2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

(a)(i) Using the parallel resistance formula:
\(R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega\).

(a)(ii) The total resistance is the sum of the parallel combination and the series resistor:
\(R_{\text{total}} = R_p + R_3 = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).

(a)(iii) Using Ohm's law:
\(I = \frac{V}{R} = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).

(b)(i) Using the power formula:
\(P = V I\)
\(I = \frac{P}{V} = \frac{1200\text{ W}}{230\text{ V}} = 5.217\text{ A} \approx 5.2\text{ A}\) (to 2 s.f.).

(b)(ii) First, convert time to seconds:
\(t = 15 \times 60 = 900\text{ s}\).
Use the energy formula:
\(E = P \times t = 1200\text{ W} \times 900\text{ s} = 1,080,000\text{ J}\) or \(1.1 \times 10^6\text{ J}\) (or \(1.08\text{ MJ}\)).

(a)(i)
- State correct formula: \(R_p = \frac{R_1 R_2}{R_1 + R_2}\) or \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) [1]
- Correct calculation: \(4.0\ (\Omega)\) [1]

(a)(ii)
- Addition of \(4.0\ (\Omega)\) (from a(i)) and \(4.0\ (\Omega)\) to get \(8.0\ \Omega\) [1]

(a)(iii)
- State formula \(I = \frac{V}{R}\) or substitute values \(\frac{12.0}{8.0}\) [1]
- Correct answer: \(1.5\text{ A}\) (accept correct unit A or Amps) [1]

(b)(i)
- State formula \(P = VI\) or rearrangement \(I = \frac{P}{V}\) [1]
- Correct calculation to 2 s.f.: \(5.2\text{ A}\) (accept \(5.22\text{ A}\) if s.f. not followed; reject \(5.2\) without units unless unit is written elsewhere) [1]

(b)(ii)
- Convert 15 minutes to seconds: \(15 \times 60 = 900\text{ s}\) [1]
- Correct calculation of energy: \(1.08 \times 10^6\text{ J}\) or \(1.1 \times 10^6\text{ J}\) (or \(1.08\text{ MJ}\)) [1]
- Correct unit: \(\text{J}\) or \(\text{Joules}\) or \(\text{MJ}\) [1]

(a)(i) The main ore is hematite, which contains iron(III) oxide. Its chemical formula is \(\text{Fe}_2\text{O}_3\).

(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) Carbon first reacts with oxygen to form carbon dioxide:
\(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\).
This carbon dioxide then reacts further with carbon (coke) at high temperature to form carbon monoxide:
\(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\).

(b)(i) The thermal decomposition of calcium carbonate is:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).

(b)(ii) Silicon dioxide (silica) is an acidic impurity. Calcium oxide is a basic oxide. They react in an acid-base neutralization-type reaction to form liquid slag, calcium silicate:
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

(a)(i)
- Chemical name: iron(III) oxide (accept ferric oxide, reject iron oxide) [1]
- Chemical formula: \(\text{Fe}_2\text{O}_3\) [1]

(a)(ii)
- Correct reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\) [1]
- Correct balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) [1]

(a)(iii)
- Description: Carbon dioxide reacts with carbon / coke to form carbon monoxide [1]
- Equation: \(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\) [1]

(b)(i)
- Correct equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]

(b)(ii)
- Explanation: Calcium oxide is basic and silica is acidic (so they react) [1]
- Products/Term: Reacts to form calcium silicate / slag [1]
- Equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]

(a)(i) The left ventricle must pump blood throughout the entire systemic circulation (to all body organs except the lungs), which has a much higher resistance and requires high pressure. The right ventricle only pumps blood to the lungs (pulmonary circulation), which are nearby and operate at much lower pressure. Therefore, the left ventricle requires thicker muscular walls to generate this high pressure.

(a)(ii) The blood vessels are the coronary arteries. If these vessels become blocked (due to fatty deposits/plaques), the heart muscle is deprived of oxygen and glucose. The muscle cells cannot perform aerobic respiration, leading to cell death, which causes coronary heart disease (CHD) and can lead to a heart attack (myocardial infarction).

