2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模擬試題連答案詳解

During pollination, pollen grains land on the sticky surface of the stigma. When the pollen tube grows down to the ovary, the male nucleus travels down and fuses with the female nucleus in the ovule, a process called fertilisation (or fertilization).

1. stigma [1];
2. fertilisation / fertilization [0.5];

As you move down Group I, the outermost electron is further from the nucleus and shielded by more electron shells, so it is lost more easily, meaning reactivity increases down the group. The reaction of an alkali metal with water produces a metal hydroxide and hydrogen gas: \(2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2\).

1. (reactivity) increases / gets more reactive (down the group) [1];
2. hydrogen / \(\text{H}_2\) [0.5];

Using Ohm's law / resistance formula:
\[R = \frac{V}{I} = \frac{6.0\text{ V}}{0.25\text{ A}} = 24\ \Omega\]
The unit of electrical resistance is the ohm (\(\Omega\)).

1. \(R = \frac{V}{I}\) or \(\frac{6.0}{0.25}\) evaluated to 24 [1];
2. \(\Omega\) / ohms [0.5];

Isotopes are defined as atoms of the same chemical element that contain identical numbers of protons (same atomic number) but differing numbers of neutrons in their nuclei (different nucleon/mass number).

1. (atoms of the same element with the) same number of protons / same atomic number [1];
2. different number of neutrons / different nucleon number [0.5];

Light waves (and all electromagnetic waves) are transverse waves because their oscillations are perpendicular to the direction of wave propagation. In the electromagnetic spectrum ordered by increasing frequency: radio, microwave, infrared, visible, ultraviolet, X-rays, gamma rays. Thus, ultraviolet sits directly between visible light and X-rays.

1. transverse [0.5];
2. ultraviolet / UV (radiation/rays/light) [1];

(a) Auxin is the plant hormone synthesized in the shoot tip.
(b) When exposed to light from one direction, auxin moves away from light and concentrates on the shaded side. Higher auxin concentration stimulates cell elongation on the shaded side, causing the shoot to bend towards the light.

(a) auxin; [0.5]
(b) auxin diffuses to / accumulates on the shaded side (of the shoot); [0.5]
(causes) cell elongation on the shaded side (more than illuminated side) / AW; [0.5]

(a) Going down Group I from lithium to francium, reactivity increases.
(b) The number of electron shells increases down the group, increasing atomic radius and electron shielding. Consequently, the single outer valence electron is less strongly attracted to the positively charged nucleus and is lost more readily.

(a) reactivity increases (down the group); [0.5]
(b) outer electron is further from the nucleus / increased shielding / greater atomic radius; [0.5]
weaker attraction between nucleus and outer electron (so electron is lost more easily); [0.5]

(a) The relationship is given by Ohm's law: \(V = IR\) or \(R = \frac{V}{I}\).
(b) \(R = \frac{12\text{ V}}{0.40\text{ A}} = 30\ \Omega\).

(a) \(V = IR\) / \(R = \frac{V}{I}\) / \(I = \frac{V}{R}\); [0.5]
(b) \(30\); [0.5]
\(\Omega\) / ohms; [0.5]

(a) An exothermic reaction is one in which thermal energy is released/transferred to the surroundings.
(b) Bond breaking is endothermic (requires energy) and bond forming is exothermic (releases energy). In an exothermic reaction, the total energy released on bond formation is greater than the total energy required for bond breaking.

(a) (reaction that) releases / transfers thermal energy / heat to the surroundings; [0.5]
(b) bond breaking absorbs / requires energy AND bond making releases energy; [0.5]
energy released in bond making is greater than energy needed for bond breaking; [0.5]

(a) An alpha particle consists of 2 protons and 2 neutrons tightly bound together (identical to a helium-4 nucleus).
(b) Alpha particles have high mass and charge (\(+2\)), making them highly ionizing but easily absorbed by a few centimeters of air or paper (low penetration). Beta particles are faster and less ionizing, with greater penetrating power.

