2025 CCEA GCSE Science Double Award 1370 模擬試題連答案詳解

A ruled table with columns 'Temperature / °C' and 'Observation of bubbling' should show: 10 °C — a few small bubbles, slow; 25 °C — a steady stream of bubbles; 40 °C — vigorous, rapid bubbling (near the enzyme's optimum); 60 °C — little or no bubbling, as the enzyme has denatured.

1 mark: correctly ruled table with headings and units (Temperature / °C; Observation of bubbling); 1 mark: observations show bubble production increasing from 10 °C to a peak around 40 °C; 1 mark: correct observation of little/no bubbling at 60 °C linked to denaturation.

A ruled table with columns 'Concentration of \(\text{H}\text{Cl}\) / mol dm⁻³' and 'Observation' should show: 0.5 — slow, gentle fizzing, ribbon takes longest to disappear; 1.0 — moderate, steady fizzing; 2.0 — rapid, vigorous fizzing with the ribbon disappearing fastest.

1 mark: correctly ruled table with headings and units; 1 mark: observations correctly show fizzing rate increasing with concentration; 1 mark: correct comparative reference to how quickly the magnesium ribbon disappears at each concentration.

A ruled table with three columns: 'Mass / g', 'Length / mm' and 'Extension / mm', with four rows for 100 g, 200 g, 300 g and 400 g. Extension for each row is calculated as the measured length minus the unstretched length of 60 mm.

1 mark: table correctly ruled with all three columns present; 1 mark: correct headings including units (Mass / g; Length / mm; Extension / mm); 1 mark: extension column correctly defined/derived as length minus 60 mm for each row.

A ruled table with columns 'Leaf' and 'Colour with iodine' should show: leaf from sunlight — blue-black, showing starch is present as a product of photosynthesis; leaf from the dark cupboard — remains orange-brown, showing no starch has been produced because no photosynthesis occurred without light.

1 mark: correctly ruled table with appropriate headings; 1 mark: correct colour recorded for each leaf (blue-black for light, orange-brown for dark); 1 mark: correct link made between blue-black colouration and the presence of starch from photosynthesis.

A ruled table with columns 'Metal ion' and 'Flame colour' should record: lithium — crimson red; sodium — yellow/orange (often obscuring other colours); potassium — lilac; copper — blue-green.

1 mark: correctly ruled table with clear headings; 1 mark: at least three of the four flame colours correctly matched to their metal ion; 1 mark: all four flame colours correctly matched (lithium crimson, sodium yellow/orange, potassium lilac, copper blue-green).

A ruled table with two columns, 'Voltage / V' and 'Current / A', and five rows corresponding to 2 V, 4 V, 6 V, 8 V and 10 V, ready to record the ammeter reading obtained at each voltage.

1 mark: table correctly ruled with two columns and five rows; 1 mark: correct headings including units (Voltage / V; Current / A); 1 mark: voltage values correctly and fully listed in ascending order (2, 4, 6, 8, 10 V).

Mean titre = \((24.60 + 24.50 + 24.55) ÷ 3 = 73.65 ÷ 3 = 24.55\) cm³. The independent variable is whatever the student deliberately changes between titrations (e.g. the acid or alkali being tested); the dependent variable, which is measured, is the titre volume of acid required.

1 mark: correct sum of the three titres (73.65 cm³) shown as working; 1 mark: correct mean titre of 24.55 cm³ (accept ecf from an incorrect sum); 1 mark: independent and dependent variables both correctly identified.

\(\text{Average speed} = \text{distance} ÷ \text{time}\) = \(1.80 ÷ 3.0 = 0.6\) m/s. The independent variable, deliberately changed by the student, is the angle of the ramp; the dependent variable, which is measured as a result, is the trolley's average speed (via the time taken to travel a fixed distance).

1 mark: correct substitution into \(\text{speed} = \text{distance} ÷ \text{time}\); 1 mark: correct answer of 0.6 m/s with correct unit; 1 mark: independent variable (ramp angle) and dependent variable (average speed/time) both correctly identified.

Because each count was already taken over exactly one minute, the rate at each distance is simply the bubble count itself: 45 bubbles/min at 10 cm, 28 bubbles/min at 20 cm, and 12 bubbles/min at 30 cm — showing the rate of photosynthesis falls as the light source moves further away. The independent variable is the distance between the lamp and the pondweed; the dependent variable is the rate of oxygen bubble production.

1 mark: correct rates stated in bubbles per minute for all three distances; 1 mark: correct trend identified (rate decreases as distance increases); 1 mark: independent variable (lamp distance) and dependent variable (bubble rate) both correctly identified.

Mass loss = \(152.40 − 150.20 = 2.20\) g, which corresponds to the mass of carbon dioxide gas that escaped from the flask during the reaction. The independent variable, changed between repeats, is the mass of marble chips used; the dependent variable, measured as an outcome, is the mass of gas lost (tracked via the flask's changing mass).

1 mark: correct subtraction shown as working (\(152.40 − 150.20\)); 1 mark: correct mass loss of 2.20 g with unit; 1 mark: independent variable (mass of marble chips) and dependent variable (mass lost as \(\text{C}\text{O}_{2}\)) both correctly identified.

Using \(R = V ÷ I\): R = \(6.0 ÷ 0.40 = 15\) Ω. If the length of wire is varied between repeats, the independent variable is the wire length, and the dependent variable is the resulting resistance (measured via voltage and current readings).

1 mark: correct substitution into \(R = V ÷ I\); 1 mark: correct answer of 15 Ω with correct unit; 1 mark: independent variable (wire length) and dependent variable (resistance) both correctly identified.

Mean resting pulse = \((68 + 72 + 70) ÷ 3 = 210 ÷ 3 = 70\) bpm. Mean pulse after exercise = \((132 + 128 + 136) ÷ 3 = 396 ÷ 3 = 132\) bpm. The independent variable is whether the pulse is measured at rest or after exercise; the dependent variable is the pulse rate in beats per minute.

1 mark: correct mean resting pulse of 70 bpm shown with working; 1 mark: correct mean pulse after exercise of 132 bpm shown with working; 1 mark: independent variable (rest/exercise state) and dependent variable (pulse rate) both correctly identified.

pH (x-axis) would be plotted against rate of reaction in mm³/min (y-axis), with a smooth bell-shaped curve drawn through the points rather than joining them with straight lines. The curve rises from a low rate at pH 2, peaks at pH 7 with a rate of 42 mm³/min, then falls away sharply to a low rate again by pH 10, showing that the enzyme has an optimum pH of 7 either side of which its activity declines because its active site is disrupted.

1 mark: correct axes identified (pH on x-axis, rate on y-axis) with a smooth curve rather than straight lines between points; 1 mark: correct identification of the optimum pH (7) and its peak rate (42 mm³/min); 1 mark: correct description of the trend as a rise then a fall either side of the optimum (bell-shaped curve).

Rate of reaction (1/time, in s⁻¹) would be plotted on the y-axis against concentration (g/dm³) on the x-axis. Converting the data gives rates of approximately 0.005, 0.010, 0.015, 0.020 and 0.025 s⁻¹, which increase in an approximately straight, proportional line through the origin as concentration rises, showing that increasing the concentration of thiosulfate increases the frequency of successful collisions and therefore the rate of reaction.

1 mark: correct axes and use of 1/time to represent rate of reaction; 1 mark: correct calculation/comparison showing rate increases as concentration increases; 1 mark: correct trend statement — rate is approximately directly proportional to concentration, linked to increased collision frequency.

Force (x-axis) would be plotted against extension in mm (y-axis). From 1 N to 4 N the points lie on a straight line through the origin, showing extension is directly proportional to force (Hooke's law). Beyond 4 N, at 5 N and 6 N, the extension increases by a larger amount than the straight-line pattern predicts, showing the spring has exceeded its limit of proportionality and Hooke's law no longer applies.

1 mark: correct axes with force on x-axis and extension on y-axis; 1 mark: correct identification of a straight-line (directly proportional) relationship from 1 N to 4 N; 1 mark: correct identification that the graph deviates from a straight line beyond 4 N, indicating the limit of proportionality has been exceeded.

A full method should: state the independent variable (concentration of hydrogen peroxide, made by diluting a stock solution with water) and the dependent variable (volume of oxygen gas produced, or time taken to collect a fixed volume); list controlled variables (temperature, mass/concentration of catalase, pH/buffer used, total reaction volume); describe apparatus (conical flask, delivery tube, inverted measuring cylinder or gas syringe over water); describe the procedure (add catalase to hydrogen peroxide, immediately start a stopwatch, record volume of gas collected every 10–15 seconds, or record the time to collect a fixed volume such as 20 cm³); state that the procedure is repeated at each concentration and a mean calculated, with anomalous results identified and repeated; explain how rate is calculated (volume ÷ time) and plotted against concentration; note safety precautions (eye protection, handling hydrogen peroxide with care).

Band A (5–6 marks): a full, logically ordered method with independent, dependent and at least two controlled variables clearly identified; apparatus and procedure described in enough detail to be repeatable; reference to repeats/means and how rate is calculated; fluent use of specialist terms with few errors. Band B (3–4 marks): most key steps present but with some gaps in detail, control of variables, or repeats; adequate use of specialist terms. Band C (1–2 marks): a basic or partial method with significant omissions; limited use of specialist terms. Band D (0 marks): no creditable response.

A full method should: state the independent variable (temperature of the sodium thiosulfate solution, controlled using a water bath) and the dependent variable (time taken for a black 'X' mark on paper beneath the flask to disappear from view); list controlled variables (concentration and volume of thiosulfate, concentration and volume of hydrochloric acid, same flask/mark used each time); describe the procedure (heat the thiosulfate to the required temperature, add the acid, start the stopwatch immediately, view the mark from above and stop the timer the instant it disappears); state that the test is repeated at each of a range of temperatures (e.g. 20 °C, 30 °C, 40 °C, 50 °C, 60 °C) with the mean time recorded; explain that rate (1/time) is calculated and plotted against temperature; note that the same person should judge when the mark disappears for consistency, and safety precautions such as eye protection should be used.

Band A (5–6 marks): a full, logically ordered method with independent, dependent and at least two controlled variables clearly identified; apparatus and procedure described in repeatable detail; reference to repeats/means and calculation of rate as 1/time; fluent use of specialist terms. Band B (3–4 marks): most key steps present with some gaps in control of variables or repeats; adequate specialist terms. Band C (1–2 marks): a basic or partial method with significant omissions; limited specialist terms. Band D (0 marks): no creditable response.

A full method should: state the independent variable (length of the pendulum string, measured from the point of suspension to the centre of the bob using a metre rule) and the dependent variable (time period of one oscillation); list controlled variables (mass of the bob, angle from which the bob is released, same stopwatch/timer used); describe the procedure (set the pendulum swinging through a small angle, use a stopwatch to time 10 complete oscillations, divide the total time by 10 to obtain the period for one oscillation, reducing the effect of reaction-time error); state that this is repeated three times at each length and a mean period calculated, with the length varied across a suitable range (e.g. 20 cm to 100 cm); explain that period is plotted against length, or period² against length to test for a straight-line relationship; note safety precautions such as keeping the swing clear of people and apparatus.

Band A (5–6 marks): a full, logically ordered method with independent, dependent and at least two controlled variables clearly identified; use of multiple oscillations to reduce timing error explicitly described; reference to repeats/means; fluent use of specialist terms. Band B (3–4 marks): most key steps present with some gaps, e.g. omitting the multiple-oscillation timing technique or repeats; adequate specialist terms. Band C (1–2 marks): a basic or partial method with significant omissions; limited specialist terms. Band D (0 marks): no creditable response.

The soda lime absorbs the carbon dioxide released by respiration, so any change in gas volume (and hence bead movement) is due only to oxygen being taken up, not to \(\text{C}\text{O}_{2}\) being added back. The control tube (identical apparatus but without seeds) accounts for any bead movement caused by changes in atmospheric pressure or temperature rather than respiration, and this movement is subtracted from the experimental reading. The distance moved by the bead over 10 minutes, multiplied by the cross-sectional area of the capillary tube, gives the volume of oxygen absorbed; dividing this volume by the 10-minute time period gives the rate of oxygen uptake (e.g. in mm³/min).

