2024 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

At room temperature, the substance is a liquid because it has a fixed volume but takes the shape of its container. When heated until it boils, it turns into a gas. In the gaseous state, the particles are far apart and move randomly at high speeds.

1 mark for the correct option C.

Isotopes of the same element have the same number of protons and electrons, so they have identical chemical properties. However, because they have different numbers of neutrons, they have different masses and therefore different physical properties (such as density).

1 mark for the correct option B.

An atom of a Group II metal (X) has two electrons in its outer shell, which it loses to form a stable \(X^{2+}\) cation. An atom of a Group VII non-metal (Y) has seven electrons in its outer shell and gains one electron to form a stable \(Y^{-}\) anion.

1 mark for the correct option B.

Copper(II) oxide is a basic metal oxide. It reacts with dilute sulfuric acid in a neutralization reaction to form copper(II) sulfate (a salt) and water. No gas is produced. Magnesium and zinc react with acid to produce hydrogen gas. Sodium carbonate reacts with acid to produce carbon dioxide gas.

1 mark for the correct option A.

During the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)), the positive lead ions (\(\text{Pb}^{2+}\)) are attracted to the negative electrode (cathode) where they gain electrons to form lead metal. The negative bromide ions (\(\text{Br}^-\)) are attracted to the positive electrode (anode) where they lose electrons to form bromine gas.

1 mark for the correct option B.

An endothermic reaction is one in which energy (heat) is absorbed from the surroundings. This intake of energy causes the temperature of the surroundings to decrease.

1 mark for the correct option D.

Alkanes are saturated hydrocarbons containing only single carbon-carbon bonds. Their boiling points increase as molecular size increases due to stronger intermolecular forces. Alkanes undergo complete combustion to produce carbon dioxide and water.

1 mark for the correct option C.

According to the flame test colors specified in the Cambridge IGCSE Chemistry syllabus: potassium (\(\text{K}^+\)) produces a lilac flame, lithium (\(\text{Li}^+\)) produces a red flame, sodium (\(\text{Na}^+\)) produces a yellow flame, and copper(II) (\(\text{Cu}^{2+}\)) produces a blue-green flame.

1 mark for the correct option C.

The rate of diffusion of a gas is inversely related to its relative molecular mass (or relative atomic mass for noble gases). The lower the mass, the faster the gas diffuses. Among the options given: Helium has a mass of 4, Nitrogen has a mass of 28, Argon has a mass of 40, and Carbon dioxide has a mass of 44. Since Helium has the smallest mass, it diffuses the fastest.

1 mark for the correct option C.

Isotopes are defined as atoms of the same element that have the same proton number but a different number of neutrons. Particle W and Particle X both have 6 protons (identifying them both as carbon), but Particle W has 6 neutrons and Particle X has 8 neutrons. Thus, W and X are isotopes of the same element.

1 mark for the correct option A.

Methane (\( \text{CH}_4 \)) consists of four single covalent C-H bonds, meaning it has 4 shared pairs of electrons. Water (\( \text{H}_2\text{O} \)) contains two single covalent O-H bonds, meaning it has 2 shared pairs of electrons (with the remaining outer electrons of oxygen forming two non-bonding lone pairs). Therefore, option A is correct.

1 mark for the correct option A.

1. Magnesium is a metal above hydrogen in the reactivity series; it reacts with dilute hydrochloric acid to produce hydrogen gas (\( \text{H}_2 \)).
2. Copper(II) oxide is a basic oxide; it reacts with hydrochloric acid to produce copper(II) chloride and water, but no gas is produced.
3. Calcium carbonate is a carbonate; it reacts with hydrochloric acid to produce carbon dioxide gas (\( \text{CO}_2 \)).
Therefore, only substances 1 and 3 react to produce a gas.

1 mark for the correct option C.

For Group I elements, hardness decreases as you go down the group, meaning the metals become softer. Since potassium is below lithium, potassium is softer than lithium, making statement B correct. Statement A is incorrect because melting points decrease down the group. Statement C is incorrect because reactivity increases down the group. Statement D is incorrect because alkali metals react vigorously with water and must be stored under oil to prevent contact with moisture and air.

1 mark for the correct option B.

An exothermic process releases thermal energy to the surroundings, which results in an increase in the temperature of the surroundings. Here, Process 1 and Process 3 show an increase in temperature, meaning they are exothermic. Processes 2 and 4 show a decrease in temperature, meaning they are endothermic.

1 mark for the correct option B.

To decrease the rate of reaction, the concentration of the reactants must be decreased, the temperature lowered, or the surface area of the solid reactant reduced. Diluting the hydrochloric acid with water decreases its concentration, which reduces the frequency of collisions between reactant particles and thus decreases the rate of reaction. Heating, using a powder (larger surface area), or using a more concentrated acid would all increase the rate.

1 mark for the correct option A.

Method 1 (fermentation) requires yeast as a catalyst to provide the necessary enzymes for the biochemical process, making option A correct. Method 1 is carried out at mild temperatures (around \( 25\text{--}35\,^\circ\text{C} \)) because higher temperatures would denature the yeast enzymes, making option B incorrect. Method 2 is a chemical hydration process, not a fermentation process, making option C incorrect. Method 2 uses ethene, which is obtained from non-renewable crude oil, making option D incorrect.

1 mark for the correct option A.

At \(-10^\circ\text{C}\), the temperature is below the melting point of \(-7^\circ\text{C}\), so the substance is a solid. At \(25^\circ\text{C}\), the temperature is between the melting point (\(-7^\circ\text{C}\)) and the boiling point (\(58^\circ\text{C}\)), so the substance is a liquid.

A: 1 mark for identifying both states correctly based on the temperature relative to the melting and boiling points.

The proton number is 11, so there are 11 protons. The number of neutrons is the nucleon number minus the proton number: \(23 - 11 = 12\). A neutral sodium atom has 11 electrons, but the sodium ion \(\text{Na}^+\) has lost one electron to form a positive ion with a 1+ charge, so it has \(11 - 1 = 10\) electrons.

A: 1 mark for the correct combination of protons (11), neutrons (12), and electrons (10).

In a molecule of methane, \(\text{CH}_4\), carbon forms four single covalent bonds. Each single covalent bond consists of a shared pair of electrons, with one electron contributed by hydrogen and one by carbon. Thus, each of the four hydrogen atoms shares a pair of electrons with the central carbon atom.

A: 1 mark for identifying that each hydrogen shares a pair of electrons with carbon.

The reaction between calcium carbonate and dilute hydrochloric acid is: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). The carbon dioxide gas is observed as bubbles (effervescence), and because the calcium chloride product is soluble, the solid calcium carbonate dissolves.

A: 1 mark for the correct combination of observations: gas produced and solid dissolves.

The rate of production of hydrogen bubbles indicates the relative reactivity of the metals. Metal X produces bubbles very rapidly (most reactive), followed by Z (moderately fast), then Y (slowly). Metal W does not react at all (least reactive). Therefore, the correct order from most to least reactive is X, Z, Y, W.

A: 1 mark for ordering the metals correctly from most reactive to least reactive.

Carbon monoxide is produced by the incomplete combustion of carbon-containing fuels (due to insufficient oxygen). Sulfur dioxide is produced by burning fossil fuels (like coal and oil) that contain sulfur impurities.

A: 1 mark for correctly identifying the primary source of both carbon monoxide and sulfur dioxide.

Alkanes are saturated hydrocarbons with the general formula \(\text{C}_n\text{H}_{2n+2}\). They do not have carbon-carbon double bonds, they do not decolourise bromine water rapidly in the dark (this requires UV light for a substitution reaction), and their boiling points increase as the number of carbon atoms (and molecular mass) increases.

A: 1 mark for selecting the correct statement regarding the general formula of alkanes.

Sodium chloride (salt) is soluble in water, but sand is insoluble. Adding water and stirring dissolves the salt. Filtering the mixture leaves sand as the residue on the filter paper, while the salt solution passes through as the filtrate. Evaporating the water from the filtrate leaves behind dry, solid salt.

A: 1 mark for identifying the correct sequence of dissolving, filtering, and evaporating.

The rate of diffusion of a gas depends on its relative molecular mass. The relative molecular mass of nitrogen (\(N_2\)) is 28, while that of carbon dioxide (\(CO_2\)) is 44. Since carbon dioxide has a larger relative molecular mass (is heavier), its molecules move slower on average, and thus it diffuses slower than nitrogen.

Award 1 mark for selecting the correct statement (option B).

The atomic number (proton number) is 15, which indicates there are 15 protons. Since the atom is neutral, the number of electrons must also be 15. The nucleon number (mass number) is 31, meaning the number of neutrons is \(31 - 15 = 16\).

Award 1 mark for the correct combination of protons, neutrons, and electrons (option A).

