \(0.5 \times 3.0\text{ s} \times 15\text{ m/s} = 22.5\text{ m}\). Total distance = \(37.5 + 150.0 + 22.5 = 210.0\text{ m}\).

1 mark for selecting C. Award 1 mark for calculating the total area under the graph as 210 m.

Resistance \(R\) is proportional to length \(L\) and inversely proportional to cross-sectional area \(A\): \(R = \rho \frac{L}{A}\). For the second wire, \(R_2 = \rho \frac{2L}{2A} = \rho \frac{L}{A} = \text{R}\). Therefore, the resistance remains \(R\).

1 mark for selecting B. 1 mark for correct use of the resistance-dimensions relationship to find the answer is R.

Both plant and animal cells contain a cell membrane, cytoplasm, and a nucleus. Only plant cells possess a cellulose cell wall, chloroplasts, and a large central permanent vacuole. Therefore, the presence of a large central vacuole and a cellulose cell wall proves it is a plant cell.

1 mark for D. 1 mark for recognizing that a cellulose cell wall and a large central vacuole are unique to plant cells in this context.

The proton number (atomic number) is \(8\), so there are \(8\) protons. The nucleon number (mass number) is \(18\), so the number of neutrons is \(18 - 8 = 10\) neutrons. The ion has a charge of \(2-\), meaning it has gained \(2\) electrons relative to the neutral atom. A neutral oxygen atom has \(8\) electrons, so the ion has \(8 + 2 = 10\) electrons.

1 mark for B. 1 mark for correctly determining protons (8), neutrons (10), and electrons (10).

All electromagnetic waves travel at the same high speed in a vacuum (approx. \(3 \times 10^8\text{ m/s}\)). The order of the electromagnetic spectrum from shortest wavelength (highest frequency) to longest wavelength (lowest frequency) is: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves. Thus, ultraviolet radiation has a shorter wavelength than infrared radiation.

1 mark for B. 1 mark for identifying that all EM waves travel at the same speed in a vacuum and comparing wavelengths correctly.

As you move from the top to the bottom of the fractionating column, the sizes of the hydrocarbon molecules increase. Consequently: the boiling point increases (stronger intermolecular forces), the viscosity increases (larger molecules tangle more, making the liquid thicker), and the flammability decreases (larger molecules are harder to ignite).

1 mark for A. 1 mark for identifying the trends of boiling point (increases), viscosity (increases), and flammability (decreases) down the column.

At temperatures above the optimum, the active site of the enzyme changes shape because the high thermal energy breaks the bonds that maintain its three-dimensional structure. This process is called denaturation. Once denatured, the substrate can no longer fit into the active site, preventing the formation of enzyme-substrate complexes and reducing the rate of reaction.

[1 mark] B is correct because above the optimum temperature, enzymes are denatured, changing the shape of their active site so the substrate no longer fits. Reject other choices: A is incorrect as kinetic energy is high; C is incorrect because enzymes do not change activation energy in this way; D is incorrect because temperature changes do not directly alter pH to cause denaturation.

The cell sap of the potato cells contains a concentration of dissolved solutes (lower water potential). When placed in a \(0.0\text{ mol/dm}^3\) (pure water) solution, which has the highest possible water potential, water moves into the potato cells down a water potential gradient by osmosis. This causes the cells to become turgid and increases the mass of the cylinder. Since the water potential gradient is steepest here, it results in the greatest percentage increase in mass.

[1 mark] A is correct because pure water (\(0.0\text{ mol/dm}^3\)) has the highest water potential, creating the steepest water potential gradient for water to enter the potato cells via osmosis. Reject B, C, and D as higher sucrose concentrations reduce the water potential of the external solution, decreasing or reversing water uptake.

The atomic (proton) number is 11, which defines the element as sodium (Na). The nucleon (mass) number is the sum of protons and neutrons: \(11 + 12 = 23\). This is represented as \({}^{23}_{11}\text{Na}\). Sodium has the electronic configuration 2,8,1, placing it in Group I of the Periodic Table, and it is a metal.

[1 mark] C is correct because the mass number is 23 (protons + neutrons), the atomic number is 11, and sodium is a Group I metal. Reject A (not a non-metal), B (incorrect mass number and group), and D (incorrect symbol and proton count).

