2023 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

At temperatures far above the optimum, the active site of the enzyme changes shape permanently. This is called denaturation. The substrate can no longer fit into the active site, so the reaction stops and the rate decreases to zero.

1 mark: Identify that high temperatures denature the enzyme and decrease the reaction rate to zero.

The angle of incidence is measured between the incident ray and the normal (which is at \(90^\circ\) to the mirror surface). Thus, the angle of incidence is \(90^\circ - 35^\circ = 55^\circ\). According to the law of reflection, the angle of reflection is equal to the angle of incidence, which is \(55^\circ\).

1 mark: Calculate the angle of incidence and apply the law of reflection to find the correct angle of \(55^\circ\).

The reaction of a metal carbonate with an acid produces carbon dioxide gas. The standard test for carbon dioxide is to bubble it through limewater, which turns cloudy (milky) due to the formation of insoluble calcium carbonate.

1 mark: Recall that carbon dioxide is produced and identify limewater turning cloudy as the correct test.

Speed is calculated using the formula: \(\text{speed} = \frac{\text{distance}}{\text{time}}\). Here, distance = 24 m and time = 8.0 s. Speed = \(24 \div 8.0 = 3.0\text{ m/s}\).

1 mark: Use the formula speed = distance / time and perform the correct calculation to get 3.0 m/s.

An ammeter measures current and must be connected in series to allow the current to flow through it. A voltmeter measures potential difference (voltage) across a component and must be connected in parallel across that component.

1 mark: Correctly identify that the ammeter is connected in series and the voltmeter in parallel.

Copper is widely used for electrical wiring because it has high electrical conductivity, allowing current to flow easily, and it is ductile, meaning it can easily be drawn into long, thin wires without breaking.

1 mark: Identify good electrical conductivity and ductility as the key properties of copper for electrical wiring.

A deficiency in calcium (or vitamin D) leads to rickets, characterized by weak and soft bones. Iron deficiency leads to anemia, while vitamin C deficiency leads to scurvy.

1 mark: Correctly pair calcium deficiency with rickets.

The proton number (atomic number) is 3, so there are 3 protons. In a neutral atom, the number of electrons equals the number of protons, which is 3. The nucleon number (mass number) is 7. The number of neutrons is the nucleon number minus the proton number: \(7 - 3 = 4\).

1 mark: Correctly determine that there are 3 protons, 4 neutrons, and 3 electrons.

The distance between two consecutive wave crests represents the wavelength, \(\lambda = 0.20\text{ m}\). The frequency of the wave is \(f = 5.0\text{ Hz}\). Using the wave equation: \(v = f \times \lambda\), we calculate: \(v = 5.0\text{ Hz} \times 0.20\text{ m} = 1.0\text{ m/s}\).

1 mark for the correct calculation of speed using the wave equation.

Using the marble chips in powdered form increases the total surface area of the reactant. This increases the frequency of collisions between reactant particles, thereby increasing the rate of reaction.

1 mark for identifying that using powdered reactant increases the rate of reaction.

The distance travelled can be calculated from the area under the speed-time graph:
1. For the first \(4.0\text{ s}\) (acceleration from rest): \(\text{distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 10\text{ m/s} = 20\text{ m}\).
2. For the next \(6.0\text{ s}\) (constant speed): \(\text{distance} = \text{base} \times \text{height} = 6.0\text{ s} \times 10\text{ m/s} = 60\text{ m}\).

Total distance = \(20\text{ m} + 60\text{ m} = 80\text{ m}\).

1 mark for summing the correct distances calculated for both stages of the motion.

Human body enzymes typically have an optimum temperature of about \(37\text{ }^\circ\text{C}\) where they function most efficiently. At high temperatures (above \(40\text{--}50\text{ }^\circ\text{C}\)), the active site of the enzyme changes shape permanently (denaturation), preventing further catalytic activity.

1 mark for identifying the correct optimum temperature and denaturation behavior.

For resistors connected in series, the total resistance is the sum of the individual resistances: \(R_{\text{total}} = R_1 + R_2 = 4.0\text{ }\Omega + 6.0\text{ }\Omega = 10.0\text{ }\Omega\).

1 mark for calculating the combined series resistance correctly.

Limestone (calcium carbonate) is added to the blast furnace. It undergoes thermal decomposition to form calcium oxide (a basic oxide), which reacts with the acidic impurity silicon dioxide to form molten slag (calcium silicate).

1 mark for identifying limestone as the material used to remove acidic impurities.

Scurvy is a disease caused by a severe deficiency of vitamin C (ascorbic acid) in the diet, which is needed to maintain healthy connective tissues.

1 mark for identifying vitamin C as the deficient nutrient.

In a neutral atom, the number of protons and the number of electrons both equal the proton number (11). The number of neutrons is the difference between the nucleon number and the proton number: \(23 - 11 = 12\text{ neutrons}\).

1 mark for correctly determining the numbers of protons (11), electrons (11) and neutrons (12).

Electromagnetic waves are transverse waves and they all travel at the speed of \(3.0 \times 10^8\text{ m/s}\) in a vacuum. Unlike sound waves, they do not require a medium to travel, and different types of electromagnetic waves (such as radio waves and gamma rays) have different wavelengths.

1 mark for the correct option A. Reject other options because electromagnetic waves are transverse and do not require a medium, and they have varying wavelengths.

A single large lump of calcium carbonate has a smaller surface area than smaller chips of the same mass. A smaller surface area reduces the frequency of collisions between reactant particles, which decreases the rate of reaction. Increasing volume does not affect the initial rate, and increasing temperature or concentration increases the rate.

1 mark for identifying that using a larger lump (smaller surface area) decreases the rate (B). Reject other options which either increase the rate or do not change the initial rate.

The total distance travelled is represented by the area under the speed-time graph. For the first phase (acceleration from 0 to 20 m/s in 10 s), the distance is represented by the area of a triangle: \(0.5 \times 10\text{ s} \times 20\text{ m/s} = 100\text{ m}\). For the second phase (constant speed for 30 s), the distance is represented by the area of a rectangle: \(30\text{ s} \times 20\text{ m/s} = 600\text{ m}\). Total distance is \(100\text{ m} + 600\text{ m} = 700\text{ m}\).

1 mark for calculating the correct total distance of 700 m (C). Reject options that fail to account for the triangular profile of the first 10 s (e.g., treating it as a rectangle, giving 800 m) or only calculate the constant speed section (600 m).

At low temperatures like 10 degrees C, molecules move slowly because they have low kinetic energy, resulting in few successful collisions between the enzyme and substrate. At high temperatures like 80 degrees C, the enzyme is denatured because the high temperature permanently alters the shape of its active site, making it unable to bind to the substrate.

1 mark for correct selection of B. Reject options suggesting low temperature causes denaturation, or high temperature causes low kinetic energy.

Using the equation \(R = V / I\), where \(V = 12\text{ V}\) and \(I = 3.0\text{ A}\), the resistance is calculated as: \(R = 12 / 3.0 = 4.0\ \Omega\).

1 mark for calculating the resistance as 4.0 ohms (B). Reject A (calculating current divided by voltage) and D (multiplying voltage and current).

The reactivity of metals with acid is indicated by the rate of production of hydrogen gas bubbles and temperature rise. An explosive reaction (Z) is more reactive than rapid bubbling (W), which is more reactive than slow bubbling (Y), and no reaction (X) indicates the least reactive metal. Thus, the correct sequence from most to least reactive is Z -> W -> Y -> X.

1 mark for the correct reactivity order Z -> W -> Y -> X (B). Reject other options that put less reactive metals before more reactive ones.

Iodine solution turns blue-black in the presence of starch. Benedict's solution remains blue if reducing sugars are absent (it would turn green, yellow, orange, or brick-red if present). Biuret solution turns purple in the presence of protein. Therefore, starch and protein are present, but reducing sugars are absent.

1 mark for identifying starch and protein only (B). Reject other options because Benedict's remaining blue means reducing sugars are absent, and purple biuret indicates protein is present.

The lower number (proton number) is 15, which represents the number of protons. For a neutral atom, the number of electrons is equal to the number of protons, which is 15. The upper number (nucleon number or mass number) is 31, which is the sum of protons and neutrons. The number of neutrons is calculated as \(31 - 15 = 16\).

1 mark for the correct configuration of 15 protons, 16 neutrons, and 15 electrons (A). Reject other options that confuse the proton and nucleon numbers.

At temperatures above \(60^\circ\text{C}\), enzyme molecules are denatured. This means their active site permanently changes shape, so the substrate (starch) can no longer fit into it. Thus, the rate of reaction drops to zero.

1 mark for B. Reject any other options.

