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(a) Insect-pollinated flowers have large, brightly coloured petals, produce nectar/scent, and have sticky/spiky pollen grains. Wind-pollinated flowers have small, dull petals, no nectar/scent, and have smooth, light pollen grains that are easily carried by wind.
(b) Fertilization occurs in the oviduct (fallopian tube). The zygote divides to form an embryo, which is moved along the oviduct by cilia and peristalsis until it reaches the uterus, where it embeds in the thick uterus lining (implantation).
(c) Estrogen stimulates the repair and thickening of the uterus lining during the first half of the menstrual cycle. Progesterone maintains the thickness of the uterus lining in preparation for a fertilized egg; a drop in progesterone levels leads to the breakdown of the lining (menstruation).

(a) Max 3 marks. Any three contrasting pairs:
- Insect-pollinated: large/brightly coloured petals VS Wind-pollinated: small/green/dull petals (1)
- Insect-pollinated: nectar/scent present VS Wind-pollinated: absent (1)
- Insect-pollinated: sticky/large/heavy pollen VS Wind-pollinated: light/smooth/abundant pollen (1)
- Insect-pollinated: stamens/stigma enclosed within petals VS Wind-pollinated: stamens/feathery stigma hanging outside (1)

(b) Max 3 marks:
- Site of fertilization: oviduct / fallopian tube (1)
- Movement: moved by cilia / peristalsis / zygote divides into embryo (1)
- Implantation: embeds / sinks into the lining of the uterus (1)

(c) Max 4 marks:
- Estrogen causes repair / thickening of the uterus lining (1)
- Estrogen is secreted by the ovaries / follicles (1)
- Progesterone maintains the uterus lining (1)
- Fall in progesterone level causes menstruation / breakdown of lining (1)

(a) First, find the equivalent resistance of the two \(6.0\ \Omega\) resistors in parallel:
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0}\)
\(R_p = 3.0\ \Omega\)
Now, add the series \(4.0\ \Omega\) resistor:
\(R_{\text{total}} = 4.0\ \Omega + 3.0\ \Omega = 7.0\ \Omega\)

(b) Using Ohm's Law:
\(I = \frac{V}{R} = \frac{12.0}{7.0} \approx 1.71\text{ A}\)

(c) Power dissipated in the \(4.0\ \Omega\) resistor:
\(P = I^2 R = (1.714)^2 \times 4.0 \approx 11.8\text{ W}\) (or \(12\text{ W}\))

(d) A fuse contains a thin wire that melts and breaks the circuit if the current exceeds its rating. Since the normal operating current is \(1.71\text{ A}\), a \(5.0\text{ A}\) fuse is suitable because it is higher than the operating current (so it won't melt during normal use) but will blow if a fault causes the current to exceed \(5.0\text{ A}\).

(a) Max 3 marks:
- Formula for parallel resistors or correct calculation of parallel part: \(R_p = 3.0\ \Omega\) (1)
- Adding series resistance: \(R_{\text{total}} = R_p + 4.0\) (1)
- Correct final answer with units: \(7.0\ \Omega\) (1)

(b) Max 2 marks:
- Formula: \(I = V / R\) (1)
- Correct calculation: \(1.7\text{ A}\) (accept \(1.71\text{ A}\)) (1)

(c) Max 2 marks:
- Formula: \(P = I^2 R\) or \(P = V I\) (with calculated voltage drop across \(4.0\ \Omega\) resistor) (1)
- Correct calculation: \(11.7\text{ W}\) to \(12\text{ W}\) (1)

(d) Max 3 marks:
- Fuse wire melts / breaks circuit (1)
- When current exceeds the fuse rating (1)
- \(5.0\text{ A}\) is higher than the operating current of \(1.7\text{ A}\) but low enough to protect against faults (1)

(a) Acceleration \(a = \frac{v - u}{t} = \frac{20 - 0}{8.0} = 2.5\text{ m/s}^2\)

(b) Force \(F = m a = 1200 \times 2.5 = 3000\text{ N}\)

(c) Distance travelled is the area under the speed-time graph. Since it is uniform acceleration from rest, the shape is a triangle:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8.0 \times 20 = 80\text{ m}\)

(d) Kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times (20)^2 = 240,000\text{ J}\) (or \(240\text{ kJ}\)). The remaining chemical energy is lost as thermal energy (heat) to the surroundings due to friction/air resistance and engine inefficiency.

