2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模擬試題連答案詳解

(a) Chloroplasts and a cell wall (or a large permanent vacuole).
(b) Magnification = Image size / Actual size = 48 mm / 0.12 mm = 400. The magnification is 400x.
(c) The function is the absorption of water and mineral ions from the soil. The cell has an elongated hair-like projection which greatly increases its surface area, allowing for rapid osmosis and active transport.
(d) Root hair cells are located underground in the soil where there is no light. Since photosynthesis cannot take place without light, chloroplasts are not needed.

(a) 1 mark for each correct structure (chloroplasts, cell wall, large permanent vacuole) up to 2 marks. Reject cytoplasm, nucleus, cell membrane.
(b) 1 mark for correct formula: Magnification = Image size / Actual size. 1 mark for correct substitution: 48 / 0.12. 1 mark for correct final answer: 400 (or x400).
(c) 1 mark for identifying the function (absorption of water / mineral ions). 1 mark for stating that it has a long projection / large surface area. 1 mark for linking surface area to increased rate of absorption / osmosis / active transport.
(d) 1 mark for stating they are underground / in darkness. 1 mark for linking this to lack of photosynthesis / no light available to absorb.

(a) Features include: thin walls (one cell thick to minimize diffusion distance), large surface area (for maximum exchange), and a close network of blood capillaries (maintaining a steep concentration gradient).
(b) Component: Tar. Effect: Tar paralyzes or damages the cilia, leading to a buildup of mucus. This leads to chronic coughing (smoker's cough) and increases the risk of lung infections or chronic bronchitis. Alternatively, Carbon Monoxide binds irreversibly to hemoglobin, reducing the oxygen-carrying capacity of the blood.
(c) Inspired air contains approximately 21% oxygen, which decreases to around 16% in expired air because oxygen is absorbed into the blood and used by body cells for aerobic respiration. Inspired air contains approximately 0.04% carbon dioxide, which increases to around 4% in expired air because carbon dioxide is produced as a metabolic waste product of aerobic respiration and excreted via the lungs.

(a) 1 mark for each valid feature (max 3): thin walls (short diffusion distance), large surface area, good blood supply / close capillary network, well-ventilated / moist surface.
(b) 1 mark for naming a correct component (e.g., tar, carbon monoxide, nicotine). 1 mark for mechanism of damage (e.g., tar blocks cilia, carbon monoxide binds to hemoglobin). 1 mark for consequences (e.g., mucus buildup / smoker's cough, reduced oxygen transport).
(c) 1 mark for stating oxygen decreases (from 21% to 16%). 1 mark for explaining oxygen is used in aerobic respiration. 1 mark for stating carbon dioxide increases (from 0.04% to 4%). 1 mark for explaining carbon dioxide is produced during aerobic respiration.

(a) A pathogen is a disease-causing organism.
(b) Lymphocytes recognize specific foreign antigens on pathogens and produce antibodies that bind to them, causing them to clump together or labeling them for destruction. They also form memory cells to recognize the antigen quickly in future infections.
(c) Vaccination introduces harmless, weakened, or dead pathogens (or their antigens) into the body. This stimulates lymphocytes to produce specific antibodies and memory cells. If the body is later exposed to the same live pathogen, these memory cells recognize the antigen instantly and produce large amounts of antibodies rapidly, neutralizing the pathogen before disease symptoms occur.
(d) Passive immunity can be acquired through breast milk (or colostrum) from the mother, or across the placenta during pregnancy.

(a) 1 mark for: disease-causing organism.
(b) 1 mark for recognizing specific foreign antigens. 1 mark for producing specific antibodies. 1 mark for producing memory cells.
(c) 1 mark for introducing harmless / weakened / dead pathogens / antigens. 1 mark for stimulating antibody and memory cell production. 1 mark for memory cells surviving in the blood long-term. 1 mark for faster and greater antibody response during a secondary infection.
(d) 1 mark for across the placenta. 1 mark for breast milk / colostrum.

(a) A pathogen is a disease-causing organism.

(b) Antibodies have a specific, complementary shape that binds to specific antigens on pathogens. This binding can cause the pathogens to clump together (agglutination), making them less active and easier for phagocytes to engulf, or it can directly label the pathogens so they are recognized and destroyed by phagocytes.

