2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

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

(b)(i) Distilled water has a higher water potential than the cell sap of the potato cells. Water moves into the potato cells by osmosis down a water potential gradient, through the partially permeable membrane, increasing the mass of the cells.

(b)(ii) The cells would lose water and become plasmolysed or flaccid. This is because the sucrose solution in Tube C has a lower water potential than the cell sap of the potato cells, so water leaves the cells by osmosis down the water potential gradient.

(b)(iii) Any one from: surface area of the potato cylinders, duration of immersion in the solutions, variety of potato used, or age/source of the potato.

(a) Max 3 marks:
- 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) Max 3 marks:
- distilled water has higher water potential than the potato cells [1]
- water moves into the potato cells by osmosis [1]
- through a partially permeable membrane, increasing mass [1]

(b)(ii) Max 3 marks:
- cells become flaccid / plasmolysed / cytoplasm shrinks away from cell wall [1]
- sucrose solution has a lower water potential than the cells [1]
- water moves out of the cells by osmosis [1]

(b)(iii) 1 mark:
- Accept: surface area / length / diameter of potato cylinder, duration of immersion, variety / source of potato. Reject: mass of potato, volume of solution, temperature.

(a) Homeostasis is the maintenance of a constant internal environment.

(b)(i) Hormone: insulin. Target organ: liver (or muscle cells).

(b)(ii) The pancreas detects the low blood glucose levels and secretes the hormone glucagon. Glucagon travels in the blood to the liver, where it binds to receptors on liver cells. This stimulates liver cells to break down glycogen into glucose and release it into the blood, increasing the blood glucose level back to normal.

(c) Negative feedback occurs when a change in a parameter (such as blood glucose) triggers a corrective response that reverses the direction of the change, returning the parameter back to its set point.

(a) Max 2 marks:
- maintenance of a constant [1]
- internal environment [1]

(b)(i) Max 2 marks:
- insulin [1]
- liver / muscles [1]

(b)(ii) Max 4 marks:
- decrease detected by pancreas [1]
- pancreas secretes glucagon [1]
- glucagon stimulates liver to convert glycogen to glucose [1]
- glucose released into the blood [1]

(c) Max 2 marks:
- a change in blood glucose level triggers a response [1]
- that opposes / reverses the change to return it to the set point [1]

(a)(i) allele

(a)(ii) heterozygous

(b)(i)
- Parental genotypes: \(Tt \times Tt\)
- Gametes: \(T\) and \(t\) (from each parent)
- Punnett square:
| | \(T\) | \(t\) |
|---|---|---|
| \(T\) | \(TT\) | \(Tt\) |
| \(t\) | \(Tt\) | \(tt\) |
- Offspring genotypes: \(TT\), \(Tt\), \(tt\)

(b)(ii) 3 tall : 1 dwarf

(c) Fertilisation is a random process, meaning the combination of gametes is due to chance. Additionally, some seeds may fail to germinate, or the sample size may be too small to reflect the probability accurately.

(d) \(tt\)

(a)(i) 1 mark:
- allele [1]

(a)(ii) 1 mark:
- heterozygous [1]

(b)(i) Max 4 marks:
- parental genotypes: \(Tt \times Tt\) [1]
- gametes shown as \(T\) and \(t\) from both parents [1]
- Punnett square correctly constructed [1]
- offspring genotypes shown as \(TT\), \(Tt\), \(tt\) [1]

(b)(ii) 1 mark:
- 3 tall : 1 dwarf (or 3 : 1) [1]

(c) Max 2 marks:
- fertilisation is random / due to chance [1]
- small sample size [1]
- differential survival of gametes or embryos / some seeds do not germinate [1]

(d) 1 mark:
- \(tt\) [1]

(a) A producer is an organism that makes its own organic nutrients (usually glucose) using energy from sunlight through photosynthesis.

(b)(i) The food web should show:
- Grass with arrows pointing to rabbits and grasshoppers.
- Rabbits with an arrow pointing to foxes.
- Grasshoppers with an arrow pointing to frogs.
- Frogs with an arrow pointing to snakes.
(Note: All arrows must point in the direction of energy flow, from the organism being eaten to the consumer.)

(b)(ii) Tertiary consumer (or 4th trophic level).

(c) Much of the energy is lost to the environment as heat from respiration in the grasshoppers. Some energy is lost in undigested waste (egestion) or excreted waste (excretion). Additionally, not all parts of the grasshoppers are consumed by the frogs.

(d) All the populations of different species living and interacting in an ecosystem at the same time.

