2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

Water moves from a region of higher water potential (inside the cell) to a region of lower water potential (the external sucrose solution) down a water potential gradient by osmosis. This net loss of water causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, which results in the cell becoming plasmolysed.

Award 1 mark for the correct option (D). Correct direction is 'out of the cell' and correct state is 'plasmolysed'.

Yeast cells undergo anaerobic respiration (fermentation) to produce ethanol and carbon dioxide. In contrast, human muscle cells during vigorous exercise respire anaerobically to produce lactic acid only, without producing carbon dioxide.

Award 1 mark for the correct option (B). Distinguish the correct end products of anaerobic pathway in both yeast and humans.

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows a greater fraction of colliding reactant particles to have energy equal to or greater than the activation energy, increasing the frequency of successful collisions.

Award 1 mark for selecting the correct explanation of catalyst action (C). Option C matches the IGCSE definition of catalysts lowering the activation energy barrier.

First, write the balanced equation: \(Mg + 2HCl \rightarrow MgCl_2 + H_2\). Determine the moles of magnesium reacting: \(\text{moles of } Mg = 3.6 / 24 = 0.15\text{ mol}\). Since the molar ratio of \(Mg\) to \(H_2\) is \(1:1\), \(0.15\text{ mol}\) of hydrogen gas is produced. Calculate the volume: \(\text{volume of } H_2 = 0.15 \times 24 = 3.6\text{ dm}^3\).

Award 1 mark for selecting 3.6 (B). Option A represents an incorrect ratio; C is double the correct amount; D represents the calculation using incorrect mass-molar conversions.

The refractive index \(n\) is given by Snell's Law: \(n = \frac{\sin i}{\sin r}\). Substituting the values: \(n = \frac{\sin(42^\circ)}{\sin(26^\circ)} = \frac{0.669}{0.438} \approx 1.53\). Rounded to two significant figures, this is \(1.5\).

Award 1 mark for selecting the correct refractive index value (B). Option A is the reciprocal (\(\sin r / \sin i\)), options C and D are incorrect calculations.

First, convert the mass from grams to kilograms: \(m = 800\text{ g} = 0.8\text{ kg}\). The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\). Substituting the values: \(KE = 0.5 \times 0.8\text{ kg} \times (5.0\text{ m/s})^2 = 0.4 \times 25 = 10\text{ J}\).

Award 1 mark for selecting 10 J (B). Option A incorrectly drops the squaring of the velocity; C uses mass in grams directly without conversion to kilograms (resulting in incorrect order of magnitude) or other calculation errors.

For resistors in parallel, the combined resistance \(R_t\) is calculated using the formula: \(\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}\). Substituting the values: \(\frac{1}{R_t} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0}\). Therefore, \(R_t = 3.0\ \Omega\).

Award 1 mark for selecting 3.0 \(\Omega\) (A). Option C is the simple sum (series connection), while other options represent standard calculation errors.

The mass number decreases by 4 (from 210 to 206) and the proton number decreases by 2 (from 84 to 82). This matches the characteristics of an alpha-particle (helium nucleus, \(^{4}_{2}\text{He}\)), which consists of 2 protons and 2 neutrons.

Award 1 mark for identifying the decay as an alpha decay (A). A beta particle emission would keep the mass number the same and increase the proton number by 1.

Excretion is defined in the syllabus as the removal from organisms of toxic materials, the waste products of metabolism and substances in excess of requirements. Growth involves increase in size, mass, and cell number/size. Respiration releases energy, not oxygen. Sensitivity does not always result in movement of the entire organism.

1 mark for the correct option B.

The stomach has a highly acidic environment (pH around 2.0) due to hydrochloric acid, making it the ideal site for an enzyme with this optimum pH. The duodenum is alkaline (pH around 8.0) because of pancreatic juice and bile. When the enzyme enters the duodenum, the alkaline pH denatures the enzyme, rapidly decreasing its activity.

1 mark for the correct option B.

Oxygen, glucose, and antibodies are transferred from the mother's blood to the fetal blood to support the development and immunity of the fetus. Waste products like carbon dioxide and urea are transferred in the opposite direction (from fetal blood to maternal blood).

1 mark for the correct option B.

The rate of diffusion of a gas depends on its relative molecular mass. Gases with lower relative molecular masses diffuse faster because, at any given temperature, their particles have a higher average speed. Helium has the lowest relative molecular mass (\(M_r = 4\)) among the choices and therefore diffuses the fastest.

1 mark for the correct option B.

Both diamond and silicon(IV) oxide have giant covalent structures with a vast three-dimensional network of strong covalent bonds. Consequently, both substances are extremely hard and have very high melting points because a large amount of energy is required to break these strong covalent bonds. They do not conduct electricity and are insoluble in water.

1 mark for the correct option C.

In the reaction \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\), iron(III) oxide is reduced to iron, and carbon monoxide (\(\text{CO}\)) is oxidized to carbon dioxide. Here, \(\text{CO}\) is a carbon compound, and it acts as the reducing agent. In option A, carbon is an element, not a compound. Option D is thermal decomposition, not a redox reaction.

1 mark for the correct option C.

