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(a) An alternating current flows through the primary coil. This produces a continuously changing magnetic field in the soft iron core. This changing magnetic field passes through and links with the secondary coil. The changing magnetic field induces an alternating electromotive force (e.m.f.) or voltage across the secondary coil.

(b)(i) Use the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(V_s = V_p \times \frac{N_s}{N_p}\)
\(V_s = 240 \times \frac{5000}{200} = 240 \times 25 = 6000\text{ V}\)

(b)(ii) Since the output voltage is higher than the input voltage (or the number of secondary turns is greater than the primary turns), this is a step-up transformer.

(c) For a given electrical power transmitted, raising the transmission voltage decreases the current flowing through the transmission lines (since \(P = IV\)). Since power lost as heat in the cables is given by \(P_{\text{loss}} = I^2R\), a lower current significantly reduces the energy wasted as thermal energy, making the transmission much more efficient and allowing thinner, cheaper cables to be used.

(a) Max [4]:
- Alternating current in the primary coil. [1]
- Creates a changing / alternating magnetic field. [1]
- Magnetic field is linked/transmitted through the iron core. [1]
- Induces an alternating e.m.f. / voltage in the secondary coil. [1]

(b)(i) [2]:
- Correct formula or substitution: \(V_s = 240 \times \frac{5000}{200}\) [1]
- Correct calculation: \(6000\text{ V}\) (must include unit) [1]

(b)(ii) [1]:
- Step-up [1]

(c) Max [3]:
- High voltage means low current (for the same power). [1]
- Less thermal energy / heat is lost in the cables. [1]
- Formula linking power loss to current, e.g., \(P = I^2R\) / allows thinner cables. [1]

(a) The three other raw materials added to the blast furnace are:
1. Coke (carbon)
2. Limestone (calcium carbonate)
3. Air (which provides oxygen)

(b)(i) Coke (carbon) reacts with carbon dioxide to form carbon monoxide:
\(\text{C(s)} + \text{CO}_2\text{(g)} \rightarrow 2\text{CO(g)}\)

(b)(ii) Carbon monoxide reduces hematite (iron(III) oxide) to molten iron:
\(\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(l)} + 3\text{CO}_2\text{(g)}\)

(c) The silicon(IV) oxide (silica) is acidic. Limestone (calcium carbonate) undergoes thermal decomposition to produce calcium oxide (a basic oxide) and carbon dioxide:
\(\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}\)
This basic calcium oxide reacts with the acidic silica impurity in a neutralization reaction to form molten slag (calcium silicate):
\(\text{CaO(s)} + \text{SiO}_2\text{(s)} \rightarrow \text{CaSiO}_3\text{(l)}\)
The molten slag is less dense than liquid iron, so it floats on top of the molten iron at the bottom of the furnace and is tapped off separately.

(a) [3]:
- Coke [1]
- Limestone / calcium carbonate [1]
- Air / oxygen [1]

(b)(i) [2]:
- \(\text{C} + \text{CO}_2 \rightarrow 2\text{CO}\) [1] for correct formulas and balancing
- Correct state symbols: \(\text{C(s)}\), \(\text{CO}_2\text{(g)}\), \(\text{CO(g)}\) [1]

(b)(ii) [2]:
- \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) [1] for correct formulas and balancing
- Correct state symbols: \(\text{Fe}_2\text{O}_3\text{(s)}\), \(\text{CO(g)}\), \(\text{Fe(l)}\) or \(\text{Fe(s)}\), \(\text{CO}_2\text{(g)}\) [1]

(c) Max [3]:
- Thermal decomposition of limestone described or represented by equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]
- Basic calcium oxide reacts with acidic silica to form slag / calcium silicate described or represented by equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]
- Slag is less dense and melts, floating on top of the molten iron where it is tapped off / separated. [1]

(a) The balanced chemical equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light and chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) To test a leaf for starch:
1. Place the leaf in boiling water for about 30 seconds to 1 minute. This kills the leaf, breaking down cell membranes and stopping all chemical metabolic reactions.
2. Boil the leaf in ethanol. This must be done using an electric water bath because ethanol is highly flammable and must not be heated with a direct Bunsen flame. This step dissolves and removes the green chlorophyll, decolorizing the leaf so colour changes can be easily seen.
3. Rinse the leaf in warm water to wash off the ethanol and soften the leaf, making it pliable.
4. Spread the leaf on a white tile and add a few drops of iodine solution. If starch is present, the iodine changes from orange-brown to blue-black.