1 mark: correct purpose of soda lime (absorbs \(\text{C}\text{O}_{2}\) so volume change reflects \(\text{O}_{2}\) uptake only); 1 mark: correct purpose of the control tube (corrects for pressure/temperature changes unrelated to respiration); 1 mark: correct method to find volume of \(\text{O}_{2}\) used (distance moved × cross-sectional area of capillary); 1 mark: correct method to obtain rate (volume ÷ time, i.e. ÷10 minutes).

The gradient of a volume–time graph represents the rate of gas production: a steep gradient at the start shows the reaction is fastest at this point, because the concentration of acid (and surface area of marble chips) is greatest, giving the highest frequency of successful collisions. As the reaction proceeds, the acid becomes more dilute (used up) and the gradient becomes progressively shallower, showing the rate is decreasing. Once the graph becomes horizontal, the gradient is zero, showing gas is no longer being produced because the reaction has finished — one of the reactants (here, the acid, since it is not stated to be in excess relative to the chips, or the marble chips if acid is in excess) has been completely used up.

1 mark: correct link between gradient of the graph and rate of reaction; 1 mark: correct explanation of the steep initial gradient (high concentration/high collision frequency = fastest rate); 1 mark: correct explanation of the decreasing gradient (concentration falling as reactant is used up); 1 mark: correct explanation of the horizontal section (rate is zero, a reactant has been used up, reaction finished).

The voltmeter must be connected in parallel across the lamp because it measures the potential difference (p.d.) between two points, which requires it to be connected across the component; a high-resistance voltmeter connected this way takes a negligible share of the current, so it does not significantly affect the current flowing through the lamp itself. The current–voltage graph curves because, as voltage and current increase, the filament heats up; the increased vibration of the metal ions in the filament increases resistance, so for each additional volt applied, a smaller additional increase in current is produced, making the graph progressively less steep (non-ohmic behaviour).

1 mark: correct reason the voltmeter is connected in parallel (measures p.d. across the component without significantly diverting current); 1 mark: correct statement that the filament heats up as current increases; 1 mark: correct link between heating and increased resistance; 1 mark: correct explanation that increasing resistance causes the graph to become progressively less steep (non-ohmic).

Before use, the colorimeter should be calibrated ('zeroed') using a blank cuvette (e.g. water or the fruit pulp with no enzyme added) so that this reading is set as zero absorbance, allowing later readings to be compared fairly. The graph falls steeply at low pectinase concentrations because more enzyme increases the rate at which pectin is broken down, releasing more clear juice and leaving less light-scattering pulp suspended, so absorbance decreases. At higher concentrations the graph levels off because enzyme is no longer the limiting factor — the amount of pectin substrate available to be broken down becomes limiting, so adding more enzyme has little further effect on absorbance.

1 mark: correct method of calibration (zeroing against a blank/control before taking readings); 1 mark: correct statement that absorbance decreases as pectinase concentration increases; 1 mark: correct explanation linking lower absorbance to more pectin broken down/less scattering pulp; 1 mark: correct explanation of the levelling off (substrate/pectin becomes the limiting factor at high enzyme concentration).

A copper calorimeter is used because copper is an excellent thermal conductor, so heat energy from the burning fuel is transferred efficiently through the container wall into the water, minimising energy losses and giving a more accurate measurement of energy released; an insulating container would trap much of the heat and prevent it reaching the water to be measured. The straight line through the origin shows that temperature rise is directly proportional to the mass of fuel burned — each additional gram of fuel releases the same amount of energy, so burning twice the mass produces twice the temperature rise, confirming a constant energy value per gram of fuel across the range tested.

1 mark: correct property of copper (good thermal conductor) identified; 1 mark: correct explanation that this ensures efficient heat transfer into the water; 1 mark: correct statement that a straight line through the origin shows direct proportionality between temperature rise and mass of fuel burned; 1 mark: correct interpretation that this means a constant amount of energy is released per gram of fuel.

The strobe light flashes at a rate that matches the wave frequency, making the moving wave pattern appear stationary so that the distance between adjacent wave crests (the wavelength) can be measured accurately with a ruler, which would otherwise be impossible on a fast-moving wave. The graph of wavelength against frequency falls as frequency increases because, for waves travelling in the same medium, wave speed remains constant; since wave speed = frequency × wavelength, if speed is fixed then wavelength must decrease as frequency increases, giving an inverse relationship between the two variables.

1 mark: correct purpose of the strobe light (makes the wave pattern appear stationary so wavelength can be measured); 1 mark: correct statement that wave speed in the tank is constant; 1 mark: correct reference to the wave equation (speed = frequency × wavelength); 1 mark: correct explanation that wavelength must fall as frequency rises to keep speed constant (inverse relationship).

As the shoot takes up water through its cut stem to replace water lost by transpiration, an air bubble introduced into the capillary tube is drawn along it; the distance the bubble moves per unit time, read against a scale on the tube, gives the rate of water uptake, which is used as an estimate of the plant's transpiration rate. The graph is steeper with a fan because increased air movement across the leaf surface removes water vapour from just outside the stomata more quickly, steepening the concentration gradient for diffusion of water vapour out of the leaf; this increases the rate of transpiration, and therefore the rate of water uptake (and bubble movement) to replace the lost water.

1 mark: correct explanation of how the bubble's movement is used to calculate a rate (distance ÷ time, read from the scale); 1 mark: correct statement that this rate estimates the transpiration rate; 1 mark: correct explanation that the fan increases the rate of diffusion of water vapour away from the leaf/stomata; 1 mark: correct link that this steeper diffusion gradient increases transpiration and therefore water uptake.

The magnetic stirrer continuously mixes the solution as acid is added, ensuring the acid is evenly distributed rather than forming a localised area of low pH near the burette tip, so the pH probe gives a reading that is representative of the whole solution at each point. The steep central region of the graph shows the equivalence point of the titration, where the alkali has just been fully neutralised; around this point, adding only a very small additional volume of acid causes a very large change in pH, because the solution has little buffering capacity left to resist the change once neutral.

1 mark: correct purpose of the magnetic stirrer (keeps the solution evenly mixed for accurate/representative pH readings); 1 mark: correct identification of the steep region as the equivalence/neutralisation point; 1 mark: correct explanation that only a small extra volume of acid is needed to cause the large pH change there; 1 mark: correct reference to the loss of buffering capacity/large change in [\(\text{H}^{+}\)] near neutrality.

As the interrupt card attached to the trolley passes through each light gate, the gate records the (short) time taken for the card to block the beam; dividing the known length of the card by this time gives the trolley's velocity at that gate. The non-zero y-intercept shows that the trolley already had a velocity greater than zero at the moment timing started (it was not released from rest at the start of recording, or was already moving when the first gate was reached). The gradient of a velocity–time graph represents acceleration (the rate of change of velocity), and because the line is straight, the trolley has a constant acceleration down the ramp.

1 mark: correct explanation of how a light gate/interrupt card gives velocity (card length ÷ time to cross the gate); 1 mark: correct interpretation of the non-zero intercept (trolley had a starting velocity when timing began); 1 mark: correct statement that the gradient represents acceleration; 1 mark: correct additional interpretation that a straight line means the acceleration is constant.

A transect line with quadrats placed at regular, fixed intervals provides a systematic sampling method that avoids the observer selecting sampling points subjectively, and allows percentage cover to be related directly and reliably to distance from the pollution source. The graph shows low percentage cover of the pollution-sensitive species close to the outflow, where pollutant concentration is highest and most inhibits growth of sensitive species; cover increases with distance as pollutant concentration falls; the graph then levels off further from the outflow once pollution has fallen enough that it is no longer the main factor limiting the species' abundance, and other factors (e.g. competition, light) become limiting instead.

1 mark: correct reason for using a transect with regularly spaced quadrats (systematic, unbiased sampling related to distance); 1 mark: correct statement that cover is low near the outflow due to higher pollution; 1 mark: correct statement that cover increases as distance from the outflow (and pollution) decreases; 1 mark: correct explanation of the levelling off (pollution no longer the limiting factor at greater distance).

The moveable crocodile clip allows the length of resistance wire connected into the circuit to be changed easily and precisely (by reading the position against a metre rule) while keeping the type and cross-sectional area of the wire, and all other circuit components, unchanged — isolating length as the only variable affecting the measured resistance (\(R = V ÷ I\) at each length). The straight line through the origin on the resistance–length graph shows that resistance is directly proportional to the length of wire: doubling the length of wire in the circuit doubles its resistance, because there is proportionally more material for charge carriers to collide with as they pass through the wire.

1 mark: correct purpose of the moveable crocodile clip (varies only the length of wire in the circuit, keeping other variables constant); 1 mark: correct statement that resistance is calculated from V and I readings (\(R = V ÷ I\)) at each length; 1 mark: correct interpretation of the straight line through the origin as direct proportionality; 1 mark: correct explanation in terms of more material/collisions for charge carriers as length increases.

Because the reaction starts producing gas as soon as the magnesium and acid are mixed, the bung fitted with the gas syringe delivery tube must be inserted quickly and sealed securely immediately after mixing, otherwise some gas escapes before it can be collected, leading to an underestimate of the total volume produced. The horizontal section of the graph, at a constant 48 cm³, shows that gas is no longer being produced because the reaction has finished — since the acid is stated to be in excess, this means all of the magnesium ribbon has been completely used up, and 48 cm³ is the total (maximum) volume of hydrogen gas that this reaction can produce.

1 mark: correct precaution (fit the bung/syringe quickly and securely immediately after mixing to prevent gas escaping unmeasured); 1 mark: correct statement that the horizontal line means gas is no longer being produced/the reaction has finished; 1 mark: correct identification that the magnesium (the limiting reactant, since acid is in excess) has been completely used up; 1 mark: correct reading of the total/maximum volume of gas produced (48 cm³).

\(\text{Rate} = \text{volume} ÷ \text{time}\) = 24 cm³ ÷ 8 min = 3.0 cm³/min.

1 mark: correct substitution into \(\text{rate} = \text{volume} ÷ \text{time}\); 1 mark: correct arithmetic (\(24 ÷ 8\)); 1 mark: correct final answer of 3.0; 1 mark: correct unit (cm³/min) given with the answer.

Moles of \(\text{Na}\text{O}\text{H}\) = mass ÷ Mr = \(4.00 ÷ 40 = 0.10\) mol. Volume = 250 cm³ = 0.250 dm³. \(\text{Concentration} = \text{moles} ÷ \text{volume}\) = \(0.10 ÷ 0.250 = 0.40\) mol/dm³.

1 mark: correct moles of \(\text{Na}\text{O}\text{H}\) calculated (0.10 mol); 1 mark: correct conversion of volume to dm³ (0.250 dm³); 1 mark: correct substitution into \(\text{concentration} = \text{moles} ÷ \text{volume}\); 1 mark: correct final answer of 0.40 mol/dm³ with unit (accept ecf from an incorrect moles value).

Volume = \(3.0 × 2.0 × 5.0 = 30\) cm³. \(\text{Density} = \text{mass} ÷ \text{volume}\) = \(135 ÷ 30 = 4.5\) g/cm³. A source of uncertainty is the precision of the ruler used to measure the block's dimensions (e.g. ±0.1 cm parallax or reading error on each measurement), which propagates into the calculated volume and hence density.

1 mark: correct volume calculated (30 cm³); 1 mark: correct substitution into \(\text{density} = \text{mass} ÷ \text{volume}\); 1 mark: correct final answer of 4.5 g/cm³ with unit; 1 mark: a valid source of measurement uncertainty correctly identified (e.g. ruler reading/parallax error, balance precision).

Change in mass = \(4.68 − 5.20\) = −0.52 g. \(\text{Percentage change} = (\text{change} ÷ \text{original mass}) × 100\) = \((−0.52 ÷ 5.20) × 100\) = −10%. The negative sign shows this is a percentage loss in mass, because water left the potato cylinder by osmosis into the more concentrated surrounding salt solution.

1 mark: correct change in mass calculated (−0.52 g); 1 mark: correct substitution into \(\text{percentage change} = (\text{change} ÷ \text{original}) × 100\); 1 mark: correct final answer of −10% (or 10% loss); 1 mark: correct statement that this represents a decrease/loss in mass, with a valid reason (water moved out by osmosis).

Energy transferred to the water = mass × specific heat capacity × temperature change = \(100 × 4.2 × 24 = 10,080\) J. Energy per gram of fuel = \(10,080 ÷ 0.60 = 16,800\) J/g (16.8 kJ/g).