First, calculate the useful work done by the motor: \(W = F \times d = 200\text{ N} \times 3.0\text{ m} = 600\text{ J}\). Next, calculate the useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{600\text{ J}}{5.0\text{ s}} = 120\text{ W}\). Finally, use the efficiency formula (\(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)) to find the power input: \(P_{\text{in}} = \frac{120\text{ W}}{0.60} = 200\text{ W}\).

1 mark for the correct option C.

First, find the equivalent resistance of the two parallel 6.0 \(\Omega\) resistors: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). Next, find the total resistance of the circuit: \(R_t = R_p + 5.0 = 3.0 + 5.0 = 8.0\ \Omega\). Use Ohm's law to find the total current leaving the battery: \(I = \frac{V}{R_t} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\). Since the 5.0 \(\Omega\) resistor is in series with the parallel combination, the total current of 1.5 A flows directly through it.

1 mark for the correct option B.

Protease enzymes in the stomach (such as pepsin) have an acidic optimum pH (around pH 1.5 - 2.0). When the pH is increased to pH 8.0, the environment is too alkaline, causing the enzyme molecules to denature. This alters the shape of their active site permanently, so the substrate can no longer fit and the reaction rate decreases significantly.

A is correct: 1 mark.
B is incorrect: increasing pH from optimum denatures the enzyme, it does not increase kinetic energy.
C is incorrect: substrates are not denatured.
D is incorrect: optimum pH is acidic (pH 2), not pH 8.

According to the kinetic theory of gases, lighter gas molecules diffuse faster than heavier gas molecules at the same temperature. Since carbon monoxide has a lower relative molecular mass (\(M_r = 28\)) than nitrogen dioxide (\(M_r = 46\)), carbon monoxide molecules will diffuse upwards more rapidly than nitrogen dioxide molecules diffuse downwards.

A is correct: 1 mark.
B is incorrect: nitrogen dioxide is heavier so it diffuses slower.
C is incorrect: gravity does not prevent diffusion.
D is incorrect: diffusion rates depend on relative molecular mass, not just temperature.

1. Find total energy input: \(E_{\text{in}} = P \times t = 100\text{ W} \times 5.0\text{ s} = 500\text{ J}\).
2. Calculate useful energy output based on 80% efficiency: \(E_{\text{useful}} = 0.80 \times 500\text{ J} = 400\text{ J}\).
3. Relate useful work done to potential energy gained: \(W = mgh = 4.0\text{ kg} \times 10\text{ N/kg} \times h = 40h\).
4. Solve for \(h\): \(40h = 400 \implies h = 10\text{ m}\).

C is correct: 1 mark for the calculation showing 10 m.
A is incorrect: calculated without multiplying by time (using power as energy).
B is incorrect: incorrect usage of efficiency.
D is incorrect: calculated assuming 100% efficiency.

In human muscle cells during vigorous exercise, anaerobic respiration breaks down glucose to produce lactic acid only (no carbon dioxide is released). This process releases much less energy per glucose molecule than aerobic respiration because glucose is only partially broken down.

B is correct: 1 mark.
A is incorrect: anaerobic respiration in muscles does not produce carbon dioxide or water.
C is incorrect: anaerobic respiration releases less energy, not more.
D is incorrect: ethanol is produced by yeast during anaerobic respiration, not by human muscle cells.

During the electrolysis of concentrated aqueous sodium chloride:
- At the anode (+), chloride ions (\(\text{Cl}^-\)) are discharged to form chlorine gas (\(\text{Cl}_2\)).
- At the cathode (-), hydrogen ions (\(\text{H}^+\)) from water are discharged to form hydrogen gas (\(\text{H}_2\)).
- The remaining solution contains sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)), forming sodium hydroxide, which is alkaline, so the pH increases.

A is correct: 1 mark.
B, C, and D are incorrect based on the standard rules of selective discharge for concentrated aqueous sodium chloride.

1. For three identical resistors in parallel:
\(\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{\text{parallel}} = \frac{R}{3}\).
Given \(R_{\text{parallel}} = 4.0\ \Omega\), we get \(\frac{R}{3} = 4.0 \implies R = 12\ \Omega\).
2. For the same three resistors in series:
\(R_{\text{series}} = R + R + R = 3R = 3 \times 12\ \Omega = 36\ \Omega\).

D is correct: 1 mark for the calculation leading to 36 ohms.
A is incorrect: represents one-third of the parallel value.
B is incorrect: equal to the parallel combination resistance.
C is incorrect: resistance of a single resistor.

Using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
\(1.41 = \frac{\sin 45^\circ}{\sin r}\)
Since \(\sin 45^\circ \approx 0.707\) and \(1.41 \approx \sqrt{2}\):
\(\sin r = \frac{\sin 45^\circ}{1.41} = \frac{0.707}{1.41} = 0.50\).
Therefore, \(r = \sin^{-1}(0.50) = 30^\circ\).

B is correct: 1 mark.
A, C, D are incorrect calculation errors.

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy, so a greater fraction of collisions are successful. Increasing the temperature increases the kinetic energy of the particles, meaning they move faster, collide more frequently, and a larger proportion of colliding particles possess energy equal to or greater than the activation energy.

C is correct: 1 mark.
A is incorrect: temperature does not lower the activation energy.
B is incorrect: catalysts do not change kinetic energy.
D is incorrect: catalysts do not primarily increase collision frequency, they increase the probability of a collision being successful.