(a) Max 2 marks:
- Formula: \(a = \Delta v / t\) (1)
- Correct answer: \(2.5\text{ m/s}^2\) (1)

(b) Max 2 marks:
- Formula: \(F = ma\) (1)
- Correct answer: \(3000\text{ N}\) (1)

(c) Max 3 marks:
- Statement: Area under the speed-time graph (1)
- Method: \(\frac{1}{2} \times \text{base} \times \text{height}\) (1)
- Correct answer: \(80\text{ m}\) (1)

(d) Max 3 marks:
- Formula: \(E_k = \frac{1}{2}mv^2\) (1)
- Correct answer: \(240,000\text{ J}\) / \(240\text{ kJ}\) (1)
- Energy loss: transferred to thermal energy of surroundings / air resistance / friction (1)

(a)(i) For Chlorine-35, number of neutrons \(\text{A} = 35 - 17 = 18\). For Chlorine-37, the number of electrons \(\text{B}\) in a neutral atom is equal to the proton number, so \(\text{B} = 17\).
(ii) They have the same chemical properties because they have the same number of outer-shell electrons (same electronic configuration).
(b) A sodium atom (electronic configuration 2,8,1) loses one outer-shell electron to achieve a stable outer shell, forming a sodium ion, \(\text{Na}^+\). A chlorine atom (electronic configuration 2,8,7) gains this one electron to complete its outer shell, forming a chloride ion, \(\text{Cl}^-\). These oppositely charged ions attract each other through strong electrostatic forces of attraction.
(c) Sodium chloride has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions, which require a large amount of thermal energy to break. Chlorine gas consists of simple covalent molecules with weak intermolecular forces between them, which require very little energy to overcome.

(a)(i) Max 2 marks:
- \(\text{A} = 18\) (1)
- \(\text{B} = 17\) (1)
(a)(ii) Max 1 mark:
- Same number of outer shell electrons / same electronic configuration (1)

(b) Max 4 marks:
- Sodium loses 1 electron (1)
- Chlorine gains 1 electron (1)
- Forms \(\text{Na}^+\) and \(\text{Cl}^-\)... (1)
- ...which have electrostatic attraction between oppositely charged ions (1)

(c) Max 3 marks:
- Giant ionic lattice (1)
- Strong forces between oppositely charged ions require much energy to overcome (1)
- Chlorine has weak intermolecular forces between molecules (1)

(a)(i) iron(III) oxide + carbon monoxide \(\rightarrow\) iron + carbon dioxide
(ii) \(\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(l)} + 3\text{CO}_2\text{(g)}\)
(b)(i) Silicon dioxide / silica / sand (\(\text{SiO}_2\)).
(ii) Limestone thermally decomposes to form calcium oxide and carbon dioxide:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).
Calcium oxide (a basic oxide) reacts with the acidic silicon dioxide to form calcium silicate (slag):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).
(c) Pure iron has a regular arrangement of identical atoms in layers. These layers can easily slide over each other when a force is applied. In steel, carbon atoms of a different size are introduced. These different-sized atoms disrupt the regular arrangement of the layers, making it much more difficult for the layers to slide past each other.

(a)(i) Max 1 mark:
- iron(III) oxide + carbon monoxide -> iron + carbon dioxide (1)
(a)(ii) Max 2 marks:
- Correct formulae and balancing: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) (1)
- Correct state symbols: (s), (g), (l)/(s), (g) (1)

(b)(i) Max 1 mark:
- Silicon dioxide / silica / sand / \(\text{SiO}_2\) (1)
(b)(ii) Max 2 marks:
- Calcium oxide formed by thermal decomposition: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) (1)
- Acid-base reaction to form slag: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) (1)

(c) Max 4 marks:
- Pure iron has regular layers of identical atoms (1)
- Layers easily slide over one another (1)
- Alloy contains different sized atoms (1)
- This disrupts the regular pattern / prevents layers sliding (1)