Molten lead(II) bromide contains lead ions (\(Pb^{2+}\)) and bromide ions (\(Br^-\)). Bromide ions migrate to the positive electrode (anode) and are oxidised to form bromine. Lead ions migrate to the negative electrode (cathode) and are reduced to form lead metal.

Award 1 mark for identifying the correct products formed at both the anode and cathode (option A).

Since the temperature of the reaction mixture decreases, the dissolving process is endothermic. In an endothermic process, thermal energy is absorbed from the surroundings.

Award 1 mark for the correct identification of the reaction type and energy change (option A).

The gas that turns limewater cloudy is carbon dioxide. Carbon dioxide is produced when a metal carbonate, such as calcium carbonate, reacts with an acid.

Award 1 mark for identifying calcium carbonate as the reactant that produces carbon dioxide (option C).

Zinc is more reactive than copper, so it will displace copper from copper(II) sulfate solution. Zinc is less reactive than magnesium, sodium, and calcium, so it cannot displace these metals from their respective solutions.

Award 1 mark for identifying that zinc only displaces metals below it in the reactivity series, which is copper (option B).

Sulfur dioxide is produced by burning fossil fuels containing sulfur impurities. It dissolves in rainwater to form sulfurous and sulfuric acids, leading to acid rain.

Award 1 mark for identifying sulfur dioxide as the pollutant that causes acid rain (option D).

Alkanes are a homologous series of saturated hydrocarbons with the general molecular formula \(C_nH_{2n+2}\). They contain only single carbon-carbon bonds and do not react rapidly with bromine water in the dark.

Award 1 mark for identifying the correct general formula for alkanes (option C).

The relative molecular mass of nitrogen, \( \text{N}_2 \), is 28, while that of sulfur dioxide, \( \text{SO}_2 \), is 64. Gases with a lower relative molecular mass diffuse faster than those with a higher relative molecular mass at the same temperature. Therefore, nitrogen diffuses faster because its molecules have a lower relative mass.

1 mark: correct option selected (A).

During the electrolysis of molten lead(II) bromide, lead ions (\( \text{Pb}^{2+} \)) migrate to the negative electrode (cathode) and are reduced to form liquid lead (observed as a grey liquid). Bromide ions (\( \text{Br}^- \)) migrate to the positive electrode (anode) and are oxidized to form bromine gas (observed as brown fumes).

1 mark: correct option selected (A).

Carbonates react with dilute acids to produce carbon dioxide gas, which turns limewater cloudy. Sodium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide, sodium chloride, and water. Copper(II) oxide and zinc hydroxide are bases and react without gas production. Magnesium reacts to produce hydrogen gas, which does not turn limewater cloudy.

1 mark: correct option selected (C).

For Group I elements (alkali metals), reactivity with water increases down the group, while their melting points decrease down the group.

1 mark: correct option selected (A).

Carbon monoxide is produced by the incomplete combustion of carbon-containing fuels. It is extremely toxic as it binds to hemoglobin in blood, reducing its capacity to carry oxygen around the body. Sulfur dioxide is from burning sulfur-containing fossil fuels and causes acid rain, not the greenhouse effect directly. Oxides of nitrogen come from combustion in car engines, not vegetation. Methane is not produced by complete combustion.

1 mark: correct option selected (A).

The rate of a reaction is slowest when the temperature is lowest, the concentration of the dissolved reactant is lowest, and the surface area of the solid reactant is lowest (using large lumps instead of fine powder). Under option A, all three variables are optimized for the slowest rate.

1 mark: correct option selected (A).

Alkenes are unsaturated hydrocarbons and react rapidly with aqueous bromine in an addition reaction, turning the solution from orange to colorless. Alkanes are saturated and do not react with bromine in the dark, so the solution remains orange.

1 mark: correct option selected (A).

Brass is an alloy, which is a mixture of copper and zinc. Helium is a noble gas and is a pure chemical element. Water is a compound made of hydrogen and oxygen atoms chemically bonded in a fixed ratio.

1 mark: correct option selected (A).

At the anode (positive electrode), hydroxide ions from water are discharged in preference to sulfate ions, forming oxygen gas. This is observed as bubbles of a colourless gas. At the cathode (negative electrode), copper(II) ions are discharged in preference to hydrogen ions, forming copper metal as a pink solid. Because copper(II) ions are removed from the solution, the blue colour of the electrolyte gradually fades and changes from blue to colourless.

1 mark: correct option A. Other options are incorrect because oxygen gas is formed at the anode, and the copper(II) ions leaving the solution causes the blue colour to fade.

Energy required to break bonds (endothermic) = (1 mol of C–H) + (1 mol of Cl–Cl) = 413 + 242 = 655 kJ. Energy released when making bonds (exothermic) = (1 mol of C–Cl) + (1 mol of H–Cl) = 339 + 431 = 770 kJ. Energy change = energy in - energy out = 655 - 770 = -115 kJ/mol.

1 mark: correct option A. Option B has the wrong sign. Options C and D are incorrect calculations.

The formula mass (M_r) of magnesium carbonate is 24 + 12 + (3 * 16) = 84. The number of moles of MgCO3 is 4.2 / 84 = 0.05 mol. According to the equation, 1 mole of MgCO3 produces 1 mole of CO2. Thus, 0.05 mol of CO2 is produced. The volume is 0.05 * 24 dm^3 = 1.2 dm^3.

1 mark: correct option A. Option B uses an incorrect molar calculation. Options C and D result from decimal errors or incorrect stoichiometric ratios.

The rate in Experiment 2 is faster because both concentration (2.0 vs 1.0 mol/dm^3) and temperature (40 vs 20 degrees Celsius) are higher. The total volume of gas depends on the number of moles of HCl. Moles of HCl in Experiment 1 = 0.050 mol. Moles of HCl in Experiment 2 = 0.050 mol. Since the number of moles of the limiting reactant is identical, the total volume of carbon dioxide gas produced is the same.

1 mark: correct option A. Reject options suggesting that the rate is slower or same, or that the total volume of gas is halved.

The forward reaction is exothermic, so a decrease in temperature shifts the equilibrium to the right to produce more heat, increasing the yield of SO3. There are 3 moles of gaseous reactants and 2 moles of gaseous products, so an increase in pressure shifts the equilibrium to the right, which has fewer gas moles. Therefore, a low temperature and a high pressure produce the highest yield at equilibrium.

1 mark: correct option C. High temperatures decrease the yield of sulfur trioxide, and low pressures decrease the yield.

The green precipitate with sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (Cr3+ also gives a green precipitate but is soluble in excess NaOH). The white precipitate formed with barium nitrate in acidic conditions confirms the presence of sulfate ions. Therefore, solid X is iron(II) sulfate.

1 mark: correct option B. Reject options containing chromium(III) (precipitate is soluble in excess NaOH), iron(III) (precipitate is red-brown), or chloride (does not form a precipitate with barium nitrate).

Bromine is higher in Group VII than iodine, making it more reactive. It can therefore displace iodine from potassium iodide solution. Chlorine cannot displace fluorine because fluorine is more reactive. The colours become darker down the group, and astatine is a solid at room temperature.

1 mark: correct option D. Other options are incorrect based on the trends in Group VII reactivity and physical states.

Hydration of ethene is a continuous industrial process. It uses a phosphoric acid catalyst, high temperature (300 degrees Celsius), and high pressure (60 atm). It does not produce carbon dioxide as a by-product (which is produced during fermentation).

1 mark: correct option B. Reject options describing fermentation characteristics (batch, renewable, produces carbon dioxide) or incorrect reaction conditions.

The \(R_f\) value is calculated as the distance moved by the substance divided by the distance moved by the solvent front: \(R_f = 9.0 / 12.0 = 0.75\). Since amino acids are colorless, a locating agent is sprayed onto the chromatogram to react with them and form visible colored spots.

1 mark for the correct answer A: 0.75 for \(R_f\) and correct definition of locating agent function.

The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass. This means lighter gas molecules diffuse faster than heavier gas molecules at the same temperature. Hydrobromic acid (\(\text{HBr}\)) has a relative molecular mass (\(M_r\)) of 81, whereas methylamine (\(\text{CH}_3\text{NH}_2\)) has an \(M_r\) of 31. Since \(\text{HBr}\) molecules are heavier, they diffuse more slowly than the lighter \(\text{CH}_3\text{NH}_2\) molecules. Therefore, the \(\text{HBr}\) travels a shorter distance than the \(\text{CH}_3\text{NH}_2\) in the same time, and the white cloud forms closer to the hydrobromic acid end.

Award 1 mark for selecting the correct option A.
- Reject option B because lighter molecules diffuse faster, not slower.
- Reject option C because diffusion rate depends on molecular mass.
- Reject option D because heavier molecules diffuse slower, not quicker.