(c) Active immunity is defense against a pathogen by antibody production within the body itself and results in the production of memory cells. It can be acquired naturally through an infection or artificially via vaccination. Passive immunity is the temporary defense against a pathogen by antibodies acquired from an external source (another individual), and does not produce memory cells. It can be acquired naturally across the placenta or via breast milk, or artificially via injection of antibodies/antitoxins.

(d) Passive immunity only provides short-term protection because the recipient's body does not produce its own memory cells. As a result, when the transferred/injected antibodies are naturally broken down and removed from the blood, no further antibodies can be produced against that specific pathogen.

(a)
- disease-causing organism [1]

(b) Max 3 marks from:
- antibodies have specific shapes / are complementary to antigens (on the pathogen) [1]
- antibodies bind to antigens / pathogens [1]
- cause agglutination / clumping of pathogens [1]
- label / target pathogens for destruction by phagocytosis / phagocytes [1]

(c) Max 4 marks from:
- active immunity involves antibody production by the body's own lymphocytes AND passive immunity involves receiving ready-made antibodies from another source [1]
- active immunity produces memory cells AND passive immunity does not produce memory cells [1]
- active immunity is acquired via natural infection OR vaccination / introduction of weakened pathogen [1]
- passive immunity is acquired via placenta / breast milk / colostrum OR injection of antibodies / antitoxins [1]

(d) Max 2 marks from:
- no memory cells are produced (by the body) [1]
- the transferred antibodies are eventually broken down / destroyed / cleared from the body [1]

**(a)(i)**
* Magnesium has an atomic number of 12, so its electronic configuration is **2,8,2**.
* Nitrogen has an atomic number of 7, so its electronic configuration is **2,5**.

**(a)(ii)**
* Each of the three magnesium atoms transfers its two valence electrons (a total of 6 electrons transferred) to the two nitrogen atoms.
* This results in three \(\text{Mg}^{2+}\) ions (each with a stable 2,8 configuration) and two \(\text{N}^{3-}\) ions (each with a stable 2,8 configuration).

**(b)**
* Magnesium nitride exists as a **giant ionic lattice**.
* There are very strong **electrostatic forces of attraction** holding the oppositely charged magnesium ions (\(\text{Mg}^{2+}\)) and nitride ions (\(\text{N}^{3-}\)) together.
* Overcoming these strong forces requires a **high amount of thermal energy**, resulting in a high melting point.

**(c)**
* A nitrogen molecule, \(\text{N}_2\), consists of two nitrogen atoms.
* Each nitrogen atom needs 3 electrons to complete its outer shell (which has 5 electrons).
* Therefore, the two atoms **share three pairs of electrons** between them, forming a **triple covalent bond**.
* Each atom keeps **two non-bonding outer electrons** (one lone pair), giving each nitrogen atom a stable octet of 8 outer-shell electrons.

**(a)(i)** [Total: 2 marks]
* 1 mark for magnesium configuration: 2,8,2 (or 1s² 2s² 2p⁶ 3s²).
* 1 mark for nitrogen configuration: 2,5 (or 1s² 2s² 2p³).

**(a)(ii)** [Total: 2 marks]
* 1 mark for describing magnesium losing 2 electrons / forming \(\text{Mg}^{2+}\).
* 1 mark for describing nitrogen gaining 3 electrons / forming \(\text{N}^{3-}\).
* *Note:* Accept description of three Mg atoms transferring electrons to two N atoms.

**(b)** [Total: 3 marks]
* 1 mark for identifying **giant ionic lattice / structure**.
* 1 mark for mentioning strong **electrostatic attraction** between **oppositely charged ions** (or between \(\text{Mg}^{2+}\) and \(\text{N}^{3-}\)).
* 1 mark for stating that **high energy** is required to break these bonds/forces.

**(c)** [Total: 3 marks]
* 1 mark for mentioning **sharing of three pairs of electrons** (or a **triple covalent bond**).
* 1 mark for stating that each nitrogen atom has a **lone pair** of electrons (or 2 non-bonding electrons).
* 1 mark for stating that this sharing allows both atoms to achieve a **full outer shell / stable octet / 8 outer electrons**.

