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

Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (against a concentration gradient), using energy released from respiration and carrier proteins in the cell membrane.

A is correct [1]. B is incorrect as active transport requires metabolic energy from respiration, not just kinetic energy, and utilizes carrier proteins rather than moving through cell walls. C is incorrect as osmosis refers specifically to the net movement of water molecules. D is incorrect as diffusion occurs down a concentration gradient.

In electrolysis, positively charged cations move to the cathode (negative electrode) where they gain electrons (reduction). Lead(II) ions (\(\text{Pb}^{2+}\)) are attracted to the cathode and gain two electrons to form molten lead: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\).

B is correct [1]. A and C are incorrect because bromine is formed at the anode via oxidation of bromide ions. D is incorrect because reduction occurs at the cathode (lead ions gain electrons, rather than lead atoms losing electrons).

Acceleration \(a = \frac{v - u}{t} = \frac{15\text{ m/s} - 0\text{ m/s}}{6.0\text{ s}} = 2.5\text{ m/s}^2\). Using Newton's second law: \(F = ma = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\).

C is correct [1]. A (\(720\text{ N}\)) arises from dividing \(1200\) by \(15\) then multiplying by \(6.0\) incorrectly. B (\(1800\text{ N}\)) is \(1200 \times 1.5\). D (\(10\,800\text{ N}\)) is \(1200 \times (15 - 6)\).

Powdered calcium carbonate has a greater surface area than chips, increasing the collision frequency between reactant particles and therefore increasing the initial rate of reaction. However, because hydrochloric acid is the limiting reactant (calcium carbonate is in excess) and the same amount of acid is used, the total yield of carbon dioxide gas remains unchanged.

B is correct [1]. A is incorrect because the final volume of gas depends on the limiting reactant (acid), which has not changed. C and D are incorrect because changing particle size changes the surface area and therefore alters the initial rate.

Total circuit resistance \(R_T = R_1 + R_2 = 6.0\text{ }\Omega + 18.0\text{ }\Omega = 24.0\text{ }\Omega\). The current in the circuit is \(I = \frac{V}{R_T} = \frac{12\text{ V}}{24.0\text{ }\Omega} = 0.50\text{ A}\). The potential difference across the \(6.0\text{ }\Omega\) resistor is \(V = I \times R = 0.50\text{ A} \times 6.0\text{ }\Omega = 3.0\text{ V}\).

B is correct [1]. A (\(2.0\text{ V}\)) is \(12 / 6\). C (\(4.0\text{ V}\)) comes from incorrect ratio calculation. D (\(9.0\text{ V}\)) is the potential difference across the \(18.0\text{ }\Omega\) thermistor.

For photosynthesis to occur, both chlorophyll (green pigment) and light are required to produce glucose, which is stored as starch.

• Area 1 contains chlorophyll and receives light, so photosynthesis occurs and starch is present (turns blue-black).
• Area 2 lacks light, so no starch is made.
• Area 3 lacks chlorophyll, so no starch is made.
• Area 4 lacks both light and chlorophyll, so no starch is made.

Therefore, only Area 1 turns blue-black.

A is correct [1].
B is incorrect as Area 3 lacks chlorophyll so cannot synthesise starch.
C is incorrect as Area 2 lacks light.
D is incorrect as Areas 2 and 4 were kept in darkness.

In the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)):

1. At the cathode (negative electrode), positive lead ions (\(\text{Pb}^{2+}\)) gain electrons (reduction) to form molten lead metal: \(\text{Pb}^{2+} + 2\text{e}^- \to \text{Pb}\).
2. At the anode (positive electrode), negative bromide ions (\(\text{Br}^-\)) lose electrons (oxidation) to form bromine gas: \(2\text{Br}^- \to \text{Br}_2 + 2\text{e}^-\).
3. Molten electrolytes conduct electricity via the movement of mobile ions, not delocalised electrons.

C is correct [1].
A is incorrect because bromine gas forms at the positive anode.
B is incorrect because lead ions undergo reduction (gain electrons) at the cathode.
D is incorrect because conduction in molten salts is due to mobile ions.

1. Calculate acceleration \(a\):
\[a = \frac{v - u}{t} = \frac{20 - 0}{8.0} = 2.5\text{ m/s}^2\]

2. Calculate the resultant force \(F_{\text{resultant}}\) using Newton's second law:
\[F_{\text{resultant}} = m \times a = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\]

3. Determine the engine's forward driving force \(F_{\text{engine}}\):
\[F_{\text{resultant}} = F_{\text{engine}} - F_{\text{resistive}}\]
\[F_{\text{engine}} = 3000\text{ N} + 400\text{ N} = 3400\text{ N}\]

C is correct [1].
A is incorrect (subtracts resistive force from resultant force instead of adding: \(3000 - 400 = 2600\text{ N}\)).
B is incorrect (gives the resultant force \(3000\text{ N}\) without accounting for resistance).
D is incorrect (calculation error).

In bright light:
• Circular muscles of the iris contract.
• Radial muscles of the iris relax.
• This causes the pupil diameter to decrease (constrict) to reduce the amount of light entering the eye and protect the retina from damage.

A is correct [1].
B is incorrect because circular muscles contract and radial muscles relax in bright light.
C and D are incorrect because pupil diameter decreases (constricts) in bright light.

1. In bright light:
\[R_{\text{total}} = 60\,\Omega + 20\,\Omega = 80\,\Omega\]
\[V_{60} = \frac{60}{80} \times 12\text{ V} = 9.0\text{ V}\]

2. In darkness:
\[R_{\text{total}} = 60\,\Omega + 180\,\Omega = 240\,\Omega\]
\[V_{60} = \frac{60}{240} \times 12\text{ V} = 3.0\text{ V}\]

3. Decrease in potential difference across the \(60\,\Omega\) resistor:
\[\Delta V = 9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\]

B is correct [1].
A is incorrect (this is the final potential difference across the fixed resistor in darkness, \(3.0\text{ V}\)).
C is incorrect (this is the initial potential difference across the fixed resistor in bright light, \(9.0\text{ V}\)).
D is incorrect (calculation error).

The work done by the resultant force equals the change in kinetic energy of the object:
\(\text{Work done} = F \times d = 18\text{ N} \times 4.0\text{ m} = 72\text{ J}\).

Since the object starts from rest:
\(\text{Kinetic energy} = \frac{1}{2} m v^2 = 72\text{ J}\)
\(\frac{1}{2} (4.0\text{ kg}) v^2 = 72\text{ J}\)
\(2.0 v^2 = 72\)
\(v^2 = 36\)
\(v = 6.0\text{ m/s}\).

C ; [1 mark for correct selection of 6.0 m/s]

Reduction is defined as the gain of electrons or a decrease in oxidation state. In equation C, the dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) gains 6 electrons to form \(\text{Cr}^{3+}\) ions, and the oxidation state of chromium decreases from +6 to +3. Equations A, B, and D all involve the loss of electrons (oxidation).

C ; [1 mark for identifying the half-equation involving electron gain / reduction]

When focusing on a distant object, the ciliary muscles relax, which increases the tension on the suspensory ligaments causing them to tighten (pull tight). This pulls the lens into a thinner, less convex shape, reducing its refractive power.

D ; [1 mark for correct combination: ciliary muscles relax, suspensory ligaments tighten]

Initial state:
\(R_{\text{total}} = 30\,\Omega + 60\,\Omega = 90\,\Omega\)
\(V_{30} = 12\text{ V} \times \frac{30}{90} = 4.0\text{ V}\)

Final state:
\(R_{\text{total}} = 30\,\Omega + 10\,\Omega = 40\,\Omega\)
\(V_{30} = 12\text{ V} \times \frac{30}{40} = 9.0\text{ V}\)

Change in potential difference:
\(\Delta V = 9.0\text{ V} - 4.0\text{ V} = +5.0\text{ V}\) (an increase of \(5.0\text{ V}\)).

C ; [1 mark for calculating correct change in potential difference (+5.0 V)]

Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is an amphoteric oxide because it reacts with both acids and bases to form salts. Calcium oxide (\(\text{CaO}\)) is a basic oxide because it is a metal oxide that reacts with acids to form a salt and water.