During the electrolysis of concentrated aqueous sodium chloride (brine):
- At the anode (+), chloride ions (\(\text{Cl}^-\)) are preferentially discharged to form chlorine gas (\(\text{Cl}_2\)).
- At the cathode (-), hydrogen ions (\(\text{H}^+\)) are preferentially discharged because hydrogen is lower in the reactivity series than sodium, forming hydrogen gas (\(\text{H}_2\)).
- As \(\text{H}^+\) and \(\text{Cl}^-\) ions are removed from the solution, sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) remain. This leaves an alkaline solution of sodium hydroxide (\(\text{NaOH}\)), which causes the pH of the electrolyte to increase.

Award 1 mark for selecting the correct option A.
- Reject option B because sodium is too reactive to be discharged in aqueous solution.
- Reject option C because chloride ions are discharged preferentially at the anode over hydroxide ions in a concentrated solution.
- Reject option D because both products and pH change are incorrect.

To find the conditions that give the highest yield of \(Z\):
1. **Temperature:** The forward reaction is exothermic (indicated by \(\Delta H = -197\text{ kJ/mol}\)). Decreasing the temperature shifts the position of equilibrium to the right (the exothermic direction) to release heat, thereby increasing the yield of \(Z\).
2. **Pressure:** On the reactant side, there are 3 moles of gas (\(2\text{ moles of } X + 1\text{ mole of } Y\)), and on the product side, there are 2 moles of gas (\(2\text{ moles of } Z\)). Increasing the pressure shifts the equilibrium position to the side with fewer moles of gas to reduce the pressure, which is the right side, increasing the yield of \(Z\).

Therefore, low temperature and high pressure produce the highest equilibrium yield.

Award 1 mark for the correct option A.
- Reject B, C, and D because they do not correctly identify both the effect of temperature (exothermic reactions favoured by lower temperature) and pressure (fewer moles of gas favoured by higher pressure).

An acid is defined as a proton (\(\text{H}^+\)) donor:
- In **A**, \(\text{HNO}_3\) donates a proton to \(\text{H}_2\text{O}\) to become \(\text{NO}_3^-\). Thus, \(\text{HNO}_3\) acts as a proton donor.
- In **B**, \(\text{NH}_3\) accepts a proton from \(\text{H}_2\text{O}\) to form \(\text{NH}_4^+\), so it acts as a base.
- In **C**, \(\text{H}_2\text{O}\) accepts a proton from \(\text{HCl}\) to form \(\text{H}_3\text{O}^+\), so it acts as a base.
- In **D**, \(\text{CO}_3^{2-}\) accepts a proton from \(\text{H}_2\text{O}\) to form \(\text{HCO}_3^-\), so it acts as a base.

Award 1 mark for the correct option A.
- Reject options B, C, and D because the underlined species in each of those reactions accept a proton (\(\text{H}^+\)) rather than donating one.

1. Write the balanced equation for the complete combustion of ethane:
\[2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(l)\]

2. Calculate the number of moles of ethane (\(\text{C}_2\text{H}_6\)) reacting:
\[\text{Moles of } \text{C}_2\text{H}_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{6.0\text{ g}}{30\text{ g/mol}} = 0.20\text{ mol}\]

3. Use the stoichiometric ratio to find the required moles of oxygen (\(\text{O}_2\)):
From the equation, \(2\text{ mol}\) of \(\text{C}_2\text{H}_6\) reacts with \(7\text{ mol}\) of \(\text{O}_2\).
\[\text{Moles of } \text{O}_2 = 0.20\text{ mol} \times \frac{7}{2} = 0.70\text{ mol}\]

4. Calculate the volume of oxygen gas at r.t.p.:
\[\text{Volume of } \text{O}_2 = 0.70\text{ mol} \times 24\text{ dm}^3/\text{mol} = 16.8\text{ dm}^3\]

Award 1 mark for the correct option A.
- Reject option B (4.8 dm3) which assumes a 1:1 mole ratio.
- Reject option C (9.6 dm3) which assumes a 1:2 mole ratio.
- Reject option D (33.6 dm3) which fails to divide the mole ratio by 2.

The repeating unit has an ester linkage (\(- \text{CO} - \text{O} -\)) between the monomer units, which means the polymer is a polyester. Polyesters are formed by condensation polymerization with the elimination of water molecules.
- The dicarboxylic acid monomer must be \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) (which provides the \(- \text{CO} - \text{C}_6\text{H}_4 - \text{CO} -\) part after losing \(-\text{OH}\) groups).
- The diol monomer must be \(\text{HO}-\text{CH}_2\text{CH}_2-\text{OH}\) (which provides the \(- \text{O} - \text{CH}_2\text{CH}_2 - \text{O} -\) part after losing \(-\text{H}\) atoms).
Therefore, the correct choice is option A.

Award 1 mark for selecting the correct option A.
- Reject option B because this is condensation, not addition polymerization.
- Reject option C because the monomers given would form a polyamide, not a polyester, and they are incorrect.
- Reject option D because the monomers are mismatched and addition polymerization is incorrect.

Let's analyse the statements:
- **Statement 1 is incorrect:** The activation energy (\(E_a\)) is the minimum energy required for a collision to be successful. It is a constant characteristic of the reaction pathway and is not affected by temperature. (Only adding a catalyst can lower the activation energy by providing an alternative pathway).
- **Statement 2 is correct:** An increase in temperature increases the average kinetic energy of the particles, so they move faster. This causes them to collide more frequently.
- **Statement 3 is correct:** Since the particles have higher average kinetic energy, a much larger proportion of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher proportion of successful collisions.

Therefore, only statements 2 and 3 are correct.

Award 1 mark for selecting correct option A.
- Reject options B, C, and D because they include statement 1, which is incorrect since temperature does not change the activation energy of a reaction.

A catalytic converter works by converting toxic exhaust gases (specifically carbon monoxide, \(\text{CO}\), and nitrogen oxides like nitrogen monoxide, \(\text{NO}\)) into less harmful gases. The reaction between carbon monoxide and nitrogen monoxide produces carbon dioxide (\(\text{CO}_2\)) and nitrogen gas (\(\text{N}_2\)):
\[2\text{CO}(g) + 2\text{NO}(g) \rightarrow 2\text{CO}_2(g) + \text{N}_2(g)\]
Both \(\text{CO}\) (a toxic gas) and \(\text{NO}\) (which contributes to acid rain and photochemical smog) are converted into \(\text{CO}_2\) and \(\text{N}_2\), which are much less harmful.

Award 1 mark for the correct option A.
- Reject option B because methane combustion is the reaction of fuel in the engine, not the catalytic conversion of pollutant gases.
- Reject option C because it produces toxic carbon monoxide, which is the opposite of the catalytic converter's purpose.
- Reject option D because it produces nitrogen monoxide, which is a harmful pollutant, and sulfur dioxide is not converted in this manner.

The \(R_f\) value is calculated by dividing the distance traveled by the substance by the distance traveled by the solvent front: \(R_f = \frac{5.2}{8.0} = 0.65\). Pencil is used because ink would dissolve and separate. The \(R_f\) value of a substance changes depending on the solvent used.

1 mark for the correct option C. Incorrect options: A is the inverse ratio (8.0/5.2), B is chemically incorrect as ink runs, D is incorrect because Rf values are solvent-dependent.

First, find the moles of \(\text{HCl}\): \(\text{moles} = \text{concentration} \times \text{volume} = 2.00 \text{ mol/dm}^3 \times 0.0500 \text{ dm}^3 = 0.100 \text{ mol}\). From the stoichiometry of the equation, 2 moles of \(\text{HCl}\) produce 1 mole of \(\text{CO}_2\). Therefore, \(\text{moles of CO}_2 = 0.100 / 2 = 0.0500 \text{ mol}\). The volume of gas is \(\text{moles} \times 24.0 \text{ dm}^3/\text{mol} = 0.0500 \times 24.0 = 1.20 \text{ dm}^3\).

1 mark for correct calculation leading to option A. B is a distractor where stoichiometric ratio is ignored. C is a distractor where moles of acid are multiplied by 2.

Increasing the temperature increases the kinetic energy of the particles, meaning a higher proportion of collisions have energy greater than or equal to the activation energy. Increasing concentration increases the number of particles per unit volume, which increases the frequency of collisions.

1 mark for the correct statement A. B is incorrect because temperature affects energy and concentration does not. C is incorrect as concentration doesn't change speed. D is incorrect because activation energy is unaffected by temperature/concentration (only by catalysts).

During the electrolysis of concentrated aqueous \(\text{NaCl}\), chloride ions (\(\text{Cl}^-\)) are discharged at the anode to form chlorine gas (\(\text{Cl}_2\)), and hydrogen ions (\(\text{H}^+\)) are discharged at the cathode to form hydrogen gas (\(\text{H}_2\)). The remaining ions in solution are \(\text{Na}^+\) and \(\text{OH}^-\), which form alkaline sodium hydroxide, raising the pH.