A negative ion has more electrons (negatively charged) than protons (positively charged). For particle X, there are 9 protons and 10 electrons, giving it a net charge of -1. W and Z are neutral atoms (protons = electrons). Y is a positive ion (more protons than electrons).

1 mark for B. Reject any other options.

Increasing the concentration of the acid increases the number of acid particles per unit volume, which increases the rate of collision between the reactants, thereby increasing the reaction rate. Larger pieces reduce the surface area, decreasing the rate. Cooling the mixture decreases the kinetic energy of particles, reducing the rate.

1 mark for C. Reject any other options.

Metal P is most reactive because it reacts with cold water. Metal Q is less reactive because it only reacts with steam. Metal R is least reactive as it does not react with water or steam and is reduced by carbon. Thus, the order of reactivity from most to least reactive is P, then Q, then R.

1 mark for A. Reject any other options.

The distance traveled is the area under the speed-time graph. In the first \(3.0\text{ s}\), the car accelerates from rest to \(6.0\text{ m/s}\). The distance is the area of a triangle: \(0.5 \times 3.0\text{ s} \times 6.0\text{ m/s} = 9.0\text{ m}\). In the next \(4.0\text{ s}\), the car moves at a constant speed of \(6.0\text{ m/s}\). The distance is the area of a rectangle: \(4.0\text{ s} \times 6.0\text{ m/s} = 24.0\text{ m}\). Total distance = \(9.0\text{ m} + 24.0\text{ m} = 33.0\text{ m}\).

1 mark for C. Reject any other options.

Sound waves are longitudinal mechanical waves, which means they propagate by the vibration of particles parallel to the direction of wave travel. Because they require particle vibrations, they cannot travel through a vacuum. They travel faster in solids and liquids than in gases.

1 mark for B. Reject any other options.

Using Ohm's Law, \(V = I \times R\). Rearranging the formula gives \(R = V / I\). Substituting the given values: \(R = 12\text{ V} / 1.5\text{ A} = 8.0\ \Omega\).

1 mark for B. Reject any other options.

Vitamin C is essential for maintaining healthy skin and gums, and its deficiency leads to scurvy. Calcium is for bones and teeth (deficiency: rickets); Iron is for hemoglobin (deficiency: anemia); Vitamin D is for bones and teeth (deficiency: rickets).

1 mark for C. Reject any other options.

All electromagnetic waves are transverse waves and travel at the same speed in a vacuum (approximately \(3 \times 10^8\text{ m/s}\)). Looking at the order of the spectrum from longest wavelength to shortest: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Therefore, X-rays have a shorter wavelength than microwaves.

1 mark for identifying the correct relationship of wavelengths in the electromagnetic spectrum (C).

Using larger lumps of calcium carbonate decreases the available surface area of the reactant. A smaller surface area reduces the frequency of collisions between reactant particles, thereby decreasing the rate of the reaction. Increasing temperature or concentration increases the rate. Changing the volume of acid without changing concentration does not affect the initial rate of reaction.

1 mark for identifying that decreasing surface area (larger lumps) decreases the rate of reaction (B).

Mass is a measure of the amount of matter in an object and remains constant regardless of the gravitational field strength, so the mass on the Moon is \(80\text{ kg}\). Weight is calculated using the formula \(W = m \times g\). On the Moon, weight = \(80\text{ kg} \times 1.6\text{ N/kg} = 128\text{ N}\).

1 mark for correctly stating that mass remains constant and calculating the weight as \(128\text{ N}\) (A).

Amylase is an enzyme (a protein catalyst) and is not a living organism, so it cannot be 'killed' at high temperatures; instead, it becomes denatured, which permanently inactivates it. Cooling down to \(10\text{ }^\circ\text{C}\) reduces kinetic energy but does not denature the enzyme. At the optimum temperature, the enzyme's catalytic activity is highest, meaning starch is broken down at the maximum rate.

1 mark for identifying that the highest rate of starch breakdown occurs at the optimum temperature (C).

Using Ohm's law, the total resistance of the circuit is \(R_{\text{total}} = \frac{V}{I} = \frac{6.0\text{ V}}{0.50\text{ A}} = 12.0\ \Omega\). In a series circuit, total resistance is the sum of individual resistances: \(R_{\text{total}} = R_1 + R_2\). Since the resistors are identical, each resistor has a resistance of \(\frac{12.0\ \Omega}{2} = 6.0\ \Omega\).

1 mark for calculating total resistance and dividing by 2 to find individual resistance of identical series resistors (B).

Metals that are less reactive than carbon, such as iron, can be extracted from their metal oxides by reduction using carbon or carbon monoxide in a blast furnace. More reactive metals like aluminium, magnesium, and sodium must be extracted using electrolysis because carbon is not reactive enough to reduce them.

1 mark for identifying that iron is extracted by heating with carbon (B).

Iron is required for the synthesis of haemoglobin in red blood cells; its deficiency leads to anaemia. Scurvy (bleeding gums) is caused by a lack of vitamin C. Rickets is caused by a lack of vitamin D or calcium.

1 mark for matching iron deficiency with anaemia (B).

In the notation \({}_{Z}^{A}\text{X}\), \(Z\) is the proton (atomic) number and \(A\) is the nucleon (mass) number. Here, the proton number \(Z = 15\), which is the number of protons. For a neutral atom, the number of electrons is equal to the number of protons, which is 15. The number of neutrons is \(A - Z = 31 - 15 = 16\).

1 mark for identifying the correct subatomic particles (protons = 15, neutrons = 16, electrons = 15) (A).

The shortest time taken for starch breakdown is at pH 7.0 (2 minutes). A shorter time means a higher rate of reaction, which indicates that the enzyme is most active at pH 7.0. Therefore, the optimum pH is close to pH 7.0.

1 mark: Correctly identifies D as the correct option.

When light travels from air into glass, it enters a more optically dense medium, which causes its speed to decrease. The frequency of a wave is determined only by its source, so it remains unchanged. Using the wave equation \(v = f \lambda\), since speed \(v\) decreases and frequency \(f\) is constant, the wavelength \(\lambda\) must also decrease.

1 mark: Correctly identifies B as the correct option.

Using powder instead of lumps increases the total surface area of calcium carbonate exposed to the acid. This increases the frequency of collisions between the reactant particles, which increases the rate of reaction. The total volume of carbon dioxide produced remains the same because the mass of calcium carbonate and concentration/volume of hydrochloric acid are unchanged, and calcium carbonate is in excess.

1 mark: Correctly identifies B as the correct option.

First, calculate the useful work done, which is equal to the gain in gravitational potential energy: \(\Delta E_p = m \times g \times h = 40\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 4800\text{ J}\). Next, calculate the power output: \(P = \frac{W}{t} = \frac{4800\text{ J}}{6.0\text{ s}} = 800\text{ W}\).

1 mark: Correctly identifies C as the correct option.

First, calculate the resistance of the two parallel resistors (\(R_p\)): \(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\), giving \(R_p = \frac{12}{5} = 2.4\ \Omega\). Since this combination is in series with the \(3.6\ \Omega\) resistor, the total resistance is \(R_{total} = R_p + 3.6\ \Omega = 2.4\ \Omega + 3.6\ \Omega = 6.0\ \Omega\).

1 mark: Correctly identifies C as the correct option.

Highly reactive metals like Y react rapidly with cold water. Very reactive metals like Z (which are more reactive than carbon) must be extracted by electrolysis. Metals of medium reactivity like X (which are less reactive than carbon) are extracted by reduction of their oxides with carbon. Unreactive metals like W are found uncombined in nature. Therefore, the order of reactivity from most to least is Y -> Z -> X -> W.

1 mark: Correctly identifies A as the correct option.

Vitamin C is essential for making collagen, which is needed for healthy skin, gums, and blood vessels. A deficiency in Vitamin C causes scurvy. Calcium is for bone/teeth (deficiency causes rickets); Iron is for haemoglobin (deficiency causes anaemia); Vitamin D is for calcium absorption (deficiency causes rickets).

1 mark: Correctly identifies C as the correct option.

The proton number is 15, so there are 15 protons. The nucleon number is 31, so the number of neutrons is \(31 - 15 = 16\). A neutral atom of Q has 15 electrons. The \(Q^{3-}\) ion has gained 3 electrons, meaning it has \(15 + 3 = 18\) electrons.

1 mark: Correctly identifies B as the correct option.

At high temperatures, the thermal energy causes the bonds that maintain the specific 3D structure of the enzyme to break. This change in shape, known as denaturation, alters the active site so that the substrate molecule can no longer bind. Substrates do not denature, kinetic energy increases with temperature, and activation energy is unaffected by denaturation.

1 mark for the correct answer B. 0 marks for incorrect options.