(a) 2 protons and 2 neutrons / helium nucleus / \(^4_2\text{He}\); [0.5]
(b) alpha has less penetrating power (than beta) / stopped by paper / short range in air (ORA for beta); [0.5]
alpha has greater / stronger ionizing power (than beta) (ORA for beta); [0.5]

1. Sepals enclose and protect the flower when it is in the bud stage prior to opening.
2. During pollination, pollen grains from the anther are transferred to the sticky surface of the stigma at the top of the carpel.

protects (flower / flower parts when in) bud [1];
stigma [0.5];
[Total: 1.5]

Bromine is one of the two liquid elements in the Periodic Table at r.t.p. In Group VII (the halogens), reactivity decreases going down the group as the atomic radius increases and incoming electrons are less strongly attracted by the nucleus.

liquid [0.5];
(reactivity) decreases (down the group) / ORA [1];
[Total: 1.5]

Electric current is defined as the rate of flow of charge (mathematically, \(I = \frac{Q}{t}\)). The fundamental SI unit of electric charge is the coulomb (symbol \(\text{C}\)).

rate of flow of (electric) charge / charge per unit time [1];
coulomb / C [0.5];
[Total: 1.5]

Phloem tissue is responsible for transporting assimilates (sucrose and amino acids) throughout the plant. This active process of distributing nutrients from sources (e.g., photosynthesizing leaves) to sinks (e.g., roots, fruits, growing tips) is called translocation.

phloem [0.5];
translocation [1];
[Total: 1.5]

When the concentration of a solution is increased, there are more reacting particles present in a given volume. This causes the particles to collide with one another more frequently (higher collision frequency), leading to an increased rate of successful collisions and thus a faster reaction rate.

more particles per unit volume / particles are closer together [1];
more frequent collisions / increased collision rate / more collisions per second (or per unit time) [0.5];
[Total: 1.5]

(a) When a conductor moves relative to a magnetic field and cuts magnetic field lines, an electromotive force is induced across it. This phenomenon is called electromagnetic induction.

(b) According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is directly proportional to the rate at which magnetic flux lines are cut. Therefore, the induced e.m.f. can be increased by:
1. Moving the conductor faster (increasing the speed of movement).
2. Using stronger magnets (increasing the magnetic field strength / flux density).
3. Using a coil with more turns of wire instead of a single straight wire (increasing the effective length of conductor cutting the field).

(a) electromagnetic induction [0.5];

(b) any two from:
- move the wire / conductor faster / increase speed (of motion) [0.5];
- stronger magnet(s) / stronger magnetic field [0.5];
- increase the length of wire in the field / use a coil (with more turns) [0.5];

[Total: 1.5]

1. Calculate the work done in lifting the container: \(\text{Work done} = \text{force} \times \text{distance} = m \times g \times h = 650\text{ kg} \times 9.8\text{ N/kg} \times 14\text{ m} = 89\,180\text{ J}\). 2. Calculate power output: \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{89\,180\text{ J}}{28\text{ s}} = 3185\text{ W}\) (or \(3.19\text{ kW}\)).

(useful work done =) \(650 \times 9.8 \times 14\) [= \(89\,180\text{ J}\)] ; (power =) \(\frac{89\,180}{28}\) ; \(3185\text{ W}\) / \(3.19\text{ kW}\) / \(3200\text{ W}\) (correct value and unit required) ;

Step 1: Calculate the gravitational potential energy gained (work done):
\[ \Delta E_p = mgh = 45\text{ kg} \times 9.8\text{ N/kg} \times 3.6\text{ m} = 1587.6\text{ J} \]

Step 2: Calculate the power output using \( P = \frac{W}{t} \):
\[ P = \frac{1587.6\text{ J}}{12\text{ s}} = 132.3\text{ W} \approx 132\text{ W} \]

award [1] for correct calculation of work done / gravitational potential energy: \( 45 \times 9.8 \times 3.6 = 1587.6\text{ [J]} \);
award [1] for using \( P = \frac{\text{work}}{\text{time}} \) or \( \frac{1587.6}{12} \);
award [1] for correct final value with unit: \( 132\text{ W} \) / \( 132.3\text{ W} \) (accept \( \text{J/s} \));
[Total: 3]

Step 1: Calculate the relative formula mass (\( M_r \)) of \( \text{Fe}_2\text{O}_3 \):
\[ M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \]

Step 2: Calculate the amount in moles of \( \text{Fe}_2\text{O}_3 \):
\[ \text{moles of } \text{Fe}_2\text{O}_3 = \frac{32.0\text{ g}}{160\text{ g/mol}} = 0.20\text{ mol} \]

Step 3: Use the stoichiometric ratio to find moles of \( \text{Fe} \):
\[ \text{moles of } \text{Fe} = 0.20\text{ mol} \times 2 = 0.40\text{ mol} \]