(c)(i) Magnesium ions are a central component of the chlorophyll molecule, which absorbs light energy for photosynthesis. A deficiency in magnesium ions results in chlorosis, where the leaves turn yellow (particularly between veins).

(c)(ii) Nitrate ions are needed to combine with glucose to produce amino acids, which are then polymerised to form proteins necessary for growth and repair.

(a) [2]:
- Correct reactants and products: \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\) [1]
- Correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) [1]

(b) Max [5]:
- Boil leaf in water to kill cells / disrupt membranes / stop reactions [1]
- Heat/boil in ethanol to extract/remove chlorophyll [1]
- Use of water bath / no naked flame explained (safety because ethanol is flammable) [1]
- Rinse in water to soften leaf [1]
- Add iodine solution, noting that blue-black indicates starch (orange-brown if negative) [1]

(c)(i) [2]:
- Required for making chlorophyll [1]
- Deficiency causes yellowing of leaves / chlorosis [1]

(c)(ii) [1]:
- To make amino acids / proteins (for growth) [1]

(a) A double circulation means that blood passes through the heart twice for each complete circuit of the body. This is divided into the pulmonary circulation (heart to lungs and back) and the systemic circulation (heart to body tissues and back). The main advantage is that blood pressure can be kept high in the systemic circuit after passing through the delicate lungs, allowing oxygenated blood to be delivered to body tissues much faster and more efficiently than in a single circulatory system.

(b) Structure:
- The left ventricle has a much thicker, more muscular wall than the right ventricle.
Role:
- The left ventricle pumps oxygenated blood to the rest of the body through the aorta (high-pressure systemic circulation).
- The right ventricle pumps deoxygenated blood only to the lungs through the pulmonary artery (low-pressure pulmonary circulation, as the lungs are nearby and delicate). The left ventricle needs a thicker muscle wall to generate the higher force/pressure required to push blood through the much longer systemic circuit.

(c) The function of the valves (such as the atrioventricular and semilunar valves) is to ensure that blood flows in one direction only, preventing the backflow of blood when the heart contracts and relaxes.

(d) Structural differences:
1. Wall thickness: Arteries have thick, muscular, and elastic walls to withstand and maintain the high pressure of blood leaving the heart, whereas veins have thin walls with less muscle/elastic tissue because blood pressure in veins is very low.
2. Lumen size: Arteries have a relatively narrow lumen to help maintain blood pressure, while veins have a wide lumen to reduce resistance to blood flow as it returns to the heart.
3. Valves: Veins contain valves to keep blood flowing in the correct direction against gravity at low pressure, while arteries do not contain valves (except at the base of the aorta/pulmonary artery).

(a) [3]:
- Blood passes through the heart twice for each complete circuit. [1]
- Mentions pulmonary (to lungs) and systemic (to body) circuits. [1]
- Advantage: keeps blood pressure high / increases speed of blood flow / faster delivery of oxygen or nutrients. [1]

(b) [3]:
- Left ventricle has thicker muscular walls than the right ventricle. [1]
- Left ventricle pumps blood to the whole body / systemic circuit (high resistance/pressure needed). [1]
- Right ventricle pumps blood to the lungs / pulmonary circuit (low resistance/delicate organ nearby). [1]

(c) [1]:
- Prevents the backflow of blood / ensures one-way flow. [1]

(d) Max [3]:
- Artery has thick wall of muscle/elastic fibers to withstand/maintain high pressure, vein has thin wall. [1]
- Artery has narrow lumen, vein has wide lumen to minimize resistance to flow. [1]
- Veins have valves to prevent backflow (of low-pressure blood), arteries do not (except at exit from heart). [1]
(Award 1 mark for each clear structural difference, and 1 mark for linking to the function, up to a maximum of 3 marks).

