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

(a) Mass increased by \(12.5\%\); because the water potential outside the potato tissue was higher than inside the potato cells, causing water to move into the cells down a water potential gradient by osmosis.
(b) Osmosis.
(c) Between \(0.2\) and \(0.4\text{ }mol/dm^3\) (specifically around \(0.36\text{ }mol/dm^3\)); because this is the concentration where there is no percentage change in mass, meaning there is no net movement of water into or out of the cells (isotonic point).
(d) Water moves out of the potato cells by osmosis down a water potential gradient; the vacuole and cytoplasm shrink, causing turgor pressure to drop; the cell membrane pulls away from the cell wall, leaving the cells flaccid or plasmolysed.
(e) Temperature / potato variety / duration of immersion / surface area of potato cylinders.

(a)
- Mass increased / gained mass [1]
- Water potential is higher outside the tissue than inside [1]
- Water moves in by osmosis [1]

(b)
- Osmosis [1]

(c)
- Estimate in the range of 0.35 to 0.37 \(mol/dm^3\) (or state between 0.2 and 0.4 \(mol/dm^3\)) [1]
- Explanation: No net movement of water / no change in mass occurs here [1]

(d)
- Water leaves cells by osmosis / vacuole shrinks [1]
- Loss of turgor pressure / cells become flaccid / flaccid state [1]
- Plasmolysis / cell membrane pulls away from cell wall [1]

(e)
- Temperature / variety of potato / duration of immersion [1]

(a) Homeostasis is the maintenance of a constant internal environment to keep conditions within set limits.
(b)(i) Arterioles constrict / narrow; this reduces the flow of blood through the capillaries near the surface of the skin; less heat is lost to the environment by radiation.
(ii) Shivering involves rapid, involuntary muscle contraction and relaxation; this increases the rate of respiration in muscle cells, which releases heat energy as a byproduct.
(c)(i) It detects changes in the temperature of the blood flowing through it.
(ii) Nerve impulses travel as electrical signals along motor neurones; they cross synapses using neurotransmitters to reach effector organs in the skin (like arterioles and muscles).

(a)
- Maintenance of constant internal environment [1]
- Within set limits / to keep conditions stable [1]

(b)(i)
- Arterioles narrow / constrict [1] (REJECT: capillaries constrict)
- Less blood flows through capillaries close to the skin surface [1]
- Less heat lost by radiation / convection [1]

(b)(ii)
- Involuntary / rapid contraction and relaxation of muscles [1]
- Increases rate of cellular respiration releasing heat energy [1]

(c)(i)
- Detects the temperature of blood flowing through the brain / hypothalamus [1]

(c)(ii)
- Signals travel as electrical impulses along motor neurones [1]
- Reaches effectors (such as muscles or arterioles) [1]

(a) Palisade mesophyll cells are situated close to the upper epidermis to receive the most sunlight; they are vertically elongated and packed tightly together to maximize cellular surface area; they contain a high density of chloroplasts to absorb light energy.
(b)(i) Carbon dioxide diffuses into the leaf through stomata; it travels through air spaces in the spongy mesophyll; it diffuses through the cell walls, membranes, and cytoplasm of palisade cells to reach chloroplasts.
(b)(ii) The completed terms are: \(6\text{H}_2\text{O}\) and \(\text{C}_6\text{H}_{12}\text{O}_6\).
(c) Glucose is used in respiration to release energy; stored as starch; used to make cellulose for cell walls; combined with nitrate ions to make amino acids/proteins.

(a)
- Positioned near upper surface / upper epidermis [1]
- Vertically elongated / column-shaped / packed tightly [1]
- Contain many / high concentration of chloroplasts [1]

(b)(i)
- Diffuses through stomata [1]
- Travels through air spaces in spongy mesophyll [1]
- Diffuses into palisade cells / chloroplasts [1]

(b)(ii)
- \(6\text{H}_2\text{O}\) [1]
- \(\text{C}_6\text{H}_{12}\text{O}_6\) [1]

(c)
- Any two valid uses: respiration / starch storage / making cellulose / amino acid synthesis [2]

(a) Genotype is \(Dd\); phenotype is normal / healthy / carrier (no cystic fibrosis).
(b)(i) Gametes are \(D\) and \(d\) from both parents. Offspring genotypes are: \(DD\), \(Dd\), \(Dd\), and \(dd\).
(b)(ii) The probability of having a child with cystic fibrosis is \(25\%\) (or \(1/4\) or \(0.25\)).
(c) A gene is a length of DNA that codes for a protein.
(d)(i) A mutation is a spontaneous change in the base sequence of DNA.
(ii) Ionising radiation (e.g., UV rays, X-rays) and chemical mutagens (e.g., chemicals in tobacco tar/smoke).