1 mark for identifying the correct combination in option C.

The repeating unit contains an ester linkage (\(\text{-O-CO-}\)). This shows it is a polyester. Polyesters are formed by the condensation polymerisation of a diol and a dicarboxylic acid.

1 mark for identifying that a polyester is made from a diol and a dicarboxylic acid (option B).

Metal Z is more reactive than X because Z reacts vigorously with acid while X reacts slowly. Metal X is more reactive than Y because X can displace Y from solution. This gives the order of reactivity: Z > X > Y.

1 mark for the correct reactivity order (Z, X, Y) in option C.

Sulfite ions (\(\text{SO}_3^{2-}\)) react with acid to produce sulfur dioxide gas, which is a reducing agent and decolourises acidified potassium manganate(VII). They also form a white precipitate of barium sulfite when barium ions are added, which is soluble in dilute acid.

1 mark for identifying the sulfite ion (option C).

Energy absorbed to break bonds = \(\text{H-H} + \text{Cl-Cl} = 436 + 242 = 678\text{ kJ/mol}\). Energy released when bonds are formed = \(2 \times \text{H-Cl} = 2 \times 431 = 862\text{ kJ/mol}\). Enthalpy change \(\Delta H = 678 - 862 = -184\text{ kJ/mol}\). Since \(\Delta H\) is negative, the reaction is exothermic.

1 mark for correct calculation and sign/direction determination (option A).

In Experiment 1, with inert carbon electrodes, hydroxide ions (from water) are discharged at the anode to produce oxygen gas, which is observed as bubbles of a colorless gas. In Experiment 2, with reactive copper electrodes, the copper anode itself is oxidized to form copper(II) ions, so the anode dissolves and decreases in mass.

1 mark for the correct option A. Reject others.

The equation for the combustion of methane is: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\). Since the mole ratio of \(CH_4\) to \(O_2\) is 1:2, \(20\text{ cm}^3\) of \(CH_4\) reacts completely with \(40\text{ cm}^3\) of \(O_2\). This leaves \(60 - 40 = 20\text{ cm}^3\) of unreacted \(O_2\). The reaction produces \(20\text{ cm}^3\) of \(CO_2\) gas. At room temperature and pressure, water is a liquid, so its volume is negligible. Therefore, the total volume of gas remaining is \(20\text{ cm}^3\) (unreacted \(O_2\)) + \(20\text{ cm}^3\) (produced \(CO_2\)) = \(40\text{ cm}^3\).

1 mark for the correct option B. Reject others.

Curve Y represents a faster reaction rate (steeper initial gradient) but the same total yield of product (levels off at the same volume). Powdered calcium carbonate has a larger surface area than chips, which increases the rate of reaction. Since the mass of calcium carbonate and the amount of limiting reactant (hydrochloric acid) are unchanged, the final volume of gas remains the same.

1 mark for the correct option A. Reject others.

Hydrochloric acid is a strong acid that completely dissociates in water, resulting in a higher concentration of hydrogen ions than the partially dissociated weak acid, ethanoic acid. Consequently, hydrochloric acid has a lower pH, higher electrical conductivity, and reacts more rapidly with magnesium.

1 mark for the correct option C. Reject others.

Since the forward reaction is exothermic, decreasing the temperature shifts the position of equilibrium to the right to produce more heat, increasing the yield of \(SO_3\). Since there are 3 moles of gas on the reactant side and 2 moles of gas on the product side, increasing the pressure shifts the equilibrium to the side with fewer moles of gas (the right), also increasing the yield of \(SO_3\).

1 mark for the correct option B. Reject others.

The repeating unit of this polymer is \(-CH(CH_3)-CH_2-\), which contains two carbon atoms in the main chain. The corresponding monomer must have a double bond between these two carbons: \(CH(CH_3)=CH_2\), which is propene (\(CH_3CH=CH_2\)).

1 mark for the correct option B. Reject others.

Energy required to break bonds = \(1 \times (H-H) + 1 \times (Cl-Cl) = 436 + 242 = 678\text{ kJ/mol}\). Energy released when making bonds = \(2 \times (H-Cl) = 2 \times 431 = 862\text{ kJ/mol}\). Enthalpy change, \(\Delta H\) = energy broken - energy made = \(678 - 862 = -184\text{ kJ/mol}\).

1 mark for the correct option A. Reject others.

During the reaction, each \(Fe^{3+}\) ion gains one electron to become a \(Fe^{2+}\) ion. A gain of electrons is reduction. Furthermore, the oxidation state of iron decreases from +3 to +2, which also defines reduction. Thus, \(Fe^{3+}\) is reduced because its oxidation number decreases.

1 mark for the correct option B. Reject others.

(a) Filtration is the standard method for separating an insoluble solid from a liquid.
(b) The residue must be washed with distilled water to remove any remaining salt solution, then dried in an oven or between filter papers.
(c) Crystallization involves heating to concentrate the solution, cooling to allow crystallization, filtering the crystals, and drying.
(d) Simple distillation is used to separate a solvent from a solute.
(e) Testing boiling point (100 °C) or freezing point (0 °C) determines the purity of water.

(a) Filtration [1 mark]
(b) Wash with distilled water [1 mark]; dry between filter papers/warm oven [1 mark]
(c) Heat/evaporate to saturation point/crystallization point [1 mark]; leave to cool [1 mark]; filter off crystals [1 mark]; dry crystals with filter paper [1 mark]
(d) Simple distillation [1 mark]
(e) Measure boiling point / freezing point [1 mark]; boils at exactly 100 °C / freezes at 0 °C [1 mark]

(a) Rate can be monitored by gas volume or mass loss over time.
(b)(i) Smaller chips increase surface area, increasing the rate.
(b)(ii) Lower temperature decreases particle energy, decreasing the rate.
(c) Large surface area increases contact area, leading to more frequent successful collisions.
(d) Decreased temperature means particles move slower and have less energy, so the frequency of collisions decreases and fewer particles have energy greater than or equal to the activation energy.

(a) Measure volume of gas produced over time [1 mark]; measure loss of mass over time [1 mark]
(b)(i) Rate increases [1 mark]
(b)(ii) Rate decreases [1 mark]
(c) Smaller chips have a larger surface area [1 mark]; more collisions per unit time / higher frequency of collisions [1 mark]; between reactant particles [1 mark]
(d) Particles have less kinetic energy / move slower [1 mark]; lower frequency of collisions [1 mark]; fewer successful/energetic collisions [1 mark]

(a) Group I metals are soft, low-density, and have low melting points compared to transition metals.
(b)(i) Typical observations include fizzing, moving on the water, melting into a ball, and eventual disappearance.
(b)(ii) Word equation: sodium + water -> sodium hydroxide + hydrogen.
(b)(iii) Alkaline solution (sodium hydroxide) has high pH (12-14).
(c) Reactivity increases down the group, so potassium reacts more vigorously than sodium.

(a) Soft/can be cut [1 mark]; low density/floats [1 mark]; low melting point [1 mark] (Any two, max 2)
(b)(i) Melts into a ball / floats / moves on water / fizzing or bubbles / disappears [2 marks, 1 for each observation]
(b)(ii) Left-hand side: sodium + water [1 mark]; Right-hand side: sodium hydroxide + hydrogen [1 mark]
(b)(iii) pH 12-14 [1 mark]; sodium hydroxide is a strong alkali/base [1 mark]
(c) Potassium is more reactive [1 mark]; reactivity increases down Group I / potassium is below sodium in the Periodic Table [1 mark]

(a)(i) Iron(II) ions form a green precipitate with sodium hydroxide that is insoluble in excess.
(a)(ii) The ionic equation is \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\).
(b)(i) Barium nitrate test is the standard test for sulfate ions, yielding a white precipitate of barium sulfate.
(c) Formula of iron(II) sulfate is \(\text{FeSO}_4\).
(d)(i) Limewater test is used for carbon dioxide.
(d)(ii) Damp blue litmus paper is bleached by chlorine.

