2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模拟试题及答案详解

The equation for the cracking process is:
\(C_{12}H_{26} \rightarrow C_8H_{18} + 2 C_xH_y\)

We balance the carbon atoms:
\(12 = 8 + 2x\)
\(2x = 4 \Rightarrow x = 2\)

We balance the hydrogen atoms:
\(26 = 18 + 2y\)
\(2y = 8 \Rightarrow y = 4\)

Therefore, the formula of the alkene is \(C_2H_4\) (ethene).

Award 1 mark for the correct option B. Reject all other options.

Using Newton's second law:
\(a = \frac{F}{m} = \frac{12\text{ N}}{4.0\text{ kg}} = 3.0\text{ m/s}^2\)

Now, using the equation of motion for constant acceleration:
\(v = u + at = 5.0\text{ m/s} + (3.0\text{ m/s}^2 \times 3.0\text{ s}) = 5.0 + 9.0 = 14.0\text{ m/s}\)

Award 1 mark for the correct option B. Reject all other options.

Let's analyze the arrangements:
- Option A (three in parallel): \(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \Rightarrow R_p = \frac{1}{3}R\).
- Option B (two in parallel connected in series with the third): \(R_{total} = \frac{R}{2} + R = \frac{3}{2}R\).
- Option C (two in series connected in parallel with the third): The two in series have resistance \(2R\). This branch is in parallel with the third resistor \(R\):
\(\frac{1}{R_{total}} = \frac{1}{2R} + \frac{1}{R} = \frac{3}{2R} \Rightarrow R_{total} = \frac{2}{3}R\).
- Option D (three in series): \(R_s = R + R + R = 3R\).

Award 1 mark for the correct option C. Reject all other options.

The polymer contains the ester linkage \(-O-CO-\), which means it is a polyester. Synthetic polyesters (such as Terylene) are formed by condensation polymerization, in which small molecules like water are eliminated.

Award 1 mark for the correct option D. Reject all other options.

With active copper electrodes, the copper anode itself dissolves. Copper atoms in the anode lose electrons (oxidation) to form copper(II) ions in the solution:
\(Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-\).

Award 1 mark for the correct option A. Reject all other options.

Using the wave equation:
\(f = \frac{v}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-2}\text{ m}} = 2.0 \times 10^{10}\text{ Hz}\).

A wavelength of \(1.5\text{ cm}\) (\(1.5 \times 10^{-2}\text{ m}\)) falls in the range of \(1\text{ mm}\) to \(10\text{ cm}\), which belongs to the microwave region.

Award 1 mark for the correct option A. Reject all other options.

First calculate the thermal energy supplied:
\(\Delta E = P \times t = 120\text{ W} \times 60\text{ s} = 7200\text{ J}\).

Now, use the specific heat capacity equation:
\(\Delta E = m c \Delta \theta\)
\(7200\text{ J} = 0.50\text{ kg} \times c \times (80\text{ °C} - 20\text{ °C})\)
\(7200 = 0.50 \times c \times 60\)
\(7200 = 30 c\)
\(c = 240\text{ J / (kg °C)}\).

Award 1 mark for the correct option B. Reject all other options.

First determine the number of half-lives that have elapsed by halving the count rate sequentially:
- Initial: 1200
- After 1 half-life: 600
- After 2 half-lives: 300
- After 3 half-lives: 150

Thus, 3 half-lives have elapsed in 30 minutes.
\(3 \times t_{1/2} = 30\text{ minutes} \Rightarrow t_{1/2} = 10\text{ minutes}\).

Award 1 mark for the correct option B. Reject all other options.

The work done to accelerate the car is equal to its gain in kinetic energy: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (20\text{ m/s})^2 = 240\,000\text{ J}\). Average power is the work done divided by time: \(P = \frac{E_k}{t} = \frac{240\,000\text{ J}}{8.0\text{ s}} = 30\,000\text{ W} = 30\text{ kW}\).

Award 1 mark for the correct option B. Reject all other options.

The ester shown is ethyl propanoate. The part derived from the carboxylic acid contains three carbons (\(\text{CH}_3\text{CH}_2\text{COO-}\)), which corresponds to propanoic acid. The part derived from the alcohol contains two carbons (\(\text{-CH}_2\text{CH}_3\)), which corresponds to ethanol.

Award 1 mark for the correct option B. Reject all other options.

First, calculate output voltage: \(V_s = V_p \times \frac{N_s}{N_p} = 230\text{ V} \times \frac{4000}{200} = 4600\text{ V}\). Next, for a \(100\%\) efficient transformer, \(I_p V_p = I_s V_s\). Therefore, \(I_s = I_p \times \frac{V_p}{V_s} = 10\text{ A} \times \frac{230\text{ V}}{4600\text{ V}} = 0.5\text{ A}\).

Award 1 mark for the correct option A. Reject all other options.

As she falls and her speed increases, air resistance increases. Because air resistance acts upwards, opposing gravity, the downward net force decreases. Since \(F = ma\), her downward acceleration decreases, but remains positive (downward), so her speed continues to increase at a slower rate.

Award 1 mark for the correct option A. Reject all other options.