A ; [1 mark for aluminium oxide = amphoteric, calcium oxide = basic]

A factor is limiting when an increase in that factor increases the rate of the reaction. At high light intensity, increasing the temperature from \(20\text{ }^\circ\text{C}\) to \(30\text{ }^\circ\text{C}\) increases the rate of photosynthesis, which demonstrates that temperature is the limiting factor at \(20\text{ }^\circ\text{C}\).

D [1 mark]: Correctly identifies temperature as the limiting factor because an increase in temperature causes an increase in rate at high light intensity.

In the electrolysis of molten lead(II) bromide, bromide ions (\(\text{Br}^-\)) migrate to the anode (positive electrode) and are oxidised to bromine gas (\(\text{Br}_2\)). In the electrolysis of concentrated aqueous sodium chloride (brine), chloride ions (\(\text{Cl}^-\)) are discharged preferentially at the anode over hydroxide ions (\(\text{OH}^-\)), producing chlorine gas (\(\text{Cl}_2\)). Therefore, row A is correct.

A [1 mark]: Correct anode products: bromine for molten lead(II) bromide and chlorine for concentrated aqueous sodium chloride.

Initial kinetic energy: \(E_{\text{k, initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.0\text{ kg} \times (3.0\text{ m/s})^2 = 9.0\text{ J}\).
Work done on the trolley: \(W = F \times d = 4.0\text{ N} \times 5.0\text{ m} = 20\text{ J}\).
By the work-energy theorem, the final kinetic energy is:
\(E_{\text{k, final}} = E_{\text{k, initial}} + W = 9.0\text{ J} + 20\text{ J} = 29\text{ J}\).

B [1 mark]: \(E_{\text{k, initial}} = 9.0\text{ J}\), \(W = 20\text{ J}\), \(E_{\text{k, final}} = 29\text{ J}\).

Acidified potassium manganate(VII) is an oxidising agent containing \(\text{MnO}_4^-\) ions, which are purple. It oxidises iron(II) ions (\(\text{Fe}^{2+}\)) to iron(III) ions (\(\text{Fe}^{3+}\)) by the loss of an electron (\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)). During this redox reaction, \(\text{MnO}_4^-\) is reduced to colourless \(\text{Mn}^{2+}\) ions, causing the solution to turn from purple to colourless.

A [1 mark]: Correctly states that \(\text{Fe}^{2+}\) is oxidised to \(\text{Fe}^{3+}\) and the manganate(VII) solution changes from purple to colourless.

First, calculate the equivalent resistance of the parallel combination of two \(4.0\text{ }\Omega\) resistors:
\(R_p = \frac{4.0 \times 4.0}{4.0 + 4.0} = 2.0\text{ }\Omega\).
Total circuit resistance: \(R_{\text{total}} = 6.0\text{ }\Omega + 2.0\text{ }\Omega = 8.0\text{ }\Omega\).
Circuit current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\text{ }\Omega} = 1.5\text{ A}\).
Potential difference across the \(6.0\text{ }\Omega\) resistor: \(V = I \times R = 1.5\text{ A} \times 6.0\text{ }\Omega = 9.0\text{ V}\).

D [1 mark]: Correct equivalent parallel resistance \(2.0\text{ }\Omega\), total resistance \(8.0\text{ }\Omega\), current \(1.5\text{ A}\), resulting in \(V = 9.0\text{ V}\).

As temperature increases up to the optimum, the enzyme and substrate molecules gain kinetic energy and move faster. This leads to a higher frequency of collisions between enzyme and substrate molecules and an increased frequency of effective (successful) collisions, raising the reaction rate.

A is correct [1]; B is incorrect as denaturation / shape change of the active site occurs mainly above the optimum temperature; C is incorrect because kinetic energy increases with temperature; D is incorrect because kinetic energy increases, not decreases.

In concentrated aqueous sodium chloride, chloride ions (\(\text{Cl}^-\)) are oxidised at the anode to form chlorine gas (\(\text{Cl}_2\)). At the cathode, hydrogen ions (\(\text{H}^+\)) are preferentially reduced compared to sodium ions, producing hydrogen gas (\(\text{H}_2\)).

A is correct [1]; B, C and D represent common errors in predicting discharge products in aqueous electrolysis.

Work done \(W = \Delta E_p = mgh = 600\text{ kg} \times 10\text{ N/kg} \times 15\text{ m} = 90\,000\text{ J}\). Power \(P = \frac{W}{t} = \frac{90\,000\text{ J}}{20\text{ s}} = 4500\text{ W}\).

B is correct [1]; A corresponds to omitting mass factor or arithmetic error (\(450\text{ W}\)); C corresponds to work done without dividing by time (\(90\,000\text{ W}\)); D corresponds to multiplying work done by time (\(180\,000\text{ W}\)).

Barium sulfate is an insoluble salt prepared by precipitation: mix solutions containing soluble barium ions (e.g. aqueous barium chloride) and sulfate ions (e.g. dilute sulfuric acid or aqueous sodium sulfate), filter off the precipitate, wash the residue with distilled water to remove soluble contaminants, and dry.

A is correct [1]; B is incorrect because \(\text{BaSO}_4\) coats unreacted \(\text{BaCO}_3\) and cannot be crystallised from solution; C is for preparing soluble salts; D will leave sodium salts mixed with barium sulfate.

For two \(6.0\text{ }\Omega\) resistors in parallel: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\text{ }\Omega\). Total resistance of the circuit: \(R_{\text{total}} = R_p + 6.0 = 3.0 + 6.0 = 9.0\text{ }\Omega\). Total current \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{9.0\text{ }\Omega} = 1.33\text{ A} \approx 1.3\text{ A}\).

B is correct [1]; A corresponds to treating all three resistors in series (\(R = 18\text{ }\Omega\), \(I = 0.67\text{ A}\)); C corresponds to ignoring the parallel branch (\(R = 6.0\text{ }\Omega\), \(I = 2.0\text{ A}\)); D corresponds to treating all three in parallel (\(R = 2.0\text{ }\Omega\), \(I = 6.0\text{ A}\)).

At temperatures above the optimum, high thermal energy causes bonds maintaining the tertiary structure of the enzyme to break. This alters the shape of the active site (denaturation), preventing the substrate from fitting to form enzyme–substrate complexes.

B is correct [1]. Denaturation occurs at high temperatures, changing the active site shape so substrates cannot bind.

In concentrated aqueous sodium chloride, \(\text{H}^+\) ions are preferentially discharged over \(\text{Na}^+\) ions at the cathode, producing hydrogen gas (\(\text{H}_2\)). Chloride ions (\(\text{Cl}^-\)) are discharged at the anode in preference to hydroxide ions because of their high concentration, forming chlorine gas (\(\text{Cl}_2\)) via the oxidation reaction \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\).

A is correct [1]. Hydrogen is produced at the cathode, chlorine at the anode via the loss of electrons (oxidation): \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\).

First, calculate the equivalent resistance of the two \(8.0\text{ }\Omega\) resistors in parallel: \(R_p = \frac{8.0 \times 8.0}{8.0 + 8.0} = 4.0\text{ }\Omega\). Next, add the series resistor to find the total resistance: \(R_{\text{total}} = 2.0\text{ }\Omega + 4.0\text{ }\Omega = 6.0\text{ }\Omega\). Finally, use Ohm's law to calculate current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}\).

C is correct [1]. \(R_p = 4.0\text{ }\Omega\), \(R_{\text{total}} = 6.0\text{ }\Omega\), \(I = \frac{12}{6.0} = 2.0\text{ A}\).

Zinc carbonate is an insoluble base. Excess zinc carbonate is added to dilute sulfuric acid to ensure all the acid reacts. The excess unreacted zinc carbonate is removed by filtration. The filtrate (aqueous zinc sulfate) is heated to the point of crystallisation and allowed to cool slowly to form crystals, which are then filtered and dried.

A is correct [1]. Excess insoluble carbonate is added to react all acid, unreacted solid is filtered off, and the solution is crystallised.

