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In Experiment 2, the student uses smaller marble chips, which have a greater surface area per unit mass. This increases the collision frequency between reactant particles, leading to a faster initial rate of reaction. The mass of the limiting reactant (calcium carbonate, 2.0 g) remains identical, and since hydrochloric acid is in excess, both experiments must eventually yield the same total volume of carbon dioxide gas.

Award 1 mark for identifying that smaller surface area/larger chips leads to a slower rate (or larger surface area/smaller chips leads to a faster rate) and that the same mass of limiting reactant ensures the same volume of gas is produced.

The repeating unit shows that monomers are linked via an ester functional group (\(-COO-\)), indicating that it is a polyester. Polyesters are synthesized via condensation polymerisation, where monomers react to form polymer chains while eliminating a small molecule (such as water).

Award 1 mark for selecting condensation polymerisation and the ester linkage.

The decomposition of calcium carbonate to form calcium oxide and carbon dioxide, \(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\), requires thermal energy from the combustion of coke, making it an endothermic reaction. Carbon monoxide acts as a reducing agent (not an oxidizing agent) for iron(III) oxide. Slag is less dense than molten iron and therefore floats on top.

Award 1 mark for identifying the thermal decomposition of calcium carbonate as an endothermic reaction.

First, calculate the moles of nitrogen gas reacting: \(n(N_2) = \frac{1.40\text{ g}}{28.0\text{ g/mol}} = 0.05\text{ mol}\). According to the balanced equation, 1 mole of \(N_2\) produces 2 moles of \(NH_3\). Therefore, \(0.05\text{ mol} \times 2 = 0.10\text{ mol}\) of \(NH_3\) gas is formed. The volume is calculated as: \(0.10\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 2.40\text{ dm}^3\).

Award 1 mark for the correct stoichiometric calculation of gas volume (2.40 dm³).

A green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(Fe^{2+}\) (chromium(III) ions produce a green precipitate that dissolves in excess NaOH to form a green solution). A white precipitate with barium nitrate in acidic conditions confirms the presence of sulfate ions, \(SO_4^{2-}\). Thus, the salt is iron(II) sulfate.

Award 1 mark for identifying the cation as iron(II) and the anion as sulfate.

Isotopes have the same atomic number (number of protons) and thus the same number of electrons. Because their electronic configuration is identical, they have the same number of outer-shell valence electrons, which determines their chemical properties. Consequently, they possess identical chemical properties, but their different numbers of neutrons cause minor differences in physical properties like density and mass.

Award 1 mark for selecting the correct explanation connecting outer-shell electrons to chemical properties.

At the cathode (negative electrode), hydrogen ions (\(H^+\)) are discharged in preference to water molecules, forming hydrogen gas. At the anode (positive electrode), since the hydrochloric acid is concentrated, chloride ions (\(Cl^-\)) are discharged in preference to hydroxide ions (\(OH^-\)), forming chlorine gas, which is observed as a pale green gas.

Award 1 mark for correctly matching cathode product (hydrogen), anode product (chlorine), and observation (pale green gas).

Energy absorbed to break reactant bonds: \(1 \times (H-H) + 1 \times (Cl-Cl) = 436 + 242 = 678\text{ kJ/mol}\). Energy released during bond formation of products: \(2 \times (H-Cl) = 2 \times 431 = 862\text{ kJ/mol}\). Energy change (\(\Delta H\)) = Energy input - Energy output = \(678 - 862 = -184\text{ kJ/mol}\). Since the energy change is negative, the reaction is exothermic.

Award 1 mark for the correct calculation of overall energy change (-184 kJ/mol) and correctly classifying it as exothermic.

Terylene is a polyester, which is formed via condensation polymerisation. The reaction occurs between a dicarboxylic acid monomer (which has two carboxylic acid \(-\text{COOH}\) groups) and a diol monomer (which has two alcohol \(-\text{OH}\) groups), eliminating water molecules in the process.

Award 1 mark for identifying the correct pair of monomers: a dicarboxylic acid and a diol.

To investigate how temperature affects the reaction rate, temperature is changed (the independent variable) and the rate of reaction is measured (the dependent variable). To ensure a fair test, all other variables, including the volumes and concentrations of the sodium thiosulfate and hydrochloric acid solutions, must be kept constant.

Award 1 mark for identifying that the volumes and concentrations of the reactants must be kept constant as controlled variables.