First, find the resultant force acting on the crate:

\(\text{Resultant Force } F = \text{Pulling Force} - \text{Frictional Force}\)
\(F = 36\text{ N} - 12\text{ N} = 24\text{ N}\)

Next, use Newton's second law, \(F = ma\), to calculate the acceleration (\(a\)):

\(a = \frac{F}{m} = \frac{24\text{ N}}{8.0\text{ kg}} = 3.0\text{ m/s}^2\)

1 mark for the correct option B.

First, find the equivalent resistance of the two parallel resistors:

\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0\ \Omega} + \frac{1}{12\ \Omega} = \frac{2}{12\ \Omega} + \frac{1}{12\ \Omega} = \frac{3}{12\ \Omega}\)

\(R_p = \frac{12}{3} = 4.0\ \Omega\)

Now, use Ohm's law (\(V = IR\)) to find the total current (\(I\)):

\(I = \frac{V}{R_p} = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\)

1 mark for the correct option C.

As the temperature increases beyond the optimum (around \( 40\ ^\circ\text{C} \)), the thermal energy breaks the bonds that maintain the active site's structure in the enzyme. This process, called denaturation, permanently changes the shape of the active site so that it is no longer complementary to the substrate. Therefore, the reaction rate drops to near zero.

1 mark for the correct option B.

The ion \( X^{2+} \) has \( 2 + 8 + 8 = 18 \) electrons. Since it has a \( 2+ \) charge, the neutral atom must have had 2 more electrons than this ion.

Number of electrons in neutral atom \( X = 18 + 2 = 20 \).

Therefore, the electronic configuration of the neutral atom is \( 2, 8, 8, 2 \).

- The number of outer-shell (valence) electrons is 2, indicating it belongs to Group II.
- The number of electron shells is 4, indicating it is in Period 4.

1 mark for the correct option B.

Increasing the concentration means there are more reactant particles in the same volume of solution. Because they are closer together, they collide more frequently, which increases the rate of reaction. Note that lowering activation energy is the role of a catalyst, and increasing kinetic energy or the proportion of collisions with energy greater than the activation energy is the effect of increasing temperature.

1 mark for the correct option C.

Since both parents are unaffected, they must carry at least one dominant normal allele, \( F \). For their child to have cystic fibrosis (\( ff \)), the child must have inherited one recessive allele \( f \) from each parent. Thus, both parents are carriers with the heterozygous genotype \( Ff \).

A monohybrid cross of \( Ff \times Ff \) produces offspring in the following genetic ratio:
- \( FF \) (unaffected, non-carrier): \( 25\% \)
- \( Ff \) (unaffected carrier): \( 50\% \)
- \( ff \) (affected with cystic fibrosis): \( 25\% \)

Therefore, the probability of the next child inheriting the disorder is \( 25\% \).

1 mark for the correct option B.

By Snell's Law, the refractive index \( n \) is given by:

\(n = \frac{\sin i}{\sin r}\)

Given:
\(i = 45^\circ \implies \sin(45^\circ) \approx 0.707\)
\(r = 28^\circ \implies \sin(28^\circ) \approx 0.469\)

Calculate \( n \):

\(n = \frac{0.707}{0.469} \approx 1.51 \approx 1.5\)

1 mark for the correct option C.

Weak, soft bones that bend easily are characteristic symptoms of rickets, which is caused by a deficiency in Vitamin D (or calcium). Vitamin D is essential for the proper absorption of calcium from the digestive tract to strengthen bones. Vitamin C deficiency causes scurvy (bleeding gums), iron deficiency causes anemia (tiredness due to lack of hemoglobin), and protein deficiency causes kwashiorkor.

1 mark for the correct option B.

Between \(20^\circ\text{C}\) and \(35^\circ\text{C}\), enzyme and substrate molecules gain kinetic energy, leading to more frequent and energetic collisions, thus increasing the rate of successful collisions. Above \(50^\circ\text{C}\), the thermal energy is high enough to disrupt the bonds holding the enzyme's tertiary structure, permanently altering the shape of the active site so that the substrate can no longer bind (denaturation).

1 mark for the correct option.
- A: correct explanation of kinetic energy and permanent active site alteration.
- B: incorrect, temperature does not alter activation energy in this way and substrate is not denatured.
- C: incorrect, denaturation is permanent and not temporary inactivation.
- D: incorrect, the concentration of complexes would drop because of denaturation, not directly because kinetic energy is 'too high' to form complexes.

According to the kinetic theory of gases, at a given temperature, lighter molecules move faster on average than heavier molecules. Since \(\text{CO}_2\) (\(M_r = 44\)) has a lower relative molecular mass than \(\text{SO}_2\) (\(M_r = 64\)), \(\text{CO}_2\) molecules move faster and diffuse at a higher rate, travelling further along the tube in a given time.

1 mark for the correct option.
- A: correct correlation between lower relative molecular mass, higher speed, and greater diffusion distance.
- B: incorrect, heavier molecules move slower at the same temperature.
- C: incorrect, diffusion rate depends on molecular mass.
- D: incorrect, carbon dioxide travels a longer distance because it moves faster.

1. Useful work done: \(W = mgh = 150 \times 10 \times 12.0 = 18,000\text{ J}\).
2. Useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{18,000}{6.0} = 3000\text{ W} = 3.0\text{ kW}\).
3. Power input: \(P_{\text{in}} = 4.0\text{ kW}\).
4. Efficiency: \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{3.0}{4.0} \times 100\% = 75\%\).

