2023 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

(a)(i) Double circulation means that blood passes through the heart twice for each complete circuit around the body (one circuit to the lungs to be oxygenated, and one circuit to deliver oxygenated blood to the rest of the body). (a)(ii) It maintains a high blood pressure to the body organs, allowing faster and more efficient delivery of oxygen and nutrients. (b) The thick muscular wall helps withstand and maintain the high pressure of blood leaving the heart, while the elastic fibers stretch as blood surges and recoil to maintain blood pressure and smooth out the flow between heartbeats. (c) Capillary walls are only one cell thick, which provides a very short diffusion distance for substances like oxygen and glucose. They also have a very narrow lumen, which slows down blood flow to allow sufficient time for exchange and forces red blood cells close to the capillary wall for rapid diffusion.

(a)(i) [2 marks] - Blood passes through the heart twice [1]; for one complete circuit of the body [1]. (a)(ii) [1 mark] - Maintain high blood pressure to body tissues / faster delivery of oxygen or glucose / faster removal of waste [1]. (b) [3 marks] - Thick muscle wall to withstand high pressure [1]; elastic fibers stretch and recoil [1]; to maintain blood pressure / smooth out blood flow [1]. (c) [3 marks] - Walls are one cell thick / very thin [1]; provides short diffusion distance / rapid diffusion [1]; narrow lumen slows down blood flow / brings red blood cells close to tissues [1] (Max 3 total, must link structure to function for full marks).

(a)(i) Primary consumer: Rabbit or Grasshopper. Tertiary consumer: Frog (in the path Grass -> Grasshopper -> Spider -> Frog) or Fox (in the path Grass -> Grasshopper -> Frog -> Fox). (a)(ii) Grass is a producer. It absorbs light energy from the Sun and converts it into chemical energy (glucose) via photosynthesis. (b)(i) Energy is lost at each trophic level because not all parts of the grasshopper are eaten or digested (lost in feces). Energy is also lost as heat during respiration, and used for metabolic processes like movement. (b)(ii) Because energy is lost at each transfer, the amount of energy remaining decreases rapidly. After five levels, there is not enough energy left to support a viable population of consumers at a higher trophic level.

(a)(i) [2 marks] - Primary consumer: Rabbit / Grasshopper [1]; Tertiary consumer: Frog / Fox [1]. (a)(ii) [2 marks] - Producer [1]; Absorbs light / solar energy [1] (accept: performs photosynthesis). (b)(i) [3 marks] - Energy lost as heat from respiration [1]; Energy lost in undigested material / feces / uneaten parts [1]; Energy used for movement / metabolic activities [1]. (b)(ii) [2 marks] - Energy is lost at each stage / trophic level [1]; Insufficient energy remains to support another trophic level [1].

(a) \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\). (b) The palisade mesophyll cells are situated near the upper surface of the leaf to receive maximum sunlight. The cells are packed with a high density of chloroplasts to absorb light energy. Additionally, the cells are elongated and tightly packed vertically, which allows a large number of cells to be exposed directly to the sunlight. (c) At low concentrations, carbon dioxide is a limiting factor, so increasing it speeds up the rate of photosynthesis. However, at higher concentrations, the rate plateaus because carbon dioxide is no longer the limiting factor. Some other factor, such as light intensity or temperature, is now in short supply and limits the rate.

(a) [2 marks] - Correct reactants and products: \(CO_2 + H_2O \rightarrow C_6H_{12}O_6 + O_2\) [1]; Correct balancing: \(6, 6 \rightarrow 1, 6\) [1]. (b) [4 marks] - Positioned near the upper epidermis/surface [1] to receive maximum light [1]; Cells contain many chloroplasts [1] to absorb maximum light energy [1]; Cells are elongated / column-shaped [1] and tightly packed to maximize absorption in a small area [1]. (Max 4 marks, must pair feature with adaptation). (c) [3 marks] - At first, carbon dioxide is the limiting factor / rate is proportional to concentration [1]; At higher concentrations, the curve levels off because carbon dioxide is no longer limiting [1]; Another factor (e.g. light intensity / temperature) has become the limiting factor [1].

