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The initial rate of reaction is determined by the concentration of the acid. Flask Q contains 2.0 mol/dm\(^{3}\) HCl, which is a higher concentration than Flask P (1.0 mol/dm\(^{3}\) HCl). Therefore, the initial rate of reaction in Flask Q is faster.

The final volume of carbon dioxide produced depends on the number of moles of the limiting reactant, which is hydrochloric acid since calcium carbonate is in excess.
- Moles of HCl in P = 0.050 dm\(^{3}\) \(\times\) 1.0 mol/dm\(^{3}\) = 0.050 mol.
- Moles of HCl in Q = 0.025 dm\(^{3}\) \(\times\) 2.0 mol/dm\(^{3}\) = 0.050 mol.

Since both flasks contain the same number of moles of HCl, they will produce the exact same final volume of carbon dioxide gas.

1 mark: B is the correct option. A, C, and D are incorrect based on rate principles and stoichiometry calculations.

A catalyst provides an alternative reaction pathway with a lower activation energy. This means a larger proportion of colliding particles have energy greater than or equal to the activation energy, resulting in a higher rate of successful collisions. Catalysts do not increase the total collision frequency (which depends on temperature, concentration, and surface area) and are chemically unchanged at the end of the reaction.

1 mark: C is the correct option. A is incorrect because a catalyst lowers activation energy. B is incorrect because catalysts do not increase overall collision frequency. D is incorrect because catalysts are not chemically changed.

The polymer contains amide linkages (-NH-CO-). It is a polyamide, which is formed by condensation polymerisation. The monomers required to make a polyamide are a diamine (containing -NH\(_{2}\) groups) and a dicarboxylic acid (containing -COOH groups), with the elimination of water molecules.

1 mark: D is the correct option. A and B are incorrect as this is a condensation polymer. C is incorrect as a diol and dicarboxylic acid form a polyester, not a polyamide.

The monomer of poly(chloroethene) is chloroethene, which has the molecular formula C\(_{2}\)H\(_{3}\)Cl. Burning PVC (combustion) releases highly toxic and acidic hydrogen chloride (HCl) gas, which is a major environmental hazard.

1 mark: B is the correct option. A and D have incorrect molecular formulas for the monomer. C has the correct formula but eutrophication is caused by nitrate/phosphate run-off, not polymer combustion.

In the electrolysis of concentrated aqueous sodium chloride (brine):
- At the anode (+), chloride ions (Cl\(^{-}\)) are discharged in preference to hydroxide ions (OH\(^{-}\)) because of their high concentration, forming chlorine gas (Cl\(_{2}\)).
- At the cathode (-), hydrogen ions (H\(^{+}\)) are discharged in preference to sodium ions (Na\(^{+}\)) because hydrogen is lower in the reactivity series, forming hydrogen gas (H\(_{2}\)).
- The remaining solution contains sodium (Na\(^{+}\)) and hydroxide (OH\(^{-}\)) ions, forming sodium hydroxide, which is highly alkaline. Therefore, the pH increases.

1 mark: B is the correct option. A, C, and D are incorrect based on selective discharge rules and the chemical nature of the remaining electrolyte.

To electroplate an object:
1. The object to be plated (the steel spoon) must be placed at the cathode (negative electrode) where metal ions are reduced and deposited.
2. The metal used for plating (silver) must be the anode (positive electrode) so that it dissolves to replenish metal ions in the solution.
3. The electrolyte must be a soluble salt containing the ions of the plating metal (aqueous silver nitrate).

1 mark: B is correct. A has the anode and cathode reversed. C and D do not use the correct plating metal or electrolyte containing silver ions.

Barium sulfate is an insoluble salt. The standard method to prepare an insoluble salt is by precipitation, which requires mixing two soluble starting materials (solutions).
- Aqueous barium chloride is soluble.
- Aqueous sodium sulfate is soluble.

Mixing them yields a precipitate of barium sulfate and a soluble salt, sodium chloride. The precipitate can be filtered, washed with distilled water, and dried.

Option B is incorrect because solid barium carbonate reacts with sulfuric acid to form an insoluble coating of barium sulfate on the unreacted carbonate, stopping the reaction. Option C involves an insoluble reactant (calcium sulfate). Option D is a reaction starting with a solid base, which is not suitable for producing a pure insoluble salt efficiently.

1 mark: A is correct. B, C, and D are incorrect because they involve insoluble reactants or are unsuitable methods for preparing a pure sample of an insoluble salt.

To prepare a soluble salt from an acid and an insoluble base:
1. Add excess insoluble base (copper(II) oxide) to warm dilute sulfuric acid to ensure all the acid is neutralized (Step Z).
2. Filter the mixture to remove the excess, unreacted copper(II) oxide (Step W).
3. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until a saturated solution is formed (the crystallization point), then allow it to cool slowly so crystals form (Step Y).
4. Filter the crystals and dry them between sheets of filter paper (Step X).

Therefore, the correct sequence is Z \(\rightarrow\) W \(\rightarrow\) Y \(\rightarrow\) X.

1 mark: A is the correct option. All other sequences represent incorrect ordering of the experimental steps.

Using a fine powder instead of chips increases the surface area of the solid reactant. This increases the collision frequency between reactant particles, resulting in a faster initial rate of reaction. Since the mass of calcium carbonate is the same and the acid remains in the same volume and concentration, the final volume of carbon dioxide gas produced is identical. The activation energy is not affected because no catalyst is used.

1 mark for correct option C. Reject other options because changing surface area does not change the total yield (A) or the activation energy (D), and it increases the rate of reaction rather than slowing it down (B).

An increase in temperature increases the average kinetic energy of the particles (Statement 2), causing them to move faster. As a result, a much greater fraction of the colliding particles possess energy equal to or greater than the activation energy (Statement 3), leading to more successful collisions per second. The activation energy itself is a constant property of the reaction pathway and does not decrease when temperature increases (Statement 1 is incorrect; only a catalyst can provide an alternative pathway with a lower activation energy).