1 mark for correct calculations leading to 75%.
- Useful output power = 3000 W (18,000 J in 6 s).
- Input power = 4000 W.
- Ratio is 0.75 or 75%.

Anaerobic respiration in yeast (fermentation) produces carbon dioxide and ethanol, and it yields much less energy per glucose molecule than aerobic respiration because glucose is only partially broken down in the absence of oxygen.

1 mark for the correct option.
- Option A correctly identifies ethanol and carbon dioxide as products and notes the low energy yield.
- Option B contains incorrect products for anaerobic respiration.
- Option C refers to lactic acid, which is produced during anaerobic respiration in mammalian muscles, not yeast.

Silicon(IV) oxide exists as a giant covalent macromolecule, where strong covalent bonds must be broken throughout the lattice to melt it. In contrast, carbon dioxide consists of simple molecules; when it melts/boils, only the weak intermolecular forces (van der Waals forces) between the molecules need to be overcome, which requires very little energy.

1 mark for correct identification of giant covalent macromolecule vs. simple covalent molecular structures.
- A: correct structural and bonding comparison.
- B: incorrect bonding type (silicon(IV) oxide is covalent, not ionic).
- C: incorrect explanation of physical state differences.
- D: incorrect description of the silicon(IV) oxide lattice (not ionic).

When a wave passes from one medium to another, its frequency remains unchanged: \(f = 6.0 \times 10^{14}\text{ Hz}\).
The wavelength in the glass is given by: \(\lambda = \frac{v}{f} = \frac{2.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 3.33 \times 10^{-7}\text{ m} \approx 3.3 \times 10^{-7}\text{ m}\).

1 mark for the correct option.
- Frequency must be unchanged: \(6.0 \times 10^{14}\text{ Hz}\).
- Wavelength calculation using \(v = 2.0 \times 10^8\text{ m/s}\) yields \(3.3 \times 10^{-7}\text{ m}\).

During the electrolysis of aqueous copper(II) sulfate:
- At the anode (+), hydroxide ions (\(\text{OH}^-\)) are discharged in preference to sulfate ions, forming oxygen gas: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\), which is seen as bubbles of a colourless gas.
- At the cathode (-), copper ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions because copper is less reactive than hydrogen: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\), forming a pink-brown copper solid.

1 mark for correct selection.
- Anode (+): oxygen gas (colourless bubbles).
- Cathode (-): copper metal deposition (pink-brown solid).

1. Find the equivalent resistance of the two \(6.0\ \Omega\) parallel resistors: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\).
2. Find the total resistance of the series-parallel circuit: \(R_{\text{total}} = 3.0\ \Omega\text{ (series)} + 3.0\ \Omega\text{ (parallel)} = 6.0\ \Omega\).
3. Find the total current drawn from the power supply: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).

1 mark for the correct answer of 2.0 A.
- Equivalent resistance of parallel branch = 3.0 ohms.
- Total resistance = 6.0 ohms.
- Ohm's law calculation: I = 12 / 6 = 2.0 A.

(a) An enzyme is defined as a protein that acts as a biological catalyst, speeding up metabolic reactions without being changed. (b) Extreme pH changes alter the ionic bonds holding the protein structure together, changing the active site shape (denaturation). (c) Temperature increases the average kinetic energy, raising the rate of collision. (d) All proteins (including enzymes) contain C, H, O, and N.

Part (a) [2 marks]: protein [1], biological catalyst / speeds up reactions [1]. Part (b) [3 marks]: optimum pH of pepsin is acidic / pH 2 [1], pH 8 changes the shape of the active site [1], enzyme is denatured / substrate can no longer fit [1]. Part (c) [3 marks]: molecules gain kinetic energy [1], move faster / collide more frequently [1], more successful collisions per unit time [1]. Part (d) [2 marks]: carbon, hydrogen, and oxygen (any two for [1]), nitrogen [1].

(a) Contrast the close, touching but irregular arrangement of liquids with the highly spaced arrangement of gases, along with their respective motions. (b)(i) Mr(O2) = 2 x 16 = 32. Mr(SO3) = 32 + (3 x 16) = 80. (ii) Rate of diffusion is inversely proportional to the square root of molecular mass; lighter molecules diffuse faster. (c) Cooling gas to liquid is condensation, bringing particles into contact.

Part (a) [4 marks]: liquid particles are close together / touching [1], gas particles are far apart [1], liquid particles slide past each other [1], gas particles move rapidly / in all directions [1]. Part (b)(i) [2 marks]: O2 = 32 [1], SO3 = 80 [1]. Part (b)(ii) [2 marks]: oxygen / O2 [1], smaller relative molecular mass [1]. Part (c) [2 marks]: condensation / condensing [1], particles become much closer together [1].

(a) A step-down transformer lowers potential difference. (b)(i) Vs = Vp x (Ns / Np) = 240 x (50 / 400) = 30 V. (ii) Is = Vs / R = 30 / 3.0 = 10 A. (iii) P = Vs x Is = 30 x 10 = 300 W. (c) Electromagnetic induction requires a changing magnetic flux, which d.c. does not provide.