1 mark: correct substitution into energy = mass × specific heat capacity × temperature change; 1 mark: correct total energy calculated (10,080 J); 1 mark: correct division by mass of fuel (÷0.60); 1 mark: correct final answer of 16,800 J/g (or 16.8 kJ/g) with correct unit.

\(\text{Work done} = \text{force} × \text{distance}\) = \(15 × 2.0 = 30\) J. \(\text{Power} = \text{work done} ÷ \text{time}\) = \(30 ÷ 5.0 = 6.0\) W.

1 mark: correct substitution into \(\text{work done} = \text{force} × \text{distance}\); 1 mark: correct work done of 30 J with unit; 1 mark: correct substitution into \(\text{power} = \text{work done} ÷ \text{time}\); 1 mark: correct power of 6.0 W with unit (accept ecf from an incorrect work-done value).

\(\text{Magnification} = \text{image size} ÷ \text{actual size}\). Both measurements must be in the same units before dividing: image size = 45 mm, actual size = 0.030 mm. Magnification = \(45 ÷ 0.030 = 1500\). Magnification has no units, because it is a ratio of two lengths measured in the same unit, so the units cancel out.

1 mark: correct formula stated or used (\(\text{magnification} = \text{image size} ÷ \text{actual size}\)); 1 mark: both values correctly identified in consistent units (mm and mm); 1 mark: correct substitution shown as working (\(45 ÷ 0.030\)); 1 mark: correct final answer of ×1500; 1 mark: correct statement that magnification has no units, as it is a ratio.

\(\text{Percentage yield} = (\text{actual yield} ÷ \text{theoretical yield}) × 100\) = \((6.80 ÷ 8.50) × 100 = 80\)%. A valid reason for the loss, other than spillage, is that some of the product remained dissolved in the solution and was not fully crystallised/recovered during filtration and drying, or that the reaction did not go to completion / some product was lost through unwanted side reactions.

1 mark: correct substitution into percentage yield = (actual ÷ theoretical) × 100; 1 mark: correct arithmetic shown (\(6.80 ÷ 8.50\)); 1 mark: correct final answer of 80%; 1 mark: a valid reason for yield loss identified (e.g. product remaining in solution, incomplete reaction); 1 mark: reason correctly explained/developed rather than just named.

\(\text{Efficiency} = (\text{useful energy output} ÷ \text{total energy input}) × 100\) = \((350 ÷ 500) × 100 = 70\)%. Energy wasted = total energy input − useful energy output = \(500 − 350 = 150\) J.

1 mark: correct substitution into \(\text{efficiency} = (\text{useful output} ÷ \text{total input}) × 100\); 1 mark: correct arithmetic shown (\(350 ÷ 500\)); 1 mark: correct final efficiency of 70%; 1 mark: correct calculation of energy wasted (\(500 − 350\)); 1 mark: correct final answer of 150 J with unit.

Plant cells have a rigid cell wall made of cellulose, which animal cells do not have; plant cells (in photosynthetic tissue) contain chloroplasts, which animal cells never have; plant cells typically have one large, permanent vacuole, whereas animal cells have only small, temporary vacuoles if any.

1 mark for each correct, valid structural difference stated, up to a maximum of 2 (e.g. cell wall, chloroplasts, large permanent vacuole).

Isotopes are atoms of the same element (same number of protons/same atomic number) that have different numbers of neutrons, and therefore different mass numbers. For chlorine-37, neutrons = mass number − atomic number = \(37 − 17 = 20\).

1 mark: correct definition of isotopes (same protons/atomic number, different neutrons); 1 mark: correct number of neutrons (20).

\(\text{Average speed} = \text{distance travelled} ÷ \text{time taken}\). When distance is measured in metres and time in seconds, the unit of speed is metres per second (m/s).

1 mark: correct equation (\(\text{speed} = \text{distance} ÷ \text{time}\)); 1 mark: correct unit (m/s).

Carbon dioxide and water react, using light energy absorbed by chlorophyll, to produce glucose and oxygen: carbon dioxide + water --light energy--> glucose + oxygen.

1 mark: correct reactants (carbon dioxide and water) and products (glucose and oxygen); 1 mark: correct reference to light energy as the energy source.

Similarity: both types of bonding involve the outer-shell (valence) electrons of atoms rearranging so that each atom achieves a full, stable outer shell. Difference: in ionic bonding, electrons are transferred between atoms, forming oppositely charged ions held together by electrostatic attraction, whereas in covalent bonding electrons are shared between atoms as pairs, with no ions formed.

1 mark: valid similarity correctly stated; 1 mark: valid difference correctly stated.

An object will remain at rest, or continue to move at a constant velocity in a straight line, unless a resultant (unbalanced) force acts upon it.

1 mark: reference to remaining at rest or moving at constant velocity; 1 mark: reference to this being true only in the absence of a resultant/unbalanced force.

A small sample of the food is treated with a few drops of iodine solution. If starch is present, the iodine solution changes colour from orange-brown to blue-black; if starch is absent, it remains orange-brown.

1 mark: correct reagent (iodine solution); 1 mark: correct positive colour change (orange-brown to blue-black).

Diamond has a giant covalent structure in which each carbon atom is covalently bonded to four other carbon atoms, forming a rigid three-dimensional lattice. To melt diamond, a very large number of these strong covalent bonds throughout the whole structure must be broken, which requires a large amount of energy, giving diamond a very high melting point.

1 mark: giant covalent structure correctly identified; 1 mark: each carbon atom bonded to four others in a rigid lattice; 1 mark: correct link between breaking many strong covalent bonds and needing a large amount of energy.

In a gas, the particles are spaced much further apart than in a liquid of the same substance. Because they are so widely spaced, gas particles collide with the walls of their container much less frequently than the closely packed particles of a liquid. Since pressure results from the frequency and force of particle collisions with a surface, fewer collisions per unit area per unit time means a gas exerts a lower pressure than the equivalent liquid.

1 mark: particles are much further apart in a gas than a liquid; 1 mark: this results in fewer collisions with the container walls per unit time/area; 1 mark: correct link between fewer collisions and lower pressure.

The active site is the specific region of an enzyme's structure that has a shape complementary to its substrate, allowing the substrate to bind and the reaction to be catalysed. At a temperature well above the enzyme's optimum, the enzyme's tertiary structure is disrupted (it denatures), permanently changing the shape of the active site so that the substrate can no longer bind, which reduces or stops the rate of reaction.

1 mark: correct definition of the active site (region with a shape complementary to the substrate); 1 mark: reference to denaturation/change of shape at high temperature; 1 mark: correct link between the changed shape and reduced binding/rate of reaction.

Going down Group 1, atoms have an increasing number of electron shells. This means the single outer-shell electron is further from the nucleus and is shielded from the nucleus's attractive pull by more inner shells of electrons. This reduces the force of attraction between the nucleus and the outer electron, so the outer electron is lost more easily, making the element more reactive.

1 mark: correct reference to an increasing number of electron shells down the group; 1 mark: correct reference to the outer electron being further from the nucleus/more shielded; 1 mark: correct link between weaker attraction and the electron being lost more easily, increasing reactivity.

As the ball falls, energy is transferred from its gravitational potential energy store to its kinetic energy store. When the ball hits the ground and bounces, some of this energy is usefully transferred back towards a kinetic/gravitational potential energy store as it rises again, but some energy is dissipated (wastefully transferred) to the thermal energy store of the ball and surroundings, and to sound, because of inelastic deformation of the ball on impact. Since the total useful energy available after the bounce is less than before it, the ball cannot rise back to its original height.

1 mark: correct initial transfer from gravitational potential energy store to kinetic energy store as it falls; 1 mark: correct reference to energy being dissipated as heat and/or sound on impact; 1 mark: correct link between this energy dissipation and the reduced bounce height.

Alveoli are adapted for efficient gas exchange because there are millions of them, giving the lungs a very large total surface area for diffusion; their walls are only one cell thick, giving a very short diffusion distance for gases; and they are surrounded by a dense network of capillaries, which maintains a steep concentration gradient for oxygen and carbon dioxide by constantly bringing blood of a different gas composition to and from the alveoli.

1 mark: large surface area correctly identified; 1 mark: thin walls/short diffusion distance correctly identified; 1 mark: good blood supply maintaining a steep concentration gradient correctly identified.

Relative formula mass (Mr) is the sum of the relative atomic masses of all the atoms shown in the formula of a compound. For \(\text{Mg}\text{C}\text{O}_{3}\): Mr = \(24 + 12 + (16 × 3) = 24 + 12 + 48 = 84\).

1 mark: correct definition of relative formula mass; 1 mark: correct method/substitution shown (\(24 + 12 + 48\)); 1 mark: correct final answer of 84.

During alpha decay, an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (equivalent to a helium nucleus). As a result, the mass number of the nucleus decreases by 4, and the atomic number decreases by 2, forming the nucleus of a new element.

1 mark: correct description of the alpha particle emitted (2 protons and 2 neutrons/helium nucleus); 1 mark: correct change in mass number (decreases by 4); 1 mark: correct change in atomic number (decreases by 2, forming a new element).

\(\text{Actual size} = \text{image size} ÷ \text{magnification}\) = 20 mm ÷ \(400 = 0.05\) mm. Converting to micrometres: 0.05 mm × \(1000 = 50\) μm.

1 mark: correct rearrangement of the magnification formula (\(\text{actual size} = \text{image size} ÷ \text{magnification}\)); 1 mark: correct division (\(20 ÷ 400 = 0.05\) mm); 1 mark: correct conversion from mm to μm (×1000); 1 mark: correct final answer of 50 μm.

Mr(\(\text{C}\text{O}_{2}\)) = \(12 + (16 × 2) = 44\). \(\text{Moles} = \text{mass} ÷ Mr\) = \(11.0 ÷ 44 = 0.25\) mol.

1 mark: correct Mr of \(\text{C}\text{O}_{2}\) calculated (44); 1 mark: correct substitution into \(\text{moles} = \text{mass} ÷ Mr\); 1 mark: correct arithmetic (\(11.0 ÷ 44\)); 1 mark: correct final answer of 0.25 mol with unit.

Acceleration = (v − u) ÷ t = \((20 − 8) ÷ 6 = 12 ÷ 6 = 2\) m/s².

1 mark: correct formula stated/used (\(a = (v − u) ÷ t\)); 1 mark: correct substitution; 1 mark: correct arithmetic (\(12 ÷ 6\)); 1 mark: correct final answer of 2 m/s² with unit.

SA:V ratio = surface area ÷ volume = \(0.008 ÷ 0.0004 = 20\), i.e. a ratio of 20:1. This large surface area to volume ratio means there is a large area for gas diffusion relative to the small volume gases must diffuse into/out of, increasing the rate of gas exchange.

1 mark: correct division shown as working; 1 mark: correct ratio of 20:1; 1 mark: valid comment that a large SA:V ratio provides more area for diffusion; 1 mark: correct link that this increases the rate of gas exchange.

\(\text{Moles} = \text{concentration} × \text{volume}\) (in dm³) = \(0.20 × 0.500 = 0.10\) mol. \(\text{Mass} = \text{moles} × Mr\) = \(0.10 × 58.5 = 5.85\) g.

1 mark: correct conversion of volume to dm³ (0.500 dm³); 1 mark: correct moles calculated (0.10 mol); 1 mark: correct substitution into \(\text{mass} = \text{moles} × Mr\); 1 mark: correct final answer of 5.85 g with unit.

\(\text{Weight} = \text{mass} × g\) = \(2000 × 10 = 20,000\) N (downward). Resultant force = thrust − weight = \(30,000 − 20,000 = 10,000\) N upward. \(\text{Acceleration} = \text{resultant force} ÷ \text{mass}\) = \(10,000 ÷ 2000 = 5\) m/s².

1 mark: correct weight calculated (20,000 N); 1 mark: correct resultant force calculated (10,000 N); 1 mark: correct substitution into \(a = F ÷ m\); 1 mark: correct final acceleration of 5 m/s² with unit.

Percentage transferred = (energy in primary consumers ÷ energy in producers) × \(100 = (1,200 ÷ 12,000) × 100 = 10\)%. This is a typical value, as usually only about 10% of the energy/biomass available at one trophic level is transferred to the next.