1 mark for option C. Statement 1 is incorrect as temperature does not alter activation energy, making options A and B incorrect. Statement 2 and 3 are correct explanations of collision theory at higher temperatures.

The repeating unit of this addition polymer is \(-CH_2-CH(CH_3)-\), which has a two-carbon backbone with a methyl group attached to one of the carbons. This corresponds to the monomer propene, \(CH_2=CHCH_3\).

1 mark for option B. Ethane is an alkane and cannot polymerise (A). But-1-ene (C) and but-2-ene (D) would result in a polymer with different repeating structures.

Nylon is a synthetic polyamide containing amide linkages, \(-CONH-\). Terylene is a synthetic polyester containing ester linkages, \(-COO-\). Both are formed by condensation polymerisation.

1 mark for option B. Option A swaps the polymer types. Options C and D are incorrect because both nylon and Terylene are condensation polymers, not addition polymers.

During the electrolysis of concentrated aqueous sodium chloride: At the anode (+), chloride ions (\(Cl^-\)) are discharged preferentially due to their high concentration, forming chlorine gas. At the cathode (-), hydrogen ions (\(H^+\)) are discharged preferentially because hydrogen is less reactive than sodium, forming hydrogen gas. The remaining sodium ions (\(Na^+\)) and hydroxide ions (\(OH^-\)) leave sodium hydroxide in solution, making the electrolyte alkaline.

1 mark for option C. Options A and D are incorrect because sodium ions are not reduced to sodium metal in aqueous solutions. Option B is incorrect because the remaining solution contains sodium hydroxide, which is alkaline, not neutral.

When active copper electrodes are used, copper atoms at the anode are oxidised to copper(II) ions, so the anode dissolves and its mass decreases (A is incorrect). At the cathode, copper(II) ions are reduced and deposited as copper metal, increasing the mass of the cathode (D is correct). Since the rate of copper dissolving at the anode equals the rate of copper depositing at the cathode, the concentration of copper(II) ions in the electrolyte remains constant, and the blue colour does not fade (B is incorrect). No oxygen gas is produced at the anode because the copper anode itself reacts (C is incorrect).

1 mark for option D. Options A, B, and C describe observations only associated with using inert electrodes (like carbon or platinum) in the same electrolyte.

An insoluble salt like barium sulfate is prepared by a precipitation reaction using two soluble starting materials. Barium chloride is a soluble salt, and dilute sulfuric acid is a soluble acid. Mixing them yields a precipitate of barium sulfate and hydrochloric acid. Barium carbonate (A) is insoluble, preventing a complete reaction. Barium hydroxide and copper(II) sulfate (C) would produce a mixture of two precipitates (barium sulfate and copper(II) hydroxide), which are difficult to separate. Barium oxide (D) reacts with hydrochloric acid to form soluble barium chloride, not barium sulfate.

1 mark for option B. Accept only soluble reactants that produce a single insoluble product.

According to Le Chatelier's principle: 1. Because the forward reaction is exothermic, a lower temperature shifts the equilibrium to the right, increasing the yield of ammonia. 2. Because there are fewer moles of gas on the product side (2 moles of \(NH_3\)) compared to the reactant side (4 moles of \(N_2\) and \(H_2\)), a higher pressure shifts the equilibrium to the right. Therefore, a low temperature and high pressure give the highest equilibrium yield of ammonia.

1 mark for option C. While low temperature and high pressure yield the most ammonia theoretically, high temperatures are used in practice as a compromise for reaction rate, but the question specifies maximum equilibrium yield.

The final volume of hydrogen gas depends on the number of moles of the limiting reactant, which is sulfuric acid (since zinc is in excess). Initial moles of \(\text{H}_2\text{SO}_4 = 0.100\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.1\text{ mol}\). Changing to \(100\text{ cm}^3\) of \(2.0\text{ mol/dm}^3\) sulfuric acid increases the moles of \(\text{H}_2\text{SO}_4\) to \(0.2\text{ mol}\), thereby increasing the final volume of gas. Since the concentration of the acid is higher, the initial rate of reaction is also higher.

Award 1 mark for B. Reject A because using powdered zinc does not change the amount of limiting reactant. Reject C because a lower temperature decreases the rate of reaction. Reject D because a catalyst does not increase the yield of product.

According to collision theory, increasing the temperature increases the average kinetic energy of the particles. This means a larger fraction of colliding particles have energy equal to or greater than the activation energy, leading to a higher rate of reaction. The activation energy remains constant.

Award 1 mark for A. Reject B because temperature does not affect activation energy. Reject C because the frequency of successful collisions decreases at lower temperatures. Reject D because particles move slower at lower temperatures.

The polymer contains ester linkages, which makes it a polyester. Polyesters are synthesized from a diol monomer and a dicarboxylic acid monomer. Thus, the diol \(\text{HO} - \square - \text{OH}\) is one of the monomers used.

Award 1 mark for B. Reject A and D because the linkages are ester bonds, not amide bonds. Reject C because amines are not used to make polyesters.

The repeating unit is \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\), which contains three carbon atoms. It is formed by the addition polymerization of propene, \(\text{CH}_2=\text{CH}-\text{CH}_3\).

Award 1 mark for D. Reject A and C because alkanes cannot undergo addition polymerization. Reject B because butene forms a different polymer chain with four-carbon segments.

At the cathode, hydrogen ions are discharged preferentially over sodium ions, producing hydrogen gas. At the anode, chloride ions are discharged preferentially over hydroxide ions due to their high concentration, producing chlorine gas. Hydroxide ions and sodium ions remain in the solution, making it alkaline and increasing the pH.

Award 1 mark for A. Reject B because oxygen is not produced when chloride is concentrated. Reject C and D because sodium is not discharged at the cathode in aqueous solution.