During the electrolysis of molten lead(II) bromide: at the cathode (negative electrode), positive lead ions (\(\text{Pb}^{2+}\)) gain electrons and are reduced to form lead metal: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\). At the anode (positive electrode), negative bromide ions (\(\text{Br}^-\)) lose electrons and are oxidised to form bromine gas: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\).

[1 mark] B is correct because lead is formed at the cathode (reduction of metal ions) and bromine gas is formed at the anode (oxidation of halide ions). Reject options where cathode half-equation is incorrect or where the products are reversed.

Alkanes are saturated hydrocarbons containing only single covalent bonds (C-C and C-H), whereas alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C). This is the key structural difference between the two homologous series.

[1 mark] D is correct because it accurately defines the difference between saturated alkanes and unsaturated alkenes. Reject A because alkenes decolourise bromine water, not alkanes. Reject B because ethene has the formula \(\text{C}_2\text{H}_4\). Reject C because only ethene reacts with steam to form ethanol.

The kinetic energy gained by the car is given by: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (20\text{ m/s})^2 = 600 \times 400 = 240\,000\text{ J}\). Power is the rate of energy transfer: \(P = \frac{\text{Energy}}{\text{time}} = \frac{240\,000\text{ J}}{8.0\text{ s}} = 30\,000\text{ W} = 30\text{ kW}\).

[1 mark] C is correct. Calculation steps: Kinetic energy gained = \(240\,000\text{ J}\); Power = \(240\,000 / 8 = 30\,000\text{ W}\) or \(30\text{ kW}\). Reject other options which result from incorrect formulae or missing factor of 1/2 in kinetic energy.

First, calculate the combined resistance \(R_p\) of the parallel combination: \(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3.0}\ \Omega^{-1}\), so \(R_p = 3.0\ \Omega\). Using Ohm's Law, the total current \(I\) is: \(I = \frac{V}{R_p} = \frac{12\text{ V}}{3.0\ \Omega} = 4.0\text{ A}\). Alternatively, the current through each resistor can be found: \(I_1 = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\) and \(I_2 = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}\). The total current is \(I_1 + I_2 = 3.0\text{ A} + 1.0\text{ A} = 4.0\text{ A}\).

[1 mark] C is correct because the combined parallel resistance is \(3.0\ \Omega\) and total current \(I = 12 / 3.0 = 4.0\text{ A}\). Reject A (current for single 12 ohm branch), B (current for single 4 ohm branch), and D (sum of resistances instead of parallel formula).

Sound waves are longitudinal waves. This means that the particles of the medium (air) vibrate back and forth parallel to the direction of energy transfer (the direction of wave propagation).

[1 mark] A is correct because sound is longitudinal and particle movement is parallel. Reject B and D because sound is not a transverse wave. Reject C because particles in longitudinal waves do not vibrate perpendicular to wave travel.

Iodine solution turns blue-black in the presence of starch. If the most starch is still present, it means the rate of amylase activity was the lowest. Amylase activity is lowest at \(5\ ^\circ\text{C}\) due to the low kinetic energy of the molecules, meaning fewer successful collisions occur between the enzyme and substrate.

A is correct because the lowest temperature results in the lowest enzyme activity, leaving the most starch undigested. (1 mark)

Palisade mesophyll cells are plant cells, so they contain a cell wall, chloroplasts, and a large permanent vacuole. Red blood cells are animal cells specialized for carrying oxygen, so they lack these structures (they also lack a nucleus). Therefore, cell wall, chloroplasts, and a large permanent vacuole are all present in the palisade cell but absent in the red blood cell.

B is correct as all three listed structures are present in the plant cell but absent in the red blood cell. (1 mark)

The number of neutrons is found by subtracting the proton number from the nucleon number: \(39 - 19 = 20\). A neutral atom with proton number 19 has 19 electrons. Since the ion has a \(1+\) charge, it has lost one electron: \(19 - 1 = 18\) electrons.

