2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模擬試題連答案詳解

(a) Asexual reproduction is a process resulting in the production of genetically identical offspring from one parent. An advantage is that it is rapid, allowing a species to colonise an area quickly without needing a partner. (b) Sperm are produced in the testes, then pass into the sperm duct (vas deferens). During ejaculation, they travel along the sperm duct, receive fluids from the prostate gland or seminal vesicle, and then enter the urethra, through which they leave the body. (c) Adaptations include: a flagellum (tail) for swimming towards the egg; many mitochondria in the midpiece to provide energy or ATP for movement; and an acrosome containing digestive enzymes in the head to penetrate the outer layer of the egg.

(a) Definition: production of genetically identical offspring [1 mark] from one parent [1 mark]. Advantage: rapid reproduction / allows rapid colonisation / only requires one parent (no mate needed) / conserves energy [1 mark]. (b) Pathway: Testes to sperm duct (vas deferens) [1 mark]; glands (prostate/seminal vesicle) add fluid [1 mark]; urethra [1 mark]; penis [1 mark]. (c) Adaptations (any three): Flagellum/tail for swimming [1 mark]; Mitochondria to release energy/for respiration [1 mark]; Acrosome containing enzymes to digest egg membrane/jelly coat [1 mark]; Haploid nucleus containing half the genetic material [1 mark]. Max 3 marks.

(a) Adrenaline is secreted by the adrenal glands directly into the blood plasma, which transports it throughout the body. Its physiological effects include increased heart rate, increased breathing rate, and widened pupils. (b) (i) Nervous transmission is extremely rapid because electrical impulses travel along neurones, while endocrine transmission is slower because hormones are carried by the blood. (ii) Nervous response is short-lived, while hormone response is typically longer-lasting. (c) Gravitropism is a response in which parts of a plant grow towards or away from gravity. A positive gravitropic response in roots causes them to grow downwards into the soil, which is beneficial because it anchors the plant firmly and allows the roots to reach water and mineral ions.

(a) Transport: dissolved in blood / blood plasma [1 mark]. Effects (any two): increased heart rate / increased breathing rate / increased blood glucose / pupil dilation [2 marks]. (b) (i) Speed: nervous transmission is fast/rapid [1 mark] whereas hormonal/endocrine transmission is slow [1 mark]. (ii) Duration: nervous effect is short-lived/instantaneous [1 mark] whereas hormonal/endocrine effect is long-lasting [1 mark]. (c) Gravitropism definition: a growth response to gravity / growing towards or away from gravity [1 mark]. Benefit (any two): anchors the plant firmly in the soil [1 mark]; grows downwards to find/absorb water [1 mark]; grows downwards to find/absorb mineral ions [1 mark]. Max 2 marks for benefits.

Part (a): Positive phototropism (or phototropism). Part (b)(i): Group C (tip covered with an opaque cap) did not bend, showing that the tip is the site of light detection. Group B (tip removed) did not grow significantly, indicating the tip produces a growth-promoting substance. Group C grew well but did not bend, showing that growth occurs below the tip, but bending requires light detection by the tip. Part (b)(ii): Group D acts as a control for the physical presence of the cap, showing that the cap itself does not physically prevent bending as long as light can penetrate. Part (c): Auxin is synthesized in the tip of the shoot. Under unilateral light, auxin diffuses downwards and accumulates on the shaded side. A higher concentration of auxin on the shaded side stimulates greater cell elongation on that side, causing the shoot to bend towards the light.

Part (a): Positive phototropism [1 mark] (do not accept gravitropism). Part (b)(i): The tip detects light because Group C (opaque cap) shows 0 curvature [1 mark]; growth occurs below the tip because Group C grows (5.8 mm) [1 mark]; the tip produces growth hormone because Group B (no tip) shows almost no growth [1 mark]. Part (b)(ii): To act as a control for the cap's presence [1 mark]; to show the cap does not mechanically prevent bending [1 mark]. Part (c): Auxin produced in tip [1 mark]; diffuses down shoot [1 mark]; accumulates on shaded side [1 mark]; causes cell elongation [1 mark] leading to bending towards light [1 mark] (max 4 marks).

Part (a): Independent variable is temperature (or water availability). Dependent variable is germination percentage. Part (b): As temperature increases from 5 to 35 degrees C, germination increases from 5 percent to 92 percent. The optimum temperature is 35 degrees C. Above 35 degrees C, germination decreases to 12 percent at 45 degrees C. Part (c): The three conditions are water, oxygen, and suitable temperature. The data supports the requirement of water because in Treatment 2 (dry), no seeds germinated (0 percent) at any temperature, while in Treatment 1 (moist) germination occurred. Part (d): High temperature (45 degrees C) denatures enzymes needed for germination, changing the shape of their active sites so they can no longer function.