(a)(i) Iron(II) / \(\text{Fe}^{2+}\) [1 mark]
(a)(ii) Correct formulae and balancing: \(\text{Fe}^{2+} + 2\text{OH}^{-} \rightarrow \text{Fe(OH)}_2\) [1 mark]; correct state symbols: (aq), (aq) -> (s) [1 mark]
(b)(i) Sulfate / \(\text{SO}_4^{2-}\) [1 mark]
(b)(ii) Barium sulfate [1 mark]
(c) \(\text{FeSO}_4\) [1 mark]
(d)(i) Bubble into limewater [1 mark]; goes cloudy/milky [1 mark]
(d)(ii) Damp blue litmus paper [1 mark]; bleached / turns white [1 mark]

(a) A hydrocarbon is defined as a molecule containing only carbon and hydrogen atoms.
(b) Alkanes have the general formula \(\text{C}_n\text{H}_{2n+2}\).
(c)(i) Propane has 3 carbon atoms in a chain, each bonded to hydrogen atoms to satisfy carbon's valency of 4.
(c)(ii) The complete combustion of propane: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\).
(d)(i) Photochemical chlorination requires ultraviolet (UV) light to break the chlorine-chlorine bond.
(d)(ii) Methane undergoes monosubstitution to form chloromethane.
(d)(iii) The hydrogen atom displaced combines with the other chlorine atom to form hydrogen chloride gas.

(a) Compound of carbon and hydrogen [1 mark]; ONLY / contains no other elements [1 mark]
(b) \(\text{C}_n\text{H}_{2n+2}\) [1 mark]
(c)(i) Three carbon chain with single bonds [1 mark]; all single C-H bonds shown (8 hydrogens) [1 mark]
(c)(ii) Correct reactants and products: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\) [1 mark]; correct balancing: 5, 3, 4 [1 mark]
(d)(i) Ultraviolet light / UV / sunlight [1 mark]
(d)(ii) Chloromethane [1 mark]
(d)(iii) Hydrogen chloride [1 mark] [Reject hydrochloric acid]

(a)(i) A process that absorbs heat, causing a temperature drop, is endothermic.
(a)(ii) Heat flows from surroundings to the reaction mixture.
(b)(i) An exothermic reaction has reactants at a higher energy level than products. \(\Delta H\) is negative, indicated by a downward arrow.
(b)(ii) Breaking bonds (H-H and Cl-Cl) requires energy. Making bonds (H-Cl) releases energy. If more energy is released than absorbed, the reaction is exothermic.
(c) Hydrogen is a clean fuel because its combustion product is only water.

(a)(i) Endothermic [1 mark]; temperature decreases / heat is taken in [1 mark]
(a)(ii) Heat transferred from surroundings to the system / reaction mixture [1 mark]
(b)(i) Horizontal reactant line higher than product line, both correctly labeled [1 mark]; downward arrow for energy change [1 mark]; arrow labeled \(\Delta H\) or enthalpy change [1 mark]
(b)(ii) Bond breaking is endothermic / absorbs energy AND bond making is exothermic / releases energy [1 mark]; more energy is released in bond making than is absorbed in bond breaking [2 marks]
(c) Hydrogen [1 mark]

(a) Water treatment involves two key steps: filtration (removes insoluble solids like sand, silt, and organic matter) and chlorination (disinfects water by killing bacteria and pathogens).
(b) Chemical additives include: 1. Chlorine (disinfectant to ensure safety from pathogens). 2. Fluoride (promotes dental health by strengthening tooth enamel).
(c) Testing for water can be done using anhydrous salts. Anhydrous copper(II) sulfate turns from white to blue, or anhydrous cobalt(II) chloride paper turns from blue to pink when water is added.

(a) Filtration [1 mark]; to remove insoluble solids/particles [1 mark]; Chlorination / addition of chlorine [1 mark]; to kill bacteria/microbes/pathogens [1 mark]
(b) Chlorine [1 mark]; to sterilize / disinfect / kill microorganisms [1 mark]; Fluoride [1 mark]; to prevent tooth decay / strengthen teeth [1 mark] (Alternatively, accept lime/calcium hydroxide [1 mark] to reduce acidity / raise pH [1 mark])
(c) Add to anhydrous copper(II) sulfate / anhydrous cobalt(II) chloride paper [1 mark]; turns from white to blue / turns from blue to pink [1 mark]

(a) Anode is the positive electrode; cathode is the negative electrode.
(b)(i) Lead ions (\(\text{Pb}^{2+}\)) migrate to the cathode and gain electrons to form lead metal.
(b)(ii) Bromide ions (\(\text{Br}^{-}\)) are oxidized at the anode, releasing reddish-brown bromine gas.
(b)(iii) The half-equation at the anode is \(2\text{Br}^{-} \rightarrow \text{Br}_2 + 2\text{e}^{-}\).
(c)(i) Electrolysis of concentrated aqueous sodium chloride (brine) yields hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide remains in solution.
(c)(ii) Hydrogen ions are discharged preferentially because sodium is highly reactive and hydrogen is lower in the reactivity series.

(a) Anode [1 mark]; Cathode [1 mark]
(b)(i) Lead [1 mark]
(b)(ii) Brown fumes / brown gas [1 mark] [Reject brown liquid/precipitate]
(b)(iii) Correct species: \(2\text{Br}^{-} \rightarrow \text{Br}_2\) [1 mark]; balanced with electrons: \(2\text{Br}^{-} \rightarrow \text{Br}_2 + 2\text{e}^{-}\) (or \(2\text{Br}^{-} - 2\text{e}^{-} \rightarrow \text{Br}_2\)) [1 mark]
(c)(i) Chlorine [1 mark]; Hydrogen [1 mark]; Sodium hydroxide [1 mark]
(c)(ii) Hydrogen is less reactive than sodium / sodium is more reactive than hydrogen (so H+ ions are preferentially discharged) [1 mark]

(a) Calculate moles of each element:
- Moles of Fe = 70.0 / 56 = 1.25 mol
- Moles of O = 30.0 / 16 = 1.875 mol
Divide by the smallest value to find the ratio:
- Fe: 1.25 / 1.25 = 1
- O: 1.875 / 1.25 = 1.5
Multiply by 2 to get whole numbers: Fe = 2, O = 3.
Therefore, the empirical formula is Fe2O3.

(b) The balanced chemical equation is:
Fe2O3 + 3CO -> 2Fe + 3CO2

(c) Step 1: Calculate the molar mass (Mr) of Fe2O3:
Mr(Fe2O3) = (2 * 56) + (3 * 16) = 112 + 48 = 160 g/mol.
Step 2: Calculate moles of Fe2O3 in 24.0 g:
moles = 24.0 / 160 = 0.15 mol.
Step 3: Determine moles of Fe produced from the equation stoichiometry (1 mol of Fe2O3 produces 2 mol of Fe):
moles of Fe = 2 * 0.15 = 0.30 mol.
Step 4: Calculate mass of Fe:
mass = 0.30 * 56 = 16.8 g.

(d) Iron is reduced because its oxidation number decreases from +3 in Fe2O3 to 0 in Fe. Carbon is oxidized because its oxidation number increases from +2 in CO to +4 in CO2. Since both oxidation and reduction occur, it is a redox reaction.

(e) Carbon monoxide is toxic because it binds strongly to haemoglobin in the blood, preventing oxygen from being transported around the body.

(f) Coke (carbon).

(g) Carbon dioxide reacts with hot coke (carbon) to form carbon monoxide:
C + CO2 -> 2CO

(a) [3 marks total]
- 1 mark for calculating moles of Fe and O (1.25 and 1.875).
- 1 mark for the ratio 1 : 1.5 (or 2 : 3).
- 1 mark for the final empirical formula Fe2O3.

(b) [2 marks total]
- 1 mark for correct reactants and products.
- 1 mark for correct balancing.

(c) [3 marks total]
- 1 mark for Mr of Fe2O3 = 160 and calculating 0.15 mol.
- 1 mark for establishing the 1:2 molar ratio of Fe2O3 to Fe (0.30 mol Fe).
- 1 mark for final mass of 16.8 g (allow ecf from incorrect moles).

(d) [2 marks total]
- 1 mark for identifying reduction as a decrease in the oxidation number of iron from +3 to 0.
- 1 mark for identifying oxidation as an increase in the oxidation number of carbon from +2 to +4.

(e) [1 mark]
- 1 mark for stating it is poisonous/toxic or that it binds to haemoglobin/reduces oxygen transport.

(f) [1 mark]
- 1 mark for coke (accept carbon).

(g) [1.33 marks]
- 1 mark for correct formula of reactants and products (C + CO2 -> CO).
- 0.33 mark for correct balancing (C + CO2 -> 2CO).

(a)(i) In dilute sulfuric acid, H+ and OH- ions are present along with SO4^2-. At the cathode, H+ is discharged to form hydrogen gas (H2). At the anode, OH- is discharged to form oxygen gas (O2).
(a)(ii) With active copper electrodes in CuSO4, copper atoms from the anode dissolve to form Cu2+ ions. At the cathode, Cu2+ ions are discharged to form copper metal (Cu).

(b) Cathode: Cu2+(aq) + 2e- -> Cu(s) (copper ions gain electrons / are reduced).
Anode: 4OH-(aq) -> O2(g) + 2H2O(l) + 4e- (hydroxide ions lose electrons / are oxidized).