(b)(i) Adaptations include:
- Biconcave disc shape: increases the surface area-to-volume ratio, allowing faster diffusion of oxygen.
- Absence of a nucleus: allows more space inside the cell for carrying hemoglobin (and thus more oxygen).
- Presence of hemoglobin: a protein that readily binds with oxygen in high concentrations (lungs) and releases it in low concentrations (tissues).

(b)(ii) Platelets are involved in blood clotting, which helps to seal wounds, stop bleeding, and prevent the entry of pathogens into the blood.

(b)(iii) Capillaries have walls that are only one cell thick (made of endothelial cells), which minimizes the diffusion distance for nutrients, oxygen, and carbon dioxide, allowing rapid and efficient exchange.

(a)(i)
- Left ventricle pumps blood to the body / further distance AND right ventricle pumps to the lungs / shorter distance [1]
- Left ventricle must generate a higher pressure (to overcome higher resistance) [1]

(a)(ii)
- Coronary arteries [1]
- Blockage restricts / stops oxygen / glucose supply to heart muscle [1]
- Muscle cannot respire (aerobically) / dies / leading to heart attack [1]

(b)(i) Any two pairs for [2] marks (1 mark per pair of feature + adaptation):
- Biconcave disc shape AND increases surface area for diffusion [1]
- No nucleus AND provides more space for hemoglobin / oxygen [1]
- Contains hemoglobin AND binds to / transports oxygen [1]
- Small / flexible AND can squeeze through narrow capillaries [1]

(b)(ii)
- Blood clotting / scab formation / prevents bleeding / prevents pathogen entry [1]

(b)(iii)
- Wall is only one cell thick [1]
- Short diffusion distance (for rapid exchange of substances) [1]

(a)(i) The formula for parallel resistance is \(1/R_p = 1/R_1 + 1/R_2\). Substituting the values: \(1/R_p = 1/6.0 + 1/12.0 = 2/12 + 1/12 = 3/12\). Therefore, \(R_p = 12/3 = 4.0\ \Omega\). (a)(ii) Using Ohm's law, \(I = V / R\). Here, \(I_{total} = 12.0\text{ V} / 4.0\ \Omega = 3.0\text{ A}\). (b)(i) The total resistance is the parallel combination in series with \(R_3\): \(R_{total} = R_p + R_3 = 4.0 + 4.0 = 8.0\ \Omega\). (b)(ii) The total power can be calculated using \(P = V^2 / R\). Substituting the values: \(P = (12.0)^2 / 8.0 = 144 / 8.0 = 18.0\text{ W}\) (or using \(I = V/R = 12/8 = 1.5\text{ A}\) and then \(P = I^2 R = 1.5^2 \times 8.0 = 18.0\text{ W}\)). (c) Before disconnecting \(R_2\), the total current in the circuit was 1.5 A. The voltage across the parallel combination was \(V_p = I_{total} \times R_p = 1.5 \times 4.0 = 6.0\text{ V}\). The current through \(R_1\) was \(I_1 = V_p / R_1 = 6.0 / 6.0 = 1.0\text{ A}\). When \(R_2\) is disconnected, the circuit becomes a simple series circuit with \(R_1\) and \(R_3\). The total resistance is now \(R = R_1 + R_3 = 6.0 + 4.0 = 10.0\ \Omega\). The new current in the circuit (which is the current flowing through \(R_1\)) is \(I = 12.0 / 10.0 = 1.2\text{ A}\). Thus, the current through \(R_1\) increases from 1.0 A to 1.2 A.

