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

(a) Osmosis is defined as the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), down a concentration gradient, through a partially permeable membrane.

(b) (i) Change in mass = \(3.78 - 4.20 = -0.42\text{ g}\).
Percentage change in mass = \(\frac{-0.42}{4.20} \times 100 = -10.00\%\).

(ii) The \(0.0\text{ mol/dm}^3\) solution (pure water) has a higher water potential than the cytoplasm/vacuole of the potato cells. Therefore, water molecules move into the potato cells by osmosis, down a water potential gradient. This net movement of water increases the mass of the potato tissue.

(c) Drying the cylinders removes surface water, which would otherwise contribute to the measured mass and cause inaccuracies in the percentage change calculation.

(d) The internal concentration is approximately \(0.4\text{ mol/dm}^3\) because at this external concentration there is no net movement of water (percentage change in mass is \(0.00\%\)), meaning the water potential inside the potato cells is equal to the water potential of the external solution.

(a)
- net movement of water molecules [1]
- from a region of higher water potential to a region of lower water potential / down a water potential gradient [1]
- through a partially permeable membrane [1]

(b) (i)
- correct calculation of mass change: \(-0.42\text{ g}\) (or shown as decrease of \(0.42\text{ g}\)) [1]
- correct percentage change calculation: \(-10.00\%\) (accept \(10.00\%\) decrease, must be 2 d.p.) [1]

(b) (ii)
- solution has a higher water potential than the potato cells [1]
- water moves into the potato cells by osmosis [1]
- net entry of water increases the mass [1]

(c)
- to remove excess liquid/surface water (which would add to the measured mass) [1]

(d)
- \(0.4\text{ mol/dm}^3\) because there is no net movement of water / zero percentage change in mass [1]

(a) The pancreas contains specialized receptor cells that monitor the glucose level of blood passing through it.

(b) When blood glucose concentration rises:
1. The pancreas detects the rise and secretes insulin.
2. Insulin is transported in the blood plasma to the liver (and muscles).
3. It stimulates liver cells to take up glucose from the blood.
4. It stimulates the conversion of soluble glucose into insoluble glycogen for storage, lowering the blood glucose levels back to normal.

(c) When blood glucose concentration falls:
1. The pancreas secretes the hormone glucagon.
2. Glucagon targets the liver cells.
3. It stimulates the breakdown of stored glycogen back into glucose.
4. Glucose is released into the bloodstream, raising blood glucose levels back to normal.

(d) (i) Symptoms of Type 1 diabetes include: presence of glucose in the urine, extreme thirst, frequent urination, rapid weight loss, and lethargy/fatigue.
(ii) Type 1 diabetes is managed and treated through regular, subcutaneous injections of insulin, alongside careful monitoring of blood glucose levels and dietary management.

(a)
- pancreas [1]

(b)
- (hormone) insulin [1]
- released from the pancreas [1]
- target organ: liver / muscle cells [1]
- effect: converts glucose to glycogen / increases glucose uptake by cells [1]

(c)
- (hormone) glucagon (secreted by pancreas) [1]
- targets liver [1]
- effect: breaks down glycogen to glucose (released into blood) [1]

(d) (i)
- glucose in urine / extreme thirst / frequent urination / fatigue / unexplained weight loss [1]

(d) (ii)
- insulin injections / insulin therapy [1]

(a) (i) Genotype is defined as the genetic makeup of an organism in terms of the alleles present.
(ii) Heterozygous means having two different alleles of a particular gene (e.g. \(Ff\)).

(b)
(i) Individual 1 is normal but has an affected son (Individual 3, genotype \(ff\)). Thus, Individual 1 must carry the recessive allele. Genotype is \(Ff\).
(ii) Individual 3 is affected by the recessive condition, so the genotype must be homozygous recessive: \(ff\).
(iii) Individual 4 is normal but has an affected daughter (Individual 6, genotype \(ff\)). Thus, she must carry the recessive allele. Genotype is \(Ff\).

