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First, calculate the work done against gravity: \(\text{Work done} = m \cdot g \cdot h = 60 \times 9.8 \times 15 = 8820\text{ J}\). Next, calculate the power: \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{8820}{30} = 294\text{ W}\).

1 mark for calculating work done (\(8820\text{ J}\)) or showing correct substitution \(P = \frac{mgh}{t}\). 1 mark for correct power with units: \(294\text{ W}\) (accept \(290\text{ W}\) from 2 significant figures calculation).

The distance travelled can be found from the area under the speed-time graph.

Distance during acceleration (triangle area): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 8.0 = 16\text{ m}\).

Distance during constant speed (rectangle area): \(\text{width} \times \text{height} = 6.0 \times 8.0 = 48\text{ m}\).

Total distance = \(16 + 48 = 64\text{ m}\).

1 mark for calculating either individual distance (\(16\text{ m}\) or \(48\text{ m}\)) or showing correct method to find the total area under graph. 1 mark for correct final total distance of \(64\text{ m}\).

First, calculate the extension \(x\) of the spring: \(x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\).

Using Hooke's Law, \(F = kx\), rearrange to solve for \(k\):
\(k = \frac{F}{x} = \frac{6.0\text{ N}}{3.0\text{ cm}} = 2.0\text{ N/cm}\).

1 mark for calculating the correct extension of \(3.0\text{ cm}\). 1 mark for correct calculation of spring constant: \(2.0\text{ N/cm}\) (or \(2\text{ N/cm}\)).

Bromine water (or aqueous bromine) is orange/brown. Ethene is an unsaturated alkene, so it will undergo an addition reaction with bromine, decolourising the bromine water. Ethane is a saturated alkane and does not react with bromine water under normal conditions, so the solution remains orange.

1 mark for naming the reagent: bromine water / aqueous bromine (accept bromine). 1 mark for the correct difference in observation: decolourises with ethene AND remains orange/brown/unchanged with ethane.

Hydrocarbon \(\text{C}_3\text{H}_6\) has the general pattern \(\text{C}_n\text{H}_{2n}\), which represents the alkenes. Therefore, the homologous series is alkenes and the general formula is \(\text{C}_n\text{H}_{2n}\).

1 mark for identifying the homologous series as alkene(s). 1 mark for stating the general formula as \(\text{C}_n\text{H}_{2n}\).

Up to the optimum temperature, an increase in temperature increases the kinetic energy of enzymes and substrates, leading to more frequent successful collisions and a faster rate of photosynthesis. Beyond the optimum temperature, the high temperature alters the active site of enzymes (denaturation), which stops the reaction and decreases the rate.

1 mark for: up to the optimum temperature, rate increases because enzymes have more kinetic energy / more successful collisions. 1 mark for: beyond the optimum temperature, rate decreases because enzymes involved are denatured.

Increasing the concentration means there are more reactant particles in a given volume (per unit volume). This increases the frequency of collisions (collisions per unit time) between reactant particles, thereby increasing the rate of reaction.

1 mark for stating there are more particles per unit volume / space. 1 mark for stating there is an increased frequency of collisions / more collisions per second.

Redox reactions involve both oxidation and reduction occurring simultaneously. In this reaction, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to form iron (\(\text{Fe}\)), so it is reduced. Carbon monoxide (\(\text{CO}\)) gains oxygen to form carbon dioxide (\(\text{CO}_2\)), so it is oxidized.

1 mark for stating that iron(III) oxide / \(\text{Fe}_2\text{O}_3\) is reduced because it loses oxygen. 1 mark for stating that carbon monoxide / \(\text{CO}\) is oxidized because it gains oxygen.

1. Find the resultant force acting on the box: \(F_{\text{net}} = \text{Applied force} - \text{Friction} = 40\text{ N} - 16\text{ N} = 24\text{ N}\). 2. Use Newton's second law \(F = ma\) to calculate acceleration: \(a = \frac{F_{\text{net}}}{m} = \frac{24\text{ N}}{12\text{ kg}} = 2.0\text{ m/s}^2\).

1 mark for calculating the resultant force of \(24\text{ N}\) (or showing \(40 - 16\)). 1 mark for the correct acceleration of \(2.0\text{ m/s}^2\) (accept \(2\text{ m/s}^2\)).

1. Calculate the work done by the motor: \(W = \text{Force} \times \text{distance} = 50\text{ N} \times 4.0\text{ m} = 200\text{ J}\). 2. Calculate the power output: \(P = \frac{\text{Work done}}{\text{time}} = \frac{200\text{ J}}{5.0\text{ s}} = 40\text{ W}\).

