2025 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

Active transport is an active process that requires energy released from aerobic respiration. Therefore, a high oxygen concentration is required to maximize the rate of respiration. Additionally, respiration is an enzyme-controlled process, meaning the rate of respiration is higher at 25 °C than at 10 °C (as 25 °C is closer to the optimum temperature for plant enzymes). High oxygen combined with higher temperature results in the fastest active transport of nitrate ions. Thus, condition **D** is correct.

1 mark for the correct option D.

When temperature increases beyond the optimum value, the active site of the enzyme changes shape due to the breaking of bonds holding its specific three-dimensional structure. Once the active site's shape is permanently altered, it is no longer complementary to the substrate, meaning the substrate cannot bind and the enzyme is denatured. Thus, the rate of reaction decreases rapidly.

1 mark for the correct option B.

1. Find the number of protons:
The ion has a 3+ charge, which means it has lost 3 electrons compared to its neutral state.
Since the ion has 10 electrons, the neutral atom must have \( 10 + 3 = 13 \) electrons.
Since a neutral atom has the same number of protons and electrons, the atomic (proton) number is 13.

2. Find the number of neutrons:
\( \text{Number of neutrons} = \text{nucleon number} - \text{proton number} = 27 - 13 = 14 \).

Therefore, row **B** is correct.

1 mark for the correct option B.

1. Activation energy (\(E_a\)) is the minimum energy required to start a reaction, which is the difference between the peak transition state energy and the reactant energy:
\( E_a = 380\text{ kJ/mol} - 250\text{ kJ/mol} = +130\text{ kJ/mol} \).

2. The overall energy change (\(\Delta H\)) is the difference between the energy of the products and the energy of the reactants:
\( \Delta H = 110\text{ kJ/mol} - 250\text{ kJ/mol} = -140\text{ kJ/mol} \).

Thus, row **A** is correct.

1 mark for the correct option A.

Alkenes, such as propene, are unsaturated hydrocarbons containing a carbon-carbon double bond (\(\text{C}=\text{C}\)). They undergo addition reactions where the double bond opens up to accept halogen atoms. Aqueous bromine is orange-brown in color, and it is decolored (turns colorless) in the presence of an alkene as the bromine reacts. Hence, row **A** is correct.

1 mark for the correct option A.

1. Calculate the acceleration of the car:
\( a = \frac{v - u}{t} = \frac{25\text{ m/s} - 10\text{ m/s}}{5.0\text{ s}} = \frac{15}{5.0} = 3.0\text{ m/s}^2 \).

2. Use Newton's second law to calculate the average resultant force:
\( F = m \times a = 1200\text{ kg} \times 3.0\text{ m/s}^2 = 3600\text{ N} \).

Therefore, the correct option is **C**.

1 mark for the correct option C.

All electromagnetic waves travel at the speed of light in a vacuum:
\( v = 3.0 \times 10^8\text{ m/s} \).

Using the wave equation:
\( v = f \lambda \implies \lambda = \frac{v}{f} \)

Substitute the values:
\( \lambda = \frac{3.0 \times 10^8\text{ m/s}}{5.0 \times 10^{14}\text{ Hz}} = 0.6 \times 10^{-6}\text{ m} = 6.0 \times 10^{-7}\text{ m} \).

Therefore, option **B** is correct.

1 mark for the correct option B.

1. Calculate the combined resistance of the parallel combination (\(R_p\)):
\( \frac{1}{R_p} = \frac{1}{4.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{3 + 1}{12.0\ \Omega} = \frac{4}{12.0\ \Omega} \)
\( R_p = \frac{12.0}{4} = 3.0\ \Omega \).

2. Calculate the total equivalent resistance of the circuit (\(R_{\text{total}}\)):
Since \(R_p\) is in series with the \(3.0\ \Omega\) resistor:
\( R_{\text{total}} = R_p + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega \).

3. Calculate the total current in the circuit (which is the current flowing through the series resistor):
\( I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A} \).

Hence, the correct option is **D**.

1 mark for the correct option D.

When a potato cylinder is placed in a solution with the same water potential as its cell sap, there is no net movement of water by osmosis, so the percentage change in mass is \(0\\%\). Since the mass increased (positive change) at \(0.2\text{ mol/dm}^3\) and decreased (negative change) at \(0.4\text{ mol/dm}^3\), the concentration at which no mass change occurs (i.e. where the water potentials are equal) must lie between \(0.2\text{ mol/dm}^3\) and \(0.4\text{ mol/dm}^3\).