**(a)**
* **Method 1 (Gas collection):** Place the hydrochloric acid and marble chips in a conical flask. Immediately seal with a delivery tube connected to a **gas syringe** (or over-water gas jar). Use a **stopwatch** to measure the **volume of carbon dioxide gas** collected at **regular time intervals** (e.g., every 10 seconds).
* **Method 2 (Mass loss):** Place the flask with hydrochloric acid on a **digital balance**, add the marble chips, and loosely plug the neck with cotton wool. Use a **stopwatch** to record the **decrease in mass** at **regular time intervals** as the \(\text{CO}_2\) gas escapes.

**(b)**
* **Effect:** The rate of reaction increases.
* **Explanation:**
* Powdering the calcium carbonate greatly increases its **surface area**.
* This exposes more reactant particles, leading to a **higher frequency of collisions** (or more collisions per unit time) between reactant particles.

**(c)**
* When temperature increases, the reactant particles gain **kinetic energy** and move faster.
* This leads to **more frequent collisions** because particles collide more often.
* More importantly, a much **greater proportion/fraction of particles** now possess energy equal to or greater than the **activation energy** (\(E_a\)).
* Therefore, a higher percentage of collisions are **successful**, significantly increasing the overall reaction rate.

**(a)** [Total: 3 marks]
* 1 mark for specifying a suitable method of collection/measurement: using a **gas syringe** to collect gas OR a **digital balance** to measure mass loss.
* 1 mark for stating that **volume of gas** OR **mass** is measured.
* 1 mark for stating that measurements are taken at **regular time intervals** (or using a **stopwatch**).

**(b)** [Total: 3 marks]
* 1 mark for stating that the rate of reaction **increases**.
* 1 mark for identifying that powder has a **larger surface area**.
* 1 mark for stating there is a **higher frequency of collisions** (accept 'more collisions per second/unit time'; reject 'more collisions' without a time reference).

**(c)** [Total: 4 marks]
* 1 mark for stating that particles gain **kinetic energy** (or move faster).
* 1 mark for stating that there are **more frequent collisions** (or higher collision frequency).
* 1 mark for mentioning **activation energy** (particles have energy \(\ge\) activation energy).
* 1 mark for stating that a **greater proportion/fraction of collisions are successful** (or lead to reaction).

**(a)(i)**
* Roasting zinc blende in air:
\(\text{ZnS} + \text{O}_2 \rightarrow \text{ZnO} + \text{SO}_2\)
Balancing the equation gives:
\(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)

**(a)(ii)**
* Chemical equation:
\(\text{ZnO(s)} + \text{CO(g)} \rightarrow \text{Zn(g/l)} + \text{CO}_2\text{(g)}\)
* **Species reduced:** Zinc oxide (\(\text{ZnO}\)).
* **Reason:** Reduction is defined as the loss of oxygen. Zinc oxide loses oxygen to form zinc metal.

**(b)(i)**
* An **alloy** is a mixture of a metal with other elements (which can be other metals or non-metals).

**(b)(ii)**
* In pure zinc, all atoms are of the **same size** and are arranged in a **regular lattice / layers**.
* These layers can **easily slide** over one another when a force is applied, making pure zinc relatively soft and malleable.
* Brass contains copper atoms, which have a **different size** compared to zinc atoms.
* The presence of these different-sized atoms **disrupts the regular arrangement/layers** of zinc atoms.
* This **prevents the layers from sliding** over each other easily, making brass harder and stronger.

**(a)(i)** [Total: 2 marks]
* 1 mark for correct reactant and product formulae: \(\text{ZnS} + \text{O}_2 \rightarrow \text{ZnO} + \text{SO}_2\).
* 1 mark for correct balancing: \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\).

**(a)(ii)** [Total: 3 marks]
* 1 mark for correct balanced chemical equation: \(\text{ZnO} + \text{CO} \rightarrow \text{Zn} + \text{CO}_2\).
* 1 mark for identifying **zinc oxide / \(\text{ZnO}\)** as the species reduced (reject zinc / Zn).
* 1 mark for explaining that it **loses oxygen**.

**(b)(i)** [Total: 1 mark]
* 1 mark for defining an alloy as a **mixture of a metal with other elements** (or mixture of two or more metals).