Part (a): Independent variable: temperature / moisture treatment [1 mark]. Dependent variable: germination percentage [1 mark]. Part (b): Germination increases from 5 to 35 degrees C [1 mark]; optimum germination is at 35 degrees C [1 mark]; germination decreases above 35 degrees C [1 mark]. Part (c): Water, oxygen, suitable temperature [1 mark for all three]; Treatment 2 had 0 percent germination [1 mark]; showing water is essential [1 mark]. Part (d): Enzymes denatured at 45 degrees C [1 mark]; active site changes shape / substrate cannot fit [1 mark].

**(a)**
Reactants: \(\text{Mg}\) and \(\text{H}_2\text{SO}_4\)
Products: \(\text{MgSO}_4\) and \(\text{H}_2\)
State symbols: \(\text{Mg(s)}\), \(\text{H}_2\text{SO}_4\text{(aq)}\), \(\text{MgSO}_4\text{(aq)}\), \(\text{H}_2\text{(g)}\)
Full balanced equation: \(\text{Mg(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{MgSO}_4\text{(aq)} + \text{H}_2\text{(g)}\)

**(b)**
1. Convert the volume of hydrogen gas to \(\text{dm}^3\):
\(240\text{ cm}^3 = \frac{240}{1000} = 0.24\text{ dm}^3\)
2. Calculate the moles of \(\text{H}_2\):
\(\text{Moles of } \text{H}_2 = \frac{0.24\text{ dm}^3}{24.0\text{ dm}^3/\text{mol}} = 0.010\text{ mol}\)
3. From the chemical equation, the mole ratio of \(\text{Mg} : \text{H}_2\) is \(1 : 1\).
Therefore, moles of \(\text{Mg} = 0.010\text{ mol}\).
4. Calculate the mass of \(\text{Mg}\):
\(\text{Mass} = 0.010\text{ mol} \times 24\text{ g/mol} = 0.24\text{ g}\)

**(c)**
1. From the equation, the mole ratio of \(\text{H}_2\text{SO}_4 : \text{Mg}\) is \(1 : 1\).
Therefore, moles of \(\text{H}_2\text{SO}_4 = 0.010\text{ mol}\).
2. Calculate the volume of sulfuric acid in \(\text{dm}^3\):
\(\text{Volume} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.010\text{ mol}}{0.50\text{ mol/dm}^3} = 0.020\text{ dm}^3\)
3. Convert volume to \(\text{cm}^3\):
\(0.020\text{ dm}^3 \times 1000 = 20\text{ cm}^3\)

**(d)**
Visible changes include: effervescence / bubbles of gas, the magnesium ribbon dissolving / disappearing, or the reaction mixture getting hot.

(a)
- Correct chemical formulas of reactants and products: \(\text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\) [1]
- Correctly balanced equation [1]
- Correct state symbols: \(\text{(s)}\), \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(g)}\) [1]

(b)
- Calculate moles of gas: \(\frac{240}{24000} = 0.010\text{ mol}\) [1]
- Deduced mole ratio of 1:1, so moles of Mg = 0.010 mol [1]
- Calculate mass of Mg: \(0.010 \times 24 = 0.24\text{ g}\) [1]

(c)
- Moles of sulfuric acid = 0.010 mol [1]
- Calculate volume in \(\text{dm}^3\): \(\frac{0.010}{0.50} = 0.020\text{ dm}^3\) [1]
- Convert to \(\text{cm}^3\): \(20\text{ cm}^3\) [1] (Allow error carried forward from previous steps)

(d)
- Any one valid visible change: effervescence / bubbles / gas evolved / magnesium dissolves / solid disappears / test tube gets hot [1]

**(a)**
The neutralization of sodium hydroxide by hydrochloric acid produces sodium chloride and water:
\(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)

**(b)**
1. Convert the volume of hydrochloric acid from \(\text{cm}^3\) to \(\text{dm}^3\):
\(22.0\text{ cm}^3 = \frac{22.0}{1000} = 0.0220\text{ dm}^3\)
2. Calculate the moles:
\(\text{Moles of HCl} = \text{concentration} \times \text{volume} = 0.15\text{ mol/dm}^3 \times 0.0220\text{ dm}^3 = 0.0033\text{ mol}\)