Step 4: Calculate the mass of \( \text{Fe} \):
\[ \text{mass of } \text{Fe} = 0.40\text{ mol} \times 56\text{ g/mol} = 22.4\text{ g} \]

award [1] for calculating \( M_r(\text{Fe}_2\text{O}_3) = 160 \) and moles of \( \text{Fe}_2\text{O}_3 = 0.20\text{ [mol]} \);
award [1] for multiplying moles by mole ratio (\( 0.20 \times 2 = 0.40\text{ [mol Fe]} \)) [ecf];
award [1] for final mass of \( 22.4\text{ g} \) (or \( 22.4 \));
[Total: 3]

Step 1: Convert image size and actual size to the same units:
\[ \text{Image length} = 28\text{ mm} = 28 \times 1000\ \mu\text{m} = 28\,000\ \mu\text{m} \]
(or actual length \( = \frac{35}{1000} = 0.035\text{ mm} \))

Step 2: Use the magnification formula:
\[ \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{28\,000\ \mu\text{m}}{35\ \mu\text{m}} = 800 \]

award [1] for stating or using \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \);
award [1] for converting units correctly (\( 28\text{ mm} = 28\,000\ \mu\text{m} \) OR \( 35\ \mu\text{m} = 0.035\text{ mm} \));
award [1] for correct final answer: \( (\times)800 \);
[Total: 3]

Step 1: Calculate the power of the heating element using \( P = \frac{V^2}{R} \):
\[ P = \frac{(230\text{ V})^2}{26.5\ \Omega} = \frac{52\,900}{26.5} = 1996.2\text{ W} \]
(Alternatively, \( I = \frac{V}{R} = \frac{230}{26.5} = 8.679\text{ A} \) and \( P = VI = 230 \times 8.679 = 1996.2\text{ W} \))

Step 2: Convert time to seconds:
\[ t = 3.0\text{ min} \times 60\text{ s/min} = 180\text{ s} \]

Step 3: Calculate energy transferred:
\[ E = P \times t = 1996.2\text{ W} \times 180\text{ s} = 359\,316\text{ J} \approx 3.6 \times 10^5\text{ J} \text{ (or } 359\text{ kJ)} \]

award [1] for calculating power \( P = \frac{V^2}{R} = 1996\text{ [W]} \) OR current \( I = 8.68\text{ [A]} \);
award [1] for conversion of time to seconds (\( 180\text{ s} \)) and multiplying \( E = P \times t \) / \( E = VIt \);
award [1] for correct numerical value and appropriate unit: \( 359\,000\text{ J} \) / \( 359\text{ kJ} \) / \( 3.6 \times 10^5\text{ J} \);
[Total: 3]

Step 1: Calculate the temperature increase (\( \Delta T \)) of the water:
\[ \Delta T = 53.5\ ^\circ\text{C} - 21.5\ ^\circ\text{C} = 32.0\ ^\circ\text{C} \]

Step 2: Use the thermal energy equation \( Q = mc\Delta T \):
\[ Q = 150\text{ g} \times 4.18\text{ J}/(\text{g}\,^\circ\text{C}) \times 32.0\ ^\circ\text{C} \]
\[ Q = 20\,064\text{ J} \approx 20.1\text{ kJ} \text{ (or } 20\,100\text{ J)} \]

award [1] for calculating \( \Delta T = 32.0\ [^\circ\text{C}] \);
award [1] for correct substitution into \( Q = mc\Delta T \) (\( 150 \times 4.18 \times 32.0 \));
award [1] for correct final value with unit: \( 20\,064\text{ J} \) / \( 20\,100\text{ J} \) / \( 20.1\text{ kJ} \);
[Total: 3]

(a) \(a = \frac{v - u}{t} = \frac{15.0 - 0}{3.50} = 4.2857\text{ m/s}^2 \approx 4.29\text{ m/s}^2\)

(b) \(F = m a = 1.40\text{ kg} \times 4.2857\text{ m/s}^2 = 6.00\text{ N}\)

(c) \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.40\text{ kg} \times (15.0\text{ m/s})^2 = 0.70 \times 225 = 157.5\text{ J} \approx 158\text{ J}\)

(a) \(4.29\text{ m/s}^2\) / \(4.3\text{ m/s}^2\) [1];
(b) \(6.00\text{ N}\) / \(6.0\text{ N}\) (allow ecf from (a)) [1];
(c) \(158\text{ J}\) / \(157.5\text{ J}\) [1]