A is correct because the number of neutrons is 20 and the number of electrons is 18. (1 mark)

Calcium carbonate decomposes thermally to form calcium oxide and carbon dioxide. Calcium oxide is a basic oxide that reacts with acidic impurities (silicon dioxide, \(\text{SiO}_2\)) to form liquid slag (calcium silicate, \(\text{CaSiO}_3\)), which can be easily separated from the molten iron.

B is correct because the decomposition of limestone produces calcium oxide, which neutralises silicon dioxide impurities to form slag. (1 mark)

Ethene is an alkene containing a carbon-carbon double bond, making it unsaturated. It can undergo addition reactions, such as reacting with steam to produce ethanol. Ethane is a saturated alkane containing only single bonds, so it does not undergo addition reactions.

C is correct as ethene undergoes an addition reaction with steam to form ethanol, which ethane cannot. (1 mark)

First, calculate the work done: \(\text{Work done} = \text{Force} \times \text{distance} = 400\text{ N} \times 5.0\text{ m} = 2000\text{ J}\). Next, calculate power: \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{2000\text{ J}}{8.0\text{ s}} = 250\text{ W}\).

A is correct. Formula for work done used (1 mark) and formula for power used correctly to find 250 W. (1 mark)

For the series combination, the total resistance is \(R_{\text{series}} = R + R = 2R\). For the parallel combination, the total resistance is \(R_{\text{parallel}} = \frac{R}{2}\). The ratio of series to parallel resistance is \(\frac{2R}{R/2} = 4\), which corresponds to a ratio of \(4 : 1\).

D is correct. Identifying series resistance is 2R and parallel resistance is 0.5R, yielding a ratio of 4:1. (1 mark)

All electromagnetic waves, regardless of their wavelength or frequency, travel at the same speed (speed of light, approximately \(3.0 \times 10^8\text{ m/s}\)) when they are in a vacuum.

C is correct as speed is the only constant property among the options for all electromagnetic waves in a vacuum. (1 mark)

At temperatures above the optimum, the increased thermal energy causes the weaker bonds maintaining the tertiary structure of the enzyme to break. This changes the specific shape of the active site, meaning the substrate is no longer complementary and can no longer fit to form an enzyme-substrate complex. This process is called denaturation.

1 mark for identifying that the active site of the enzyme changes shape so that the substrate can no longer fit (B).

Inspired air consists of approximately 21% oxygen, 0.04% carbon dioxide, and a variable amount of water vapour. Expired air has less oxygen (approx. 16%) as it has been absorbed for respiration, more carbon dioxide (approx. 4%) as it is excreted as a waste product of respiration, and is saturated with water vapour due to evaporation from the moist lining of the respiratory tract.

1 mark for selecting the correct row containing the accurate percentages and water vapour content for both inspired and expired air (A).

During the electrolysis of molten lead(II) bromide, the negative bromide ions (\(Br^-\)) are attracted to the positive electrode (anode). At the anode, they lose electrons (oxidation) to form bromine gas: \(2Br^- \rightarrow Br_2 + 2e^-\). Lead ions (\(Pb^{2+}\)) are reduced at the negative electrode (cathode) to form lead metal.

1 mark for identifying that oxidation of bromide ions occurs at the positive electrode (anode) (C).

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond (\(C=C\)). They undergo an addition reaction with aqueous bromine, decolourising it from orange to colourless. Alkanes are saturated hydrocarbons and do not react with bromine water under standard conditions. Thus, ethene (an alkene) is X, and ethane (an alkane) is Y.

1 mark for correctly identifying X as ethene and Y as ethane (B).

In Group I (alkali metals), reactivity increases down the group because the outer shell electron is further from the nucleus, meaning there is weaker electrostatic attraction and it is lost more easily. In Group VII (halogens), reactivity decreases down the group because the outer shell is further from the nucleus, making it harder to attract and gain an incoming electron.

1 mark for identifying that Group I reactivity increases and Group VII reactivity decreases down the groups (C).

