2023 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

At high temperatures, such as \(80\ ^\circ\text{C}\), the thermal energy breaks the bonds that maintain the specific three-dimensional shape of the enzyme amylase. This causes the enzyme to denature, permanently changing the shape of its active site. As a result, the substrate (starch) can no longer fit into the active site, no enzyme-substrate complexes can form, and no starch is broken down into reducing sugars. Because no reducing sugars are present, the Benedict's test remains negative (blue).

Award 1 mark for B (identifying that the amylase has denatured, altering its active site so starch cannot bind).

To prepare a pure, dry sample of a soluble salt from an insoluble base and an acid:
1. Add excess copper(II) oxide solid to hot dilute sulfuric acid to ensure all the acid is completely reacted and neutralised.
2. Filter the mixture to remove the unreacted copper(II) oxide (residue).
3. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until a saturated solution is formed (the crystallisation point).
4. Allow the saturated solution to cool and crystallise slowly.
5. Filter the crystals, rinse them with a small amount of cold distilled water to remove impurities, and dry them with filter paper.

Award 1 mark for B (correct sequence of adding excess base, filtering, heating filtrate to a saturated solution, and cooling to crystallise).

All electromagnetic waves travel at the same high speed in a vacuum (approximately \(3 \times 10^8\text{ m/s}\)). The order of the electromagnetic spectrum in order of increasing frequency (and decreasing wavelength) is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Gamma rays have a higher frequency than ultraviolet waves, and both travel at the same speed in a vacuum, making statement C correct.

Award 1 mark for C (correctly identifying the relative frequency of gamma rays and ultraviolet waves, and noting that their vacuum speed is identical).

1. First, calculate the work done against gravity (which is equal to the increase in gravitational potential energy):
\(W = F \times d = m \times g \times h\)
\(W = 20 \text{ kg} \times 10 \text{ N/kg} \times 5.0 \text{ m} = 1000 \text{ J}\)

2. Next, calculate the power output by dividing the work done by the time taken:
\(P = \frac{W}{t} = \frac{1000 \text{ J}}{4.0 \text{ s}} = 250 \text{ W}\).

Therefore, the correct option is C.

[1 mark] Award for correct calculation of power output: 250 W.
- Correct formula for power: \(P = \frac{W}{t}\)
- Correct work done: 1000 J
- Correct answer: C

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond. They react readily with aqueous bromine, causing the orange-brown solution to decolorize (become colorless). Therefore, compound Y must be an alkene and is unsaturated.

Alkanes are saturated hydrocarbons containing only single bonds. They do not react with aqueous bromine under normal conditions, so the mixture remains orange-brown. Therefore, compound X must be an alkane and is saturated.

[1 mark] Correct choice is C.
- Award mark for identifying that decolorization indicates Y is an alkene and is unsaturated.

At 80 °C, the temperature is far above the optimum temperature for amylase. The enzyme is denatured because the active site changes shape, meaning it can no longer bind to starch.

As a result, no digestion of starch occurs:
1. Starch remains present in the test-tube, so the iodine test gives a positive result (blue-black).
2. No reducing sugars are produced, so the Benedict's test gives a negative result (remains blue).

Therefore, the correct option is A.

[1 mark] Correct option is A.
- Award mark for identifying that high temperature (80 °C) denatures amylase, keeping starch present (blue-black with iodine) and preventing sugar formation (blue with Benedict's).

The distance traveled during the constant speed phase is calculated as: \(d_1 = 12\text{ m/s} \times 8.0\text{ s} = 96\text{ m}\). The distance traveled during the deceleration phase is represented by the area of a triangle: \(d_2 = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\). The total distance is the sum of both phases: \(96\text{ m} + 24\text{ m} = 120\text{ m}\).

1 mark for the correct calculation of total distance (120 m). Award 1 mark for Option B.

The atomic number (bottom number) represents the number of protons, which is 17. The nucleon number (top number) is 37, so the number of neutrons is \(37 - 17 = 20\). Since the ion has a charge of \(1-\), it has gained 1 electron compared to its neutral atomic state, so it has \(17 + 1 = 18\) electrons.

1 mark for identifying 17 protons, 20 neutrons, and 18 electrons. Award 1 mark for Option A.

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (\(C=C\)). This reactive double bond allows them to undergo an addition reaction with aqueous bromine, which converts the orange-brown bromine into a colorless dibromoalkane. Alkanes, being saturated hydrocarbons with only single bonds, do not react with bromine water under standard conditions.

1 mark: D is the correct option.
- Reject A and B because alkanes are saturated and do not decolorize bromine water.
- Reject C because alkenes must contain at least one double bond, not only single bonds.

First, calculate the total resistance of the series circuit:

\(R_{\text{total}} = 4.0\ \Omega + 8.0\ \Omega = 12.0\ \Omega\)

Next, calculate the total circuit current using Ohm's law:

\(I = \frac{V}{R_{\text{total}}} = \frac{6.0\text{ V}}{12.0\ \Omega} = 0.50\text{ A}\)

In a series circuit, the current is the same at every point. Therefore, the current flowing through the \(4.0\ \Omega\) resistor is \(0.50\text{ A}\).

Finally, calculate the potential difference across the \(8.0\ \Omega\) resistor:

\(V = I \times R = 0.50\text{ A} \times 8.0\ \Omega = 4.0\text{ V}\)

1 mark: A is the correct option.
- B is incorrect because \(2.0\text{ V}\) is the p.d. across the \(4.0\ \Omega\) resistor, not the \(8.0\ \Omega\) resistor.
- C and D are incorrect because they use an incorrect current calculation.

Enzymes are proteins with a specific three-dimensional shape, including an active site that fits a specific substrate. High temperatures (such as \(65\ ^\circ\text{C}\)) break the bonds holding the protein structure together, causing the enzyme to denature. This permanently changes the shape of the active site, meaning the substrate can no longer fit, and the reaction cannot resume even after cooling.

1 mark: A is the correct option.
- B is incorrect because at high temperatures, kinetic energy is high, not low.
- C is incorrect because temperature changes, not pH, caused this inactivation, and denaturation is permanent.
- D is incorrect because enzymes affect the activation energy pathway, but denaturation refers specifically to the loss of active site shape.