1 mark for the correct option B.

As temperature increases up to the optimum temperature, the kinetic energy of both enzyme and substrate molecules increases. This causes them to move faster, resulting in more frequent collisions, and a higher proportion of these collisions have sufficient energy to overcome the activation energy barrier, leading to a higher rate of reaction.

1 mark for the correct option C.

Arteries (P) have thick muscular walls to withstand high blood pressure, a narrow lumen, and no valves. Veins (Q) have thinner walls, a wide lumen, and valves to prevent the backflow of slow-moving blood. Capillaries (R) have walls that are only one cell thick (no muscular layer) to facilitate exchange of materials, a very narrow lumen, and no valves.

1 mark for the correct option A.

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Because they have the same proton number, neutral atoms of isotopes have the same number of electrons and the same electronic configuration (including the number of outer-shell electrons). Since chemical properties depend on the arrangement of outer-shell electrons, isotopes have the same chemical properties.

1 mark for the correct option B.

In this reaction, iodide ions (\(\text{I}^-\)) lose electrons to form iodine (\(\text{I}_2\)). Oxidation is defined as the loss of electrons, so iodide ions are oxidized. Chlorine (\(\text{Cl}_2\)) gains electrons to form chloride ions (\(\text{Cl}^-\)), so it is reduced. Therefore, chlorine acts as the oxidizing agent, and iodide ions act as the reducing agent.

1 mark for the correct option A.

First, calculate the work done (change in gravitational potential energy, \(\Delta E_p\)):
\[W = mgh = 500\text{ kg} \times 9.8\text{ m/s}^2 \times 20\text{ m} = 98000\text{ J}\]
Next, calculate the power output:
\[P = \frac{W}{t} = \frac{98000\text{ J}}{8.0\text{ s}} = 12250\text{ W}\]

1 mark for the correct option B.

When light enters an optically denser medium like glass from air, its speed decreases. The frequency of a wave is determined by the source and remains unchanged during refraction. Since speed is given by the wave equation \(v = f\lambda\), and \(f\) is constant while \(v\) decreases, the wavelength \(\lambda\) must also decrease.

1 mark for the correct option A.

Let \(R\) be the resistance of each individual resistor.
1. The resistance of the two parallel resistors is:
\[R_p = \frac{R \times R}{R + R} = 0.5R\]
2. The total resistance of the arrangement (series combination of the parallel pair and the third resistor) is:
\[R_{\text{total}} = R_p + R = 0.5R + R = 1.5R\]
3. Since \(R_{\text{total}} = 9.0\\ \Omega\):
\[1.5R = 9.0\\ \Omega \implies R = \frac{9.0}{1.5} = 6.0\\ \Omega\]

1 mark for the correct option C.

Osmosis is the net movement of water molecules from a region of higher water potential (inside the cell sap) to a region of lower water potential (the concentrated salt solution) through a partially permeable membrane. Since water leaves the plant cells, they lose turgor pressure and become flaccid, leading to plasmolysis where the cell membrane pulls away from the cell wall.

1 mark for selecting D. Correctly identifying that water moves out of the cells down a water potential gradient and causes them to become plasmolysed.

At high temperatures (such as \(60\ ^\circ\text{C}\)), enzymes (which are proteins) are denatured. This means the bonds maintaining their specific three-dimensional structure are broken, causing the active site to change shape. The substrate can no longer fit into the modified active site, and the reaction ceases.

1 mark for selecting C. Correctly explaining that denaturation refers to the change in shape of the enzyme's active site, preventing substrate binding.

As Group VII (the halogens) is descended, the atoms get larger, meaning the outer shell is further from the nucleus, so it is harder to attract an incoming electron, which decreases reactivity. At the same time, the diatomic molecules become larger, resulting in stronger intermolecular forces of attraction, which increases the melting point.

1 mark for selecting C. Correctly identifying both down-group trends for Group VII halogens.

In an endothermic reaction, energy is absorbed from the surroundings, which causes the temperature of the surroundings to decrease. Because the system absorbs energy, the products have a higher energy level than the reactants.

1 mark for selecting A. Correctly identifying that endothermic reactions cause a temperature decrease in the surroundings and result in products with higher energy than reactants.

Metal R is the most reactive because it reacts vigorously with cold water (typical of alkali metals like potassium or sodium). Metal S is the second most reactive because it reacts slowly with cold water. Metal P is less reactive because it does not react with cold water but does react with dilute acid (typical of metals like zinc or iron). Metal Q is the least reactive as it does not react with either cold water or dilute acid (typical of unreactive metals like copper or gold). This gives the order: R to S to P to Q.