In copper purification, the cathode is pure copper and the anode is impure copper. The rate of copper dissolution at the anode equals the rate of copper deposition at the cathode, keeping the concentration of copper(II) ions in the electrolyte constant. Hence, the blue color remains unchanged.

Award 1 mark for C. Reject A because the blue color does not fade. Reject B and D because the impure copper must be at the anode and the pure copper at the cathode.

Barium sulfate is an insoluble salt. The standard preparation method for an insoluble salt is precipitation, which requires mixing two soluble starting materials. Aqueous barium chloride and dilute sulfuric acid are both soluble. Mixing them produces a precipitate of barium sulfate, which is then filtered, washed with distilled water, and dried.

Award 1 mark for B. Reject A and C because reacting an insoluble carbonate or metal with sulfuric acid is prevented by the formation of an insoluble barium sulfate coating. Reject D because titration is used to prepare soluble salts.

The oxidation state of carbon in carbon monoxide is +2, and in carbon dioxide it is +4. Because the oxidation state increases, carbon monoxide is oxidized. Iron(III) oxide is reduced because the oxidation state of iron decreases from +3 in \(\text{Fe}_2\text{O}_3\) to 0 in \(\text{Fe}\).

Award 1 mark for A. Reject B because the loss of oxygen is reduction. Reject C because carbon monoxide acts as the reducing agent. Reject D because the oxidation state of iron decreases from +3 to 0.

The initial rate of reaction depends on the concentration of the acid. A higher concentration (\(2.0\text{ mol/dm}^3\) compared to \(1.0\text{ mol/dm}^3\)) will increase the initial rate of reaction.

The volume of hydrogen gas collected depends on the number of moles of the limiting reactant, which is sulfuric acid (since zinc is in excess).

- Original moles of \(\text{H}_2\text{SO}_4 = 0.025\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.025\text{ mol}\).
- Moles of \(\text{H}_2\text{SO}_4\) in option C = \(0.010\text{ dm}^3 \times 2.0\text{ mol/dm}^3 = 0.020\text{ mol}\).

Since \(0.020\text{ mol} < 0.025\text{ mol}\), a smaller volume of hydrogen gas is collected. Option C satisfies both conditions.

Correct answer: C

- 1 mark for identifying the correct option (C).

The polymer shown is an addition polymer, recognizable by its continuous carbon-carbon single bond backbone with no heteroatoms (like O or N) in the main chain. The repeating unit is \(-\text{CH}_2-\text{CH}(\text{CN})-\), which is formed from the monomer acrylonitrile (propenenitrile), \(\text{CH}_2=\text{CH}(\text{CN})\). This monomer contains 3 carbon atoms, 3 hydrogen atoms, and 1 nitrogen atom (\(\text{C}_3\text{H}_3\text{N}\)), and contains a carbon-carbon double bond.

Correct answer: B

- 1 mark for identifying the correct option (B).

At the positive electrode (anode), hydroxide ions (\(\text{OH}^-\)) from the water are selectively discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)) to produce oxygen gas: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\). This is observed as bubbles of a colourless gas.

At the negative electrode (cathode), copper(II) ions (\(\text{Cu}^{2+}\)) are selectively discharged in preference to hydrogen ions (\(\text{H}^+\)) because copper is lower in the reactivity series: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\). This is observed as a pink/brown solid deposit.

As \(\text{Cu}^{2+}\) and \(\text{OH}^-\) ions are discharged and removed, \(\text{H}^+\) and \(\text{SO}_4^{2-}\) ions remain in the solution, making it acidic and causing the pH to decrease.

Correct answer: A

- 1 mark for identifying the correct option (A).

Barium sulfate is an insoluble salt. The correct preparation method for insoluble salts is precipitation. This involves mixing two soluble salt solutions (such as aqueous barium chloride and aqueous sodium sulfate) to form the precipitate. The precipitate is then filtered off as the residue, washed with distilled water to remove any soluble impurities (such as sodium chloride), and dried.

Correct answer: A

- 1 mark for identifying the correct option (A).

At higher temperatures, particles have more kinetic energy (Statement 1) and move faster, leading to a higher frequency of collisions (Statement 4). Additionally, because the particles are more energetic, a larger fraction of them have energy greater than or equal to the activation energy, meaning a greater proportion of collisions are successful (Statement 3). Temperature does not lower the activation energy; only a catalyst can do this (Statement 2 is incorrect).

Correct answer: B

- 1 mark for identifying the correct option (B).

The polymer contains the ester linkage \(-\text{O}-\text{CO}-\), which means it is a polyester (specifically Terylene). It is formed by condensation polymerisation of a diol and a dicarboxylic acid, which eliminates a water molecule (\(\text{H}_2\text{O}\)) during the formation of each ester linkage.

Correct answer: B

- 1 mark for identifying the correct option (B).

During electroplating:
1. The object to be plated (the steel key) must be the cathode (negative electrode) so that positive metal ions in solution are attracted to it and reduced to metal atoms.
2. The anode (positive electrode) must be made of the metal being plated (copper sheet) so that it dissolves to replenish the metal ions in the electrolyte: \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\).
3. The electrolyte must be an aqueous solution containing the ions of the metal being plated (aqueous copper(II) sulfate).

Correct answer: B

- 1 mark for identifying the correct option (B).

Potassium is a highly reactive Group I alkali metal. Adding potassium metal directly to dilute hydrochloric acid is extremely violent and explosive, making it highly dangerous and unsafe as a laboratory method. Soluble potassium salts (like potassium chloride) are instead prepared using acid-base titration involving potassium hydroxide and hydrochloric acid.

Correct answer: B

- 1 mark for identifying the correct option (B).

In Experiment 2, magnesium powder is used, which has a larger surface area than the large pieces in Experiment 1. Also, the concentration of hydrochloric acid is higher in Experiment 2 (\( 2.0\text{ mol/dm}^3 \) compared to \( 1.0\text{ mol/dm}^3 \)). Both of these changes increase the frequency of effective collisions, so the initial rate of reaction is faster in Experiment 2.