(a)
- Genotype: \(Dd\) [1]
- Phenotype: Normal / unaffected / carrier [1]

(b)(i)
- Correct parental gametes: \(D\) and \(d\) from both [1]
- Correct heterozygous offspring genotypes: \(Dd\) and \(Dd\) [1]
- Correct homozygous offspring genotypes: \(DD\) and \(dd\) [1]

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

(c)
- Length of DNA that codes for a protein [1]

(d)(i)
- Spontaneous change in the base sequence of DNA [1]

(d)(ii)
- Any two mutagenic factors: UV radiation, X-rays, gamma rays, mustard gas, tobacco tar [2]

(a) Anode: Effervescence / bubbles of colourless gas. Cathode: Pink-brown solid / copper metal deposited on the electrode. (b) Gas: Oxygen. Test: Place a glowing splint in the gas; it relights. (c) Copper(II) ions (Cu2+) are responsible for the blue colour. As copper(II) ions are discharged at the cathode to form copper atoms, their concentration in the solution decreases, causing the blue colour to fade. (d) Cu2+(aq) + 2e- -> Cu(s). (e) Mass of copper anode decreases. Copper atoms at the anode lose electrons and dissolve into the solution as copper(II) ions: Cu(s) -> Cu2+(aq) + 2e-.

(a) 1 mark for correct anode observation (effervescence/bubbles), 1 mark for correct cathode observation (pink/brown solid). (b) 1 mark for identifying oxygen, 1 mark for correct test (glowing splint relights). (c) 1 mark for linking blue colour to copper(II) ions / Cu2+, 1 mark for stating concentration decreases as ions are discharged to form copper atoms. (d) 1 mark for correct reactants and products (Cu2+ and Cu), 1 mark for correct balancing of electrons (2e- on reactant side). (e) 1 mark for stating mass decreases / anode dissolves, 1 mark for explanation (copper atoms are oxidised to form copper(II) ions).

(a)(i) C10H22 -> C8H18 + C2H4. (a)(ii) Formula of X: C2H4. Homologous series: alkene. (b) Reagent: aqueous bromine / bromine water. Octane: remains orange / yellow / brown (no reaction). Hydrocarbon X (ethene): turns colourless / is decolourised. (c)(i) 25 O2 and 18 H2O. (c)(ii) Carbon monoxide. Harmful effect: reduces the capacity of blood to carry oxygen by binding irreversibly to haemoglobin / causes suffocation.

(a)(i) 1 mark for C10H22 -> C8H18 + C2H4. (a)(ii) 1 mark for formula C2H4, 1 mark for homologous series alkene. (b) 1 mark for specifying bromine water / aqueous bromine. 1 mark for octane observation (stays orange/brown/no change), 1 mark for X/ethene observation (decolourises / turns colourless). (c)(i) 1 mark for balancing O2 with coefficient 25, 1 mark for balancing H2O with coefficient 18. (c)(ii) 1 mark for naming carbon monoxide, 1 mark for describing its toxic effect (reduces oxygen-carrying capacity of blood / binds to haemoglobin).

(a) Volume of hydrochloric acid, concentration of hydrochloric acid, temperature of the mixture, starting mass of calcium carbonate. (b)(i) Rate of reaction increases. Powder has a larger surface area, leading to more frequent collisions between reactant particles per unit time. (b)(ii) The total volume of carbon dioxide remains unchanged because the same mass of calcium carbonate and same volume/concentration of acid are used (the quantities of reactants are unchanged). (c)(i) Kinetic energy of particles increases (particles move faster). (c)(ii) More reactant particles have energy greater than or equal to the activation energy, resulting in a higher frequency of successful collisions.

(a) 1 mark for each correct control variable (any two from: volume of acid, concentration of acid, starting temperature, mass of calcium carbonate) [max 2]. (b)(i) 1 mark for rate increases, 1 mark for mentioning larger surface area of powder, 1 mark for more frequent collisions. (b)(ii) 1 mark for stating total volume of gas remains unchanged, 1 mark for explanation (same quantity of limiting reactant / mass of reactants is identical). (c)(i) 1 mark for stating kinetic energy of particles increases / move faster. (c)(ii) 1 mark for stating more particles have energy greater than/equal to activation energy, 1 mark for stating higher frequency of successful collisions.