(c) Electrolyte: The intensity of the blue color of the solution remains unchanged because the rate at which Cu2+ ions are discharged at the cathode is exactly equal to the rate at which Cu2+ ions are formed at the anode.
Anode: The anode dissolves and decreases in mass/size because copper atoms lose electrons to form soluble Cu2+ ions.

(d) This process is used industrially for the refining/purification of copper.

(a) [4 marks total]
- (i) 1 mark for oxygen at the anode; 1 mark for hydrogen at the cathode.
- (ii) 1 mark for copper dissolves / copper(II) ions at the anode; 1 mark for copper metal at the cathode.

(b) [4 marks total]
- Cathode: 1 mark for Cu2+ + 2e- -> Cu; 1 mark for correct state symbols: Cu2+(aq) and Cu(s).
- Anode: 1 mark for 4OH- -> O2 + 2H2O + 4e- (or 2H2O -> O2 + 4H+ + 4e-); 1 mark for correct state symbols.

(c) [4 marks total]
- 1 mark for stating that the electrolyte remains blue / no change in colour intensity.
- 1 mark for explaining that the concentration of Cu2+ ions remains constant / rate of formation equals rate of discharge.
- 1 mark for stating that the copper anode decreases in mass / dissolves.
- 1 mark for explaining that copper atoms are oxidized to Cu2+ ions (Cu -> Cu2+ + 2e-).

(d) [1.33 marks]
- 1.33 marks for stating 'purification of copper' or 'refining copper'.

(a) A suitable method is to measure the volume of carbon dioxide gas produced over time using a gas syringe connected to the reaction flask. Alternatively, the reaction flask can be placed on a digital balance to monitor the decrease in mass as carbon dioxide gas escapes over time.

(b)(i) The initial rate of reaction increases.
(b)(ii) Increasing the concentration of hydrochloric acid means there are more reactant particles (H+ ions) per unit volume. This increases the frequency of collisions between the reacting particles, leading to a higher rate of successful collisions per unit time.

(c) Increasing the temperature increases the kinetic energy of the particles, causing them to move faster and collide more frequently. More importantly, a much larger proportion of the colliding particles now possess energy greater than or equal to the activation energy (Ea). This significantly increases the proportion of successful collisions per unit time, which raises the rate of reaction.

(d) Step 1: Calculate the number of moles of HCl used:
moles of HCl = concentration * volume in dm³ = 1.00 mol/dm³ * (50.0 / 1000) dm³ = 0.050 mol.

Step 2: Determine moles of CO2 produced from the stoichiometry of the equation:
According to the equation, 2 moles of HCl produce 1 mole of CO2.
moles of CO2 = 0.050 / 2 = 0.025 mol.

Step 3: Convert moles of CO2 to volume at r.t.p.:
volume of CO2 = moles * molar volume = 0.025 mol * 24.0 dm³/mol = 0.60 dm³ (or 600 cm³).

(a) [2 marks total]
- 1 mark for describing the measurement of gas volume (or mass loss) over time.
- 1 mark for specifying the apparatus: gas syringe (or digital balance).

(b) [4 marks total]
- (i) 1 mark for stating that the rate increases.
- (ii) 1 mark for stating that higher concentration means more particles per unit volume.
- 1 mark for mentioning more frequent collisions (or greater collision frequency).
- 1 mark for linking this to more successful collisions per unit time.

(c) [4 marks total]
- 1 mark for stating that particles gain kinetic energy / move faster.
- 1 mark for mentioning more frequent collisions.
- 1 mark for stating that more particles have energy greater than or equal to the activation energy.
- 1 mark for linking this to a higher proportion of successful collisions.

(d) [3.33 marks total]
- 1 mark for calculating moles of HCl = 0.050 mol.
- 1 mark for calculating moles of CO2 = 0.025 mol (using 2:1 ratio).
- 1.33 marks for calculating volume of CO2 = 0.60 dm³ (or 600 cm³), including correct units.

(a)(i) Nitrogen is obtained from the fractional distillation of liquid air.
(a)(ii) Hydrogen is obtained by reacting methane (from natural gas) with steam, or by the catalytic cracking of hydrocarbons.

(b) The typical reaction conditions are a temperature of approximately 450 °C and a pressure of about 200 atmospheres (or 20,000 kPa).

(c) The catalyst used in the process is finely divided iron.

(d)(i) Increasing the temperature decreases the equilibrium yield of ammonia. This is because the forward reaction is exothermic (ΔH is negative). According to Le Chatelier's principle, the system will shift in the endothermic direction (to the left) to absorb the added heat, thereby decreasing the yield of ammonia.
(d)(ii) Increasing the pressure increases the equilibrium yield of ammonia. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium in the direction of fewer moles of gas. There are 4 moles of gaseous reactants on the left and only 2 moles of gaseous products on the right, so the equilibrium shifts to the right, increasing the yield.

(e) At a low temperature like 100 °C, the rate of reaction is extremely slow because the reactant molecules have very low kinetic energy. Consequently, the reaction would take too long to reach equilibrium, making the process commercially unviable.

(a) [2 marks total]
- (i) 1 mark for fractional distillation of liquid air (accept air).
- (ii) 1 mark for steam reforming of methane / natural gas / cracking of hydrocarbons.

(b) [2 marks total]
- 1 mark for temperature: 450 °C (accept range 400-500 °C).
- 1 mark for pressure: 200 atm (accept range 150-250 atm).

(c) [1 mark]
- 1 mark for iron (Fe).

(d) [6 marks total]
- (i) 1 mark for predicting yield decreases; 1 mark for identifying the forward reaction is exothermic; 1 mark for explaining that equilibrium shifts to the left to absorb heat.
- (ii) 1 mark for predicting yield increases; 1 mark for identifying fewer moles of gas on the right side (2 mol vs 4 mol); 1 mark for explaining that equilibrium shifts to the right to oppose the pressure increase.

(e) [2.33 marks total]
- 1 mark for identifying that the rate of reaction is too slow at low temperatures.
- 1.33 marks for explaining that it takes too long to reach equilibrium, which is economically inefficient.

(a)(i) Addition polymerization.
(a)(ii) The structure of poly(ethene) showing two repeat units is drawn as a chain of four carbon atoms linked by single bonds, with each carbon atom bonded to two hydrogen atoms. Open single bonds at each end of the chain indicate continuation:
H H H H
| | | |
-[C - C - C - C]-
| | | |
H H H H

(b)(i) The two functional groups are the amine group (-NH2) and the carboxylic acid group (-COOH).
(b)(ii) The amide linkage (peptide link) is: -C(=O)-N(H)-. It must show the double bond between carbon and oxygen and the single bond between nitrogen and hydrogen.
(b)(iii) It is condensation polymerization. The small molecule eliminated is water (H2O).

(c)(i) The ester linkage is: -C(=O)-O-. It must show the double bond between carbon and oxygen and the single bond between carbon and oxygen.
(c)(ii) Terylene is used for clothing, synthetic fibers, ropes, and sails.

(d)(i) Amino acids.
(d)(ii) Proteins are hydrolysed by heating with concentrated hydrochloric acid (or by using protease enzymes).

(a) [3 marks total]
- (i) 1 mark for addition polymerization.
- (ii) 2 marks for drawing -C-C-C-C- with single bonds, correct hydrogens on each carbon, and open extension bonds at both ends. Deduct 1 mark if carbon-carbon double bonds are present or if extension bonds are missing.

(b) [6 marks total]
- (i) 2 marks (1 mark for amine / amino, 1 mark for carboxylic acid / carboxyl).
- (ii) 2 marks for drawing the amide linkage: 1 mark for C=O, 1 mark for N-H and linking C-N single bond.
- (iii) 2 marks (1 mark for condensation polymerization, 1 mark for water / H2O).

(c) [2 marks total]
- (i) 1 mark for drawing the ester link -C(=O)-O-.
- (ii) 1 mark for stating clothing / textiles / synthetic fibers / sails.

(d) [2.33 marks total]
- (i) 1 mark for amino acids.
- (ii) 1.33 marks for heating with concentrated hydrochloric acid (accept heating with aqueous acid/alkali or protease enzymes).

(a) An acid is a proton (H+) donor.

(b) A strong acid completely dissociates/ionizes in aqueous solution to produce H+ ions (e.g., hydrochloric acid, HCl). A weak acid only partially dissociates/ionizes in aqueous solution (e.g., ethanoic acid, CH3COOH).