Acceleration: \(a = \frac{v - u}{t} = \frac{16 - 0}{4.0} = 4.0\text{ m/s}^2\). Resultant force: \(F = ma = 0.50\text{ kg} \times 4.0\text{ m/s}^2 = 2.0\text{ N}\). Distance travelled is the area under the speed-time graph: \(\text{distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 16\text{ m/s} = 32\text{ m}\).

A is correct [1]. \(F = ma = 0.50 \times 4.0 = 2.0\text{ N}\); \(\text{distance} = \text{average speed} \times t = 8.0 \times 4.0 = 32\text{ m}\).

In a shoot exposed to unilateral light, auxin diffuses towards the shaded side of the shoot tip. This results in a higher concentration of auxin on the shaded side. Auxin stimulates cell elongation in shoots, so the cells on the shaded side elongate more rapidly than those on the illuminated side, causing the shoot to bend towards the light.

A [1]

During the electrolysis of concentrated aqueous sodium chloride (brine):
- At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged preferentially over hydroxide ions, producing chlorine gas (\(\text{Cl}_2\)).
- At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged preferentially over sodium ions, producing hydrogen gas (\(\text{H}_2\)).
- As \(\text{H}^+\) ions are removed, an excess of hydroxide ions (\(\text{OH}^-\)) and sodium ions (\(\text{Na}^+\)) remain in solution, forming aqueous sodium hydroxide (an alkali). Hence, the pH increases.

A [1]

Work done on the trolley is given by:
\(W = F \times d = 6.0\text{ N} \times 3.0\text{ m} = 18\text{ J}\).

By the conservation of energy, this work done is transferred into kinetic energy (\(E_k\)):
\(E_k = \frac{1}{2}mv^2\)
\(18 = \frac{1}{2} \times 4.0 \times v^2 = 2.0 v^2\)
\(v^2 = \frac{18}{2.0} = 9.0\)
\(v = \sqrt{9.0} = 3.0\text{ m/s}\).

B [1]

1. Effervescence producing a gas that turns limewater milky confirms the presence of carbonate ions (\(\text{CO}_3^{2-}\)).
2. With aqueous \(\text{NaOH}\), \(\text{Al}^{3+}\) and \(\text{Zn}^{2+}\) both give a white precipitate soluble in excess.
3. With aqueous \(\text{NH}_3\), \(\text{Al}^{3+}\) forms a white precipitate insoluble in excess, whereas \(\text{Zn}^{2+}\) forms a white precipitate soluble in excess.
Therefore, compound \(X\) contains \(\text{Al}^{3+}\) and \(\text{CO}_3^{2-}\).

A [1]

1. Initial total resistance of the series circuit:
\(R_{\text{total}} = 30\,\Omega + 60\,\Omega = 90\,\Omega\).

2. Potential difference across the \(30\,\Omega\) resistor:
\(V = \frac{30}{90} \times 12\text{ V} = \frac{1}{3} \times 12\text{ V} = 4.0\text{ V}\).

3. When the resistance of the variable resistor decreases, the total resistance of the circuit decreases, causing the circuit current to increase. Because the voltage across the fixed resistor is \(V = I R\), the larger current causes the potential difference across the \(30\,\Omega\) resistor to increase.

A [1]

Amylase has an optimum pH of around 6.8. Lowering the pH significantly to 2.5 causes the enzyme to denature because extreme pH alters the shape of the active site so that the substrate can no longer bind, thus decreasing the rate of reaction. Increasing substrate or enzyme concentration (B and C) increases the rate of effective collisions, and increasing the temperature from \(35\ ^\circ\text{C}\) towards optimum (D) increases kinetic energy and reaction rate.

A ; [1 mark for correct identification of condition that denatures the enzyme / decreases reaction rate]

An oxidising agent oxidises another substance while itself being reduced (gaining electrons or losing oxygen). In equation A, \(\text{CuO}\) loses oxygen to form \(\text{Cu}\) (or \(\text{Cu}^{2+}\) gains 2 electrons to form \(\text{Cu}\)), meaning \(\text{CuO}\) is reduced and acts as the oxidising agent. In B, C, and D, the underlined substances (\(\text{Mg}\), \(\text{CO}\), and \(\text{Zn}\)) are oxidised, so they act as reducing agents.

A ; [1 mark for identifying the substance undergoing reduction / acting as oxidising agent]

1. Calculate useful work done: \(W = \Delta E_p = mgh = 60\text{ kg} \times 10\text{ N/kg} \times 8.0\text{ m} = 4800\text{ J}\).
2. Calculate useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{4800\text{ J}}{12\text{ s}} = 400\text{ W}\).
3. Calculate efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{400}{600} \times 100\% = 66.7\% \approx 67\%\).

C ; [1 mark for correct calculation of useful power output / work done and percentage efficiency]

In the electrolysis of concentrated aqueous \(\text{NaCl}\):
- At the cathode (negative electrode), \(\text{H}^+\) ions are discharged preferentially over \(\text{Na}^+\) ions because hydrogen is less reactive, producing hydrogen gas (\(\text{H}_2\)).
- At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are present in high concentration and are discharged in preference to hydroxide ions (\(\text{OH}^-\)), producing chlorine gas (\(\text{Cl}_2\)).
- The anode half-equation is the oxidation of chloride ions: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\).

A ; [1 mark for correct cathode product, anode product, and anode half-equation]

Using the potential divider formula:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2} = 12\text{ V} \times \frac{60\,\Omega}{30\,\Omega + 60\,\Omega} = 12 \times \frac{60}{90} = 12 \times \frac{2}{3} = 8.0\text{ V}\).
Alternatively, total resistance \(R = 30 + 60 = 90\,\Omega\), current \(I = \frac{12}{90} = \frac{2}{15}\text{ A}\), and \(V = I \times R_2 = \frac{2}{15} \times 60 = 8.0\text{ V}\).

C ; [1 mark for correct calculation of potential difference across the variable resistor]

Use the formula for gravitational potential energy:
$$\Delta E_p = mgh$$
$$\Delta E_p = 450\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 52\,920\text{ J}$$
Rounded to two significant figures, this is \(53\,000\text{ J}\) (or \(53\text{ kJ}\)).

(\Delta E_p =) mgh \text{ OR } 450 \times 9.8 \times 12 [1];
52920 / 53000 (\text{J}) [1]

To test for sulfate ions in solution:
1. Acidify with dilute nitric acid (or dilute hydrochloric acid) to remove any carbonate ions.
2. Add aqueous barium nitrate (or aqueous barium chloride).
3. A positive result is the formation of a white precipitate of barium sulfate.

add dilute nitric acid / dilute hydrochloric acid AND add barium nitrate (solution) / barium chloride (solution) [1];
white precipitate [1]

The central nervous system (CNS) consists of the brain and the spinal cord.

brain [1];
spinal cord [1]

Use the electrical power equation:
$$P = I \times V$$
$$P = 0.25\text{ A} \times 6.0\text{ V} = 1.5\text{ W}$$
The correct unit for power is watts (W).

(P =) IV \text{ OR } 0.25 \times 6.0 [1];
1.5 AND W / watts / \text{J/s} [1]

Rusting is the corrosion of iron and requires both oxygen (present in air) and water (or water vapour/moisture).

oxygen / air [1];
water / moisture / water vapour [1]

1. Determine the extension \(x\) of the spring:
\[x = 18.0\text{ cm} - 12.0\text{ cm} = 6.0\text{ cm} = 0.060\text{ m}\]

2. Use Hooke's law, \(F = kx\), to calculate the spring constant \(k\):
\[k = \frac{F}{x} = \frac{4.5\text{ N}}{0.060\text{ m}} = 75\text{ N/m}\]

(Alternatively, \(k = \frac{4.5\text{ N}}{6.0\text{ cm}} = 0.75\text{ N/cm}\)).

• \(x = 6.0\text{ cm}\) (or \(0.060\text{ m}\)) AND correct formula substitution \(k = \frac{4.5}{0.060}\) or \(k = \frac{4.5}{6.0}\) [1]
• \(75\text{ N/m}\) OR \(0.75\text{ N/cm}\) (correct value with matching unit) [1]

When concentration increases, there are more reactant particles present in a given volume (or particles are closer together). This results in a higher frequency of collisions (more collisions per unit time/second) between the reacting particles, which increases the rate of reaction.

