2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模拟试题及答案详解

During inhalation (inspiration):
1. The external intercostal muscles contract (pulling ribs up and out).
2. The diaphragm contracts and flattens (moves downwards).
3. These combined movements increase the volume inside the thorax (thoracic cavity), reducing internal pressure below atmospheric pressure so air rushes in.

contract / tighten ; [0.5]
flattens / moves down / descends ; [0.5]
increase / get bigger / expand ; [0.5]

(i) \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_{2}\text{H}_{4} + \text{C}_{8}\text{H}_{18}\). Therefore, hydrocarbon \(\text{X}\) is \(\text{C}_{8}\text{H}_{18}\).
(ii) Aqueous bromine is orange (or brown/yellow). Alkenes undergo an addition reaction with bromine, decolourising it so the final solution is colourless.

(i) \(\text{C}_{8}\text{H}_{18}\) ; [0.5]
(ii) (from) orange / brown / red-brown / yellow ; [0.5]
(to) colourless / decolourised ; [0.5] (reject: 'clear' alone)

Use Ohm's Law formula:
\(R = \frac{V}{I}\)
\(R = \frac{12.0\text{ V}}{2.5\text{ A}} = 4.8\)

The SI unit of electrical resistance is the ohm (\(\Omega\)).

\(4.8\) ; [1.0] (award [0.5] for formula \(R = \frac{V}{I}\) or substitution \(\frac{12}{2.5}\) if final value incorrect)
\(\Omega\) / ohm(s) ; [0.5]

1. Water evaporates from the cell walls of the spongy mesophyll layer into the interconnecting air spaces within the leaf.
2. The water vapour diffuses down its concentration gradient through the stomata (singular: stoma) into the drier surrounding air.

spongy ; [0.75]
stomata / stoma / stomatal pores ; [0.75]

(i) Nucleon number conservation: \(14 = A + 0 \implies A = 14\).
Proton number conservation: \(6 = Z + (-1) \implies Z = 7\).
(ii) Gamma (\(\gamma\)) radiation is high-energy electromagnetic radiation and has the greatest penetrating power compared to alpha and beta radiation.

(i) (A =) 14 ; [0.5]
(Z =) 7 ; [0.5]
(ii) gamma / \(\gamma\) ; [0.5]

When the diaphragm contracts and flattens (moves downwards), the volume of the thoracic cavity increases. According to Boyle's law, as volume increases, the internal gas pressure inside the thorax decreases below atmospheric pressure, allowing air to rush into the lungs.

Volume increases [1];
Pressure decreases / falls below atmospheric pressure [0.5];

Aqueous bromine (bromine water) is originally orange or red-brown. When shaken with an unsaturated hydrocarbon such as propene (\(\text{C}_3\text{H}_6\)), the bromine adds across the double bond to form 1,2-dibromopropane (\(\text{CH}_3\text{CHBrCH}_2\text{Br}\)), resulting in a colourless solution (decolourisation).

(Colour change from) orange / brown / red-brown to colourless / decolourised [1];
1,2-dibromopropane / dibromopropane [0.5];
(Reject: 'clear' instead of 'colourless')

Using Ohm's law:
$$R = \frac{V}{I}$$
$$R = \frac{12\text{ V}}{0.75\text{ A}} = 16\,\Omega$$

Formula or substitution: \(R = \frac{V}{I}\) OR \(\frac{12}{0.75}\) [1];
Correct value with unit: 16 \(\mathbf{\Omega}\) / 16 ohms [0.5];

Xylem vessels consist of dead, hollow cells joined end-to-end with no end walls, forming a continuous tube for uninterrupted water flow. Their cell walls are reinforced with lignin, which provides structural rigidity and prevents the vessels from collapsing under negative pressure (tension).