First, calculate the weight (gravitational force) of the load: \(F = m \times g = 25\text{ kg} \times 10\text{ m/s}^2 = 250\text{ N}\). Next, calculate the work done in lifting the load: \(W = F \times d = 250\text{ N} \times 8.0\text{ m} = 2000\text{ J}\). Finally, calculate the power output: \(P = \frac{W}{t} = \frac{2000\text{ J}}{4.0\text{ s}} = 500\text{ W}\).

Award 1 mark for the correct answer (B). [1 mark]

At high temperatures, the thermal energy causes the enzyme molecule's shape to change permanently (denaturation). Consequently, the active site is no longer complementary in shape to the substrate, preventing the substrate from fitting and reacting. Option A is incorrect because enzymes are non-living molecules and cannot be 'killed'. Option B describes what happens at lower temperatures, not high temperatures. Option D is incorrect because denaturation affects the enzyme's active site, not the substrate.

Award 1 mark for the correct answer (C). [1 mark]

Metal \(W\) is the most reactive because it reacts very rapidly with cold water. Metal \(Z\) is the next most reactive because it reacts slowly with cold water. Metal \(X\) is less reactive than \(W\) and \(Z\) because it only reacts with dilute acid, not cold water. Metal \(Y\) is the least reactive as it does not react with either cold water or dilute acid. Thus, the order of decreasing reactivity is \(W \rightarrow Z \rightarrow X \rightarrow Y\).

Award 1 mark for the correct answer (A). [1 mark]

Ethene is an unsaturated alkene containing a double carbon-carbon bond (\(\text{C}=\text{C}\)). It reacts with aqueous bromine (which is orange) in an addition reaction where bromine atoms add across the double bond to form colorless 1,2-dibromoethane. Thus, the solution turns from orange to colorless.

[1 mark] for selecting option A, indicating that an addition reaction occurs and the color change is from orange to colorless.

All electromagnetic waves are transverse waves and they all travel at the same speed in a vacuum (approximately \(3.0 \times 10^8 \text{ m/s}\)).

[1 mark] for selecting option D, identifying that electromagnetic waves are transverse and share the same speed in a vacuum.

In this reaction, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become metallic iron (\(\text{Fe}\)). By definition, loss of oxygen is reduction, so iron(III) oxide is reduced. Carbon monoxide (\(\text{CO}\)) gains oxygen to form carbon dioxide (\(\text{CO}_2\)), which is oxidation.

[1 mark] for selecting option D, representing the correct identification of reduction based on the loss of oxygen.

At \( 37^\circ\text{C} \) (near human body temperature), amylase functions at or near its optimum rate. High kinetic energy results in a high frequency of successful collisions between enzyme active sites and starch molecules, while the enzyme remains in its native, non-denatured state. Starch is rapidly broken down into reducing sugars (maltose), which react with Benedict's reagent upon heating to form a red precipitate. At \( 5^\circ\text{C} \), the kinetic energy is too low, so the reaction is extremely slow. At \( 60^\circ\text{C} \) and \( 85^\circ\text{C} \), the amylase enzyme is denatured, destroying its active site so no starch is digested.

[1 mark] B is selected.
- Reject other choices as they describe either incorrect temperatures (A and C) or incorrect mechanisms of enzyme action (D).

As you descend Group VII (from chlorine to bromine to iodine):
1. The colours of the elements become darker: chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid.
2. The reactivity of the elements decreases: chlorine is more reactive than bromine, which is more reactive than iodine. Therefore, as you descend the group, their ability to displace other halides from solution (reactivity with aqueous potassium halides) decreases.

[1 mark] C is selected.
- Reject A: Reactivity with aqueous potassium halides decreases, not increases.
- Reject B and D: The elements become darker (not lighter) and their melting points increase.

Sound is a mechanical wave that requires a medium to travel. It travels much faster in solids (such as wood) than in gases (such as air) because the particles in a solid are much closer together, allowing vibrations to be transmitted more rapidly.

When a wave transitions from one medium to another, its frequency (\( f \)) remains constant because it is determined solely by the source of the sound.

Using the wave equation:
\( v = f \lambda \)

Since speed (\( v \)) increases and frequency (\( f \)) remains constant, the wavelength (\( \lambda \)) must also increase.

[1 mark] A is selected.
- Reject B, C, and D because the speed of sound must increase in the solid, and according to \( v = f \lambda \), the wavelength must also increase when frequency is constant.

Inspired air contains approximately 21% oxygen and 0.04% carbon dioxide. During gas exchange in the alveoli, oxygen diffuses into the blood and carbon dioxide diffuses out of the blood into the lungs. Therefore, expired air has a lower percentage of oxygen (about 16%) and a higher percentage of carbon dioxide (about 4%). This matches option A.

1 mark for identifying the correct row (Option A).

The atomic number is 17, so there are 17 protons. The nucleon number is 37, so there are 37 - 17 = 20 neutrons. Since it is a negative chloride ion with a charge of -1, it has gained 1 electron, making the number of electrons 17 + 1 = 18. This matches option A.

1 mark for identifying the correct number of protons (17), neutrons (20), and electrons (18).

First, find the acceleration of the car using the formula: acceleration = (final speed - initial speed) / time = (25 - 15) / 5.0 = 2.0 m/s^2. Next, calculate the resultant force using Newton second law: force = mass x acceleration = 800 kg x 2.0 m/s^2 = 1600 N. This matches option B.

1 mark for calculating the acceleration of 2.0 m/s^2 and finding the force of 1600 N.

Aqueous bromine (bromine water) is used to test for unsaturation (alkenes). Bromine water is naturally orange-brown. When added to an alkene like hexene, an addition reaction occurs, and the solution turns from orange to colorless (decolorizes). Alkanes do not react with bromine water under room conditions, so the solution remains orange.

1 mark for the correct option A.

The formula for wave speed is \(v = f \lambda\). First, convert the wavelength from centimeters to meters: \(\lambda = 12\text{ cm} = 0.12\text{ m}\). Then, calculate the wave speed: \(v = 5.0\text{ Hz} \times 0.12\text{ m} = 0.60\text{ m/s}\).

1 mark for the correct option A.