The motion can be divided into two parts on a speed-time graph: 1) Uniform acceleration from \(0\) to \(6.0\text{ m/s}\) in \(4.0\text{ s}\). The distance is the area of the triangle: \(\frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ m/s} = 12.0\text{ m}\). 2) Constant speed of \(6.0\text{ m/s}\) for \(6.0\text{ s}\) (from \(t = 4.0\text{ s}\) to \(t = 10.0\text{ s}\)). The distance is the area of the rectangle: \(6.0\text{ s} \times 6.0\text{ m/s} = 36.0\text{ m}\). The total distance is the sum of these areas: \(12.0\text{ m} + 36.0\text{ m} = 48.0\text{ m}\).

1 mark for calculating the total distance as \(48\text{ m}\) by finding the sum of the triangular and rectangular areas under the speed-time profile (B).

All electromagnetic waves travel at the same high speed in a vacuum (approximately \(3.0 \times 10^8\text{ m/s}\)). Gamma rays have much shorter wavelengths than radio waves. Sound waves are longitudinal mechanical waves, not electromagnetic waves. Ultraviolet radiation has a higher frequency than infrared radiation.

1 mark for identifying that all electromagnetic waves travel at the same speed in a vacuum (B).

First, calculate the equivalent resistance (\(R_p\)) of the two parallel resistors: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3.0\ \Omega} + \frac{1}{6.0\ \Omega} = \frac{2}{6.0} + \frac{1}{6.0} = \frac{3}{6.0}\). This gives \(R_p = 2.0\ \Omega\). Next, apply Ohm's law to find the total current: \(I = \frac{V}{R_p} = \frac{6.0\text{ V}}{2.0\ \Omega} = 3.0\text{ A}\).

1 mark for calculating the combined resistance of \(2.0\ \Omega\) and dividing the total voltage by this resistance to find the total current of \(3.0\text{ A}\) (C).

Part (a): The enzyme amylase breaks down starch (substrate) into maltose (product). Part (b): Low temperature reduces kinetic energy, causing molecules to move slowly, leading to fewer successful enzyme-substrate collisions. At 37 °C (optimum), kinetic energy is higher, maximizing collisions. Part (c): High temperatures above the optimum disrupt the chemical bonds maintaining the protein structure, changing the active site shape. The enzyme is denatured and cannot bind to starch.

Part (a) [2 marks]: 1 mark for starch, 1 mark for maltose. Part (b) [4 marks]: 1 mark for low kinetic energy at 10 °C, 1 mark for fewer successful collisions, 1 mark for higher kinetic energy at 37 °C, 1 mark for higher frequency of successful collisions. Part (c) [3 marks]: 1 mark for stating the enzyme is denatured, 1 mark for explaining that the active site changes shape, 1 mark for explaining that the substrate can no longer bind/fit.

Part (a): Copper ions are discharged at the cathode to form metallic copper (pink/brown solid). Hydroxide ions are discharged at the anode to form oxygen gas (colorless bubbles). Part (b): The cathode half-equation shows reduction: copper(II) ions gain two electrons to form neutral copper atoms. Part (c): Prolonged electrolysis removes \( \text{Cu}^{2+} \) and \( \text{OH}^- \) from solution, leaving a solution of sulfuric acid (containing \( \text{H}^+ \) and \( \text{SO}_4^{2-} \)), which lowers the pH.

Part (a) [3 marks]: 1 mark for pink/brown solid at cathode, 1 mark for bubbles/effervescence at anode, 1 mark for identifying the gas as oxygen. Part (b) [2 marks]: 1 mark for correct formula of reactants and products, 1 mark for correct balancing with 2 electrons. Part (c) [4 marks]: 1 mark for connecting blue color to \( \text{Cu}^{2+} \) ions, 1 mark for concentration of copper ions decreasing, 1 mark for hydrogen ions remaining in solution, 1 mark for explaining increased concentration of hydrogen ions decreases pH.

Part (a): Use formula \( a = \frac{v - u}{t} = \frac{8.0 - 0}{4.0} = 2.0\ \text{m/s}^2 \). Part (b): Use formula \( \text{KE} = \frac{1}{2} m v^2 = 0.5 \times 15 \times (8.0)^2 = 0.5 \times 15 \times 64 = 480\ \text{J} \). Part (c): Power is work done per unit time. Assuming no losses, work done equals kinetic energy gained: \( P = \frac{W}{t} = \frac{480}{4.0} = 120\ \text{W} \).