The concentrated sucrose solution has a lower water potential than the inside of the potato cells. Water therefore moves out of the potato cells by osmosis (down the water potential gradient). As a result, the vacuole shrinks and the cell membrane pulls away from the cell wall, leaving the cells plasmolysed.

1 mark for the correct answer B. 0 marks for incorrect options.

In terms of oxygen transfer, oxidation is the gain of oxygen and reduction is the loss of oxygen. Iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become iron (\(\text{Fe}\)), so it is reduced. Carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)), so it is oxidised.

1 mark for the correct answer B. 0 marks for incorrect options.

Metals that react with cold water (such as sodium or calcium) are highly reactive. Metals that react with steam but not cold water (such as magnesium or zinc) are moderately reactive. Metals whose oxides are reduced by carbon (like copper or iron) but do not react with steam are lower in reactivity. Metals found uncombined (native, like gold) are extremely unreactive. Thus, the order of reactivity from most to least is W -> X -> Y -> Z.

1 mark for the correct answer A. 0 marks for incorrect options.

The initial kinetic energy of the car is calculated using \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5\text{ kg} \times (2.0\text{ m/s})^2 = 1.0\text{ J}\). The work done on the car by the force is \(W = F \times d = 2.0\text{ N} \times 4.0\text{ m} = 8.0\text{ J}\). Since there are no resistive forces, the final kinetic energy is the sum of the initial kinetic energy and the work done: \(1.0\text{ J} + 8.0\text{ J} = 9.0\text{ J}\).

1 mark for correct calculation leading to option C (Method: Work done = F x d, KE = 0.5 m v^2. Total KE = initial KE + work done. Accuracy: 9.0 J).

First convert the wavelength to meters: \(3.0\text{ cm} = 0.03\text{ m}\). Use the wave equation \(v = f\lambda\) to calculate frequency: \(f = \frac{v}{\lambda} = \frac{1500\text{ m/s}}{0.03\text{ m}} = 50\ 000\text{ Hz}\).

1 mark for correct calculation leading to option D (Method: v = f lambda, convert cm to m, f = v/lambda. Accuracy: 50 000 Hz).

The equivalent resistance of the two \(6.0\ \Omega\) resistors in parallel is \(1/R_p = 1/6.0 + 1/6.0 = 2/6.0\) which gives \(R_p = 3.0\ \Omega\). The total resistance of the circuit is this parallel resistance in series with the \(5.0\ \Omega\) resistor: \(R_{total} = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\). Using Ohm's Law, the total current is \(I = V/R = 12\text{ V} / 8.0\ \Omega = 1.5\text{ A}\).

1 mark for correct calculation leading to option B (Method: Rp = 3.0 ohms, Rtotal = 8.0 ohms, I = V/Rtotal. Accuracy: 1.5 A).

Bleeding gums and painful joints (along with poor wound healing) are classic symptoms of scurvy, which is caused by a deficiency of Vitamin C in the diet.

1 mark for the correct answer A. 0 marks for incorrect options.

At temperatures above the optimum (such as \(45^\circ\text{C}\)), the enzyme's high thermal energy disrupts the bonds holding its tertiary structure together, denaturing the enzyme. This alters the shape of the active site so that the substrate is no longer complementary and cannot bind. Note that denaturation does not break covalent peptide bonds to form free amino acids, nor does heating decrease kinetic energy.

1 mark: Correctly identifies that at \(45^\circ\text{C}\) the enzyme's active site is denatured and no longer fits the substrate.

Using a fine powder increases the surface area of the reactant, which increases the frequency of successful collisions and thus increases the initial rate of reaction. Since the mass of calcium carbonate is unchanged (and it is already in excess), and the volume and concentration of the limiting reactant (hydrochloric acid) are kept the same, the total moles of reactant do not change, so the total volume of carbon dioxide gas produced remains the same.

1 mark: Correctly identifies that increasing surface area increases rate without changing the yield when the limiting reactant is unchanged.

First, calculate the useful work done (increase in gravitational potential energy): \(\Delta E_p = F \times h = 50\text{ N} \times 8.0\text{ m} = 400\text{ J}\).

Next, calculate the efficiency: \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\% = \frac{400\text{ J}}{600\text{ J}} \times 100\% \approx 66.7\%\) (which rounds to \(67\%\)).

1 mark: Correctly calculates efficiency by dividing useful work (400 J) by total input energy (600 J).

First, calculate the combined resistance of the two parallel \(8.0\ \Omega\) resistors:
\(1/R_p = 1/8.0 + 1/8.0 = 2/8.0 \implies R_p = 4.0\ \Omega\).

Next, calculate the total resistance of the series combination:
\(R_{\text{total}} = 4.0\ \Omega + R_p = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).

Finally, calculate the current using Ohm's law:
\(I = V / R_{\text{total}} = 6.0\text{ V} / 8.0\ \Omega = 0.75\text{ A}\).

1 mark: Correctly determines the total equivalent resistance to be 8.0 ohms and calculates the current as 0.75 A.

When a wave passes from one medium to another, its frequency remains constant because it is determined by the source. Therefore, the frequency in shallow water remains \(5.0\text{ Hz}\).
Using the wave equation \(v = f \lambda\) in shallow water:
\(\lambda = v / f = 1.5\text{ m/s} / 5.0\text{ Hz} = 0.30\text{ m}\).

1 mark: Correctly identifies that frequency is unchanged (5.0 Hz) and applies the wave equation to find the wavelength (0.30 m).

In the mouth, salivary glands secrete saliva containing salivary amylase, which digests starch. In the stomach, protein is digested by pepsin, which is produced by the stomach wall (not pancreas). In the small intestine, starch is digested by pancreatic amylase, not stomach. Fat is digested by pancreatic lipase (not liver; liver produces bile, which is an emulsifier, not an enzyme).

1 mark: Correctly identifies the mouth, starch, and salivary glands.

1. Metal X reacts with cold water, indicating very high reactivity.
2. Metal Y reacts with dilute acid but not water, so it is less reactive than X but still more reactive than hydrogen.
3. Metal Z's oxide can be reduced by carbon, meaning it is less reactive than carbon. Since Y reacts rapidly with acid, it sits above carbon and Z.
4. Metal W does not react with dilute acid, placing it below hydrogen (least reactive).

Thus, the correct order from most to least reactive is X \(\rightarrow\) Y \(\rightarrow\) Z \(\rightarrow\) W.

1 mark: Correctly deduces the order of reactivity as X > Y > Z > W based on the observations.

For the ion \({}_{24}^{52}\text{E}^{2+}\):
- The atomic number (subscript 24) is the number of protons, so protons = 24.
- The mass number (superscript 52) is the sum of protons and neutrons. Thus, neutrons = 52 - 24 = 28.
- The charge of \(2+\) means the neutral atom has lost two electrons. The neutral atom would have 24 electrons, so this ion has 24 - 2 = 22 electrons.

Therefore, Protons = 24, Neutrons = 28, Electrons = 22.

1 mark: Correctly identifies the number of protons (24), neutrons (28), and electrons (22).

Increasing temperature from 10 °C to 37 °C increases the kinetic energy of the reactants, resulting in a higher rate of successful collisions. Above 45 °C, the high thermal energy disrupts the bonds holding the enzyme's three-dimensional structure together, denaturing the enzyme. This changes the shape of its active site, meaning the substrate can no longer fit, and the reaction rate decreases rapidly.

1 mark: correct option selected. Reject all other options.

First, calculate the maximum speed achieved: \(v = a \times t = 2.5\text{ m/s}^2 \times 6.0\text{ s} = 15\text{ m/s}\). Then, calculate the distance for each phase:
1) Accelerating phase: \(\text{distance}_1 = \frac{1}{2} \times 15\text{ m/s} \times 6.0\text{ s} = 45\text{ m}\).
2) Constant speed phase: \(\text{distance}_2 = 15\text{ m/s} \times 12.0\text{ s} = 180\text{ m}\).
3) Decelerating phase: \(\text{distance}_3 = \frac{1}{2} \times 15\text{ m/s} \times 4.0\text{ s} = 30\text{ m}\).
Total distance = \(45 + 180 + 30 = 255\text{ m}\).

1 mark: correct option selected. Reject all other options.

The frequency of a wave depends only on its source and does not change when transitioning between mediums. In deep water: \(f = \frac{v}{\lambda} = \frac{24\text{ cm/s}}{4.0\text{ cm}} = 6.0\text{ Hz}\). Therefore, the frequency in shallow water remains \(6.0\text{ Hz}\). In shallow water, using the same frequency: \(\lambda = \frac{v}{f} = \frac{18\text{ cm/s}}{6.0\text{ Hz}} = 3.0\text{ cm}\).