(a)(i) 1 mark for correct formula or working: \(1/R = 1/6 + 1/12\). 1 mark for correct value and unit: \(4.0\ \Omega\). (a)(ii) 1 mark for formula \(I = V/R\) or substitution. 1 mark for correct value with unit: \(3.0\text{ A}\). (b)(i) 1 mark for adding parallel resistance to series resistance: \(4.0 + 4.0\). 1 mark for correct value with unit: \(8.0\ \Omega\). (b)(ii) 1 mark for correct formula \(P = V^2 / R\) or \(P = VI\) after finding correct current. 1 mark for correct value with unit: \(18.0\text{ W}\). (c) 1 mark for stating that the current increases (or changes from 1.0 A to 1.2 A). 1 mark for explanation involving calculation of new resistance (10 ohms) and new current (1.2 A), comparing it to previous current through R1 (1.0 A).

(a)(i) Carbon monoxide (CO) is the main reducing agent. It reduces iron(III) oxide to iron: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). (a)(ii) Inside the hot furnace, calcium carbonate decomposes thermally to form calcium oxide and carbon dioxide: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\). Calcium oxide is basic and reacts with the acidic silicon dioxide (silica) impurity to form calcium silicate (slag): \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\). (b)(i) Pure iron consists of a regular arrangement of metallic ions in layers that slide over each other easily. Carbon atoms in steel are of a different size to iron atoms, which disrupts this regular lattice arrangement. This prevents the layers of atoms from sliding past each other easily, making the alloy much harder and stronger. (b)(ii) Copper is a much better electrical conductor than steel, ensuring less energy is lost as heat. Copper is also highly ductile, making it easier to draw into thin wires without breaking.

(a)(i) 1 mark for naming carbon monoxide. 1 mark for correct reactant and product formulas in the equation. 1 mark for correct balancing of the equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). (a)(ii) 1 mark for explaining thermal decomposition of limestone or writing \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\). 1 mark for describing calcium oxide reacting with silicon dioxide (neutralisation of acidic oxide by basic oxide) or writing \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\). 1 mark for identifying the product as slag / calcium silicate. (b)(i) 1 mark for mentioning different sizes of atoms disrupting the regular structure/layers. 1 mark for explaining that this prevents layers of atoms from sliding over each other. (b)(ii) 1 mark for stating that copper is a better electrical conductor. 1 mark for stating that copper is more ductile / easily drawn into wires.

(a) Double circulation means that for every complete circuit of the body, blood passes through the heart twice. This consists of the pulmonary circulation (heart to lungs and back) and the systemic circulation (heart to rest of the body and back). (b) The left ventricle must pump blood through the systemic circulation (all around the body), which requires a very high pressure to overcome the high resistance of the systemic blood vessels. The right ventricle only pumps blood to the lungs (pulmonary circulation), which are nearby, delicate, and require much lower pressure to prevent damage. (c) Arteries have thick, muscular, and elastic walls to withstand and maintain the high pressure of blood coming from the heart, and a narrow lumen to maintain pressure. Veins have thin walls because the blood is under low pressure, a wide lumen to reduce resistance to blood flow, and possess valves to prevent the backflow of blood. (d)(i) The pulmonary artery. (d)(ii) The pulmonary vein.

(a) 1 mark for stating blood flows through the heart twice for each complete circuit of the body. 1 mark for identifying the two circuits as pulmonary and systemic. (b) 1 mark for stating left ventricle pumps blood to the whole body / systemic circuit at high pressure. 1 mark for stating right ventricle only pumps blood to the lungs / pulmonary circuit at lower pressure. (c) 1 mark for stating artery has a thicker muscular/elastic wall OR narrower lumen than a vein. 1 mark for relating artery wall/lumen to withstanding/maintaining high pressure. 1 mark for stating vein has a thinner wall OR wider lumen than an artery. 1 mark for relating vein structure to low pressure flow OR mentioning valves to prevent backflow of blood. (d)(i) 1 mark for pulmonary artery. (d)(ii) 1 mark for pulmonary vein.