**(c)**
1. Identify the stoichiometry of the reaction:
From the balanced equation, \(1\text{ mol}\) of \(\text{NaOH}\) reacts with \(1\text{ mol}\) of \(\text{HCl}\).
Therefore, moles of \(\text{NaOH}\) in \(25.0\text{ cm}^3 = 0.0033\text{ mol}\).
2. Calculate the concentration of \(\text{NaOH}\):
\(\text{Volume of NaOH in dm}^3 = \frac{25.0}{1000} = 0.0250\text{ dm}^3\)
\(\text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.0033\text{ mol}}{0.0250\text{ dm}^3} = 0.132\text{ mol/dm}^3\) (or \(0.13\text{ mol/dm}^3\))

**(d)**
1. Use the conversion formula:
\(\text{Concentration in g/dm}^3 = \text{Concentration in mol/dm}^3 \times M_r\)
2. Substitute the values:
\(0.132\text{ mol/dm}^3 \times 40.0\text{ g/mol} = 5.28\text{ g/dm}^3\)

(a)
- Correct chemical formulas of reactants: \(\text{NaOH} + \text{HCl}\) [1]
- Correct chemical formulas of products: \(\text{NaCl} + \text{H}_2\text{O}\) [1]

(b)
- Conversion of volume to \(\text{dm}^3\): \(\frac{22.0}{1000} = 0.0220\text{ dm}^3\) [1]
- Calculation of moles of \(\text{HCl}\): \(0.15 \times 0.0220 = 0.0033\text{ mol}\) [1]

(c)
- Deducing 1:1 mole ratio, so moles of NaOH = 0.0033 mol [1]
- Conversion of sodium hydroxide volume to \(\text{dm}^3\): \(\frac{25.0}{1000} = 0.0250\text{ dm}^3\) [1]
- Calculate concentration of sodium hydroxide: \(\frac{0.0033}{0.0250} = 0.132\text{ mol/dm}^3\) [1] (Accept \(0.13\text{ mol/dm}^3\))

(d)
- Recognize formula: \(\text{conc in g/dm}^3 = \text{conc in mol/dm}^3 \times M_r\) [1]
- Correct calculation process: \(0.132 \times 40.0\) [1]
- Correct final answer with correct units: \(5.28\text{ g/dm}^3\) [1] (Allow ECF from (c): e.g., if concentration is \(0.13\text{ mol/dm}^3\), then answer is \(5.2\text{ g/dm}^3\))

(a) The balanced chemical equation with state symbols is:
\(\text{CuCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CuCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

(b) Step 1: Calculate the relative formula mass (\(M_r\)) of \(\text{CuCO}_3\) and \(\text{CuCl}_2\):
\(M_r(\text{CuCO}_3) = 64 + 12 + (3 \times 16) = 124\)
\(M_r(\text{CuCl}_2) = 64 + (2 \times 35.5) = 135\)

Step 2: Calculate the number of moles of \(\text{CuCO}_3\) reacted:
\(\text{Moles of CuCO}_3 = \frac{\text{mass}}{\text{formula mass}} = \frac{6.20}{124} = 0.050\text{ mol}\)

Step 3: Determine the moles of \(\text{CuCl}_2\) formed:
From the balanced equation, \(1\text{ mol}\) of \(\text{CuCO}_3\) produces \(1\text{ mol}\) of \(\text{CuCl}_2\).
Therefore, \(\text{moles of CuCl}_2 = 0.050\text{ mol}\).

Step 4: Calculate the theoretical mass of \(\text{CuCl}_2\):
\(\text{Mass} = 0.050\text{ mol} \times 135\text{ g/mol} = 6.75\text{ g}\).

(c) Percentage yield calculation:
\(\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% = \frac{5.40}{6.75} \times 100\% = 80.0\%\)

(d) Acceptable reasons include: loss of product during transfer, filtration, or evaporation; incomplete reaction; or side reactions occurring.

(a) [3 marks total]
- Correct chemical formulae of reactants and products: \(\text{CuCO}_3 + 2\text{HCl} \rightarrow \text{CuCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) [1]
- Equation fully balanced [1]
- All state symbols correct: \(\text{CuCO}_3\text{(s)}\), \(\text{HCl(aq)}\), \(\text{CuCl}_2\text{(aq)}\), \(\text{CO}_2\text{(g)}\), \(\text{H}_2\text{O(l)}\) [1]

(b) [4 marks total]
- Correct \(M_r\) values calculated (\(\text{CuCO}_3 = 124\) and \(\text{CuCl}_2 = 135\)) [1]
- Correct conversion of 6.20 g of \(\text{CuCO}_3\) to 0.050 mol [1]
- Deduces 0.050 mol of \(\text{CuCl}_2\) is produced (1:1 ratio) [1]
- Correct theoretical mass of 6.75 g [1]

(c) [2 marks total]
- Correct formula for percentage yield and substitution of values: \(\frac{5.40}{6.75} \times 100\) [1]
- Correct answer: 80.0% (or 80%) [1]

(d) [1 mark total]
- Accept any one valid reason: some product left on the filter paper or beaker / loss during transfer / evaporation / incomplete reaction. Reject: general 'experimental error' / 'human error' (unless specified as spilling).