The total volume of hydrogen gas is determined by the number of moles of the limiting reactant, which is the acid (since magnesium is in excess in both experiments):
- Moles of \( \text{HCl} \) in Experiment 1 = \( 0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol} \)
- Moles of \( \text{HCl} \) in Experiment 2 = \( 0.025\text{ dm}^3 \times 2.0\text{ mol/dm}^3 = 0.050\text{ mol} \)

Since the number of moles of acid is identical in both cases, the total volume of hydrogen gas produced remains the same.

1 mark for selecting B.
- Reject A: The total volume is determined by moles of the limiting reactant, which is the same for both, not greater.
- Reject C: Powder and higher acid concentration increase the rate, they do not decrease it.
- Reject D: The initial rate is faster, not the same, and the total volume is the same, not smaller.

The repeating unit contains the ester linkage, \( \text{–O–CO–} \). This identifies the polymer as a polyester. Polyesters are formed by condensation polymerization of two monomers: a diol (which contains two alcohol groups, \( \text{–OH} \)) and a dicarboxylic acid (which contains two carboxylic acid groups, \( \text{–COOH} \)), eliminating a molecule of water for each linkage formed.

1 mark for selecting C.
- Reject A: This is a condensation polymer, not an addition polymer.
- Reject B: It contains ester linkages, not amide linkages (like nylon).
- Reject D: Acid hydrolysis of this polymer yields two different monomers (a diol and a dicarboxylic acid), and neither monomer contains a carbon-carbon double bond.

During the electrolysis of concentrated aqueous sodium chloride (brine):
- At the cathode, hydrogen ions (\( \text{H}^+ \)) are discharged in preference to sodium ions (\( \text{Na}^+ \)) because hydrogen is lower in the reactivity series, forming hydrogen gas (\( \text{H}_2 \)).
- At the anode, chloride ions (\( \text{Cl}^- \)) are in high concentration and are discharged in preference to hydroxide ions (\( \text{OH}^- \)), forming chlorine gas (\( \text{Cl}_2 \)).
- As \( \text{H}^+ \) and \( \text{Cl}^- \) are selectively discharged, the sodium ions (\( \text{Na}^+ \)) and hydroxide ions (\( \text{OH}^- \)) remain in solution, forming an alkaline solution of sodium hydroxide (\( \text{NaOH} \)). Thus, the pH of the solution increases.

1 mark for selecting C.
- Reject A, D: Sodium is not discharged at the cathode in aqueous solution.
- Reject B: Chlorine is discharged at the anode instead of oxygen due to its high concentration.

Barium sulfate (\( \text{BaSO}_4 \)) is an insoluble salt. Insoluble salts are prepared using precipitation, which requires mixing two soluble starting salts. Barium chloride (\( \text{BaCl}_2 \)) and sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) are both soluble in water. When mixed, they react to produce insoluble barium sulfate and soluble sodium chloride. The precipitate is separated from the reaction mixture by filtration, washed with distilled water to remove soluble impurities, and dried.

1 mark for selecting B.
- Reject A: Barium metal is too reactive, and the insoluble salt formed would coat the metal and stop the reaction.
- Reject C: Titration is used to prepare soluble salts from soluble starting materials.
- Reject D: Insoluble barium carbonate reacts poorly with sulfuric acid because the insoluble barium sulfate coats the carbonate; also, because barium sulfate is insoluble, it would not be present in the filtrate.

Increasing the temperature increases the kinetic energy of the particles, which makes them move faster. This increases the collision frequency slightly. However, the most significant effect is that a much larger proportion of the colliding particles now possess energy equal to or greater than the activation energy of the reaction. This greatly increases the frequency of successful collisions, accelerating the reaction rate.

1 mark for selecting B.
- Reject A: The activation energy is a constant and is only lowered by adding a catalyst.
- Reject C: Temperature changes do not affect the concentration of the reactants.
- Reject D: Changing the reaction pathway is a property of a catalyst, not temperature.

Let's evaluate each statement:
- Statement 1 is correct: The structure shows different amino acid residues linked by amide (or peptide) linkages (\( \text{–CO–NH–} \)), which is the characteristic structure of a protein.
- Statement 2 is incorrect: It contains amide linkages, not ester linkages (which are \( \text{–CO–O–} \)).
- Statement 3 is correct: Acid or enzymatic hydrolysis breaks the amide linkages in a protein, converting it back to its constituent amino acid monomers.
- Statement 4 is incorrect: It is formed by condensation polymerization (accompanied by the elimination of water molecules), not addition polymerization.
Therefore, statements 1 and 3 are correct.

1 mark for selecting A.
- Reject B, C, D: Since statements 2 and 4 are incorrect, any choice containing them is incorrect.

During the extraction of aluminium by electrolysis:
- Oxides ions (\( \text{O}^{2-} \)) are oxidized at the graphite (carbon) anodes to form oxygen gas (\( \text{O}_2 \)). At high operating temperatures (around \( 950^\circ\text{C} \)), this oxygen reacts with the carbon anodes to form carbon dioxide gas (\( \text{CO}_2 \)). Thus, the anodes burn away and must be replaced regularly.
- Cryolite is added to lower the melting point of aluminium oxide from \( 2000^\circ\text{C} \) to \( 950^\circ\text{C} \) and to improve conductivity, making option A incorrect.
- Aluminium ions gain electrons (are reduced, not oxidized) at the cathode, making option C incorrect.
- The electrolyte is molten, not aqueous, because water would be preferentially reduced at the cathode, producing hydrogen gas instead of aluminium, making option D incorrect.