Part (a) [1 mark]: decreases alternating voltage [1]. Part (b)(i) [2 marks]: use of transformer equation Vs / Vp = Ns / Np [1], 30 (V) [1]. Part (b)(ii) [2 marks]: use of I = V / R [1], 10 (A) [1]. Part (b)(iii) [2 marks]: use of P = V x I or P = I^2 x R [1], 300 (W) [1]. Part (c) [3 marks]: output voltage is zero [1], d.c. produces a constant/steady magnetic field [1], induction requires a changing magnetic field [1].

(a) Anaerobic respiration of glucose in yeast produces ethanol and carbon dioxide: C6H12O6 -> 2C2H5OH + 2CO2. (b) Yeast fermentation yields ethanol and CO2, while mammalian muscles yield lactic acid only. (c)(i) Strenuous activity exceeds the aerobic respiration limit. (ii) Lactic acid is metabolized in the liver with oxygen.

Part (a) [3 marks]: C6H12O6 as reactant [1], C2H5OH and CO2 as products [1], balanced equation [1]. Part (b) [2 marks]: yeast produces ethanol, muscle produces lactic acid [1], yeast produces carbon dioxide, muscle does not [1]. Part (c)(i) [2 marks]: oxygen supply is inadequate [1], to meet the rapid / high energy demand [1]. Part (c)(ii) [3 marks]: transported by blood to liver [1], oxidized to CO2 and H2O / converted to glucose [1], requires oxygen (paying oxygen debt) [1].

(a) A closed reaction system connected to a graduated gas collector. (b) Catalysts lower the energy barrier (Ea), allowing more molecular collisions to result in reaction. (c) High temperature raises both collision frequency and the energy of colliding particles.

Part (a) [3 marks]: flask with stopper [1], delivery tube [1], gas syringe or inverted measuring cylinder over water [1]. Part (b) [3 marks]: alternative reaction pathway [1], lower activation energy [1], increased frequency of successful collisions [1]. Part (c) [4 marks]: particles gain kinetic energy / move faster [1], increased collision frequency [1], greater proportion of particles have energy >= Ea [1], increased frequency of successful collisions [1].

(a) Weight = m x g = 15 x 10 = 150 N. (b) Work = force x distance = 150 x 8.0 = 1200 J. (c) Useful Power = Work / time = 1200 / 12 = 100 W. (d)(i) Efficiency = (Useful Power Output / Power Input) x 100 = (100 / 150) x 100 = 66.7%. (ii) Energy is lost as heat. (e) Loss in GPE = Gain in KE -> mgh = 0.5 x m x v^2 -> v = sqrt(2gh) = sqrt(2 x 10 x 8.0) = sqrt(160) = 12.65 m/s.

Part (a) [1 mark]: 150 (N) [1]. Part (b) [2 marks]: use of W = F x d or mgh [1], 1200 (J) [1]. Part (c) [2 marks]: use of P = W / t [1], 100 (W) [1]. Part (d)(i) [2 marks]: use of efficiency = output / input [1], 66.7% or 0.67 [1]. Part (d)(ii) [1 mark]: wasted as thermal energy to surroundings [1]. Part (e) [2 marks]: equating KE and PE (v = sqrt(2gh)) [1], 12.6 or 13 (m/s) [1].

(a)(i) Phenotype is physical appearance. (ii) Heterozygous means possessing alleles R and r. (b) A test cross Rr x rr results in 50% Rr (red) and 50% rr (white). (c) Rr x RR yields RR and Rr, both of which are red; thus, no white flowers are possible.

Part (a)(i) [1 mark]: physical/observable features of an organism [1]. Part (a)(ii) [1 mark]: possess two different alleles of a gene [1]. Part (b) [5 marks]: parent genotypes Rr and rr [1], correct gametes R, r, and r [1], correct Punnett square setup [1], offspring genotypes Rr and rr [1], ratio 1 red : 1 white [1]. Part (c) [3 marks]: 0 / 0% [1], homozygous red parent only passes on dominant R allele [1], all offspring will receive R / be red [1].

(a) The decay equation preserves mass number (210 = 206 + 4) and atomic number (84 = 82 + 2). (b) Half-life is defined by exponential decay. (c)(i) Number of half-lives = 414 / 138 = 3. Remaining atoms = 8.0 x 10^15 x (0.5)^3 = 1.0 x 10^15. (ii) Decayed fraction = 1 - 0.125 = 0.875 = 87.5%. (d) Alpha has +2 charge and greater mass, making it much more ionizing than beta.

Part (a) [3 marks]: correct helium/alpha nuclide 4/2 He or 4/2 alpha [1], lead nuclide as 206/82 Pb [1], fully balanced equation [1]. Part (b) [2 marks]: time taken for half the radioactive atoms to decay / activity to halve [2]. Part (c)(i) [2 marks]: determining 3 half-lives [1], 1.0 x 10^15 atoms [1]. Part (c)(ii) [2 marks]: fraction decayed is 7/8 or 100% - 12.5% [1], 87.5% [1]. Part (d) [1 mark]: alpha is more ionizing than beta [1].

(a) (i) Velocity is speed in a specified direction (vector quantity).
(ii) Acceleration is calculated using: \( a = \frac{v - u}{t} = \frac{12 - 0}{4.0} = 3.0\text{ m/s}^2 \).