1 mark: correct option selected. Reject all other options.

In experiment X, calcium carbonate is in excess, so the limiting reactant is the hydrochloric acid. The total volume of carbon dioxide produced depends on the number of moles of acid, which is \(0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.05\text{ mol}\). Using powdered marble chips increases the surface area of the solid reactant, which increases the rate of collision and therefore produces a faster initial rate of reaction. Keeping the volume and concentration of hydrochloric acid the same ensures the same number of moles of limiting reactant, yielding the exact same volume of carbon dioxide.

1 mark: correct option selected. Reject all other options.

1) Calculate the combined resistance of the parallel branch, \(R_p\): \(R_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = 2.0\ \Omega\).
2) Calculate total resistance, \(R_{total}\), by adding the series resistor: \(R_{total} = R_p + 4.0\ \Omega = 2.0 + 4.0 = 6.0\ \Omega\).
3) Use Ohm's law to calculate total current, \(I\): \(I = \frac{V}{R_{total}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).

1 mark: correct option selected. Reject all other options.

- From 'W + Oxide of X → Reaction', W is more reactive than X (\(W > X\)).
- From 'Y + Oxide of W → Reaction', Y is more reactive than W (\(Y > W\)).
- From 'Z + Oxide of Y → No reaction', Y is more reactive than Z (\(Y > Z\)).
- From 'Z + Oxide of W → Reaction', Z is more reactive than W (\(Z > W\)).
Combining these findings: \(Y > Z > W > X\).

1 mark: correct option selected. Reject all other options.

Vitamin C is essential for maintaining healthy connective tissues, skin, and gums, and its deficiency leads to scurvy. Calcium is needed for strong bones/teeth (rickets); iron is needed for hemoglobin (anemia); and Vitamin D is needed for calcium absorption (rickets).

1 mark: correct option selected. Reject all other options.

The atomic (proton) number is 17, which identifies the element as chlorine. The nucleon (mass) number is the total number of protons and neutrons: \(17 + 18 = 35\). Since there are 18 electrons and only 17 protons, there is an excess of one electron, giving it a single negative charge (\(\text{Cl}^-\)). Hence, it is a negatively charged ion of chlorine with a nucleon number of 35.

1 mark: correct option selected. Reject all other options.

In the electromagnetic spectrum, ultraviolet waves have higher frequencies and therefore shorter wavelengths than infrared waves. All electromagnetic waves travel at the same speed (the speed of light, \(3 \times 10^8\) m/s) in a vacuum.

1 mark for the correct option C.

In this reaction, carbon monoxide (\(CO\)) gains oxygen to become carbon dioxide (\(CO_2\)), which means it is oxidized. Since \(CO\) causes the iron(III) oxide to be reduced to iron, \(CO\) acts as the reducing agent. Iron(III) oxide loses oxygen and is reduced, acting as the oxidizing agent.

1 mark for identifying the correct statement regarding redox role (option B).

At temperatures above the optimum (e.g., at 60 °C), the active site of the enzyme changes shape permanently due to denaturation. This prevents the substrate from binding to the active site, stopping the reaction.

1 mark for explaining enzyme denaturation at high temperatures (option B).

The work done on the object is equal to Force \(\times\) distance: \(W = 8.0\text{ N} \times 10\text{ m} = 80\text{ J}\). Since the surface is frictionless and it starts from rest, all work done is converted into kinetic energy, so \(E_k = 80\text{ J}\). To find the speed: \(E_k = \frac{1}{2}mv^2 \Rightarrow 80 = \frac{1}{2} \times 4.0 \times v^2 \Rightarrow 80 = 2v^2 \Rightarrow v^2 = 40 \Rightarrow v = \sqrt{40} \approx 6.3\text{ m/s}\).

1 mark for calculating kinetic energy as 80 J and speed as 6.3 m/s (option B).

A more reactive metal displaces a less reactive metal from its compound. W is the most reactive because it displaces X, Y, and Z. Z is more reactive than Y and X (as it reacts with both their nitrates) but less reactive than W. Y reacts only with the nitrate of X, so Y is more reactive than X but less reactive than W and Z. X is the least reactive as it cannot displace any of the other metals. The correct order is W > Z > Y > X.

1 mark for deducing the correct reactivity order from displacement data (option A).

First, find the total resistance (\(R_p\)) of the parallel combination: \(1/R_p = 1/12 + 1/6 = 3/12 = 1/4\), so \(R_p = 4\ \Omega\). Then, use Ohm's law to find the total current: \(I = V / R_p = 6.0\text{ V} / 4\ \Omega = 1.5\text{ A}\).

1 mark for calculating total resistance and total current (option C).

Bleeding gums and poor wound healing are classic symptoms of scurvy, which is caused by a deficiency in Vitamin C. Vitamin C is highly abundant in citrus fruits such as oranges and lemons.

1 mark for correctly matching the symptoms of scurvy to Vitamin C deficiency and citrus fruits (option B).

In concentrated aqueous NaCl, the ions present are \(Na^+\), \(H^+\), \(Cl^-\), and \(OH^-\). At the cathode, \(H^+\) ions are reduced to hydrogen gas because hydrogen is lower than sodium in the reactivity series. At the anode, because the chloride ion concentration is high, \(Cl^-\) ions are oxidized to chlorine gas.

1 mark for correctly identifying the gaseous products of concentrated aqueous NaCl electrolysis (option C).

(a) An enzyme is a protein that functions as a biological catalyst.
(b) As temperature increases towards the optimum temperature, the rate of reaction increases because particles have more kinetic energy and collide more frequently. Above the optimum temperature, the rate of reaction decreases rapidly because the enzyme is denatured, changing the shape of its active site so the substrate can no longer fit.
(c) Carbon, hydrogen, oxygen, nitrogen (any three).

(a) [2 marks] - biological catalyst (1), - protein (1). (b) [4 marks] - rate increases up to optimum (1), - particles have more kinetic energy / more successful collisions (1), - above optimum, rate decreases (1), - enzyme is denatured / active site changes shape (1). (c) [3 marks] - carbon (1), - hydrogen (1), - oxygen or nitrogen (1).

(a) The angle of reflection is equal to the angle of incidence, which is 35 degrees.
(b) The normal is an imaginary line drawn perpendicular (at 90 degrees) to the reflecting surface at the point of incidence.
(c) Regions with shorter wavelengths than visible light are ultraviolet, X-rays, and gamma rays.
(d) Hazards of ultraviolet radiation include sunburn, premature aging of skin, skin cancer, or eye damage.
(e) (i) Sound waves are longitudinal waves.
(ii) A compression is a region in a longitudinal wave where the particles are closest together (high pressure).

(a) [1 mark] - 35 degrees (1). (b) [1 mark] - line perpendicular / at 90 degrees to the mirror surface (1). (c) [2 marks] - any two from: ultraviolet, X-rays, gamma rays (1 mark each). (d) [1 mark] - skin cancer / sunburn / cataracts / cell damage (1). (e) [3 marks] - (i) longitudinal (1), - (ii) region of high pressure / where air particles are pushed close together (2).

(a) The unit of electric current is the ampere (or amp), symbol A.
(b) Formula: \( R = \frac{V}{I} \)
Working: \( R = \frac{6.0}{0.40} = 15 \)
Unit: \( \Omega \) (ohms)
(c) The two lamps must be connected in parallel with each other. This ensures that if one lamp breaks, the circuit for the other lamp remains complete. The switch and the battery must be in series in the main branch so that opening the switch stops the current to both parallel branches simultaneously.
(d) A fuse is a safety device containing a thin wire that melts and breaks the circuit when the current exceeds a specified limit, preventing overheating and risk of fire.

(a) [1 mark] - ampere / amp / A (1). (b) [3 marks] - formula: R = V / I (1), - calculation: 15 (1), - unit: ohm / \u03a9 (1). (c) [3 marks] - lamps in parallel with each other (1), - switch in series with battery/power supply (1), - correct explanation of independent paths (1). (d) [2 marks] - melts / breaks the circuit if current is too high (1), - prevents damage/fire (1).

(a) Limestone (calcium carbonate) is added to react with acidic silicon dioxide impurities to form molten slag (calcium silicate).
(b) (i) iron(III) oxide + carbon monoxide -> iron + carbon dioxide
(ii) Iron(III) oxide is reduced because it loses oxygen during the reaction (forming iron).
(c) (i) Magnesium is the most reactive as it reacts with acid, followed by hydrogen, and copper is the least reactive as it is below hydrogen and does not react with acid. Order: magnesium, hydrogen, copper.
(ii) Test: Insert a burning/lighted splint into a test tube of the gas. Positive result: The gas burns with a squeaky pop sound.