1 mark: correct substitution into percentage = \((1,200 ÷ 12,000) × 100\); 1 mark: correct arithmetic; 1 mark: correct final answer of 10%; 1 mark: valid comment that around 10% is a typical efficiency for energy transfer between trophic levels.

\(\text{Moles of Mg} = \text{mass} ÷ Ar\) = \(2.4 ÷ 24 = 0.10\) mol. From the equation, the mole ratio of Mg : \(\text{Mg}\text{O}\) is 1 : 1, so moles of \(\text{Mg}\text{O}\) formed = 0.10 mol. Mass of \(\text{Mg}\text{O}\) = moles × Mr = \(0.10 × 40 = 4.0\) g.

1 mark: correct moles of Mg calculated (0.10 mol); 1 mark: correct 1:1 mole ratio applied from the balanced equation; 1 mark: correct moles of \(\text{Mg}\text{O}\) obtained (0.10 mol); 1 mark: correct final mass of 4.0 g with unit.

\(\text{Density} = \text{mass} ÷ \text{volume}\) = \(0.0032 ÷ 2.0 = 0.0016\) kg/m³.

1 mark: correct formula stated/used (\(\text{density} = \text{mass} ÷ \text{volume}\)); 1 mark: correct substitution; 1 mark: correct arithmetic; 1 mark: correct final answer of 0.0016 kg/m³ with unit.

\(\text{Rate} = \text{volume} ÷ \text{time}\) = \(15 ÷ 5 = 3.0\) cm³/min. Converting to per second: \(3.0 ÷ 60 = 0.05\) cm³/s.

1 mark: correct substitution into \(\text{rate} = \text{volume} ÷ \text{time}\); 1 mark: correct rate of 3.0 cm³/min; 1 mark: correct method to convert minutes to seconds (÷60); 1 mark: correct final answer of 0.05 cm³/s.

For the 2 nm cube: surface area = \(6 × 2² = 24\) nm²; volume = \(2³ = 8\) nm³; SA:V = \(24 ÷ 8 = 3\), i.e. 3:1. For the 20 nm cube: surface area = \(6 × 20² = 2400\) nm²; volume = \(20³ = 8000\) nm³; SA:V = \(2400 ÷ 8000 = 0.3\), i.e. 0.3:1. This confirms that smaller particles have a much larger surface area to volume ratio.

1 mark: correct surface area and volume for the 2 nm cube; 1 mark: correct SA:V ratio of 3:1 for the 2 nm cube; 1 mark: correct surface area and volume for the 20 nm cube; 1 mark: correct SA:V ratio of 0.3:1 for the 20 nm cube, showing the smaller particle has a larger ratio.

Gravitational potential energy lost = mgh = \(0.50 × 10 × 4.0 = 20\) J. Since no energy is transferred to the surroundings, all of this is transferred to the kinetic energy store of the ball, so kinetic energy just before impact = 20 J.

1 mark: correct substitution into \(GPE = \text{mgh}\); 1 mark: correct GPE value of 20 J; 1 mark: correct statement that this equals the kinetic energy gained (conservation of energy); 1 mark: correct final answer of 20 J with unit.

t = √(\(2 × 0.20 ÷ 10\)) = √(\(0.40 ÷ 10\)) = √\(0.04 = 0.2\) s.

1 mark: correct substitution into the given equation; 1 mark: correct value calculated inside the square root (0.04); 1 mark: correct square root taken; 1 mark: correct final answer of 0.2 s with unit.

\(Rf = \text{distance moved by the spot} ÷ \text{distance moved by the solvent front}\) = \(6.4 ÷ 8.0 = 0.80\).

1 mark: correct formula stated/used (\(Rf = \text{distance moved by spot} ÷ \text{distance moved by solvent}\)); 1 mark: correct substitution; 1 mark: correct arithmetic; 1 mark: correct final answer of 0.80 (no units).

Enzymes: amylase (produced in the mouth and pancreas) breaks down starch into maltose/sugars; protease (produced in the stomach as pepsin, and in the pancreas) breaks down proteins into amino acids; lipase (produced in the pancreas) breaks down fats into fatty acids and glycerol, with bile from the liver first emulsifying fats to increase the surface area available to lipase. Small intestine adaptations: the inner lining is covered in villi and microvilli, which greatly increase the surface area for absorption; villi walls are only one cell thick, giving a short diffusion distance; villi are supplied by a dense network of capillaries (and a lacteal for fats), maintaining a concentration gradient for efficient absorption; the small intestine is also very long, increasing the time available for digestion and absorption.

Band A (5–6 marks): at least two enzymes correctly named with their substrates and products, AND at least two small-intestine adaptations clearly explained (not just listed); fluent, accurate use of specialist terms. Band B (3–4 marks): some correct enzyme and adaptation detail present but less complete, e.g. enzymes named without full detail, or adaptations listed without explanation. Band C (1–2 marks): a basic or fragmentary description with limited accurate detail. Band D (0 marks): no creditable response.

Elements in the Periodic Table are arranged in order of increasing atomic (proton) number. Elements in the same period have the same number of electron shells, with the outer shell filling progressively with electrons across the period. Elements in the same group have the same number of electrons in their outer shell, which is why they show similar chemical properties. Going down Group 7, reactivity decreases: atoms have an increasing number of electron shells, so the outer shell (which needs to gain one electron to become full) is further from the nucleus and shielded by more inner shells; this weakens the attraction between the nucleus and an incoming electron, making it harder for the atom to gain an electron, so reactivity decreases down the group.

Band A (5–6 marks): correct explanation of arrangement by proton number, correct link between periods/electron shells and groups/outer electrons, AND a correct, fully explained trend in Group 7 reactivity (more shells, more shielding, weaker attraction, harder to gain an electron); fluent use of specialist terms. Band B (3–4 marks): general structure of the table explained correctly but the Group 7 trend only partially explained, or vice versa. Band C (1–2 marks): a basic or fragmentary account with limited accurate detail. Band D (0 marks): no creditable response.

In nuclear fission, a large, unstable nucleus such as uranium-235 absorbs a neutron and splits into two smaller daughter nuclei, releasing a large amount of energy along with two or three further neutrons. These released neutrons can go on to be absorbed by other uranium-235 nuclei, causing them to split in turn — a self-sustaining chain reaction. In a power station, a moderator (e.g. graphite or water) slows down the fast neutrons so they are more easily absorbed by further uranium nuclei, sustaining the reaction; control rods (e.g. made of boron) are inserted or withdrawn to absorb excess neutrons and regulate the rate of the chain reaction, preventing it running out of control. The energy released as heat is used to heat water into steam, which drives turbines connected to generators, producing electricity.

Band A (5–6 marks): accurate, detailed description of fission (neutron absorbed, nucleus splits, energy and further neutrons released) AND correct explanation of the chain reaction AND correct explanation of at least one control mechanism (control rods/moderator) AND correct link to electricity generation (steam, turbine, generator); fluent use of specialist terms. Band B (3–4 marks): the basic fission/chain reaction process is described but control mechanisms or the electricity-generation link are only partially explained. Band C (1–2 marks): a basic or fragmentary account with limited accurate detail. Band D (0 marks): no creditable response.

(a) Light intensity decreases as distance from the light source increases (following an inverse square relationship — intensity is proportional to 1 ÷ distance²). (b) At low light intensities, light is the limiting factor: increasing it increases the rate of the light-dependent reaction, providing more energy (as ATP and reduced NADP) for the light-independent reaction, so the overall rate of photosynthesis increases. Beyond a certain intensity, light is no longer limiting, and another factor such as carbon dioxide concentration or temperature becomes limiting instead, so further increases in light intensity produce no further increase in rate and the graph plateaus. (c) Temperature should be controlled, e.g. using a water bath around the beaker, because temperature also affects the rate of the enzyme-controlled reactions of photosynthesis; if it were not controlled, a change in temperature could increase or decrease the rate independently of light intensity, making it impossible to draw a valid conclusion about the effect of light alone.

(a) 2 marks: 1 mark for correctly stating intensity decreases with distance; 1 mark for reference to the inverse square relationship. (b) 3 marks: 1 mark for explaining the initial increase in rate with light intensity; 1 mark for correct use of the concept of a limiting factor; 1 mark for correctly explaining the plateau (another factor becomes limiting). (c) 2 marks: 1 mark for a valid controlled variable (e.g. temperature or \(\text{C}\text{O}_{2}\) concentration); 1 mark for a valid, developed reason why it must be controlled.

(a) Amylase; it converts starch into maltose (a reducing sugar). (b) Pepsin has an optimum pH that is strongly acidic (around pH 2); the stomach produces hydrochloric acid, which provides this acidic environment. At a pH far from this optimum, the enzyme's active site changes shape (its activity is reduced/it may denature), so pepsin only works effectively within the acidic conditions the stomach provides. (c) An enzyme's active site has a specific shape that is complementary to the shape of one particular substrate molecule, like a lock and key; only a substrate with a matching, complementary shape can bind to the active site to form an enzyme-substrate complex and be catalysed. A differently shaped substrate does not fit the active site, so no enzyme-substrate complex forms and no reaction is catalysed.

(a) 2 marks: 1 mark for amylase; 1 mark for maltose as the product. (b) 2 marks: 1 mark for reference to pepsin's low/acidic optimum pH provided by stomach \(\text{H}\text{Cl}\); 1 mark for explaining reduced activity away from this optimum. (c) 3 marks: 1 mark for the complementary-shape concept; 1 mark for correct reference to formation of an enzyme-substrate complex; 1 mark for explaining why a differently shaped substrate cannot bind/react.

(a) Glucose + oxygen → carbon dioxide + water (with energy released). (b) Regular exercise can increase lung/vital capacity, allowing more air (and so more oxygen) to be taken in with each breath, and can increase the density of capillaries around the alveoli, increasing the surface area available for gas exchange. Regular exercise can also strengthen the heart muscle, increasing its stroke volume (the volume of blood pumped per beat), so the same amount of oxygenated blood can be delivered to the body with fewer heartbeats, lowering resting heart rate over time.

(a) 2 marks: 1 mark for correct reactants (glucose and oxygen); 1 mark for correct products (carbon dioxide and water, with energy released). (b) 4 marks: up to 2 marks for each of two valid, clearly explained ways exercise improves respiratory/circulatory efficiency (e.g. increased lung capacity/alveolar surface area; strengthened heart muscle/increased stroke volume and reduced resting heart rate), 1 mark for identification and 1 mark for explanation/development of each.

(a) A stimulus is detected by a receptor; an electrical impulse travels along a sensory neurone to the central nervous system (spinal cord), where it passes across a synapse to a relay neurone; the impulse then passes across another synapse to a motor neurone, which carries it to an effector (a muscle or gland), producing the response. (b) Because a reflex action only involves the spinal cord and does not require conscious processing by the brain, it occurs much faster than a consciously controlled response; this rapid response helps to protect the body from harm, for example by causing a hand to be withdrawn from a hot object before pain is even consciously registered.

(a) 3 marks: 1 mark for receptor detecting the stimulus and impulse travelling along a sensory neurone; 1 mark for a synapse/relay neurone in the spinal cord (CNS); 1 mark for a motor neurone carrying the impulse to an effector. (b) 3 marks: 1 mark for correctly stating that a reflex bypasses conscious brain processing; 1 mark for correctly linking this to a faster response; 1 mark for a valid, developed explanation of the protective benefit (e.g. named example).

(a) The pyramid of biomass would be widest at the bottom, representing grass (the producer), becoming progressively narrower for the rabbit (primary consumer) and narrower again for the fox (secondary consumer/top predator), since the total biomass present decreases at each higher trophic level. (b) Not all of the biomass/energy present at one trophic level is passed on to the next: some biomass is used by the organism in respiration and lost from the food chain as heat energy; some biomass is lost in waste materials such as faeces and urine, which are not eaten by the next consumer; and some parts of an organism (e.g. bones, roots) are not eaten or digested by the next consumer. As a result, typically only around 10% of the biomass at one level is transferred to the next, so biomass decreases at each successive trophic level.

(a) 2 marks: 1 mark for a correctly shaped pyramid narrowing from a wide base; 1 mark for the trophic levels in the correct order (grass, rabbit, fox). (b) 4 marks: up to 3 marks for named, explained sources of biomass loss (e.g. respiration/heat loss, waste materials, uneaten/undigested parts — 1 mark each, max 3); 1 mark for the overall conclusion that this results in progressively less biomass being available at each higher trophic level.