(a)(i) zinc oxide + sulfuric acid -> zinc sulfate + water. (a)(ii) Step 1: Add excess zinc oxide to dilute sulfuric acid and warm. Step 2: Filter the mixture to remove the unreacted zinc oxide. Step 3: Heat the filtrate to evaporate some water / until the crystallization point is reached. Step 4: Allow the hot solution to cool so crystals form, filter to collect crystals, and dry them with filter paper or in a warm oven. (b)(i) A strong acid is an acid that completely ionises / dissociates in aqueous solution. (b)(ii) From red (in acid) to yellow (in excess sodium hydroxide/alkali). (c) Mass of ZnSO4 = 65 + 32 + (16 * 4) = 161. Mass of 7H2O = 7 * (2 * 1 + 16) = 126. Total relative formula mass = 161 + 126 = 287.

(a)(i) 1 mark for correct word equation (zinc oxide + sulfuric acid -> zinc sulfate + water). (a)(ii) 1 mark for adding excess zinc oxide to acid (and warming), 1 mark for filtering to remove unreacted solid, 1 mark for heating filtrate to crystallization point / evaporating some water, 1 mark for cooling to crystallize and drying with filter paper. (b)(i) 1 mark for complete ionisation/dissociation in water. (b)(ii) 1 mark for red in acid, 1 mark for yellow in alkali. (c) 1 mark for calculating mass of ZnSO4 (161) or 7H2O (126), 1 mark for correct final relative formula mass of 287.

(a) (i) The drone is moving with a constant speed (or uniform velocity) of 6.0 m/s.

(ii) Acceleration is calculated by finding the gradient of the graph during the first 4.0 s:
$$\text{acceleration} = \frac{\text{change in speed}}{\text{time taken}}$$
$$\text{acceleration} = \frac{6.0 \text{ m/s} - 0}{4.0 \text{ s}} = 1.5 \text{ m/s}^2$$

(iii) Distance is the area under the speed-time graph. We can divide the area into a triangle, a rectangle, and another triangle:
- Area 1 (0 to 4.0 s) = $\frac{1}{2} \times 4.0 \times 6.0 = 12 \text{ m}$
- Area 2 (4.0 to 10.0 s) = $6.0 \times 6.0 = 36 \text{ m}$
- Area 3 (10.0 to 15.0 s) = $\frac{1}{2} \times 5.0 \times 6.0 = 15 \text{ m}$
$$\text{Total distance} = 12 + 36 + 15 = 63 \text{ m}$$
Alternatively, using the area of a trapezium:
$$\text{Area} = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (6.0 + 15.0) \times 6.0 = 63 \text{ m}$$

(b) (i) Weight is given by:
$$\text{Weight} = m \times g = 4.5 \text{ kg} \times 10 \text{ N/kg} = 45 \text{ N}$$

(ii) Resultant force is given by Newton's second law:
$$F = m \times a = 4.5 \text{ kg} \times 1.5 \text{ m/s}^2 = 6.75 \text{ N}$$

(a) (i) constant speed / uniform velocity / speed of 6.0 m/s [1]

(ii) formula used: $a = \frac{\Delta v}{t}$ OR $\frac{6}{4}$ [1]
- 1.5 [1]
- $\text{m/s}^2$ OR $\text{m } \text{s}^{-2}$ [1]

(iii) evidence of calculating area under the graph [1]
- correct calculated values of individual parts (12, 36, 15) OR correct use of trapezium formula [1]
- 63 (m) [1]

(b) (i) 45 (N) [1]

(ii) formula used: $F = m \times a$ OR $4.5 \times 1.5$ [1]
- 6.75 (N) [1] (accept 6.8)

(a) (i) Humans cannot hear ultrasound because its frequency (40 kHz = 40,000 Hz) is above the upper limit of the human audible frequency range, which is 20,000 Hz (or 20 kHz).

(ii) Wavelength (\\lambda) is calculated using the wave equation:
$$v = f \times \lambda \implies \lambda = \frac{v}{f}$$
$$\lambda = \frac{340 \text{ m/s}}{40000 \text{ Hz}} = 0.0085 \text{ m} \text{ (or } 8.5 \text{ mm)}$$

(b) Sound travels to the water surface and back to the sensor. The total distance travelled is twice the distance to the water surface ($2d$):
$$2d = v \times t \implies d = \frac{v \times t}{2}$$
$$d = \frac{340 \text{ m/s} \times 0.12 \text{ s}}{2} = 20.4 \text{ m}$$

(c) (i) Infrared radiation is located between visible light and microwaves on the electromagnetic spectrum (it has a longer wavelength than visible light but a shorter wavelength than microwaves).

(ii) All electromagnetic waves travel at the same high speed in a vacuum (approximately $3.0 \times 10^8 \text{ m/s}$) / are transverse waves.

(iii) Infrared waves are electromagnetic waves and do not require a medium to propagate. Sound waves are longitudinal mechanical waves that require a physical medium (particles) to vibrate and transmit energy, which is absent in a vacuum.