As temperature increases up to the optimum temperature, molecules gain more kinetic energy, move faster, and collide more frequently and with more energy. This increases the rate of reaction. Low temperatures do not denature enzymes (they only decrease kinetic energy), and temperatures above the optimum decrease the rate of reaction because the active site denatures (changes shape permanently).

1 mark for the correct option B.

At extreme pH values (such as pH 2.0 and pH 12.0), enzymes become denatured. This means the specific three-dimensional shape of the protein, including its active site, is permanently altered. Consequently, the substrate can no longer fit or bind to the active site, and the rate of reaction drops to zero. Kinetic energy is affected by temperature, not pH, making option B incorrect.

1 mark: A is the correct answer. Other options represent incorrect biological mechanisms for pH denaturation.

As we go down Group VII, the colors of the elements become progressively darker: chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid. Therefore, option D is correct. Option A is incorrect because chlorine is a gas at room temperature and pressure. Option B is incorrect because reactivity decreases down the group (chlorine is more reactive than bromine). Option C is incorrect because iodine is less reactive than chlorine and cannot displace it.

1 mark: D is the correct answer.

Using the wave equation \(v = f \lambda\), we can rearrange it to find the wavelength: \(\lambda = \frac{v}{f}\). Substituting the given values: \(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{1.0 \times 10^8\text{ Hz}} = 3.0\text{ m}\). Therefore, option B is correct.

1 mark: B is the correct calculation. Options A, C, and D represent incorrect rearrangements or calculation errors.

As you go down Group I (with increasing atomic number): 1. Reactivity with water increases. 2. Density generally increases. 3. Melting point decreases. 4. They form basic oxides when reacted with oxygen. Therefore, the statement that their melting point decreases is correct.

1 mark: C. Reject A: Density generally increases down the group. Reject B: Reactivity with water increases down the group. Reject D: Alkali metals form basic oxides, not acidic oxides.

Step 1: Calculate the total resistance of the series circuit: \( R = 10\ \Omega + 20\ \Omega = 30\ \Omega \). Step 2: Calculate the total current using Ohm's Law: \( I = \frac{V}{R} = \frac{6.0\text{ V}}{30\ \Omega} = 0.20\text{ A} \). Step 3: Since it is a series circuit, the current is the same at all points, so the current in the \( 10\ \Omega \) resistor is also \( 0.20\text{ A} \).

1 mark: A. Reject B: calculated as \( 6.0\text{ V} / 20\ \Omega = 0.30\text{ A} \). Reject C: calculated as \( 6.0\text{ V} / 10\ \Omega = 0.60\text{ A} \).

Enzymes are protein molecules (not carbohydrates) that act as biological catalysts. They are not used up in the reactions they catalyse. Low temperatures decrease the kinetic energy of the molecules and rate of reaction but do not denature the enzymes (denaturation occurs at high temperatures, usually above \( 45^\circ\text{C} \)). The active site of an enzyme has a complementary shape to its specific substrate molecule, which fits into it. Thus, statement C is correct.

1 mark: C. Reject A: Enzymes are proteins, not carbohydrates. Reject B: Low temperatures inactivate enzymes but do not denature them. Reject D: Catalysts are not used up in the reactions they catalyse.

At extreme pH values (far below or far above the optimum), enzymes undergo denaturation. Denaturation alters the shape of the enzyme's active site, preventing the substrate from fitting. Therefore, the reaction rate drops to zero because the enzyme can no longer bind to its substrate.

Award 1 mark for identifying that denaturation changes the shape of the active site so that the substrate cannot fit.

In Group VII, as you go down the group: 1) the elements become darker in colour (fluorine/chlorine are pale gases, bromine is a reddish-brown liquid, iodine is a dark grey solid); 2) their melting and boiling points increase; 3) their density increases; and 4) their reactivity decreases. Therefore, statement D is correct.

Award 1 mark for selecting the correct trends of increasing boiling points and darker colours down Group VII.

Radio waves are electromagnetic waves, which travel at the speed of light: \(v = 3.0 \times 10^8\text{ m/s}\). Using the wave equation: \(v = f \lambda\), we rearrange to find wavelength: \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{9.4 \times 10^7\text{ Hz}} \approx 3.191\text{ m}\), which rounds to \(3.2\text{ m}\).

Award 1 mark for recalling the speed of electromagnetic waves in air/vacuum (\(3.0 \times 10^8\text{ m/s}\)) and correctly applying the wave equation to calculate the wavelength.

To find the total distance travelled from a description of motion with constant acceleration, constant speed, and constant deceleration, we can find the area under the speed-time graph representing this motion.

1. For the first phase (acceleration from rest to \( 6\text{ m/s} \) in \( 4\text{ s} \)):
\(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\text{ s} \times 6\text{ m/s} = 12\text{ m} \)

2. For the second phase (constant speed of \( 6\text{ m/s} \) for \( 6\text{ s} \)):
\(\text{Distance}_2 = \text{width} \times \text{height} = 6\text{ s} \times 6\text{ m/s} = 36\text{ m} \)

3. For the third phase (deceleration to rest in \( 2\text{ s} \)):
\(\text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\text{ s} \times 6\text{ m/s} = 6\text{ m} \)

Total distance = \( 12\text{ m} + 36\text{ m} + 6\text{ m} = 54\text{ m} \).

Award 1 mark for the correct calculation of total distance (54 m), which corresponds to option C.

As Group VII is descended (from chlorine to iodine):
- Boiling points increase because the molecular size increases, leading to stronger intermolecular forces.
- The colour of the elements becomes darker (chlorine is a pale green gas, bromine is a red-brown liquid, iodine is a grey-black solid).
- Chemical reactivity decreases because the outer shell is further from the nucleus, making it harder to attract and gain an electron.

Award 1 mark for identifying that boiling point increases, colour becomes darker, and chemical reactivity decreases (option A).

Amylase is a protein that acts as an enzyme. At high temperatures like \( 80^{\circ}\text{C} \), the thermal energy breaks bonds maintaining the enzyme's three-dimensional shape. This changes the shape of the active site (denaturation), meaning the substrate (starch) can no longer fit into the active site to react. Starch is a carbohydrate, not a protein, and does not denature in this way.