(a) A sodium atom (electron arrangement 2,8,1) loses its single outer-shell electron to form a positively charged \(\text{Na}^{+}\) ion with a stable, full outer shell (2,8). A chlorine atom (2,8,7) gains this electron to form a negatively charged \(\text{Cl}^{-}\) ion with a stable, full outer shell (2,8,8). The oppositely charged ions are then held together by strong electrostatic forces of attraction. (b) Solid sodium chloride has a giant ionic lattice structure, in which each ion is surrounded by, and strongly electrostatically attracted to, oppositely charged ions in a regular, repeating three-dimensional arrangement. Its high melting point results because a very large amount of energy is needed to overcome these numerous strong electrostatic forces of attraction throughout the whole lattice. (c) In the solid, the ions are held in fixed positions within the rigid lattice and cannot move freely, so there are no mobile charge carriers and it cannot conduct electricity. When molten, the lattice structure breaks down and the ions become free to move, allowing them to carry electric charge through the liquid, so molten sodium chloride does conduct.

(a) 3 marks: 1 mark for sodium losing an electron to form \(\text{Na}^{+}\); 1 mark for chlorine gaining an electron to form \(\text{Cl}^{-}\); 1 mark for reference to both ions achieving a full/stable outer shell. (b) 2 marks: 1 mark for a correctly described giant ionic lattice structure; 1 mark for correctly explaining the high melting point in terms of many strong electrostatic forces requiring much energy to overcome. (c) 2 marks: 1 mark for correctly explaining that ions are fixed in the solid and cannot carry charge; 1 mark for correctly explaining that ions are free to move and carry charge when molten.

(a) A glowing splint is inserted into the gas; if oxygen is present, the splint relights/glows more brightly. (b) The gas is bubbled through limewater (calcium hydroxide solution); if carbon dioxide is present, the limewater turns cloudy/milky. (c) Obtaining two spots with different Rf values shows that the mixture contains at least two different substances/dyes, because each moves a different distance relative to the solvent front depending on its own solubility in the solvent and its attraction to the paper. The identity of each dye can be confirmed by comparing its Rf value, measured under the same solvent and conditions, with the Rf values of known reference dyes run on the same chromatogram.

(a) 2 marks: 1 mark for the correct method (glowing splint); 1 mark for the correct positive result (relights/glows more brightly). (b) 2 marks: 1 mark for the correct method (bubble through limewater); 1 mark for the correct positive result (turns cloudy/milky). (c) 3 marks: 1 mark for correctly interpreting two spots as showing at least two substances present; 1 mark for correctly explaining that Rf differs due to differing solubility/attraction to the paper; 1 mark for a correct method to confirm identity (compare Rf with known references under identical conditions).

(a) Acid + metal carbonate → salt + water + carbon dioxide. (b) The mixture is filtered to remove the excess, unreacted solid copper carbonate, leaving a blue copper sulfate solution as the filtrate. This filtrate is then gently heated/evaporated to reduce its volume until it reaches saturation point (crystals begin to form), then left to cool so crystals form; the crystals are then filtered off and dried (e.g. between sheets of filter paper) to obtain pure, dry copper sulfate crystals. (c) In any acid, the reacting species is the hydrogen ion, \(\text{H}^{+}\text{(aq)}\); in any alkali, the reacting species is the hydroxide ion, \(\text{O}\text{H}^{-}\text{(aq)}\). Neutralisation always involves the same reaction between these two ions to form water: \(\text{H}^{+}\text{(aq)} + \text{O}\text{H}^{-}\text{(aq)} \rightarrow \text{H}_{2}\text{O}\text{(l)}\), regardless of which specific acid or alkali is used, so this same ionic equation applies to every acid–alkali neutralisation.

(a) 1 mark: fully correct word equation. (b) 3 marks: 1 mark for filtration to remove excess solid; 1 mark for evaporation/heating the filtrate to the point of saturation; 1 mark for cooling/crystallising and drying the crystals. (c) 3 marks: 1 mark for identifying \(\text{H}^{+}\text{(aq)}\) as the reacting species from acids; 1 mark for identifying \(\text{O}\text{H}^{-}\text{(aq)}\) as the reacting species from alkalis; 1 mark for the correct ionic equation \(\text{H}^{+}\text{(aq)} + \text{O}\text{H}^{-}\text{(aq)} \rightarrow \text{H}_{2}\text{O}\text{(l)}\) and explanation that this is common to all such reactions.

(a) Graphite has a giant covalent structure arranged in layers of hexagonally-bonded carbon atoms, with each carbon atom covalently bonded to only three others within its layer. The layers are held together only by weak intermolecular forces, allowing them to slide over each other easily; this is why graphite is soft and slippery and can be used as a lubricant (and in pencils). (b) Each carbon atom in graphite uses only three of its four outer-shell electrons in covalent bonding, leaving one delocalised electron per atom that is free to move throughout the structure between the layers; this allows graphite to conduct electricity. In diamond, every carbon atom uses all four outer-shell electrons in covalent bonds, so there are no delocalised electrons free to move, and diamond cannot conduct electricity.

(a) 3 marks: 1 mark for a correctly described layered giant covalent structure with each carbon bonded to three others; 1 mark for weak forces between layers allowing them to slide; 1 mark for correctly linking this to use as a lubricant. (b) 3 marks: 1 mark for identifying delocalised electrons present in graphite and free to move; 1 mark for correctly linking this to electrical conduction; 1 mark for correctly contrasting with diamond, which has no delocalised/free electrons.

(a) Across a period, elements have an increasing number of electrons in their outer shell, from 1 in Group 1 up to a full outer shell in Group 0. It is the number of outer-shell electrons that determines an element's chemical properties (how readily and in what way it reacts), so chemical properties change progressively across a period even though the number of electron shells stays the same. (b) Atoms of noble gases already have a full outer shell of electrons, which is a stable electron configuration. This means they have no tendency to lose, gain or share electrons in order to bond with other atoms, so they exist as very unreactive, stable single atoms.

(a) 3 marks: 1 mark for correctly stating outer-shell electron number increases across a period; 1 mark for correctly linking outer-shell electrons to chemical properties/reactivity; 1 mark for a correct overall conclusion (properties change across a period as a result). (b) 3 marks: 1 mark for correctly identifying a full outer shell; 1 mark for correctly linking this to no tendency to lose/gain/share electrons; 1 mark for the correct conclusion (very unreactive).

(a) The graph is a straight line rising from the origin (0, 0) to (4, 8), showing uniform acceleration; then a horizontal straight line from (4, 8) to (14, 8), showing constant velocity; then a straight line falling from (14, 8) to (16, 0), showing uniform deceleration to rest. (b) Acceleration = (v − u) ÷ t = \((8 − 0) ÷ 4 = 2\) m/s². (c) Total distance = area under the graph = area of the first triangle + area of the rectangle + area of the second triangle = (\(½ × 4 × 8\)) + (\(10 × 8\)) + (\(½ × 2 × 8\)) = \(16 + 80 + 8 = 104\) m.

(a) 2 marks: 1 mark for the correct overall shape (rising, then flat, then falling to zero); 1 mark for correctly consistent time/velocity values at each stage. (b) 2 marks: 1 mark for correct substitution into \(a = (v − u) ÷ t\); 1 mark for the correct final answer of 2 m/s² with unit. (c) 3 marks: 1 mark for the correct method (splitting the area under the graph into a triangle, rectangle and triangle); 1 mark for correct calculation of each of the three areas; 1 mark for the correct total distance of 104 m.

(a) For a system in equilibrium (balanced), the sum of the clockwise moments about a pivot equals the sum of the anticlockwise moments about the same pivot. (b) \(\text{Moment} = \text{force} × \text{distance}\). Child's moment = \(300 × 1.5 = 450\) Nm. For balance, adult's moment must also equal 450 Nm: \(450 = 450\) × d, so d = \(450 ÷ 450 = 1.0\) m. (c) At 0.6 m, the adult's moment would be \(450 × 0.6 = 270\) Nm, which is less than the child's moment of 450 Nm. Since the moments would no longer be equal, the see-saw would be unbalanced and would tip down on the child's side, as the child's (larger) moment would dominate.

(a) 1 mark: correct statement of the principle of moments. (b) 3 marks: 1 mark for correctly calculating the child's moment (450 Nm); 1 mark for correct rearrangement to find distance; 1 mark for the correct final answer of 1.0 m with unit. (c) 3 marks: 1 mark for correctly calculating the adult's new moment (270 Nm); 1 mark for correctly comparing this with the child's moment and identifying they are unequal; 1 mark for the correct conclusion that the see-saw tips down on the child's side.

(a) In a solid, particles are closely packed in a fixed, regular arrangement and can only vibrate about fixed positions. In a liquid, particles are still close together but arranged randomly, and are able to move and slide past one another. In a gas, particles are far apart, arranged randomly, and move rapidly and randomly in all directions, with very weak forces between them. (b) Heating the gas increases the average kinetic energy of its particles, so they move faster on average. These faster-moving particles collide with the container walls more frequently, and each collision exerts a greater force on the wall. Both a higher frequency of collisions and a greater force per collision increase the total force acting on a given area of the container walls, increasing the gas pressure.

(a) 3 marks: 1 mark each for a correctly described state (solid, liquid, gas), covering both particle arrangement and movement. (b) 3 marks: 1 mark for particles gaining kinetic energy/moving faster when heated; 1 mark for more frequent collisions with the container walls; 1 mark for a greater force per collision, both increasing pressure.

(a) The kinetic energy store of the car decreases as it slows down; the thermal energy store of the brakes (and surrounding air) increases, as work done by friction between the brake pads and discs transfers energy to heat. (b) KE = \(½ × 1200 × 20² = ½ × 1200 × 400 = 240,000\) J. (c) The calculation assumes that all of the car's kinetic energy is transferred to the thermal energy store of the brakes, with no energy 'lost' to other stores such as sound or overcoming air resistance during braking — in reality, some energy is dissipated in these other ways.

(a) 2 marks: 1 mark for the correct decreasing store (kinetic energy store of the car); 1 mark for the correct increasing store (thermal energy store of the brakes). (b) 4 marks: 1 mark for correct substitution into \(KE = ½\text{mv}²\); 1 mark for correctly squaring the velocity (400); 1 mark for correct arithmetic; 1 mark for the correct final answer of 240,000 J with unit. (c) 1 mark: a valid assumption correctly stated (e.g. no energy lost to sound/air resistance, all KE transferred to the brakes).

(a) In beta decay, a neutron within the nucleus converts into a proton and an electron; the high-energy electron (the beta particle) is emitted from the nucleus, while the proton remains. As a result, the atomic number of the nucleus increases by 1 (forming a new element), while the mass number stays the same, since the total number of protons plus neutrons is unchanged. (b) 24 days ÷ 8 days per half-life = 3 half-lives. Activity after each half-life: 640 → 320 (1st) → 160 (2nd) → 80 Bq (3rd).

(a) 3 marks: 1 mark for a neutron converting into a proton and an electron; 1 mark for the electron (beta particle) being emitted from the nucleus; 1 mark for the correct effect on atomic number (+1) and mass number (unchanged). (b) 3 marks: 1 mark for correctly calculating 3 half-lives have elapsed; 1 mark for showing the repeated halving as working; 1 mark for the correct final answer of 80 Bq.

Osmosis is the net movement of water molecules from a region of higher water potential (a more dilute solution) to a region of lower water potential (a more concentrated solution), through a partially permeable membrane.

1 mark: correct direction of net water movement (high to low water potential/dilute to concentrated); 1 mark: correct reference to a partially permeable membrane.

Arteries have thick, muscular and elastic walls to withstand and maintain the high pressure of blood being pumped away from the heart. Veins have thinner walls, carry blood at lower pressure back to the heart, and often contain valves to prevent the backflow of blood.

1 mark for each correct, valid difference stated, up to a maximum of 2 (e.g. wall thickness, pressure/direction of flow, presence of valves).

The menstrual cycle is controlled by several hormones, including follicle-stimulating hormone (FSH), luteinising hormone (LH), oestrogen and progesterone, which act on the ovaries and uterus lining.

1 mark for each correctly named hormone, up to a maximum of 2 (accept FSH, LH, oestrogen, progesterone).

In order of decreasing reactivity: magnesium, then iron, then copper. Magnesium and iron are both more reactive than hydrogen, so both react with dilute hydrochloric acid; copper is less reactive than hydrogen, so it does not react with the acid.