1 mark for selecting A. Correctly deducing the reactivity order from the experimental observations.

Alkenes are unsaturated hydrocarbons containing a \(C=C\) double bond, which allows them to undergo addition reactions, such as reacting with aqueous bromine to turn it from orange/brown to colourless. Alkanes are saturated (only containing \(C-C\) single bonds) and do not react with aqueous bromine under normal conditions, so the orange colour remains.

1 mark for selecting B. Correctly identifying that alkenes decolourise bromine water, while alkanes do not.

Plastic is optically denser than air. When light travels from a less optically dense medium (air) to a more optically dense medium (plastic), its speed decreases (it travels slower). This reduction in speed causes the ray to bend towards the normal, making the angle of refraction less than the angle of incidence (\(40\ ^\circ\)).

1 mark for selecting D. Correctly identifying that light slows down in plastic and bends towards the normal, resulting in a smaller angle of refraction.

According to the Combined Science syllabus, the combined resistance of two resistors in parallel is always less than the resistance of either resistor by itself. Here, the individual resistors are \(6.0\ \Omega\) and \(12\ \Omega\). The combined resistance must be less than the smaller resistance, which is \(6.0\ \Omega\). Among the options, only \(4.0\ \Omega\) is less than \(6.0\ \Omega\). (Alternatively, using the formula: \(1/R_{\text{total}} = 1/6.0 + 1/12 = 3/12\), which gives \(R_{\text{total}} = 4.0\ \Omega\).)

1 mark for selecting A. Correctly applying the rule that combined resistance in parallel is less than the individual resistances, or correctly calculating the equivalent resistance.

Water moves from a region of higher water potential (less negative, \(-100\text{ kPa}\)) to a region of lower water potential (more negative, \(-400\text{ kPa}\)) by osmosis. Therefore, the net movement of water molecules is into the cell. Since it is a plant cell, it has a strong, rigid cellulose cell wall that prevents it from bursting. Instead, the cell becomes turgid as water enters.

Award 1 mark for the correct option A.

At \(50^\circ\text{C}\), which is above the optimum temperature for most human enzymes, the active site of the enzyme changes shape because the enzyme molecule denatures. As a result, the substrate molecule is no longer complementary to the active site, and the rate of reaction decreases.

Award 1 mark for the correct option C.

In Group I (alkali metals), as you go down the group:
1. The melting point decreases, so Period 3 \(X\) has a higher melting point than Period 4 \(Y\).
2. The reactivity increases, so Period 4 \(Y\) reacts more rapidly with water than Period 3 \(X\). This means \(X\) reacts less rapidly with water than \(Y\).

Award 1 mark for the correct option B.

First, calculate the combined resistance \(R_p\) of the two resistors in parallel:

\(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\ \Omega^{-1}\)

\(R_p = \frac{12}{5} = 2.4\ \Omega\)

Next, use Ohm's law to find the total current \(I\):

\(I = \frac{V}{R_p} = \frac{12\text{ V}}{2.4\ \Omega} = 5.0\text{ A}\)

Alternatively, calculate the individual currents:

\(I_1 = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\)

\(I_2 = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\)

\(I_{\text{total}} = 3.0\text{ A} + 2.0\text{ A} = 5.0\text{ A}\).

Award 1 mark for the correct option C.

A mature plant palisade mesophyll cell contains a cell wall, a nucleus, and chloroplasts. A mature human red blood cell has no cell wall, no chloroplasts, and loses its nucleus during maturation to maximize space for haemoglobin. Option B is incorrect because cell membranes and cytoplasm are present in red blood cells. Option C is incorrect because cytoplasm is present in red blood cells. Option D is incorrect because cell membranes are present in red blood cells.

Award 1 mark for the correct option A.

Incomplete combustion of carbon-containing fuels produces carbon monoxide (\(\text{CO}\)). Carbon monoxide is highly toxic because it binds irreversibly to haemoglobin in red blood cells, thereby reducing the oxygen-carrying capacity of the blood.

Award 1 mark for the correct option B.

To prepare an insoluble salt like lead(II) sulfate by precipitation, we must mix two soluble salts: lead(II) nitrate and sodium sulfate. Once mixed, a precipitate of lead(II) sulfate forms. The mixture is filtered, the residue (lead(II) sulfate) is washed with distilled water, and then dried. Evaporating to dryness (Option B) would leave soluble sodium nitrate mixed with the lead(II) sulfate. Lead(II) oxide (Option C) and lead(II) carbonate (Option D) are insoluble bases/salts and do not react effectively with sulfuric acid to make a pure insoluble salt because the reaction stops once the insoluble sulfate coats the starting material.