• More particles per unit volume / particles closer together [1]
• Higher frequency of collisions / more collisions per second / more collisions per unit time [1]
(Note: Reject 'more collisions' without a reference to time/rate/frequency)

Adrenaline is secreted by the adrenal glands in response to danger or stress. It prepares the body for action by increasing heart rate, increasing breathing rate/depth, dilating pupils, and stimulating the conversion of glycogen to glucose to increase blood glucose concentration.

Award [1] mark for each correct physiological effect (max 2):
• Increased heart rate / pulse rate [1]
• Increased breathing rate / deeper breathing [1]
• Dilation of pupils [1]
• Increased blood glucose concentration [1]
• Diversion of blood flow towards skeletal muscles / away from digestive system [1]

1. Convert time to seconds:
\[t = 3.0\text{ min} \times 60 = 180\text{ s}\]

2. Use the electrical energy formula \(E = V \times I \times t\):
\[E = 6.0\text{ V} \times 0.45\text{ A} \times 180\text{ s} = 486\text{ J}\]

• Correct time conversion \(t = 180\text{ s}\) AND substitution \(E = 6.0 \times 0.45 \times 180\) [1]
• \(486\text{ J}\) OR \(0.486\text{ kJ}\) [1]

A strong acid completely ionises (dissociates) into ions when dissolved in water, releasing all available hydrogen ions (\(\text{H}^+\)). In contrast, a weak acid only partially ionises (dissociates) in aqueous solution, establishing an equilibrium where only a small fraction of the molecules release \(\text{H}^+\) ions.

• Strong acid: completely / fully ionised / dissociated (in aqueous solution) [1]
• Weak acid: partially / incompletely ionised / dissociated (in aqueous solution) [1]

Adrenaline prepares the body for vigorous action:
(i) It increases the heart rate to supply more oxygen and glucose to respiring muscles.
(ii) It causes the pupil to dilate (increase in diameter) to allow more light into the eye and enhance vision.

(i) increases / goes up / beats faster [1];
(ii) increases / dilates / widens / gets larger [1];

Reject: 'contracts' for (ii).

When the concentration of the acid is decreased:
1. The rate of the reaction decreases.
2. In terms of collision theory, a lower concentration means fewer reacting particles per unit volume, which reduces the collision frequency (fewer successful collisions per unit time).

rate (of reaction) decreases / reaction is slower [1];
fewer particles per unit volume / particles further apart AND lower collision frequency / fewer collisions per second (or unit time) [1];

Note: Do not accept 'fewer collisions' without a reference to time or unit volume for the second mark.

Use the electrical power equation:
\[ P = V \times I \]
\[ P = 12\text{ V} \times 0.75\text{ A} = 9.0\text{ W} \]

9.0 / 9 [1];
\text{W} / \text{watts} / \text{J/s} [1];

During the electrolysis of molten lead(II) bromide:
- Negative bromide ions (\(\text{Br}^-\)) migrate to the anode where they lose electrons to form bromine (\(\text{Br}_2\)).
- Positive lead ions (\(\text{Pb}^{2+}\)) migrate to the cathode where they gain electrons to form lead metal (\(\text{Pb}\)).

(i) bromine / \(\text{Br}_2\) [1];
(ii) lead / \(\text{Pb}\) [1];

Reject: bromide for (i), lead ions for (ii).

1. An \(\alpha\)-particle consists of two protons and two neutrons (identical to a helium-4 nucleus).
2. Because of its relatively large mass and \(+2\) charge, it has a high ionising power, meaning it readily collides with and ionises air molecules, rapidly losing its kinetic energy.

(helium nucleus /) 2 protons and 2 neutrons [1];
(strongly / highly) ionising / causes many ionisations / loses energy quickly (through frequent collisions with air molecules) [1];

Reject: 'helium atom' for mark 1.

Electromotive force (\(\text{e.m.f.}\)) is defined as the electrical energy transferred (or work done) by an electrical source per unit charge in driving that charge around a complete circuit.

Mathematically, \(\text{e.m.f.} = \frac{W}{Q}\), where \(W\) is work done / energy supplied and \(Q\) is charge.

work done / energy transferred (by a source) per unit charge [1] ;
in driving / moving charge around a complete circuit [1] ;

[Total: 2]

1. Calculate the work done (change in gravitational potential energy):
\[ \Delta E_p = mgh = 350\text{ kg} \times 9.8\text{ N/kg} \times 12.0\text{ m} = 41\,160\text{ J} \]

2. Calculate the useful power output:
\[ P = \frac{W}{t} = \frac{41\,160\text{ J}}{14.0\text{ s}} = 2940\text{ W} \]

(\Delta E_p =) mgh \text{ OR } 350 \times 9.8 \times 12.0 \text{ [1]};
(P =) \frac{\text{work done}}{\text{time}} \text{ OR } \frac{41\,160}{14.0} \text{ [1]};
2940 \text{ (or } 2.94\text{)} \text{ AND } \text{W} \text{ (or } \text{J/s} \text{ / } \text{kW if } 2.94\text{)} \text{ [1]}

1. Calculate the amount in moles of magnesium reacted:
\[ n(\text{Mg}) = \frac{\text{mass}}{A_r} = \frac{3.60\text{ g}}{24\text{ g/mol}} = 0.150\text{ mol} \]

2. Use the molar ratio from the equation:
\[ \text{ratio of Mg : H}_2 = 1 : 1 \implies n(\text{H}_2) = 0.150\text{ mol} \]

3. Calculate the volume of \( \text{H}_2 \) gas:
\[ V = n \times 24.0\text{ dm}^3/\text{mol} = 0.150\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 3.60\text{ dm}^3 \]

\text{moles of Mg} = \frac{3.60}{24} = 0.15(0) \text{ (mol)} \text{ [1]};
\text{moles of } \text{H}_2 = 0.15(0) \text{ (mol)} \text{ [1]};
(V =) 0.150 \times 24.0 = 3.60 \text{ (dm}^3\text{)} \text{ [1]}

Method 1 (Combined resistance):
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30.0} + \frac{1}{20.0} = \frac{5}{60} = \frac{1}{12.0\ \Omega} \implies R_p = 12.0\ \Omega \]
\[ I = \frac{V}{R_p} = \frac{12.0\text{ V}}{12.0\ \Omega} = 1.00\text{ A} \]

Method 2 (Sum of branch currents):
\[ I_1 = \frac{12.0\text{ V}}{30.0\ \Omega} = 0.40\text{ A}, \quad I_2 = \frac{12.0\text{ V}}{20.0\ \Omega} = 0.60\text{ A} \]
\[ I_{\text{total}} = 0.40\text{ A} + 0.60\text{ A} = 1.00\text{ A} \]

\frac{1}{R_p} = \frac{1}{30} + \frac{1}{20} \text{ OR } R_p = 12(.0) \text{ (}\Omega\text{)} \text{ OR } I_1 = \frac{12}{30} (= 0.4\text{ A}) \text{ AND } I_2 = \frac{12}{20} (= 0.6\text{ A}) \text{ [1]};
I = \frac{12.0}{12.0} \text{ OR } I = 0.40 + 0.60 \text{ [1]};
1.0(0) \text{ AND } \text{A} \text{ (or amperes)} \text{ [1]}

1. Convert the image size and actual size to the same units:
\[ \text{Image size } (I) = 54\text{ mm} = 54\,000\ \mu\text{m} \]
\[ \text{Actual size } (A) = 75\ \mu\text{m} \]

2. Use the magnification formula:
\[ \text{Magnification } (M) = \frac{\text{Image size } (I)}{\text{Actual size } (A)} \]
\[ M = \frac{54\,000\ \mu\text{m}}{75\ \mu\text{m}} = 720 \text{ (or } \times 720 \text{)} \]