Because the gas is highly soluble in water, it cannot be collected over water. Since the gas is denser than air, it must be collected by downward delivery (upward displacement of air) rather than upward delivery. To obtain a dry sample, the gas must first pass through a drying agent such as anhydrous calcium chloride.

Award 1 mark for selecting the correct combination of a drying agent and downward delivery.

The combustion of carbon (coke) in oxygen to form carbon dioxide, \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\), is highly exothermic. This reaction is crucial because it provides the heat required to maintain the high temperatures needed for the reduction of iron ore in the blast furnace.

Award 1 mark for identifying the reaction of coke with oxygen as the main highly exothermic heat-producing reaction.

Hydrochloric acid is a strong acid and is fully ionized in aqueous solution, resulting in a higher concentration of hydrogen ions than in ethanoic acid (a weak acid, which is only partially ionized). Because of this high concentration of mobile ions, hydrochloric acid has a lower pH, reacts faster with metals, and conducts electricity significantly better than ethanoic acid.

Award 1 mark for identifying that the strong acid (hydrochloric acid) conducts electricity better due to full dissociation/higher ion concentration.

Chemical properties of elements are determined by their electronic configuration, specifically the number of electrons in their outer shells. Since isotopes of the same element have the same number of protons, they also have the same number of electrons and identical electronic configurations, resulting in identical chemical properties.

Award 1 mark for identifying that identical chemical properties are due to isotopes having the same electronic configuration.

The reaction of the salt with aqueous sodium hydroxide to form a green precipitate that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The reaction with barium nitrate in acidic conditions to form a white precipitate indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the salt \(X\) is iron(II) sulfate.

Award 1 mark for correctly identifying the cation as iron(II) and the anion as sulfate to deduce the formula of salt \(X\) as iron(II) sulfate.

A mixture consists of two or more substances that are physically blended but not chemically combined, meaning they can be separated by physical methods (e.g., filtration, distillation). A compound consists of two or more elements chemically combined in a fixed ratio, meaning its components can only be separated using chemical processes (e.g., electrolysis).

Award 1 mark for identifying the correct comparative property regarding separation methods.

Statement 1 is correct because the polymer contains ester linkages ($-\text{O}-\text{CO}-$) which are formed by condensation polymerisation.
Statement 2 is correct because the repeating unit is made from ethane-1,2-diol ($\text{HO}-\text{CH}_2\text{CH}_2-\text{OH}$) and butanedioic acid ($\text{HOOC}-\text{CH}_2\text{CH}_2-\text{COOH}$).
Statement 3 is correct because during condensation polymerisation of polyesters, a water molecule is eliminated for each ester link formed.
Statement 4 is incorrect because it is a condensation polymer, not an addition polymer.

1 mark for selecting the correct option (B).

To obtain half the volume of carbon dioxide, the moles of the limiting reactant (hydrochloric acid) must be halved. The initial number of moles of acid is $0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol}$. Halving this gives $0.025\text{ mol}$.
In option A, the moles of acid is $0.025\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.025\text{ mol}$, which is correct.
Furthermore, Curve Y is steeper, meaning the initial rate of reaction is faster. Using calcium carbonate powder instead of chunks increases the surface area, which increases the rate of reaction.

1 mark for the correct option (A).

Ammonia ($\text{NH}_3$) is an alkaline gas, so it cannot be dried using an acidic drying agent like concentrated sulfuric acid, as they would react to form ammonium sulfate. Therefore, calcium oxide (a basic oxide) must be used as the drying agent.
Ammonia has a lower relative molecular mass ($M_r = 17$) than air (average $M_r \approx 29$), making it less dense than air. It must be collected by upward delivery (downward displacement of air) in an inverted gas jar.

1 mark for the correct option (D).

The main acidic impurity in the iron ore is silicon(IV) oxide, $\text{SiO}_2$ (sand). Calcium carbonate decomposes to produce calcium oxide, $\text{CaO}$, which is a basic oxide. The basic calcium oxide reacts with the acidic silicon(IV) oxide to form liquid slag, calcium silicate ($\text{CaSiO}_3$). This neutralisation reaction represents the removal of acidic impurities.

1 mark for the correct option (D).