Award 1 mark for the correct option A.

The electromagnetic spectrum in order of decreasing wavelength is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. The radiation with a wavelength longer than visible light but shorter than microwaves is infrared radiation. A common use of infrared radiation is in remote controls and optical fibre communication.

Award 1 mark for the correct option A.

As the temperature increases up to the optimum, the kinetic energy of both the enzyme and substrate molecules increases. This causes them to move faster, leading to more frequent collisions per unit time and thus a higher rate of enzyme-substrate complex formation.

C is the correct answer because an increase in temperature directly increases the kinetic energy of the molecules, resulting in more frequent collisions. A, B, and D are scientifically incorrect statements in this context.

When the concentration of carbon dioxide inside the sealed jar becomes constant, the plant has reached its compensation point. At this point, the rate of carbon dioxide uptake by photosynthesis is exactly equal to the rate of carbon dioxide release by respiration.

B is correct because equal rates of photosynthesis and respiration result in no net change in carbon dioxide concentration. A, C, and D are incorrect as they do not describe the biological balance responsible for a constant carbon dioxide level.

Down Group VII, the molecular size increases, which leads to stronger intermolecular forces of attraction. Therefore, the melting and boiling points increase down the group. Bromine is a red-brown liquid at room temperature (not a solid), reactivity decreases down the group (so fluorine is more reactive than chlorine), and a less reactive halogen (iodine) cannot displace a more reactive one (bromine).

D is correct because boiling points increase down Group VII. A is incorrect because bromine is a liquid. B is incorrect because reactivity decreases down the group. C is incorrect because iodine is less reactive than bromine and cannot displace it.

The total resistance \(R\) of the two parallel resistors is calculated as: \(\frac{1}{R} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3 + 1}{12.0} = \frac{4}{12.0}\), which gives \(R = 3.0\ \Omega\). Using Ohm's Law: \(I = \frac{V}{R} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\). Alternatively, the currents in the individual branches are \(\frac{6.0}{4.0} = 1.5\text{ A}\) and \(\frac{6.0}{12.0} = 0.5\text{ A}\), summing to \(2.0\text{ A}\).

D is correct because the parallel combination equivalent resistance is \(3.0\ \Omega\), yielding \(2.0\text{ A}\). A is incorrect (uses series resistance instead). B and C only identify single-branch currents.

The reaction of ethene (unsaturated) with steam to produce ethanol is an hydration reaction, which is a type of addition reaction. This reaction is catalysed by phosphoric acid (an acid catalyst). Yeast is used in the fermentation of glucose, which is a different method of ethanol production.

B is correct because the hydration of ethene is an addition reaction catalyzed by an acid (such as phosphoric acid). A is incorrect because yeast is not the catalyst for this reaction. C and D are incorrect because it is not a substitution reaction.

Using the wave equation: \(v = f \lambda\), we have \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 5.0 \times 10^{-7}\text{ m}\). This wavelength (equivalent to \(500\text{ nm}\)) is within the range of visible light (approximately \(400\text{ nm}\) to \(700\text{ nm}\)).

A is correct because the computed wavelength is \(5.0 \times 10^{-7}\text{ m}\) and this corresponds to the visible light spectrum. B is incorrect because this wavelength is not in the infrared region. C and D are incorrect because they use the inverted formula.

Leached fertilisers supply nutrients causing rapid algae growth (algal bloom). This layer of algae blocks light from reaching submerged aquatic plants, causing them to die. Aerobic bacteria decompose the dead plants, replicating quickly and consuming dissolved oxygen. The resulting oxygen depletion causes fish and other aquatic animals to suffocate and die.

A is correct because it correctly describes the biological chain of events in eutrophication. B, C, and D contain incorrect ecological relationships and outcomes.

In any reaction, bond-breaking absorbs energy (endothermic) and bond-making releases energy (exothermic). For a reaction to be overall exothermic, the energy released when new bonds are formed must be greater than the energy absorbed to break existing bonds. Consequently, net energy is transferred to the surroundings, raising their temperature.

B is correct because exothermic reactions release more energy during bond-making than is absorbed during bond-breaking. A, C, and D describe endothermic processes or properties.