(a) Xylem vessels transport water and mineral ions from the roots to the leaves (one-way flow). They are made of dead cells with lignified walls and have no end walls (hollow tubes). Phloem vessels transport sucrose and amino acids from sources to sinks (two-way flow / translocation). They are made of living cells, have sieve plates, and are supported by companion cells.
(b)(i) Water evaporates from the leaves (transpiration), which creates a transpiration pull. Cohesion (water molecules sticking together due to hydrogen bonding) and adhesion (sticking to xylem walls) allow a continuous column of water to be pulled up the xylem.
(ii) An increase in wind speed increases the rate of transpiration. This is because wind blows away the water vapour accumulating near the stomata, maintaining a steep concentration gradient of water vapour between the inside and outside of the leaf.
(iii) An increase in humidity decreases the rate of transpiration. High humidity means there is a high concentration of water vapour in the air outside the leaf, which reduces the concentration gradient, slowing down the diffusion of water vapour out of the stomata.

(a) Max 4 marks:
- Xylem: transports water / mineral ions (1)
- Xylem: dead cells / hollow / lignin (1)
- Phloem: transports sucrose / amino acids (1)
- Phloem: living cells / sieve plates / companion cells (1)
(Note: Max 2 marks for functions, Max 2 marks for structures)

(b)(i) Max 2 marks:
- Transpiration pull / tension created by evaporation (1)
- Cohesion between water molecules / continuous water column (1)

(b)(ii) Max 2 marks:
- Wind speed increases transpiration (1)
- Blows water vapour away / maintains steep concentration gradient (1)

(b)(iii) Max 2 marks:
- Humidity decreases transpiration (1)
- Air outside has more water vapour / reduces concentration gradient (1)

(a)(i) Copper
(ii) Oxygen (gas)
(b)(i) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)
(ii) \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\)
(c) 1. A pink/brown solid (copper metal) deposits on the cathode. 2. Bubbles of a colourless gas (oxygen) are produced at the anode. 3. The blue colour of the copper(II) sulfate solution gradually fades (becomes lighter blue / colourless) because \(\text{Cu}^{2+}\) ions are removed from the solution.
(d) If copper electrodes are used, the anode would dissolve / decrease in mass (\(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\)) instead of producing oxygen gas. The intensity of the blue colour of the electrolyte would remain constant because the rate of \(\text{Cu}^{2+}\) ions entering the solution at the anode equals the rate of \(\text{Cu}^{2+}\) ions depositing at the cathode.

(a) Max 2 marks:
- (i) Cathode product: copper (1)
- (ii) Anode product: oxygen (1)

(b) Max 4 marks:
- (i) Cathode half-equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) (1) + correct state symbols (1)
- (ii) Anode half-equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (1) + correct state symbols (1)

(c) Max 2 marks:
- Pink/brown solid depositing on cathode (1)
- Effervescence / bubbles of colourless gas at anode (1)
- Blue colour of solution fades (1)

(d) Max 2 marks:
- Anode dissolves / decreases in mass (no bubbles) (1)
- Blue colour of solution does not fade / remains constant (1)

(a) 1. Thin walls (one cell thick) providing a short diffusion pathway. 2. Large surface area for maximum diffusion of gases. 3. Good blood supply / dense capillary network to maintain a steep concentration gradient. 4. Moist surface to allow gases to dissolve.
(b) During inspiration, the external intercostal muscles contract, pulling the ribcage upwards and outwards. At the same time, the diaphragm contracts and flattens. This increases the volume of the thorax (chest cavity), which decreases the air pressure inside the lungs below atmospheric pressure. Consequently, air is drawn into the lungs.
(c)(i) Inspired air has approximately \(21\%\) oxygen, while expired air has approximately \(16\%\) oxygen (decreases).
(ii) Inspired air has approximately \(0.04\%\) carbon dioxide, while expired air has approximately \(4\%\) carbon dioxide (increases).
(iii) Inspired air has variable water vapour, while expired air is saturated with water vapour (increases).