(a) (i) Effervescence / bubbles of a colorless gas (oxygen) are formed.
(ii) A pink / brown / red-brown solid (copper metal) is deposited on the electrode.

(b) (i) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)
(ii) \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\)
(Alternatively: \(2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-\))

(c) The blue color of the solution is due to the presence of aqueous copper(II) ions, \(\text{Cu}^{2+}\text{(aq)}\). As these ions are discharged at the cathode to form copper atoms, they are removed from the solution. Since the carbon anodes are inert, they do not replace these copper ions, so the concentration of \(\text{Cu}^{2+}\) ions decreases, causing the blue color to fade.

(d) (i) The copper anode dissolves / becomes smaller / decreases in mass, and no bubbles of gas are produced.
(ii) The intensity of the blue color remains constant (unchanged).

(a) [2 marks total]
(i) Bubbles / effervescence / gas released [1]
(ii) Pink / brown / red-brown solid deposit [1]

(b) [4 marks total]
(i) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)
- Correct species (reactants and products) [1]
- Fully balanced with state symbols [1]
(ii) \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\) (or equivalent)
- Correct species (reactants and products) [1]
- Fully balanced with state symbols [1]

(c) [2 marks total]
- Identifies that blue color is due to \(\text{Cu}^{2+}\) ions [1]
- Explains that the concentration of \(\text{Cu}^{2+}\) decreases as they are discharged at the cathode / not replaced [1]

(d) [2 marks total]
(i) Anode dissolves / decreases in mass / no bubbles of gas observed [1]
(ii) Color intensity remains the same / does not change [1]

For (a): Use the acceleration formula: \(a = \frac{v - u}{t}\). Substituting the values: \(a = \frac{4.0 - 0}{2.0} = 2.0\text{ m/s}^2\). For (b): The total distance is the sum of distance in the acceleration phase and distance in the constant velocity phase. Distance 1 = area under the speed-time graph = \(0.5 \times 2.0\text{ s} \times 4.0\text{ m/s} = 4.0\text{ m}\). Distance 2 = \(5.0\text{ s} \times 4.0\text{ m/s} = 20.0\text{ m}\). Total distance = \(4.0 + 20.0 = 24.0\text{ m}\). For (c): Use the kinetic energy formula: \(E_k = \frac{1}{2}mv^2\). Substituting the values: \(E_k = 0.5 \times 0.50\text{ kg} \times (4.0\text{ m/s})^2 = 4.0\text{ J}\). For (d): Power is work done divided by time taken: \(P = \frac{W}{t}\). Here, work done is equal to the increase in kinetic energy: \(W = 4.0\text{ J}\) over a time of \(2.0\text{ s}\). Therefore, \(P = \frac{4.0\text{ J}}{2.0\text{ s}} = 2.0\text{ W}\).

(a) 1 mark for formula or substitution: \(a = (4.0 - 0)/2.0\); 1 mark for correct value with unit: \(2.0\text{ m/s}^2\). (b) 1 mark for calculating distance during acceleration: \(4.0\text{ m}\); 1 mark for calculating distance during constant speed: \(20\text{ m}\); 1 mark for adding the distances; 1 mark for correct total: \(24\text{ m}\). (c) 1 mark for formula or substitution: \(E_k = 0.5 \times 0.50 \times 4.0^2\); 1 mark for correct value: \(4.0\text{ J}\). (d) 1 mark for power formula: \(P = E/t\) or substitution; 1 mark for correct value: \(2.0\text{ W}\).