(a) (i) Using Ohm's Law:
\(R = \frac{V}{I}\)
\(R = \frac{12\text{ V}}{1.5\text{ A}} = 8.0\ \Omega\)

(ii) In a series circuit, total resistance is the sum of individual resistances:
\(R_{\text{total}} = R_{\text{resistor}} + R_{\text{lamp}}\)
\(8.0\ \Omega = 4.0\ \Omega + R_{\text{lamp}}\)
\(R_{\text{lamp}} = 4.0\ \Omega\)

(iii) First, calculate the power of the lamp using \(P = I^2 R\) or \(P = V_{\text{lamp}} I\):
\(V_{\text{lamp}} = I \times R_{\text{lamp}} = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\)
\(P = 6.0\text{ V} \times 1.5\text{ A} = 9.0\text{ W}\)
Convert time to seconds:
\(t = 5.0\text{ mins} \times 60\text{ s/min} = 300\text{ s}\)
Calculate energy transferred:
\(E = P \times t = 9.0\text{ W} \times 300\text{ s} = 2700\text{ J}\) (or \(2.7\text{ kJ}\))

(b) A step-up transformer has more turns on the secondary coil than on the primary coil (\(N_s > N_p\)). When an alternating current (a.c.) flows through the primary coil, it produces a constantly changing magnetic field. This magnetic field is linked to the secondary coil by the soft iron core. The changing magnetic field cuts through the secondary coil, inducing an alternating electromotive force (voltage) across its ends by electromagnetic induction. Because there are more turns on the secondary coil, the induced voltage is greater than the input voltage.

(a) (i)
- Formula \(R = V / I\) used or implied: [1 mark]
- Correct calculation to give \(8.0\ \Omega\) (accept 8): [1 mark]

(ii)
- Subtraction of 4.0 from the answer to (a)(i) to give \(4.0\ \Omega\) (allow error carried forward from (a)(i)): [1 mark]

(iii)
- Conversion of time to seconds: \(300\text{ s}\): [1 mark]
- Use of \(E = I^2 R t\) or \(E = V I t\) (using correct voltage across lamp of \(6.0\text{ V}\)): [1 mark]
- Correct final answer: \(2700\text{ J}\) or \(2.7\text{ kJ}\) with correct unit: [1 mark]

(b)
- Mentions that there are more turns on the secondary coil than the primary coil: [1 mark]
- States that alternating current in the primary coil creates a changing/alternating magnetic field: [1 mark]
- Explains that the core transfers this magnetic field to the secondary coil: [1 mark]
- Describes that the changing magnetic field induces a voltage/e.m.f. in the secondary coil (electromagnetic induction): [1 mark]

(a) (i) Coke is the raw material that provides carbon. (Coal is also accepted).

(ii) The balanced equation is:
\(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

(iii) Redox is a reaction where both reduction and oxidation happen simultaneously. In terms of oxygen transfer:
- \(\text{Fe}_2\text{O}_3\) is reduced because it loses oxygen to become \(\text{Fe}\).
- \(\text{CO}\) is oxidized because it gains oxygen to become \(\text{CO}_2\).

(b) (i) Pure iron has a regular arrangement of metallic ions in layers. When a force is applied, these layers slide over each other easily, making pure iron soft. Steel is an alloy containing carbon atoms of different sizes. These different-sized atoms disrupt the regular arrangement of the layers. As a result, the layers cannot slide over each other as easily, making steel harder.

(ii) Magnesium is higher in the reactivity series than iron, meaning it is more reactive and oxidizes more readily. It loses electrons preferentially compared to iron. This prevents the iron from reacting with water and oxygen to form rust.

(a) (i)
- Coke / coal: [1 mark] (Reject: carbon / charcoal)

(ii)
- Correct formulas for reactants and products: [1 mark]
- Correct balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\): [1 mark]

(iii)
- Identifies that reduction is the loss of oxygen (iron(III) oxide loses oxygen) AND oxidation is the gain of oxygen (carbon monoxide gains oxygen): [2 marks]
- (Award [1 mark] if only one of these definitions/processes is correctly described in context)

(b) (i)
- Describes steel containing atoms of different sizes / carbon atoms of different sizes: [1 mark]
- Explains that this disrupts the regular arrangement / lattice of iron layers: [1 mark]
- Concludes that layers cannot slide / slip over each other easily: [1 mark]

(ii)
- States that magnesium is more reactive than iron: [1 mark]
- Explains that magnesium oxidizes / corrodes / loses electrons preferentially (instead of iron): [1 mark]

(a) (i) The balanced chemical equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
(Light and chlorophyll can be written over/under the arrow but are not required as reactants/products).