The cracking equation is \(\text{C}_{12}\text{H}_{26} \rightarrow \text{C}_6\text{H}_{14} + 2X\). Balancing the atoms: the remaining atoms are \(12 - 6 = 6\) carbon atoms and \(26 - 14 = 12\) hydrogen atoms. Since two molecules of X are formed, each molecule of X has the formula \(\text{C}_3\text{H}_6\). This fits the general formula \(\text{C}_n\text{H}_{2n}\), so X is an alkene (propene).

Award 1 mark for the correct option A. Reject all other options.

Using Fleming's Left-Hand Rule: align your first finger (field) pointing downwards, and your second finger (current) pointing to the right. Your thumb (force) will point into the page.

Award 1 mark for the correct option A. Reject all other options.

At the anode, concentrated chloride ions are oxidised to form chlorine gas (\(\text{Cl}_2\)). At the cathode, hydrogen ions from water are reduced to form hydrogen gas (\(\text{H}_2\)). The remaining solution contains sodium (\(\text{Na}^+\)) and hydroxide (\(\text{OH}^-\)) ions, forming sodium hydroxide, which is alkaline, so the pH of the electrolyte increases.

Award 1 mark for the correct option A. Reject all other options.

The change in pressure is calculated using \(\Delta P = \rho g \Delta h\), where \(\Delta h = 150\text{ m} - 50\text{ m} = 100\text{ m}\). So, \(\Delta P = 1025\text{ kg/m}^3 \times 10\text{ N/kg} \times 100\text{ m} = 1\,025\,000\text{ Pa} \approx 1.0 \times 10^6\text{ Pa}\).

Award 1 mark for the correct option B. Reject all other options.

The cracking equation is C10H22 -> C3H6 (propene) + C4H8 (but-1-ene) + X. Balancing the carbon atoms: 10 - 3 - 4 = 3 carbon atoms. Balancing the hydrogen atoms: 22 - 6 - 8 = 8 hydrogen atoms. Therefore, the formula of alkane X is C3H8.

1 mark for the correct option C.

The center of mass of the uniform 3.0 m plank is at its midpoint, 1.5 m from the left end, which is 0.5 m to the right of the pivot. The clockwise moment due to the plank weight about the pivot is 120 N * 0.5 m = 60 N m. The anticlockwise moment due to the block of weight W placed at the left edge (1.0 m from pivot) is W * 1.0 m. For horizontal equilibrium, the clockwise and anticlockwise moments are equal: W * 1.0 = 60, which gives W = 60 N.

1 mark for the correct option B.

First, calculate the resistance of the two 6.0 ohm resistors connected in parallel: R_p = (6.0 * 6.0) / (6.0 + 6.0) = 3.0 ohms. Next, add the resistance of the third resistor connected in series: R_total = 6.0 + 3.0 = 9.0 ohms.

1 mark for the correct option C.

When an alcohol reacts with a carboxylic acid, an ester is formed. The first part of the ester's name comes from the alcohol (ethanol gives ethyl-) and the second part comes from the carboxylic acid (propanoic acid gives -propanoate). Therefore, the ester formed is ethyl propanoate.

1 mark for the correct option A.

The initial kinetic energy is 0.5 * 800 kg * (20 m/s)^2 = 160,000 J = 160 kJ. The final kinetic energy is 0.5 * 800 kg * (30 m/s)^2 = 360,000 J = 360 kJ. The increase in kinetic energy is 360 kJ - 160 kJ = 200 kJ.

1 mark for the correct option C.

The voltage in the secondary coil is V_s = V_p * (N_s / N_p) = 12 V * (800 / 200) = 48 V. The current in the secondary coil is I_s = V_s / R = 48 V / 4.0 ohms = 12 A. For an ideal transformer, the input power equals the output power, so V_p * I_p = V_s * I_s, which gives 12 V * I_p = 48 V * 12 A, resulting in I_p = 48 A.

1 mark for the correct option C.

The general formula of a saturated monohydric alcohol is CnH2n+1OH. Its relative molecular mass is given by 12n + (2n+1) + 16 + 1 = 14n + 18. Setting this equal to 60 gives 14n + 18 = 60, so 14n = 42, which means n = 3 (propanol). The alkene Y that reacts with steam to form propanol must be propene (C3H6).

1 mark for the correct option B.

At constant terminal velocity, the parachutist is not accelerating, so the resultant force on them is 0 N. This means the upward air resistance force must exactly balance the downward weight of the parachutist. The weight is W = m * g = 70 kg * 10 N/kg = 700 N. Therefore, the air resistance force is 700 N and the resultant force is 0 N.

1 mark for the correct option B.

During addition polymerisation, the double bond of the monomer opens up to form single covalent bonds. The repeating unit of the polymer is -[CH(CH3)-CH(CH3)]-. Reintroducing the double bond between these two carbons gives CH3-CH=CH-CH3, which is but-2-ene.

1 mark for the correct option B.

Propanoic acid (CH3CH2COOH) reacts with ethanol (CH3CH2OH) to form ethyl propanoate and water. The structure of ethyl propanoate is CH3CH2COOCH2CH3.

1 mark for the correct option A.

When the magnet is exactly at the centre of the solenoid, the rate of change of magnetic flux linkage is zero, so no EMF is induced and the voltmeter reads zero. As the magnet exits, the flux decreases, causing an induced EMF in the opposite direction (Lenz's Law), which deflects the needle to the right.

1 mark for the correct option A.