(c) Individual 4 is phenotypically normal, meaning she must possess at least one dominant allele (\(F\)). Her daughter (Individual 6) has cystic fibrosis, which is a recessive condition, meaning the daughter's genotype is \(ff\). Since offspring receive one allele from each biological parent, Individual 4 must have passed on a recessive allele (\(f\)) to her daughter. Therefore, Individual 4 must be heterozygous with the genotype \(Ff\).

(d) (i) Both Individual 4 and Individual 5 are unaffected but have an affected child (6, \(ff\)), so both must be heterozygous carriers (\(Ff\)).

**Parental Genotypes:** \(Ff \times Ff\)

**Gametes:** \(F\) and \(f\) from both parents.

**Punnett Square:**

| | **F** | **f** |
| :---: | :---: | :---: |
| **F** | \(FF\) | \(Ff\) |
| **f** | \(Ff\) | \(ff\) |

**Offspring Phenotypes:**
- \(FF\): Normal lung function
- \(Ff\): Normal lung function (carrier)
- \(ff\): Affected by cystic fibrosis

(ii) From the Punnett square, there is 1 box out of 4 that results in the affected genotype (\(ff\)). Therefore, the probability of having an affected child is \(1\text{ in }4\), \(0.25\), or \(25\%\).

(a) (i)
- genetic makeup of an organism in terms of the alleles present [1]
(a) (ii)
- having two different alleles of a gene / alleles are different [1]

(b)
- (i) \(Ff\) [1]
- (ii) \(ff\) [1]
- (iii) \(Ff\) [1]

(c)
- Individual 4 is normal so must have at least one dominant allele / \(F\) [1]
- Individual 4 has an affected daughter / Individual 6 / \(ff\), so she must pass on a recessive allele / \(f\) (making her \(Ff\)) [1]

(d) (i)
- correct parental genotypes shown as \(Ff \times Ff\) [1]
- correct Punnett square with offspring genotypes: \(FF, Ff, Ff, ff\) [1]
- identification of phenotypes corresponding to genotypes (normal for \(FF, Ff\) and cystic fibrosis/affected for \(ff\)) [1]

(d) (ii)
- \(1/4\) / \(25\%\) / \(0.25\) / 1 in 4 [1]

(a) (i) Homeostasis is defined as the maintenance of a constant internal environment.
(ii) The pancreas is the endocrine gland that detects changes in blood glucose levels and secretes the regulatory hormones.

(b) When blood glucose levels rise above normal (e.g., after a meal):
1. The pancreas detects this rise and secretes the hormone insulin into the bloodstream.
2. Insulin travels through the blood to the liver and muscle cells (target organs).
3. This stimulates the liver and muscle cells to take up more glucose from the blood.
4. Inside these cells, glucose is converted into insoluble glycogen for storage, thereby lowering the blood glucose concentration back to its set-point.

(c) Completed Table 1.1:
- Speed of transmission: Fast / rapid (Nervous) vs. Slow / slower (Endocrine)
- Duration of response: Short-lived / brief (Nervous) vs. Long-lasting (Endocrine)
- Nature of signal: Electrical impulses (Nervous) vs. Chemical / hormones (Endocrine)
- Pathway of transmission: Along neurones / nerve cells (Nervous) vs. In the blood / blood plasma (Endocrine)

(a) (i) [1 mark]
- Maintenance of a constant internal environment.

(ii) [1 mark]
- Pancreas (reject: liver).