Part (a) [3 marks]: 1 mark for formula or substitution, 1 mark for correct value (2.0), 1 mark for correct unit (\( \text{m/s}^2 \)). Part (b) [2 marks]: 1 mark for correct formula or substitution, 1 mark for correct calculation (480 J). Part (c) [3 marks]: 1 mark for equating work done to kinetic energy (480 J), 1 mark for power formula, 1 mark for correct calculation (120 W).

Part (a): Rate of diffusion is maximized by maximizing surface area (vast number of alveoli), minimizing thickness of the barrier (alveolar wall is one cell thick), and maintaining a high concentration gradient (continuous blood flow through dense capillary networks). Part (b): In the airway lining, goblet cells produce protective mucus. Cilia on ciliated cells sweep this trapped material out of the respiratory tract.

Part (a) [6 marks]: Award 1 mark for naming each feature (up to 3) and 1 mark for its corresponding explanation. Feature 1: Large surface area (1) -> more area for diffusion (1). Feature 2: Thin walls/one cell thick (1) -> short diffusion distance (1). Feature 3: Good blood supply/capillaries (1) -> maintains concentration gradient (1). Part (b) [3 marks]: 1 mark for goblet cells secreting mucus, 1 mark for mucus trapping dust/pathogens, 1 mark for ciliated cells/cilia sweeping mucus away from lungs.

Part (a): An acid reacts with a metal oxide (base) to produce salt and water, which is a neutralisation reaction. The balanced equation is \( \text{MgO} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O} \). Part (b): Filtration separates insoluble excess base. Heating evaporates water to reach crystallization point. Cooling allows regular crystals to precipitate. Secondary filtration separates crystals, and drying yields pure product.

Part (a) [3 marks]: 1 mark for identifying neutralisation, 2 marks for balanced equation (1 mark for correct reactants and products, 1 mark for correct balancing). Part (b) [5 marks]: 1 mark for filtering excess solid, 1 mark for heating/evaporating filtrate to crystallization/saturation point, 1 mark for cooling to allow crystallization, 1 mark for filtering crystals, 1 mark for drying with filter paper.

Part (a): Use the parallel resistance formula: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0} \), so \( R_p = 3.0\ \Omega \). Part (b): Use Ohm's law for the whole circuit: \( I = \frac{V}{R_p} = \frac{12.0}{3.0} = 4.0\ \text{A} \). Part (c): In parallel, potential difference across each resistor is equal to the battery voltage (12.0 V). Current through \( R_1 \): \( I_1 = \frac{V}{R_1} = \frac{12.0}{4.0} = 3.0\ \text{A} \). Current through \( R_2 \): \( I_2 = \frac{V}{R_2} = \frac{12.0}{12.0} = 1.0\ \text{A} \). Verification: \( I_1 + I_2 = 3.0\ \text{A} + 1.0\ \text{A} = 4.0\ \text{A} \).

Part (a) [3 marks]: 1 mark for parallel resistance formula, 1 mark for correct substitution, 1 mark for correct value (3.0 ohms). Part (b) [2 marks]: 1 mark for substitution into \( I = V/R \), 1 mark for correct value (4.0 A). Part (c) [3 marks]: 1 mark for calculating \( I_1 = 3.0\ \text{A} \), 1 mark for calculating \( I_2 = 1.0\ \text{A} \), 1 mark for verifying sum equals total current.

Part (a): Insect-pollinated flowers have structures evolved to attract insects (colorful petals, sticky pollen). Wind-pollinated flowers are adapted to catch air currents (exposed hanging anthers, aerodynamic pollen). Part (b): Pollination leads to fertilization. The pollen tube serves as a pathway for male gametes to travel down the style to reach the female gamete in the ovule.

Part (a) [6 marks]: 2 marks for petals (1 mark for insect-pollinated description, 1 mark for wind-pollinated description), 2 marks for anthers (1 mark for insect-pollinated, 1 mark for wind-pollinated), 2 marks for pollen grains (1 mark for insect-pollinated, 1 mark for wind-pollinated). Part (b) [3 marks]: 1 mark for pollination definition (transfer from anther to stigma), 1 mark for tube growth down style, 1 mark for entering ovary/ovule.