In a parallel circuit, the potential difference across each branch is equal to the supply voltage (12 V). Current through 4.0 Ohm resistor: I1 = 12 / 4.0 = 3.0 A. Current through 6.0 Ohm resistor: I2 = 12 / 6.0 = 2.0 A. Total current I = 6.0 A, so the current through R is I3 = 6.0 - (3.0 + 2.0) = 1.0 A. Resistance R = 12 / 1.0 = 12 Ohms.

1 mark for the correct option C.

We can use the impulse-momentum theorem: Impulse = F * t = Change in momentum = m * vf - m * vi. Impulse = 3.0 N * 2.0 s = 6.0 Ns. Initial momentum: pi = 0.50 kg * 4.0 m/s = 2.0 kg m/s. Final momentum: pf = pi + Impulse = 2.0 + 6.0 = 8.0 kg m/s. Final velocity: vf = 8.0 kg m/s / 0.50 kg = 16 m/s. Alternatively, acceleration a = F / m = 3.0 / 0.50 = 6.0 m/s^2. vf = vi + a * t = 4.0 + 6.0 * 2.0 = 16 m/s.

1 mark for the correct option D.

Useful work output (gain in GPE) = m * g * h = 80 kg * 10 N/kg * 15 m = 12,000 J = 12 kJ. Efficiency = (Useful work output / Total energy input) * 100% = 60%. Total electrical energy input = 12 kJ / 0.60 = 20 kJ.

1 mark for the correct option C.

In a spinal reflex arc, the stimulus is detected by a receptor, which generates a nerve impulse. This impulse travels along the sensory neurone to the spinal cord, across a synapse to a relay neurone, and then across another synapse to a motor neurone, which carries the impulse to the effector (a muscle or gland).

1 mark for the correct option A.

At high temperatures, the thermal energy breaks the bonds that maintain the specific three-dimensional shape of the enzyme's protein structure. This leads to denaturation, causing the active site to change shape irreversibly so that the substrate molecule is no longer complementary and cannot bind.

1 mark for the correct option B.

Propanol (\(C_3H_7OH\)) provides the propyl group (\(-CH_2CH_2CH_3\)) for the alkyl part of the ester. Butanoic acid (\(C_3H_7COOH\)) provides the butanoyl group (\(CH_3CH_2CH_2CO-\)) for the acyl part of the ester. Therefore, the ester formed is propyl butanoate, with the structure \(CH_3CH_2CH_2COOCH_2CH_2CH_3\).

1 mark for the correct option.

Fermentation uses glucose from plants (renewable), while hydration uses ethene from crude oil (non-renewable). Hydration is much faster, produces highly pure ethanol directly, and operates at a higher temperature (around 300°C compared to 30°C for fermentation).

1 mark for the correct option.

Using conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the values: \(0.50 \times 4.0 + 1.50 \times 0 = (0.50 + 1.50) v \implies 2.0 = 2.0 v \implies v = 1.0\text{ m/s}\).

1 mark for the correct option.

Useful energy output (Work done) = \(mgh = 40 \times 10 \times 15 = 6000\text{ J}\). Total energy input = \(VIt = 240 \times 5.0 \times 10 = 12000\text{ J}\). Efficiency = \(\frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\% = \frac{6000}{12000} \times 100\% = 50\%\).

1 mark for the correct option.

Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies \frac{12}{240} = \frac{50}{N_s} \implies N_s = 1000\text{ turns}\). For a 100% efficient transformer: \(V_p I_p = V_s I_s \implies 12 \times I_p = 240 \times 0.20 \implies I_p = 4.0\text{ A}\).

1 mark for the correct option.

The parallel combination of three identical resistors has resistance \(R_p = \frac{R}{3}\). When this is connected in series with the fourth resistor of resistance \(R\), the total combined resistance is \(R_{total} = R_p + R = \frac{R}{3} + R = \frac{4}{3}R\).

1 mark for the correct option.

During addition polymerization, the double bond of propene opens to form the repeating unit \(-CH_2-CH(CH_3)-\). Connecting two of these units together yields \(-CH_2-CH(CH_3)-CH_2-CH(CH_3)-\).

1 mark for the correct option.

Using Fleming's Left Hand Rule: the first finger points in the direction of the magnetic field (into the page), the second finger points in the direction of the conventional current (left to right, since the alpha particle is positively charged), so the thumb points in the direction of the force (upwards, towards the top of the page).

1 mark for the correct option.

(a)(i) Decane has 10 carbons and octane has 8, so the remaining molecule must have 2 carbons. Hydrogens: \(22 - 18 = 4\). Hence, \(\text{C}_2\text{H}_4\).

(ii) \(\text{C}_2\text{H}_4\) is ethene. The structure is \(\text{H}_2\text{C}=\text{CH}_2\) with a double bond between the two carbon atoms and single bonds to four hydrogen atoms.

(iii) Industrial cracking requires high temperatures of around \(450^\circ\text{C}\) to \(800^\circ\text{C}\) and a catalyst of silicon dioxide (silica) and aluminum oxide (alumina).

(b)(i) Ethene undergoes addition polymerization.

(ii) The repeating unit shows a single carbon-carbon bond with two hydrogen atoms on each carbon and extension bonds on either side: \(-[\text{CH}_2-\text{CH}_2]-\).

(c) Add bromine water to both samples. With decane, there is no colour change (remains orange/brown). With ethene (alkene **X**), the mixture turns from orange to colourless.