(a) The gas produced at the positive anode is chlorine. It can be identified using damp blue litmus paper, which turns red and then bleaches white. (b) At the negative cathode, bubbles of a colourless gas (hydrogen) are observed. Since hydrogen is lower in the reactivity series than sodium, hydrogen ions (\(H^+\)) are preferentially discharged to form hydrogen gas, while sodium ions (\(Na^+\)) remain in the solution. (c) As hydrogen ions from water are discharged, hydroxide ions (\(OH^-\)) are left behind. The accumulation of these hydroxide ions increases the pH, making the solution alkaline.

(a) Chlorine gas [1]; damp blue litmus paper [1]; turns red then bleaches white [1]. (b) Bubbles / effervescence / gas produced [1]; H+ and Na+ are present but hydrogen is less reactive than sodium [1]; H+ is preferentially discharged to form hydrogen gas [1]. (c) Hydrogen ions are discharged / removed [1]; leaving an excess of hydroxide (OH-) ions in solution [1]; concentration of OH- increases the pH [0.67].

(a) High temperature (around 600-700 degrees Celsius) and a catalyst (such as alumina, silica, or zeolites) are required. (b)(i) Subtracting octane (C8H18) from decane (C10H22) gives C2H4, which is ethene. (ii) Ethene is drawn with a carbon-carbon double bond: H2C=CH2 with all individual C-H bonds explicitly shown. (c) Add bromine water (which is orange). Ethene will decolourise the bromine water (turn from orange to colourless), while octane will cause no colour change (remains orange).

(a) High temperature [1]; Catalyst / alumina / silica / zeolite [1]. (b)(i) C2H4 [1]; ethene [1]. (b)(ii) Fully displayed structure of ethene showing C=C double bond and four C-H single bonds [1.67]. (c) Bromine water / aqueous bromine [1]; starting colour orange/brown/yellow [1]; colourless with ethene AND stays orange with octane [1].

(a)(i) The main reducing agent is carbon monoxide. The balanced equation is \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\). (a)(ii) Limestone undergoes thermal decomposition to produce calcium oxide and carbon dioxide. The calcium oxide (a basic oxide) reacts with silicon dioxide (sand, an acidic impurity) to form molten calcium silicate (slag, \(CaSiO_3\)), which is removed from the furnace. (b)(i) Brass is composed of copper and zinc. (b)(ii) Pure copper has regular layers of identical atoms that slide over each other easily when force is applied. In brass, the introduction of zinc atoms of a different size disrupts this regular lattice, preventing layers from sliding easily, making it harder.

(a)(i) Carbon monoxide [1]; equation reactants and products correct [1]; balancing correct [1]. (a)(ii) Calcium carbonate decomposes to calcium oxide [1]; calcium oxide reacts with silicon dioxide / sand [1]; calcium silicate / slag is formed [1]. (b)(i) Copper and zinc [1] (both needed). (b)(ii) Zinc atoms are a different size to copper atoms [0.67]; layers of atoms are disrupted and cannot slide over each other easily [1].

(a) Acceleration is change in speed divided by time:
\( a = \frac{v - u}{t} = \frac{8.0 - 0}{4.0} = 2.0 \text{ m/s}^2 \)

(b) Distance can be found from the area under the speed-time graph:
- Distance during acceleration (0 to 4 s) = area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 8.0 = 16 \text{ m} \)
- Distance during constant speed (4 to 10 s) = area of rectangle = \( \text{width} \times \text{height} = 6.0 \times 8.0 = 48 \text{ m} \)
- Total distance = \( 16 + 48 = 64 \text{ m} \)

(c) First, calculate the useful kinetic energy gained by the car:
\( E_k = \frac{1}{2} m v^2 \)
Convert mass to kg: \( 120 \text{ g} = 0.12 \text{ kg} \)
\( E_k = \frac{1}{2} \times 0.12 \times 8.0^2 = 0.06 \times 64 = 3.84 \text{ J} \)
Then, calculate the input energy using the efficiency formula:
\( \text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\\% \)
\( 0.75 = \frac{3.84}{E_{\text{in}}} \)
\( E_{\text{in}} = \frac{3.84}{0.75} = 5.12 \text{ J} \)

(a)
- Use of \( a = \Delta v / t \) or equivalent: [1 mark]
- Correct answer with unit (2.0 m/s^2): [1 mark]