**(b)(ii)** [Total: 4 marks]
* 1 mark for stating that pure zinc has atoms of the **same size** in a **regular arrangement / layers**.
* 1 mark for stating that these layers **slide easily** over one another.
* 1 mark for stating that brass contains atoms of **different sizes** (copper atoms are a different size).
* 1 mark for stating that this **disrupts the regular layers/structure**, preventing them from sliding easily.

**Part (a)**
(i) Bromine water (or aqueous bromine) is added to ethene. The orange/brown solution turns colourless (is decolourised).
(ii) The repeating unit of poly(ethene) is \(-\text{CH}_2-\text{CH}_2-\). It has a single carbon-to-carbon bond with continuation bonds on both ends:

H H
| |
-[C - C]-
| |
H H

**Part (b)**
(i) Condensation polymerisation forms both the polymer and a small molecule (usually water or hydrogen chloride), whereas addition polymerisation only forms the polymer.
(ii) Amide linkage showing all atoms and bonds:
O H
|| |
-C - N-
(iii) Proteins (or polypeptides) contain the same amide (peptide) linkages.
(iv) Terylene is a polyester.
(v) Ester linkage showing all atoms and bonds:
O
||
-C - O-

**Part (c)**
(i) Non-biodegradable means the substance cannot be broken down or decomposed by microorganisms (such as bacteria or fungi).
(ii)
- Landfill: They do not decay and take up landfill space for an extremely long time.
- Incineration: Burning them releases greenhouse gases (such as carbon dioxide) or toxic gases (such as hydrogen chloride) into the atmosphere.

**Part (a)**
(i)
- Reagent: Bromine water / aqueous bromine [1]
- Result: Turns from orange / brown / yellow to colourless [1] (Do not accept 'clear')
(ii)
- Correct structure of repeating unit showing a single C-C bond, 4 hydrogen atoms, and continuation bonds through brackets or extending outwards [1]

**Part (b)**
(i)
- Condensation polymerisation produces a small molecule (like water) as well as the polymer, whereas addition polymerisation only produces the polymer [1]
(ii)
- Correct amide linkage showing carbon double-bonded to oxygen and single-bonded to nitrogen, which is single-bonded to hydrogen: \(-\text{C}(=\text{O})-\text{NH}-\) with continuation bonds [1]
(iii)
- Protein / polypeptide [1]
(iv)
- Polyester [1]
(v)
- Correct ester linkage showing carbon double-bonded to oxygen and single-bonded to oxygen: \(-\text{C}(=\text{O})-\text{O}-\) with continuation bonds [1]

**Part (c)**
(i)
- Cannot be broken down / decomposed by microbes / bacteria / fungi / living organisms [1]
(ii)
- Landfill: takes up space / remains there indefinitely / visual pollution [1]
- OR Incineration: produces toxic gases / greenhouse gases (e.g., carbon dioxide) [1]

*(a) Life cycle of a high-mass star:*
After the stable main sequence phase, a massive star expands and cools to become a red supergiant. Eventually, it undergoes a rapid collapse and explodes in a supernova. The remaining core collapses further to become either a neutron star or, if the mass is extremely large, a black hole.

*(b)(i) Meaning of redshift:*
Redshift is the increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation emitted by a source (like a galaxy) because it is moving away from the observer.

*(b)(ii) Evidence for the Big Bang:*
- Redshift shows that almost all distant galaxies are moving away from us.
- The further away a galaxy is, the greater its redshift, meaning it is moving away faster.
- This indicates that the Universe is expanding.
- Extrapolating this expansion backwards in time suggests that the entire Universe originated from a single, extremely hot and dense point (the Big Bang) in the past.

*(b)(iii) Calculation of distance:*
Using Hubble's Law:
\(v = H_0 d\)

Rearranging for \(d\):
\(d = \frac{v}{H_0}\)

Substitute the values:
\(d = \frac{1.1 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 5.0 \times 10^{24}\text{ m}\)

*(a) Life cycle of a high-mass star:* (Max 3 marks)
- Red supergiant (or expands and cools) [1]
- Supernova (or explodes) [1]
- Neutron star OR black hole [1]

*(b)(i) Meaning of redshift:* (Max 2 marks)
- Increase in observed wavelength / decrease in frequency [1]
- Due to the source moving away from the observer [1]

*(b)(ii) Evidence for the Big Bang:* (Max 3 marks)
- Redshift shows galaxies are moving away from us / Earth [1]
- More distant galaxies have greater redshift / are moving faster [1]
- This indicates the Universe is expanding from a single point [1]

*(b)(iii) Distance calculation:* (Max 2 marks)
- Correct formula or substitution: \(d = \frac{v}{H_0}\) [1]
- Correct final calculation: \(5.0 \times 10^{24}\text{ m}\) [1]
- Accept correct unit (meters or m).