Any two from:
- Hollow / no cytoplasm / no cell contents [0.75];
- End walls broken down / continuous tube [0.75];
- Lignified / contains lignin / thickened cell walls [0.75];
- Narrow lumen (to sustain capillary action / water column) [0.75];
(Maximum 1.5 marks)

First determine the number of half-lives elapsed:
$$\text{Number of half-lives} = \frac{60\text{ min}}{20\text{ min}} = 3$$
Halve the initial quantity 3 times:
$$\text{After 1 half-life (20 min)} = 2.4 \times 10^{12}$$
$$\text{After 2 half-lives (40 min)} = 1.2 \times 10^{12}$$
$$\text{After 3 half-lives (60 min)} = 6.0 \times 10^{11}$$

Identification of 3 half-lives elapsed OR dividing initial count by 8 / halving three times [1];
6.0 \times 10^{11} / 6 \times 10^{11} [0.5];

During inhalation (inspiration), the muscular diaphragm contracts and moves downwards (flattens). Simultaneously, the external intercostal muscles contract to pull the ribs upwards and outwards, which increases the volume of the thorax and reduces the pressure inside the lungs below atmospheric pressure.

downwards / down / flattens / lower [1];
outwards / out [0.5];
(Total: 1.5 marks)

(i) Aqueous bromine is initially orange/brown/red-brown and becomes colourless (is decolourised) when it reacts with an alkene.
(ii) The reaction across the carbon-carbon double bond (\(\text{C}=\text{C}\)) is an addition reaction (bromination).

(i) orange / brown / red-brown to colourless / decolourised [1]; (reject: clear instead of colourless)
(ii) addition [0.5];
(Total: 1.5 marks)

Using Ohm's law / resistance formula:
\[R = \frac{V}{I}\]
\[R = \frac{12.0\text{ V}}{0.75\text{ A}} = 16\ \Omega\]

correct calculation \(R = \frac{12.0}{0.75} = 16\) [1];
correct unit: \(\Omega\) / ohms [0.5];
(Total: 1.5 marks)

(i) Magnesium is more reactive than copper, displacing copper ions. Visual changes: the blue solution fades/turns colourless as \(\text{Cu}^{2+}\) ions react, and a brown/pink-brown/reddish deposit of solid copper forms.
(ii) The ionic equation is \(\text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)}\).

(i) (blue) solution becomes colourless / fades OR pink / brown / reddish solid forms [0.5];
(ii) \(\text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)}\) balanced formulae [0.5], correct state symbols [0.5];
(Total: 1.5 marks)

Number of half-lives elapsed \(= \frac{45\text{ min}}{15\text{ min}} = 3\).
After 1 half-life (15 min): \(4.8 \times 10^{12} \div 2 = 2.4 \times 10^{12}\)
After 2 half-lives (30 min): \(2.4 \times 10^{12} \div 2 = 1.2 \times 10^{12}\)
After 3 half-lives (45 min): \(1.2 \times 10^{12} \div 2 = 6.0 \times 10^{11}\)

identifying that 3 half-lives have elapsed [0.5];
correct final answer \(6.0 \times 10^{11}\) (or \(6 \times 10^{11}\)) [1];
(Total: 1.5 marks)

Alveoli and gas exchange surfaces possess adaptations to maximize the rate of diffusion according to Fick's Law: 1. A large total surface area. 2. A thin surface (one cell thick) providing a short diffusion distance. Other valid features include a moist lining or a rich capillary network/good blood supply to maintain a steep concentration gradient.

Award [0.75] marks for each correct feature up to a maximum of [1.5]:
• large surface area [0.75];
• thin walls / one cell thick / short diffusion distance [0.75];
• good blood supply / well vascularised [0.75];
• good ventilation [0.75];
• moist surface [0.75].