1 mark for correct calculation of work done (\(200\text{ J}\)) or stating the correct formula (\(P = \frac{Fd}{t}\)). 1 mark for the correct power of \(40\text{ W}\) (accept \(40\text{ J/s}\)).

1. Calculate the extension (\(x\)) of the spring: \(x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\) (or \(0.030\text{ m}\)). 2. Use Hooke's Law \(F = kx\) to find the spring constant \(k\): \(k = \frac{F}{x} = \frac{6.0\text{ N}}{3.0\text{ cm}} = 2.0\text{ N/cm}\). Alternatively, using SI units: \(k = \frac{6.0\text{ N}}{0.030\text{ m}} = 200\text{ N/m}\).

1 mark for calculating the extension as \(3.0\text{ cm}\) or \(0.03\text{ m}\). 1 mark for correct value and unit of spring constant: \(2.0\text{ N/cm}\) or \(200\text{ N/m}\).

Bromine water (aqueous bromine) is orange/brown. When added to an alkene, an addition reaction occurs, and the solution becomes colourless. Alkanes do not react with bromine water in the absence of UV light, so the solution remains orange/brown.

1 mark for stating the test reagent: bromine water / aqueous bromine. 1 mark for the observation with alkene: turns from orange / yellow / brown to colourless (accept 'decolourises', reject 'becomes clear').

1. The cracking equation is: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\). By subtracting the atoms of octane from decane, the molecular formula of the other product is found to be \(\text{C}_2\text{H}_4\). 2. Since \(\text{C}_2\text{H}_4\) has the general formula \(\text{C}_n\text{H}_{2n}\), it belongs to the alkene family.

1 mark for the correct molecular formula: \(\text{C}_2\text{H}_4\). 1 mark for identifying the homologous series: alkene (or alkenes).

Under high light intensity, the rate of photosynthesis is no longer limited by light. Instead, the concentration of carbon dioxide is the limiting factor. Therefore, increasing the carbon dioxide concentration will increase the rate of photosynthesis.

1 mark for stating that the rate of photosynthesis increases / goes up. 1 mark for explaining that carbon dioxide was the limiting factor.

1. At higher temperatures, particles have more kinetic energy and move faster, leading to more frequent collisions (more collisions per unit time). 2. A greater proportion of particles have energy equal to or greater than the activation energy, meaning a higher percentage of collisions are successful.

1 mark for explaining that particles have more kinetic energy / move faster, leading to more frequent collisions (or more collisions per unit time). 1 mark for explaining that a greater proportion of collisions are successful / a greater proportion of particles have energy equal to or greater than the activation energy.

Reduction is defined as the loss of oxygen. In this reaction, copper(II) oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)), so it is the substance that is reduced.

1 mark for identifying copper(II) oxide / \(\text{CuO}\) as the substance reduced (reject 'copper' or 'Cu'). 1 mark for explaining that it loses oxygen during the reaction.

To find the speed of the toy car, we use the formula for kinetic energy:
\(E_k = \frac{1}{2}mv^2\)

Rearranging the formula to solve for speed (\(v\)):
\(v^2 = \frac{2E_k}{m}\)
\(v^2 = \frac{2 \times 16\text{ J}}{0.50\text{ kg}} = 64\text{ m}^2/\text{s}^2\)
\(v = \sqrt{64} = 8.0\text{ m/s}\)

Award 1 mark for correct substitution or rearrangement of the formula: \(v = \sqrt{\frac{2 \times 16}{0.50}}\)
Award 1 mark for correct value with unit: \(8.0\text{ m/s}\) (accept \(8\text{ m/s}\))

During a cracking reaction, the total number of carbon and hydrogen atoms remains constant.

Total carbon atoms initially = 10
Carbon atoms in ethene (\(\text{C}_2\text{H}_4\)) and propene (\(\text{C}_3\text{H}_6\)) = \(2 + 3 = 5\)
Carbon atoms in alkane \(X\) = \(10 - 5 = 5\)

Total hydrogen atoms initially = 22
Hydrogen atoms in ethene and propene = \(4 + 6 = 10\)
Hydrogen atoms in alkane \(X\) = \(22 - 10 = 12\)

Therefore, the molecular formula of alkane \(X\) is \(\text{C}_5\text{H}_{12}\).

Award 1 mark for showing working, such as calculating the number of carbon atoms (5) or hydrogen atoms (12), or writing the unbalanced or partially balanced equation.
Award 1 mark for the correct molecular formula: \(\text{C}_5\text{H}_{12}\) (reject 'pentane' unless formula is also provided).