*(a) Light entering along the normal:*
The angle of incidence at the boundary is \(0^\circ\) (along the normal). Therefore, the angle of refraction is also \(0^\circ\), meaning the ray does not change direction.

*(b)(i) Calculate the angle of refraction:*
When light travels from glass to air, Snell's Law can be written as:
\(n = \frac{\sin(r)}{\sin(i)}\)
where \(i\) is the angle in glass (\(38^\circ\)) and \(r\) is the angle in air.

Rearranging for \(\sin(r)\):
\(\sin(r) = n \times \sin(i)\)
\(\sin(r) = 1.52 \times \sin(38^\circ)\)
\(\sin(38^\circ) \approx 0.6157\)
\(\sin(r) = 1.52 \times 0.6157 = 0.9359\)
\(r = \sin^{-1}(0.9359) \approx 69.4^\circ\)

*(b)(ii) Calculate the critical angle:*
\(\sin(c) = \frac{1}{n}\)
\(\sin(c) = \frac{1}{1.52} \approx 0.6579\)
\(c = \sin^{-1}(0.6579) \approx 41.1^\circ\) (or \(41^\circ\))

*(b)(iii) Angle of incidence increased to \(45^\circ\):*
Since the new angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light cannot refract out of the glass. Instead, it undergoes total internal reflection, reflecting back into the glass at an angle of \(45^\circ\).

*(c) Application in communications:*
- Application: Optical fibers (or fiber optic cables).
- Explanation: Information is transmitted as light pulses that travel down the core of the fiber by undergoing continuous total internal reflection at the boundary between the core and the cladding.

*(a) Light entering along the normal:* (Max 1 mark)
- Angle of incidence is zero / ray strikes perpendicular to surface / along the normal, so it does not bend. [1]

*(b)(i) Angle of refraction:* (Max 3 marks)
- Correct formula: \(n = \frac{\sin(r)}{\sin(i)}\) or \(\sin(r) = n \sin(i)\) [1]
- Correct substitution: \(\sin(r) = 1.52 \times \sin(38^\circ)\) [1]
- Correct answer: \(69.4^\circ\) (Accept \(69^\circ\) or \(69.3^\circ - 69.5^\circ\)) [1]

*(b)(ii) Critical angle:* (Max 2 marks)
- Correct formula: \(\sin(c) = \frac{1}{n}\) [1]
- Correct calculation: \(41.1^\circ\) (Accept \(41^\circ\)) [1]

*(b)(iii) Angle of incidence increased to \(45^\circ\):* (Max 2 marks)
- Total internal reflection occurs [1]
- Because the angle of incidence is greater than the critical angle [1]

*(c) Application:* (Max 2 marks)
- Optical fibers / fiber optics / endoscope [1]
- Light travels down/within the fiber/tube by undergoing continuous total internal reflection [1]

*(a) Nuclear Equation:*
Carbon-14 decays to Nitrogen-14 by emitting a beta particle. A beta particle is an electron, represented as \(^{0}_{-1}\beta\) or \(^{0}_{-1}\text{e}\).
\(^{14}_{~6}\text{C} \rightarrow ^{14}_{~7}\text{N} + ^{~0}_{-1}\beta\)

*(b)(i) Calculation of remaining atoms:*
First, find the number of half-lives that have elapsed:
Number of half-lives \(= \frac{\text{Total time}}{\text{Half-life}} = \frac{17190\text{ years}}{5730\text{ years}} = 3\text{ half-lives}\)

After 3 half-lives, the number of remaining atoms is:
\(N = 8.0 \times 10^9 \times \left(\frac{1}{2}\right)^3\)
\(N = 8.0 \times 10^9 \times \frac{1}{8} = 1.0 \times 10^9\text{ atoms}\)

*(b)(ii) Radioactive detector:*
A Geiger-Müller (GM) tube (or detector).