(a) [1 mark] - limestone / calcium carbonate (1). (b) [3 marks] - (i) iron + carbon dioxide (1), - (ii) iron(III) oxide (1) because it loses oxygen (1). (c) [4 marks] - (i) magnesium > hydrogen > copper (2) (1 mark for magnesium most reactive, 1 mark for hydrogen then copper), - (ii) burning / lighted splint (1), - squeaky pop (1).

(a) Increasing concentration increases the rate of reaction. This is because there are more reactant particles in a given volume, leading to a greater frequency of successful collisions between particles.
(b) (i) Using a single large lump decreases the rate of reaction.
(ii) A large lump has a smaller total surface area than the same mass of small pieces. This means fewer calcium carbonate particles are exposed to the acid, resulting in a lower frequency of collisions.
(c) Test: Bubble the gas into limewater. Positive result: The limewater turns cloudy (or milky).

(a) [3 marks] - rate increases (1), - more particles per unit volume (1), - greater frequency of collisions (1). (b) [3 marks] - (i) rate decreases (1), - (ii) smaller surface area (1), - fewer collisions per unit time / lower frequency of collisions (1). (c) [2 marks] - bubble into limewater (1), - turns cloudy/milky (1).

(a) (i) Carbohydrates provide the primary source of energy for cellular respiration.
(ii) Proteins are needed for the growth and repair of cells and tissues, and to make enzymes.
(iii) Fats act as an energy store, provide thermal insulation, and protect internal organs.
(b) (i) Scurvy is the deficiency disease caused by a lack of Vitamin C.
(ii) Vitamin D is required for the absorption of calcium from food, which is essential for developing strong bones and teeth (preventing rickets).
(c) (i) Iron is required to produce haemoglobin, the protein in red blood cells that binds to and transports oxygen throughout the body.
(ii) Good dietary sources of iron include red meat, liver, dark green vegetables (such as spinach), and legumes.

(a) [3 marks] - (i) source of energy (1), - (ii) growth / tissue repair (1), - (iii) energy storage / insulation (1). (b) [2 marks] - (i) scurvy (1), - (ii) absorption of calcium / bone health (1). (c) [3 marks] - (i) production of haemoglobin (1) for oxygen transport / in red blood cells (1), - (ii) red meat / spinach / liver (1).

(a) Formula: \( \text{average speed} = \frac{\text{distance}}{\text{time}} \)
Working: \( \text{speed} = \frac{1800}{120} = 15 \)
Unit: m/s (or metres per second)

(b) Formula: \( W = m \times g \)
Working: \( W = 80 \times 10 = 800 \)
Unit: N (or Newtons)

(c) As the bicycle is braked, kinetic energy is transferred to thermal (heat) energy in the brake pads and wheels, causing them to warm up.

(a) [3 marks] - formula: speed = distance / time (1), - calculation: 15 (1), - unit: m/s or m/s^-1 (1). (b) [3 marks] - formula: W = m * g (1), - calculation: 800 (1), - unit: N or Newtons (1). (c) [2 marks] - kinetic energy (1), - transferred to thermal / heat energy (1).

(a) (i) The proton number is 11, so there are 11 protons.
(ii) The number of neutrons is the nucleon number minus the proton number: \( 23 - 11 = 12 \).
(iii) For a neutral atom, the number of electrons equals the number of protons, which is 11.
(b) Most of the mass of an atom is concentrated in the nucleus (protons and neutrons have much greater mass than electrons).
(c) (i) Proton: +1
(ii) Electron: -1
(d) An element is a substance made up of only one type of atom, which cannot be broken down into simpler substances by chemical means.
(e) There are 4 different elements present in \( \text{NaHCO}_3 \): sodium (Na), hydrogen (H), carbon (C), and oxygen (O).

(a) [3 marks] - (i) 11 (1), - (ii) 12 (1), - (iii) 11 (1). (b) [1 mark] - nucleus / center of the atom (1). (c) [2 marks] - (i) +1 / positive (1), - (ii) -1 / negative (1). (d) [2 marks] - substance containing only one type of atom (1), - cannot be broken down chemically (1). (e) [1 mark] - 4 elements (1).

(a) (i) In a longitudinal wave, the particles of the medium vibrate/oscillate back and forth parallel to the direction of wave propagation (energy transfer). In a transverse wave, the particles vibrate perpendicular to the direction of wave propagation.

(ii) The sound wave travels to the cliff and back, covering a total distance of \(d_{\text{total}}\):
\(d_{\text{total}} = \text{speed} \times \text{time} = 340\text{ m/s} \times 0.60\text{ s} = 204\text{ m}\).
Since the echo travels to the cliff and back, the distance to the cliff is half of the total distance:
\(\text{distance} = \frac{204\text{ m}}{2} = 102\text{ m}\).

(b) (i) The normal is an imaginary line perpendicular (at \(90^\circ\)) to the mirror surface.
Angle of incidence \(i = 90^\circ - 40^\circ = 50^\circ\).

(ii) By the law of reflection, the angle of reflection is equal to the angle of incidence:
Angle of reflection \(r = 50^\circ\).

(iii) Characteristics of an image in a plane mirror include: virtual, upright, same size as the object, laterally inverted, and the same distance behind the mirror as the object is in front of it.

(a)(i)
- Vibrations in longitudinal wave are parallel to energy travel AND vibrations in transverse wave are perpendicular to energy travel. [1 mark]

(a)(ii)
- Recall/use of: distance = speed \u00d7 time (e.g., \(340 \times 0.60 = 204\text{ m}\)) [1 mark]
- Division by 2 to find distance to cliff (\(204 / 2 = 102\)) [1 mark]
- Correct unit: \(\text{m}\) or metres [1 mark]

(b)(i)
- Subtraction from \(90^\circ\) shown: \(90 - 40\) [1 mark]
- Correct value: \(50^\circ\) [1 mark]

(b)(ii)
- Correct value: \(50^\circ\) (allow follow-through from (b)(i)) [1 mark]

(b)(iii)
- Any two from: virtual / upright / same size / same distance behind mirror / laterally inverted. [2 marks, 1 mark for each correct feature]

(a) Frequency remains unchanged / constant. Speed decreases.

(b) \( n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.50} = 2.0 \times 10^8 \text{ m/s} \).

(c) As the wave meets the boundary at an angle, different parts of a wavefront reach the boundary at different times. The part of the wavefront that enters the denser glass block first slows down, while the part still in the air continues at the higher speed. This difference in speed across the wavefront causes the wavefront to pivot/bend, changing the direction of travel towards the normal.

Part (a):
- Frequency remains constant [1]
- Speed decreases [1]

Part (b):
- Use of \( v = \frac{c}{n} \) [1]
- Calculation to give \( 2.0 \times 10^8 \) [1]
- Correct unit \( \text{m/s} \) [1]

Part (c):
- Wavefronts approach the boundary at an angle so different parts reach the boundary at different times [1]
- The part entering the glass slows down first (while other part is still in air) [1]
- This difference in speed causes the wavefronts to pivot / bend / change direction [1]

(a) Connect the reaction flask to a gas syringe using a delivery tube. Measure the volume of carbon dioxide gas collected at regular time intervals (e.g., every 10 seconds) using a stopwatch.

(b) Increasing concentration increases the rate of reaction. This is because there are more reactant particles (hydrogen ions / acid particles) per unit volume. Therefore, the frequency of collisions between reactant particles increases, leading to a higher rate of successful collisions.

(c) The total volume of carbon dioxide gas produced remains the same. Calcium carbonate is the limiting reactant because the acid is in excess in both experiments. Since the mass of calcium carbonate remains unchanged, the total moles of product formed must be the same.

Part (a):
- Collect gas in a gas syringe (or by displacement of water in a measuring cylinder) [1]
- Measure volume of gas at regular time intervals / use a stopwatch [1]

Part (b):
- Rate of reaction increases [1]
- More particles per unit volume [1]
- More frequent collisions (or higher collision frequency) [1]
- More frequent successful/effective collisions [1]

Part (c):
- Volume of carbon dioxide is the same [1]
- Calcium carbonate is the limiting reactant / acid is in excess [1]

(a) Gain in GPE = \( mgh = 1.2 \text{ kg} \times 9.8 \text{ m/s}^2 \times 5.0 \text{ m} = 58.8 \text{ J} \).

(b)(i) Useful power output = \( \frac{\text{Useful energy output}}{\text{time}} = \frac{58.8 \text{ J}}{4.0 \text{ s}} = 14.7 \text{ W} \).