\text{magnification} = \frac{\text{image size}}{\text{actual size}} \text{ (seen or implied)} \text{ [1]};
\text{conversion of units: } 54\text{ mm} = 54\,000\ \mu\text{m} \text{ OR } 75\ \mu\text{m} = 0.075\text{ mm} \text{ [1]};
(\times) 720 \text{ [1]}

1. Calculate the total energy supplied by the heater:
\[ E = P \times t = 450\text{ W} \times 160\text{ s} = 72\,000\text{ J} \]

2. Calculate the temperature change:
\[ \Delta T = 66.0^\circ\text{C} - 18.0^\circ\text{C} = 48.0^\circ\text{C} \]

3. Calculate the specific heat capacity:
\[ c = \frac{E}{m \Delta T} = \frac{72\,000\text{ J}}{0.60\text{ kg} \times 48.0^\circ\text{C}} = 2500\text{ J}/(\text{kg}^\circ\text{C}) \]

(E =) P \times t = 450 \times 160 = 72\,000 \text{ (J)} \text{ [1]};
(c =) \frac{E}{m \Delta T} \text{ OR } \frac{72\,000}{0.60 \times 48(.0)} \text{ [1]};
2500 \text{ AND } \text{J}/(\text{kg}^\circ\text{C}) \text{ (or } \text{J}/(\text{kg K})\text{)} \text{ [1]}

(a) Using kinetic energy formula: \(E_k = \frac{1}{2}mv^2\)
\(E_k = 0.5 \times 1.80 \times (14.0)^2 = 0.5 \times 1.80 \times 196 = 176.4\text{ J}\) (or \(176\text{ J}\)).

(b) Using power formula: \(P = \frac{W}{t} = \frac{E_k}{t}\)
\(P = \frac{176.4}{3.50} = 50.4\text{ W}\).

(a) \(\frac{1}{2} \times 1.80 \times 14.0^2\) / substitution into \(E_k = \frac{1}{2}mv^2\); [1]
\(176.4\) / \(176\text{ (J)}\); [1]

(b) \(\frac{176.4}{3.50}\) / \(50.4\text{ (W)}\) (allow ecf from (a)); [1]

Step 1: Calculate moles of \(\text{CaCO}_3\):
\(\text{moles} = \frac{\text{mass}}{M_r} = \frac{15.0}{100} = 0.150\text{ mol}\)

Step 2: Mole ratio of \(\text{CaCO}_3 : \text{CO}_2\) is \(1 : 1\), so \(\text{moles of }\text{CO}_2 = 0.150\text{ mol}\).

Step 3: Calculate volume of \(\text{CO}_2\):
\(\text{volume} = \text{moles} \times 24.0\text{ dm}^3/\text{mol} = 0.150 \times 24.0 = 3.60\text{ dm}^3\).

\(\text{moles of }\text{CaCO}_3 = \frac{15.0}{100} = 0.150\text{ (mol)}\); [1]
\(\text{volume of }\text{CO}_2 = 0.150 \times 24.0\); [1]
\(3.60\text{ (dm}^3\text{)}\); [1]

Method 1:
Calculate total equivalent resistance \(R_T\):
\(\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30.0} + \frac{1}{20.0} = \frac{2 + 3}{60.0} = \frac{5}{60.0}\)
\(R_T = \frac{60.0}{5} = 12.0\ \Omega\)
Calculate total current: \(I = \frac{V}{R_T} = \frac{12.0}{12.0} = 1.00\text{ A}\)

Method 2:
Calculate current through each branch:
\(I_1 = \frac{12.0}{30.0} = 0.40\text{ A}\)
\(I_2 = \frac{12.0}{20.0} = 0.60\text{ A}\)
Total current \(I = I_1 + I_2 = 0.40 + 0.60 = 1.00\text{ A}\).

Correct formula for combined parallel resistance \(\frac{1}{R_T} = \frac{1}{30} + \frac{1}{20}\) OR individual currents \(I_1 = \frac{12}{30}\) and \(I_2 = \frac{12}{20}\); [1]
\(R_T = 12.0\ (\Omega)\) OR \(I_1 = 0.40\text{ (A)}\) and \(I_2 = 0.60\text{ (A)}\); [1]
\(1.00\text{ (A)}\); [1]

Step 1: Calculate energy supplied by heater:
\(E = P \times t = 450\text{ W} \times 120\text{ s} = 54000\text{ J}\)

Step 2: Calculate temperature change:
\(\Delta T = 54.0 - 22.0 = 32.0\ ^\circ\text{C}\)

Step 3: Rearrange \(E = mc\Delta T\) to solve for \(c\):
\(c = \frac{E}{m\Delta T} = \frac{54000}{0.800 \times 32.0} = \frac{54000}{25.6} = 2109.375\text{ J}/(\text{kg}\ ^\circ\text{C}) \approx 2110\text{ J}/(\text{kg}\ ^\circ\text{C})\).

\(\text{Energy } E = 450 \times 120 = 54000\text{ (J)}\); [1]
\(\Delta T = 32.0\text{ (}^\circ\text{C)}\) and substitution \(c = \frac{54000}{0.800 \times 32.0}\); [1]
\(2110\text{ (J}/(\text{kg}\ ^\circ\text{C}))\) (accept \(2109\) or \(2109.4\)); [1]

(a) \(\text{Magnification } M = \frac{\text{Image size } (I)}{\text{Actual size } (A)} = \frac{48.0\text{ mm}}{0.060\text{ mm}} = 800\) (or \(\times 800\)).

(b) Convert chloroplast actual length to mm:
\(4.5\ \mu\text{m} = 0.0045\text{ mm}\)
Image length \(I = M \times A = 800 \times 0.0045\text{ mm} = 3.6\text{ mm}\).

(a) \(\frac{48.0}{0.060} = (\times)800\); [1]

(b) Conversion of \(4.5\ \mu\text{m}\) to \(0.0045\text{ mm}\) OR \(4.5 \times 10^{-3}\text{ mm}\); [1]
\(3.6\text{ (mm)}\); [1]

1. Auxin is produced in the shoot tip (apex) and diffuses downwards into the elongation zone.
2. Light coming from one side (unilateral light) causes auxin to move across to the shaded side of the shoot, resulting in an unequal distribution of auxin.
3. The higher concentration of auxin on the shaded side stimulates cells in that region to elongate more rapidly.
4. Because the shaded side grows and elongates faster than the illuminated side, the unequal growth causes the shoot to bend towards the light source (positive phototropism).

1. auxin produced / synthesised in the tip / apex (of the shoot) [1];
2. auxin moves / diffuses to / accumulates on the shaded side / side away from light [1];
3. (higher concentration of auxin causes) cell elongation / cells become longer on the shaded side [1];
4. shaded side grows / elongates faster (than illuminated side) causing the shoot to bend towards the light [1]

1. At higher temperatures, particles absorb thermal energy, increasing their kinetic energy and speed.
2. Faster moving particles collide more frequently per unit time (increased collision frequency).
3. More importantly, a much higher fraction of colliding particles have energy equal to or exceeding the activation energy (\(E_a\)).
4. This leads to a greater frequency of successful/effective collisions, resulting in a higher reaction rate.

1. particles gain (kinetic) energy / move faster [1];
2. particles collide more frequently / higher collision frequency [1];
3. greater proportion / more of the particles have energy greater than / equal to the activation energy (\(E_a\)) [1];
4. higher frequency of successful / effective collisions (per unit time) [1]

1. Water molecules in a liquid state exist with a wide range of kinetic energies.
2. The most energetic (fastest) molecules near the surface overcome intermolecular forces of attraction and escape into the gas phase.
3. The removal of these high-energy molecules reduces the average kinetic energy of the remaining liquid molecules.
4. Since temperature is a measure of the average kinetic energy of particles, the liquid's temperature drops, and thermal energy is transferred away from the skin, resulting in a cooling sensation.