(a) \(R = \frac{V}{I} = \frac{24.0\text{ V}}{3.50\text{ A}} = 6.857\ Ω \approx 6.86\ Ω\)

(b) \(P = V \times I = 24.0\text{ V} \times 3.50\text{ A} = 84.0\text{ W}\)

(c) Time in seconds \(t = 5.00 \times 60\text{ s} = 300\text{ s}\)
\(E = P \times t = 84.0\text{ W} \times 300\text{ s} = 25\,200\text{ J} = 25.2\text{ kJ}\)

(a) \(6.86\ Ω\) / \(6.9\ Ω\) [1];
(b) \(84.0\text{ W}\) / \(84\text{ W}\) [1];
(c) \(25\,200\text{ J}\) / \(25.2\text{ kJ}\) (allow ecf from (b)) [1]

1. As temperature increases, the particles gain kinetic energy and move faster.
2. This results in a higher frequency of collisions between reactant particles.
3. A greater proportion of the particles have energy greater than or equal to the activation energy (≥ \(E_a\)), so a higher percentage of collisions are successful per unit time.

Award 1 mark for each of the following points (max 3):
- particles gain kinetic energy / move faster;
- increased collision frequency / particles collide more often / more collisions per unit time (ignore 'more collisions' without time reference);
- greater proportion / fraction of particles have energy equal to or greater than activation energy / more collisions have \(E \ge E_a\) / more successful collisions per unit time;

1. The ciliary muscles contract.
2. This causes the suspensory ligaments to loosen / slacken.
3. The lens becomes thicker / more convex / more rounded, which increases the refraction of light rays so they converge onto the retina.

Award 1 mark for each point (max 3):
- ciliary muscles contract;
- suspensory ligaments slacken / loosen / become less taut;
- lens becomes fatter / thicker / more convex / more rounded (so light is refracted more strongly onto the retina);

1. Down Group I, atoms have more electron shells / atomic radius increases.
2. There is more shielding between the nucleus and the valence electron, and the outer electron is further from the nucleus.
3. The electrostatic attraction between the positively charged nucleus and the outer electron is weaker, so the outer electron is lost more easily.

Award 1 mark for each point (max 3):
- atomic radius increases / number of occupied electron shells increases;
- increased shielding / outer electron is further from the nucleus;
- weaker electrostatic attraction between nucleus and outer (valence) electron, so electron is lost more readily / easily;

1. An alternating current in the primary coil generates an alternating (changing) magnetic field.
2. The soft iron core concentrates and transfers this changing magnetic field to the secondary coil.
3. The changing magnetic field cuts through the secondary coil, inducing an alternating voltage / e.m.f. across it by electromagnetic induction.

Award 1 mark for each point (max 3):
- alternating current in primary coil creates a changing / alternating magnetic field;
- magnetic field is transferred / linked through the (iron) core to the secondary coil;
- changing magnetic field cuts the secondary coil / changing magnetic flux linkage induces an alternating e.m.f. / voltage;

1. Transpiration occurs as water evaporates from the surfaces of mesophyll cells and diffuses out through stomata.
2. This loss of water produces a transpiration pull (tension/negative pressure) in the xylem.
3. Water molecules are held together in an unbroken column by cohesion (hydrogen bonding), drawing water up from roots to leaves.

Award 1 mark for each point (max 3):
- evaporation of water at mesophyll surface / diffusion of water vapour through stomata (transpiration);
- creates a tension / transpiration pull / negative pressure at the top of xylem;
- cohesion between water molecules (maintaining a continuous unbroken column) / adhesion of water molecules to xylem walls;

1. Transpiration occurs when water evaporates from the surfaces of mesophyll cells inside the leaf and diffuses out through open stomata into the atmosphere.
2. This loss of water creates a negative pressure / tension / transpiration pull at the top of the plant.
3. Water molecules are attracted to one another by cohesion (forming a continuous, unbroken column of water in xylem vessels) and to the cellulose walls by adhesion, drawing water upwards continuously from root to leaf.

1. Evaporation of water from mesophyll cell surfaces / diffusion of water vapour out through stomata (transpiration) [1];
2. Creates transpiration pull / suction / negative pressure gradient (in xylem) [1];
3. Water molecules drawn up in a continuous column due to cohesion (between water molecules) / adhesion (to xylem walls) [1].