(a) (i) Double circulation is a system where blood travels through the heart twice during one complete cycle through the body (pulmonary and systemic circuits).
(ii) Exercise increases the rate of cellular respiration in muscles. To support this, the heart pumps faster (increased heart rate) to deliver oxygen and glucose to muscle tissues and remove carbon dioxide.
(b) Capillaries are adapted by having walls that are only one cell thick (minimising diffusion distance) and a highly branched network with a narrow lumen (slowing down blood to allow time for diffusion and providing a large total surface area).

(a) (i) Blood passes through the heart twice [1]; for one complete circuit of the body [1].
(ii) Heart rate increases [1]; to deliver more oxygen/glucose to muscles [1]; for increased rate of respiration [1].
(b) Wall is one cell thick [1]; reduces diffusion distance [1]; Highly branched / large surface area / narrow lumen [1]; slows blood flow / increases diffusion rate [1].

(a) (i) Draw three carbon atoms with one double bond between two of them. Ensure the first carbon has 2 hydrogens, the middle has 1, and the end carbon has 3 (total 6 hydrogens, all single bonds drawn). [2 marks]
(ii) General formula for alkenes is \( C_nH_{2n} \). [1 mark]
(b) An addition reaction occurs where bromine atoms add across the double bond, decolourising the bromine water from orange-brown to colourless. [3 marks]
(c) Alkanes are saturated and do not have double bonds, meaning they cannot undergo addition reactions. They require UV light to undergo substitution reactions with halogens instead. [3 marks]

(a) (i) Correct structure showing C=C and C-C bonds [1]; all 6 hydrogen atoms with single bonds drawn [1].
(ii) \( C_nH_{2n} \) [1].
(b) Addition [1]; from orange / red-brown / yellow [1]; to colourless [1] (reject: clear).
(c) Propane is saturated / an alkane [1]; contains only single C-C bonds [1]; no C=C double bond to undergo addition [1].

(a) Acceleration \( a = \frac{v - u}{t} = \frac{6.0 \text{ m/s} - 0 \text{ m/s}}{3.0 \text{ s}} = 2.0 \text{ m/s}^2 \).
(b) Resultant force \( F = m \times a = 0.50 \text{ kg} \times 2.0 \text{ m/s}^2 = 1.0 \text{ N} \).
(c) Kinetic energy \( K.E. = \frac{1}{2} m v^2 = 0.5 \times 0.50 \text{ kg} \times (6.0 \text{ m/s})^2 = 0.25 \times 36 = 9.0 \text{ J} \).
(d) The law of conservation of energy states that the total energy of an isolated system remains constant; energy can change forms but cannot be created or destroyed.

(a) Formula \( a = \frac{\Delta v}{t} \) or correct numbers used [1]; 2.0 m/s² (with correct unit) [1].
(b) Formula \( F = ma \) or correct numbers [1]; 1.0 [1]; unit: N / Newtons [1].
(c) Formula \( K.E. = \frac{1}{2}mv^2 \) or correct numbers [1]; 9.0 J [1].
(d) Energy cannot be created or destroyed [1]; only transformed/transferred from one form to another [1].

(a) Enzymes are biological catalysts that speed up biochemical reactions without being consumed in the process. They are made of protein.
(b) As temperature increases towards the optimum, molecules gain kinetic energy and move faster. This increases the collision frequency between starch molecules and the active site of amylase, resulting in more enzyme-substrate complexes per second.
(c) At 75 °C, the excess thermal energy vibrates the protein structure, breaking crucial hydrogen and ionic bonds. This denatures the enzyme, altering the active site's shape permanently, so starch can no longer bind.

(a) Biological catalyst [1]; made of protein [1].
(b) Molecules gain kinetic energy / move faster [1]; higher frequency of collisions [1]; more successful collisions / more enzyme-substrate complexes formed [1].
(c) Enzyme / active site is denatured [1]; hydrogen / internal bonds broken [1]; active site shape changes permanently [1]; substrate can no longer fit/bind [1] (reject: enzyme is killed).

(a) Zinc carbonate reacts with sulfuric acid to produce zinc sulfate, carbon dioxide gas, and water.
(b) Excess means that all acid has reacted, so no more gas is produced (fizzing stops) and solid remains visible.
(c) The steps are: filtration (to remove unreacted solid), partial evaporation (heating to concentration), cooling (to allow crystallisation), and filtration/drying (drying between filter papers).
(d) Zinc metal reacts very vigorously with acid, releasing flammable hydrogen gas. Zinc carbonate reacts more safely at a controlled rate, producing harmless, non-flammable carbon dioxide.