(b) [4 marks max]
- Pancreas secretes insulin; [1]
- Insulin travels in the blood; [1]
- Targets the liver / muscle cells; [1]
- Stimulates uptake of glucose from the blood; [1]
- Glucose converted into glycogen; [1]

(c) [4 marks]
- 1 mark for each correct row filled in Table 1.1:
- Row 1: Fast / rapid AND Slow / slower; [1]
- Row 2: Short-lived / short duration AND Long-lasting / long duration; [1]
- Row 3: Electrical impulse(s) AND Chemical / hormone(s); [1]
- Row 4: Along neurones (or axons/nerves) AND In the blood (plasma); [1]

(a) As the reaction proceeds, reactant particles (hydrogen ions in the acid) are used up, so their concentration decreases. This leads to a lower frequency of collisions (fewer collisions per unit time) between reactant particles, resulting in fewer successful collisions per second.
(b) Any two from: increase the temperature of the acid, use powdered calcium carbonate to increase the surface area, or use a catalyst.
(c) First, calculate the relative formula mass of \(\text{CaCO}_3\): \(M_r = 40 + 12 + (3 \times 16) = 100\).
Next, find the moles of \(\text{CaCO}_3\): \(\text{moles} = 0.25 / 100 = 0.0025\text{ mol}\).
From the equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, so 0.0025 mol of \(\text{CO}_2\) is produced.
Finally, calculate the volume: \(\text{Volume} = 0.0025 \times 24 = 0.06\text{ dm}^3\) (or \(60\text{ cm}^3\)).
(d) Bubble the gas through limewater. If carbon dioxide is present, the limewater turns cloudy or milky.

(a) Max [3] marks:
- Reactant particles/acid ions are used up / concentration decreases [1]
- Frequency of collisions decreases / fewer collisions per unit time [1]
- Fewer successful collisions per unit time [1]

(b) Max [2] marks:
- Increase temperature [1]
- Use powdered calcium carbonate / increase surface area [1]
(Accept: add a catalyst [1])

(c) Max [3] marks:
- Correct calculation of Mr of CaCO3 as 100 and moles as 0.0025 mol [1]
- State 1:1 ratio or moles of CO2 is 0.0025 mol [1]
- Correct final volume of 0.06 dm3 (or 60 cm3) with unit [1]

(d) Max [2] marks:
- Test: Bubble gas through limewater [1]
- Result: Turns cloudy / milky [1]

(a)(i) A sodium atom transfers its single outer shell electron to a chlorine atom. This forms a sodium ion (\(\text{Na}^+\)) with a stable full outer shell, and a chloride ion (\(\text{Cl}^-\)) with a stable full outer shell.
(a)(ii) Sodium chloride has a giant ionic lattice structure with very strong electrostatic forces of attraction between oppositely charged ions. Breaking these strong electrostatic bonds requires a very large amount of thermal energy.
(b)(i) Covalent bonding, because both phosphorus and chlorine are non-metal elements that share outer shell electrons to achieve stable full outer shells.
(b)(ii) 1. There are three covalent bonds in a \(\text{PCl}_3\) molecule, which means there are 3 shared pairs of electrons.
2. Phosphorus is in Group 15 (Group V) and has 5 outer electrons. Since 3 are shared with chlorine atoms, 2 non-bonding outer shell electrons (1 lone pair) remain on the phosphorus atom.
3. Each chlorine atom is in Group 17 (Group VII) and has 7 outer electrons. Each chlorine shares 1 electron, leaving 6 non-bonding outer electrons per chlorine. For three chlorine atoms, the total is \(3 \times 6 = 18\) non-bonding outer electrons.

(a)(i) Max [3] marks:
- Sodium atom loses one electron [1]
- Chlorine atom gains one electron [1]
- Ions with full outer shells are formed / Na+ and Cl- formed [1]

(a)(ii) Max [2] marks:
- Giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions [1]
- Requires a large amount of energy to overcome / break [1]

(b)(i) Max [2] marks:
- Covalent [1]
- Non-metals share electrons [1]

(b)(ii) Max [3] marks:
- 1. 3 [1]
- 2. 2 [1]
- 3. 18 [1]