Statement 1: A pH of 1.0 corresponds to a hydrogen ion concentration of $10^{-1} = 0.10\text{ mol/dm}^3$. Since the acid concentration is $0.10\text{ mol/dm}^3$, acid X is completely ionised (it is a strong acid).
Statement 2: A lower pH means a higher concentration of hydrogen ions. Thus, acid X has a higher concentration of hydrogen ions than acid Y ($10^{-1}$ vs $10^{-3}\text{ mol/dm}^3$).
Statement 3: Both are monoprotic acids of the same concentration ($0.10\text{ mol/dm}^3$) and volume ($25\text{ cm}^3$), containing the same total amount of ionisable hydrogen atoms. Since zinc is in excess, both will produce the same volume of hydrogen gas upon completion.
Statement 4: Acid X has a higher concentration of ions, making it a better conductor of electricity than acid Y.

1 mark for the correct option (B).

Isotopes are atoms of the same element that contain the same number of protons but different numbers of neutrons. Since they are neutral atoms, the number of electrons must also equal the number of protons.
- Option A has 17 protons (same element), 20 neutrons (different number from 18), and 17 electrons (neutral atom). This is an isotope of X.
- Option B is an anion of the same isotope of X ($X^-$).
- Option C represents an atom of a different element (argon, 18 protons).
- Option D represents an atom of a different element (sulfur, 16 protons).

1 mark for the correct option (A).

1. The reaction with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. This confirms the presence of iron(II) ions, $\text{Fe}^{2+}$. (Chromium(III) ions, $\text{Cr}^{3+}$, also form a green precipitate but it is soluble in excess sodium hydroxide to give a green solution).
2. The reaction with barium nitrate in the presence of nitric acid produces a white precipitate, which confirms the presence of sulfate ions, $\text{SO}_4^{2-}$.
Combining these ions gives iron(II) sulfate, $\text{FeSO}_4$.

1 mark for the correct option (A).

1. Calculate the number of moles of each reactant:
- $\text{moles of CaCO}_3 = \frac{10.0\text{ g}}{100\text{ g/mol}} = 0.10\text{ mol}$.
- $\text{moles of HCl} = 0.100\text{ dm}^3 \times 1.5\text{ mol/dm}^3 = 0.15\text{ mol}$.

2. Determine the limiting reactant:
According to the balanced chemical equation, $1\text{ mol}$ of $\text{CaCO}_3$ reacts with $2\text{ mol}$ of $\text{HCl}$.
To completely react $0.10\text{ mol}$ of $\text{CaCO}_3$, $0.20\text{ mol}$ of $\text{HCl}$ is needed. Since only $0.15\text{ mol}$ of $\text{HCl}$ is present, $\text{HCl}$ is the limiting reactant.

3. Calculate the moles of $\text{CO}_2$ produced:
- $\text{moles of CO}_2 = \frac{1}{2} \times \text{moles of HCl} = \frac{0.15\text{ mol}}{2} = 0.075\text{ mol}$.

4. Calculate the volume of $\text{CO}_2$ at r.t.p.:
- $\text{Volume of CO}_2 = 0.075\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.8\text{ dm}^3$.

1 mark for the correct option (B).

The repeating unit contains an amide linkage (-\text{CONH}-). It is formed by a condensation reaction between a dicarboxylic acid and a diamine. The dicarboxylic acid block has 4 carbon atoms between the carbonyl groups: \(\text{HOOC}-(CH_2)_4-\text{COOH}\). The diamine block has 6 carbon atoms between the amine groups: \(\text{H}_2\text{N}-(CH_2)_6-\text{NH}_2\). Therefore, option A is the correct monomer pair.

[1 mark] - Correctly identifies the two monomers (dicarboxylic acid and diamine) with correct carbon chain lengths.

In Experiment 2, using powder instead of lumps increases the surface area, which increases the initial rate of reaction. Since the number of moles of the limiting reactant (hydrochloric acid) is the same (\(0.05\text{ mol}\)), the final volume of carbon dioxide gas produced remains the same. In Experiment 3, a lower concentration of acid (\(0.5\text{ mol/dm}^3\)) is used, which decreases the initial rate of reaction. The number of moles of acid is still \(0.10 \times 0.5 = 0.05\text{ mol}\), so the final volume of gas is also the same as in Experiment 1.

[1 mark] - Correctly identifies that increasing surface area increases rate but does not change the yield of gas, and selects option A.

Dilute hydrochloric acid and copper(II) carbonate react to produce carbon dioxide gas. The gas can be dried by bubbling it through concentrated sulfuric acid (which is acidic and does not react with \(\text{CO}_2\)). The dry gas can then be collected in a gas syringe. Collecting over water (option D) would re-wet the gas. Anhydrous copper(II) sulfate (option B) is used as a chemical test for water, not as a standard drying agent for preparing gases.