1 mark for selecting B.
- Reject A: Cryolite is added to increase conductivity, not decrease it.
- Reject C: Aluminium ions undergo reduction (gain of electrons) at the cathode.
- Reject D: The process must use a molten electrolyte because aluminium is highly reactive.

The correct experimental procedure is:
- Step 2: Add excess copper(II) oxide to warm dilute sulfuric acid to ensure all the acid is neutralized.
- Step 3: Filter the mixture to remove the unreacted, excess solid copper(II) oxide.
- Step 1: Heat the filtrate (copper(II) sulfate solution) to evaporate some water until a hot saturated solution is obtained.
- Step 5: Cool the hot saturated solution slowly to allow copper(II) sulfate crystals to grow.
- Step 4: Separate the crystals from the remaining liquid by filtration, and dry them using filter paper.
This matches the sequence: 2 \(\rightarrow\) 3 \(\rightarrow\) 1 \(\rightarrow\) 5 \(\rightarrow\) 4.

1 mark for selecting A.
- Reject B: The filtrate must be filtered (Step 3) before heating (Step 1) to remove excess copper(II) oxide.
- Reject C: Filtration (Step 3) cannot take place before the chemical reaction (Step 2).
- Reject D: Heating (Step 1) cannot happen before the reactants are mixed.

(a) Two observations during the reaction:
1. Effervescence / bubbling / fizzing.
2. The magnesium ribbon dissolves / disappears.

(b) (i) As the reaction proceeds, the reactants (hydrochloric acid and magnesium) are used up. The concentration of acid reactant particles decreases, meaning there are fewer particles per unit volume, which results in a lower frequency of successful collisions.
(ii) A higher concentration of hydrochloric acid means there are more acid particles per unit volume. This leads to a higher frequency of collisions (more collisions per unit time) between the reacting particles, resulting in a higher frequency of successful collisions and thus an increased initial rate of reaction.

(c) (i) The initial rate of reaction increases.
(ii) Magnesium powder has a larger surface area than magnesium ribbon. This exposes more magnesium atoms to the hydrochloric acid, which increases the frequency of collisions between the reactants and therefore increases the rate of reaction.
(iii) The final volume of hydrogen gas collected remains the same. Magnesium is the limiting reactant and its mass is unchanged. Since the number of moles of the limiting reactant is identical, the same amount of product is formed.

Total Marks: 13

(a) [2 marks total]
- Any two observations from: effervescence/fizzing [1 mark], magnesium disappears/dissolves [1 mark], heat is evolved/test tube gets warm [1 mark].

(b)(i) [2 marks total]
- Concentration of reactant particles / acid decreases [1 mark].
- Lower frequency of collisions / fewer successful collisions per unit time [1 mark].

(b)(ii) [3 marks total]
- More particles per unit volume [1 mark].
- Higher frequency of collisions / more collisions per second [1 mark].
- Higher frequency of successful collisions [1 mark].

(c)(i) [1 mark total]
- Rate increases [1 mark].

(c)(ii) [2 marks total]
- Greater surface area [1 mark].
- More collisions per unit time / greater collision frequency [1 mark].

(c)(iii) [3 marks total]
- Volume remains the same [1 mark].
- Mass / moles of magnesium are the same [1 mark].
- Magnesium is the limiting reactant [1 mark].

(a) (i) A catalyst is a substance that increases the rate of a chemical reaction, but remains chemically unchanged at the end of the reaction.
(ii) A catalyst provides an alternative reaction pathway with a lower activation energy, meaning more particles have energy greater than or equal to the activation energy.
(iii) Filter the reaction mixture at the end to retrieve the manganese(IV) oxide. Wash the solid with distilled water, dry it, and weigh it. The mass of the dry manganese(IV) oxide recovered should be identical to the starting mass.

(b) (i) Activation energy is the minimum energy that colliding particles must have in order to react.
(ii) Bond breaking is an endothermic process (takes in energy) and bond making is an exothermic process (releases energy). The energy released during the making of bonds in the products (\(\text{H}_2\text{O}\) and \(\text{O}_2\)) is greater than the energy absorbed in breaking the bonds of the reactants (\(\text{H}_2\text{O}_2\)).

(c) At \(80\ ^\circ\text{C}\), the rate of reaction will decrease significantly or stop. This is because catalase is an enzyme (protein), and high temperatures denature enzymes by altering the shape of their active sites, rendering them inactive.

Total Marks: 13

(a)(i) [2 marks total]
- Substance that increases rate of reaction [1 mark].
- Remains chemically unchanged / mass unchanged at the end [1 mark].

(a)(ii) [2 marks total]
- Alternative reaction pathway [1 mark].
- Lower activation energy [1 mark].

(a)(iii) [3 marks total]
- Filter the mixture to retrieve the solid catalyst [1 mark].
- Wash and dry the solid [1 mark].
- Show that the mass of the dry solid remains unchanged [1 mark].

(b)(i) [1 mark total]
- Minimum energy required for colliding particles to react [1 mark].

(b)(ii) [3 marks total]
- Bond breaking absorbs energy AND bond making releases energy [1 mark].
- More energy is released during bond making than absorbed during bond breaking [1 mark].
- Correctly references reactants and products [1 mark].

(c) [2 marks total]
- Rate decreases / reaction stops [1 mark].
- Enzyme / catalase is denatured [1 mark]. (Do NOT accept 'killed')

(a) (i) A polymer is a large molecule / macromolecule built up from many small repeating units (monomers) linked together by covalent bonds.
(ii) It contains a carbon-carbon double bond (\(\text{C}=\text{C}\)) / is unsaturated.
(iii) The repeating unit consists of a single-bonded carbon-carbon backbone: \(\text{-[CH}_2\text{-CH(C}_6\text{H}_5)]\text{-}\). In drawn form: two carbon atoms joined by a single covalent bond. Each carbon has two open continuation bonds on either end of the unit. The first carbon has two hydrogen atoms attached, and the second carbon has one hydrogen atom and one phenyl group (\(\text{C}_6\text{H}_5\)) attached.