Award 1 mark for explaining that the active site of the amylase changed shape so starch can no longer fit (option B).

The atomic number (bottom number, 15) represents the number of protons, so there are 15 protons. The nucleon number (top number, 31) represents the total number of protons and neutrons, so the number of neutrons is 31 - 15 = 16. The 3- charge indicates that the phosphorus atom has gained 3 electrons, so the number of electrons is 15 + 3 = 18. This matches option B.

Award 1 mark for selecting the correct option (B) showing 15 protons, 16 neutrons, and 18 electrons.

(a) State and use the formula:
\(\Delta E_p = m \times g \times h\)
\(\Delta E_p = 250\text{ kg} \times 9.8\text{ m/s}^2 \times 20\text{ m} = 49000\text{ J}\) (or \(49\text{ kJ}\))

(b) Assuming no energy loss, potential energy lost = kinetic energy gained:
\(E_k = \frac{1}{2} m v^2 = 49000\text{ J}\)
\(\frac{1}{2} \times 250\text{ kg} \times v^2 = 49000\)
\(125 \times v^2 = 49000\)
\(v^2 = 392\)
\(v = \sqrt{392} \approx 19.8\text{ m/s}\)
(If \(g = 10\text{ m/s}^2\) is used: \(E_p = 50000\text{ J}\), \(v^2 = 400\), \(v = 20\text{ m/s}\))

(c) In a real system, resistive forces such as friction between wheels and track, as well as air resistance, oppose the motion. Work is done against these forces, transferring some of the initial gravitational potential energy into thermal energy and sound energy, which are dissipated to the surroundings. Thus, less kinetic energy is gained. According to the law of conservation of energy, the total energy remains constant, but it is not all converted to kinetic energy.

(a) [Total: 3 marks]
- 1 mark for correct formula: \(\Delta E_p = mgh\)
- 1 mark for correct calculation: \(250 \times 9.8 \times 20 = 49000\) (allow \(50000\) if \(g = 10\text{ m/s}^2\) is specified)
- 1 mark for correct unit: \(\text{J}\) or \(\text{kJ}\)

(b) [Total: 3 marks]
- 1 mark for equating kinetic energy to potential energy: \(\frac{1}{2}mv^2 = 49000\) (or \(50000\))
- 1 mark for rearranging and solving for \(v^2\): \(v^2 = 392\) (or \(400\))
- 1 mark for correct final speed with unit: \(19.8\text{ m/s}\) (accept range \(19.7\text{--}19.8\text{ m/s}\), or \(20\text{ m/s}\) if \(g=10\) is used)

(c) [Total: 2.88 marks]
- 1 mark for identifying friction / air resistance opposing the motion.
- 1 mark for explaining that energy is transferred to thermal energy (heat) and/or sound to the surroundings.
- 0.88 marks for stating that the total energy is conserved (the sum of kinetic, thermal, and sound energy equals the initial potential energy).

(a) (i) The balanced equation representing the cracking of decane is:
\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\)
(ii) The essential conditions for catalytic cracking are:
1. High temperature (approx. \(450^\circ\text{C}\) to \(800^\circ\text{C}\)).
2. A catalyst (such as alumina / silicon dioxide / zeolite).

(b) The reagent used is aqueous bromine (bromine water), which is orange/brown.
- When added to octane (an alkane), there is no reaction in the absence of UV light, so the mixture remains orange/brown.
- When added to ethene (an alkene), an addition reaction occurs, and the bromine water is decolourised (turns from orange/brown to colourless).

(c) The type of polymerisation is addition polymerisation.
The repeat unit of poly(ethene) is drawn with two carbon atoms connected by a single covalent bond, each bonded to two hydrogen atoms, and single bonds extending through brackets on both sides to show linkage:
\(\text{-(CH}_2\text{-CH}_2\text{)-}\)

(a) [Total: 4 marks]
- (i) [2 marks]: 1 mark for correct reactants and products (\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\)); 1 mark for correct balancing.
- (ii) [2 marks]: 1 mark for specifying a high temperature (accept range \(450\text{--}800^\circ\text{C}\)); 1 mark for mentioning a catalyst (accept alumina, silica, or zeolite; reject general 'catalyst' if not specified or described as a mineral catalyst).

(b) [Total: 3 marks]
- 1 mark for correct reagent: bromine water / aqueous bromine (reject liquid bromine).
- 1 mark for octane observation: remains orange/brown/yellow.
- 1 mark for ethene observation: decolourises / turns colourless (reject 'clear').

(c) [Total: 1.88 marks]
- 1 mark for identifying 'addition polymerisation'.
- 0.88 marks for drawing the correct repeat unit (must show a single bond between carbon atoms, four C-H single bonds, and extension bonds passing through the brackets/parentheses).

(a) As temperature increases, the salivary amylase (enzyme) and starch (substrate) molecules gain kinetic energy. They move faster, which increases the frequency of collisions between the enzyme and substrate molecules. Additionally, a larger fraction of colliding molecules have sufficient energy to overcome the activation energy barrier, leading to a higher rate of successful collisions and thus a faster rate of reaction.

(b) At \(60^\circ\text{C}\), the excessive heat energy causes the weak bonds (hydrogen bonds) maintaining the specific three-dimensional shape of the enzyme's active site to vibrate and break. The active site permanently changes shape, meaning the substrate (starch) is no longer complementary and cannot fit into the active site. The enzyme is denatured, causing the rate of reaction to drop rapidly to zero.

(c) Two key variables to keep constant are:
1. pH of the reaction mixture (using a buffer solution).
2. Concentration / volume of salivary amylase (enzyme) solution.
3. Concentration / volume of the starch (substrate) solution.

(a) [Total: 3 marks]
- 1 mark for stating that molecules gain kinetic energy and move faster.
- 1 mark for stating that the frequency of collisions increases.
- 1 mark for stating that a higher proportion of collisions are successful / have energy greater than or equal to the activation energy.

(b) [Total: 3 marks]
- 1 mark for stating that the enzyme denatures / active site changes shape.
- 1 mark for explaining that the substrate (starch) can no longer fit or bind to the active site / they are no longer complementary.
- 1 mark for stating that the rate of reaction decreases rapidly / drops to zero.