(b)(ii) \( \text{Efficiency} = \frac{\text{Useful power output}}{\text{Power input}} \times 100 \% \Rightarrow 0.60 = \frac{14.7}{\text{Power input}} \Rightarrow \text{Power input} = \frac{14.7}{0.60} = 24.5 \text{ W} \).

(c) The energy that is not usefully transferred is dissipated / lost to the surroundings, primarily as thermal energy (due to friction in bearings) and sound energy.

Part (a):
- Use of \( \Delta E_p = mgh \) [1]
- Correct calculation: \( 58.8 \text{ J} \) [1]

Part (b)(i):
- Use of \( P = \frac{W}{t} \) [1]
- Correct calculation: \( 14.7 \text{ W} \) (accept \( \text{J/s} \)) [1]

Part (b)(ii):
- Use of \( \text{Efficiency} = \frac{\text{Power output}}{\text{Power input}} \) [1]
- Correct calculation: \( 24.5 \text{ W} \) [1]

Part (c):
- Dissipated / transferred to surroundings [1]
- Form of energy specified: thermal / heat / sound [1]

(a) Amylase is a protein with a specific three-dimensional shape. It has an active site. The shape of the active site is complementary only to the shape of the starch molecule (substrate). This is known as the lock and key mechanism, meaning only starch can fit into the active site of amylase to form an enzyme-substrate complex.

(b) As temperature increases from 20 °C to 40 °C, the amylase and starch molecules gain more kinetic energy. This causes them to move faster, leading to a higher frequency of collisions between the substrate molecules and the active site. More collisions have sufficient energy, resulting in a higher rate of enzyme-substrate complex formation and a higher reaction rate.

(c) At 80 °C, the high temperature disrupts the bonds maintaining the three-dimensional structure of the enzyme protein. This changes the shape of the active site (denaturation). The starch substrate can no longer fit into the modified active site, so no enzyme-substrate complexes can form, and the reaction stops.

Part (a):
- Enzyme has an active site [1]
- Active site has a complementary shape to the substrate (starch) [1]
- Reference to 'lock and key' model / enzyme-substrate complex [1]

Part (b):
- Molecules gain more kinetic energy [1]
- Move faster / higher frequency of collisions [1]
- More frequent successful collisions / more enzyme-substrate complexes formed per unit time [1]

Part (c):
- Enzyme is denatured [1]
- Shape of active site is changed so substrate can no longer fit [1]

(a) For two resistors in parallel:

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

Therefore, \( R_p = 6.0 \ \Omega \).

(b) Since the parallel combination is in series with the 4.0 \(\Omega\) resistor:

\( R_{\text{total}} = R_p + 4.0 = 6.0 + 4.0 = 10.0 \ \Omega \).

(c) Using Ohm's law:

\( I = \frac{V}{R_{\text{total}}} = \frac{6.0 \text{ V}}{10.0 \ \Omega} = 0.60 \text{ A} \).

(d) Power dissipated in the 4.0 \(\Omega\) resistor:

\( P = I^2 R = (0.60 \text{ A})^2 \times 4.0 \ \Omega = 0.36 \times 4.0 = 1.44 \text{ W} \).

Alternatively, potential difference across 4.0 \(\Omega\) resistor is \( V = I R = 0.60 \times 4.0 = 2.4 \text{ V} \).

\( P = I V = 0.60 \times 2.4 = 1.44 \text{ W} \).

Part (a):
- Use of \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \) or \( R_p = \frac{R_1 R_2}{R_1 + R_2} \) [1]
- Correct calculation to show 6.0 \(\Omega\) [1]

Part (b):
- Correct addition: \( 6.0 + 4.0 = 10.0 \ \Omega \) [1]

Part (c):
- Use of \( I = \frac{V}{R} \) [1]
- Calculation of 0.60 A [1]

Part (d):
- Use of \( P = I^2 R \) or \( P = I V \) [1]
- Correct calculation to give 1.44 [1]
- Unit: W / Watts [1]

(a) The three raw materials are hematite (or iron ore), coke (or carbon), and limestone (or calcium carbonate).

(b) The balanced equation is:

\( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)

(c) It is a redox reaction because both oxidation and reduction occur simultaneously. \( \text{Fe}_2\text{O}_3 \) is reduced because it loses oxygen to become iron (\( \text{Fe} \)), while \( \text{CO} \) is oxidized because it gains oxygen to become \( \text{CO}_2 \).

(d) Calcium oxide (a basic oxide) reacts with silicon dioxide (an acidic impurity) to form calcium silicate (slag):

\( \text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \)

Thus, the chemical name is calcium silicate.

Part (a):
- Hematite / iron ore [1]
- Coke / carbon [1]
- Limestone / calcium carbonate [1]

Part (b):
- Correct formulas of reactants and products: \( \text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2 \) [1]
- Correct balancing: 1, 3, 2, 3 [1]

Part (c):
- Iron(III) oxide is reduced because it loses oxygen [1]
- Carbon monoxide is oxidized because it gains oxygen [1]

Part (d):
- Calcium silicate (accept \( \text{CaSiO}_3 \)) [1.88]

(a)(i) Vitamin C is required for maintaining healthy skin, gums, blood vessels, and tissue repair (collagen synthesis).

(a)(ii) Vitamin D is required for calcium absorption and maintaining strong bones and teeth.

(a)(iii) Iron is required for the production of hemoglobin in red blood cells to transport oxygen.

(b)(i) Scurvy

(b)(ii) Rickets

(c) Mechanical digestion is the physical breakdown of food into smaller pieces (e.g., chewing by teeth, churning in stomach). This increases the surface area of the food. It does not alter the chemical nature of the molecules. In contrast, chemical digestion is the breakdown of large, insoluble food molecules into small, soluble ones, catalyzed by enzymes, which changes the chemical structure of the food.

Part (a):
- (i) Healthy skin / gums / healing wounds / making collagen [1]
- (ii) Calcium absorption / healthy bones / teeth [1]
- (iii) Making hemoglobin / red blood cells (for oxygen transport) [1]

Part (b):
- (i) Scurvy [1]
- (ii) Rickets [1]

Part (c):
- Mechanical digestion involves physical breakdown / chewing / churning / breaking large pieces to smaller pieces [1]
- Increases surface area of food [1]
- Chemical digestion changes chemical structure / uses enzymes / breaks down large insoluble molecules into small soluble molecules [1.88]

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

(b)(i) Chlorine-35 atom:
- Protons = 17
- Neutrons = 35 - 17 = 18
- Electrons = 17

(b)(ii) Chloride-37 ion (\( \text{Cl}^- \)):
- Protons = 17 (since it is still chlorine)
- Neutrons = 37 - 17 = 20
- Electrons = 18 (since it has a charge of -1, meaning it has gained 1 electron, so 17 + 1 = 18)

(c) Chlorine has 7 outer shell electrons. To gain a full stable outer shell of 8 electrons, two chlorine atoms share one pair of electrons. Each chlorine atom contributes one electron to the shared pair, forming a single covalent bond.

Part (a):
- Atoms with the same number of protons / same atomic number [1]
- Different number of neutrons / different nucleon number [1]

Part (b)(i):
- Protons = 17, Electrons = 17 [1]
- Neutrons = 18 [1]

Part (b)(ii):
- Protons = 17, Neutrons = 20 [1]
- Electrons = 18 [0.88]

Part (c):
- Chlorine atoms need to share electrons to obtain a full outer shell / 8 electrons [1]
- One pair of electrons is shared (one from each atom) [1]

(a)(i) In a parallel circuit, the potential difference across each branch is equal to the supply voltage, so the potential difference across the \(6.0\ \Omega\) resistor is \(12.0\text{ V}\).
Using Ohm's law:
\(I = \frac{V}{R} = \frac{12.0\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).

(a)(ii) The total current is the sum of currents in the parallel branches:
\(I_{\text{lamp}} = I_{\text{total}} - I_{\text{resistor}} = 3.5\text{ A} - 2.0\text{ A} = 1.5\text{ A}\).

(a)(iii) The voltage across the lamp is also \(12.0\text{ V}\).
Using Ohm's law:
\(R_{\text{lamp}} = \frac{V}{I_{\text{lamp}}} = \frac{12.0\text{ V}}{1.5\text{ A}} = 8.0\ \Omega\).

(b) The combined resistance can be calculated using either the parallel resistance formula or Ohm's law on the total circuit:
\(R_{\text{total}} = \frac{V}{I_{\text{total}}} = \frac{12.0\text{ V}}{3.5\text{ A}} \approx 3.4\ \Omega\) (or \(3.43\ \Omega\)).
Alternatively:
\(\frac{1}{R_{\text{total}}} = \frac{1}{6.0} + \frac{1}{8.0} = \frac{7}{24}\),
\(R_{\text{total}} = \frac{24}{7} \approx 3.4\ \Omega\).