(a) zinc carbonate + sulfuric acid [1] -> zinc sulfate + carbon dioxide + water [1] (all reactants and products must be correct for the marks).
(b) Fizzing / bubbling stops OR solid remains undissolved at the bottom [1].
(c) Filter the mixture (to remove excess zinc carbonate) [1]; heat the filtrate / solution to saturation point / evaporate some water [1]; leave to cool / crystallise [1]; dry crystals between pieces of filter paper / in a warm oven [1].
(d) Reacting zinc metal with acid produces hydrogen gas [1]; which is highly flammable / explosive / the reaction is too violent [1].

(a) Combined resistance of parallel resistors: \( \frac{1}{R_p} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \), so \( R_p = 6.0 \ \Omega \).
(b) Total resistance \( R_T = R_{\text{series}} + R_p = 4.0 + 6.0 = 10.0 \ \Omega \).
(c) Using Ohm's Law: \( I = \frac{V}{R_T} = \frac{24 \text{ V}}{10.0 \ \Omega} = 2.4 \text{ A} \).
(d) The 4.0 ohm resistor experiences the total current of 2.4 A. Since the two parallel resistors are identical, the current splits equally, so each 12 ohm resistor has 1.2 A, which is half of the total current.

(a) Correct formula used (\( R_p = \frac{R_1 R_2}{R_1 + R_2} \) or equivalent) [1]; 6.0 \( \Omega \) [1].
(b) Adding series resistor (\( 4.0 + R_p \)) [1]; 10.0 \( \Omega \) [1].
(c) Formula \( I = \frac{V}{R} \) [1]; calculation shown with correct numbers [1]; 2.4 A (with unit) [1].
(d) Current through the 4.0 ohm resistor is the total current [1]; which splits equally so it is twice the current through each 12 ohm resistor [1].

(a) Insect-pollinated pollen is sticky/spiky (clings to insects) and larger/heavier (keeps it from drifting in wind). Wind-pollinated pollen is smooth, light, and produced in huge numbers.
(b) Fertilization occurs specifically inside the ovule inside the ovary of the flower.
(c) Pollination is the transfer of pollen from the male anther to the female stigma. Following this, the pollen grain grows a pollen tube down the style, carrying the male nucleus to the ovary and entering the ovule through the micropyle.

(a) Sticky / spiky outer wall [1]; adheres easily to insect bodies [1]; Larger / heavier grains [1]; prevents wind dispersal / ensures transport by insects [1].
(b) Ovule [1] (accept: ovary).
(c) Definition: Transfer of pollen from anther to stigma [1]; pollen tube grows down the style [1]; enters the ovary / ovule [1]; through the micropyle [1].

(a) (i) Protons = 11, Electrons = 11, Neutrons = 23 - 11 = 12.
(ii) Protons = 17, Neutrons = 35 - 17 = 18, Electrons = 17 + 1 = 18 (since it has a 1- charge).
(b) Sodium (2,8,1) loses 1 valence electron to achieve a stable octet, forming \( \text{Na}^+ \). Chlorine (2,8,7) gains this electron to form \( \text{Cl}^- \). The electrostatic attraction between the positive sodium ions and negative chloride ions forms strong ionic bonds in a giant lattice.
(c) Ionic compounds typically have high melting/boiling points, are soluble in water, and conduct electricity in liquid/aqueous states but not as solids.

(a) (i) Protons = 11, Neutrons = 12, Electrons = 11 [1.5] (all three must be correct for full credit, or 0.5 per correct value).
(ii) Protons = 17, Neutrons = 18, Electrons = 18 [1.5] (all three must be correct, or 0.5 per correct value).
(b) Sodium atom loses one electron [1]; chlorine atom gains one electron [1]; forms oppositely charged ions (\( \text{Na}^+ \) and \( \text{Cl}^- \)) [1]; held together by strong electrostatic attraction [1].
(c) High melting / boiling point [1]; conducts electricity when molten or in aqueous solution [1].

**(a)(i)**
Using the formula for acceleration:
\(a = \frac{v - u}{t} = \frac{12\text{ m/s} - 0\text{ m/s}}{3.0\text{ s}} = 4.0\text{ m/s}^2\).

**(a)(ii)**
The total mass of the system \(m = 4.0\text{ kg} + 1.0\text{ kg} = 5.0\text{ kg}\).
Using the formula for kinetic energy:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 5.0\text{ kg} \times (12\text{ m/s})^2 = 0.5 \times 5.0 \times 144 = 360\text{ J}\).