(b) (i) Monomer formula: \(\text{CH}_2=\text{CHCH}_3\) (or structural: propene); Name: propene.
(ii) Addition polymerization.

(c) (i) They contain only single covalent bonds (carbon-carbon \(\text{C-C}\) and carbon-hydrogen \(\text{C-H}\) bonds) which are very strong, stable, and non-polar, making them highly unreactive.
(ii) 1. They fill up landfill sites, taking up valuable space because they do not decompose.
2. In oceans, marine animals can ingest them or get entangled, leading to suffocation or death.
(iii) Addition polymers have a strong, non-polar carbon-carbon backbone that is highly resistant to chemical attack. In contrast, condensation polymers contain polar ester or amide linkages, which can be readily broken down via hydrolysis by acids, bases, or specialized microbes.

Total Marks: 13

(a)(i) [1 mark total]
- Large molecule / macromolecule made of many small repeating units / monomers [1 mark].

(a)(ii) [1 mark total]
- Carbon-carbon double bond / \(\text{C}=\text{C}\) / unsaturation [1 mark].

(a)(iii) [2 marks total]
- Carbon-carbon backbone with a single bond and open continuation bonds shown [1 mark].
- Correct substituents attached: two \(\text{-H}\) atoms on one carbon, and one \(\text{-H}\) and one \(\text{-C}_6\text{H}_5\) on the second carbon [1 mark].

(b)(i) [2 marks total]
- Formula: \(\text{CH}_2=\text{CHCH}_3\) / \(\text{C}_3\text{H}_6\) [1 mark].
- Name: propene [1 mark].

(b)(ii) [1 mark total]
- Addition (polymerization) [1 mark].

(c)(i) [1 mark total]
- Contain only strong, stable, non-polar \(\text{C-C}\) and \(\text{C-H}\) single bonds [1 mark].

(c)(ii) [2 marks total]
- Fill up landfill sites [1 mark].
- Harm to marine life / ingestion / entanglement [1 mark].

(c)(iii) [3 marks total]
- Addition polymers have a carbon-carbon single bond backbone which cannot be easily broken [1 mark].
- Condensation polymers contain ester / amide linkages [1 mark].
- These linkages are polar and can undergo hydrolysis / biological breakdown [1 mark].

(a) Two differences:
1. Condensation polymerization produces a small molecule (usually water) as a byproduct, whereas addition polymerization forms only the polymer product.
2. Addition polymerization requires monomers with unsaturated carbon-carbon double bonds (\(\text{C}=\text{C}\)), while condensation polymerization requires monomers with reactive functional groups at both ends (such as -COOH and -OH).

(b) (i) Structure of dicarboxylic acid: \(\text{HO-C(=O)-R-C(=O)-OH}\). Every bond within the \(\text{-COOH}\) groups must be clearly shown, particularly the double bond to oxygen and the single bond to the hydroxyl group.
(ii) Structure of diol: \(\text{H-O-R'-O-H}\). All single covalent bonds in the hydroxyl groups (\(\text{O-H}\)) must be fully drawn out.
(iii) Structure of Terylene repeating unit: \(\text{-[O-R'-O-C(=O)-R-C(=O)]-}\) with open bonds at both ends.

(c) (i) Amino acids.
(ii) Amide linkage (or peptide link).
(iii) Hydrolysis [1]; using dilute hydrochloric acid under reflux conditions (or using protease enzymes) [1].

Total Marks: 13

(a) [2 marks total]
- Condensation forms a small molecule byproduct (like \(\text{H}_2\text{O}\)) but addition does not [1 mark].
- Condensation requires reactive functional groups at monomer ends, while addition requires unsaturated carbon-carbon double bonds [1 mark].

(b)(i) [2 marks total]
- Correct carboxylic acid groups shown as \(\text{-C(=O)-OH}\) showing all bonds [1 mark].
- Rest of the molecule showing \(\text{R}\) connecting the two acid groups [1 mark].

(b)(ii) [2 marks total]
- Correct hydroxyl groups shown as \(\text{-O-H}\) showing the O-H bond [1 mark].
- Rest of the molecule showing \(\text{R}'\) connecting the two -OH groups [1 mark].

(b)(iii) [3 marks total]
- Correct ester linkage shown as \(\text{-O-C(=O)-}\) with all bonds [1 mark].
- Correct linkage of \(\text{R}\) and \(\text{R}'\) through the ester linkage [1 mark].
- Continuation bonds shown on both ends [1 mark].

(c)(i) [1 mark total]
- Amino acids [1 mark].

(c)(ii) [1 mark total]
- Amide link / peptide link [1 mark].

(c)(iii) [2 marks total]
- Hydrolysis [1 mark].
- Acid catalyst / dilute HCl / heat / enzymes [1 mark].

(a) (i)
Cathode product: Zinc (\(\text{Zn}\))
Anode product: Chlorine gas (\(\text{Cl}_2\))
(ii)
Cathode reaction: \(\text{Zn}^{2+}(\text{l}) + 2\text{e}^- \rightarrow \text{Zn(l)}\)
Anode reaction: \(2\text{Cl}^-(\text{l}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\)

(b) (i) Anode product: Chlorine gas (\(\text{Cl}_2\)).
Observation: Effervescence / bubbles of pale-green gas / sharp, choking smell.
(ii) Zinc is more reactive than hydrogen (hydrogen is lower in the reactivity series). In aqueous solution, hydrogen ions (\(\text{H}^+\)) are more easily reduced than zinc ions (\(\text{Zn}^{2+}\)), so hydrogen gas is liberated at the cathode.
(iii) The pH increases / becomes alkaline. Both hydrogen ions (\(\text{H}^+\)) and chloride ions (\(\text{Cl}^-\)) are discharged and removed from the solution. This leaves a high relative concentration of hydroxide ions (\(\text{OH}^-\)) and zinc ions (\(\text{Zn}^{2+}\)) in solution, making the mixture alkaline (or forming zinc hydroxide).