(a)(i) \(\text{C}_2\text{H}_4\) [1]
(ii) ethene [1], fully correct displayed structure of ethene showing double bond and all H atoms [1]
(iii) temperature in range \(450\text{--}800^\circ\text{C}\) [1], catalyst: silica / alumina / zeolite / clay [1]
(b)(i) addition (polymerisation) [1]
(ii) single carbon-carbon bond with extension bonds [1], correct groups attached to carbons (all H atoms) [1]
(c) add bromine water/aqueous bromine [1], decane remains orange AND ethene/alkene **X** turns colourless [1]

(a)(i) Acceleration \(a = \frac{v-u}{t} = \frac{12 - 0}{8.0} = 1.5\text{ m/s}^2\).

(ii) Total distance is the area under the speed-time graph:
Distance in acceleration phase = \(\frac{1}{2} \times 8.0 \times 12 = 48\text{ m}\).
Distance in constant speed phase = \(15 \times 12 = 180\text{ m}\).
Distance in deceleration phase = \(\frac{1}{2} \times 5.0 \times 12 = 30\text{ m}\).
Total distance = \(48 + 180 + 30 = 258\text{ m}\).

(b)(i) Resultant force \(F = m \times a = 85\text{ kg} \times 1.5\text{ m/s}^2 = 127.5\text{ N}\).
Forward force = Resultant force + Opposing force = \(127.5\text{ N} + 35\text{ N} = 162.5\text{ N}\).

(ii) Work done = Force \(\times\) distance = \(35\text{ N} \times 180\text{ m} = 6300\text{ J}\) (or \(6.3\text{ kJ}\)).

(a)(i) formula used: \(a = \frac{\Delta v}{t}\) [1], \(1.5\) [1], \(\text{m/s}^2\) [1]
(ii) area under acceleration phase = \(48\text{ m}\) OR area under deceleration phase = \(30\text{ m}\) [1], area under constant speed phase = \(180\text{ m}\) [1], addition of all three phases to give \(258\text{ m}\) [1]
(b)(i) calculation of resultant force: \(85 \times 1.5 = 127.5\text{ N}\) [1], addition of opposing force to give \(162.5\text{ N}\) [1]
(ii) work done formula: \(W = F \times d\) OR \(35 \times 180\) [1], \(6300\text{ J}\) (or \(6.3\text{ kJ}\)) [1]

(a)(i) Resistance \(R = \frac{V}{I} = \frac{8.0}{2.5} = 3.2\ \Omega\).

(ii) Power \(P = V \times I = 8.0\text{ V} \times 2.5\text{ A} = 20\text{ W}\).

(iii) Time \(t = 10\text{ minutes} = 600\text{ s}\).
Energy \(E = P \times t = 20\text{ W} \times 600\text{ s} = 12\ 000\text{ J}\) (or \(12\text{ kJ}\)).

(b)(i) The brightness of the lamp decreases. Increasing the resistance of the variable resistor increases the total resistance of the circuit, which decreases the current flowing through the lamp (reducing its power).

(ii) The magnetic field pattern consists of concentric circles centred on the wire.

(a)(i) formula: \(R = \frac{V}{I}\) [1], \(3.2\ \Omega\) [1]
(ii) formula: \(P = VI\) [1], \(20\text{ W}\) [1]
(iii) conversion of time to \(600\text{ s}\) [1], formula: \(E = Pt\) OR \(20 \times 600\) [1], \(12\ 000\text{ J}\) (or \(12\text{ kJ}\)) [1]
(b)(i) brightness decreases [1], explanation: total circuit resistance increases so current through the lamp decreases [1]
(ii) concentric circles [1]

(a)(i) A sensory neurone carries impulses from receptors to the CNS.

(ii) An electrical impulse arrives at the pre-synaptic membrane. This triggers the release of chemical neurotransmitter molecules from vesicles. The neurotransmitter diffuses across the synaptic cleft/gap and binds to specific receptors on the post-synaptic membrane, generating a new electrical impulse.

(b)(i) Secreted by the adrenal glands.

(ii) Physiological effects include: increased heart rate, increased breathing rate, widening of pupils (pupil dilation), conversion of glycogen to glucose (increasing blood glucose concentration), and diversion of blood flow to the muscles.

(c) To focus on a distant object, the ciliary muscles relax, which causes the suspensory ligaments to tighten (become taut). This pulls the lens, causing it to become thinner, flatter, and less refractive.

(a)(i) sensory (neurone) [1]
(ii) any three from: neurotransmitter released from vesicles [1], neurotransmitter diffuses across synaptic gap [1], binds to receptors [1], triggers impulse in next/post-synaptic neurone [1]
(b)(i) adrenal (glands) [1]
(ii) any three from: increased heart rate [1], increased breathing rate [1], pupils dilate [1], increased blood glucose [1], diversion of blood to muscles [1]
(c) ciliary muscles relax AND suspensory ligaments tighten/pull [1], lens becomes thinner / flatter / less convex [1]

(a)(i) Energy needed to break bonds = \(436\text{ (H-H)} + 242\text{ (Cl-Cl)} = 678\text{ kJ}\).

(ii) Energy released when bonds are made = \(2 \times 431\text{ (H-Cl)} = 862\text{ kJ}\).