(a) Max 2 marks:
- makes its own organic nutrients / food [1]
- using energy from sunlight / photosynthesis [1]

(b)(i) Max 3 marks:
- grass at base with arrows pointing to rabbits and grasshoppers [1]
- arrow from rabbits to foxes [1]
- arrow from grasshoppers to frogs and frogs to snakes [1]
- Note: Deduct 1 mark if arrow directions are reversed / incorrect.

(b)(ii) 1 mark:
- tertiary consumer / fourth trophic level [1]

(c) Max 3 marks:
- energy lost as heat during respiration [1]
- energy lost in excretion / egestion / faeces / waste [1]
- not all parts of the organism are eaten / digested [1]

(d) 1 mark:
- all the populations of different species living and interacting in an ecosystem [1]

(a)(i) Thermal decomposition of copper(II) carbonate yields copper(II) oxide and carbon dioxide gas: \(CuCO_3(s) \rightarrow CuO(s) + CO_2(g)\). (b) Moles of \(CuCO_3 = 6.20 / 124 = 0.05\text{ mol}\). Since the stoichiometry is 1:1, moles of \(CO_2\) produced is also 0.05 mol. Volume of \(CO_2 = 0.05 \times 24 = 1.20\text{ dm}^3\) (or \(1200\text{ cm}^3\)). (c)(i) Observations include: the green solid dissolves/disappears, effervescence/fizzing, and the solution turns blue. (c)(ii) Using warm acid increases the kinetic energy of the particles, so they move faster and collide more frequently. Additionally, a greater proportion of particles have energy greater than or equal to the activation energy, resulting in a higher rate of successful collisions.

(a)(i) 1 mark for correct reactants and products; 1 mark for correct balancing; 1 mark for correct state symbols. (b) 1 mark for calculating moles of CuCO3 (0.05 mol); 1 mark for stating mole ratio is 1:1; 1 mark for calculating volume of gas (1.20 dm3 or 1200 cm3). (c)(i) 1 mark for each of two correct observations: green solid disappears, effervescence, blue solution forms (max 2). (c)(ii) 1 mark for stating particles have more kinetic energy / collide more frequently; 1 mark for stating more particles have energy greater than or equal to activation energy.

(a)(i) Catalytic cracking requires high temperature (typically 450-800 °C) and a catalyst (such as alumina, silica, or zeolite). (a)(ii) The missing product is \(C_5H_{10}\) (pentene), as the total number of carbon and hydrogen atoms must balance on both sides of the equation. (b)(i) Propene has a double bond between two carbon atoms and a single bond to the third carbon. All atoms must be explicitly shown with their bonds: H-C(H)=C(H)-C(H)(H)H. (b)(ii) The reagent used is aqueous bromine (or bromine water). With propene (unsaturated), the orange-brown colour disappears/decolourises. With propane (saturated), the solution remains orange-brown. (c) The repeat unit of poly(propene) has a backbone of two single-bonded carbon atoms with extension bonds, containing three hydrogen atoms and one methyl (\(-CH_3\)) group attached.

(a)(i) 1 mark for high temperature (or range 450-800 °C); 1 mark for a catalyst (or alumina/silica/zeolite). (a)(ii) 1 mark for C5H10. (b)(i) 1 mark for a correct carbon skeleton with one double and one single C-C bond; 1 mark for correctly showing 6 C-H bonds and all atoms labeled. (b)(ii) 1 mark for specifying bromine water / aqueous bromine; 1 mark for propene decolourising; 1 mark for propane showing no change / remaining orange. (c) 1 mark for drawing a two-carbon backbone with single bond and open-ended bonds at the sides; 1 mark for correctly showing three hydrogen atoms and one methyl group attached.

(a)(i) Lead(II) ions (\(Pb^{2+}\)) are positively charged and move to the negative electrode (cathode), where they gain electrons to form grey liquid lead. Bromide ions (\(Br^-\)) are negatively charged and move to the positive electrode (anode), where they lose electrons to form red-brown bromine gas. (a)(ii) At the cathode: \(Pb^{2+}(l) + 2e^- \rightarrow Pb(l)\). At the anode: \(2Br^-(l) \rightarrow Br_2(g) + 2e^-\). (b)(i) At the anode, chloride ions are preferentially discharged to form chlorine gas. At the cathode, hydrogen ions are discharged to form hydrogen gas. (b)(ii) Water dissociates into \(H^+\) and \(OH^-\). As \(H^+\) ions are discharged at the cathode and \(Cl^-\postive\) ions are discharged at the anode, a high concentration of \(Na^+\) and \(OH^-\postive\) ions remains in the solution, forming sodium hydroxide which is highly alkaline.