(c) [Total: 2.88 marks]
- 1.44 marks for each correct variable named (max 2): pH, concentration of enzyme, volume of enzyme, concentration of starch, volume of starch. (Reject 'amount of enzyme/starch' unless qualified by volume or concentration).

(a) Acceleration is calculated using the formula:
\(a = \frac{v - u}{t}\)
Given: \(u = 0\text{ m/s}\), \(v = 3.0\text{ m/s}\), \(t = 1.5\text{ s}\).
\(a = \frac{3.0 - 0}{1.5} = 2.0\text{ m/s}^2\).

(b) Resultant force is calculated using Newton's second law:
\(F = m \times a\)
Given: \(m = 0.5\text{ kg}\), \(a = 2.0\text{ m/s}^2\).
\(F = 0.5 \times 2.0 = 1.0\text{ N}\).

(c) (i) Since the car travels at a constant speed, there is no acceleration. According to Newton's first law, the resultant force must be \(0\text{ N}\).
(ii) The total distance is the sum of the distance travelled down the slope (\(d_1\)) and the distance on the flat track (\(d_2\)).
- Distance on slope (\(d_1\)): Since it accelerates uniformly from rest to \(3.0\text{ m/s}\):
\(d_1 = \text{average speed} \times t = \frac{0 + 3.0}{2} \times 1.5 = 1.5 \times 1.5 = 2.25\text{ m}\).
- Distance on flat track (\(d_2\)):
\(d_2 = \text{speed} \times t = 3.0 \times 4.0 = 12.0\text{ m}\).
Total distance = \(d_1 + d_2 = 2.25 + 12.0 = 14.25\text{ m}\).

(d) As the car goes down the slope, its height decreases and its speed increases. Thus, gravitational potential energy is transferred to kinetic energy.

(a) [2 marks]
- Award 1 mark for correct formula \(a = \frac{\Delta v}{t}\) or correct substitution \(3.0 / 1.5\).
- Award 1 mark for correct calculation: \(2.0\) (accept \(2\)).

(b) [2 marks]
- Award 1 mark for correct calculation: \(1.0\) (accept \(1\)) [must follow from (a) or use \(F = m \times a\)].
- Award 1 mark for correct unit: \(N\) or Newtons.

(c) (i) [1 mark]
- Award 1 mark for stating \(0\) / \(0\text{ N}\) / zero.
(ii) [3 marks]
- Award 1 mark for calculating distance on slope: \(2.25\text{ m}\).
- Award 1 mark for calculating distance on flat track: \(12.0\text{ m}\).
- Award 1 mark for correct total distance: \(14.25\text{ m}\) (accept \(14.3\text{ m}\)).

(d) [1 mark]
- Award 1 mark for: gravitational potential energy to kinetic energy (accept GPE to KE).

(a) (i) The reaction is between copper(II) oxide and carbon. Heating them produces copper metal and carbon dioxide gas (or carbon monoxide gas is also acceptable):
\(\text{copper(II) oxide} + \text{carbon} \rightarrow \text{copper} + \text{carbon dioxide}\)

(ii) Redox reactions involve both oxidation (gain of oxygen) and reduction (loss of oxygen). Here, copper(II) oxide loses oxygen to form copper, so it is reduced. Carbon gains oxygen to form carbon dioxide, so it is oxidized.

(b) (i) Oxygen gas (\(\text{O}_2\)) from the air blast reacts with hot coke to form carbon dioxide (\(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\)). This carbon dioxide then reacts with more hot coke to form carbon monoxide (\(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\)).

(ii) Carbon monoxide reduces iron(III) oxide to iron, forming carbon dioxide as a product:
\(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

(c) (i) Brass is an alloy of copper and zinc.

(ii) In pure copper, all atoms are of the same size and are arranged in a regular lattice. This allows the layers of atoms to easily slide over each other when a force is applied, making it relatively soft/malleable. In brass, the introduction of zinc atoms, which are larger/different in size than copper atoms, disrupts this regular arrangement. Consequently, the layers cannot slide past each other easily, making the alloy much harder.

(a) (i) [1 mark]
- Award 1 mark for correct word equation: copper(II) oxide + carbon \(\rightarrow\) copper + carbon dioxide (accept carbon monoxide as product). Reject symbols/formulae for a word equation.
(ii) [2 marks]
- Award 1 mark for explaining copper(II) oxide is reduced because it loses oxygen.
- Award 1 mark for explaining carbon is oxidized because it gains oxygen.

(b) (i) [1 mark]
- Award 1 mark for oxygen (accept air).
(ii) [2 marks]
- Award 1 mark for correct formulae of reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\).
- Award 1 mark for correct balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).

(c) (i) [1 mark]
- Award 1 mark for zinc.
(ii) [2 marks]
- Award 1 mark for mentioning that pure copper has layers of same-sized atoms that slide over each other.
- Award 1 mark for explaining that zinc atoms are of a different size and disrupt the regular layers/structure, preventing them from sliding.

(a) An enzyme is defined as a protein that acts as a biological catalyst. It speeds up the rate of metabolic reactions without being used up or permanently altered in the process.

(b) (i) The optimum pH is the pH at which the enzyme activity is at its highest (maximum rate of reaction). According to the results, this is pH 2.0.
(ii) Enzymes are proteins whose tertiary structure and shape of the active site are maintained by chemical bonds. At a pH far from the optimum (such as pH 5.0), the changes in hydrogen ion concentration disrupt these bonds. This permanently changes the shape of the active site, meaning the enzyme is denatured. Because the active site has changed shape, the substrate can no longer fit into it, meaning no enzyme-substrate complexes can form.

(c) (i) Proteins are polymers made of amino acid monomers. Thus, digestion of proteins produces amino acids.
(ii) Under the lock and key hypothesis, the enzyme is like a lock and the substrate is like a key. The active site of the protease enzyme has a specific 3D shape that is complementary only to the shape of a protein molecule. Starch molecules have a completely different shape, so they cannot fit into the active site of the protease enzyme.

(a) [2 marks]
- Award 1 mark for stating that it is a "protein".
- Award 1 mark for stating that it is a "biological catalyst" (or speeds up chemical reactions).