**(b)(i)**
Using the formula for gravitational potential energy:
\(\Delta E_p = mgh = 5.0\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 735\text{ J}\).

**(b)(ii)**
Using the formula for efficiency:
\(\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%\)
\(\text{efficiency} = \frac{735\text{ J}}{4800\text{ J}} \times 100\% = 15.3125\% \approx 15.3\%\) (or \(15\%\)).

**(a)(i)**
- 1 mark for showing correct formula or substitution: \(\frac{12}{3.0}\)
- 1 mark for correct value: \(4.0\) (accept \(4\))
- 1 mark for correct unit: \(\text{m/s}^2\) (or \(\text{m/s/s}\) or \(\text{m s}^{-2}\))

**(a)(ii)**
- 1 mark for correct substitution using total mass \(5.0\text{ kg}\): \(\frac{1}{2} \times 5.0 \times 12^2\)
- 1 mark for correct final value: \(360\)

**(b)(i)**
- 1 mark for correct substitution: \(5.0 \times 9.8 \times 15\)
- 1 mark for correct final value: \(735\)

**(b)(ii)**
- 1 mark for correct formula/substitution: \(\frac{735}{4800} \times 100\) (allow error carried forward from (b)(i))
- 1 mark for correct final percentage: \(15.3\%\) (accept \(15\%\) or \(15.31\%\) or \(0.153\))

An exemplar planning response should address all prompt requirements:

1. **Apparatus**:
- Beaker or cup to hold the hydrochloric acid.
- Forceps or plastic spoon to safely transfer the agar blocks into and out of the acid.
- Paper towels to dry the blocks.

2. **Method**:
- Use the ruler, cutting tile, and scalpel to cut the large pink agar block into at least three cubes of different dimensions, such as: \(1.0\text{ cm} \times 1.0\text{ cm} \times 1.0\text{ cm}\), \(2.0\text{ cm} \times 2.0\text{ cm} \times 2.0\text{ cm}\), and \(3.0\text{ cm} \times 3.0\text{ cm} \times 3.0\text{ cm}\).
- Measure out equal volumes of dilute hydrochloric acid into three separate beakers.
- Place one cube into each beaker and start the stop-watch immediately.
- Record the time taken for each cube to turn completely colourless.

3. **Measurements**:
- Measure the dimensions of each cube to calculate their surface area, volume, and surface area to volume (SA:V) ratio.
- Measure the time taken (\(t\)) for the pink colour to completely disappear.
- Alternatively, leave the cubes in the acid for a fixed time (e.g. 10 minutes), cut them in half, measure the distance of the colourless zone, and calculate the rate of diffusion as: \(\text{rate} = \frac{\text{distance of penetration}}{\text{time}}\).

4. **Constant Variables**:
- Temperature of the acid and agar blocks.
- Concentration of the hydrochloric acid used.
- Volume of acid in each beaker (ensuring the cubes are completely submerged).

5. **Processing Results**:
- Plot a graph of the rate of diffusion (or time taken for complete neutralisation) on the vertical axis against the surface area to volume ratio on the horizontal axis to draw a conclusion.

**Apparatus (Max 2 marks)**
- [1] Beaker / container to hold the acid.
- [1] Forceps / plastic spoon to transfer blocks OR paper towels to blot blocks dry.

**Method (Max 3 marks)**
- [1] Cut at least three different sizes of agar cubes (e.g. \(1.0\text{ cm}\), \(2.0\text{ cm}\), and \(3.0\text{ cm}\) sides) using the ruler and scalpel.
- [1] Place cubes in the dilute hydrochloric acid.
- [1] Observe the colour change and record the time taken for the pink colour to completely disappear.

**Measurements (Max 2 marks)**
- [1] Measure the time using a stop-watch.
- [1] Calculate the surface area, volume, and SA:V ratio of each cube OR measure the depth of penetration of acid in a fixed time.

**Constant Variables (Max 2 marks)**
- [1] Keep the concentration of the hydrochloric acid constant.
- [1] Keep the temperature of the solutions constant.

**Processing Results (Max 1 mark)**
- [1] Plot a graph of the rate of diffusion (or time taken) against the surface area to volume ratio.