Total Marks: 13

(a)(i) [2 marks total]
- Cathode: Zinc [1 mark].
- Anode: Chlorine [1 mark].

(a)(ii) [4 marks total]
- Cathode equation: \(\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}\) [1 mark].
- State symbols for cathode: \(\text{Zn}^{2+}(\text{l})\) and \(\text{Zn(l)}\) or \(\text{Zn(s)}\) [1 mark].
- Anode equation: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) [1 mark].
- State symbols for anode: \(\text{Cl}^-(\text{l})\) and \(\text{Cl}_2(\text{g})\) [1 mark].

(b)(i) [2 marks total]
- Chlorine [1 mark].
- Green gas / bubbles / effervescence [1 mark].

(b)(ii) [2 marks total]
- Hydrogen is lower in reactivity series / zinc is more reactive [1 mark].
- Hydrogen ions / \(\text{H}^+\) discharged preferentially [1 mark].

(b)(iii) [3 marks total]
- pH increases / solution becomes alkaline [1 mark].
- \(\text{H}^+\) and \(\text{Cl}^-\) ions are removed [1 mark].
- Leaving behind excess hydroxide ions / \(\text{OH}^-\) ions [1 mark].

(a) (i) Balanced equation with state symbols:
\(\text{CuCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{H}_2\text{O(l)} + \text{CO}_2(\text{g})\)

(ii)
To ensure all acid has reacted: Add excess copper(II) carbonate to the warm sulfuric acid until no more dissolves / no more carbon dioxide bubbles are given off.
To obtain crystals: Filter the mixture to remove the excess unreacted copper(II) carbonate. Heat the copper(II) sulfate solution (filtrate) in an evaporating dish until the crystallization point is reached (or when a glass rod dipped in the solution forms crystals upon cooling). Allow the hot saturated solution to cool slowly to form large crystals. Filter the crystals from the remaining liquid, wash them with a small amount of cold distilled water, and dry them using filter paper.

(b) (i) Any soluble barium salt, e.g., barium chloride or barium nitrate, and any soluble sulfate salt, e.g., sodium sulfate, potassium sulfate, or ammonium sulfate (or dilute sulfuric acid).
(ii) Ionic equation with state symbols:
\(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

Total Marks: 13

(a)(i) [3 marks total]
- Correct reactants and products: \(\text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2\) [2 marks].
- Correct state symbols: \(\text{CuCO}_3\text{(s)}\), \(\text{H}_2\text{SO}_4\text{(aq)}\), \(\text{CuSO}_4\text{(aq)}\), \(\text{H}_2\text{O(l)}\), \(\text{CO}_2\text{(g)}\) [1 mark].

(a)(ii) [5 marks total]
- Add copper(II) carbonate in excess [1 mark].
- Until no more dissolves / no more effervescence occurs [1 mark].
- Filter off the excess copper(II) carbonate [1 mark].
- Heat filtrate to crystallization point / saturate solution [1 mark].
- Cool, filter crystals, wash with cold distilled water, and dry between filter papers [1 mark].

(b)(i) [2 marks total]
- Soluble barium salt: Barium chloride / barium nitrate [1 mark].
- Soluble sulfate: Sodium sulfate / potassium sulfate / ammonium sulfate / dilute sulfuric acid [1 mark].

(b)(ii) [3 marks total]
- Correct ions and product: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\) [2 marks].
- Correct state symbols: \(\text{Ba}^{2+}\text{(aq)}\), \(\text{SO}_4^{2-}\text{(aq)}\), \(\text{BaSO}_4\text{(s)}\) [1 mark].

(a) The diagram should show a conical flask containing the reaction mixture (zinc, dilute sulfuric acid, and catalyst) fitted with a stopper and a delivery tube connected to a gas syringe (or an inverted measuring cylinder filled with water). All apparatus must be sealed and labeled correctly.

(b) Two constant variables: temperature of the acid, volume of dilute sulfuric acid, concentration of the sulfuric acid, mass/surface area of zinc granules.

(c) Anomalous result: 21 cm³ at 4.0 g/dm³ (expected volume should be higher than 45 cm³). Possible practical errors: gas leak in the apparatus, stopper not inserted quickly enough, or incorrect measurement read from the syringe.

(d) The reaction stops because one of the reactants (either zinc or sulfuric acid) has been completely used up.

(e) Draw a tangent to the curve at 15 seconds, and then calculate the gradient of this tangent (change in volume divided by change in time).

(f) Test: Insert a lighted splint into a test-tube of the gas. Observation: Gas burns with a squeaky pop.

Total: 10 marks

(a) [2 marks]
- 1 mark: Conical flask connected via gas-tight delivery tube to a gas syringe or inverted measuring cylinder over water.
- 1 mark: All major apparatus labeled correctly (flask, delivery tube, gas syringe/measuring cylinder, and reaction mixture).

(b) [2 marks]
- 2 marks: Any two correct variables (e.g., volume of acid, concentration of acid, mass of zinc, surface area of zinc, temperature). [1 mark for each]

(c) [2 marks]
- 1 mark: Correctly identifying 21 cm³ (at 4.0 g/dm³) as anomalous.
- 1 mark: Reasonable practical explanation (e.g., loose stopper/gas escaped, delayed fitting of stopper, or error in reading syringe).

(d) [1 mark]
- 1 mark: Reactant used up / zinc or acid is limiting reactant.

(e) [1 mark]
- 1 mark: Draw a tangent to the curve at 15 s and find its gradient / slope.

(f) [2 marks]
- 1 mark: Lighted splint.
- 1 mark: Squeaky pop / pops.