Step 1: Convert the operating time from minutes to seconds:
\(t = 45 \times 60\text{ s} = 2700\text{ s}\)

Step 2: Calculate the energy transferred in joules using the equation \(E = V \times I \times t\):
\(E = 240\text{ V} \times 8.5\text{ A} \times 2700\text{ s} = 5\,508\,000\text{ J}\)

Step 3: Convert the energy from joules (J) to megajoules (MJ):
\(E = \frac{5\,508\,000}{10^6}\text{ MJ} = 5.508\text{ MJ} \approx 5.51\text{ MJ}\)

conversion of time to seconds: \(45 \times 60 = 2700\text{ (s)}\) ;
substitution into \(E = VIt\) or \(E = Pt\): \(240 \times 8.5 \times 2700\) or \(5\,508\,000\text{ (J)}\) ;
final answer in MJ: 5.51 / 5.508 (MJ) ;

Step 1: Calculate relative formula masses (\(M_r\)):
\(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\)
\(M_r(\text{CaO}) = 40 + 16 = 56\)

Step 2: Calculate moles of \(\text{CaCO}_3\):
\(\text{moles of }\text{CaCO}_3 = \frac{10.0\text{ g}}{100\text{ g/mol}} = 0.100\text{ mol}\)

Step 3: Determine mass of \(\text{CaO}\) formed (mole ratio is 1:1):
\(\text{moles of }\text{CaO} = 0.100\text{ mol}\)
\(\text{mass of }\text{CaO} = 0.100\text{ mol} \times 56\text{ g/mol} = 5.60\text{ g}\)

calculation of \(M_r\) values: \(M_r(\text{CaCO}_3) = 100\) AND \(M_r(\text{CaO}) = 56\) ;
calculation of moles of \(\text{CaCO}_3\): \(\frac{10.0}{100} = 0.10\text{ (mol)}\) ;
final mass of \(\text{CaO}\): 5.6(0) (g) ;

Step 1: State the formula linking magnification, image size, and actual size:
\(\text{magnification} = \frac{\text{image size}}{\text{actual size}}\)

Step 2: Ensure units match (both are given in mm):
\(\text{Image size} = 48\text{ mm}\)
\(\text{Actual size} = 0.080\text{ mm}\)

Step 3: Calculate magnification:
\(\text{magnification} = \frac{48}{0.080} = 600\)

formula stated or implied: \(\text{magnification} = \frac{\text{image size}}{\text{actual size}}\) / \(M = \frac{I}{A}\) ;
correct substitution: \(\frac{48}{0.080}\) ;
correct calculation: (\(\times\)) 600 ;

Step 1: Calculate the gain in kinetic energy (\(E_k\)):
\(E_k = \frac{1}{2}mv^2 = 0.5 \times 1.4\text{ kg} \times (6.5\text{ m/s})^2\)
\(E_k = 0.7 \times 42.25 = 29.575\text{ J}\)

Step 2: Calculate the average useful power using \(P = \frac{E}{t}\):
\(P = \frac{29.575\text{ J}}{3.0\text{ s}} = 9.858\text{ W} \approx 9.9\text{ W}\)

use of \(E_k = \frac{1}{2}mv^2\): \(0.5 \times 1.4 \times (6.5)^2\) OR \(29.6\text{ (J)}\) / \(29.575\text{ (J)}\) ;
use of \(P = \frac{E}{t}\) or \(P = \frac{W}{t}\): \(\frac{29.575}{3.0}\) ;
correct final power: 9.9 / 9.86 (W) ;

Step 1: Convert the frequency from megahertz (MHz) to hertz (Hz):
\(f = 98.0\text{ MHz} = 98.0 \times 10^6\text{ Hz}\)

Step 2: Use the wave speed equation \(v = f \lambda\) rearranged for wavelength:
\(\lambda = \frac{v}{f}\)

Step 3: Substitute the known values:
\(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{98.0 \times 10^6\text{ Hz}} = 3.0612\text{ m} \approx 3.06\text{ m}\)

frequency conversion: \(98.0 \times 10^6\text{ (Hz)}\) ;
rearrangement and substitution: \(\lambda = \frac{3.0 \times 10^8}{98.0 \times 10^6}\) ;
correct wavelength: 3.06 (m) [accept 3.1 (m) or 3.061 (m)] ;

1. Convert time to seconds:
\[t = 12 \times 60 = 720\text{ s}\]

2. Use the electrical energy formula:
\[E = V \times I \times t\]
\[E = 230\text{ V} \times 4.5\text{ A} \times 720\text{ s} = 745\,200\text{ J}\]