[1 mark] - Correctly identifies concentrated sulfuric acid as the drying agent and a gas syringe for gas collection.

Roasting zinc blende (mainly \(\text{ZnS}\)) in air produces zinc oxide and sulfur dioxide gas: \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\). In the furnace, the zinc oxide is reduced to zinc by carbon monoxide: \(\text{ZnO} + \text{CO} \rightarrow \text{Zn} + \text{CO}_2\) (or directly by carbon: \(\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\)). Carbon dioxide cannot act as a reducing agent.

[1 mark] - Correctly identifies sulfur dioxide as the gaseous product of roasting and carbon monoxide as the reducing agent.

1. Correct: pH is a logarithmic measure of hydrogen ion concentration. A lower pH represents a higher concentration of hydrogen ions (\([\text{H}^+] = 10^{-2}\text{ mol/dm}^3\) for X and \(10^{-5}\text{ mol/dm}^3\) for Y).
2. Incorrect: Since both are strong monoprotic acids of equal volume, the one with the lower pH (Solution X) has a much higher concentration and thus contains more moles of hydrogen ions, producing a larger total volume of hydrogen gas.
3. Correct: The higher concentration of hydrogen ions in Solution X increases the frequency of effective collisions, resulting in a faster initial rate of reaction.
Therefore, statements 1 and 3 are correct.

[1 mark] - Correctly evaluates the relationship between pH, hydrogen ion concentration, reaction rate, and total volume of product.

Let the abundance of \(^{65}\text{E}\) be \(x\) (expressed as a fraction) and the abundance of \(^{63}\text{E}\) be \(1-x\).
The relative atomic mass is given by:
\(65x + 63(1-x) = 63.5\)
\(65x + 63 - 63x = 63.5\)
\(2x = 0.5\)
\(x = 0.25\)
Converting the fraction to a percentage: \(0.25 \times 100\% = 25\%\).

[1 mark] - Set up and solve the weighted average abundance equation correctly to find 25%.

1. The green precipitate with NaOH indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\).
2. Heating with NaOH produces a gas that turns damp red litmus blue (ammonia, \(\text{NH}_3\)), indicating the presence of ammonium ions, \(\text{NH}_4^+\).
3. The white precipitate with acidified barium chloride indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, the mixture must contain iron(II) ions, ammonium ions, and sulfate ions. Iron(II) sulfate and ammonium chloride contains all of these.

[1 mark] - Identifies Fe2+ from the green precipitate, NH4+ from the ammonia gas, and SO42- from the barium sulfate precipitate.

According to Le Chatelier's principle:
1. The forward reaction is exothermic (\(\Delta H < 0\)), so decreasing the temperature shifts the equilibrium to the right to produce more heat, increasing the yield.
2. There are 4 moles of gas on the left (\(1 + 3\)) and 2 moles of gas on the right. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas (the right), increasing the yield.
Therefore, low temperature and high pressure yield the highest percentage of ammonia at equilibrium.

[1 mark] - Explains effect of temperature on exothermic reaction and pressure on gas moles correctly to select low temp and high pressure.

The polymer shown is a polyester (Terylene), characterized by ester linkages (\(-\text{O}-\text{CO}-\)). It is produced via condensation polymerisation of a diol (\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\)) and a dicarboxylic acid (\(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\)) with the elimination of water molecules. Therefore, option C is correct.

1 mark for the correct option C.

The total volume of carbon dioxide gas depends on the total moles of the limiting reactant, which is hydrochloric acid (since calcium carbonate is in excess). Original moles of HCl = \(0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.05\text{ mol}\). In option A, moles of HCl = \(0.025\text{ dm}^3 \times 2.0\text{ mol/dm}^3 = 0.05\text{ mol}\). Because the concentration is higher (2.0 mol/dm³ compared to 1.0 mol/dm³), the initial rate of reaction is higher. Since the total moles of HCl is unchanged, the total volume of carbon dioxide produced remains the same. Thus, option A is correct.

1 mark for the correct option A.

The reaction is: \(\text{CuCO}_3(\text{s}) + 2\text{HNO}_3(\text{aq}) \rightarrow \text{Cu(NO}_3)_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). Option A is suitable because carbon dioxide gas escapes, causing a detectable loss in mass. Option B is suitable because carbon dioxide gas can be collected and measured using a gas syringe. Option C is suitable because hydrogen ions (acidic) are consumed, which continuously changes the pH of the mixture. Option D is NOT suitable because the reaction takes place in an aqueous solution where water is the solvent; the small volume of water produced cannot be measured separately. Therefore, option D is the correct answer.