When viewing a near object such as a textbook:
1. The ciliary muscles contract.
2. This causes the suspensory ligaments to slacken / become loose.
3. The lens becomes thicker / more convex / more rounded due to reduced pulling force.
4. This increases the refraction (bending) of incoming light rays so they converge sharply on the retina.

Any three points from:
- ciliary muscles contract ;
- suspensory ligaments slacken / loosen / become less taut ;
- lens becomes fatter / thicker / more convex / more spherical ;
- (lens) refracts / bends light rays more (strongly onto retina) ;

[Max 2.65 marks / 3 scoring points credited proportionally]

When the temperature is increased:
1. Particles gain thermal energy, which is converted to kinetic energy, so they move faster.
2. Particles collide more frequently per unit time.
3. A greater proportion of particles possess energy equal to or greater than the activation energy (\(E_a\)).
4. Therefore, the frequency of successful / effective collisions increases, leading to a higher reaction rate.

Any three points from:
- particles gain kinetic energy / move faster ;
- frequency of collisions increases / particles collide more often (per unit time) ;
- more / greater proportion of particles have energy \(\ge\) activation energy / \(E_a\) ;
- higher frequency / rate of successful / effective collisions ;

[Max 2.65 marks]

1. When the parachute opens, the large surface area causes a sudden increase in upward air resistance / drag.
2. Upward air resistance is now greater than downward weight (gravity), creating an unbalanced / resultant force upwards, which causes the skydiver to decelerate.
3. As speed decreases, air resistance decreases until it balances the downward weight.
4. When air resistance equals weight, resultant force becomes zero, and the skydiver falls at a steady, lower terminal velocity.

Any three points from:
- air resistance / drag increases (due to larger surface area) and is greater than weight / gravity ;
- (upward) resultant / net force causes deceleration / slows skydiver down ;
- as speed decreases, air resistance decreases ;
- air resistance becomes equal to weight / forces balance / resultant force is zero (so constant speed reached) ;

[Max 2.65 marks]

1. Water evaporates from the damp cell walls of spongy mesophyll cells into air spaces and diffuses out of the leaf through stomata (transpiration).
2. This loss of water lowers the water potential in leaf cells, creating a tension / negative pressure / transpiration pull in the xylem vessels.
3. Cohesion (attractive forces between water molecules) and adhesion (attraction between water molecules and xylem walls) maintain a continuous, unbroken column of water pulled up from the roots.

Any three points from:
- evaporation of water from (mesophyll) cell surfaces / diffusion of water vapour out through stomata ;
- creates transpiration pull / tension / negative pressure in xylem (ORA: water potential gradient) ;
- water molecules are held together by cohesion / forces of attraction (between water molecules) ;
- forms a continuous / unbroken column of water (pulled up xylem) ;

[Max 2.65 marks]

To construct the real, inverted image formed by a thin converging lens:
1. Draw a ray from the top of object \(O\) parallel to the principal axis up to the centre line of the lens, then refract it straight through the principal focus \(F\) on the right-hand side of the lens.
2. Draw a second ray from the top of object \(O\) passing straight through the optical centre of the lens without undergoing deviation.
3. Identify the intersection point of these two rays on the right side of the lens.
4. Draw an inverted arrow from the principal axis down to this intersection point and label it \(I\).

1. Ray drawn from top of \(O\) parallel to principal axis and refracted through focus \(F\) on the opposite side [1];
2. Ray drawn from top of \(O\) straight through the optical centre of the lens without bending [1];
3. Inverted image \(I\) drawn correctly from the principal axis to the point where the two rays intersect, with an arrowhead [1].

1. At the point of incidence on the top surface, construct a dashed line perpendicular (at \(90^\circ\)) to the surface to represent the normal.
2. In entering a denser medium (glass), light slows down and refracts towards the normal. Draw the refracted ray inside the block such that the angle of refraction \(r < 45^\circ\).
3. When the ray hits the bottom surface, construct a second normal. In exiting back into air, light speeds up and bends away from the normal.
4. Draw the emergent ray exiting the block, parallel to the incident ray but laterally displaced.

1. Normal drawn at \(90^\circ\) to the top boundary at the point of incidence [1];
2. Refracted ray drawn inside the glass bending towards the normal such that angle of refraction is less than the angle of incidence [1];
3. Emergent ray drawn exiting the bottom boundary bending away from the normal and parallel to the original incident ray [1].