(b) (i) [1 mark]
- Award 1 mark for pH 2.0 (accept 2).
(ii) [3 marks]
- Award 1 mark for stating that the enzyme/active site changes shape.
- Award 1 mark for stating that the enzyme is denatured.
- Award 1 mark for stating that the substrate can no longer fit into the active site / no enzyme-substrate complexes can form.

(c) (i) [1 mark]
- Award 1 mark for amino acids.
(ii) [2 marks]
- Award 1 mark for explaining that the active site has a specific shape complementary only to the protein substrate (or lock and key reference).
- Award 1 mark for explaining that starch has a different shape that does not fit into the protease active site.

(a) \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

(b) Limestone decomposes at high temperatures to form calcium oxide and carbon dioxide:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
Calcium oxide is a basic oxide, which reacts with acidic silicon dioxide (sandy impurities) to form calcium silicate (slag):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)
This liquid slag is tapped off at the bottom of the furnace.

(c) (i) Red-brown solid deposit forms on the zinc; blue solution fades or becomes colorless.
(ii) Zinc is more reactive than copper, so zinc displaces copper ions from the solution.

(a)
- Correct formulae of reactants and products [1]
- Fully balanced equation [1]

(b)
- Thermal decomposition of limestone to produce calcium oxide OR \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]
- CaO is a basic oxide that neutralises/reacts with acidic silicon dioxide impurities [1]
- Reaction to form slag OR \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1.88]

(c) (i)
- Red-brown solid/deposit forms on zinc [1]
- Blue color of solution fades/becomes paler/colorless [1]

(c) (ii)
- Zinc is more reactive than copper (so displaces it) [1]

(a) (i) Acceleration \(a = \frac{v - u}{t} = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\).
(ii) Kinetic energy \(E_k = \frac{1}{2} m v^2 = 0.5 \times 1200 \times 15^2 = 600 \times 225 = 135\,000\text{ J}\) (or \(135\text{ kJ}\)).

(b) (i) Forward force is \(800\text{ N}\). Because the car is moving at a constant speed, the acceleration is zero, which means the resultant force is zero. Therefore, the forward engine force must balance the resistive forces.
(ii) \(\text{Power} = \text{force} \times \text{speed} = 800\text{ N} \times 15\text{ m/s} = 12\,000\text{ W}\) (or \(12\text{ kW}\)).

(a) (i)
- Formula/working: \(15 / 6.0\) [1]
- Correct value \(2.5\) and unit \(\text{m/s}^2\) [1] (Accept \(\text{m s}^{-2}\))

(a) (ii)
- Formula \(E_k = \frac{1}{2} m v^2\) or substitution \(0.5 \times 1200 \times 15^2\) [1]
- Correct calculation to \(135\,000\) [1]
- Correct unit \(\text{J}\) or \(\text{kJ}\) (matched with \(135\)) [0.88]

(b) (i)
- Magnitude: \(800\text{ N}\) [1]
- Explanation: constant speed means forces are balanced / resultant force is zero [1]

(b) (ii)
- Formula \(P = F \times v\) or substitution \(800 \times 15\) [1]
- Correct answer \(12\,000\text{ W}\) or \(12\text{ kW}\) [1]

(a) Any three from:
1. Large surface area: provides more space over which gas molecules can diffuse at any given moment.
2. One-cell-thick walls / thin walls: creates a very short diffusion distance for the gases.
3. Good blood supply / dense capillary network: continuously transports gases away, maintaining a steep concentration gradient.
4. Well-ventilated lungs: brings in fresh air high in oxygen and low in carbon dioxide, maintaining a steep concentration gradient.
5. Moist lining: gases dissolve in the moisture, which allows them to diffuse across the cell membrane more easily.

(b) Aerobic respiration occurring in the body cells produces carbon dioxide as a waste product. This carbon dioxide is transported by the blood to the lungs, where it diffuses into the alveoli and is exhaled. Thus, expired air contains more carbon dioxide than inspired air.

(c) Goblet cells secrete sticky mucus which traps inhaled dust, pollen, and pathogens (such as bacteria and viruses). Ciliated cells have tiny hair-like structures called cilia which beat rhythmically to sweep the mucus upwards, away from the lungs towards the back of the throat where it can be swallowed or coughed out.

(a)
- First feature + correct explanation [1.88]
- Second feature + correct explanation [1]
- Third feature + correct explanation [1]
(Accept: large surface area, thin walls / one cell thick, good blood supply / capillary network, ventilation / moist surface)

(b)
- Respiration (in cells) produces carbon dioxide as a waste product [1]
- Carbon dioxide is carried to the lungs by blood and released into alveoli / exhaled [1]

(c)
- Goblet cells produce/secrete mucus [1]
- Mucus traps dust/pathogens/bacteria [1]
- Ciliated cells sweep mucus up/away from the lungs/to throat [1]

To investigate the effect of pH on catalase activity:
1. Identify variables:
- Independent variable: pH of the buffer solution (pH 4, 5, 6, 7, and 8).
- Dependent variable: Volume of oxygen gas produced in a set time (e.g., 2 minutes) or the rate of reaction calculated as \(\text{volume} / \text{time}\).
- Control variables: Volume and concentration of hydrogen peroxide (e.g., 10 cm\(^3\) of 3% solution), volume and concentration of yeast suspension (e.g., 5 cm\(^3\)), and the temperature of the reactants (controlled using a water bath at 25\(^\circ\)C).

2. Experimental method:
- Set up a side-arm flask connected by delivery tubing to a gas syringe.
- Use a syringe/measuring cylinder to add 5 cm\(^3\) of yeast suspension and 5 cm\(^3\) of pH 4 buffer solution to the side-arm flask.
- Swirl gently and leave the mixture in a water bath at 25\(^\circ\)C for 2 minutes to equilibrate.
- Add 10 cm\(^3\) of hydrogen peroxide solution to the flask, immediately place the stopper on the flask, and start the stopwatch.
- Measure and record the volume of oxygen gas collected in the gas syringe after exactly 2 minutes.
- Rinse the flask thoroughly and repeat the exact procedure for buffer solutions of pH 5, 6, 7, and 8.