(a)(i) 1 mark for Pb2+ moving to the cathode; 1 mark for observing grey liquid lead at the cathode; 1 mark for Br- moving to the anode; 1 mark for observing red-brown gas at the anode. (a)(ii) 1 mark for correct cathode half-equation with state symbols; 1 mark for correct anode half-equation with state symbols. (b)(i) 1 mark for chlorine at the anode; 1 mark for hydrogen at the cathode. (b)(ii) 1 mark for noting H+ and Cl- are discharged / removed; 1 mark for explaining that Na+ and OH- remain in solution to form sodium hydroxide.

(a)(i) Nitrogen is obtained from the fractional distillation of liquid air. Hydrogen is obtained by reacting methane (from natural gas) with steam. (a)(ii) The chemical equation is \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\). The essential industrial conditions are a temperature of 450 °C, a pressure of 200 atmospheres, and an iron catalyst. (b)(i) First, perform a titration with an indicator to find the exact volume of dilute sulfuric acid needed to neutralize a set volume of aqueous ammonia. Second, repeat the reaction using these exact volumes but without the indicator to ensure the product is uncontaminated. Third, heat the solution to evaporate most of the water, leave to cool so crystals form, then filter the crystals, wash them with a small amount of cold distilled water, and dry them between filter papers. (b)(ii) Calcium hydroxide is alkaline and reacts with ammonium salts to release ammonia gas. This gas escapes into the air, thereby reducing the nitrogen content remaining in the soil.

(a)(i) 1 mark for nitrogen from liquid air; 1 mark for hydrogen from natural gas/methane/steam. (a)(ii) 1 mark for the correct balanced reversible equation; 1 mark for temperature of 450 °C (accept 350-500 °C); 1 mark for pressure of 200 atm (accept 150-300 atm); 1 mark for iron catalyst. (b)(i) 1 mark for using titration with indicator to determine exact neutralizing volumes; 1 mark for repeating without indicator using the same volumes; 1 mark for heating to point of crystallization, filtering, washing, and drying the crystals. (b)(ii) 1 mark for explaining that calcium hydroxide reacts with ammonium sulfate to produce ammonia gas which escapes (or reduces soil nitrogen).

For (a)(i), the acceleration is given by the change in speed divided by time: \(a = \frac{v - u}{t} = \frac{6.0 - 0}{4.0} = 1.5\text{ m/s}^2\). For (a)(ii), the total distance is the area under the speed-time graph: distance during acceleration (0 to 4.0 s) is \(0.5 \times 4.0 \times 6.0 = 12\text{ m}\), and distance during constant speed (4.0 to 9.0 s, duration of 5.0 s) is \(5.0 \times 6.0 = 30\text{ m}\). Total distance is \(12 + 30 = 42\text{ m}\). For (b)(i), kinetic energy is given by \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.5 \times 6.0^2 = 27\text{ J}\). For (b)(ii), work done is given by \(W = F \times d\). The distance travelled during this constant speed phase is \(30\text{ m}\). \(W = 4.5\text{ N} \times 30\text{ m} = 135\text{ J}\).

(a)(i) Correct substitution of speed and time: a = 6.0 / 4.0 [1 mark]. Correct final acceleration value with unit: 1.5 m/s^2 [1 mark]. (a)(ii) Calculation of distance during acceleration phase: 12 m [1 mark]. Calculation of distance during constant speed phase: 30 m [1 mark]. Correct sum with unit: 42 m [1 mark]. (b)(i) Correct formula or substitution: E_k = 0.5 * 1.5 * (6.0^2) [1 mark]. Correct kinetic energy with unit: 27 J [1 mark]. (b)(ii) Recalling correct distance of 30 m [1 mark]. Correct formula and calculation of work done: W = 4.5 * 30 [1 mark]. Correct work done value with unit: 135 J [1 mark].

For (a)(i), the critical angle \(c\) is calculated using \(\sin(c) = \frac{1}{n}\). \(\sin(c) = \frac{1}{1.52} \approx 0.6579\), giving \(c = \arcsin(0.6579) \approx 41.1^\circ\). For (a)(ii), since \(35^\circ\) is less than the critical angle, the light refracts out of the glass block. Since \(50^\circ\) is greater than the critical angle, total internal reflection occurs. For (b)(i), the speed of light in glass is \(v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.52} \approx 1.97 \times 10^8\text{ m/s}\). For (b)(ii), the wavelength is \(\lambda = \frac{v}{f}\). The frequency is unchanged in the glass block. \(\lambda = \frac{1.97 \times 10^8}{5.1 \times 10^{14}} \approx 3.86 \times 10^{-7}\text{ m}\). Expressed to 2 significant figures: \(3.9 \times 10^{-7}\text{ m}\).