(a)(i) Industrial catalytic cracking requires high temperatures (typically around 500 °C) and a catalyst (such as zeolite, alumina, or silica).
(a)(ii) The total atoms on the reactant side are 10 carbons and 22 hydrogens. The products have \(2 \times 2 = 4\) carbons and \(2 \times 4 = 8\) hydrogens in the two ethene molecules. Subtracting these gives: \(10 - 4 = 6\) carbons and \(22 - 8 = 14\) hydrogens. The missing product is hexane, \(\text{C}_6\text{H}_{14}\).
(b)(i) Unsaturated means the compound contains one or more carbon-to-carbon double bonds (\(\text{C}=\text{C}\)).
(b)(ii) Add bromine water (aqueous bromine) to both substances. Ethene (unsaturated) will rapidly decolourise the bromine water, changing it from orange/brown to colourless. Ethane (saturated) will not react, so the bromine water remains orange/brown.
(c)(i) Addition polymerisation is the process where unsaturated monomers join together to form a polymer without forming any other products.
(c)(ii) The monomer ethene contains a carbon-to-carbon double bond (\(\text{C}=\text{C}\)). During polymerisation, this double bond breaks / opens up, meaning the polymer poly(ethene) contains only carbon-to-carbon single bonds (\(\text{C}-\text{C}\)) in its backbone chain.

(a)(i) Max [2] marks:
- High temperature (accept 450 to 800 °C) [1]
- Catalyst / zeolite / silica / alumina [1]
(a)(ii) Max [1] mark:
- C6H14 [1]
(b)(i) Max [1] mark:
- Contains a carbon-carbon double bond / C=C [1]
(b)(ii) Max [3] marks:
- Test: Bromine water / aqueous bromine [1]
- Ethene result: turns colourless / decolourises [1]
- Ethane result: remains orange / yellow / brown / no change [1]
(c)(i) Max [1] mark:
- Addition (polymerisation) [1]
(c)(ii) Max [2] marks:
- Monomer (ethene) has a double bond (C=C) [1]
- Polymer (poly(ethene)) has single bonds (C-C) [1]

(a) Magnesium metal reacts with aqueous hydrochloric acid to produce aqueous magnesium chloride and hydrogen gas: \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\). (b)(i) At the start, the concentration of hydrochloric acid particles is at its highest, meaning there are more reactant particles per unit volume, resulting in the highest frequency of successful collisions. (ii) As the acid is used up, its concentration decreases, so the rate of reaction decreases. (c) Step 1: Calculate moles of magnesium: \(\text{moles of Mg} = 0.18\text{ g} / 24\text{ g/mol} = 0.0075\text{ mol}\). Step 2: Determine moles of hydrogen: Since the mole ratio of \(\text{Mg}\) to \(\text{H}_2\) is 1:1, moles of \(\text{H}_2\) produced = \(0.0075\text{ mol}\). Step 3: Calculate volume of gas: \(\text{Volume} = 0.0075\text{ mol} \times 24\text{ dm}^3/\text{mol} = 0.18\text{ dm}^3\). (d) Increasing temperature increases the kinetic energy of the particles, causing them to move faster. This results in a higher frequency of collisions. Furthermore, a greater proportion of colliding particles possess energy equal to or greater than the activation energy, leading to a higher frequency of successful collisions and thus an increased rate.

(a) [2 marks] 1 mark for correct balanced chemical equation: \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\). 1 mark for all correct state symbols: \(\text{(s)}\), \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(g)}\). (b)(i) [2 marks] 1 mark for stating concentration of acid/reactant particles is highest at the start. 1 mark for stating this results in the highest frequency of successful collisions (or most successful collisions per unit time). (b)(ii) [1 mark] 1 mark for stating that the rate of reaction decreases / slows down. (c) [3 marks] 1 mark for calculating moles of Mg: \(0.18 / 24 = 0.0075\text{ mol}\). 1 mark for using 1:1 ratio to identify moles of \(\text{H}_2 = 0.0075\text{ mol}\). 1 mark for calculating volume of gas: \(0.0075 \times 24 = 0.18\text{ dm}^3\) (allow error carried forward from incorrect moles). (d) [2 marks] 1 mark for stating that the rate of reaction increases because particles have more kinetic energy / move faster. 1 mark for stating that there is a higher frequency of successful collisions / more collisions have energy greater than or equal to the activation energy.