(iii) Overall energy change = \(678 - 862 = -184\text{ kJ}\). Since the value is negative, the reaction is exothermic.

(b) The reaction is exothermic because more energy is released during the formation of new bonds (bond making) in \(\text{HCl}\) than is taken in to break the original bonds (bond breaking) in \(\text{H}_2\) and \(\text{Cl}_2\).

(c) In the energy level diagram:
1. The energy level of the reactants (\(\text{H}_2 + \text{Cl}_2\)) is drawn higher than the energy level of the products (\(2\text{HCl}\)).
2. A curve rises from the reactant level to a maximum peak (the transition state) and then falls to the product level.
3. The activation energy is represented by an upward arrow from the reactants to the peak of the curve.
4. The overall energy change (\(\Delta H\)) is shown by a downward arrow from the reactants level to the products level.

(a)(i) \(678\text{ (kJ)}\) [1]
(ii) \(862\text{ (kJ)}\) [1]
(iii) \(-184\text{ (kJ)}\) [1], exothermic [1]
(b) more energy is released when making bonds / in products [1] than is absorbed/taken in when breaking bonds / in reactants [1]
(c) reactants at a higher energy level than products [1], curve showing energy peak between reactants and products [1], activation energy shown as vertical distance from reactants to peak [1], energy change shown as vertical distance between reactants and products [1]

(a)(i) The word equation is: glucose \(\rightarrow\) lactic acid.

(ii) Anaerobic respiration is less efficient because glucose is only partially broken down (incomplete oxidation), releasing far less energy per molecule of glucose compared to aerobic respiration.

(iii) The accumulation of lactic acid causes muscle fatigue, pain, or cramps.

(b) This is called the oxygen debt (or excess post-exercise oxygen consumption, EPOC). The elevated breathing and heart rate supply extra oxygen to transport lactic acid from muscles to the liver, where it is oxidized/broken down into carbon dioxide and water, or converted back into glucose.

(c)(i) The word equation in yeast is: glucose \(\rightarrow\) carbon dioxide + ethanol (alcohol).

(ii) Industrial processes include: bread-making (baking, where carbon dioxide makes dough rise) and brewing (production of beer, wine, or biofuel ethanol).

(a)(i) glucose \(\rightarrow\) lactic acid [1]
(ii) incomplete breakdown of glucose [1], releases much less energy / fewer ATP molecules [1]
(iii) causes muscle fatigue / pain / cramps [1]
(b) oxygen debt [1], extra oxygen needed to break down/oxidize lactic acid [1], which occurs in the liver / converts lactic acid to carbon dioxide and water/glucose [1]
(c)(i) glucose \(\rightarrow\) carbon dioxide + ethanol [1]
(ii) any two from: bread-making / baking [1], brewing of beer/wine [1], production of biofuels / ethanol [1]

(a)(i) Moles of \(\text{Mg} = \frac{\text{mass}}{\text{RAM}} = \frac{0.36}{24} = 0.015\text{ mol}\).

(ii) The molar ratio is \(1\text{ Mg} : 1\text{ H}_2\). Therefore, \(0.015\text{ mol}\) of \(\text{H}_2\) is produced.

(iii) Volume of gas = moles \(\times\) 24 = \(0.015 \times 24 = 0.36\text{ dm}^3\) (or \(360\text{ cm}^3\)).

(b)(i) 1. Moles of \(\text{HCl}\) required = \(2 \times \text{moles of Mg} = 2 \times 0.015 = 0.030\text{ mol}\).
2. \(\text{Volume in dm}^3 = \frac{\text{moles}}{\text{concentration}} = \frac{0.030}{2.0} = 0.015\text{ dm}^3\).
3. \(\text{Volume in cm}^3 = 0.015 \times 1000 = 15\text{ cm}^3\).

(ii) Replacing ribbon with powder increases the rate of reaction. This is because powder has a larger surface area per unit mass, which increases the frequency of collisions between reactant particles.

(a)(i) \(0.015\text{ (mol)}\) [1]
(ii) \(0.015\text{ (mol)}\) [1]
(iii) formula: \(\text{moles} \times 24\) [1], \(0.36\text{ dm}^3\) (or \(360\text{ cm}^3\)) [1]
(b)(i) moles of \(\text{HCl}\) needed = \(0.030\text{ mol}\) [1], formula: \(\text{volume} = \frac{\text{moles}}{\text{concentration}}\) [1], volume in \(\text{dm}^3\) = \(0.015\text{ dm}^3\) [1], volume in \(\text{cm}^3\) = \(15\text{ cm}^3\) [1]
(ii) rate increases due to larger surface area of powder [1], higher frequency of collisions between reactant particles [1]

(a)(i) Similarity: both transfer energy (or both undergo reflection/refraction). Difference: sound waves are longitudinal (require a medium) while light waves are transverse (can travel in a vacuum).

(ii) Speed \(v = f \times \lambda = 450\text{ Hz} \times 0.75\text{ m} = 337.5\text{ m/s}\).

(b)(i) Refractive index \(n = \frac{\sin i}{\sin r} = \frac{\sin 40^\circ}{\sin 25^\circ} = \frac{0.6428}{0.4226} \approx 1.52\).

(ii) Speed decreases because glass is optically denser than air. Frequency remains unchanged/constant as it is determined by the source.

(c)(i) Ultraviolet (UV) has a higher frequency than infrared (IR).