(ii)
- Carbon dioxide: Diffuses from the atmosphere into the leaf through stomata.
- Water: Absorbed from the soil into root hair cells by osmosis, then transported through xylem vessels.

(b) (i) Light intensity is varied by placing the light source at different measured distances from the beaker containing the aquatic plant (or using different power/wattage light bulbs).

(ii) To ensure a fair test, variables that must be kept constant include:
- Temperature (e.g., using a water bath/heat shield to block heat from the lamp).
- Carbon dioxide concentration (e.g., using a fixed concentration of sodium hydrogencarbonate solution).
- The same piece of aquatic plant.

(c) Palisade mesophyll cell adaptations include:
- Packed with a large number of chloroplasts containing chlorophyll to absorb maximum light energy.
- Elongated, column-shaped cells arranged vertically near the upper surface of the leaf to allow many cells to pack closely together and receive direct sunlight.
- Large central vacuole which pushes chloroplasts towards the periphery (edge) of the cell, reducing the diffusion distance for carbon dioxide.

(a) (i)
- Correct formulas for reactants and products: [1 mark]
- 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 mark]

(ii)
- Carbon dioxide enters via diffusion through stomata: [1 mark]
- Water enters via osmosis through root hair cells: [1 mark]

(b) (i)
- Move the lamp to different distances from the plant / vary the distance of the light source: [1 mark]

(ii)
- Any two from: temperature, carbon dioxide concentration / concentration of hydrogencarbonate, size/mass of plant, wavelength/color of light: [2 marks]

(c)
- Contains many chloroplasts (to capture light): [1 mark]
- Columnar / elongated shape: [1 mark]
- Positioned near the upper surface of the leaf / arranged vertically to fit more cells: [1 mark]

**Part (a)(i)**
Using the parallel resistor formula:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0}\)
\(R_p = \frac{12.0}{4} = 3.0\ \Omega\)

**Part (a)(ii)**
Since the parallel pair is in series with \(R_3\):
\(R_{\text{total}} = R_p + R_3 = 3.0 + 5.0 = 8.0\ \Omega\)

**Part (a)(iii)**
The current through \(R_3\) is the total current of the circuit, as \(R_3\) is in series with the power supply.
Using Ohm's Law:
\(I = \frac{V}{R_{\text{total}}}\)
\(I = \frac{16.0\text{ V}}{8.0\ \Omega} = 2.0\text{ A}\)

**Part (b)**
The potential difference across the parallel combination (\(V_p\)) is:
\(V_p = I \times R_p = 2.0\text{ A} \times 3.0\ \Omega = 6.0\text{ V}\)
(Alternatively, using the potential divider rule: \(V_p = 16.0 \times \frac{3.0}{8.0} = 6.0\text{ V}\))

**Part (c)**
The power dissipated by \(R_3\) (\(P_3\)) is calculated using:
\(P = I^2 \times R_3\) or \(P = V_3 \times I\)
Since \(V_3 = I \times R_3 = 2.0 \times 5.0 = 10.0\text{ V}\):
\(P = 2.0^2 \times 5.0 = 4.0 \times 5.0 = 20.0\text{ W}\) (or \(P = 10.0 \times 2.0 = 20.0\text{ W}\))
The unit of power is Watts (\(\text{W}\)).