1 mark: correct order (magnesium, iron, copper); 1 mark: correct identification that magnesium and iron react with the acid but copper does not.

Iron rusts (is oxidised) only when both water (moisture) and oxygen are present together; if either is absent, rusting does not occur.

1 mark: water/moisture correctly identified; 1 mark: oxygen correctly identified.

The rate of reaction between a solid and a solution can be increased by increasing the temperature, increasing the concentration of the solution, increasing the surface area of the solid (e.g. using smaller pieces or a powder), or by adding a suitable catalyst.

1 mark for each correct, valid method stated, up to a maximum of 2.

Wave speed = frequency × wavelength. Frequency is measured in hertz (Hz).

1 mark: correct equation (speed = frequency × wavelength); 1 mark: correct unit of frequency (Hz).

The law of reflection states that the angle of incidence is equal to the angle of reflection, both measured from the normal (a line perpendicular to the reflecting surface at the point of incidence).

1 mark: correct statement that the angle of incidence equals the angle of reflection; 1 mark: correct reference to both angles being measured from the normal.

In a series circuit, the current is the same at every point around the circuit. In a parallel circuit, the current splits between the branches, so the total current supplied is equal to the sum of the currents flowing through each individual branch.

1 mark: correct statement for series (current the same throughout); 1 mark: correct statement for parallel (current splits/total equals sum of branch currents).

A genome is the entire genetic material (all of the DNA/genes) of an organism. Knowledge of the human genome can be used in medicine to identify genes associated with inherited disorders, allowing genetic screening or testing so that individuals at risk can be identified and offered appropriate advice, monitoring or treatment.

1 mark: correct definition of genome; 1 mark: a valid medical use identified (e.g. genetic screening for inherited disorders); 1 mark: the use correctly developed/explained.

Continuous variation shows a full, unbroken range of values between two extremes, with no distinct categories; it is usually controlled by several genes (and often influenced by environment), for example human height. Discontinuous variation falls into a small number of distinct, separate categories with no intermediate values; it is usually controlled by a single gene, for example human blood group.

1 mark: correct description of continuous variation with a valid example; 1 mark: correct description of discontinuous variation with a valid example; 1 mark: correct reference to genetic control (many genes for continuous, a single gene for discontinuous).

Phagocytes are a type of white blood cell that engulf and digest pathogens directly (phagocytosis). Lymphocytes are a type of white blood cell that recognise specific antigens on the surface of a pathogen and produce complementary antibodies, which bind to the pathogen and help to destroy it or mark it for destruction.

1 mark: phagocytosis correctly described; 1 mark: antibody production by lymphocytes correctly described; 1 mark: correct additional detail/development of either mechanism (e.g. reference to antigens).

A saturated hydrocarbon contains only single carbon-carbon bonds (e.g. an alkane); an unsaturated hydrocarbon contains at least one carbon-carbon double bond (e.g. an alkene). Bromine water can be used to test for unsaturation: an unsaturated hydrocarbon rapidly decolourises bromine water (from orange to colourless), while a saturated hydrocarbon does not.

1 mark: correct definitions of saturated and unsaturated; 1 mark: correct test reagent (bromine water); 1 mark: correct results distinguishing the two (decolourises for unsaturated, stays orange for saturated).

At the cathode (negative electrode), positively charged lead ions (\(\text{Pb}^{2+}\)) are attracted, gain electrons and are reduced to form molten lead metal. At the anode (positive electrode), negatively charged bromide ions (\(\text{Br}^{-}\)) are attracted, lose electrons and are oxidised to form bromine.

1 mark: correct cathode process and product (\(\text{Pb}^{2+}\) gains electrons/reduced to Pb); 1 mark: correct anode process and product (\(\text{Br}^{-}\) loses electrons/oxidised to \(\text{Br}_{2}\)); 1 mark: correct reference to ions being attracted to the oppositely charged electrode.

To test for carbon dioxide, the gas is bubbled through limewater (calcium hydroxide solution); a positive result is the limewater turning cloudy/milky. To test for hydrogen, a lit splint is held at the mouth of the test tube; a positive result is a squeaky pop sound as the hydrogen ignites and burns rapidly.

1 mark: correct \(\text{C}\text{O}_{2}\) test and result (limewater turns cloudy); 1 mark: correct hydrogen test and result (lit splint gives a squeaky pop); 1 mark: correct distinction between using a lit splint for hydrogen (not a glowing splint, which tests for oxygen).

The strength of a solenoid's magnetic field can be increased by increasing the current flowing through it, by increasing the number of turns (coils) of wire, and by adding a soft iron core inside the coil.

1 mark for each correct factor stated, up to a maximum of 3 (increasing current; increasing number of turns; adding an iron core).

An orbit is the curved path followed by one body (e.g. a satellite) around another (e.g. a planet), maintained by the gravitational force of attraction between them. Because this gravitational force acts towards the centre of the orbit — perpendicular to the satellite's direction of motion at every point — it continually changes the satellite's direction (keeping it moving in a circle) rather than doing work on it to speed it up or slow it down, so its speed remains constant without needing any additional forward-driving force.

1 mark: correct definition of an orbit (curved path due to gravitational attraction); 1 mark: correct reference to gravity providing the centripetal force; 1 mark: correct explanation that a force perpendicular to motion changes direction, not speed.

Potential difference (voltage) is the energy transferred per unit charge passed between two points in a circuit. It is linked to current and resistance by the equation \(V = I × R\) (potential difference = current × resistance).

1 mark: correct definition of potential difference (energy transferred per unit charge); 1 mark: correct equation (\(V = I × R\)); 1 mark: correct identification of the unit of potential difference (volt, V).

Parent Tt produces gametes T and t; parent tt produces only gametes t. Combining these gives offspring genotypes Tt, Tt, tt, tt — a ratio of 1 Tt : 1 tt. Since Tt is tall (T dominant) and tt is short, half of the offspring (2 out of 4) are expected to be tall, giving a 50% probability.

1 mark: correct gametes identified for each parent (T,t and t,t); 1 mark: correct genotype combinations shown (Tt, Tt, tt, tt); 1 mark: correct genotype ratio (1:1); 1 mark: correct percentage probability of tall offspring (50%).

\(\text{Cardiac output} = \text{stroke volume} × \text{heart rate}\) = \(70 × 72 = 5040\) cm³/min.

1 mark: correct formula stated/used (\(\text{cardiac output} = \text{stroke volume} × \text{heart rate}\)); 1 mark: correct substitution; 1 mark: correct arithmetic; 1 mark: correct final answer of 5040 cm³/min with unit.

Percentage success rate = (successful cycles ÷ total cycles) × \(100 = (68 ÷ 340) × 100 = 20\)%.

1 mark: correct substitution into percentage = \((68 ÷ 340) × 100\); 1 mark: correct arithmetic; 1 mark: correct final answer of 20%; 1 mark: percentage sign/unit correctly given.

\(\text{Moles of Zn} = \text{mass} ÷ Ar\) = \(6.5 ÷ 65 = 0.10\) mol. The equation shows a 1:1 mole ratio of Zn to Cu, so moles of Cu formed = 0.10 mol. \(\text{Mass of Cu} = \text{moles} × Ar\) = \(0.10 × 64 = 6.4\) g.

1 mark: correct moles of zinc calculated (0.10 mol); 1 mark: correct 1:1 mole ratio applied; 1 mark: correct moles of copper obtained (0.10 mol); 1 mark: correct final mass of 6.4 g with unit.

Volume = moles × molar volume at rtp = \(0.50 × 24 = 12\) dm³.

1 mark: correct relationship used (volume = moles × 24 dm³); 1 mark: correct substitution; 1 mark: correct arithmetic; 1 mark: correct final answer of 12 dm³ with unit.

\(\text{Moles of copper deposited} = \text{mass} ÷ Ar\) = \(0.32 ÷ 64 = 0.005\) mol. The anode (made of impure copper) loses mass during electrorefining, as copper atoms at the anode lose electrons and dissolve into solution as \(\text{Cu}^{2+}\) ions, which then travel to the cathode to be deposited as pure copper.

1 mark: correct substitution into \(\text{moles} = \text{mass} ÷ Ar\); 1 mark: correct final answer of 0.005 mol with unit; 1 mark: correct statement that the anode loses mass; 1 mark: correct reason (copper anode dissolves into solution as ions).

\(\text{sin}(C) = 1 ÷ n\) = \(1 ÷ 1.5 = 0.667\). C = sin⁻¹\((0.667) = 41.8\)°, which rounds to approximately 42°.

1 mark: correct substitution into \(\text{sin}(C) = 1 ÷ n\); 1 mark: correct value of sin(C) (0.667); 1 mark: correct method (inverse sine) to find the angle; 1 mark: correct final answer of approximately 42° (accept 41–42°).

\(Vs ÷ Vp = Ns ÷ Np\), so \(Vs = Vp × (Ns ÷ Np)\) = \(12 × (1000 ÷ 200) = 12 × 5 = 60\) V.

1 mark: correct transformer equation stated/used; 1 mark: correct turns ratio calculated (\(1000 ÷ 200 = 5\)); 1 mark: correct substitution/rearrangement; 1 mark: correct final answer of 60 V with unit.

Change in mass = \(5.28 − 4.80 = 0.48\) g. \(\text{Percentage change} = (\text{change} ÷ \text{original mass}) × 100\) = \((0.48 ÷ 4.80) × 100\) = +10%. Since the potato cylinders gained mass, water must have moved into the potato cells by osmosis, which means the surrounding sucrose solution had a higher water potential (was more dilute/less concentrated) than the cytoplasm of the potato cells.

1 mark: correct change in mass calculated (0.48 g); 1 mark: correct substitution into \(\text{percentage change} = (\text{change} ÷ \text{original}) × 100\); 1 mark: correct final answer of +10%; 1 mark: correct statement that the solution was less concentrated/more dilute than the cytoplasm; 1 mark: correct reasoning linking mass gain to net water movement into the cells by osmosis.

Percentage increase = (\((150 − 20) ÷ 20\)) × \(100 = (130 ÷ 20) × 100 = 650\)%. The data shows the number of new cases rising steeply through Weeks 1–4, but the increase from Week 4 to Week 5 (150 to 160) is much smaller than in earlier weeks, showing the rate of new infections beginning to level off/slow down, possibly due to increasing immunity in the population or the effect of control measures.

1 mark: correct change in cases calculated (130); 1 mark: correct substitution into percentage increase formula; 1 mark: correct final answer of 650%; 1 mark: correct description of the overall rising trend; 1 mark: correct identification that the rate of increase slows/levels off by Week 5.

Rate (0–10 s) = volume ÷ time = \(40 ÷ 10 = 4\) cm³/s. Rate (40–60 s) = \((80 − 70) ÷ (60 − 40) = 10 ÷ 20 = 0.5\) cm³/s. These values show the rate of reaction decreases considerably as the reaction proceeds, because the concentration of acid falls as it is used up, reducing the frequency of successful collisions between acid and marble particles per unit time.

1 mark: correct rate calculated for 0–10 s (4 cm³/s); 1 mark: correct rate calculated for 40–60 s (0.5 cm³/s); 1 mark: correct comparison stating the rate has decreased; 2 marks: correct explanation in terms of falling acid concentration reducing the frequency of successful collisions.

Moles of C = \(2.4 ÷ 12 = 0.2\). Moles of H = \(0.6 ÷ 1 = 0.6\). Moles of O = \(3.2 ÷ 16 = 0.2\). Dividing each by the smallest value (0.2): C = 1, H = 3, O = 1. Empirical formula = \(\text{C}\text{H}_{3}\text{O}\).

1 mark: correct moles of carbon (0.2); 1 mark: correct moles of hydrogen (0.6); 1 mark: correct moles of oxygen (0.2); 1 mark: correct simplified whole-number ratio (1:3:1); 1 mark: correct final empirical formula (\(\text{C}\text{H}_{3}\text{O}\)).

Energy to break bonds (reactants) = \((4 × 413) + (2 × 498) = 1652 + 996 = 2648\) kJ/mol. Energy released forming bonds (products) = \((2 × 805) + (4 × 464) = 1610 + 1856 = 3466\) kJ/mol. Overall energy change = energy in − energy out = \(2648 − 3466\) = −818 kJ/mol. Since more energy is released forming bonds than is needed to break them, the reaction is exothermic.