(c)(i) 0.10 mol/dm³ hydrochloric acid (strong acid) has a pH of approximately 1. 0.10 mol/dm³ ethanoic acid (weak acid) has a pH of approximately 3 to 5.
(c)(ii) Experiment: Add an equal mass/length of magnesium ribbon (or calcium carbonate) to equal volumes of both 0.10 mol/dm³ hydrochloric acid and 0.10 mol/dm³ ethanoic acid in separate test tubes.
Observations: Hydrochloric acid will show rapid effervescence/bubbling, and the magnesium will dissolve quickly. Ethanoic acid will show slow/gentle effervescence, and the magnesium will dissolve much more slowly.
Conclusion: The faster rate of reaction in hydrochloric acid indicates a higher concentration of H+(aq) ions, showing that it has dissociated to a greater extent and is therefore the stronger acid.

(d) The neutralization reaction involves hydrogen ions reacting with hydroxide ions to form water:
H+(aq) + OH-(aq) -> H2O(l)

(a) [1 mark]
- 1 mark for stating that an acid is a proton (H+) donor.

(b) [4 marks total]
- 1 mark for stating that a strong acid completely dissociates/ionizes in water.
- 1 mark for a correct example of a strong acid (e.g. hydrochloric acid / sulfuric acid / nitric acid).
- 1 mark for stating that a weak acid partially dissociates/ionizes in water.
- 1 mark for a correct example of a weak acid (e.g. ethanoic acid / citric acid / carbonic acid).

(c) [6 marks total]
- (i) 2 marks: 1 mark for hydrochloric acid pH = 1 (accept 0 to 2); 1 mark for ethanoic acid pH = 3 to 5 (accept 3 to 5).
- (ii) 4 marks:
- 1 mark for describing a valid reaction (adding magnesium/zinc or calcium carbonate).
- 1 mark for stating the observation with hydrochloric acid (rapid bubbling / fast rate).
- 1 mark for stating the observation with ethanoic acid (slow bubbling / slow rate).
- 1 mark for linking the difference in rate to the concentration of H+ ions in solution.

(d) [2.33 marks total]
- 1 mark for correct formulas of reactants and products: H+ + OH- -> H2O.
- 1 mark for correct state symbols: (aq) for ions, (l) for water.
- 0.33 mark for ensuring charges are balanced.

Part (a): The mass of carbon dioxide lost is calculated by subtracting each mass reading from the starting mass (124.50 g): At 0 s: 124.50 - 124.50 = 0.00 g; At 30 s: 124.50 - 124.15 = 0.35 g; At 60 s: 124.50 - 123.90 = 0.60 g; At 90 s: 124.50 - 123.75 = 0.75 g; At 120 s: 124.50 - 123.65 = 0.85 g; At 150 s: 124.50 - 123.60 = 0.90 g; At 180 s: 124.50 - 123.60 = 0.90 g. Part (b): The cotton wool plug allows carbon dioxide gas to escape but prevents liquid droplets (acid spray) from spitting out, which would cause an incorrect, artificially high calculated mass loss. Part (c): The reaction was completed at 150 seconds because the mass of the flask stops decreasing and remains constant at 123.60 g from this point onwards. Part (d): Experimental Plan: 1. Measure a fixed volume (e.g., 25 cm3) of aqueous hydrogen peroxide of a known concentration into a conical flask. 2. Weigh out a fixed mass (e.g., 0.5 g) of MnO2 powder. 3. Connect the flask to a gas syringe. 4. Add the MnO2 powder to the flask, immediately replace the stopper, and start a stopwatch. 5. Measure the volume of oxygen gas produced in a set time (e.g., 2 minutes) or measure the time taken to collect a specific volume (e.g., 50 cm3) of gas. 6. Repeat the entire procedure using the same volume and concentration of hydrogen peroxide, at the same temperature, and with the exact same mass (0.5 g) of CuO powder and then Fe2O3 powder. 7. Compare the rates: the catalyst that produces the largest volume of gas in the set time (or takes the shortest time to collect the fixed volume) is the most effective catalyst.

Part (a) [2 marks total]: 1 mark for any three correct values, 2 marks for all six correct values (0.35, 0.60, 0.75, 0.85, 0.90, 0.90). Part (b) [1 mark]: State that cotton wool prevents loss of liquid/spray/droplets (Reject: 'prevents gas escaping'). Part (c) [2 marks]: 1 mark for identifying 150 seconds. 1 mark for explaining that the mass becomes constant / stops decreasing. Part (d) [6.33 marks]: 1 mark for keeping the volume and concentration of hydrogen peroxide constant. 1 mark for keeping the mass and particle size (powdered state) of the catalysts constant. 1 mark for measuring the volume of gas collected (using a gas syringe or over water) OR measuring the loss in mass over time. 1 mark for recording the time using a stopwatch. 1 mark for specifying that the temperature must be kept constant. 1.33 marks for explaining the criteria for concluding which catalyst is best (e.g., the one producing the fastest rate of gas release / largest volume in a set time).

Part (a): Pencil lead (graphite) is insoluble in chromatography solvents, so it will not dissolve or run up the paper and interfere with the results. Pen ink is soluble and would separate into colours. Part (b): The Rf value is calculated using the formula: Rf = (distance moved by spot) / (distance moved by solvent front). For Spot X: Rf = 5.2 / 8.0 = 0.65. For Spot Y: Rf = 2.4 / 8.0 = 0.30. Part (c): Water is a highly polar solvent; some organic dyes are non-polar and insoluble in water, meaning they would remain on the baseline and not separate. A suitable alternative organic solvent is ethanol or propanone. Part (d): Extraction and separation procedure: 1. Place fresh spinach leaves into a mortar, add a small volume of propanone (or ethanol) to act as a solvent, and grind thoroughly using a pestle to break open the cells and dissolve the pigments. 2. Filter the mixture through filter paper in a funnel to remove solid leaf debris and collect the deep green filtrate in a small beaker. 3. Draw a pencil baseline about 2 cm from the bottom of a strip of chromatography paper. 4. Use a capillary tube or fine dropper to put a small spot of the green filtrate on the pencil line. Allow it to dry, and repeat several times to build up a concentrated, dark spot. 5. Pour a small volume of a suitable solvent (such as propanone or cyclohexane) into a chromatography tank or beaker, ensuring the liquid level is below the pencil line. 6. Suspend the paper in the beaker, cover it with a lid to prevent evaporation, and allow the solvent to ascend. 7. Remove the paper before the solvent reaches the top, immediately mark the solvent front with a pencil, and allow the chromatogram to dry to observe the separated yellow-green and blue-green pigment spots.

Part (a) [1 mark]: State that pencil is insoluble / does not run / does not contaminate the chromatogram. Part (b) [3 marks]: 1 mark for the formula Rf = (distance moved by substance) / (distance moved by solvent). 1 mark for Spot X = 0.65 (must be exactly 0.65). 1 mark for Spot Y = 0.30 (must be exactly 0.30). Part (c) [3 marks]: 2 marks for explaining that some dyes are insoluble in water and will not move/separate. 1 mark for naming ethanol, propanone, or ethyl ethanoate as an alternative solvent. Part (d) [6.33 marks]: 1 mark for crushing leaves using a mortar and pestle. 1 mark for adding an organic solvent (e.g. propanone or ethanol) to extract pigments. 1 mark for filtering the mixture to obtain a clear liquid filtrate. 1 mark for applying a concentrated spot of filtrate onto a pencil line on chromatography paper. 1 mark for placing the paper in a container with a solvent level below the spot. 1.33 marks for covering the container, letting the solvent run, and marking the solvent front.

Part (a): Test 1: A lilac flame indicates the presence of potassium ions, K+. Test 2: Addition of nitric acid followed by silver nitrate yields a cream precipitate, which confirms the presence of bromide ions, Br-. Test 3: The chemical formula of Solid E is KBr. Part (b): (i) Adding sodium hydroxide to a solution containing iron(III) ions produces a red-brown precipitate of iron(III) hydroxide, which is insoluble in excess sodium hydroxide. (ii) Adding nitric acid and silver nitrate to a solution containing chloride ions (from iron(III) chloride) produces a white precipitate of silver chloride. Part (c): Titration Procedure: 1. Pipette exactly 25.0 cm3 of the 0.100 mol/dm3 sodium hydroxide solution into a clean conical flask. 2. Add 2 to 3 drops of phenolphthalein indicator to the flask (the solution turns pink). 3. Fill a clean burette with the dilute hydrochloric acid (Solution G), ensuring the jet space is filled, and record the initial reading. 4. Run Solution G slowly into the flask while swirling continuously until the pink colour just disappears (turns colourless). This is the end-point. 5. Record the final burette reading to find the titre volume. 6. Repeat the titration until at least two concordant results (within 0.1 cm3 of each other) are obtained, and calculate the average titre volume, V cm3. Calculation: Moles of NaOH used = (25.0 / 1000) * 0.100 = 0.0025 moles. Since HCl and NaOH react in a 1:1 ratio (HCl + NaOH -> NaCl + H2O), the moles of HCl in V cm3 of Solution G = 0.0025 moles. Molarity of HCl = (0.0025 / V) * 1000 = 2.5 / V mol/dm3. The relative formula mass of HCl is 1 + 35.5 = 36.5 g/mol. Therefore, the concentration in g/dm3 = (2.5 / V) * 36.5 = 91.25 / V g/dm3.