**Part (a)**
(i)
- Use of \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) or \(R_p = \frac{R_1 \times R_2}{R_1 + R_2}\) [1]
- Correct calculation leading to \(3.0\ \Omega\) [1]

(ii)
- \(8.0\ \Omega\) [1]

(iii)
- Use of \(I = \frac{V}{R}\) [1]
- Correct current: \(2.0\text{ A}\) (accept \(2\text{ A}\)) [1]

**Part (b)**
- Use of \(V = I \times R_p\) or potential divider formula [1]
- Correct potential difference: \(6.0\text{ V}\) (accept \(6\text{ V}\)) [1]

**Part (c)**
- Use of formula \(P = I^2 R\) or \(P = V I\) or \(P = \frac{V^2}{R}\) [1]
- Correct numerical value: \(20.0\) (or \(20\)) [1]
- Correct unit: \(\text{W}\) or Watts [1]

**Part (a)**
The three materials added to the top or bottom of the blast furnace are:
1. Coke (carbon, which acts as fuel and reducing agent)
2. Limestone (calcium carbonate, which removes silica impurities)
3. Air (oxygen, which reacts with coke to heat the furnace and form carbon monoxide)

**Part (b)(i)**
The balanced chemical equation for the reduction of hematite by carbon monoxide is:
\(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

**Part (b)(ii)**
Iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) is the substance reduced. Reduction is the loss of oxygen; the iron(III) oxide loses oxygen to form iron (\(\text{Fe}\)).

**Part (c)**
- Calcium carbonate (limestone) thermally decomposes in the furnace to form calcium oxide and carbon dioxide:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
- Calcium oxide (\(\text{CaO}\)), a basic oxide, reacts with the acidic impurity silicon(IV) oxide (\(\text{SiO}_2\)) to form calcium silicate (\(\text{CaSiO}_3\)), commonly called slag:
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)
- Slag is molten at these high temperatures and, being less dense than molten iron, floats on top of the liquid iron, where it is periodically tapped off.

**Part (a)**
- Coke [1]
- Limestone / calcium carbonate [1]
- Air / oxygen [1]
(Accept in any order)

**Part (b)**
(i)
- Correct chemical formulas of all reactants and products [1]
- Correct balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) [1]

(ii)
- Iron(III) oxide / \(\text{Fe}_2\text{O}_3\) [1]
- because it loses oxygen (to form iron) [1] (Reject: iron is reduced; it must be iron(III) oxide)

**Part (c)**
- Statement that limestone decomposes to form calcium oxide / \(\text{CaO}\) [1]
- Correct equation for slag formation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]
- Slag is less dense than iron / floats on molten iron and is tapped off / separated [1]

**Part (a)(i)**
The balanced chemical equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
(Accept the word equation only if requested, but syllabus specifies balanced chemical equation. Light and chlorophyll can be written over the arrow but are not required as reactants/products.)

**Part (a)(ii)**
- Source of energy: Sunlight / light
- Pigment: Chlorophyll

**Part (b)(i)**
As the distance increases, the light intensity decreases (it follows an inverse square relationship, but 'decreases' is sufficient for the mark).

**Part (b)(ii)**
To keep water temperature constant, the student can use a glass heat shield or screen between the light source and the beaker to block thermal radiation, use an LED bulb (which releases very little heat), or place the container of pondweed in a larger, thermostatically-controlled water bath.

**Part (b)(iii)**
At very high light intensities, light is no longer the factor that limits the rate of photosynthesis. The rate has reached a maximum plateau because another factor, such as carbon dioxide concentration or temperature, is now in short supply and acts as the limiting factor.

**Part (c)**
- Magnesium ions are required by plants to manufacture chlorophyll.
- If grown in magnesium-deficient soil, the plant cannot make sufficient chlorophyll, resulting in a yellowing of the leaves (chlorosis), particularly between the veins.

**Part (a)**
(i)
- Correct formulas for 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]
- Correctly balanced equation: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) [1]

(ii)
- Light / sunlight [1]
- Chlorophyll [1]

**Part (b)**
(i)
- Decreases [1]

(ii)
- Place a transparent glass barrier / heat shield between the light and the plant OR use LED lamp OR put the plant in a water bath [1]

(iii)
- Light is no longer the limiting factor [1]
- Another factor (such as carbon dioxide concentration or temperature) is limiting / at its maximum rate [1]

**Part (c)**
- Role: making chlorophyll [1]
- Appearance: yellowing leaves / chlorosis [1]