3. Convert joules to megajoules (\(\text{MJ}\)):
\[E = \frac{745\,200}{10^6} = 0.7452\text{ MJ} \approx 0.75\text{ MJ}\]

conversion of time to seconds: \(12 \times 60 = 720\text{ s}\) [1];
use of \(E = VIt\) or \(P = VI\) and \(E = Pt\) (e.g. \(230 \times 4.5 \times 720\) or \(1035 \times 720\)) [1];
final answer \(0.7452\text{ MJ}\) / \(0.745\text{ MJ}\) / \(0.75\text{ MJ}\) [1]

1. Calculate relative formula masses (\(M_r\)):
\[M_r(\text{MgCO}_3) = 24 + 12 + (3 \times 16) = 84\]
\[M_r(\text{MgO}) = 24 + 16 = 40\]

2. Calculate moles of \(\text{MgCO}_3\):
\[\text{moles of } \text{MgCO}_3 = \frac{12.6}{84} = 0.15\text{ mol}\]

3. Determine mass of \(\text{MgO}\) using the 1:1 mole ratio:
\[\text{moles of } \text{MgO} = 0.15\text{ mol}\]
\[\text{mass of } \text{MgO} = 0.15\text{ mol} \times 40\text{ g/mol} = 6.0\text{ g}\]

correct \(M_r\) calculation: \(M_r(\text{MgCO}_3) = 84\) AND \(M_r(\text{MgO}) = 40\) [1];
calculating moles of \(\text{MgCO}_3\): \(\frac{12.6}{84} = 0.15\text{ mol}\) [1];
calculating mass of \(\text{MgO}\): \(0.15 \times 40 = 6.0\text{ g}\) [1]

1. Convert the measured image length into micrometres (\(\mu\text{m}\)):
\[\text{Image length} = 54\text{ mm} = 54 \times 1000\,\mu\text{m} = 54\,000\,\mu\text{m}\]

2. State or apply the magnification formula:
\[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\]

3. Calculate the magnification:
\[\text{Magnification} = \frac{54\,000\,\mu\text{m}}{75\,\mu\text{m}} = 720\]

conversion of units so image and actual size are in the same unit (e.g. \(54\text{ mm} = 54\,000\,\mu\text{m}\) or \(75\,\mu\text{m} = 0.075\text{ mm}\)) [1];
use of formula \(\text{Magnification} = \frac{\text{image size}}{\text{actual size}}\) [1];
final answer \(720\) or \(\times 720\) [1]

1. Use the wave speed equation:
\[v = f \times \lambda\]
\[\lambda = \frac{v}{f}\]

2. Substitute the values:
\[\lambda = \frac{3.0 \times 10^8\text{ m/s}}{9.6 \times 10^7\text{ Hz}}\]
\[\lambda = 3.125\text{ m}\]

3. Round to 2 significant figures:
\[\lambda \approx 3.1\text{ m}\]

formula \(v = f\lambda\) or \(\lambda = \frac{v}{f}\) [1];
correct substitution: \(\frac{3.0 \times 10^8}{9.6 \times 10^7}\) [1];
correct value to 2 significant figures: \(3.1\text{ m}\) (accept \(3.13\text{ m}\) only if working shows unrounded \(3.125\) before rounding instruction) [1]

1. Calculate the number of moles of propane:
\[\text{moles of } \text{C}_3\text{H}_8 = \frac{\text{mass}}{M_r} = \frac{6.6\text{ g}}{44\text{ g/mol}} = 0.15\text{ mol}\]

2. Determine the moles of oxygen gas needed:
From the balanced equation, \(1\text{ mol}\) of \(\text{C}_3\text{H}_8\) reacts with \(5\text{ mol}\) of \(\text{O}_2\).
\[\text{moles of } \text{O}_2 = 0.15\text{ mol} \times 5 = 0.75\text{ mol}\]