(a) The balanced equation for the thermal decomposition of calcium carbonate is:
\(\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}\)

(b)(i) \(M_r(\text{CaCO}_3) = A_r(\text{Ca}) + A_r(\text{C}) + 3 \times A_r(\text{O}) = 40 + 12 + (3 \times 16) = 100\)

(b)(ii) \(M_r(\text{CaO}) = A_r(\text{Ca}) + A_r(\text{O}) = 40 + 16 = 56\)

(c)(i) \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{25.0\text{ g}}{100} = 0.25\text{ mol}\)

(c)(ii) From the balanced equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CaO}\) is \(1:1\).
Therefore, \(0.25\text{ mol}\) of \(\text{CaCO}_3\) produces \(0.25\text{ mol}\) of \(\text{CaO}\).
\(\text{mass of CaO} = \text{moles} \times M_r = 0.25\text{ mol} \times 56 = 14.0\text{ g}\)

(c)(iii) From the balanced equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is also \(1:1\).
So \(0.25\text{ mol}\) of \(\text{CO}_2\) gas is produced.
\(\text{volume of gas} = \text{moles} \times 24.0\text{ dm}^3/\text{mol}\)
\(\text{volume} = 0.25 \times 24.0 = 6.0\text{ dm}^3\) (or \(6\text{ dm}^3\))

(a) [1]:
- \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]

(b)(i) [1]:
- \(100\) [1]

(b)(ii) [1]:
- \(56\) [1]

(c)(i) [2]:
- Use of formula: \(\text{moles} = \frac{\text{mass}}{M_r}\) [1]
- Correct calculation: \(0.25\text{ mol}\) [1]

(c)(ii) [2]:
- Identifies \(1:1\) mole ratio so moles of \(\text{CaO} = 0.25\) [1]
- Correct calculation of mass: \(0.25 \times 56 = 14.0\text{ g}\) [1]

(c)(iii) [3]:
- Identifies moles of \(\text{CO}_2 = 0.25\) [1]
- Multiplies moles by \(24.0\) [1]
- Correct volume: \(6.0\text{ dm}^3\) (allow correct unit of \(\text{dm}^3\) or \(\text{litres}\)) [1]

(a) In solid lead(II) bromide, the ions are held in fixed positions in a giant ionic lattice and cannot move to carry electrical charge. When molten, the lattice breaks down, and the ions (\(\text{Pb}^{2+}\) and \(\text{Br}^-\)) become free to move towards the electrodes.

(b)(i) At the negative electrode (cathode), positive lead ions (\(\text{Pb}^{2+}\)) are attracted. They gain electrons (reduction) to form lead metal. The observation is a bead or deposit of shiny grey liquid/metal at the bottom. Equation:
\(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\)

(b)(ii) At the positive electrode (anode), negative bromide ions (\(\text{Br}^-\)) are attracted. They lose electrons (oxidation) to form bromine gas. The observation is red-brown gas or fumes. Equation:
\(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\) (or \(2\text{Br}^- - 2e^- \rightarrow \text{Br}_2\))

(c)(i) In concentrated aqueous sodium chloride, both \(\text{Na}^+\) and \(\text{H}^+\) ions are attracted to the cathode. Because hydrogen is less reactive than sodium, hydrogen ions are preferentially discharged to form hydrogen gas, \(\text{H}_2\).

(c)(ii) At the anode, both \(\text{Cl}^-\) and \(\text{OH}^-\) ions are attracted. Because it is a concentrated solution, chloride ions are preferentially discharged to form chlorine gas, \(\text{Cl}_2\).

(a) [2]:
- Ions in solid are fixed / cannot move. [1]
- Ions in molten state are free to move / mobile. [1]
(Do not accept 'electrons' are free to move for ionic conduction)

(b)(i) [3]:
- Observation: shiny / grey / silvery liquid or bead of metal. [1]
- Correct species in equation: \(\text{Pb}^{2+}\), \(e^-\), and \(\text{Pb}\) [1]
- Correctly balanced equation: \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\) [1]

(b)(ii) [3]:
- Observation: red-brown / brown gas / vapour. [1]
- Correct species in equation: \(\text{Br}^-\), \(\text{Br}_2\), and \(e^-\) [1]
- Correctly balanced equation: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\) [1]