(a)(i) Correct formula or substitution: sin(c) = 1 / 1.52 [1 mark]. Correct critical angle: 41 degrees or 41.1 degrees [1 mark]. (a)(ii) State that at 35 degrees the ray refracts out into air [1 mark]. State that at 50 degrees total internal reflection occurs [1 mark]. (b)(i) Correct formula or substitution: v = (3.0 * 10^8) / 1.52 [1 mark]. Correct speed with unit: 1.97 * 10^8 m/s (accept 2.0 * 10^8 m/s) [1 mark]. (b)(ii) State that frequency remains unchanged [1 mark]. Correct formula or substitution: lambda = (1.97 * 10^8) / (5.1 * 10^14) [1 mark]. Correct calculated wavelength: 3.86 * 10^-7 m [1 mark]. Correct rounding to 2 significant figures: 3.9 * 10^-7 m [1 mark].

For (a)(i), the current through the \(6.0\ \Omega\) resistor is equal to the total current leaving the supply because they are connected in series. For (a)(ii), the resistance of the parallel combination \(R_p\) is calculated using \(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0}\). Therefore, \(R_p = \frac{12.0}{4} = 3.0\ \Omega\). For (a)(iii), the total resistance is the sum of the series resistor and the parallel combination: \(R_{\text{total}} = 6.0 + 3.0 = 9.0\ \Omega\). For (b)(i), the total current is \(I = \frac{V}{R_{\text{total}}} = \frac{12.0}{9.0} = 1.33\text{ A}\) (or \(\frac{4}{3}\text{ A}\)). For (b)(ii), the potential difference across the parallel combination is \(V_p = I \times R_p = 1.33 \times 3.0 = 4.0\text{ V}\). For (b)(iii), the power dissipated in the \(6.0\ \Omega\) resistor is \(P = I^2 R = (1.33^2) \times 6.0 = 10.7\text{ W}\) (or using \(V_{\text{series}} = 12.0 - 4.0 = 8.0\text{ V}\), \(P = \frac{V^2}{R} = \frac{8.0^2}{6.0} = 10.7\text{ W}\)).

(a)(i) State that currents are equal [1 mark]. Explain that the resistor is in series with the supply so all current must pass through it [1 mark]. (a)(ii) Correct parallel resistance formula: 1/R_p = 1/4 + 1/12 [1 mark]. Correct calculation leading to 3.0 ohm [1 mark]. (a)(iii) Correct total resistance: 9.0 ohm (accept ecf from a-ii) [1 mark]. (b)(i) Correct formula or substitution: I = 12.0 / 9.0 [1 mark]. Correct current with unit: 1.33 A or 1.3 A [1 mark]. (b)(ii) Correct formula or substitution: V = I * R_p = 1.33 * 3.0 [1 mark]. Correct potential difference with unit: 4.0 V [1 mark]. (b)(iii) Correct calculation of power with unit: 10.7 W or 11 W [1 mark].

For (a)(i), an alpha particle contains 2 protons and 2 neutrons. For (a)(ii), conservation of mass number: \(212 = A + 4\), so \(A = 208\). Conservation of atomic number: \(83 = Z + 2\), so \(Z = 81\). For (b)(i), half-life is the time taken for the activity or number of radioactive nuclei in a sample to fall to half of its initial value. For (b)(ii), the number of half-lives elapsed is \(\frac{242}{60.5} = 4\). The activity halves 4 times: \(3200 \rightarrow 1600 \rightarrow 800 \rightarrow 400 \rightarrow 200\text{ Bq}\). For (b)(iii), two safety precautions are using tongs to maintain distance, and storing radioactive sources in lead-lined containers.

(a)(i) Mention 2 protons [1 mark] and 2 neutrons [1 mark]. (a)(ii) Correct value for A: 208 [1 mark]. Correct value for Z: 81 [1 mark]. (b)(i) Mention of 'time taken' [1 mark]. Mention 'for activity/number of nuclei to halve' [1 mark]. (b)(ii) Determine that 4 half-lives have elapsed: 242 / 60.5 = 4 [1 mark]. Correct final activity with unit: 200 Bq [1 mark]. (b)(iii) First valid precaution (e.g., use tongs, lead shielding, protective clothing) [1 mark]. Second different valid precaution [1 mark].