(a) Max 3 marks:
- Thin walls / one-cell thick -> short diffusion distance (1)
- Large surface area (many alveoli) -> faster diffusion rate (1)
- Rich blood supply / network of capillaries -> maintains steep concentration gradient (1)
- Moist lining -> allows gases to dissolve (1)

(b) Max 4 marks:
- External intercostal muscles contract AND ribs move up/out (1)
- Diaphragm contracts AND flattens (1)
- Volume of thorax increases (1)
- Pressure inside lungs decreases below atmospheric pressure (air enters) (1)

(c) Max 3 marks:
- (i) Oxygen: inspired \(\approx 21\%\) VS expired \(\approx 16\%\) (or decreases) (1)
- (ii) Carbon dioxide: inspired \(\approx 0.04\%\) VS expired \(\approx 4\%\) (or increases) (1)
- (iii) Water vapour: inspired is variable VS expired is saturated (or increases) (1)

(a) Water moves from the soil (higher water potential) into the root hair cells (lower water potential) by osmosis down a water potential gradient through a partially permeable cell membrane.
(b) Xylem vessels consist of dead cells aligned end-to-end with no end walls, forming a continuous tube. The absence of cytoplasm and organelles (hollow tubes) allows uninterrupted flow of water. The cell walls are thickened with lignin, which provides structural support, preventing the vessels from collapsing under tension.
(c)(i) Any two from: increased temperature, decreased humidity (dry air), increased wind speed, or increased light intensity.
(c)(ii) Example 1 (Humidity): A lower humidity reduces the water vapour concentration in the air surrounding the leaf. This increases the concentration gradient between the air spaces inside the leaf and the external air, increasing the rate of diffusion of water vapour out through the stomata.
Example 2 (Wind speed): Wind blows away water vapour accumulating near the leaf surface, keeping the external humidity low. This maintains a steep water vapour concentration gradient, increasing the rate of transpiration.

(a)
- Osmosis mentioned: [1]
- Down a water potential gradient (from high to low water potential): [1]
- Across a partially permeable membrane: [1]

(b)
- Hollow / no cytoplasm / dead cells to allow free flow: [1]
- Lignified / thickened walls to prevent collapse / withstand tension: [1]
- No end walls / continuous tube: [1]

(c)(i)
- Any two correct factors (e.g., temperature, wind speed, light intensity, low humidity): [2] (1 mark per correct factor)

(c)(ii)
- Stating how the chosen factor alters the external environment (e.g., wind moves water vapour away / low humidity means dry air): [1]
- Linking this to a steeper/increased water vapour concentration gradient and faster diffusion of water vapour: [1]

(a) During the electrolysis of concentrated aqueous sodium chloride:
- Chloride ions (\(\text{Cl}^-\)) are discharged at the anode to produce chlorine gas, \(\text{Cl}_2\).
- Hydrogen ions (\(\text{H}^+\)) from water are discharged preferentially over sodium ions (\(\text{Na}^+\)) at the cathode to produce hydrogen gas, \(\text{H}_2\).
(b) The ionic half-equation at the cathode is: \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\).
(c) Water dissociates into \(\text{H}^+\) and \(\text{OH}^-\). Since \(\text{H}^+\) ions are continuously discharged at the cathode, \(\text{OH}^-\)(aq) ions remain behind in the solution. These, along with the remaining \(\text{Na}^+\)(aq) ions, form sodium hydroxide (\(\text{NaOH}\)), which is a strong alkali.
(d)(i) Bubbles of colourless gas are observed at both electrodes. The volume of gas produced at the cathode (hydrogen) is twice the volume of gas produced at the anode (oxygen).
(d)(ii) At the anode, hydroxide ions (\(\text{OH}^-\)) from water are discharged: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O}\text{(l)} + 4\text{e}^-\) (or \(2\text{H}_2\text{O}\text{(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-\)).

(a)
- Anode: chlorine / \(\text{Cl}_2\): [1]
- Cathode: hydrogen / \(\text{H}_2\): [1]
(b)
- Correct species (\(\text{H}^+\), \(\text{e}^-\), \(\text{H}_2\)): [1]
- Correct balancing and state symbols: [1]
(c)
- \(\text{H}^+\) ions are discharged / removed from solution: [1]
- Leaving behind excess hydroxide (\(\text{OH}^-\)) ions / forming sodium hydroxide: [1]
(d)(i)
- Bubbles of colourless gas at both electrodes: [1]
- Twice the volume of gas at the cathode compared to the anode (ratio 2:1): [1]
(d)(ii)
- Correct reactants and products (\(\text{OH}^-\), \(\text{O}_2\), \(\text{H}_2\text{O}\), \(\text{e}^-\)): [1]
- Correct balancing: [1]