1. molecules (in liquid) have a range / distribution of kinetic energies [1];
2. molecules with the highest kinetic energy / fastest molecules overcome attractive forces / escape from surface [1];
3. average kinetic energy of the remaining molecules decreases [1];
4. temperature (of remaining water) decreases / thermal energy is absorbed from the skin [1]

1. In molten lead(II) bromide, ions are free to move. \(\text{Pb}^{2+}\) cations are attracted to the negative cathode, while \(\text{Br}^-\eval{}\) anions are attracted to the positive anode.
2. At the cathode: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\). Lead ions gain electrons (reduction) to form silvery molten lead.
3. At the anode: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\). Bromide ions lose electrons (oxidation) to form orange/brown bromine gas.
4. Overall redox reaction separates the compound into its constituent elements via electron transfer at the electrodes.

1. \(\text{Pb}^{2+}\) (cations) move to the cathode / negative electrode AND \(\text{Br}^-\eval{}\) (anions) move to the anode / positive electrode [1];
2. at the cathode, \(\text{Pb}^{2+}\) ions gain electrons / undergo reduction to produce lead (metal) [1];
3. at the anode, \(\text{Br}^-\eval{}\) ions lose electrons / undergo oxidation to produce bromine (gas / vapor) [1];
4. correct ionic half-equations: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\) AND \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\) [1]

1. An alternating current in the primary winding produces a continuously changing (varying) magnetic field.
2. The soft iron core provides a high-permeability magnetic pathway that channels and concentrates the magnetic flux to the secondary winding.
3. The secondary coil experiences a continually changing magnetic flux linkage (the magnetic field lines cut through the secondary coil).
4. According to Faraday's law of electromagnetic induction, a changing magnetic flux induces an alternating electromotive force (e.m.f. / voltage) across the terminals of the secondary coil.

1. alternating current in primary coil creates a changing / alternating magnetic field [1];
2. (soft) iron core transfers / channels / concentrates the magnetic field to the secondary coil [1];
3. magnetic field / flux cuts through / links with the secondary coil [1];
4. (changing magnetic field) induces an alternating e.m.f. / voltage in the secondary coil (by electromagnetic induction) [1]

1. At higher temperatures (\(40\text{ }^\circ\text{C}\)), the reactant particles gain thermal energy which is converted into kinetic energy / particles move faster.
2. This causes particles to collide more frequently / higher frequency of collisions.
3. A greater proportion / fraction of colliding particles have energy equal to or exceeding the activation energy (\(E_a\)).
4. Therefore, there is a higher frequency of successful / effective collisions per unit time.

Any four from:
- (particles have) greater kinetic energy / move faster (at \(40\text{ }^\circ\text{C}\)); [1]
- collisions are more frequent / higher collision rate / collide more often; [1]
- more particles have energy greater than or equal to the activation energy / \(E_a\); [1]
- a greater proportion / fraction of collisions are successful / effective; [1]
- higher frequency of successful collisions / more successful collisions per unit time / per second; [1]

Auxin is synthesized in the shoot tip and transported downwards along the stem. When the shoot is positioned horizontally, gravity causes auxin to accumulate in higher concentrations on the lower side of the shoot. In shoots, higher auxin concentration stimulates cell elongation. Consequently, cells on the lower side elongate more rapidly than cells on the upper side, creating an unequal growth rate that causes the shoot to bend upwards.

- auxin is made / produced in the shoot tip; [1]
- auxin accumulates / moves to the lower side (of the shoot) (due to gravity); [1]
- (higher concentration of) auxin stimulates cell elongation / cell expansion in shoots; [1]
- lower side elongates / grows faster than the upper side / unequal growth causes the shoot to bend upwards (negative gravitropism); [1]

Molecules in the layer of water on the skin possess a range of kinetic energies. The most energetic molecules (those with the highest kinetic energy) are able to overcome the attractive forces between neighbouring water molecules and escape the liquid surface as water vapour. As the fastest particles leave, the mean / average kinetic energy of the remaining water molecules decreases, which lowers the temperature of the water. Thermal energy is then conducted / transferred from the warmer skin to the cooler water, resulting in a cooling sensation.

- particles in liquid have a range / distribution of kinetic energies / speeds; [1]
- particles with the highest kinetic energy / fastest particles escape / leave the liquid (surface); [1]
- (escape requires overcoming) attractive / intermolecular forces; [1]
- average kinetic energy of remaining particles decreases; [1]
- thermal energy is transferred / conducted from the skin to the water (lowering skin temperature); [1]
[Max 4 marks]

In aqueous solution, both \(\text{H}^+\) and \(\text{Na}^+\) ions are attracted to the negative cathode. Because hydrogen is less reactive than sodium (it has a lower tendency to remain as an ion), \(\text{H}^+\) ions are preferentially discharged by gaining electrons (reduction) to form \(\text{H}_2\) gas.

The ionic half-equation is:
\[2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\]

(a)
- \(\text{H}^+\) / hydrogen ions and \(\text{Na}^+\) / sodium ions are both present (at cathode); [1]
- hydrogen is less reactive than sodium / \(\text{H}^+\) ions are discharged more readily / gain electrons more easily than \(\text{Na}^+\); [1]
(b)
- \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (allow \(2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-\)); [1]

1. At the start (instant of jump), the upward air resistance is zero; the only force acting is weight downwards, so the skydiver accelerates at maximum acceleration (\(g = 9.8\text{ m/s}^2\)).
2. As speed increases, air resistance (drag) opposing the motion increases.
3. The resultant downward force (\(\text{weight} - \text{air resistance}\)) decreases, so acceleration decreases (non-uniform acceleration).
4. Eventually, air resistance equals weight (forces become balanced / resultant force = 0), so acceleration is zero and terminal velocity is reached.

- initially weight is the only force / downward force is greater than upward force, so skydiver accelerates (downwards); [1]
- as speed / velocity increases, air resistance / drag increases; [1]
- resultant downward force decreases, so acceleration decreases / rate of increase of velocity decreases; [1]
- air resistance equals weight / forces become balanced / resultant force is zero; [1]
- moves at constant speed / zero acceleration / terminal velocity reached; [1]
[Max 4 marks]

(a) Reading the thermometer scale: At \(t = 0\text{ s}\), the liquid column reaches \(82.5\,^\circ\text{C}\). At \(t = 180\text{ s}\), the liquid column reaches \(64.0\,^\circ\text{C}\). Both values are recorded to one decimal place.
(b) Temperature decrease \(\Delta \theta = 82.5 - 64.0 = 18.5\,^\circ\text{C}\).
(c) A standard experimental precaution when reading a liquid-in-glass thermometer is positioning the line of sight perpendicular/level with the meniscus to eliminate parallax error, or stirring the water gently to ensure uniform temperature distribution.

(a) \(\theta_0 = 82.5\) [1]; \(\theta_{180} = 64.0\) (allow 64) [1];
(b) \(18.5\,^\circ\text{C}\) (ecf from candidate's table values) [1];
(c) View scale at eye level / perpendicular to scale / avoid parallax error OR stir liquid before taking reading OR keep bulb fully immersed / not touching walls of container [1].

(a) Reading the scales: \(V_{30} = 36\text{ cm}^3\) and \(V_{90} = 66\text{ cm}^3\).
(b) Rate between \(0\) and \(30\text{ s}\) is \(\frac{36\text{ cm}^3 - 0\text{ cm}^3}{30\text{ s}} = 1.2\text{ cm}^3/\text{s}\).
(c) Between \(30\text{ s}\) and \(90\text{ s}\) (an interval of \(60\text{ s}\)), only \(66 - 36 = 30\text{ cm}^3\) is produced (rate = \(0.5\text{ cm}^3/\text{s}\)), which is less gas produced per unit time than in the first \(30\text{ s}\).

(a) \(36\) (\(\text{cm}^3\)) [1]; \(66\) (\(\text{cm}^3\)) [1];
(b) \(1.2\) with correct unit \(\text{cm}^3/\text{s}\) or \(\text{cm}^3\,\text{s}^{-1}\) [1];
(c) Less gas is produced per second / volume increase is only \(30\text{ cm}^3\) in \(60\text{ s}\) (or \(0.5\text{ cm}^3/\text{s}\)) compared to \(36\text{ cm}^3\) in the first \(30\text{ s}\) (\(1.2\text{ cm}^3/\text{s}\)) [1].