Part (a) [4 marks total]: Test 1: 1 mark for potassium / K+. Test 2: 1 mark for bromide / Br- (1 additional mark for correct ionic reasoning). Test 3: 1 mark for KBr. Part (b) [3 marks total]: (i) 1 mark for red-brown precipitate, 1 mark for stating it is insoluble in excess. (ii) 1 mark for white precipitate. Part (c) [6.33 marks]: 1 mark for pipetting a fixed volume (e.g. 25.0 cm3) of NaOH into a conical flask and adding phenolphthalein. 1 mark for filling a burette with Solution G and recording the start volume. 1 mark for titrating until the colour changes from pink to colourless. 1 mark for repeating to obtain concordant titres and calculating the average titre, V. 1 mark for showing the correct mole ratio calculation: moles of HCl = moles of NaOH = 0.0025. 1.33 marks for demonstrating the correct final calculation steps to convert to g/dm3 (Molarity of HCl = 2.5 / V mol/dm3, then multiplying by 36.5 to get 91.25 / V g/dm3).

Part (a): Conical flask (for the reaction mixture) and a gas syringe or an inverted measuring cylinder filled with water to collect and measure the gas. Part (b): The temperature of the acid, or the surface area/length of the magnesium ribbon, or the total volume of the acid mixture must be kept constant. Part (c): Test with a lighted splint. Hydrogen gas ignites with a characteristic squeaky 'pop' sound. Part (d): To study the effect of concentration: 1. Prepare different concentrations of HCl (e.g., 2.0, 1.5, 1.0, 0.5 mol/dm³) by mixing appropriate ratios of 2.0 mol/dm³ acid and distilled water (e.g., 50 cm³ acid + 0 cm³ water for 2.0 mol/dm³, 25 cm³ acid + 25 cm³ water for 1.0 mol/dm³). 2. Pour a constant volume (e.g., 50 cm³) of the first acid concentration into a conical flask. 3. Add a fixed mass or length (e.g., 2 cm) of magnesium ribbon to the flask. 4. Immediately seal with a stopper connected to a gas syringe and start a stopwatch. 5. Record the volume of gas collected at fixed intervals (e.g., every 10 seconds) until the reaction stops. 6. Repeat steps 2-5 for each diluted acid concentration. 7. Compare the initial rates of reaction (gradient of the curves) or the time taken to collect a fixed volume of gas.

Part (a) [2 marks]: 1 mark for conical flask / reaction flask. 1 mark for gas syringe / inverted measuring cylinder over water. Part (b) [1 mark]: 1 mark for identifying a valid controlled variable (temperature / surface area of magnesium / volume of acid). Part (c) [2 marks]: 1 mark for lighted splint. 1 mark for 'pop' sound / squeaky pop. (Reject glowing splint). Part (d) [5 marks]: 1 mark for explaining a method of dilution (using different volumes of acid and water to vary concentration). 1 mark for using a constant volume of acid and a constant mass/length of magnesium. 1 mark for measuring the volume of gas. 1 mark for measuring time / using a stopwatch. 1 mark for repeating the experiment with different concentrations and comparing the rates.

Part (a): Iron(II) ions react with hydroxide ions to form insoluble iron(II) hydroxide. The observation is a green precipitate that remains insoluble when excess sodium hydroxide is added. Part (b): Heating ammonium ions with sodium hydroxide releases ammonia gas (\(NH_3\)), which is alkaline. The observation is effervescence and the damp red litmus paper turning blue. Part (c): Sulfate ions react with barium ions to form insoluble barium sulfate. Dilute nitric acid is added first to prevent other anions (like carbonate) from precipitating. The observation is a white precipitate. Part (d)(i): Copper(II) ions react with ammonia to form a light blue precipitate of copper(II) hydroxide. In excess ammonia, this precipitate dissolves to form a deep blue soluble complex. Part (d)(ii): The standard test for chloride ions involves acidifying the solution with dilute nitric acid to remove carbonates, then adding aqueous silver nitrate. A white precipitate of silver chloride confirms chloride.

Part (a) [2 marks]: 1 mark for green precipitate. 1 mark for insoluble in excess. Part (b) [2 marks]: 1 mark for gas/effervescence (or pungent smell). 1 mark for damp red litmus turning blue. Part (c) [2 marks]: 1 mark for white precipitate. 1 mark for indicating the precipitate does not dissolve in acid (or stating white precipitate remains). Part (d)(i) [2 marks]: 1 mark for light blue precipitate. 1 mark for dissolves in excess to form a deep blue solution. Part (d)(ii) [2 marks]: 1 mark for adding dilute nitric acid followed by aqueous silver nitrate. 1 mark for white precipitate.

Part (a): Pencil lead consists of graphite, which is insoluble in common solvents and will remain on the baseline. Ink is soluble and contains dyes that would dissolve and separate, interfering with the results. Part (b): The solvent level must be below the baseline. If it is above, the dye samples will dissolve directly into the reservoir of solvent rather than traveling up the paper. Part (c): The retention factor (\(R_f\)) is the ratio of the distance moved by the solute to the distance moved by the solvent front. Formula: \(R_f = \frac{\text{distance traveled by substance}}{\text{distance traveled by solvent front}}\). Part (d): Yellow \(R_f = 3.0 \text{ cm} / 7.5 \text{ cm} = 0.40\) (or 0.4). Blue \(R_f = 6.0 \text{ cm} / 7.5 \text{ cm} = 0.80\) (or 0.8). Part (e): Colorless substances (such as amino acids or sugars) cannot be seen directly. They are visualized by spraying the chromatogram with a locating agent (like ninhydrin) which reacts with the substances to form colored products, or by placing the paper under ultraviolet (UV) light where the spots fluoresce.

Part (a) [2 marks]: 1 mark for stating pencil/graphite is insoluble (or does not dissolve/run). 1 mark for stating ink is soluble and would separate/run (interfering with the chromatogram). Part (b) [2 marks]: 1 mark for stating solvent level must be below the baseline. 1 mark for explaining that spots would wash off/dissolve into the solvent. Part (c) [1 mark]: 1 mark for correct formula: distance moved by spot / distance moved by solvent front. Part (d) [2 marks]: 1 mark for correct yellow Rf of 0.4 / 0.40. 1 mark for correct blue Rf of 0.8 / 0.80. Part (e) [3 marks]: 1 mark for using a locating agent. 1 mark for spraying/applying the agent. 1 mark for heating/drying or viewing under UV light to make the spots visible.

Part (a): A volumetric pipette is used to deliver a highly accurate and fixed volume of 25.0 cm³ of sodium hydroxide. A burette is used to add varying amounts of the acid precisely until the end-point is reached. Part (b): In alkaline solution, methyl orange is yellow. As acid is added and neutralization occurs, the solution becomes neutral/slightly acidic, turning orange at the end-point (further acid would turn it red). Part (c): First calculate the three titre volumes: Titre 1 = 22.80 - 0.00 = 22.80 cm³. Titre 2 = 23.15 - 1.00 = 22.15 cm³. Titre 3 = 22.40 - 0.20 = 22.20 cm³. Compare the values: 22.15 cm³ and 22.20 cm³ are within 0.20 cm³ of each other, making them concordant. Titre 1 is not concordant. Average = \((22.15 + 22.20) / 2 = 22.175 \text{ cm}^3\) or \(22.18 \text{ cm}^3\). Part (d): For safety: use a funnel and fill the burette below eye level (e.g., clamp it lower down or stand on a stool) to prevent acid spilling onto the face or eyes. For accuracy: ensure there are no air bubbles trapped in the space below the tap (jet), and remove the funnel before commencing the titration so extra drops do not fall into the burette during the experiment.

Part (a) [2 marks]: 1 mark for pipette / volumetric pipette. 1 mark for burette. Part (b) [2 marks]: 1 mark for yellow (initial color). 1 mark for orange / peach / pink (end-point color). (Accept: yellow to orange. Reject: yellow to red, as red indicates excess acid). Part (c) [3 marks]: 1 mark for correctly calculating all three titres (22.80, 22.15, 22.20 cm³). 1 mark for identifying 22.15 cm³ and 22.20 cm³ as the concordant results. 1 mark for calculating the correct average: 22.18 (or 22.175) cm³. Part (d) [2 marks]: 1 mark for a safety precaution with explanation (e.g., fill below eye level to protect eyes / use a funnel to avoid spills). 1 mark for an accuracy precaution with explanation (e.g., remove funnel to prevent drips / run solution through the tap to remove air bubbles).