(c)(i) [1]:
- Hydrogen (gas) / \(\text{H}_2\) [1]

(c)(ii) [1]:
- Chlorine (gas) / \(\text{Cl}_2\) [1]

(a)(i) Kinetic Energy:
\(\text{K.E.} = \frac{1}{2} m v^2\)
\(\text{K.E.} = \frac{1}{2} \times 1200\text{ kg} \times (15\text{ m/s})^2\)
\(\text{K.E.} = 600 \times 225 = 135\,000\text{ J}\) (or \(135\text{ kJ}\))

(a)(ii) Work done is the change in kinetic energy:
\(\text{Work done} = 135\,000\text{ J}\)

(a)(iii) Power:
\(P = \frac{W}{t} = \frac{\text{work done}}{\text{time}}\)
\(P = \frac{135\,000\text{ J}}{6.0\text{ s}} = 22\,500\text{ W}\) (or \(22.5\text{ kW}\))

(b)(i) Forward force = 400 N. According to Newton's first law, since the car is traveling at a constant speed, its acceleration is zero, which means there is no resultant force. The forward thrust must exactly balance the total resistive force.

(b)(ii) Distance:
\(d = v \times t = 15\text{ m/s} \times 20\text{ s} = 300\text{ m}\)
Work done:
\(W = F \times d = 400\text{ N} \times 300\text{ m} = 120\,000\text{ J}\) (or \(120\text{ kJ}\))

(a)(i) [3]:
- Correct formula: \(\text{K.E.} = \frac{1}{2}mv^2\) [1]
- Substitution and calculation: \(\frac{1}{2} \times 1200 \times 15^2 = 135\,000\) [1]
- Correct unit: \(\text{J}\) or \(\text{kJ}\) [1]

(a)(ii) [1]:
- \(135\,000\text{ J}\) (or same value as calculated in (a)(i)) [1]

(a)(iii) [2]:
- Correct formula: \(P = \frac{W}{t}\) [1]
- Correct calculation: \(22\,500\text{ W}\) or \(22.5\text{ kW}\) [1]

(b)(i) [2]:
- \(400\text{ N}\) [1]
- Reason: forces must be balanced / resultant force is zero (at constant speed) [1]

(b)(ii) [2]:
- Distance = \(300\text{ m}\) [1]
- Work = \(120\,000\text{ J}\) (or \(120\text{ kJ}\)) [1]

(a)(i) When the hot plate (stimulus) is touched:
1. Temperature/pain receptors in the skin of the hand detect the heat and generate nerve impulses.
2. The nerve impulses travel along a sensory neurone to the spinal cord (part of the central nervous system).
3. In the spinal cord, the impulse passes across a microscopic gap (synapse) to a relay neurone.
4. The impulse then travels across another synapse to a motor neurone, which carries the impulse out of the spinal cord to the effector (biceps muscle in the arm).
5. The muscle contracts, pulling the hand away.

(a)(ii) Reflex actions are automatic, rapid, and bypass the conscious parts of the brain. This ensures an immediate response that minimizes damage or prevents serious injury to tissues.

(b)(i) Speed of transmission:
- Nervous system: very fast / almost instantaneous (via electrical impulses).
- Hormonal system: slower (depends on blood flow to transport hormones).

(b)(ii) Duration of response:
- Nervous system: short-lived / transient (stops as soon as impulses stop).
- Hormonal system: longer-lasting (hormones remain in the blood until broken down by the liver or excreted).

(b)(iii) Transmission medium and method:
- Nervous system: signals are transmitted as electrical impulses along specialized cells called neurones.
- Hormonal system: signals are chemical substances called hormones, which are secreted by endocrine glands directly into the bloodstream and carried dissolved in the blood plasma.