Part (a): Hydrocarbon contains ONLY carbon and hydrogen. Decane cracking: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_6\text{H}_{14} + 2\text{C}_2\text{H}_4 \). Part (b): Bromine water reacts with the unsaturated double bond of ethene, decolorizing it. Hexane is saturated, so it does not react without UV light. Part (c): The structure of ethene must show the double covalent bond between the carbon atoms and the four single bonds to hydrogen.

Part (a) [3 marks]: 1 mark for definition of hydrocarbon (carbon and hydrogen ONLY), 2 marks for balanced equation (1 mark for correct formulas, 1 mark for correct balancing with 2 moles of ethene). Part (b) [3 marks]: 1 mark for naming bromine water, 1 mark for correct hexane result, 1 mark for correct ethene result. Part (c) [2 marks]: 1 mark for drawing carbon-carbon double bond, 1 mark for showing four correct carbon-hydrogen single bonds.

(a) (i) At the cathode, positive copper ions (\(\text{Cu}^{2+}\)) are attracted and gain electrons (reduction) to form copper metal: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\).
(ii) At the anode, hydroxide ions (\(\text{OH}^-\)) from water are discharged preferentially over sulfate ions to form oxygen gas and water: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\). The visual observation is bubbles of gas.

(b) (i) The blue colour of aqueous copper(II) sulfate is due to hydrated \(\text{Cu}^{2+}\) ions. As these ions are discharged at the cathode to form copper metal, their concentration in the solution decreases, causing the blue colour to fade and eventually become colourless.
(ii) As \(\text{OH}^-\)(aq) is discharged at the anode and \(\text{Cu}^{2+}\)(aq) at the cathode, \(\text{H}^+\) and \(\text{SO}_4^{2-}\) ions remain behind in the solution, forming dilute sulfuric acid. The concentration of \(\text{H}^+\) ions relative to \(\text{OH}^-\) ions increases, which lowers the pH (making it acidic).

(c) (i) When copper electrodes are used, the anode is active. Copper atoms lose electrons and dissolve into solution as copper ions: \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\). Thus, the mass of the anode decreases.
(ii) This process is used industrially for refining/purifying copper or for electroplating objects with copper.

Part (a)
(i) Copper [1] (Accept: Cu)
(ii) Bubbles / effervescence / gas given off [1]; oxygen (gas) [1] (Reject: oxide / hydroxide)

Part (b)
(i) Blue colour is due to copper ions / \(\text{Cu}^{2+}\) [1];
Copper ions are discharged / removed from solution / turn into copper atoms [1]
(ii) pH decreases / becomes acidic [1];
Hydrogen ions / \(\text{H}^+\) remain in solution (or hydroxide ions / \(\text{OH}^-\)) are discharged [1]

Part (c)
(i) Mass decreases / dissolves / becomes smaller [1]
(ii) Refining / purification of copper OR electroplating [1] (Accept: extraction of copper)

(a) Iodine solution turns blue-black in the presence of starch, but remains orange/brown/yellow when starch is fully digested. (b) 1 minute and 15 seconds = \(1 \times 60 + 15 = 75\) seconds. (c) Controlling temperature (using a water bath) and the concentration/volume of both reactants ensures a fair test where only pH varies. (d) To find the optimum temperature, run the investigation at a range of temperatures (e.g., 20 °C to 60 °C) using a buffer solution to keep pH constant. The rate of starch breakdown is calculated as 1/time, and the optimum temperature corresponds to the minimum time taken for starch to disappear.

a) Yellow / brown / orange [1 mark]. Reject: blue-black / clear. b) 75 (seconds) [1 mark]. c) Any two from: temperature [1], concentration of amylase [1], volume of starch solution [1], volume of amylase solution [1] [Max 2 marks]. d) Experimental plan: 1. Use water baths at at least 3-5 different temperatures (e.g. 20 °C to 60 °C) [1 mark]. 2. Describe mixing of amylase and starch after allowing them to equilibrate to the water bath temperature [1 mark]. 3. Describe method of testing: add drops of mixture to iodine solution in a spotting tile at regular intervals (e.g. 30 seconds) [1 mark]. 4. Identify independent variable (temperature) and dependent variable (time taken for starch to be completely digested / iodine to stay orange) [1 mark]. 5. State control variables: constant pH (using a buffer), constant volumes/concentrations of reactants [1 mark]. 6. Explain how to find optimum: the temperature with the shortest time for starch to disappear / fastest rate [1 mark].