(a) (i) frequency is higher than the upper limit of human hearing (20,000 Hz / 20 kHz) [1]

(ii) formula used: $v = f \times \lambda$ OR $\lambda = \frac{340}{40000}$ [1]
- 0.0085 [1]
- m (or metres) [1]

(b) formula used: $2d = v \times t$ OR $d = \frac{v \times t}{2}$ [1]
- $340 \times 0.12$ (= 40.8 m) OR dividing total distance by 2 [1]
- 20.4 (m) [1]

(c) (i) between visible light and microwaves [1]

(ii) any one from: travel at speed of light / $3 \times 10^8$ m/s in vacuum / transverse waves / can travel through a vacuum [1]

(iii) electromagnetic waves do not need a medium / sound waves are mechanical and require particles/medium [1]

(a) (i) Electrical power ($P$) is calculated by:
$$P = V \times I$$
$$P = 12 \text{ V} \times 1.5 \text{ A} = 18 \text{ W}$$

(ii) Electrical energy ($E$) transferred is:
$$E = P \times t$$
First, convert minutes to seconds: $5.0 \text{ minutes} = 5.0 \times 60 = 300 \text{ s}$.
$$E = 18 \text{ W} \times 300 \text{ s} = 5400 \text{ J} \text{ (or } 5.4 \text{ kJ)}$$

(b) (i) In parallel, if one lamp fails/blows, the other remains lit (or the lamps can be controlled independently by separate switches, or both get the full 12 V and shine with normal brightness).

(ii) For two resistors in parallel:
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12}$$
$$R_p = 12 \text{ } \Omega$$
Alternatively, $R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{24 \times 24}{24 + 24} = 12 \text{ } \Omega$.

(c) (i) The strength of an electromagnet can be increased by increasing the current in the coil, increasing the number of turns on the coil, or using a soft iron core.

(ii) Iron (or soft iron) is used because it is magnetically soft, meaning it is easily magnetised and quickly loses its magnetism when the current is turned off.

(a) (i) formula used: $P = V \times I$ OR $12 \times 1.5$ [1]
- 18 (W) [1]

(ii) formula used: $E = P \times t$ OR $18 \times 5 \times 60$ [1]
- 5400 [1]
- J (or Joules) [1]

(b) (i) if one lamp fails, the other remains lit / lamps can be controlled independently / both lamps receive the full supply voltage [1]

(ii) formula used: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$ OR $R_p = \frac{R_1 \times R_2}{R_1 + R_2}$ [1]
- 12 ($\Omega$) [1]

(c) (i) increase current OR increase number of turns on the coil OR use a soft iron core [1]

(ii) (soft) iron because it loses its magnetism quickly when current is switched off / easily demagnetised [1]

(a) (i) Isotopes are atoms of the same element with the same number of protons (atomic number) but a different number of neutrons (mass number).

(ii) From the symbol $^{99}_{43}\text{Tc}$:
- Atomic number (bottom) = number of protons = 43
- Mass number (top) = protons + neutrons = 99
- Number of neutrons = $99 - 43 = 56$

(b) (i) When a nucleus emits gamma radiation, there is no change in the number of protons or neutrons. The nucleus simply loses excess energy and becomes more stable.

(ii) Gamma radiation is highly penetrating, allowing it to easily pass through body tissues and be detected outside the body. It is also weakly ionising, meaning it causes significantly less damage/harm to internal healthy cells than highly ionising alpha radiation (which would be completely absorbed by the body tissues, causing extreme local radiation damage).

(c) (i) Calculate the number of half-lives ($n$) that have passed in 24 hours:
$$n = \frac{24 \text{ hours}}{6.0 \text{ hours}} = 4 \text{ half-lives}$$
After each half-life, the number of radioactive atoms is halved:
$$\text{Atoms remaining} = 8.0 \times 10^{12} \times \left(\frac{1}{2}\right)^4$$
$$\text{Atoms remaining} = \frac{8.0 \times 10^{12}}{16} = 5.0 \times 10^{11}$$

(ii) The half-life is the time taken for half of the radioactive nuclei in a sample to decay (or the time taken for the activity of the sample to halve).

(a) (i) atoms of the same element / same proton number [1]
- different number of neutrons / different nucleon number [1]

(ii) protons = 43 [1]
- neutrons = 56 [1]

(b) (i) no change in proton or neutron number / atomic mass and atomic number remain unchanged (just loses energy) [1]

(ii) gamma is highly penetrating / can escape the body to be detected [1]
- gamma is weakly ionising / causes less cell damage [1]

(c) (i) 4 half-lives [1]
- $5.0 \times 10^{11}$ [1] (accept $5 \times 10^{11}$)

(ii) time taken for half the radioactive nuclei to decay OR time taken for activity / count rate to halve [1]