(a) The digital balance display shows \(2.13\text{ g}\). Change in mass = \(\text{final mass} - \text{initial mass} = 2.13 - 2.50 = -0.37\text{ g}\).
(b) \(\text{Percentage change} = \left(\frac{-0.37}{2.50}\right) \times 100\% = -14.8\%\).
(c) Calculating percentage change accounts for slight differences in the initial starting mass of each cylinder, enabling valid and direct comparison.

(a) \(2.13\) [1]; \(-0.37\) (accept \(0.37\text{ g decrease}\), minus sign required if given as raw value) [1];
(b) \(-14.8\) (allow \(-15\) or \(14.8\%\text{ decrease}\)) (ecf from (a)) [1];
(c) Initial masses of cylinders are different / not identical / to allow valid comparison [1].

(a) Reading the meter dials: \(V = 2.40\text{ V}\) (or \(2.4\text{ V}\)) and \(I = 0.60\text{ A}\) (or \(0.6\text{ A}\)).
(b) \(R = \frac{V}{I} = \frac{2.40}{0.60} = 4.0\,\Omega\).
(c) Resistance is directly proportional to length. Doubling the length from \(40.0\text{ cm}\) to \(80.0\text{ cm}\) doubles the resistance: \(4.0 \times 2 = 8.0\,\Omega\).

(a) \(V = 2.4(0)\) [1]; \(I = 0.6(0)\) [1];
(b) \(4.0\) (allow \(4\)) [1];
(c) \(8.0\) (allow \(8\), ecf candidate's value in (b) \(\times 2\)) [1].

(a) Burette scales read downwards: initial reading = \(1.40\text{ cm}^3\), final reading = \(24.10\text{ cm}^3\).
(b) Titre 1 = \(24.10 - 1.40 = 22.70\text{ cm}^3\).
(c) The three titres are \(22.70\text{ cm}^3\), \(22.65\text{ cm}^3\), and \(22.70\text{ cm}^3\). All three lie within a range of \(0.05\text{ cm}^3\) (well within \(0.20\text{ cm}^3\)). Average titre = \(\frac{22.70 + 22.65 + 22.70}{3} = 22.68\text{ cm}^3\) (or \(22.7\text{ cm}^3\)).

(a) Initial reading \(1.40\) and final reading \(24.10\) [2] (1 mark for each correct reading recorded to 2 decimal places or 1 decimal place if ending in .0);
(b) \(22.70\) (ecf from candidate readings) [1];
(c) Identifies concordant titres (all three / 1, 2 and 3 OR 1 and 3 OR 2 and 3) AND calculates average correctly: \(22.68\) or \(22.70\) (\(\text{cm}^3\)) [1].

(a) Graph plotting [4 marks]:
- Axes (A): \(y\)-axis labelled 'volume of gas / \(\text{cm}^3\)' and \(x\)-axis labelled 'time / \(\text{s}\)'.
- Scale (S): Linear scales chosen so that the plotted data occupies more than half of the available grid in both directions (e.g., \(x\)-axis: \(2\text{ cm} = 20\text{ s}\); \(y\)-axis: \(2\text{ cm} = 10\text{ cm}^3\)).
- Plotting (P): All 7 points \((0,0), (20,24), (40,41), (60,54), (80,62), (100,66), (120,66)\) plotted accurately to within half a small square.
- Line (L): A single, smooth, continuous curve of best fit drawn passing through the origin and levelling off at \(66\text{ cm}^3\).

(b)(i) [1 mark]:
- Reading at \(t = 50\text{ s}\) from the curve: \(48\text{ cm}^3\) (acceptable range: \(47.0\text{ cm}^3\) to \(49.0\text{ cm}^3\)).
- Clear dashed or solid construction lines shown on the graph at \(t = 50\text{ s}\) projected to the curve and across to the \(y\)-axis.

(b)(ii) [1 mark]:
- The rate decreases (or becomes zero) as shown by the decreasing gradient / curve becoming less steep / flattening out.

(a)
- A (Axes): labelled with quantity and unit: volume (of gas) / \(\text{cm}^3\) and time / \(\text{s}\) ; [1]
- S (Scale): linear, not awkward, plotted points occupy \(\ge 50\%\) of grid in both dimensions ; [1]
- P (Plotting): all points plotted correctly to \(\pm 0.5\) small square ; [1]
- L (Line): smooth best-fit curve drawn through points with no double lines / feathering ; [1]

(b)(i)
- Correct reading from candidate's graph at \(t = 50\text{ s}\) (expected: \(48.0 \pm 1.0\text{ cm}^3\)) AND evidence of construction lines shown on graph ; [1]

(b)(ii)
- (Rate) decreases / reaction slows down AND gradient / slope decreases / curve becomes less steep / levels off ; [1]

(a) Graph plotting [4 marks]:
- Axes (A): \(y\)-axis labelled 'resistance / \(\Omega\)' and \(x\)-axis labelled 'length / \(\text{cm}\)'.
- Scale (S): Both axes starting from \((0,0)\), linear scales chosen such that the plotted points occupy \(\ge 50\%\) of grid in both dimensions (e.g., \(x\)-axis: \(2\text{ cm} = 10.0\text{ cm}\); \(y\)-axis: \(2\text{ cm} = 1.0\text{ }\Omega\) or \(2.0\text{ }\Omega\)).
- Plotting (P): All 6 points \((10.0, 1.4), (20.0, 2.9), (30.0, 4.3), (40.0, 5.8), (50.0, 7.2), (60.0, 8.7)\) plotted accurately to \(\pm 0.5\) small square.
- Line (L): Single, sharp, straight line of best fit drawn with a ruler passing through origin.

(b)(i) [1 mark]:
- Gradient \(m = \frac{\Delta R}{\Delta L}\).
- Using points from the line spanning at least half the length of the drawn line, e.g., \((0,0)\) and \((60.0, 8.7)\):
$$\text{gradient} = \frac{8.7 - 0}{60.0 - 0} = 0.145\text{ }\Omega / \text{cm}$$
- Range of acceptable values: \(0.142\text{ to } 0.148\text{ }\Omega / \text{cm}\).
- Clear indication of coordinates / triangle used on graph.

(b)(ii) [1 mark]:
- \(R = \text{gradient} \times 80.0\text{ cm} = 0.145 \times 80.0 = 11.6\text{ }\Omega\).
- Accept \(11.4\text{ }\Omega\) to \(11.8\text{ }\Omega\) or \(\text{ecf}\) from candidate's value in (b)(i).

(a)
- A (Axes): labelled with quantity and unit: resistance / \(\Omega\) and length / \(\text{cm}\) ; [1]
- S (Scale): linear, starts at \((0,0)\), data occupies \(\ge 50\%\) of grid in both directions ; [1]
- P (Plotting): all 6 points plotted accurately to within \(\pm 0.5\) small square ; [1]
- L (Line): straight line of best fit drawn with a ruler, well balanced ; [1]

(b)(i)
- Correct calculation of gradient using values from graph spanning at least half the drawn line with indication shown on graph (expected: \(0.145 \pm 0.003\text{ }\Omega / \text{cm}\)) ; [1]

(b)(ii)
- Correct calculation of resistance using candidate's gradient \(\times 80.0\) (e.g. \(11.6\text{ }\Omega\), accept \(11.4 - 11.8\text{ }\Omega\)) ; [1]

Apparatus:
- Conical flask with delivery tube and bung
- Gas syringe (or inverted measuring cylinder filled with water in a trough)
- Stopwatch / timer
- Measuring cylinder (to measure acid and water volumes)
- Ruler and scissors (to measure and cut magnesium ribbon) / balance

Method:
1. Prepare at least 5 different concentrations of hydrochloric acid (e.g., \(2.0\text{ mol/dm}^3\), \(1.6\text{ mol/dm}^3\), \(1.2\text{ mol/dm}^3\), \(0.8\text{ mol/dm}^3\), and \(0.4\text{ mol/dm}^3\)) by diluting the \(2.0\text{ mol/dm}^3\) stock acid with appropriate measured volumes of distilled water (e.g., \(40\text{ cm}^3\) acid + \(10\text{ cm}^3\) water for \(1.6\text{ mol/dm}^3\)).
2. Measure a constant total volume (e.g. \(50\text{ cm}^3\)) of the chosen acid concentration into a conical flask.
3. Cut a piece of magnesium ribbon to a fixed length (e.g. \(3.0\text{ cm}\)) or measure a fixed mass.
4. Add the magnesium ribbon to the flask, immediately insert the bung connected to the gas syringe, and start the stopwatch.
5. Record the volume of hydrogen gas collected after a fixed time (e.g. \(60\text{ s}\)) OR measure the time taken to collect a fixed volume of gas (e.g. \(30\text{ cm}^3\)) / time taken for the magnesium to completely dissolve.
6. Repeat the procedure with the remaining acid concentrations.