(b) The net vertical force is: \( F_{\text{net}} = F - W = ma \).
First, calculate the weight: \( W = mg = 1.8 \times 10 = 18\text{ N} \).
Then, substitute into the equation of motion: \( F - 18 = 1.8 \times 3.0 \implies F - 18 = 5.4 \implies F = 23.4\text{ N} \).

(c) Power is related to force and velocity by: \( P = F_{\text{res}} \times v \implies F_{\text{res}} = \frac{P}{v} = \frac{240}{12} = 20\text{ N} \).

(d) The energy stored in the battery is chemical (potential) energy. During horizontal flight at a constant speed, this energy is ultimately transferred to thermal (internal) energy of the surrounding air due to work done against air resistance.

(a) (i) speed in a given direction / rate of change of displacement [1]
(ii) substitution: \( a = \frac{12}{4} \) [1]; correct value with units: \( 3.0\text{ m/s}^2 \) [1]

(b) calculation of weight: \( W = 18\text{ N} \) [1]; correct equation: \( F - 18 = 1.8 \times 3.0 \) or net force of \( 5.4\text{ N} \) [1]; final value: \( F = 23.4\text{ N} \) [1]

(c) formula used: \( P = Fv \) or substitution: \( F = \frac{240}{12} \) [1]; final value: \( 20\text{ N} \) [1]

(d) chemical (potential) energy [1]; thermal / internal energy (of air/surroundings) [1]

(a) The balanced equation is: \( 2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} \).

(b) (i) Increasing the concentration increases the rate of reaction.
(ii) A higher concentration means there are more reactant particles per unit volume. This increases the frequency of collisions (more collisions per second) between reactant particles, resulting in a higher rate of successful collisions per unit time.

(c) On a coordinate system of Potential Energy vs. Reaction Coordinate: the reactants are represented by a horizontal line at a higher level than the products. A curve rises from the reactant level to a maximum peak and then falls to the product level. The activation energy \(E_a\) is the upward difference from the reactants to the peak. The enthalpy change \(\Delta H\) is the downward difference from the reactants to the products.

(a) correct formulae of reactants and products: \( \text{H}_2\text{O}_2 \), \( \text{H}_2\text{O} \), \( \text{O}_2 \) [1]; balancing: \( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \) [1]; correct state symbols: \( \text{(aq)} \), \( \text{(l)} \), \( \text{(g)} \) [1]

(b) (i) rate increases / reaction is faster [1]
(ii) more particles per unit volume / space [1]; higher frequency of collisions / more collisions per second [1]; more successful collisions per unit time [1]

(c) reactants line drawn higher than products line [1]; activation energy, \(E_a\), shown from reactants to the peak [1]; enthalpy change, \(\Delta H\), shown between reactants and products [1]

(a) (i) Pepsin is the protease secreted in the stomach. It breaks down large proteins into smaller peptides and amino acids.
(ii) Hydrochloric acid maintains a highly acidic environment (pH 1.5 - 2.0). This provides the optimum pH for the pepsin enzyme to function. Additionally, it kills harmful bacteria/pathogens ingested with food by denaturing their cellular proteins.

(b) Villi are finger-like projections found on the inner surface of the small intestine. They are adapted for absorption because:
1. They significantly increase the total surface area available for absorption.
2. Their walls are only one cell thick (thin epithelium), shortening the diffusion distance.
3. They have a rich network of blood capillaries to quickly transport absorbed amino acids away, maintaining a steep concentration gradient.

(c) Mechanical digestion physically breaks down food into smaller pieces without changing its chemical identity (e.g., teeth chewing, stomach wall churning), which increases the surface area for enzymes. Chemical digestion breaks large, insoluble molecules into small, soluble molecules by breaking covalent chemical bonds using enzymes (e.g., amylase breaking down starch to maltose).

(a) (i) pepsin [1]; amino acids / peptides [1]
(ii) Any two from: provides acidic / low pH / optimum pH for pepsin [1]; kills bacteria / pathogens in food [1]; denatures proteins to make them easier to digest [1]

(b) Any three from: finger-like projections increase surface area [1]; thin wall / single-layer epithelium for short diffusion distance [1]; rich capillary network to carry nutrients and maintain concentration gradient [1]; presence of microvilli (further increases surface area) [1]

(c) Mechanical digestion: physical breakdown of food into smaller pieces / no change in chemical structure [1]; plus correct example: teeth chewing / stomach churning [1]; Chemical digestion: chemical breakdown of large insoluble to small soluble molecules / involves enzymes [1]

(a) (i) Ultrasound is defined as sound with a frequency higher than the upper limit of human hearing, which is 20 kHz (or 20000 Hz).
(ii) Wavelength is calculated using: \( \lambda = \frac{v}{f} \).
Convert frequency to Hz: \( 40\text{ kHz} = 40000\text{ Hz} \).
\( \lambda = \frac{1500}{40000} = 0.0375\text{ m} \) (or 3.75 cm).
(iii) Ultrasound is a longitudinal wave. The water particles vibrate back and forth parallel to the direction of wave travel (energy transfer).

(b) (i) The speed of radio waves (and all electromagnetic waves) in a vacuum is \( 3.0 \times 10^8\text{ m/s} \).
(ii) Two major differences are:
1. Sound waves are longitudinal, whereas EM waves are transverse.
2. Sound waves require a physical medium to propagate, whereas EM waves can travel through a vacuum.
(iii) The region of the EM spectrum between infrared and ultraviolet is visible light.