When water waves move from deep water to shallow water, their speed decreases while their frequency remains constant (\(v = f\lambda\)), causing the wavelength \(\lambda\) to decrease.
1. The wavefronts inside the shallow water must be drawn with a consistently smaller spacing (shorter wavelength) than in the deep water.
2. The wavefronts must be tilted (refracted) so that the wave ray / direction of propagation bends towards the normal to the boundary.

1. Wavelength in shallow water shown clearly shorter (wavefronts closer together) than in deep water [1];
2. Wavefronts refracted at the boundary such that the direction of propagation bends towards the normal [1].

1. Locate the image \(P'\) directly behind the mirror along the perpendicular line from \(P\) to the mirror plane, such that the perpendicular distance of \(P\) to the mirror equals the perpendicular distance of \(P'\) to the mirror.
2. Draw two straight lines from \(P'\) towards the observer's eye; make the sections behind the mirror dashed (virtual rays) and solid from the mirror surface to the eye (reflected rays).
3. Draw solid incident rays from object \(P\) to the points on the mirror where the reflected rays originate.
4. Add correct arrowheads showing the direction of light travelling from \(P\) to the mirror and then reflecting into the eye.

1. Image \(P'\) positioned along the normal to the mirror at the exact same distance behind the mirror as object \(P\) is in front [1];
2. Two reflected rays drawn diverging from the mirror into the eye, with dashed construction lines tracing back to \(P'\) [1];
3. Two incident rays drawn from \(P\) to the mirror surface with correct arrowheads showing light direction [1].

1. Because the light ray is travelling in an optically denser medium (core) towards a less dense medium (cladding) and the angle of incidence \(i\) is greater than the critical angle \(c\) (\(i > c\)), no light refracts into the cladding.
2. The ray undergoes total internal reflection: draw the reflected ray bouncing off the core-cladding interface back into the core, ensuring the angle of reflection equals the angle of incidence (\(r = i\)).
3. The phenomenon is total internal reflection.

1. Reflected ray drawn reflected back into the core with angle of reflection equal to angle of incidence (by eye) and no refracted ray in the cladding [1];
2. Total internal reflection (TIR) stated [1].

1. Ray 1: Drawn parallel to the principal axis from the top of object \( O \) to the centreline of the lens, then refracted straight through the principal focus \( F_2 \) on the right.
2. Ray 2: Drawn straight through the optical centre \( C \) of the lens without deviation.
3. The intersection point of the two rays marks the top of the inverted image \( I \).
4. An arrow pointing downwards from the principal axis to the intersection point represents the image \( I \).
5. Since light rays physically converge at the point of the image, the image is real.

• Ray drawn from top of object parallel to principal axis and passing through \( F_2 \) after lens [0.75]
• Undeviated ray drawn from top of object passing through optical centre of lens [0.75]
• Clearly drawn arrow labelled \( I \) at the point of intersection AND state 'real' [0.75]

1. Since the ray enters face \( AB \) at normal incidence (angle of incidence \( = 0^\circ \)), it passes straight through without deviation to face \( AC \).
2. At face \( AC \), the angle between the ray and the normal is \( 45^\circ \). Since \( 45^\circ > 42^\circ \) (the critical angle), total internal reflection occurs.
3. The reflected ray obeys the law of reflection (angle of reflection \( = 45^\circ \)), directing the ray vertically downwards towards face \( BC \).
4. The reflected ray meets face \( BC \) at an angle of \( 90^\circ \) to the boundary (normal incidence) and emerges straight out into the air without refraction.

• Undeviated straight ray continuing through face \( AB \) to face \( AC \) [0.75]
• Angle of incidence stated as \( 45^\circ \) (or calculation showing \( 45^\circ > 42^\circ \)) [0.75]
• Total internal reflection shown at \( AC \) with angle of reflection \( = 45^\circ \) AND ray leaving face \( BC \) at \( 90^\circ \) without bending [0.75]

1. Draw two distinct rays from point \( S \) to the reflective side of mirror \( M \).
2. Draw perpendicular normal lines (dashed) at each point of incidence.
3. Construct reflected rays such that the angle of reflection equals the angle of incidence (\( r = i \)) for each ray, with arrows pointing away from the mirror.
4. Project both reflected rays backward behind the mirror using dashed straight lines until they intersect.
5. Label the point of intersection as image \( I \), which is located at equal perpendicular distance behind the mirror line.