*(c) Safety precautions:*
1. Handle the source with long-handled tongs to maximize distance from the body.
2. Do not point the source directly at anyone (especially eyes) and keep it pointed away.
3. Store the source in a lead-lined container immediately when not in use.
(Any two valid precautions)

*(d) Relative ionizing power:*
- Beta vs Gamma: A beta particle is a charged particle (charge \(-1\)) with mass, whereas a gamma ray is an uncharged, massless electromagnetic wave. Charged, massive particles are much more likely to interact electromagnetically with atomic electrons, knocking them out (ionizing them).
- Beta vs Alpha: An alpha particle has a much larger charge (\(+2\)) and mass (4 units) than a beta particle (charge \(-1\), negligible mass). This larger charge and size make collisions and strong electrostatic interactions far more frequent for alpha particles, making alpha much more strongly ionizing than beta.

*(a) Nuclear Equation:* (Max 3 marks)
- Correct symbol for Nitrogen with mass number 14: \(^{14}\text{N}\) [1]
- Correct atomic number for Nitrogen: 7 (i.e. \(^{14}_{~7}\text{N}\)) [1]
- Correct beta particle notation: \(^{~0}_{-1}\beta\) or \(^{~0}_{-1}\text{e}\) [1]

*(b)(i) Atoms remaining:* (Max 2 marks)
- Correct number of half-lives identified (3) [1]
- Correct final answer: \(1.0 \times 10^9\) (or \(1 \times 10^9\)) [1]

*(b)(ii) Detector:* (Max 1 mark)
- Geiger-Müller (GM) tube / GM counter (Accept cloud chamber / spark counter) [1]

*(c) Safety precautions:* (Max 2 marks)
- Use tongs / do not touch directly [1]
- Point source away from people [1]
- Store in lead-lined box when not in use [1]
- Limit exposure time [1]
(Any two points, 1 mark each)

*(d) Relative ionizing power:* (Max 2 marks)
- Beta is more ionizing than gamma because beta has charge / mass (while gamma has no charge / no mass) [1]
- Beta is less ionizing than alpha because beta has less charge / less mass than alpha (alpha has \(+2\) charge and mass of 4) [1]

(a)(i) The proton number is defined as the number of protons in the nucleus of an atom. (ii) The nucleon number is defined as the total number of protons and neutrons (nucleons) in the nucleus of an atom. (b)(i) In beta decay, a neutron in the nucleus decays into a proton and an electron (beta particle). The equation is written as: \(^{14}_{6}\text{C} \rightarrow\ ^{14}_{7}\text{N} + \ ^{0}_{-1}\text{e}\). Both nucleon number (14 = 14 + 0) and charge (6 = 7 - 1) balance. (b)(ii) Because a neutron changes into a proton, the number of protons increases by 1, and the number of neutrons decreases by 1. (c) First, determine the number of half-lives that have elapsed: \(n = \frac{17100\text{ years}}{5700\text{ years}} = 3\). Next, calculate the remaining number of atoms after 3 half-lives: Remaining atoms = \(8.0 \times 10^9 \times (\frac{1}{2})^3 = 8.0 \times 10^9 \times \frac{1}{8} = 1.0 \times 10^9\) atoms.

(a)(i) 1 mark: number of protons (in the nucleus). (a)(ii) 1 mark: total number of protons and neutrons (in the nucleus). (b)(i) 1 mark: correct reactant and product symbols (\(^{14}_{6}\text{C}\) and \(^{14}_{7}\text{N}\)). 1 mark: correct beta particle symbol (\(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\)). 1 mark: correctly balanced equation. (b)(ii) 1 mark: number of protons increases by 1. 1 mark: number of neutrons decreases by 1. (c) 1 mark: calculates that 3 half-lives have passed (\(17100 / 5700 = 3\)). 1 mark: clear halving process shown (e.g. \(8.0 \times 10^9 \rightarrow 4.0 \times 10^9 \rightarrow 2.0 \times 10^9 \rightarrow 1.0 \times 10^9\)). 1 mark: correct final answer of \(1.0 \times 10^9\) (allow correct equivalent values, e.g., \(1 \times 10^9\)).