(ii) Applications of infrared include: remote controls, thermal imaging, cooking/heaters, and optical fibre communication.

(a)(i) similarity: both transfer energy / reflect / refract [1], difference: sound is longitudinal OR requires a medium AND light is transverse OR can travel in a vacuum [1]
(ii) formula: \(v = f\lambda\) [1], \(337.5\text{ m/s}\) (or \(338\text{ m/s}\)) [1]
(b)(i) formula: \(n = \frac{\sin i}{\sin r}\) [1], \(1.52\) [1]
(ii) speed decreases [1], frequency remains constant/unchanged [1]
(c)(i) ultraviolet / UV [1]
(ii) any one from: remote controls / thermal imaging / heaters / cooking / optical fibres [1]

(a) Decane is cracked to produce octene and ethene: \(C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_4\). (b) Add aqueous bromine (bromine water). Decane is an alkane and does not react, so the solution remains orange/brown. Octene is an alkene and reacts, decolourising the bromine water. (c)(i) This is an addition polymerization reaction. (ii) Ethene has a carbon-carbon double bond, whereas poly(ethene) consists of a long chain of carbon atoms joined only by single covalent bonds. (d) Unsaturated means containing at least one carbon-carbon double bond. Hydrocarbon means a compound consisting of hydrogen and carbon atoms only.

(a) \(C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_4\) [1]

(b) Test: Add bromine water / aqueous bromine [1]
Observation with decane: Stays orange / brown / yellow (no change) [1]
Observation with octene: Decolourises / turns colourless [1] (REJECT 'clear')

(c) (i) Addition (polymerization) [1]
(ii) Poly(ethene) contains only single carbon-carbon bonds / ethene has carbon-carbon double bond [1]; many monomer units linked together / form a long chain [1]

(d) Unsaturated: Contains a carbon-carbon double bond / \(C=C\) [1]
Hydrocarbon: Contains carbon and hydrogen [1] only [1]

(a) As the coil rotates, it cuts through magnetic field lines, inducing an electromotive force (e.m.f.) across the coil. The slip rings and brushes maintain contact with the external circuit, reversing the direction of the induced current every half-turn to produce alternating current. (b)(i) Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging gives \(N_s = N_p \times \frac{V_s}{V_p} = 120 \times \frac{4000}{240} = 2000\) turns. (b)(ii) At a higher voltage, the current required to transmit the same power is reduced. This smaller current minimizes resistive heating losses in the transmission lines, making transmission much more efficient. (c) To increase the induced e.m.f., one can: 1. Rotate the coil faster. 2. Use stronger magnets / increase magnetic field strength. 3. Increase the number of turns on the coil.

(a) Coil cuts magnetic field lines / experiences changing magnetic field [1]
Electromotive force / e.m.f. / voltage is induced [1]
Slip rings and brushes allow the alternating current to flow to the external circuit / reverse current direction every half-turn [1]

(b) (i) Correct formula used: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) or substitution \(\frac{240}{4000} = \frac{120}{N_s}\) [1]
\(N_s = 2000\) [1]
(ii) Higher voltage means lower current (for same power) [1]
Less thermal energy / heat is lost in the cables (improving efficiency) [1] (REJECT 'no energy lost')

(c) Increase speed of rotation [1]
Use stronger magnets / increase magnetic field strength [1]
Increase the number of turns on the coil [1]

(a)(i) Acceleration is calculated using \(a = \frac{v - u}{t} = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\). (a)(ii) Distance travelled is the area under the speed-time graph: \(\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 6.0 \times 15 = 45\text{ m}\). (b)(i) Kinetic energy is \(E_k = \frac{1}{2} m v^2 = 0.5 \times 1200 \times 15^2 = 135\text{ }000\text{ J}\) (or 135 kJ). (b)(ii) Useful power is \(P = F \times v = 800 \times 15 = 12\text{ }000\text{ W}\) (or 12 kW). (c) The car travels at a constant speed because the forward force of 800 N is exactly balanced by equal and opposite resistive forces (friction and air resistance) totaling 800 N, making the net resultant force zero.

(a) (i) \(a = \frac{v}{t}\) or substitution \(\frac{15}{6.0}\) [1]
\(a = 2.5\text{ m/s}^2\) (allow correct unit if omitted elsewhere) [1]
(ii) \(\text{Distance} = \frac{1}{2} \times b \times h\) or substitution \(0.5 \times 6.0 \times 15\) [1]
\(\text{Distance} = 45\text{ m}\) [1]

(b) (i) \(E_k = \frac{1}{2} m v^2\) or substitution \(0.5 \times 1200 \times 15^2\) [1]
\(135\text{ }000\text{ J}\) / \(135\text{ kJ}\) [1]
(ii) \(P = F \times v\) or substitution \(800 \times 15\) [1]
\(12\text{ }000\text{ W}\) / \(12\text{ kW}\) [1]

(c) Opposing resistive forces / friction / air resistance equal 800 N [1]
The forces are balanced / resultant force is zero [1]

(a) A temperature/pain receptor in the skin detects the heat and initiates an electrical impulse. This impulse travels along a sensory neurone to the central nervous system (spinal cord). It crosses a synapse to a relay neurone, then another synapse to a motor neurone. The impulse travels along the motor neurone to the effector muscle, which contracts to pull the hand away. (b)(i) The hormone is adrenaline, secreted by the adrenal glands. (ii) Other physiological changes include pupil dilation, increased breathing rate, increased blood glucose concentration, and blood being diverted away from the digestive system to the muscles. (c) Nervous responses are extremely rapid (fractions of a second) and short-lived, whereas hormonal responses are slower to act (seconds to minutes) but produce a more prolonged effect.