(a) (i) sound with frequency greater than 20 kHz / 20000 Hz [1]
(ii) substitution: \( \lambda = \frac{1500}{40000} \) [1]; final value with correct unit: 0.0375 m or 3.75 cm [1]
(iii) longitudinal [1]; particles vibrate parallel to / along the direction of wave propagation / energy transfer [1]

(b) (i) \( 3.0 \times 10^8\text{ m/s} \) (accept \( 3 \times 10^8 \)) [1]
(ii) Any two from: EM waves do not require a medium / sound waves do [1]; EM waves are transverse / sound waves are longitudinal [1]; EM waves travel at the speed of light / sound waves travel much slower [1]
(iii) visible light [1]; explanation or description that it lies between infrared and ultraviolet in the EM spectrum [1]

(a) The timer reads 1 minute and 40 seconds. 1 minute = 60 seconds, so 60 + 40 = 100 seconds.
(b) Rate = 1000 / time. At 40^\circ\text{C}, the time is 50 s. Rate = 1000 / 50 = 20.0.
(c) The optimum temperature is the temperature at which the rate of reaction is the highest, which is 40^\circ\text{C} with a rate of 20.0.
(d) Controlled variables include the volume and concentration of starch, the volume and concentration of amylase, and the pH of the mixture.
(e) Equilibrating the solutions in the water bath ensures they are exactly at the target temperature at the moment of mixing.
(f) Starch is detected using iodine solution. A positive result is a change from yellow-brown/orange-brown to blue-black.

(a) 1 mark: 100 (seconds)
(b) 1 mark: 20 / 20.0
(c) 1 mark: 40 (^\circ\text{C}) [Accept: temperature with the shortest time / highest rate]
(d) 2 marks: Any two from: volume of starch, concentration of starch, volume of amylase, concentration of amylase, pH [Reject: temperature]
(e) 1 mark: to allow the solutions to reach the target/constant temperature before mixing
(f) 2 marks: use iodine solution (1 mark); colour change to blue-black (1 mark)

(a) Change in mass = Final mass - Initial mass = 3.36 - 3.20 = +0.16 g.
Percentage change = (0.16 / 3.20) * 100 = +5.0%.
(b) The internal concentration of the cells corresponds to the sucrose concentration where there is no net movement of water (0% change in mass). This lies between 0.4\text{ mol/dm}^3 (positive change) and 0.6\text{ mol/dm}^3 (negative change).
(c) Osmosis is the diffusion of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane.
(d) Excess solution clinging to the outside of the potato cylinder adds extra mass that is not part of the tissue, introducing error.
(e) Repeating the trials allows calculation of mean values and identification of anomalies.
(f) Using a scalpel or cork borer requires cutting downwards onto a tile, keeping hands clear of the cutting edge.

(a) 2 marks: Correct calculation of mass change (0.16 g) and percentage change (+5.0%) with working [1 mark for correct method, 1 mark for correct answer with '+' sign or indicating increase]
(b) 2 marks: Between 0.4 and 0.6 (mol/dm^3) (1 mark); explanation: the concentration at which there is no change in mass / where the curve crosses the x-axis (1 mark)
(c) 1 mark: Osmosis
(d) 1 mark: to remove liquid on the surface / ensure only the mass of the potato tissue is measured
(e) 1 mark: repeat the experiment (for each concentration) / use more potato cylinders per solution
(f) 1 mark: cut on a cutting board / cut away from fingers / wear safety goggles

(a) A conical flask is used to contain the reacting mixture, and a gas syringe is used to collect and measure the volume of gas.
(b) Directly reading the syringe values gives 24\text{ cm}^3 at 20 s and 42\text{ cm}^3 at 40 s.
(c) Rate of gas production = (Volume at 40 s - Volume at 20 s) / (40 - 20) = (42 - 24) / 20 = 18 / 20 = 0.9\text{ cm}^3/\text{s}.
(d) The concentration of acid decreases as it is consumed, and eventually one or both reactants are completely used up.
(e) Smaller marble chips have a larger total surface area exposed to the acid. This increases the frequency of collisions between reactant particles, resulting in a higher rate of reaction.

(a) 2 marks: Conical flask / reaction vessel (1 mark) and gas syringe / graduated cylinder (1 mark)
(b) 2 marks: 24 (cm^3) (1 mark) and 42 (cm^3) (1 mark)
(c) 2 marks: evidence of volume difference (18 cm^3) divided by time difference (20 s) (1 mark); 0.9 (cm^3/s) (1 mark)
(d) 1 mark: reactants are used up / concentration of acid decreases
(e) 2 marks: rate increases (1 mark); because of larger surface area / more frequent collisions per unit time (1 mark)

(a) A yellow flame in a flame test is the characteristic test result for sodium ions (Na^+).
(b) (i) Barium nitrate test giving a white precipitate in acidic conditions confirms the presence of sulfate ions (SO_4^{2-}).
(ii) Barium ions react with sulfate ions to form insoluble barium sulfate, BaSO_4.
(iii) Carbonate ions would react with barium ions to form a white precipitate of barium carbonate, BaCO_3. Adding nitric acid decomposes any carbonate present, preventing a false positive.
(c) (i) Copper(II) ions react with a few drops of aqueous ammonia to form a light blue precipitate of copper(II) hydroxide.
(ii) In excess ammonia, this precipitate dissolves to form a deep/dark blue soluble complex.
(d) Wearing personal protective equipment like safety goggles and lab coats is a standard safety measure.