• Two straight incident rays from \( S \) to mirror with correct arrowheads [0.75]
• Reflected rays accurately drawn with angle of reflection equal to angle of incidence (within \( \pm 2^\circ \)) [0.75]
• Reflected rays projected backwards as dashed lines to intersect at point \( I \), positioned symmetrically opposite \( S \) [0.75]

(a) Balancing carbon atoms: 4 carbons on the left require \(4\text{ CO}_2\) on the right (or 8 for 2 butane molecules).
Balancing hydrogen atoms: 10 hydrogens on the left require \(5\text{ H}_2\text{O}\) on the right (or 10 for 2 butane molecules).
Balancing oxygen atoms: \((4 \times 2) + 5 = 13\) oxygen atoms, which requires \(6.5\text{ O}_2\) (or 13 for 2 butane molecules).
Using lowest whole numbers gives: \(2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}\).

(b) Carbon dioxide is a gas at room temperature and pressure, so its state symbol is \(\text{(g)}\).

(a) Correct balancing of \(\text{C}\) and \(\text{H}\) (e.g. \(4\text{ CO}_2\) and \(5\text{ H}_2\text{O}\) for \(1\text{ C}_4\text{H}_{10}\)) [1]; fully balanced equation with correct \(\text{O}_2\) coefficient [0.5];
(b) \(\text{(g)}\) / gas [1]

(a) The reactants are solid calcium carbonate, \(\text{CaCO}_3\text{(s)}\), and aqueous hydrochloric acid, \(\text{HCl(aq)}\). The products are aqueous calcium chloride, \(\text{CaCl}_2\text{(aq)}\), gaseous carbon dioxide, \(\text{CO}_2\text{(g)}\), and liquid water, \(\text{H}_2\text{O(l)}\).
Balancing chlorine requires \(2\text{HCl}\):
\[ \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \]

(b) In aqueous solution, \(\text{HCl}\) dissociates into \(\text{H}^+\) and \(\text{Cl}^-\), and \(\text{CaCl}_2\) into \(\text{Ca}^{2+}\) and \(\text{Cl}^-\). Since \(\text{CaCO}_3\) is solid, it remains intact. Eliminating the spectator chloride ions, \(\text{Cl}^-\), gives:
\[ \text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \]

(a) Correct formulae of all reactants and products [1]; correct balancing and all correct state symbols [0.5];
(b) \(\text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) [1]

(a) Oxygen is on the right as \(\text{CO}_2\) (2 atoms), so \(2\text{ PbO}\) is required on the left. This requires \(2\text{ Pb}\) on the right:
\[ 2\text{PbO} + \text{C} \rightarrow 2\text{Pb} + \text{CO}_2 \]

(b) Reactants: \(\text{Fe}_2\text{O}_3\) and \(\text{CO}\). Products: \(\text{Fe}\) and \(\text{CO}_2\).
Balancing iron: \(\text{Fe}_2\text{O}_3\) gives \(2\text{Fe}\).
Balancing oxygen and carbon: Each \(\text{CO}\) accepts 1 oxygen from \(\text{Fe}_2\text{O}_3\) to become \(\text{CO}_2\). Since there are 3 oxygen atoms in \(\text{Fe}_2\text{O}_3\), \(3\text{ CO}\) molecules form \(3\text{ CO}_2\) molecules:
\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]

(a) \(2\text{PbO} + \text{C} \rightarrow 2\text{Pb} + \text{CO}_2\) (both '2' coefficients correct) [1];
(b) Correct formulae of reactants and products (\(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\)) [1]; correct balancing (\(1, 3 \rightarrow 2, 3\)) [0.5]

(a) To balance the 2 sodium atoms in \(\text{Na}_2\text{SO}_4\), place a 2 in front of \(\text{NaOH}\). There are then 4 hydrogen atoms on the left (2 from \(2\text{NaOH}\) and 2 from \(\text{H}_2\text{SO}_4\)), requiring a 2 in front of \(\text{H}_2\text{O}\):
\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

(b) In an aqueous neutralisation between a strong acid and a strong alkali, hydrogen ions react with hydroxide ions to form water:
\[ \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} \]

(a) \(2\text{NaOH}\) and \(2\text{H}_2\text{O}\) (both numbers correct) [1];
(b) \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\) [1]; all state symbols correct: \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(l)}\) [0.5]