3. Reliability and Data Analysis:
- Repeat the entire experiment at each pH at least three times. Calculate the mean volume of gas produced at each pH, excluding any anomalous results.
- Present the data in a table with headings: pH, Volume of oxygen collected in 2 minutes (cm\(^3\)) for Trials 1, 2, 3, and the Mean Volume (cm\(^3\)).
- Plot a line graph of the Mean Volume of oxygen (y-axis) against pH (x-axis). Draw a smooth curve of best fit. The peak of this curve represents the optimum pH for catalase activity.

4. Safety:
- Wear eye protection (safety goggles) because hydrogen peroxide is an irritant and can cause eye damage.

1 mark: Identifies the independent variable (pH 4 to 8).
1 mark: Identifies the dependent variable (volume of oxygen gas produced in a fixed time OR time taken to collect a fixed volume of gas).
2 marks: Identifies two relevant control variables (volume/concentration of yeast suspension, volume/concentration of hydrogen peroxide, or temperature of the mixture) [1 mark for each].
1 mark: Mentions suitable apparatus to collect and measure gas (e.g., a gas syringe or a measuring cylinder inverted over water).
1 mark: Describes a clear step-by-step mixing sequence, including sealing the flask quickly with a stopper after adding the hydrogen peroxide.
1 mark: Explains the need for repeating each pH at least three times and calculating a mean to ensure reliability.
1 mark: Describes how to process the data (plotting a graph of volume of gas or rate of reaction against pH).
1 mark: Explains how to identify the optimum pH from the graph (the peak of the curve/highest rate value).
1 mark: Identifies a sensible safety precaution with a valid reason (e.g., wearing goggles because hydrogen peroxide is corrosive/an irritant).

To investigate the effect of sodium thiosulfate concentration on the rate of reaction:
1. Varying the Independent Variable:
- Create at least five different concentrations of sodium thiosulfate by diluting the stock solution with distilled water. Keep the total volume of the thiosulfate-water mixture constant (e.g., 50 cm\(^3\)). For example, mix stock thiosulfate and water in ratios of 50:0, 40:10, 30:20, 20:30, and 10:40 (in cm\(^3\)).

2. Experimental Setup and Endpoint Determination:
- Draw a bold black cross on a piece of white paper and place a clean conical flask directly over it.
- Use a measuring cylinder to measure and add 50 cm\(^3\) of the prepared sodium thiosulfate solution (or thiosulfate-water mixture) to the flask.
- Measure 10 cm\(^3\) of dilute hydrochloric acid using a separate measuring cylinder.
- Pour the acid into the flask, start the stopwatch immediately, and swirl the flask once to mix.
- Look vertically downwards through the mouth of the flask at the black cross. Stop the stopwatch the exact moment the cross is no longer visible (the endpoint).

3. Control Variables:
- Keep the volume of hydrochloric acid constant (10 cm\(^3\)).
- Keep the concentration of hydrochloric acid constant.
- Keep the temperature constant (perform all trials at room temperature).
- Keep the total volume of the reaction mixture constant (60 cm\(^3\)).

4. Data Processing and Analysis:
- Repeat the experiment for each concentration at least three times and calculate the mean time (\(t_{\text{mean}}\)) taken for the cross to disappear.
- Calculate the rate of reaction using the formula: \(\text{Rate} = \frac{1}{t_{\text{mean}}}\) (or \(\frac{1000}{t_{\text{mean}}}\)).
- Plot a graph of rate (y-axis) against the concentration of sodium thiosulfate (x-axis) to analyze the relationship.

5. Safety Precaution:
- Conduct the experiment in a well-ventilated area or fume cupboard. This is because the reaction produces sulfur dioxide (\(\text{SO}_2\)) gas, which is toxic and can trigger asthma attacks or cause respiratory irritation.

1 mark: Explains how to vary the concentration of sodium thiosulfate by adding different volumes of distilled water while keeping the total volume constant.
1 mark: Describes the 'disappearing cross' method (placing the flask over a paper cross and observing from above until it is obscured).
1 mark: Identifies the dependent variable as the time taken for the cross to disappear.
1 mark: Identifies that the rate of reaction is calculated as \(1 / \text{time}\).
2 marks: Lists two controlled variables (volume of acid, concentration of acid, or temperature of solutions) [1 mark each].
1 mark: Mentions using separate, clean measuring cylinders for measuring the different reactants to avoid cross-contamination.
1 mark: Mentions starting the stopwatch immediately upon mixing the reactants.
1 mark: Mentions repeating the experiment at each concentration to calculate a mean and identify anomalies.
1 mark: Identifies that sulfur dioxide gas is toxic/harmful and suggests carrying out the experiment in a fume cupboard or well-ventilated room.

To investigate how the resistance of a wire depends on its length:
1. Circuit Design:
- Connect the power supply, the switch, the ammeter, and the test wire in series to form a closed loop.
- Connect the voltmeter in parallel across the portion of the resistance wire being measured (specifically across the two crocodile clips making contact with the wire).

2. Experimental Procedure:
- Attach one crocodile clip to the start of the resistance wire (at the 0.0 cm mark on the ruler).
- Attach the second mobile crocodile clip to the wire at a distance of 20.0 cm (measured using the meter ruler).
- Close the switch, quickly read and record the current (\(I\)) on the ammeter and the potential difference (\(V\)) on the voltmeter, then immediately open the switch to prevent the wire from heating up.
- Move the second crocodile clip to 40.0 cm, 60.0 cm, 80.0 cm, and 100.0 cm. Repeat the measurement of current and potential difference for each length.

3. Calculating Resistance:
- For each length, calculate the resistance (\(R\)) using Ohm's Law: \(R = \frac{V}{I}\), where \(V\) is potential difference in volts and \(I\) is current in amperes.

4. Control Variables:
- Temperature of the wire: Keep the current low and switch off the circuit between readings so the wire does not heat up, as heating increases resistance.
- Use the exact same wire throughout the experiment to ensure the cross-sectional area (thickness) and material (resistivity) remain constant.

5. Data Analysis and Conclusion:
- Repeat the measurements to obtain average current and potential difference readings for each length, and calculate the mean resistance.
- Plot a graph of mean resistance \(R\) (y-axis) against length \(L\) (x-axis).
- Draw a straight line of best fit. If the line is straight and passes through the origin, it proves that resistance is directly proportional to length.