(a) See marking scheme for drawing criteria. (b) Halfway between 30 and 40 cm³ is 35 cm³. The reading at 120 seconds is 64 cm³. (c) As the reaction progress, acid reactant particles are used up, decreasing the concentration and rate of effective collisions. (d) Multiple trials (repeats) and calculating average values reduces the impact of random experimental error. (e) Bubbling CO2 into limewater produces a white precipitate of calcium carbonate, rendering the solution cloudy.

a) Flask containing acid and marble chips [1 mark]; bung and delivery tube leading to syringe without leaks [1 mark]; gas syringe clearly labelled [1 mark]. b) 35 cm³ [1 mark] and 64 cm³ [1 mark]. c) Volume collected per unit time (rate) decreases [1 mark] because the concentration of acid decreases as it is used up [1 mark]. d) Repeat the experiment and find the average / mean of the volume at each time [1 mark]. e) Test: Bubble gas through limewater (calcium hydroxide solution) [1 mark]. Result: Limewater turns cloudy / milky [1 mark].

(a) A ruler or meter rule is used. To avoid parallax errors, look straight/perpendicularly at the ruler marking. (b) Extension \(x = L - L_0 = 6.1\text{ cm} - 2.5\text{ cm} = 3.6\text{ cm}\). (c) Hooke's Law states force (load) is directly proportional to extension, meaning a plot of load against extension yields a straight line through the origin. (d) Safety goggles protect against injury if the spring snaps under tension. (e) Both wires must be tested using the same independent variable increments (load). The wire with the smaller gradient (extension/load) or smaller absolute extension for equivalent loads is stiffer.

a) Ruler / half-metre rule [1 mark]. Read scale with eye level directly perpendicular to the ruler / level with the bottom of the spring [1 mark]. b) \(6.1 - 2.5 = 3.6\) cm [1 mark]. c) Relationship: Load is directly proportional to extension [1 mark]. Graph shape: Straight line passing through the origin [1 mark]. d) Wear safety goggles / eye protection (in case the spring/wire snaps or flies off) [1 mark]. e) Planning comparison: 1. Clamp wire A and measure its original length [1 mark]. 2. Add known masses / loads to the wire and measure the new length to find the extension [1 mark]. 3. Repeat the exact same procedure for wire B using the same set of loads [1 mark]. 4. Compare the extensions: the wire with the smaller extension for the same load is the stiffer wire (or compare gradients of load-extension graphs, steeper gradient = stiffer wire) [1 mark].

(a) Oxygen supports combustion, hence it relights a glowing splint. (b) \(\text{Rate} = 48 \div 60\text{ s} = 0.8\text{ bubbles/s}\). (c) Temperature and concentration of sodium hydrogencarbonate (CO2 source) must be kept constant. (d) Counting bubbles assumes constant bubble volume, which is untrue. Collecting gas over water in a graduated tube or syringe yields true volumetric measurements. (e) Lights emit heat which can warm the plant's water, changing the temperature. Since temperature affects enzyme activity and rate, a water jacket block prevents temperature fluctuations from confounding light intensity results.

a) Test: Use a glowing splint [1 mark]. Result: The splint relights [1 mark]. b) \(48 \div 60 = 0.8\) bubbles per second [1 mark]. c) Any two from: temperature, carbon dioxide concentration (concentration of sodium hydrogencarbonate), species/mass/length of pondweed, wavelength of light [2 marks]. d) Reason: Bubbles can be of different sizes / bubbles are produced too fast to count accurately [1 mark]. Improvement: Collect the gas in an inverted graduated tube / measuring cylinder / gas syringe to measure actual volume [1 mark]. e) The lamp emits thermal energy/heat [1 mark]. Temperature is a variable that affects the rate of photosynthesis / enzyme activity [1 mark]. The glass beaker of water absorbs heat from the lamp while letting light through, acting as a heat shield to keep temperature constant [1 mark].