For (a): The formula for parallel resistors is \(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}\). Substituting the values: \(\frac{1}{R_p} = \frac{1}{12} + \frac{1}{6.0} = \frac{3}{12}\), so \(R_p = 4.0\ \Omega\). For (b): The total resistance is the sum of R1 and the parallel combination: \(R_{total} = R_1 + R_p = 2.0 + 4.0 = 6.0\ \Omega\). For (c): The current through R1 is the total current of the circuit, given by Ohm's Law: \(I = \frac{V}{R_{total}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). For (d): The potential difference across R1 is \(V_1 = I \times R_1 = 2.0\text{ A} \times 2.0\ \Omega = 4.0\text{ V}\). For (e): First convert time to seconds: \(10\text{ minutes} = 600\text{ s}\). Energy converted is \(E = I^2 \times R_1 \times t = (2.0)^2 \times 2.0 \times 600 = 8.0 \times 600 = 4800\text{ J}\) (or \(4.8\text{ kJ}\)).

(a) 1 mark for formula or substitution: \(1/R = 1/12 + 1/6\); 1 mark for correct value: \(4.0\ \Omega\). (b) 1 mark for adding resistances: \(2.0 + 4.0\); 1 mark for correct total: \(6.0\ \Omega\). (c) 1 mark for Ohm's law: \(I = V/R\); 1 mark for correct value: \(2.0\text{ A}\). (d) 1 mark for potential difference formula or substitution: \(2.0 \times 2.0\); 1 mark for correct value: \(4.0\text{ V}\). (e) 1 mark for converting time to seconds (600 s) and choosing energy formula \(E = Pt\) or \(E = I^2 R t\); 1 mark for correct value: \(4800\text{ J}\) (or \(4.8\text{ kJ}\)).

For (a), the voltmeter must be connected in parallel across the resistor \(R\), and the ammeter must be connected in series with the resistor \(R\). For (b), a variable resistor is represented by a standard rectangle with a diagonal arrow pointing through it, whereas a thermistor is represented by a rectangle with a diagonal line that has a flat horizontal section at its bottom-left end. For (c), we use the Ohm's law formula: \(R = \frac{V}{I}\). Substituting the given values: \(R = \frac{3.0\text{ V}}{0.50\text{ A}} = 6.0\ \Omega\). For (d)(i), the graph is a curve with a decreasing gradient as the potential difference increases (forming an S-shape through the origin). This occurs because a higher current increases the temperature of the filament, which increases the lattice vibrations of the metal atoms and consequently increases its resistance. For (d)(ii), the graph is a straight line of constant positive gradient passing through the origin, indicating that current is directly proportional to potential difference.

(a) Voltmeter connected in parallel across the resistor \(R\) [1 mark]; Ammeter connected in series with the resistor \(R\) [1 mark].
(b) Variable resistor described as a rectangle with a diagonal arrow through it [1 mark]; Thermistor described as a rectangle with a diagonal line having a horizontal step/flat section at the bottom [1 mark].
(c) State formula: \(R = \frac{V}{I}\) [1 mark]; Correct calculation to yield \(6.0\ \Omega\) (accept 6 or 6.0 ohms) [1 mark].
(d)(i) S-shaped curve passing through the origin / curve with a decreasing gradient [1 mark]; current increases the temperature of the filament [1 mark]; higher temperature increases resistance [1 mark].
(d)(ii) Straight line passing through the origin [1 mark].

For (a), a scalar quantity has magnitude only, whereas a vector quantity has both magnitude and direction. Velocity is a vector quantity. For (b), the acceleration is represented by the gradient (slope) of the speed-time graph, and the distance travelled is represented by the total area under the speed-time graph. For (c), the weight is calculated using the formula \(W = m \times g\). Substituting the values: \(W = 15\text{ kg} \times 9.8\text{ m/s}^2 = 147\text{ N}\). For (d), the weight acts downwards (\(147\text{ N}\)) and the air resistance acts upwards (\(50\text{ N}\)). The resultant force is: \(147\text{ N} - 50\text{ N} = 97\text{ N}\) acting downwards. For (e), as the box falls and its speed increases, the upward air resistance force acting on it also increases. Eventually, the magnitude of the upward air resistance becomes equal to the downward weight of the box. At this point, the forces are balanced, the resultant force is zero, and the box stops accelerating, falling at a constant terminal velocity.

(a) Scalar has magnitude only, vector has magnitude and direction [1 mark]; Velocity is a vector [1 mark].
(b) Acceleration is the gradient (slope) [1 mark]; Distance travelled is the area under the graph [1 mark].
(c) Correct formula used: \(W = m \times g\) [1 mark]; Correct calculation: \(15 \times 9.8 = 147\text{ N}\) [1 mark].
(d) Correct subtraction of forces (\(147 - 50\)) to give \(97\text{ N}\) [1 mark]; Direction stated as downwards [1 mark].
(e) Air resistance increases as speed increases [1 mark]; Air resistance becomes equal to weight, so the resultant force is zero (and acceleration becomes zero) [1 mark].