(c) First, convert time to seconds:
\(t = 5.0\text{ minutes} \times 60\text{ s/minute} = 300\text{ s}\).

Now, calculate the power of the \(6.0\ \Omega\) resistor:
\(P = V \times I = 12.0\text{ V} \times 2.0\text{ A} = 24.0\text{ W}\) (or \(P = I^2 R = 2.0^2 \times 6.0 = 24.0\text{ W}\)).

Calculate the energy transferred:
\(E = P \times t = 24.0\text{ W} \times 300\text{ s} = 7200\text{ J}\).

Convert to kilojoules:
\(E = \frac{7200}{1000} = 7.2\text{ kJ}\).

(a)(i) [2 Marks]
- State that potential difference across the parallel resistor is \(12.0\text{ V}\) [1]
- Calculation: \(I = \frac{12.0}{6.0} = 2.0\text{ A}\) [1]

(a)(ii) [1 Mark]
- \(3.5 - 2.0 = 1.5\text{ A}\) [1]

(a)(iii) [2 Marks]
- State/use formula: \(R = \frac{V}{I}\) [1]
- Substitution and calculation: \(R = \frac{12.0}{1.5} = 8.0\ \Omega\) [1]

(b) [2 Marks]
- Use of \(R_{\text{total}} = \frac{V}{I_{\text{total}}}\,\) OR \(\,\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\) [1]
- Correct calculation: \(3.4\ \Omega\) (accept \(3.43\ \Omega\) or \(\frac{24}{7}\ \Omega\)) [1]

(c) [2 Marks]
- Conversion of time: \(5.0\text{ minutes} = 300\text{ s}\) [1]
- Correct formula usage and conversion to \(\text{kJ}\): \(E = 12.0 \times 2.0 \times 300 = 7200\text{ J} = 7.2\text{ kJ}\) [1]
(Accept alternative calculation using \(P = I^2 R = 2.0^2 \times 6.0 = 24\text{ W}\))

(a) reading Trial 1: exactly halfway between 14 and 15 gives 14.5 \(\text{cm}^3\). Trial 2 is exactly on 28.0 \(\text{cm}^3\).
(b) The independent variable is the parameter changed by the experimenter (concentration of hydrogen peroxide). The dependent variable is the parameter measured (volume of oxygen gas produced).
(c) Control variables include the temperature of the reactants (controlled using a water bath) and the concentration or volume of the yeast suspension to ensure enzyme concentration remains constant.
(d) Taking regular interval readings allows the initial rate of reaction (the steepest part of the curve) to be calculated. At later times, the rate slows down as substrate is depleted.
(e) Hydrogen peroxide is a chemical irritant. Safety goggles must be worn to protect eyes from splashes.
(f) The standard test for oxygen is that it relights a glowing splint.

[Total: 10 marks]
(a) [2 marks] 1 mark for each correct volume reading: 14.5 \(\text{cm}^3\) and 28.0 \(\text{cm}^3\) (must include units or correct decimals).
(b) [2 marks] 1 mark for identifying independent variable as hydrogen peroxide concentration, and 1 mark for dependent variable as volume of gas.
(c) [2 marks] 1 mark for each valid control variable (e.g., yeast volume, yeast concentration, temperature, or pH).
(d) [2 marks] 1 mark for identifying that it allows the initial/maximum rate to be plotted, 1 mark for stating that rate changes/slows down over time due to substrate depletion.
(e) [1 mark] 1 mark for identifying a valid hazard (hydrogen peroxide is an irritant/corrosive) and its matching precaution (goggles/gloves).
(f) [1 mark] 1 mark for glowing splint relights.

(a) Mass loss = \( 156.45\text{ g} - 155.65\text{ g} = 0.80\text{ g} \).
(b) Sealing the flask with a rubber bung would prevent gas from escaping, building up pressure and keeping the mass constant. Using cotton wool allows gas to exit while blocking liquid droplets/spray.
(c) Limewater (calcium hydroxide solution) turns cloudy when carbon dioxide is bubbled through it due to the formation of insoluble calcium carbonate.
(d) (i) Dividing a solid into a powder increases its surface area to volume ratio, increasing the rate of reaction.
(ii) A larger surface area means more reactant particles are exposed, leading to a higher frequency of successful collisions.
(e) Mass of product is determined by the limiting reactant. Since the initial mass of calcium carbonate is identical and acid is in excess, the same mass of gas is evolved at completion.

[Total: 10 marks]
(a) [1 mark] Correct subtraction showing mass loss of 0.80 g (accept 0.8 g).
(b) [2 marks] 1 mark for stating that it allows gas to escape (preventing pressure build-up), 1 mark for stating that it prevents acid spray/droplets from escaping.
(c) [2 marks] 1 mark for limewater test, 1 mark for correct observation (turns cloudy / milky / white precipitate).
(d) [4 marks]
- (i) [1 mark] Correctly states that the rate of reaction increases / reaction is faster.
- (ii) [3 marks] 1 mark for larger surface area (of powder), 1 mark for more frequent collisions, 1 mark for referencing "per unit time" or "rate of collisions".
(e) [1 mark] 1 mark for stating that the same starting mass of reactants yields the same quantity of product / the stoichiometry/limiting reactant is unchanged.

(a) To measure current, an ammeter must be placed in series with the component. To measure potential difference, a voltmeter must be placed in parallel across the component. Therefore, the ammeter is in series with the cell, switch, and resistance wire, while the voltmeter is connected in parallel across the test wire length.
(b) (i) Using Ohm's Law: \( R = \frac{V}{I} = \frac{2.20}{0.40} = 5.5\ \Omega \). The unit symbol is \( \Omega \) (ohms).
(ii) Resistance per unit length is calculated by dividing the total resistance by the length: \( \frac{5.5\ \Omega}{50.0\text{ cm}} = 0.11\ \Omega/\text{cm} \).
(c) (i) As the temperature of a metal wire increases, metal ions vibrate more, which increases the resistance of the wire. This alters the independent relationship between length and resistance.
(ii) A simple way to prevent heating is to only close the switch long enough to take a rapid reading of the meters, then immediately open the switch to stop current flow.

[Total: 10 marks]
(a) [3 marks]
- 1 mark for stating ammeter connected in series.
- 1 mark for stating voltmeter connected in parallel across the test wire.
- 1 mark for mentioning a complete, closed circuit including cell/power supply and switch.
(b) [3 marks]
- (i) [2 marks] 1 mark for correct formula \( R = V/I \) and calculation yielding 5.5; 1 mark for correct unit symbol \( \Omega \) (or ohms).
- (ii) [1 mark] Correct calculation of 0.11 (accept 11 \( \Omega/\text{m} \) if clearly converted, but question specifies \( \Omega/\text{cm} \)).
(c) [4 marks]
- (i) [2 marks] 1 mark for stating that resistance depends on temperature / temperature is a variable that affects resistance; 1 mark for identifying that it is no longer a fair test of length vs resistance.
- (ii) [2 marks] 1 mark for turning off the power/circuit between readings; 1 mark for explanation (allows wire to cool down to room temperature / minimizes current heating effect).

To construct a successful plan, address each prompt systematically:
1. **Apparatus**: You need containers (beakers) to hold the water, insulating materials (bubble wrap, paper, cotton wool), a method of heating water (kettle), a device to measure volume (measuring cylinder), a device to measure temperature (thermometer), and a timer (stopwatch).
2. **Procedure**: Wrap three identical beakers in equal layers of the respective materials. Boil water and measure equal volumes (e.g., 150 \(\text{cm}^3\)) into each. Place a thermometer in each. Record the starting temperature (aiming for the same initial temperature, e.g., 80\(^\circ\text{C}\)). Start the timer and record temperature at set intervals (e.g., every minute) for 10 minutes.
3. **Variables to control**: Volume of hot water used, starting temperature, layers/thickness of wrap, beaker size/shape, whether a lid is used, and room draft/ambient temperature.
4. **Measurements**: Initial and final temperatures, time.
5. **Analysis**: Determine the temperature drop: \( \Delta T = T_{\text{initial}} - T_{\text{final}} \). Compare values; the best insulator has the lowest temperature drop. Alternatively, describe plotting temperature vs. time graphs.