3. Calculate the volume of \(\text{O}_2\) at r.t.p.:
\[\text{Volume} = \text{moles} \times 24\text{ dm}^3/\text{mol} = 0.75 \times 24 = 18\text{ dm}^3\]

moles of propane \(= \frac{6.6}{44} = 0.15\text{ mol}\) [1];
moles of oxygen \(= 0.15 \times 5 = 0.75\text{ mol}\) [1];
volume of oxygen \(= 0.75 \times 24 = 18\text{ dm}^3\) (or \(18\,000\text{ cm}^3\)) [1]

Step 1: Calculate the acceleration of the vehicle using the formula \(a = \frac{v - u}{t}\).
\(a = \frac{15\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = 3.0\text{ m/s}^2\)

Step 2: Calculate the resultant force using Newton's second law \(F = m a\).
\(F = 650\text{ kg} \times 3.0\text{ m/s}^2 = 1950\text{ N}\)

calculation of acceleration: \(a = \frac{15}{5.0} = 3.0\text{ (m/s}^2\text{)}\) [1];
use of \(F = ma\) or \(F = 650 \times 3.0\) [1];
correct numerical answer (\(1950\) / \(1.95 \times 10^3\)) AND correct unit \(\text{N}\) / newtons [1]

Step 1: Calculate the relative formula mass (\(M_\text{r}\)) of \(\text{MgCO}_3\) and \(\text{MgO}\).
\(M_\text{r}(\text{MgCO}_3) = 24 + 12 + (3 \times 16) = 84\)
\(M_\text{r}(\text{MgO}) = 24 + 16 = 40\)

Step 2: Calculate the amount in moles of \(\text{MgCO}_3\).
\(\text{Moles of } \text{MgCO}_3 = \frac{\text{mass}}{M_\text{r}} = \frac{16.8\text{ g}}{84} = 0.20\text{ mol}\)

Step 3: Determine the moles of \(\text{MgO}\) formed and calculate its mass.
From the 1:1 mole ratio, \(\text{moles of } \text{MgO} = 0.20\text{ mol}\).
\(\text{Mass of } \text{MgO} = 0.20\text{ mol} \times 40\text{ g/mol} = 8.0\text{ g}\)

correct \(M_\text{r}\) values: \(M_\text{r}(\text{MgCO}_3) = 84\) AND \(M_\text{r}(\text{MgO}) = 40\) [1];
\(\text{moles of } \text{MgCO}_3 = \frac{16.8}{84} = 0.20\text{ (mol)}\) [1];
\(\text{mass of } \text{MgO} = 0.20 \times 40 = 8.0\text{ (g)}\) [1]

The alveoli have several adaptations that maximize the rate of gas exchange according to Fick's Law:
1. A large overall surface area provided by millions of microscopic spherical alveoli, allowing more gas molecules to diffuse simultaneously.
2. Very thin walls (a single layer of flattened epithelial cells) which minimizes the diffusion distance for oxygen and carbon dioxide.
3. A rich network of surrounding blood capillaries (and continuous ventilation) that rapidly carries oxygen away and brings carbon dioxide, maintaining a steep concentration gradient across the exchange surface.

1 mark for each valid structural adaptation with its linked explanation (up to max 3 marks):
- large surface area / folded membrane (1) AND allows more particles to diffuse at once / increases diffusion rate (1);
- thin walls / one cell thick / thin epithelium (1) AND provides a short diffusion pathway / distance (1);
- rich capillary network / good blood supply / continuous blood flow (1) AND maintains a steep concentration gradient (1);
- moist lining (1) AND allows gases to dissolve before diffusing (1).

When the concentration of hydrochloric acid is increased, there are more hydrogen ions / acid particles in a given volume of solution. Because the particles are closer together, the frequency of collisions between the acid particles and the surface of the calcium carbonate increases. Consequently, there are more successful collisions per unit time, resulting in a higher rate of reaction.

Award 1 mark for each of the following points:
- more particles per unit volume / particles are closer together (1);
- increased frequency of collisions / more collisions per second / particles collide more often (1);
- more successful / effective collisions per unit time / per second (1).