(a)(i) The total distance travelled by the sound wave to the wall and back is 2d. Using the formula: distance = speed x time, we get 2d = 330 m/s x 0.80 s = 264 m. Therefore, d = 264 / 2 = 132 m. (a)(ii) Sound is a longitudinal wave. (b) Convert wavelength to meters: lambda = 1.5 cm = 0.015 m. Using the wave equation: v = f x lambda = 8.0 Hz x 0.015 m = 0.12 m/s. (c)(i) The speed of all electromagnetic waves in a vacuum is 3.0 x 10^8 m/s. (c)(ii) Ultraviolet has a higher frequency (and shorter wavelength) than infrared. A common application of ultraviolet is sterilising medical equipment or security marking. A common application of infrared is remote controls or thermal imaging.

(a)(i) 1 mark for recall of formula/working (e.g. 2d = speed x time or 330 x 0.80), 1 mark for correct final answer with unit [132 m]. (a)(ii) 1 mark for longitudinal. (b) 1 mark for converting wavelength to m [0.015 m], 1 mark for using v = f x lambda, 1 mark for correct final answer with unit [0.12 m/s] (accept 12 cm/s only if unit changed). (c)(i) 1 mark for 3.0 x 10^8 m/s (accept 3 x 10^8 m/s or 300,000,000 m/s). (c)(ii) 1 mark for identifying ultraviolet as higher frequency, 1 mark for a valid use of ultraviolet, 1 mark for a valid use of infrared.

(a) The diagram must show a battery, a switch, and an ammeter connected in series with the main loop. The main loop then splits into two parallel branches, one containing a 6.0 ohm resistor and the other containing a 12 ohm resistor. (b) For parallel resistors: 1/Rp = 1/R1 + 1/R2 = 1/6.0 + 1/12 = 2/12 + 1/12 = 3/12. Therefore, Rp = 12 / 3 = 4.0 ohms. (c) The total resistance is 4.0 ohms. Using Ohm's Law: I = V / R = 12 V / 4.0 ohms = 3.0 A. (d) First, determine the power dissipated by the 6.0 ohm resistor: P = V^2 / R = 12^2 / 6.0 = 144 / 6.0 = 24 W. Convert time to seconds: t = 5.0 minutes = 5.0 x 60 = 300 s. Energy E = P x t = 24 W x 300 s = 7200 J (or 7.2 kJ).

(a) 1 mark for correct symbols for battery, ammeter, switch, and resistors; 1 mark for parallel arrangement of the two resistors; 1 mark for placing the switch and ammeter in the main loop of the circuit. (b) 1 mark for correct formula 1/Rp = 1/R1 + 1/R2 or equivalent, 1 mark for correct calculation [4.0 ohms]. (c) 1 mark for recalling V = I x R or I = V/R, 1 mark for correct substitution and showing 3.0 A. (d) 1 mark for calculating power of the 6.0 ohm resistor [24 W] or using E = (V^2 / R) * t, 1 mark for converting time to seconds [300 s], 1 mark for final correct energy with unit [7200 J or 7.2 kJ].

(a) Acceleration is given by: a = (v - u) / t = (6.0 - 0) / 4.0 = 1.5 m/s^2. (b) The total distance is the area under the speed-time graph. First part (triangle, 0 to 4 s): distance1 = 0.5 x base x height = 0.5 x 4.0 x 6.0 = 12 m. Second part (rectangle, 4 s to 9 s): distance2 = base x height = 5.0 s x 6.0 m/s = 30 m. Total distance = 12 + 30 = 42 m. (c) Kinetic energy: Ek = 0.5 x m x v^2 = 0.5 x 0.50 kg x (6.0 m/s)^2 = 0.25 x 36 = 9.0 J. (d) The work done by the braking force to stop the car is equal to the change in its kinetic energy. Work Done = Kinetic Energy lost = 9.0 J. Work Done = Force x distance => 9.0 J = Force x 3.0 m => Force = 9.0 / 3.0 = 3.0 N.