1 mark: correct total energy to break bonds (2648 kJ/mol); 1 mark: correct total energy released forming bonds (3466 kJ/mol); 1 mark: correct method (energy in − energy out); 1 mark: correct final value of −818 kJ/mol; 1 mark: correct conclusion that the reaction is exothermic, with a valid reason.

Wave speed = frequency × wavelength = \(340 × 1.0 = 340\) m/s, which is consistent with the known speed of sound in air (~340 m/s). Because the speed of sound in a given medium remains constant, if frequency increases, wavelength must decrease proportionally (since speed = frequency × wavelength is fixed), keeping the product the same.

1 mark: correct formula used (speed = frequency × wavelength); 1 mark: correct substitution and arithmetic; 1 mark: correct answer of 340 m/s with a valid comparison to the known speed of sound; 2 marks: correct explanation that wavelength decreases as frequency increases because wave speed in the same medium is constant.

\(1 ÷ R(\text{total}) = 1 ÷ \text{R1} + 1 ÷ \text{R2}\) = \(1 ÷ 4 + 1 ÷ 12 = 3 ÷ 12 + 1 ÷ 12 = 4 ÷ 12 = 1 ÷ 3\). Therefore R(total) = 3 Ω.

1 mark: correct formula stated (1 ÷ Rtotal = 1 ÷ R\(1 + 1\) ÷ R2); 1 mark: correct substitution; 1 mark: correct combination using a common denominator (\(3/12 + 1/12 = 4/12\)); 1 mark: correct simplification to \(1/3\); 1 mark: correct final answer of 3 Ω with unit.

Circumference of orbit = 2 × π × radius = \(2 × 3.14 × 4.2 × 10⁷\) ≈ \(2.64 × 10⁸\) m. \(\text{Speed} = \text{circumference} ÷ \text{time}\) = \(2.64 × 10⁸ ÷ 86,400\) ≈ 3053 m/s, which rounds to approximately 3050 m/s.

1 mark: correct formula stated/used; 1 mark: correct substitution of values; 1 mark: correct calculation of the orbit's circumference (≈\(2.64 × 10⁸\) m); 1 mark: correct division by time; 1 mark: correct final answer of approximately 3050 m/s (accept 3000–3100 m/s).

A vaccine contains a small, harmless quantity of a dead or inactive pathogen, or antigens from it. This stimulates white blood cells (lymphocytes) to produce specific antibodies against the pathogen, and importantly, long-lived memory cells are also produced. If the vaccinated person is later exposed to the live pathogen, these memory cells enable a fast, strong secondary immune response, producing antibodies quickly enough to destroy the pathogen before the person becomes ill, giving long-term individual immunity. Herd immunity occurs at the population level: if a sufficiently high proportion of the population is vaccinated (immune), the pathogen has far fewer susceptible hosts to infect, so it cannot spread easily through the population; this indirectly protects individuals who cannot be vaccinated (e.g. for medical reasons) or in whom vaccination was less effective, because they are unlikely to come into contact with an infected person.

Band A (5–6 marks): accurate description of how a vaccine stimulates antibody and memory cell production, a correct explanation of the rapid secondary response memory cells enable, AND a correct, developed explanation of herd immunity (reduced transmission when a high proportion of the population is immune, protecting unvaccinated individuals); fluent, accurate use of specialist terms. Band B (3–4 marks): the individual immunity mechanism or herd immunity is well explained but the other is only partially covered. Band C (1–2 marks): a basic or fragmentary account with limited accurate detail. Band D (0 marks): no creditable response.

According to collision theory, a chemical reaction can only occur when reacting particles collide with each other with sufficient energy — at least the activation energy — and in the correct orientation; not all collisions are successful. Increasing temperature increases the average kinetic energy of the particles, so they move faster; this increases both the frequency of collisions between particles and, more significantly, the proportion of collisions that have enough energy to be successful (equal to or greater than the activation energy), substantially increasing the rate of reaction. Increasing the concentration of a solution increases the number of particles present in a given volume; this means reacting particles are closer together on average, so they collide more frequently, increasing the rate of successful collisions and therefore the rate of reaction — but, unlike a temperature increase, it does not change the energy of individual collisions or the proportion of collisions that are successful.

Band A (5–6 marks): correct statement of collision theory (successful collisions require sufficient energy/activation energy), a correct, developed explanation of the effect of temperature (faster particles, more frequent AND more energetic/successful collisions), AND a correct, developed explanation of the effect of concentration (more particles per volume, more frequent collisions); fluent, accurate use of specialist terms. Band B (3–4 marks): collision theory and at least one of the two factors are explained correctly, but coverage of the other factor is partial or the explanations lack full development. Band C (1–2 marks): a basic or fragmentary account with limited accurate detail. Band D (0 marks): no creditable response.

In the Sun's core, hydrogen nuclei fuse together under conditions of extremely high temperature and pressure to form helium nuclei, releasing very large amounts of energy in the process; this energy is what powers the Sun and is eventually radiated as light and heat. Nuclear fusion has not yet been used to generate electricity commercially on Earth because achieving and sustaining the extremely high temperatures and pressures needed to force positively charged nuclei close enough together to fuse (overcoming their strong electrostatic repulsion) is technologically very difficult; current experimental fusion reactors require more energy to create and maintain these conditions (e.g. through magnetic confinement) than the fusion reaction itself currently releases, so a sustained, net energy-positive commercial reactor has not yet been achieved.

Band A (5–6 marks): accurate description of hydrogen fusing into helium under extreme temperature/pressure releasing energy, AND a correct, developed explanation of why commercial fusion has not been achieved (extreme conditions needed to overcome nuclear repulsion; energy input currently exceeds output); fluent, accurate use of specialist terms. Band B (3–4 marks): the fusion process is described correctly but the explanation of the barriers to commercial use is only partially developed, or vice versa. Band C (1–2 marks): a basic or fragmentary account with limited accurate detail. Band D (0 marks): no creditable response.

(a) Parent 1 (Ff) produces gametes F and f; parent 2 (Ff) also produces gametes F and f. Combining these gives offspring genotypes FF, Ff, Ff and ff. (b) Genotype ratio: 1 FF : 2 Ff : 1 ff. Phenotype ratio: 3 unaffected : 1 affected (since FF and Ff are both unaffected, and only ff shows the disorder). Probability of a child having cystic fibrosis = 1 in 4 (25%). (c) Both parents are unaffected because they each carry one dominant allele (F), which masks the effect of the recessive allele (f) in their own phenotype — they are carriers. However, each parent can still pass on either allele to their offspring; if a child inherits the recessive f allele from both parents, becoming ff, they will have cystic fibrosis, even though neither parent shows the disorder themselves.

(a) 2 marks: 1 mark for correctly showing the gametes F and f from each parent; 1 mark for the four correct genotype combinations (FF, Ff, Ff, ff). (b) 2 marks: 1 mark for the correct genotype/phenotype ratios; 1 mark for the correct probability (\(1/4\) or 25%). (c) 3 marks: 1 mark for correctly stating both parents are unaffected carriers; 1 mark for explaining that the dominant allele masks the recessive allele in the parents; 1 mark for correctly explaining that a child needs to inherit f from both parents to be affected.

(a) Deoxygenated blood enters the right atrium from the vena cava; it passes through the tricuspid valve into the right ventricle, which pumps it through the pulmonary valve into the pulmonary artery towards the lungs. Oxygenated blood returns from the lungs via the pulmonary vein into the left atrium; it passes through the bicuspid (mitral) valve into the left ventricle, which pumps it through the aortic valve into the aorta and around the body. (b) The left ventricle must pump blood at high pressure all the way around the entire body (the systemic circulation), a much longer route with greater overall resistance than the right ventricle's task of pumping blood only the short distance to the nearby lungs (the pulmonary circulation). It therefore needs a thicker, more muscular wall to generate the greater force required to maintain this higher pressure.

(a) 4 marks: 1 mark for vena cava into the right atrium; 1 mark for the tricuspid valve into the right ventricle and onward to the pulmonary artery; 1 mark for the pulmonary vein into the left atrium; 1 mark for the bicuspid valve into the left ventricle and onward to the aorta. (b) 3 marks: 1 mark for correctly contrasting pumping to the whole body versus only the lungs; 1 mark for reference to the greater distance/pressure required; 1 mark for correctly linking this to the need for a thicker, more muscular wall.

(a) The combined pill contains the hormones oestrogen and progesterone. These hormones inhibit the release of FSH and LH from the pituitary gland, which prevents the maturation and release of an egg (ovulation); the hormones can also thicken cervical mucus, making it harder for sperm to reach an egg. (b) An advantage of hormonal contraception is that, if taken correctly, it is highly effective at preventing pregnancy and does not require any action to be taken at the time of intercourse itself. A disadvantage is that hormonal contraception provides no protection against sexually transmitted infections, unlike a barrier method such as a condom, and it may also cause side effects for some users.

(a) 3 marks: 1 mark for correctly naming oestrogen and progesterone; 1 mark for correctly explaining inhibition of FSH/LH preventing ovulation; 1 mark for an additional correct mechanism (e.g. thickened cervical mucus). (b) 4 marks: 2 marks for a valid, explained advantage; 2 marks for a valid, explained disadvantage (1 mark for identification, 1 mark for explanation, in each case).

(a) Within a bacterial population there is genetic variation, and by chance a small number of bacteria may already carry a mutation giving them some resistance to a particular antibiotic. When the antibiotic is used, non-resistant bacteria are killed, but resistant bacteria survive and are able to reproduce, passing the resistance allele on to their offspring. Over many generations, the proportion of resistant bacteria in the population increases, until eventually most or all of the population is resistant. (b) The resistance allele already existed in some bacteria by chance mutation before the antibiotic was ever used — exposure to the antibiotic did not cause bacteria to develop resistance in response; it simply acted as a selection pressure that favoured the survival and reproduction of bacteria that already happened to carry the resistant allele.

(a) 4 marks: 1 mark for reference to existing genetic variation/mutation; 1 mark for the antibiotic (selection pressure) killing non-resistant bacteria; 1 mark for resistant survivors reproducing and passing on the allele; 1 mark for the resistant proportion increasing over generations. (b) 2 marks: 1 mark for correctly stating resistance pre-existed by chance mutation rather than being caused by exposure; 1 mark for correctly distinguishing the antibiotic as a selection pressure rather than an instructive cause.

(a) A vaccine contains a small, harmless quantity of a dead or inactivated pathogen, or antigens from it. This triggers white blood cells (lymphocytes) to produce specific antibodies against the pathogen, and memory cells are also produced. If the person is later exposed to the live pathogen, these memory cells allow a rapid secondary immune response, producing antibodies quickly enough to destroy the pathogen before the person becomes ill. (b) The concentration of antibodies (and sometimes memory cells) in the blood can decline over time after the original vaccination, so the strength of immunity may weaken. A booster dose re-exposes the immune system to the antigen, restimulating memory cells to produce antibodies again and renewing/boosting long-term immunity.

(a) 4 marks: 1 mark for the vaccine containing a dead/inactive pathogen or antigen; 1 mark for stimulating antibody production by white blood cells; 1 mark for memory cells being produced; 1 mark for a correctly explained rapid secondary response on future exposure. (b) 2 marks: 1 mark for correct reference to declining antibody levels/immunity over time; 1 mark for correct explanation of a booster restimulating memory cells to renew protection.

(a) Both Generation 1 parents must be Ff. Neither parent is affected, so each must carry at least one dominant allele (F); however, since they have an affected (ff) daughter, each parent must also carry a copy of the recessive allele (f). (b) The disorder must be recessive because two unaffected (phenotypically normal) parents were able to produce a child with the disorder. This is only possible if the allele is recessive and both parents are unaffected carriers (heterozygous, Ff) — if the allele were dominant, at least one parent would have had to show the disorder themselves in order to pass it on to a child.

(a) 2 marks: 1 mark for correctly identifying both parents as Ff; 1 mark for correct reasoning linking this to having an affected child despite being unaffected. (b) 4 marks: 1 mark for the observation that two unaffected parents produced an affected child; 1 mark for stating this is only possible for a recessive allele; 1 mark for a correct explanation via carrier (heterozygous) status; 1 mark for a correct contrast with what would be required if the allele were dominant.