[Note: Do NOT credit "more collisions" alone without reference to time / frequency / per second].

1. An alternating potential difference applied to the primary coil causes an alternating current to flow through it.
2. This produces a continuously changing (alternating) magnetic field in the soft iron core.
3. The magnetic field passes through the core and cuts through the secondary coil, inducing an alternating electromotive force / potential difference across the secondary coil by electromagnetic induction.

Award 1 mark for each point:
- alternating current / p.d. in primary coil produces a changing / alternating magnetic field (1);
- the (iron) core channels / transfers this changing magnetic field to the secondary coil (1);
- the changing magnetic field cuts the secondary coil / links through it, inducing a potential difference / voltage (1).

A magnesium atom has two valence electrons and loses both electrons to attain a stable outer shell, forming a magnesium ion with a charge of \(+2\) (\(\text{Mg}^{2+}\)). Two chlorine atoms, each needing one electron to complete their octet, each gain one of these lost electrons to form two chloride ions, each with a charge of \(-1\) (\(\text{Cl}^-\)). The oppositely charged \(\text{Mg}^{2+}\) and \(\text{Cl}^-\dots\) ions are strongly held together by electrostatic forces of attraction to form the ionic compound \(\text{MgCl}_2\).

Award 1 mark for each point:
- magnesium (atom) loses two electrons to form a \(\text{Mg}^{2+}\) ion (1);
- two chlorine atoms each gain one electron (or one electron transferred to each of two Cl atoms) to form two \(\text{Cl}^-\dots\) / chloride ions (1);
- strong electrostatic attraction between oppositely charged ions / between positive and negative ions (1).

When the temperature increases, the thermal energy is converted into kinetic energy, causing the gas particles to move at higher average speeds. Because the particles travel faster, they collide with the inner walls of the container more frequently. Furthermore, each collision involves a greater change in momentum, producing a larger force per collision. Since pressure is force per unit area (\(P = \frac{F}{A}\)), the total pressure exerted on the container walls increases.

Award 1 mark for each point:
- particles gain kinetic energy / move faster / have higher average speed (1);
- particles collide with the walls more frequently / more collisions per second (1);
- greater change in momentum per collision / collisions exert a greater force (hence greater force per unit area / pressure) (1).

1. Large total surface area: millions of microscopic alveoli provide an extensive surface area over which oxygen and carbon dioxide can diffuse simultaneously.
2. Very thin walls / one cell thick: this provides a short diffusion distance for gases passing between the alveolar air and the blood.
3. Good blood supply / surrounded by dense capillary network: the continuous flow of blood removes oxygen and brings carbon dioxide, maintaining a steep concentration gradient across the alveolar membrane.
(Alternative: Moist surface/lining allows gases to dissolve before diffusing).

Any three from:
- large surface area (to volume ratio) / many alveoli (for rapid diffusion) [1];
- thin walls / one cell thick / thin epithelium (giving a short diffusion pathway) [1];
- rich capillary network / good blood supply (maintaining steep concentration gradient) [1];
- moist lining / layer of moisture (allowing gases to dissolve) [1];
- good ventilation / airflow (maintaining concentration gradient) [1].

When the temperature is increased:
1. Particles gain kinetic energy and move faster.
2. Particles collide more frequently / higher collision frequency.
3. A greater fraction of colliding particles have energy greater than or equal to the activation energy (\(E_a\)), resulting in a higher rate/frequency of successful (effective) collisions.

Award marks for:
- particles gain kinetic energy / move faster [1];
- higher collision frequency / particles collide more often / more collisions per unit time [1];
- more particles have energy greater than / equal to the activation energy \(OR\) higher frequency of successful / effective collisions [1].

1. For a given electrical power transmitted, \(P = IV\), raising the voltage (\(V\)) decreases the electric current (\(I\)).
2. The power lost as thermal energy in the transmission lines depends on the square of the current: \(P_{\text{loss}} = I^2R\).
3. Because current \(I\) is significantly reduced, the thermal energy generated per second due to the cable resistance is much lower, improving transmission efficiency.