1 mark: Describes/draws a complete series circuit containing power supply, switch, ammeter, and the test wire.
1 mark: Connects the voltmeter in parallel across the section of the test wire being measured.
1 mark: Describes how to vary the length of the wire using a movable crocodile clip/sliding contact.
1 mark: Explains that the length is measured using the meter ruler attached to the wire.
1 mark: Identifies the primary measurements needed (current from the ammeter, potential difference from the voltmeter).
1 mark: States the correct formula to calculate resistance: \(R = V / I\).
1 mark: Identifies wire temperature as a control variable and explains how to control it (e.g., turning off the switch between readings or using a low current).
1 mark: Identifies wire thickness/cross-sectional area or wire material as a control variable.
1 mark: Mentions repeating measurements at each length to find the average current/p.d. or resistance.
1 mark: Describes using a graph of resistance against length to show the relationship (specifying that a straight line through the origin indicates direct proportionality).

### Experimental Plan

**1. Additional Apparatus Required:**
- Conical flask (reaction vessel)
- Rubber bung fitted with a delivery tube
- Gas syringe (or an inverted measuring cylinder in a trough of water)
- Measuring cylinders (for measuring acid and water volumes)
- Stopwatch or digital timer
- Thermometer (to monitor temperature)

**2. Method and Dilution of Acid:**
To vary the concentration, prepare at least five different concentrations of hydrochloric acid by diluting the stock \(2.0\text{ mol/dm}^3\) acid with distilled water, keeping the total volume of liquid constant at \(20\text{ cm}^3\) for each trial:
- Trial 1: \(20\text{ cm}^3\) acid + \(0\text{ cm}^3\) water (Concentration = \(2.0\text{ mol/dm}^3\))
- Trial 2: \(16\text{ cm}^3\) acid + \(4\text{ cm}^3\) water (Concentration = \(1.6\text{ mol/dm}^3\))
- Trial 3: \(12\text{ cm}^3\) acid + \(8\text{ cm}^3\) water (Concentration = \(1.2\text{ mol/dm}^3\))
- Trial 4: \(8\text{ cm}^3\) acid + \(12\text{ cm}^3\) water (Concentration = \(0.8\text{ mol/dm}^3\))
- Trial 5: \(4\text{ cm}^3\) acid + \(16\text{ cm}^3\) water (Concentration = \(0.4\text{ mol/dm}^3\))

**3. Experimental Procedure:**
- Measure \(20\text{ cm}^3\) of the prepared acid mixture using a measuring cylinder and pour it into the conical flask.
- Add one \(2\text{ cm}\) strip of magnesium ribbon to the flask.
- Immediately insert the rubber bung connected to the gas syringe and start the stopwatch.
- Record the volume of gas collected in the syringe every \(10\text{ seconds}\) for a total of \(60\text{ seconds}\).
- Repeat the entire procedure for the remaining four acid concentrations.

**4. Control Variables:**
- **Temperature:** Conduct all experiments in the same room at the same ambient temperature, or use a constant-temperature water bath.
- **Magnesium Ribbon:** Ensure each piece of magnesium ribbon is exactly \(2\text{ cm}\) long and comes from the same source roll to keep its surface area and mass constant.
- **Total Volume:** Keep the total volume of the reaction mixture constant at \(20\text{ cm}^3\).

**5. Data Analysis:**
- Plot a graph of volume of gas collected (y-axis) against time (x-axis) for each concentration.
- Determine the initial rate of reaction by calculating the gradient of the steep, linear starting portion of each curve (\(\text{gradient} = \frac{\Delta y}{\Delta x}\)).
- Alternatively, calculate the average rate of reaction over the first \(30\text{ seconds}\) using: \(\text{Rate} = \frac{\text{Volume of gas}}{\text{Time taken}}\).

**6. Results Table:**

| Concentration of acid / \(\text{mol/dm}^3\) | Volume of gas at \(10\text{ s}\) / \(\text{cm}^3\) | Volume of gas at \(20\text{ s}\) / \(\text{cm}^3\) | Volume of gas at \(30\text{ s}\) / \(\text{cm}^3\) | Volume of gas at \(40\text{ s}\) / \(\text{cm}^3\) | Volume of gas at \(50\text{ s}\) / \(\text{cm}^3\) | Volume of gas at \(60\text{ s}\) / \(\text{cm}^3\) |
|---|---|---|---|---|---|---|
| 2.0 | | | | | | |
| 1.6 | | | | | | |
| 1.2 | | | | | | |
| 0.8 | | | | | | |
| 0.4 | | | | | | |

**Apparatus (Max 2 marks):**
- **MP1:** Gas syringe OR inverted measuring cylinder in a trough of water (for gas collection).
- **MP2:** Conical flask with delivery tube/bung AND a stopwatch/timer.

**Method & Dilution (Max 2 marks):**
- **MP3:** Describes a logical method to dilute the stock acid to obtain at least 5 different concentrations (by varying the relative volumes of acid and water).
- **MP4:** States that the total volume of the acid mixture must be kept constant (e.g., always \(20\text{ cm}^3\)).

**Procedure & Measurements (Max 2 marks):**
- **MP5:** Explains that magnesium is added and the flask is immediately sealed with a bung to start gas collection.
- **MP6:** Describes measuring the volume of gas at regular time intervals (e.g., every \(10\text{ s}\)) OR measures the time taken to collect a fixed volume of gas.

**Control Variables (Max 2 marks):**
- **MP7:** Identifies keeping the temperature constant (e.g., using a water bath or monitoring with a thermometer).
- **MP8:** Identifies keeping the surface area/mass/length of the magnesium ribbon constant.

**Data Analysis (Max 1 mark):**
- **MP9:** Explains how rate is determined, e.g., by finding the gradient of a volume-time graph, or calculating \(\frac{\text{volume}}{\text{time}}\).

**Results Table (Max 1 mark):**
- **MP10:** Draws a clear table containing appropriate headings with units, e.g., concentration in \(\text{mol/dm}^3\) and volume of gas in \(\text{cm}^3\) or time in \(\text{s}\).