(b)
- Calculation of distance during first 4.0 s (16 m) OR area under first part of graph: [1 mark]
- Calculation of distance during remaining 6.0 s (48 m) OR area under second part of graph: [1 mark]
- Correct total distance (64 m): [1 mark]

(c)
- Conversion of mass to kg (0.12 kg) AND use of \( E_k = \frac{1}{2} m v^2 \): [1 mark]
- Calculation of useful kinetic energy (3.84 J): [1 mark]
- Correct use of efficiency formula (dividing kinetic energy by 0.75): [1 mark]
- Correct final answer (5.12 J): [1 mark]

(a) Sound is a longitudinal wave. The particles/molecules of the medium vibrate or oscillate back and forth parallel to the direction of energy transfer (wave propagation), forming alternating regions of high pressure (compressions) and low pressure (rarefactions).

(b) (i) The pulse travels to the seabed and back, so the total distance is twice the depth:
\( 2 \times \text{depth} = \text{speed} \times \text{time} \)
\( 2 \times \text{depth} = 1500 \text{ m/s} \times 0.40 \text{ s} = 600 \text{ m} \)
\( \text{depth} = 300 \text{ m} \)

(ii) Use the wave equation \( v = f \lambda \):
Convert frequency to Hz: \( 30 \text{ kHz} = 30\\,000 \text{ Hz} \)
\( \lambda = \frac{v}{f} = \frac{1500}{30\\,000} = 0.05 \text{ m} \) (or 5.0 cm)

(c) (i) The frequency of a wave is determined by the source and remains the same when crossing a boundary between media.
(ii) Use \( \lambda = \frac{v}{f} \) with the speed in air:
\( \lambda = \frac{340}{30\\,000} \approx 0.0113 \text{ m} \) (or 1.1 cm)

(a)
- Mention of molecules vibrating/oscillating back and forth parallel to the direction of wave/energy travel: [1 mark]
- Mention of compressions and rarefactions (or regions of high and low pressure): [1 mark]

(b)(i)
- Correct formula or understanding that distance is halved: [1 mark]
- Correct calculation (300 m): [1 mark]

(b)(ii)
- Conversion of frequency to 30,000 Hz and use of \( v = f \lambda \): [1 mark]
- Correct calculation (0.05 m or 5.0 cm): [1 mark]

(c)(i)
- Correct statement: remains the same: [1 mark]

(c)(ii)
- Use of same frequency (30,000 Hz) with speed of 340 m/s: [1 mark]
- Correct calculation (0.0113 m or 1.1 cm): [1 mark]

(a) The circuit diagram should show a loop containing the 6.0 V d.c. power supply. The ammeter and the switch must be placed in series in the main branch of the circuit so that the ammeter measures the total current and the switch controls the entire circuit. The circuit branch must then split into two parallel branches, one containing the 12 \( \Omega \) resistor and the other containing the 6.0 \( \Omega \) resistor, before recombining back to the power supply.

(b) For parallel resistors, the formula is:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \)
\( \frac{1}{R_p} = \frac{1}{12} + \frac{1}{6.0} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4.0} \)
\( R_p = 4.0 \\ \Omega \)

(c) Use Ohm's Law: \( I = \frac{V}{R_p} \)
\( I = \frac{6.0 \text{ V}}{4.0 \\ \Omega} = 1.5 \text{ A} \)

(d) Convert time to seconds: \( 5.0 \text{ minutes} = 5.0 \times 60 = 300 \text{ s} \)
Use the electrical energy formula: \( E = V I t \)
\( E = 6.0 \text{ V} \times 1.5 \text{ A} \times 300 \text{ s} = 2700 \text{ J} \) (or 2.7 kJ)

(a)
- Correct symbols for power supply/battery, switch, ammeter, and two resistors: [1 mark]
- Switch and ammeter placed in the main branch: [1 mark]
- Both resistors connected in parallel with each other: [1 mark]

(b)
- Correct formula used, e.g., \( 1/R = 1/R_1 + 1/R_2 \) or product-over-sum: [1 mark]
- Correct calculation showing equivalent resistance is 4.0 \( \Omega \): [1 mark]

(c)
- Correct use of \( I = V/R \) with total resistance: [1 mark]
- Correct current (1.5 A): [1 mark]

(d)
- Conversion of time to seconds (300 s): [1 mark]
- Correct energy calculation (2700 J): [1 mark]