[Total: 10 marks]
- **Apparatus** [2 marks]: 1 mark for thermometer and stopwatch; 1 mark for beakers, measuring cylinder, and insulation sheets.
- **Method/Procedure** [3 marks]:
- 1 mark for wrapping beakers with different materials and adding hot water.
- 1 mark for using a measuring cylinder to control the volume of water.
- 1 mark for recording initial temperature and starting a timer to record subsequent temperatures.
- **Control Variables** [2 marks]: 1 mark for each valid control variable (e.g., same volume of water, same initial temperature, same thickness/layers of insulating material, same size/type of beaker, use of lids on all or none) - maximum 2 marks.
- **Measurements** [1 mark]: 1 mark for stating that temperature is measured at regular intervals (e.g., every minute) or comparing initial and final temperatures.
- **Analysis & Conclusion** [2 marks]:
- 1 mark for stating how to process data (e.g., calculate temperature change \( \Delta T \) or plot cooling curves of temperature against time).
- 1 mark for identifying that the best insulator has the lowest temperature drop / slowest rate of cooling.

Part (a): When starch is present, iodine solution changes from its natural orange-brown color to blue-black. When all the starch is digested, the mixture no longer contains starch, so the iodine solution remains orange-brown. Part (b): A clear, ruled table with correct headers and units is required: Column 1: Temperature / \(^\circ\text{C}\) (values: 20, 30, 40, 50, 60); Column 2: Time taken to reach endpoint / s (values: 300, 180, 90, 240, > 600 or 'not digested'). Part (c): The pH must be kept constant using a buffer solution, or the concentration of starch/amylase must be kept constant to ensure a fair test. Part (d): Pre-heating the reactants ensures that they are already at the exact target temperature when mixed, preventing temperature lag. Part (e): At \(60^\circ\text{C}\), high thermal energy breaks the bonds maintaining the tertiary structure of the amylase protein. The active site changes shape (denatures), meaning the starch substrate is no longer complementary and cannot bind. Part (f): Testing at shorter intervals (e.g., every 10 seconds) reduces the uncertainty of the exact moment the starch is fully digested, making the time measurement more precise.

(a) 1 mark for blue-black (starch present). 1 mark for yellow/brown/orange (starch digested). [2] (b) 1 mark for correct column headings with units (Temperature / \(^\circ\text{C}\) and Time / s). 1 mark for entering all 5 data points correctly, including noting that at \(60^\circ\text{C}\) it was not digested / exceeded 600 s. [2] (c) 1 mark for pH / concentration of amylase / concentration of starch. [1] (d) 1 mark for ensuring the reactants are at the target temperature before mixing. [1] (e) 1 mark for enzyme is denatured. 1 mark for active site changes shape / substrate no longer fits. [2] (f) 1 mark for testing at shorter time intervals. 1 mark for suggesting a specific shorter interval (e.g., every 10 or 15 s). [2]

Part (a): The diagram must be clear and functional, featuring: 1. A reaction vessel (conical flask) containing the reactants (calcium carbonate chips and hydrochloric acid). 2. A tight-fitting rubber bung/stopper with a single delivery tube. 3. The delivery tube connects directly to a graduated gas syringe. 4. All parts must be clearly labelled. Part (b)(i): The rate of reaction is fastest at the start and progressively decreases. This is supported by the data: from 0 to 30 s, the volume increase is \(24\text{ cm}^3\), whereas from 120 to 150 s, the volume increase is only \(2\text{ cm}^3\). Part (b)(ii): The reaction stops because the limiting reactant (hydrochloric acid) is completely depleted. Part (c): The chemical test for carbon dioxide gas involves bubbling it through limewater (aqueous calcium hydroxide). A positive result is indicated when the limewater turns cloudy/milky due to the formation of insoluble calcium carbonate precipitate. Part (d)(i): Using a powder instead of large chips increases the rate of reaction. Part (d)(ii): Powders have a much larger total surface area exposed to the acid. This increases the frequency of collisions between hydrogen ions in the acid and the calcium carbonate particles per unit time, thereby increasing the reaction rate.

(a) 1 mark for drawing a closed reaction flask containing solid and liquid. 1 mark for showing a gas syringe connected via a delivery tube. 1 mark for correct labels. [3] (b)(i) 1 mark for stating that the rate decreases over time. 1 mark for citing values showing a decreasing volume change per unit time. [2] (b)(ii) 1 mark for stating that the acid (limiting reactant) is completely used up. [1] (c) 1 mark for bubbling the gas through limewater. 1 mark for stating that the limewater turns cloudy or milky. [2] (d)(i) 1 mark for stating the rate increases. [1] (d)(ii) 1 mark for explaining that powder has a larger surface area, leading to a higher frequency of collisions. [1]

Part (a): The normal line is drawn perpendicular (\(90^\circ\)) to the boundary of the glass block. The student should use a protractor aligned with the edge of the block outline at the exact point of incidence to mark and draw this line. Part (b)(i): A protractor is used to measure angles. Part (b)(ii): Using the given values: \(\sin(45^\circ) = 0.7071\), \(\sin(28^\circ) = 0.4695\), \(n = \frac{0.7071}{0.4695} \approx 1.51\). Part (c)(i): Since \(n = \frac{\sin i}{\sin r}\), multiplying both sides by \(\sin r\) yields \(\sin i = n \cdot \sin r\). Comparing this to the equation of a straight line, \(y = mx\), if \(\sin i\) is on the y-axis and \(\sin r\) is on the x-axis, the gradient \(m\) will equal the refractive index \(n\). Part (c)(ii): To improve accuracy, the student should use a thin, sharp pencil to trace the lines, use a narrow slit on the ray box to get a thin beam, and align the pencil marks with the center of the light beam. Part (d): The optical pin method requires: 1. Pinning down the paper, tracing the block, and drawing an incident line. Place two pins (P1 and P2) at least 5 cm apart along this incident line. 2. Looking through the opposite side of the block at the images of P1 and P2. Place a third pin (P3) such that it covers/aligns with the images of P1 and P2. 3. Place a fourth pin (P4) aligned with P3 and the images of P1 and P2. 4. Remove the block and pins, then draw a line through P3 and P4 to the block boundary. Join this point to the incident point to find the path of the refracted ray inside the block.

(a) 1 mark for stating that the normal must be drawn at \(90^\circ\) to the surface at the point of incidence. [1] (b)(i) 1 mark for protractor. [1] (b)(ii) 1 mark for calculating correct sines (\(\sin 45^\circ = 0.71\) and \(\sin 28^\circ = 0.47\)). 1 mark for final answer of 1.5 or 1.51. [2] (c)(i) 1 mark for stating \(\sin i\) should be on the y-axis. 1 mark for explaining that this makes the gradient of the graph equal to the refractive index \(n\). [2] (c)(ii) 1 mark for any one of: use a thin ray of light / use a sharp pencil / mark the center of the ray. [1] (d) 1 mark for placing two pins along the incident line. 1 mark for viewing from the opposite side and placing two more pins so they appear to align with the first two. 1 mark for drawing a line through the pin holes to determine the exit ray and joining it to the incident point. [3]

Part (a): The circuit diagram should feature: 1. A cell or power supply, a closed switch, an ammeter, and the resistance wire connected in series. 2. A sliding contact (represented as an arrow pointing to the resistance wire). 3. A voltmeter connected in parallel with the length of the resistance wire being measured (across the fixed end of the wire and the sliding contact). 4. Standard circuit symbols must be used correctly throughout. Part (b)(i): The voltmeter reading is exactly halfway between 1.0 V and 2.0 V, which is \(1.5\text{ V}\). The current \(I\) is read directly as \(0.40\text{ A}\). Part (b)(ii): Using Ohm's Law: \(R = \frac{V}{I} = \frac{1.5}{0.40} = 3.75\ \Omega\). The unit is the ohm (represented by the symbol \(\Omega\)). Part (c)(i): Resistance is directly proportional to the length of the wire (a straight-line graph through the origin is expected). Part (c)(ii): Current flowing through a wire causes heating due to electrical resistance. Since resistance increases with temperature, switching off the current between readings prevents temperature rise and ensures a fair test. Part (c)(iii): A variable resistor can be adjusted to limit the current flowing through the circuit, keeping it low to minimize heating effects, or to keep the current constant across different tests.

(a) 1 mark for power supply, switch, and ammeter in series with the test wire. 1 mark for voltmeter connected in parallel across the test wire. 1 mark for correct symbols used for all components, including a clear representation of the sliding contact. [3] (b)(i) 1 mark for \(V = 1.5\text{ V}\). 1 mark for \(I = 0.40\text{ A}\). [2] (b)(ii) 1 mark for calculating \(R = 3.75\) (allow ecf from b(i)). 1 mark for stating the correct unit, \(\Omega\) or ohms. [2] (c)(i) 1 mark for stating that resistance increases as length increases / resistance is directly proportional to length. [1] (c)(ii) 1 mark for explaining that keeping the switch open prevents the wire from heating up (which would change its resistance). [1] (c)(iii) 1 mark for explaining that it limits the current / prevents overheating / allows adjustment of current. [1]