(a) (i) The light blue precipitate insoluble in excess sodium hydroxide indicates the presence of copper(II) ions, \(Cu^{2+}\).
(ii) The white precipitate formed with nitric acid and silver nitrate indicates the presence of chloride ions, \(Cl^-\).
(iii) The combination of \(Cu^{2+}\) and \(Cl^-\)

(a) (i) [1] copper(II) / \(Cu^{2+}\)
(ii) [1] chloride / \(Cl^-\)
(iii) [1] \(CuCl_2\)

(b) (i) [2] diagram showing:
- [1] filter funnel with filter paper correctly positioned
- [1] beaker or conical flask positioned below the funnel to collect the filtrate, both parts clearly labelled.
(ii) [2]
- [1] Rinse the beaker with distilled water and pour the washings through the filter paper.
- [1] Dry the precipitate in a warm oven / incubator / desiccator before weighing.
(iii) [1] \(1.84\text{ g} - 1.18\text{ g} = 0.66\text{ g}\)

(c) [2] Any two from:
- volume of solution Y used (e.g. \(50.0\text{ cm}^3\))
- concentration of solution Y
- concentration of the sodium hydroxide used
- temperature of the solutions

(a) (i) A protractor is used to measure angles.
(ii) To ensure accuracy, the student should use a sharp pencil to draw thin rays, or mark the centre of the ray precisely.
(b) (i) Graph plotting:
- Label axes with units: Angle of refraction \(r / ^{\circ}\) on the y-axis, and Angle of incidence \(i / ^{\circ}\) on the x-axis.
- Use scales that cover at least half of the grid.
- Plot points accurately.
(ii) Draw a smooth curve of best fit passing close to all the points.
(iii) Read the y-value corresponding to \(x = 50\). This is typically around \(32^{\circ}\).
(c) The ray box lamp can become very hot during use. To minimise the risk of burns, the student should avoid touching the metal housing or the bulb directly, and turn off the ray box when it is not in use.

(a) (i) [1] protractor
(ii) [1] Any one from:
- Use a sharp pencil to draw the lines / normal.
- Mark the centre of the light ray with small dots.
- Align the baseline of the protractor accurately with the boundary/normal.

(b) (i) [3]
- [1] Correctly labelled axes with quantities and units (\(r / ^{\circ}\) and \(i / ^{\circ}\)).
- [1] Linear scales covering at least half of the grid.
- [1] All 6 points plotted accurately to within half a small square.
(ii) [1] Smooth, single-line curve of best fit drawn.
(iii) [1] Correct value read from the plotted curve at \(i = 50^{\circ}\) (accept range \(31.0^{\circ}\) to \(33.0^{\circ}\) depending on graph).

(c) [2]
- [1] Hazard: Ray box gets hot / bulb can burn fingers.
- [1] Precaution: Do not touch the bulb / switch off when not taking readings.

(a) (i) Polystyrene is an excellent thermal insulator, which minimises heat transfer/loss to the surrounding air, making the temperature measurement more accurate.
(ii) A thermometer is used to measure temperature.
(b) (i) \(\Delta T = T_{\text{max}} - T_{\text{initial}} = 35.5 - 21.0 = 14.5^{\circ}\text{C}\).
(ii) Stirring ensures that the reactants are thoroughly mixed for a complete reaction and that the heat is distributed evenly throughout the solution.
(c) (i) As the mass of zinc increases from \(0.2\text{ g}\) to \(0.8\text{ g}\), the temperature rise increases proportionally. Above \(0.8\text{ g}\), the temperature rise remains constant at \(14.5^{\circ}\text{C}\).
(ii) Zinc is in excess because adding more of it (above \(0.8\text{ g}\)) does not cause any further increase in temperature, indicating that all of the copper(II) sulfate has already been fully reacted.
(iii) Exothermic reactions release thermal energy, causing the temperature to rise.
(iv) An electronic balance is used to measure mass.

(a) (i) [1] Polystyrene is a (good) thermal insulator / reduces heat loss to surroundings.
(ii) [1] thermometer

(b) (i) [1] \(14.5\text{ (}^{\circ}\text{C)}\)
(ii) [1] To ensure reactants are mixed / heat is distributed evenly / reaction goes to completion.

(c) (i) [2]
- [1] As mass of zinc increases, temperature rise increases (up to \(0.8\text{ g}\)).
- [1] Above \(0.8\text{ g}\) the temperature rise remains constant.
(ii) [2]
- [1] Zinc is in excess.
- [1] Explanation: No further temperature increase occurs after adding more than \(0.8\text{ g}\) of zinc (showing all copper(II) sulfate has reacted).
(iii) [1] exothermic
(iv) [1] (electronic) balance / scales