(a) Plan:
1. Cut strips of the three different polymer sheets to the exact same dimensions (length and width) using a ruler and scissors.
2. Clamp one end of the first polymer strip (e.g., LDPE) securely to a stand.
3. Attach a mass hanger to the free end of the strip.
4. Add weights (masses) of known value (e.g., 10g or 50g at a time) to the hanger sequentially and carefully.
5. Record the total mass required to make the strip break.
6. Repeat the process for the HDPE and starch-based polymer strips.
7. Keep the temperature and rate of adding masses constant (fair test variables).
8. The polymer strip that holds the greatest total mass before breaking is the strongest.

(b) Hazard: Falling weights could injure feet. Precaution: Place a soft cushion, tray of sand, or cardboard box directly underneath the suspended weights, or keep feet well clear of the area below the weights.

(c) Chemical test: Add a few drops of aqueous iodine solution to a sample of each polymer. Positive result: The starch-based polymer will turn blue-black, while the polyethylene will remain brown/orange.

Total: 10 marks

(a) [6 marks]
- 1 mark: Cut strips of the same dimensions (length and width) / same thickness.
- 1 mark: Securely clamp one end of the strip and hang weights on the other end.
- 1 mark: Add weights/masses gradually / one by one.
- 1 mark: Record the mass at which the strip breaks.
- 1 mark: Repeat for all three polymers.
- 1 mark: State that the strongest polymer is the one that supports the greatest mass before breaking.

(b) [2 marks]
- 1 mark: Hazard: falling masses/weights injuring feet/toes.
- 1 mark: Precaution: use a box/cushion to catch falling masses OR keep feet clear of the drop zone.

(c) [2 marks]
- 1 mark: Add iodine solution / aqueous iodine.
- 1 mark: Starch-based polymer turns blue-black (accept dark blue / black; ignore change for polyethylene).

(a) The diagram must include:
- A beaker containing aqueous copper(II) sulfate.
- Two electrodes dipping into the solution.
- A complete d.c. circuit with a power supply (battery/cell) connected to both electrodes.
- Correct labels: anode (+), cathode (-), and copper(II) sulfate electrolyte.

(b) Setup A observations:
(i) Anode (+): Bubbles of a colorless gas (oxygen) are evolved.
(ii) Cathode (-): A pink-brown / reddish-brown solid deposit (copper) forms on the electrode.
(iii) Electrolyte: The blue color of the solution gradually fades / becomes lighter / colorless.

(c) Setup B observations and explanations:
(i) At the anode, instead of gas bubbles forming, the copper anode dissolves / decreases in mass / becomes smaller. This is because copper atoms are oxidized to copper(II) ions (\(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\)) as copper is not inert.
(ii) The blue color intensity of the electrolyte remains constant. This is because copper(II) ions are discharged at the cathode to form copper metal at the same rate that copper(II) ions are produced at the anode by dissolving copper atoms, maintaining a constant concentration of \(\text{Cu}^{2+}\) ions in solution.

Total: 10 marks

(a) [3 marks]
- 1 mark: Beaker with liquid labeled as aqueous copper(II) sulfate / copper(II) sulfate solution.
- 1 mark: Two electrodes immersed in the liquid, connected to a d.c. power source (battery or cell symbols).
- 1 mark: Correctly labels anode (connected to +) and cathode (connected to -).

(b) [3 marks]
- 1 mark: (i) bubbles of gas / effervescence.
- 1 mark: (ii) pink / brown / pink-brown / red-brown solid deposit.
- 1 mark: (iii) blue solution becomes lighter / fades / colorless.

(c) [4 marks]
- (i) [2 marks]
- 1 mark: Anode dissolves / loses mass / no gas bubbles.
- 1 mark: Copper anode reacts/is oxidized to form \(\text{Cu}^{2+}\) ions.
- (ii) [2 marks]
- 1 mark: Blue color remains the same / constant.
- 1 mark: Copper ions are added to solution at the same rate they are removed.

(a) Steps to prepare the saturated solution:
1. Measure a known volume of dilute sulfuric acid into a beaker and warm/heat it gently (to increase the rate of reaction).
2. Add nickel(II) carbonate powder a little at a time, stirring continuously, until no more dissolves (excess solid is visible at the bottom of the beaker and effervescence stops).
3. Filter the mixture using a funnel and filter paper to remove the unreacted excess nickel(II) carbonate solid.
4. Collect the filtrate (which is the pure nickel(II) sulfate solution).

(b) Adding an excess of nickel(II) carbonate ensures that all of the dilute sulfuric acid has completely reacted, so the final salt crystals are not contaminated with acid.

(c) Obtaining crystals:
1. Heat the filtrate in an evaporating basin until the crystallization point is reached (indicated by crystals forming on a cold glass rod dipped into the solution).
2. Leave the hot, saturated solution to cool and crystallize slowly.
3. Filter off the crystals from the remaining liquid, and dry them by patting gently with filter paper / paper towel.

(d) The solution changes from colorless to green (accept light green / emerald green).

(e) Heating to dryness would remove the water of crystallization, turning the hydrated nickel(II) sulfate crystals into an anhydrous powder / ruining the crystal structure.

Total: 10 marks

(a) [4 marks]
- 1 mark: Add nickel(II) carbonate to (warm) dilute sulfuric acid.
- 1 mark: Add until in excess / until no more dissolves / effervescence stops.
- 1 mark: Stir.
- 1 mark: Filter to remove excess nickel(II) carbonate / collect filtrate.

(b) [1 mark]
- 1 mark: To ensure all sulfuric acid is completely reacted/used up.

(c) [3 marks]
- 1 mark: Heat/evaporate filtrate to crystallization point / evaporate some water.
- 1 mark: Cool (slowly) to form crystals.
- 1 mark: Filter crystals and dry using filter paper (reject heat to dry).

(d) [1 mark]
- 1 mark: Colorless to green.

(e) [1 mark]
- 1 mark: Water of crystallization would be lost / would produce anhydrous powder / crystals would break.