Award marks for:
- (for a constant power) higher voltage results in a lower / smaller current (\(P = IV\)) [1];
- power / thermal energy loss depends on current squared / \(P = I^2R\) / current causes heating in cables [1];
- lower current reduces heating / less thermal energy wasted to the surroundings [1].

1. The crude oil is heated and vaporised before entering the fractionating column.
2. The fractionating column has a temperature gradient (hotter at the base, cooler at the top).
3. Hydrocarbons with different carbon chain lengths have different boiling points (due to different strengths of intermolecular forces), so they condense at different levels/heights where the temperature equals their boiling point and are collected separately.

Award marks for:
- crude oil is heated / vaporised [1];
- column is hotter at the bottom and cooler at the top / temperature gradient in the column [1];
- fractions / hydrocarbons condense at different heights / temperatures because they have different boiling points / different molecular sizes [1].

1. Alpha particles have a large mass and a relatively large charge (\(+2e\)) compared to beta particles or gamma rays.
2. This strong electric field / large charge allows them to easily attract and knock electrons out of neutral air atoms/molecules (high ionising ability).
3. Every ionisation event causes the alpha particle to transfer kinetic energy; because it causes many ionisations per unit distance, it loses its kinetic energy rapidly, leading to a short penetration range.

Award marks for:
- alpha particles have a large charge (\(+2\)) / large mass / slow speed compared to beta [1];
- easily pull / remove electrons from atoms / molecules due to strong electrostatic attraction [1];
- (rapid) loss of kinetic energy per collision / each ionisation removes energy, so the particle stops quickly / over a short distance [1].

1. During exercise, working muscles contract more vigorously and require increased energy/ATP, which is supplied by an increased rate of aerobic respiration.
2. Increased respiration produces more carbon dioxide as a waste product, which dissolves in the blood and decreases blood pH (making it more acidic).
3. This increase in carbon dioxide concentration is detected by receptors/the brain, which signals the diaphragm and intercostal muscles to increase the rate and depth of ventilation to remove carbon dioxide rapidly and supply more oxygen to the muscles.

Any three from:
- (increased) muscle contraction / activity requires more energy / ATP (from respiration) ;
- (increased) rate of aerobic respiration produces more carbon dioxide / lowers blood pH ;
- higher carbon dioxide / lower pH detected by receptors / brain (respiratory centre) ;
- increased ventilation removes \(\text{CO}_2\) faster / supplies \(\text{O}_2\) faster (to meet metabolic demand / repay oxygen debt) ;

1. Ray 1 starts at the tip of the object, travels parallel to the principal axis to the lens axis, and refracts downwards passing directly through the principal focus \(F\) on the right side of the lens. 2. Ray 2 travels from the tip of the object straight through the optical centre of the lens without bending. 3. The real image is located where the two rays intersect beyond \(F\). An arrow is drawn with its tail on the principal axis and its head at the intersection point, pointing downwards to show an inverted image.

(i) Straight ray from top of object parallel to axis, refracted through \(F\) on right of lens [1]; (ii) Straight undeviated ray from top of object passing through optical centre [1]; (iii) Inverted arrow drawn from axis to ray intersection [1].

(a) In the alveoli, the concentration of oxygen is higher than in the deoxygenated blood arriving at the capillary, so oxygen diffuses down its concentration gradient from the alveolus into the capillary. (b) The concentration of carbon dioxide in the capillary blood is higher than in the alveolus, so carbon dioxide diffuses from the capillary into the alveolus. (c) Oxygen binds reversibly to haemoglobin inside red blood cells (erythrocytes) to form oxyhaemoglobin.

(a) Arrow labelled O pointing from alveolus lumen into capillary [1]; (b) Arrow labelled C pointing from capillary into alveolus lumen [1]; (c) Red blood cells / erythrocytes / haemoglobin [1].