(a) Extension \(x = 17.0\text{ cm} - 12.0\text{ cm} = 5.0\text{ cm}\) (or \(0.05\text{ m}\)).
Using Hooke's Law: \(F = kx\)
\(k = \frac{F}{x} = \frac{4.0\text{ N}}{5.0\text{ cm}} = 0.8\text{ N/cm}\).
In SI units: \(k = \frac{4.0\text{ N}}{0.05\text{ m}} = 80\text{ N/m}\).

(b) For a load \(F = 12.0\text{ N}\):
Extension \(x = \frac{F}{k} = \frac{12.0\text{ N}}{0.8\text{ N/cm}} = 15.0\text{ cm}\).
New length = unstretched length + extension = \(12.0\text{ cm} + 15.0\text{ cm} = 27.0\text{ cm}\).
(Alternatively using SI units: \(x = \frac{12.0}{80} = 0.15\text{ m}\). New length = \(0.12\text{ m} + 0.15\text{ m} = 0.27\text{ m}\)).

(c)(i) Work done = Force \(\times\) distance = \(m \times g \times h\)
\(W = 15\text{ kg} \times 10\text{ N/kg} \times 4.5\text{ m} = 675\text{ J}\).

(c)(ii) Power = \(\frac{\text{Work done}}{\text{time taken}}\)
\(P = \frac{675\text{ J}}{3.0\text{ s}} = 225\text{ W}\).

(a)
- Calculating the extension as 5.0 cm (or 0.05 m): [1]
- Calculating the value of spring constant: 0.8 (or 80): [1]
- Correct unit matching the calculated value (N/cm or N/m): [1]

(b)
- Stating Hooke's law or ratio method: [1]
- Calculating the new extension as 15.0 cm (or 0.15 m): [1]
- Finding the new length as 27.0 cm (or 0.27 m): [1]

(c)(i)
- Formula \(W = mgh\) or substitution \(15 \times 10 \times 4.5\): [1]
- Correct final value and unit (675 J or Nm): [1]

(c)(ii)
- Formula \(P = W/t\) or substitution \(675 / 3\): [1]
- Correct final value and unit (225 W or J/s): [1]

(a)(i) The main reducing agent in the blast furnace is carbon monoxide (\(\text{CO}\)), formed by the reaction of carbon dioxide with hot coke.
(a)(ii) The balanced equation is: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).
(b) Limestone (\(\text{CaCO}_3\)) decomposes thermally in the blast furnace to form calcium oxide (\(\text{CaO}\)) and carbon dioxide:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).
Calcium oxide is a basic oxide. It reacts with the acidic silicon dioxide (\(\text{SiO}_2\)) impurities (sand) to form molten slag (calcium silicate, \(\text{CaSiO}_3\)):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).
(c)(i) Brass consists of copper and zinc.
(c)(ii) In a pure metal, atoms are of the same size and are arranged in a regular lattice pattern. This allows layers of atoms to slide past one another easily when a force is applied, making pure metals malleable and soft. In an alloy, atoms of different elements have different sizes. This disrupts the regular layered structure of the host metal, making it much harder for the layers to slide over each other, increasing the strength and hardness of the material.

(a)(i)
- Carbon monoxide / \(\text{CO}\): [1] (Reject: carbon / carbon dioxide)
(a)(ii)
- Correct reactants and products: [1]
- Correct balancing: [1]

(b)
- Explaining thermal decomposition of \(\text{CaCO}_3\) to \(\text{CaO}\) OR basic \(\text{CaO}\) reacting with acidic impurity \(\text{SiO}_2\) to form slag: [1]
- Equation for thermal decomposition: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\): [1]
- Equation for slag formation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\): [1]

(c)(i)
- Copper and zinc (both required): [1]
(c)(ii)
- Mentioning that pure metals have regularly sized atoms arranged in layers that slide over each other easily: [1]
- Alloys contain atoms of different sizes: [1]
- This disrupts the regular layers / structure, preventing them from sliding easily: [1]