(a) 1 mark: sodium (ion) / Na^+
(b) 3 marks:
(i) sulfate (ion) / SO_4^{2-} (1 mark)
(ii) barium sulfate / BaSO_4 (1 mark)
(iii) to remove/destroy carbonate ions / prevent false positive (1 mark)
(c) 2 marks:
(i) light blue precipitate / blue ppt. (1 mark)
(ii) dissolves / dark blue solution (1 mark)
(d) 1 mark: wear safety goggles / gloves / use a fume cupboard if hazardous gases are evolved

(a) Reading the meters directly gives 0.35 A and 1.40 V.
(b) Using Ohm's Law: R = V / I = 1.40 V / 0.35 A = 4.0 \Omega.
(c) Current passing through a wire causes heating due to Joule heating. Resistance of metals increases with temperature, so keeping the switch closed would introduce temperature-related errors.
(d) A graph of R against l is a straight line through the origin, indicating direct proportionality.
(e) Parallax error occurs if the scale is viewed from an angle. To minimize it, align the eye perpendicular to the mark on the rule.
(f) Since resistance R is inversely proportional to cross-sectional area (A = \pi d^2 / 4), doubling the diameter increases the area by a factor of 4, thus reducing the resistance to one-quarter (1/4) of its original value.

(a) 2 marks: Current = 0.35 (A) (1 mark) and Voltage = 1.4(0) (V) (1 mark)
(b) 2 marks: correct calculation (4.0) (1 mark) and unit (\Omega / ohms) (1 mark)
(c) 1 mark: to prevent the wire from warming up / overheating (which changes resistance)
(d) 1 mark: resistance is directly proportional to length (or resistance increases as length increases)
(e) 2 marks: parallax error / rule not aligned with wire (1 mark); view perpendicularly / tape wire flat to rule (1 mark)
(f) 1 mark: resistance decreases (by a factor of 4)

(a) The mass of the bolt is directly read from the electronic balance as 78.4 g.
(b) (i) Initial volume of water, V1 = 50.0\text{ cm}^3. Final volume of water, V2 = 58.0\text{ cm}^3.
(ii) Volume of the bolt, V3 = V2 - V1 = 58.0 - 50.0 = 8.0\text{ cm}^3.
(c) Density = mass / volume = 78.4 g / 8.0\text{ cm}^3 = 9.8\text{ g/cm}^3.
(d) Looking at the measuring cylinder from an angle causes parallax error. The reading must be taken with the eye at the same horizontal level as the bottom of the curved water surface (the meniscus).
(e) Soluble solids like salt dissolve in water. This means water molecules fit into spaces between salt particles, resulting in a total volume increase that is much less than the actual volume of the salt block.

(a) 1 mark: 78.4 (g)
(b) 3 marks:
(i) V1 = 50.0 (cm^3) and V2 = 58.0 (cm^3) (2 marks)
(ii) V3 = 8.0 (cm^3) (1 mark)
(c) 2 marks: evidence of density formula (mass / volume) (1 mark); 9.8 (g/cm^3) (1 mark)
(d) 1 mark: read at the bottom of the meniscus / view at eye level / view horizontally
(e) 1 mark: salt dissolves in water (giving inaccurate volume)

To conduct a fair test, we must insulate the sides of the beakers using bubble wrap so that heat loss only occurs through the top exposed water surface. The independent variable is the surface area of the water, which is varied by using beakers of different diameters.

1. Measure the internal diameter (d) of each beaker using a ruler and calculate the surface area (A = \pi d^2 / 4).
2. Insulate the curved vertical sides of each beaker equally.
3. Use a measuring cylinder to pour the same volume of hot water into each beaker.
4. Insert a thermometer into the water and record the initial temperature once it stabilizes.
5. Start a stopwatch and record the temperature of the water at 1-minute intervals for 10 minutes.
6. Control variables such as starting temperature, water volume, room temperature, and drafts.
7. Calculate the temperature drop (\Delta T) or rate of cooling (\Delta T / \Delta t) for each beaker.
8. Plot a graph of rate of cooling against exposed surface area. A steeper temperature decline for larger diameters shows that cooling rate increases with exposed surface area.

9 marks total, awarded for points from the following categories:
- Apparatus (max 2 marks): thermometer (1 mark) and stopwatch (1 mark)
- Method (max 3 marks):
- Pour equal volumes of hot water into beakers of different diameters (1 mark)
- Insulate sides of the beakers (to ensure heat is only lost from the top) (1 mark)
- Record temperature at regular intervals / after a set time (1 mark)
- Variables to control (max 2 marks):
- Same volume of water (1 mark)
- Same initial temperature of water / same room temperature (1 mark)
- Data processing and conclusion (max 2 marks):
- Calculate the temperature change (\Delta T) / plot a graph of temperature against time or rate of cooling against surface area (1 mark)
- The beaker with the largest surface area will have the fastest rate of cooling / largest temperature drop (1 mark)