1. At the start of the jump, the only major force is weight acting downwards, causing the skydiver to accelerate downwards. 2. As speed increases, the upward air resistance acting on the skydiver increases. 3. Eventually, the air resistance becomes equal and opposite to the weight. The resultant force becomes zero, so the acceleration is zero, and the skydiver travels at a constant speed known as terminal velocity.

1 mark: State that weight causes downward acceleration or is initially greater than air resistance. 1 mark: State that air resistance increases as speed increases. 1 mark: State that terminal velocity is reached when air resistance equals weight, resulting in zero resultant force or zero acceleration.

1. The limit of proportionality is the point beyond which the extension of a material is no longer directly proportional to the applied load. 2. Before this limit, the relationship between load and extension is linear, meaning double the force results in double the extension. 3. Once this limit is exceeded, the relationship becomes non-linear, with the extension increasing much more rapidly for each unit of additional load, and the wire may suffer permanent deformation.

1 mark: Definition of limit of proportionality as the point where extension is no longer proportional to load. 1 mark: Describing the linear or proportional relationship before this limit. 1 mark: Describing the non-linear relationship after this limit (extension increases disproportionately or plastic deformation occurs).

1. The monomer molecules must be unsaturated, meaning they contain a carbon-to-carbon double bond (\(C=C\)). 2. During the reaction, these double bonds open up, allowing adjacent monomer molecules to join together end-to-end to form a long, continuous chain (the polymer). 3. It is classified as an addition polymerisation because the monomer units add together to form a single product, with no other chemical species (such as water) being produced or lost.

1 mark: State that monomers are unsaturated or contain \(C=C\) double bonds. 1 mark: Describe the opening up of double bonds to link monomers into a long chain. 1 mark: Explain that addition polymerisation produces only one product or no other molecules are lost.

1. Add aqueous bromine (bromine water) to each gas sample. 2. With ethane, there is no reaction, so the solution remains orange, yellow, or brown. 3. With ethene, an addition reaction occurs across the double bond, which decolourises the bromine water, turning it from orange/yellow to colourless.

1 mark: Identify the reagent as aqueous bromine or bromine water. 1 mark: State that ethane remains orange/yellow/brown (no change). 1 mark: State that ethene turns the bromine water colourless or decolourises it.

1. As the temperature rises from \(10\ ^\circ\text{C}\) towards the optimum temperature (typically around \(30\ ^\circ\text{C}\)), the kinetic energy of the enzymes and substrate molecules increases. This leads to more frequent, successful collisions and an increased rate of photosynthesis. 2. At the optimum temperature, the rate of photosynthesis reaches its maximum. 3. As the temperature rises further towards \(40\ ^\circ\text{C}\), the high thermal energy causes the enzymes to denature. The active sites change shape, so they can no longer bind to substrates, causing the rate of photosynthesis to decrease rapidly.

1 mark: State that increasing temperature increases kinetic energy, leading to more frequent successful collisions (increasing rate). 1 mark: Mention optimum temperature where the rate of photosynthesis is at its maximum. 1 mark: Explain that at higher temperatures (approaching \(40\ ^\circ\text{C}\)), enzymes denature (active site changes shape), rapidly decreasing the rate.

1. The cells in the spongy mesophyll are loosely packed with large intercellular air spaces between them. This creates a large internal surface area and allows gases (carbon dioxide and oxygen) to diffuse rapidly throughout the leaf tissue. 2. The cell walls are thin, which shortens the diffusion distance for gases entering and leaving the cells. 3. The cells are coated with a thin film of moisture, which allows carbon dioxide to dissolve before diffusing across the cell membrane into the cytoplasm.

1 mark: Mention the presence of large air spaces or loosely packed cells for rapid gas diffusion. 1 mark: Mention the thin cell walls for a short diffusion distance. 1 mark: Mention the thin film of moisture on cell surfaces which dissolves gases to aid diffusion.

1. Increasing the concentration of a solution increases the number of reactant particles in a given volume (particles are closer together). 2. This increases the frequency of collisions between reactant particles (they collide more often / more collisions per unit time). 3. Consequently, there is an increased rate of successful collisions (collisions with energy greater than or equal to the activation energy), which increases the overall rate of reaction.

1 mark: State that there are more particles per unit volume. 1 mark: State that this increases the collision frequency or number of collisions per unit time (do not accept just 'more collisions' without a time reference). 1 mark: State that this leads to a higher rate of successful collisions.

1. Oxidation is the gain of oxygen. In this reaction, carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)), so carbon monoxide is oxidised. 2. Reduction is the loss of oxygen. In this reaction, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become iron (\(\text{Fe}\)), so iron(III) oxide is reduced. 3. Since both oxidation and reduction processes take place simultaneously, the overall reaction is a redox reaction.