(a)(i) [4]:
- Receptors (in skin) detect stimulus and generate impulse. [1]
- Impulse travels along sensory neurone to central nervous system / spinal cord. [1]
- Passes across synapses via relay neurone. [1]
- Passes along motor neurone to the effector (muscle) causing contraction. [1]

(a)(ii) [1]:
- They are rapid / automatic and protect the body from harm / prevent injury. [1]

(b)(i) [1]:
- Nervous is very fast, whereas hormonal is slower. [1]

(b)(ii) [1]:
- Nervous response is short-lived, whereas hormonal is long-lasting. [1]

(b)(iii) [3]:
- Nervous signals are electrical impulses, whereas hormonal signals are chemical molecules. [1]
- Nervous signals travel along neurones / nerve fibers. [1]
- Hormonal signals travel through the blood / bloodstream / blood plasma. [1]

(a)(i) For parallel resistors, the combined resistance \(R_p\) is calculated using: \(1/R_p = 1/R_1 + 1/R_2 = 1/8.0 + 1/24.0 = 3/24.0 + 1/24.0 = 4/24.0\). Therefore, \(R_p = 24.0 / 4 = 6.0\ \Omega\). (a)(ii) The total resistance \(R_{\text{total}}\) of the series-parallel combination is: \(R_{\text{total}} = 4.0\ \Omega + R_p = 4.0\ \Omega + 6.0\ \Omega = 10.0\ \Omega\). (a)(iii) Using Ohm's Law, total current \(I = V / R_{\text{total}} = 12\text{ V} / 10.0\ \Omega = 1.2\text{ A}\). (a)(iv) The potential difference across the 4.0 \(\Omega\) resistor is \(V = I \times R = 1.2\text{ A} \times 4.0\ \Omega = 4.8\text{ V}\). (b) Resistance is directly proportional to length, so doubling the length doubles the resistance (\(2 \times R\)). Resistance is inversely proportional to cross-sectional area, so halving the area doubles the resistance again (\(2 \times 2 \times R = 4R\)). Thus, the overall resistance increases by a factor of 4, changing from 24.0 \(\Omega\) to 96.0 \(\Omega\).

(a)(i) [2 marks total]: 1 mark for correct formula/working (e.g., \((8 \times 24)/(8 + 24)\) or \(1/8 + 1/24\)), 1 mark for correct value 6.0 \(\Omega\). (a)(ii) [1 mark total]: 1 mark for 10.0 \(\Omega\) (or ecf from a(i) + 4.0). (a)(iii) [2 marks total]: 1 mark for correct formula/working (\(I = V/R\)), 1 mark for 1.2 with correct unit (A / Ampere). (a)(iv) [2 marks total]: 1 mark for correct formula/working (\(V = I \times R\)), 1 mark for 4.8 V (accept ecf from a(iii)). (b) [3 marks total]: 1 mark for stating that doubling the length doubles the resistance, 1 mark for stating that halving the cross-sectional area doubles the resistance, 1 mark for stating that the overall resistance quadrupled / becomes 96 \(\Omega\).

(a)(i) Coke is the raw material that provides carbon in the blast furnace. (a)(ii) The balanced equation is: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). (a)(iii) Reduction is the loss of oxygen; hematite (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become iron. Oxidation is the gain of oxygen; carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)). Since both oxidation and reduction occur simultaneously, it is a redox reaction. (b)(i) Limestone (calcium carbonate) thermally decomposes to form calcium oxide (a basic oxide) and carbon dioxide. The calcium oxide then undergoes a neutralization reaction with the acidic silicon dioxide impurity to produce calcium silicate (slag). Equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\). (b)(ii) Aluminum is higher in the reactivity series than carbon. Carbon is not reactive enough to reduce aluminum oxide (it cannot displace aluminum from its oxide).

(a)(i) [1 mark]: Coke / coal. (a)(ii) [2 marks]: 1 mark for correct reactants and products, 1 mark for correct balancing. (a)(iii) [2 marks]: 1 mark for stating hematite/iron oxide loses oxygen (is reduced), 1 mark for stating carbon monoxide gains oxygen (is oxidized). (b)(i) [3 marks total]: 1 mark for mentioning that limestone decomposes to calcium oxide (CaO), 1 mark for stating that basic CaO reacts with acidic silicon dioxide (SiO2) to form slag, 1 mark for correct balanced equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\). (b)(ii) [2 marks]: 1 mark for stating aluminum is more reactive than carbon (or higher in the reactivity series), 1 mark for stating carbon cannot reduce aluminum oxide / displace aluminum.