1. Initial acceleration: From time \(t = 0\text{ s}\), \(v = 0\text{ m/s}\) to \(t = 4.0\text{ s}\), \(v = 6.0\text{ m/s}\), draw a straight line. 2. Constant speed: From \(t = 4.0\text{ s}\) to \(t = 4.0 + 6.0 = 10.0\text{ s}\), the speed remains constant at \(6.0\text{ m/s}\), represented by a horizontal line. 3. Deceleration: From \(t = 10.0\text{ s}\) to \(t = 10.0 + 2.0 = 12.0\text{ s}\), the speed decreases uniformly to \(0\text{ m/s}\), represented by a straight line down to the time axis at \((12.0, 0)\).

Straight line drawn from \((0, 0)\) to \((4.0, 6.0)\) [1]; horizontal straight line drawn from \((4.0, 6.0)\) to \((10.0, 6.0)\) [1]; straight line drawn from \((10.0, 6.0)\) to \((12.0, 0)\) [1].

(i) In an exothermic reaction, energy is released to the surroundings, so the products have lower chemical energy than the reactants. A horizontal line must be drawn below the level of the reactants and labelled 'products'. (ii) \(\Delta H\) represents the overall energy change, shown by a vertical arrow starting at the level of the reactants and pointing downwards to the level of the products. (iii) The activation energy \(E_{\text{a}}\) is the minimum energy required to start the reaction, shown by an upward arrow extending from the reactant energy level to the top peak of the energy barrier.

(i) Horizontal line drawn below the reactants line and clearly labelled products [1]; (ii) Vertical arrow drawn from reactant level to product level labelled \(\Delta H\) / enthalpy change [1]; (iii) Vertical arrow from reactant level to highest point of curve labelled \(E_{\text{a}}\) / activation energy [1].

1. The normal is defined as a line perpendicular (at \(90^\circ\)) to the surface of the block at the point of incidence.
2. When light travels from a less optically dense medium (air) into a more optically dense medium (glass), it slows down and refracts towards the normal line. Therefore, the refracted ray inside the glass must make an angle with the normal that is smaller than the angle of incidence (\(r < 45^\circ\)).

(i) normal drawn as a straight line perpendicular to the glass-air boundary at the point of incidence (dashed or solid) [1];
(ii) refracted ray drawn inside the block bent towards the normal (angle of refraction clearly smaller than angle of incidence) [1]

1. Draw a ray from the top of object \(O\) travelling parallel to the principal axis to the centre line of the lens, then refract it downwards so that it passes straight through the principal focus \(F\) on the right side of the lens.
2. Draw a second ray from the top of object \(O\) travelling straight through the optical centre \(C\) of the lens without any deviation.
3. At the point where the two refracted rays intersect on the right of the lens, draw an inverted vertical arrow from the principal axis to the intersection point and label it \(I\).

one ray from top of \(O\) parallel to principal axis refracted through \(F\) on the right-hand side OR one ray from top of \(O\) passing straight through the optical centre \(C\) without bending;
second correct ray drawn and inverted image labelled \(I\) at the intersection of the two rays;

1. Nitrogen is in Group V and has 5 outer shell electrons. Hydrogen is in Group I and has 1 outer shell electron.
2. Nitrogen forms 3 single covalent bonds by sharing one electron with each of the three hydrogen atoms. In each of the three overlap areas between the nitrogen circle and a hydrogen circle, draw one dot (\(\bullet\)) and one cross (\(\times\)).
3. This accounts for 3 of nitrogen's 5 outer electrons. Place the remaining 2 non-bonding electrons as a lone pair (two dots, \(\bullet\bullet\)) on the nitrogen atom outside the bond overlaps.
4. Verify that each hydrogen atom has 2 shared electrons and nitrogen has a complete octet of 8 outer electrons.

one shared pair of electrons consisting of one dot and one cross (\(\bullet\times\)) in each of the three \(\text{N}-\text{H}\) overlap regions;
one non-bonding pair of dots (\(\bullet\bullet\)) on the nitrogen atom and no additional electrons on any hydrogen atom;