1 mark: Identify that carbon monoxide is oxidised because it gains oxygen. 1 mark: Identify that iron(III) oxide is reduced because it loses oxygen. 1 mark: State that redox is a reaction where both oxidation and reduction occur together.

Palisade mesophyll cells are adapted for photosynthesis in three key ways: 1. Position: They are located directly beneath the upper epidermis, exposing them to the maximum amount of sunlight entering the leaf. 2. Cellular structure: They contain a high concentration of chloroplasts, which are the sites of photosynthesis, ensuring high light absorption. 3. Arrangement: They are vertically elongated (columnar) and packed closely together, which maximizes the number of cells that can absorb the incoming light.

Award 1 mark for each of the following points, up to a maximum of 3 marks:
- Mentioning that they are positioned near the top / upper surface of the leaf (to receive maximum light) [1]
- Mentioning that they contain a high density / large number of chloroplasts [1]
- Mentioning that they are closely packed together / arranged vertically / column-shaped (to maximize light absorption) [1]

1. Purpose of cracking: Fractional distillation produces an excess of larger, less useful hydrocarbon molecules (high boiling point, low flammability) compared to the market demand. Cracking is used to break these long-chain alkanes into smaller, more valuable molecules.
2. Product 1 (Alkanes): It produces shorter-chain alkanes, which make excellent fuels (like petrol) because they flow and ignite more easily.
3. Product 2 (Alkenes): It produces alkenes (unsaturated hydrocarbons), which contain a carbon-carbon double bond and are highly useful chemical feedstocks, specifically used to manufacture plastics and polymers.

Award 1 mark for each of the following points, up to a maximum of 3 marks:
- Explaining that cracking breaks down larger, less useful hydrocarbons into smaller, more useful molecules (or matches supply with demand) [1]
- Identifying that shorter-chain alkanes are produced (which are useful as fuels) [1]
- Identifying that alkenes are produced (which are used to make plastics / polymers / chemical feedstocks) [1]

Use the formula for kinetic energy: \(E_k = \frac{1}{2}mv^2\). Substituting the given values: \(E_k = 0.5 \times 0.5\text{ kg} \times (4.0\text{ m/s})^2 = 0.25 \times 16 = 4.0\text{ J}\).

1 mark for the correct formula or substitution: \(0.5 \times 0.5 \times 4.0^2\). 1 mark for the correct answer with appropriate units (4.0 J or 4 J).

First, calculate the relative formula masses: \(M_r(\text{CH}_4) = 12 + (4 \times 1) = 16\), and \(M_r(\text{CO}_2) = 12 + (2 \times 16) = 44\). According to the balanced equation, \(16\text{ g}\) of \(\text{CH}_4\) produces \(44\text{ g}\) of \(\text{CO}_2\). Therefore, \(8.0\text{ g}\) of \(\text{CH}_4\) produces: \\frac{8.0}{16} \\times 44 = 22\\text{ g}\) of \(\text{CO}_2\).

1 mark for calculating the relative molecular masses correctly (16 and 44) or identifying the molar ratio (0.5 mol of methane produces 0.5 mol of carbon dioxide). 1 mark for the correct final mass of 22 g.

Use the formula for work done: \(W = F \times d\). Substitute the given values: \(W = 15\text{ N} \times 4.0\text{ m} = 60\text{ J}\).

1 mark for the correct formula or substitution: \(15 \times 4.0\). 1 mark for the correct answer with unit (60 J, accept 60).

Use Ohm's Law to calculate the resistance: \(R = \frac{V}{I}\). Substitute the values: \(R = \frac{6.0\text{ V}}{0.15\text{ A}} = 40\\ \Omega\).

1 mark for the correct formula or substitution: \(6.0 / 0.15\). 1 mark for the correct calculated resistance with appropriate unit (40 Γ̱ or 40 ohms).

Use the wave equation: \(v = f \lambda\). Substitute the given values: \(v = 15\text{ Hz} \times 0.20\text{ m} = 3.0\text{ m/s}\).

1 mark for the correct formula or substitution: \(15 \times 0.20\). 1 mark for the correct answer with unit (3.0 m/s or 3 m/s).

Use the formula for average rate of reaction: \\text{Rate} = \\frac{\\text{Volume of gas produced}}{\\text{Time taken}}\). Substituting the values: \\text{Rate} = \\frac{36\\text{ cm}^3}{45\\text{ s}} = 0.8\\text{ cm}^3/\\text{s}\).

1 mark for correct calculation formula or substitution: \(36 / 45\). 1 mark for correct final answer with unit (0.8 cm^3/s, accept 0.8).