(a) Receptor detects the stimulus (heat / pain) [1]
Impulse travels along a sensory neurone [1]
Impulse passes via relay neurone / spinal cord / CNS [1]
Impulse travels along a motor neurone to the effector / muscle (causing contraction) [1]

(b) (i) Adrenaline [1], adrenal gland(s) [1]
(ii) Any two from: pupil dilation / increased breathing rate / increased blood glucose level / diversion of blood to muscles [2]

(c) Speed of response: Nervous is faster / hormonal is slower [1]
Duration of effect: Nervous has shorter duration / hormonal has longer duration [1]

(a) Iodine solution turns from yellow-brown to blue-black in the presence of starch. (b)(i) The independent variable is the pH of the buffer solution. (b)(ii) Controlled variables include the temperature of the reaction, the concentration of the amylase, the concentration of the starch, and the volume of both amylase and starch solutions. (c)(i) The time taken for starch digestion decreases as pH increases from 5 to 7 (reaching a minimum of 60 s at pH 7), showing the rate of reaction is fastest at pH 7, which is near the optimum pH. Above pH 7, the time taken increases (120 s at pH 8), indicating a slower rate of reaction. (c)(ii) At pH 3, the reaction will take a very long time, or there will be no reaction at all, because the highly acidic environment denatures the amylase enzyme, altering its active site so it can no longer bind to starch.

(a) blue-black [1 mark]. (b)(i) pH / hydrogen ion concentration [1 mark]. (b)(ii) Any two of: temperature, concentration of amylase, concentration of starch, volume of amylase, volume of starch [2 marks]. (c)(i) Digestion is fastest / time is shortest at pH 7 [1 mark]; because pH 7 is the optimum pH for amylase activity [1 mark]. (c)(ii) Much longer time / infinite time / no digestion [1 mark]; because the highly acidic environment denatures the amylase enzyme [1 mark].

(a) X is hexene because it reacts with aqueous bromine (decolourisation), which is a characteristic test for unsaturated hydrocarbons (alkenes). Y is hexane because alkanes are saturated and do not react with bromine water under normal conditions. (b)(i) Acidified potassium manganate(VII) is an oxidising agent that turns from purple to colourless when it oxidises ethanol. (b)(ii) The oxidation of ethanol produces ethanoic acid. (c) Adding anhydrous copper(II) sulfate to both liquids will distinguish them; water hydrates the salt, turning it from white to blue, while hexane produces no change. (d)(i) Fermentation requires yeast (as a catalyst), anaerobic conditions (absence of oxygen), and a suitable temperature (typically around 30 to 40 degrees Celsius). (d)(ii) Carbon dioxide gas is produced; when bubbled through limewater, it turns the limewater cloudy or milky.

(a) X is hexene and Y is hexane [1 mark]; hexene is an alkene / unsaturated and reacts with bromine [1 mark]. (b)(i) Purple solution turns colourless / decolourises [1 mark]. (b)(ii) Ethanoic acid [1 mark]. (c) Add anhydrous copper(II) sulfate or anhydrous cobalt(II) chloride [1 mark]; water turns copper(II) sulfate from white to blue (or cobalt(II) chloride from blue to pink) but hexane does not [1 mark]. (d)(i) Yeast [1 mark]; anaerobic conditions / absence of oxygen / temperature of 30-40 degrees Celsius [1 mark]. (d)(ii) Carbon dioxide, turns limewater cloudy/milky [1 mark].

(a) The circuit diagram should feature a power source, a switch, an ammeter connected in series with the test wire, and a voltmeter connected in parallel across the length of the test wire being measured. (b)(i) Using Ohm's Law, R = V / I. For 40.0 cm, R = 1.8 V / 0.30 A = 6.0 ohms. For 80.0 cm, R = 1.8 V / 0.15 A = 12.0 ohms. (b)(ii) Resistance increases linearly with length, indicating that resistance is directly proportional to the length of the wire. (c) A suitable safety precaution is opening the switch to turn off the current between readings. This prevents the wire from becoming hot, which could cause burns if touched. (d) As temperature increases, the resistance of metals increases. This means the measured resistance values would be artificially high and inaccurate. This heating effect can be minimized by using a variable resistor to keep currents low and only closing the switch long enough to take a reading.

(a) Ammeter connected in series with the wire [1 mark]; Voltmeter connected in parallel across the wire [1 mark]. (b)(i) 6.0 ohms at 40.0 cm [1 mark]; 12.0 ohms at 80.0 cm [1 mark]. (b)(ii) Resistance is directly proportional to length [1 mark]. (c) Switch off current / open switch between readings [1 mark]; to prevent the wire from overheating / causing burns / changing resistance [1 mark]. (d) Heating increases the resistance / changes the value [1 mark]; minimize by using low currents / keeping current on only briefly [1 mark].