Variables to Control (Fair Test):
- Length / mass / surface area of magnesium ribbon.
- Total volume of liquid in the flask (e.g. \(50\text{ cm}^3\)).
- Initial temperature of the acid / reaction mixture.

Processing and Conclusion:
- Repeat each concentration at least twice and calculate a mean value.
- Calculate the rate of reaction (e.g., \(\text{rate} = \frac{\text{volume of gas}}{\text{time}}\) or \(\text{rate} = \frac{1}{\text{time}}\)).
- Plot a graph of reaction rate (\(y\)-axis) against concentration of hydrochloric acid (\(x\)-axis).
- A direct proportionality / positive correlation indicates that increasing concentration increases the rate of reaction.

Safety Precaution:
- Wear safety goggles / eye protection because hydrochloric acid is corrosive / an irritant.

Award 1 mark for each of the following points (up to a maximum of 7 marks):

- MP1 (Apparatus): Suitable apparatus to measure rate specified (gas syringe / inverted measuring cylinder over water / balance to measure mass loss / stopwatch to time until magnesium dissolves);
- MP2 (Independent variable / Dilution): Preparing at least 4 different concentrations of acid by mixing measured volumes of acid and distilled water (or testing at least 4 specific concentrations);
- MP3 (Control variable - Magnesium): Keeping the length / mass / surface area of magnesium constant for each test;
- MP4 (Control variable - Solution/Conditions): Keeping the total volume of solution constant OR keeping starting temperature constant;
- MP5 (Measurements): Measuring the volume of gas collected in a specified fixed time period OR measuring the time taken for a fixed volume of gas to be produced / time taken for magnesium to completely react/disappear;
- MP6 (Data processing): Repeating trials for each concentration to calculate a mean OR calculating rate (e.g., \(\frac{\text{volume}}{\text{time}}\) or \(\frac{1}{\text{time}}\)) AND plotting a graph of rate (or volume or \(1/\text{time}\)) against concentration;
- MP7 (Safety): Wear eye protection / safety goggles because acid is corrosive/an irritant OR ensure no naked flames because hydrogen is flammable.

(a)(i) A \( 25\text{ cm}^3 \) or \( 50\text{ cm}^3 \) measuring cylinder (or a \( 25\text{ cm}^3 \) volumetric pipette/burette) is suitable to accurately measure \( 15\text{ cm}^3 \).
(a)(ii) A green precipitate insoluble in excess aqueous sodium hydroxide indicates the presence of iron(II) ions (\( \text{Fe}^{2+} \)).
(b)(i) Sulfate ions react with barium ions (\( \text{Ba}^{2+} \)) to produce insoluble barium sulfate, observed as a white precipitate.
(b)(ii) Dilute nitric acid is added to react with and eliminate any interfering anions, such as carbonate (\( \text{CO}_3^{2-} \)) or sulfite (\( \text{SO}_3^{2-} \)), which would otherwise form an insoluble white precipitate of barium carbonate or barium sulfite.
(c) Aqueous sodium hydroxide is corrosive and caustic; wearing chemical-resistant gloves or washing hands immediately after contact prevents chemical burns.

(a)(i) 1 mark: measuring cylinder / pipette / burette (reject: beaker / flask / dropping pipette).
(a)(ii) 1 mark: iron(II) / \( \text{Fe}^{2+} \) (reject: iron / \( \text{Fe}^{3+} \) / iron(III)).
(b)(i) 1 mark: white precipitate / white ppt / solid.
(b)(ii) 1.25 marks: to react with / eliminate / destroy carbonate (ions) / sulfite (ions) (1 mark); so they do not give a false positive / white precipitate with barium ions (0.25 mark).
(c) 1 mark: wear (nitrile/protective) gloves / wash hands immediately if spilled.

(a) Carbon dioxide gas passes through limewater (aqueous calcium hydroxide), forming insoluble calcium carbonate which turns the limewater cloudy/milky.
(b)(i) The gas evolved is carbon dioxide (\( \text{CO}_2 \)).
(b)(ii) Copper(II) carbonate decomposes upon heating to form copper(II) oxide, which is a black solid (\( \text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2 \)).
(c) If the heat is removed while the delivery tube is submerged, the air in the boiling tube contracts, causing cold limewater to suck back into the hot tube and crack the glass. To avoid this, remove the delivery tube from the liquid before extinguishing the burner.
(d) Black copper(II) oxide reacts with sulfuric acid to form copper(II) sulfate solution, which is blue (\( \text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} \)).

(a) 1 mark: (colourless to) cloudy / milky / white precipitate.
(b)(i) 1 mark: carbon dioxide / \( \text{CO}_2 \).
(b)(ii) 1 mark: copper(II) oxide / \( \text{CuO} \) (accept: copper oxide).
(c) 1.25 marks: suck-back / cold water drawn into hot tube / tube cracks or breaks (0.75 mark); remove delivery tube from limewater before removing heat / disconnecting apparatus before extinguishing flame (0.5 mark).
(d) 1 mark: blue.

(a) Zinc ions (\( \text{Zn}^{2+} \)) form a white precipitate of zinc hydroxide with aqueous ammonia which dissolves in excess ammonia to form a colourless complex solution.
(b)(i) The formation of a cream precipitate with aqueous silver nitrate acidified with dilute nitric acid indicates bromide ions (\( \text{Br}^- \)).
(b)(ii) The precipitation reaction is between silver ions and bromide ions: \( \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s}) \).
(c) A dropping pipette (teat pipette) is the standard laboratory apparatus for adding reagents dropwise.
(d) Safety goggles or safety glasses protect the eyes from splashes.

(a) 1 mark: zinc / \( \text{Zn}^{2+} \).
(b)(i) 1 mark: bromide / \( \text{Br}^- \) (reject: bromine).
(b)(ii) 1.25 marks: \( \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr} \) (1 mark); correct state symbols \( (\text{aq}) \) and \( (\text{s}) \) (0.25 mark).
(c) 1 mark: dropping pipette / teat pipette / dropper / Pasteur pipette (reject: beaker / cylinder).
(d) 1 mark: (safety) goggles / (safety) glasses / eye protection / face shield (reject: lab coat / gloves as it specifies chemical splashes to eyes/face).

(a)(i) Benedict's test requires heating in a hot water bath (typically 80–100 °C) for 3–5 minutes.
(a)(ii) In the presence of a high concentration of reducing sugar, the solution changes from blue to a brick-red precipitate.
(b) Biuret reagent contains copper(II) sulfate in an alkaline solution; in the presence of peptide bonds (protein), it changes from blue to purple/lilac/violet.
(c)(i) Iodine in potassium iodide solution (iodine solution) is used to test for starch.
(c)(ii) If starch is absent, the iodine solution remains yellow/brown/orange-brown (does not turn blue-black).
(d) A test-tube holder (or test-tube tongs/clamp) is used to safely hold the tube during heating.

(a)(i) 1 mark: heat / (hot) water bath / temperature above 70 °C (reject: warm water without qualification).
(a)(ii) 1 mark: brick-red / red / brown-red (precipitate/solid) (reject: green/yellow alone for high concentration).
(b) 1 mark: (blue to) purple / lilac / violet / mauve.
(c)(i) 1 mark: iodine (solution) / aqueous iodine / iodine in KI.
(c)(ii) 0.5 mark: yellow / brown / orange-brown / remains yellow-brown (reject: colourless / no change without specifying initial colour).
(d) 0.75 mark: test-tube holder / test-tube tongs / clamp.