(a) Both graphs rise and then level off (become horizontal) at the same final volume of gas, but the graph for the powdered marble chips rises much more steeply at the start and reaches its maximum volume in a much shorter time than the graph for the large lumps. (b) Powdered marble has a much greater total surface area exposed to the acid than the same mass in large lumps. A greater surface area means more particles of marble are in contact with the acid at any moment, increasing the frequency of successful collisions between acid and marble particles per unit time, and therefore increasing the rate of reaction. (c) The same mass of marble chips is used in both cases (just in a different physical form), so the same number of moles of calcium carbonate reacts; with the acid in excess, the same total amount of carbon dioxide gas is therefore produced regardless of the rate at which it forms — surface area affects only the rate of reaction, not the total quantity of product formed.

(a) 2 marks: 1 mark for correctly describing the steeper initial gradient for the powder; 1 mark for correctly stating both reach the same final volume. (b) 3 marks: 1 mark for greater surface area for the powder; 1 mark for more marble particles exposed/in contact with acid; 1 mark for the correct link to increased collision frequency and rate. (c) 2 marks: 1 mark for correct reference to the same mass/moles of reactant in both cases; 1 mark for correctly explaining that surface area affects rate, not total yield.

(a) Dynamic equilibrium is reached in a closed system when the rate of the forward reaction becomes equal to the rate of the backward reaction, so the concentrations of reactants and products remain constant over time, even though both reactions continue to occur. (b) There are 4 moles of gas on the reactant side (1 mole \(\text{N}_{2}\) + 3 moles \(\text{H}_{2}\)) but only 2 moles of gas on the product side (2 moles \(\text{N}\text{H}_{3}\)). Increasing pressure shifts the position of equilibrium towards the side with fewer gas moles — the product side — in order to reduce the pressure, in line with Le Chatelier's principle; this increases the yield of ammonia. (c) Because the forward reaction is exothermic, increasing temperature shifts the equilibrium position towards the endothermic (backward) direction, which absorbs the extra heat energy supplied; this favours the reactants and so decreases the yield of ammonia, even though it increases the rate at which equilibrium is reached.

(a) 2 marks: 1 mark for correct reference to forward and backward rates being equal; 1 mark for correct reference to concentrations remaining constant. (b) 3 marks: 1 mark for correctly comparing the moles of gas on each side; 1 mark for correctly applying Le Chatelier's principle (shift towards fewer moles to reduce pressure); 1 mark for the correct conclusion (yield of ammonia increases). (c) 2 marks: 1 mark for correctly applying Le Chatelier's principle to temperature (shift towards the endothermic/reverse direction); 1 mark for the correct conclusion that ammonia yield decreases.

(a) Cracking involves heating long-chain, saturated hydrocarbon molecules, often in the presence of a catalyst, to break the strong carbon-carbon bonds within them, producing a mixture of smaller molecules; this typically produces a shorter, still-saturated alkane and a smaller, unsaturated alkene. (b) Cracking is important because fractional distillation of crude oil produces more long-chain hydrocarbons than the market demands, while short-chain hydrocarbons (e.g. for petrol) are in higher demand than can be obtained directly; cracking converts less useful long-chain fractions into more marketable, shorter-chain products, and also produces alkenes, which are needed as a feedstock for making polymers/plastics. (c) Bromine water is added to a sample of the hydrocarbon; an alkene rapidly decolourises the bromine water (from orange to colourless), because its carbon-carbon double bond reacts with (adds) the bromine, whereas an alkane does not decolourise bromine water, as it has no double bond to react with the bromine.

(a) 3 marks: 1 mark for correctly describing the process (heat/catalyst breaking C–C bonds in a long-chain hydrocarbon); 1 mark for a shorter alkane as one product; 1 mark for an alkene as the other product. (b) 2 marks: 1 mark for correctly explaining the supply/demand mismatch for long vs short chains; 1 mark for correctly referencing alkenes as feedstock for polymers. (c) 2 marks: 1 mark for the correct test (bromine water) and observation for the alkene (decolourises); 1 mark for the correct observation for the alkane (no change/remains orange).

(a) Aluminium is more reactive than carbon, so carbon is not reactive enough to displace/reduce aluminium ions from aluminium oxide by removing the oxygen. Because a simple chemical reduction reaction with carbon will not work, aluminium must instead be extracted using electrolysis, which uses electrical energy to separate the metal from its ore. (b) Iron is less reactive than carbon, so carbon is able to displace iron from iron oxide in a reduction reaction — carbon removes the oxygen from the iron oxide (reducing it to iron metal) while carbon itself is oxidised to carbon dioxide (or carbon monoxide); this leaves molten iron, which is a cheaper and simpler method than electrolysis.

(a) 3 marks: 1 mark for correctly stating aluminium is more reactive than carbon; 1 mark for correctly explaining that carbon therefore cannot reduce/displace it; 1 mark for the correct conclusion that electrolysis is needed instead. (b) 3 marks: 1 mark for correctly stating iron is less reactive than carbon; 1 mark for correctly explaining that carbon can reduce/displace iron from its oxide; 1 mark for correct reference to this being a cheaper/simpler method than electrolysis.

(a) Breaking bonds in the reactants requires an input of energy, while forming new bonds in the products releases energy. In an exothermic reaction, the energy released when new bonds are formed is greater than the energy required to break the original bonds, so there is a net release of energy overall, usually transferred to the surroundings as heat. (b) On the reaction profile, the reactants are shown at a higher energy level than the products, showing that overall energy has been released by the reaction. Between them there is a 'hump' representing the activation energy — the minimum energy needed for the reaction to begin — with the peak of the hump above the reactants' energy level. The vertical difference between the reactants' level and the products' (lower) level represents the overall energy released by the reaction.

(a) 3 marks: 1 mark for correct reference to energy needed to break bonds; 1 mark for correct reference to energy released forming bonds; 1 mark for the correct explanation that more energy is released than absorbed, giving a net release. (b) 3 marks: 1 mark for correctly positioning products lower than reactants; 1 mark for a correctly shown/labelled activation energy hump above the reactants; 1 mark for correctly identifying the overall energy change as the difference between reactant and product levels.

(a) Because L1 and L2 are identical and connected in parallel, the total current from the battery splits equally between the two branches, so the current through L1 equals the current through L2. (b) L2 continues to work/stay lit, because L1 and L2 are on separate parallel branches; if L1's branch breaks, current can still flow around the separate, complete loop containing L2 and the battery, so L2 is unaffected by the fault in L1's branch. (c) In a parallel circuit, each branch is connected directly across the same two points (the same battery terminals), so the same amount of energy is transferred to a unit of charge passing through either branch between those two points. Since potential difference is defined as the energy transferred per unit charge between two points, components connected in parallel across the same two points must always have the same potential difference across them.

(a) 2 marks: 1 mark for correctly describing current splitting between parallel branches; 1 mark for the correct conclusion that the currents through identical lamps are equal. (b) 2 marks: 1 mark for correctly concluding L2 keeps working; 1 mark for the correct explanation (separate, complete parallel loop for L2). (c) 3 marks: 1 mark for correct reference to both branches connecting across the same two points; 1 mark for the correct definition/application of potential difference as energy transferred per unit charge; 1 mark for the correct conclusion that this must be equal across parallel branches.

(a) In series (Circuit A), the total resistance is the sum of both lamps' resistances, higher than for a single lamp, so a smaller current flows for the same battery voltage; additionally, the battery's total voltage is shared between the two lamps in series rather than each lamp receiving the full voltage as it would in parallel. Both effects mean each series lamp receives less power than each parallel lamp (which is connected directly across the full battery voltage), so the series lamps are dimmer. (b) Wiring appliances in parallel means each appliance is connected directly across the full supply voltage, so each operates at its normal, rated voltage and can be switched on or off independently without affecting the others. In series, all appliances would share the supply voltage (so none would receive the full rated voltage) and would depend on each other, meaning if one failed or was switched off, the whole circuit would break and every appliance would stop working. (c) A fuse contains a thin wire that melts and breaks the circuit if the current becomes too large (exceeds the fuse's rating), protecting the appliance and its wiring from overheating and reducing the risk of an electrical fire.

(a) 3 marks: 1 mark for correctly identifying higher total resistance/lower current in series; 1 mark for correctly identifying the voltage/energy being shared between series components; 1 mark for the correct conclusion linking this to dimmer lamps compared with parallel. (b) 3 marks: 1 mark for each appliance receiving the full supply voltage independently in parallel; 1 mark for appliances being able to operate/be switched independently; 1 mark for a correct contrast with the problems of a series arrangement. (c) 1 mark: a valid advantage correctly stated (e.g. protection against excessive current/overheating/reduced fire risk).

(a) A coil of insulated wire is wound around a soft iron core and connected to a battery/power supply. When current flows through the coil, a magnetic field is created around it, which is concentrated and strengthened by the iron core, magnetising it. When the current is switched off, the magnetic field disappears almost immediately, since the soft iron core loses its induced magnetism once no current is flowing to sustain it. (b) In a relay, a small current in the low-current (input) circuit flows through the relay's electromagnet coil; this creates a magnetic field that attracts an armature/switch contact, closing (or opening) a separate switch in the high-current (output) circuit. This allows a small control current to switch a much larger current on or off in a separate circuit, without the two circuits being electrically connected to each other. (c) A valid advantage: it allows a low-power control circuit (e.g. from a sensor or a low-voltage switch) to safely switch a high-power/high-current appliance without needing thick, high-current wiring throughout the control circuit, improving safety and allowing sensitive control components to be used.

(a) 3 marks: 1 mark for correctly describing a coil wound around an iron core connected to a current source; 1 mark for correctly explaining that current creates/strengthens the magnetic field via the core; 1 mark for correctly stating the field disappears when current is switched off. (b) 3 marks: 1 mark for the small current energising the electromagnet coil; 1 mark for the resulting magnetic field operating a mechanical switch; 1 mark for the correct conclusion that this allows a small current to control a separate, larger current circuit. (c) 1 mark: any valid, correctly explained advantage.

(a) A transverse wave is one in which the oscillations of the medium/field are perpendicular (at right angles) to the direction the wave travels, for example light waves or water waves. A longitudinal wave is one in which the oscillations are parallel to (along) the direction of travel, for example sound waves. (b) Sound waves are mechanical waves that require particles of a medium (solid, liquid or gas) to vibrate and pass the disturbance on from particle to particle; a vacuum contains no particles, so sound cannot be transmitted through it. Light waves are electromagnetic waves, which do not require a medium to travel — they can travel through the vacuum of space, as demonstrated by sunlight reaching Earth.

(a) 3 marks: 1 mark for a correct definition of a transverse wave with a valid example; 1 mark for a correct definition of a longitudinal wave with a valid example; 1 mark for correctly using the terms 'perpendicular'/'parallel' to distinguish the direction of oscillation from the direction of travel. (b) 3 marks: 1 mark for correctly explaining that sound requires particles of a medium to vibrate/transmit the wave; 1 mark for correctly explaining that a vacuum has no particles, so cannot transmit sound; 1 mark for correctly explaining that light (an electromagnetic wave) does not require a medium and so can cross a vacuum.

(a) A star forms from a nebula (a cloud of dust and gas) that contracts under the pull of gravity, heating up until nuclear fusion of hydrogen begins in its core, forming a stable main sequence star like the Sun. After billions of years, once the hydrogen fuel begins to run low, the star expands to become a red giant. Eventually the outer layers of the star drift away into space as a planetary nebula, leaving behind a small, dense white dwarf, which then cools very slowly over an extremely long period of time. (b) During its main sequence lifetime, a star exists in equilibrium between two opposing effects: the inward pull of gravity, which continually tends to compress the star, and the outward force/pressure produced by the energy released from nuclear fusion reactions occurring in its core (radiation and thermal pressure), which tends to make the star expand. Because these two effects are balanced, the star's overall size remains stable for as long as fusion continues steadily in its core.

(a) 3 marks: 1 mark for correct formation from a nebula/gravitational contraction into a main sequence star; 1 mark for correct reference to expansion into a red giant once hydrogen fuel runs low; 1 mark for correct reference to a white dwarf (via a planetary nebula) as the final stage. (b) 3 marks: 1 mark for correct reference to the inward gravitational force; 1 mark for correct reference to the outward force/pressure from fusion/radiation; 1 mark for a correct explanation that a stable size results from these two effects being balanced.