(a)(i) The reaction produces carbon dioxide gas, which escapes from the open flask, causing a decrease in total mass. (a)(ii) The cotton wool plug allows carbon dioxide gas to escape freely while preventing any liquid acid droplets or spray from being carried out of the flask, ensuring the mass loss is due to gas evolution only. (b)(i) Mass loss = initial mass (250.00 g) - mass at time t. For 60 s, loss = 250.00 - 249.40 = 0.60 g. For 120 s, loss = 250.00 - 249.15 = 0.85 g. For 180 s, loss = 250.00 - 249.10 = 0.90 g. (b)(ii) The reaction is complete at 150 s because the mass remains constant at 249.10 g from 150 s to 180 s, indicating no further gas is being produced. (c)(i) Using powdered calcium carbonate increases the rate of reaction. (c)(ii) Powdered calcium carbonate has a much larger surface area than marble chips. This exposes more reactant particles to the acid, resulting in a higher frequency of successful collisions per unit time.

(a)(i) Carbon dioxide gas is released / escapes from the flask [1 mark]. (a)(ii) Allows gas to escape but prevents loss of acid spray/mist [1 mark]. (b)(i) 0.60 g and 0.85 g [1 mark]; 0.90 g [1 mark]. (b)(ii) 150 s [1 mark]; mass remains constant/no further mass loss after 150 s [1 mark]. (c)(i) Rate of reaction increases [1 mark]. (c)(ii) Powder has a larger surface area [1 mark]; leading to a higher frequency of collisions / more collisions per second [1 mark].

(a)(i) The mass of the rock is 48.6 g. (a)(ii) To read the volume accurately, look horizontally at the measuring cylinder so your line of sight is level with the water surface, and read the scale at the bottom of the meniscus. (b)(i) The volume of the rock is the difference in volume: V = V2 - V1 = 73 cm3 - 55 cm3 = 18 cm3. (b)(ii) Density = mass / volume = 48.6 g / 18 cm3 = 2.7 g/cm3. (c) This displacement method is not suitable for sugar because sugar dissolves in water, meaning the volume of water would not increase by the true volume of the sugar. (d) Forgetting to dry the rock means the measured mass will be too high because it includes the mass of the adhering water. Since density = mass / volume, a higher mass value with the same volume results in an overestimation of the rock's density.

(a)(i) 48.6 g [1 mark]. (a)(ii) Read at eye level [1 mark]; to the bottom of the meniscus [1 mark]. (b)(i) Volume = 18 cm3 [1 mark]. (b)(ii) Density = mass / volume [1 mark]; 48.6 / 18 = 2.7 [1 mark]; unit of g/cm3 [1 mark]. (c) Sugar dissolves in water [1 mark]. (d) Density is higher/larger [1 mark]; because measured mass includes the water [1 mark].

(a)(i) The image formed on a screen by a converging lens when u is between f and 2f is real, inverted (both vertically and horizontally), and magnified. (a)(ii) Substituting values: f = (24.0 * 48.0) / (24.0 + 48.0) = 1152 / 72.0 = 16.0 cm. (b) To find the sharpest image, the screen should be moved back and forth slowly across the focus point, or the experiment should be performed in a darkened room to increase contrast. (c)(i) As the object distance increases, the image distance decreases, so the screen must be moved closer to the lens. (c)(ii) Increasing the object distance decreases the magnification, so the image becomes smaller. (d) A thicker converging lens has more highly curved surfaces, which bend light rays more strongly. This stronger refraction brings the rays to a focus closer to the lens, resulting in a shorter focal length.

(a)(i) Inverted [1 mark]; magnified / larger [1 mark]. (a)(ii) Substitution: (24.0 * 48.0) / (24.0 + 48.0) [1 mark]; 16.0 (cm) [1 mark]. (b) Move screen slowly back and forth / use a dark room / align object, lens, and screen [1 mark]. (c)(i) Move screen closer to the lens [1 mark]. (c)(ii) Image is smaller / diminished [1 mark]. (d) Focal length is shorter/smaller [1 mark]; because a thicker lens bends light more strongly [1 mark].

(a)(i) Placing the leaf in boiling water kills the cells, destroying the cell membranes and stopping all metabolic processes. This makes the membranes permeable so iodine can enter the cells. (a)(ii) Boiling in ethanol extracts the green pigment chlorophyll, leaving the leaf white so that colour changes with iodine can be observed easily. Ethanol is highly flammable and its vapours can easily ignite, so a water bath (electric or hot water) is used for heating instead of an open flame. (b)(i) The covered part remains yellow-brown because the black paper blocked light, preventing photosynthesis and starch production. (b)(ii) The uncovered part turns blue-black because it received sunlight, allowing photosynthesis to take place and produce starch, which reacts with iodine. (c) The carbohydrate produced during photosynthesis is converted into sucrose for transport, which travels through the phloem tissue to other parts of the plant.

(a)(i) Kills the leaf / stops reactions / makes cell membranes permeable [1 mark]. (a)(ii) To remove chlorophyll / decolourise the leaf [1 mark]; ethanol is flammable and can catch fire with an open flame [1 mark]. (b)(i) Yellow-brown / orange-brown [1 mark]; no starch was produced because light was absent / no photosynthesis [1 mark]. (b)(ii) Blue-black [1 mark]; starch is present because photosynthesis occurred in the light [1 mark]. (c) Sucrose (accept soluble sugar) [1 mark]; phloem [1 mark].