(a)(i) Double circulation means that blood passes through the heart twice for every complete circuit of the body. It consists of two separate circuits: the pulmonary circuit (to and from the lungs) and the systemic circuit (to and from the rest of the body). (a)(ii) An advantage is that blood can be pumped at a higher pressure to the body tissues, ensuring faster and more efficient delivery of oxygen and nutrients. (b)(i) Arteries have a thick wall containing muscle and elastic fibers to withstand and maintain high pressure. The elastic fibers stretch and recoil to smooth out blood flow. They also have a relatively narrow lumen to keep blood flowing under high pressure. (b)(ii) Two differences are: 1. Veins have valves to prevent backflow, whereas capillaries do not have valves. 2. Capillaries have walls that are only one cell thick, while veins have multiple-layered walls (tunicas). (c) During physical exercise, the heart rate increases. This is necessary to deliver oxygen and glucose to the respiring muscle cells at a faster rate, and to remove carbon dioxide (a waste product of respiration) more rapidly.

(a)(i) [2 marks]: 1 mark for blood passing through the heart twice for each complete circuit, 1 mark for mentioning pulmonary and systemic circuits. (a)(ii) [1 mark]: 1 mark for maintaining higher pressure / faster delivery of oxygen or nutrients. (b)(i) [3 marks]: 1 mark for thick outer wall (withstands high pressure), 1 mark for elastic fibers (allows stretch and recoil), 1 mark for narrow lumen (maintains high pressure). (b)(ii) [2 marks]: Any two of: Veins have valves vs capillaries do not (1 mark); Capillaries have walls one cell thick vs veins have thicker walls (1 mark); Capillaries have a microscopic lumen vs veins have a large lumen (1 mark). (c) [2 marks]: 1 mark for heart rate increasing, 1 mark for explanation (to transport more oxygen/glucose to muscles for increased respiration / to remove CO2 faster).

(a)(i) Moles of \(\text{Mg} = \text{mass} / A_r = 4.8 / 24 = 0.20\text{ mol}\). (a)(ii) According to the equation, 1 mole of \(\text{Mg}\) reacts with 2 moles of \(\text{HCl}\). Therefore, \(0.20\text{ mol} \times 2 = 0.40\text{ mol}\) of \(\text{HCl}\) is required. (a)(iii) The mole ratio of \(\text{Mg}\) to \(\text{H}_2\) is 1:1, so 0.20 mol of \(\text{H}_2\) gas is produced. Volume of \(\text{H}_2 = \text{moles} \times 24\ \text{dm}^3 = 0.20 \times 24 = 4.8\ \text{dm}^3\). (b)(i) The mole ratio of \(\text{HCl}\) to \(\text{MgCl}_2\) is 2:1. Moles of \(\text{MgCl}_2\) produced \(= 0.15 / 2 = 0.075\text{ mol}\). The relative formula mass (\(M_r\)) of \(\text{MgCl}_2 = 24 + (2 \times 35.5) = 24 + 71 = 95\). Mass of \(\text{MgCl}_2 = \text{moles} \times M_r = 0.075 \times 95 = 7.125\text{ g}\) (accept 7.13 g). (b)(ii) Percentage yield \(= (\text{actual yield} / \text{theoretical yield}) \times 100\% = (6.12 / 7.125) \times 100\% = 85.89\%\) (accept 85.9% or 86%).

(a)(i) [2 marks]: 1 mark for correct working (\(4.8 / 24\)), 1 mark for 0.20 mol. (a)(ii) [1 mark]: 1 mark for 0.40 mol (or ecf of \(2 \times \text{moles from a(i)}\)). (a)(iii) [2 marks]: 1 mark for 0.20 mol of H2, 1 mark for 4.8 with unit \(\text{dm}^3\). (b)(i) [3 marks total]: 1 mark for calculating moles of \(\text{MgCl}_2 = 0.075\text{ mol}\) (from 0.15 / 2), 1 mark for calculating \(M_r\) of \(\text{MgCl}_2 = 95\), 1 mark for final mass 7.125 g (accept 7.13 g). (b)(ii) [2 marks]: 1 mark for setting up percentage yield calculation \((6.12 / 7.125) \times 100\), 1 mark for 85.9% (accept 85.8% to 86.0% depending on rounding, allow ecf from b(i)).