(a) 1 mark for formula or substitution (6.0 / 4.0), 1 mark for correct final answer with unit [1.5 m/s^2]. (b) 1 mark for calculating distance of first stage [12 m], 1 mark for calculating distance of second stage [30 m], 1 mark for correct total sum with unit [42 m]. (c) 1 mark for formula Ek = 1/2 m v^2 or substitution, 1 mark for correct final answer with unit [9.0 J]. (d) 1 mark for equating kinetic energy to work done (or using kinematic equations to find acceleration a = -6 m/s^2), 1 mark for recalling W = F x d or F = m x a, 1 mark for correct final answer with unit [3.0 N].

(a)
Using the formula for change in gravitational potential energy:
\(\Delta E_p = mgh\)
\(\Delta E_p = 0.80\text{ kg} \times 10\text{ m/s}^2 \times 0.45\text{ m} = 3.6\text{ J}\)

(b)(i)
Using the kinetic energy formula:
\(E_k = \frac{1}{2}mv^2\)
\(E_k = 0.5 \times 0.80\text{ kg} \times (2.4\text{ m/s})^2\)
\(E_k = 0.40 \times 5.76 = 2.304\text{ J}\) (or \(2.3\text{ J}\) to 2 significant figures)

(b)(ii)
Work done against friction is equal to the mechanical energy lost:
\(\text{Energy lost} = E_p - E_k\)
\(\text{Energy lost} = 3.6\text{ J} - 2.304\text{ J} = 1.296\text{ J}\) (or \(1.3\text{ J}\) to 2 significant figures)

(c)(i)
Using equations of motion:
Average speed during uniform deceleration = \(\frac{u + v}{2} = \frac{2.4 + 0}{2} = 1.2\text{ m/s}\)
Time taken to stop, \(t = \frac{\text{distance}}{\text{average speed}} = \frac{1.8\text{ m}}{1.2\text{ m/s}} = 1.5\text{ s}\)
Deceleration \(a = \frac{u - v}{t} = \frac{2.4\text{ m/s} - 0\text{ m/s}}{1.5\text{ s}} = 1.6\text{ m/s}^2\)
(Alternatively, using \(v^2 = u^2 + 2as\):
\(0 = 2.4^2 + 2 \times a \times 1.8\)
\(3.6a = -5.76\)
\(a = -1.6\text{ m/s}^2\), so deceleration is \(1.6\text{ m/s}^2\).)

(c)(ii)
The kinetic energy has been transferred to thermal energy (or internal energy / heat).

(c)(iii)
As the materials warm up, the molecules gain kinetic energy and vibrate or move faster.

(a)
- 1 mark for formula or substitution: \(0.80 \times 10 \times 0.45\)
- 1 mark for correct value with unit: \(3.6\text{ J}\) (accept joules/J, reject lower case j without capital symbol)

(b)(i)
- 1 mark for substitution: \(0.5 \times 0.80 \times 2.4^2\)
- 1 mark for correct value: \(2.3\text{ J}\) or \(2.30\text{ J}\) or \(2.304\text{ J}\)

(b)(ii)
- 1 mark for recognizing energy difference: \(\Delta E_p - E_k\) (e.g. \(3.6 - \text{their (b)(i)}\))
- 1 mark for correct value: \(1.3\text{ J}\) or \(1.296\text{ J}\) (allow ecf from (a) and (b)(i))

(c)(i)
- 1 mark for choosing a valid method (either finding time \(t = 1.5\text{ s}\) first or substituting into \(v^2 = u^2 + 2as\))
- 1 mark for showing fully worked step-by-step calculation leading to \(1.6\text{ m/s}^2\)

(c)(ii)
- 1 mark for: thermal / internal energy (accept heat)

(c)(iii)
- 1 mark for: molecules vibrate faster / move faster / gain kinetic energy