1 mark for the correct option D.

In the blast furnace, carbon monoxide (\(\text{CO}\)) acts as the reducing agent, reducing iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) to molten iron: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). Option B is incorrect because limestone (\(\text{CaCO}_3\)) thermally decomposes to form calcium oxide and carbon dioxide, not oxygen. Option C is incorrect because calcium oxide reacts with silicon dioxide (sand/silica impurities), not coke, to form calcium silicate slag. Option D is incorrect because molten iron is denser than molten slag and thus sinks to the bottom, while the slag floats on top. Therefore, option A is correct.

1 mark for the correct option A.

Hydrochloric acid is a strong acid which fully ionises in water, resulting in a high concentration of hydrogen ions (\(\text{H}^+\)). Ethanoic acid is a weak acid which only partially ionises in water, resulting in a low concentration of hydrogen ions. Option A is incorrect because a higher concentration of \(\text{H}^+\) in Solution X means it has a lower pH than Solution Y. Option B is correct because the higher concentration of free-moving ions in Solution X makes it a much better electrical conductor. Option C is incorrect because Solution X has a higher hydrogen ion concentration than Solution Y. Option D is incorrect because the rate of reaction depends on the concentration of \(\text{H}^+\) ions; therefore, Solution X will react much faster than Solution Y.

1 mark for the correct option B.

Isotopes are atoms of the same element with the same number of protons (and thus the same number of electrons and electronic configuration) but different numbers of neutrons. Chemical properties are determined entirely by the outer-shell electronic configuration. Therefore, isotopes have identical chemical properties because they have the same number of electrons in their outer shell (same electronic configuration). Thus, option D is correct.

1 mark for the correct option D.

In Test 1, the formation of a green precipitate that is insoluble in excess sodium hydroxide is characteristic of iron(II) ions, \(\text{Fe}^{2+}\). (Note that chromium(III) ions, \(\text{Cr}^{3+}\), also form a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution). In Test 2, the formation of a white precipitate with barium nitrate in acidic conditions confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these ions, salt W is iron(II) sulfate (\(\text{FeSO}_4\)). Therefore, option B is correct.

1 mark for the correct option B.

1. Calculate the number of moles of \(\text{CO}_2\): \(\text{moles of CO}_2 = 4.4\text{ g} / 44\text{ g/mol} = 0.10\text{ mol}\). 2. Determine the number of atoms per molecule of \(\text{CO}_2\): Each molecule of \(\text{CO}_2\) contains 1 carbon atom and 2 oxygen atoms, which is 3 atoms in total. 3. Calculate the total moles of atoms: \(0.10\text{ mol} \times 3 = 0.30\text{ mol}\). 4. Calculate the total number of atoms: \(0.30\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 1.8 \times 10^{23}\). Therefore, option C is correct.

1 mark for the correct option C.

Detailed explanation of solutions:
(a)(i) Since methyl methacrylate contains a carbon-carbon double bond, it undergoes addition polymerisation.
(ii) The monomer's structure is \(\text{CH}_2=\text{C}(\text{CH}_3)\text{COOCH}_3\). Draw all bonds explicitly: a double bond between the two central carbons; one carbon has two C-H single bonds; the other carbon has a C-C single bond to a methyl group (\(-\text{CH}_3\)) and a C-C single bond to an ester group (\(-\text{C}(=\text{O})-\text{O}-\text{CH}_3\)).
(iii) The polymer repeat unit is obtained by opening the C=C double bond into a single bond and adding open extension bonds at each side: \(\text{—[CH}_2\text{—C}(\text{CH}_3)(\text{COOCH}_3)\text{]—}\).
(b)(i) A condensation reaction involves two functional groups reacting together to join molecules with the loss of a small molecule, typically \(\text{H}_2\text{O}\) or \(\text{HCl}\).
(ii) The monomers are a diol (\(\text{HO—Block A—OH}\)) and a dicarboxylic acid (\(\text{HOOC—Block B—COOH}\)). When they react, they form ester linkages (\(\text{—O—CO—}\)). Two repeat units would be: \(\text{—O—Block A—O—CO—Block B—CO—O—Block A—O—CO—Block B—CO—}\).
(c)(i) Biodegradable polymers are broken down naturally by microorganisms, meaning they do not accumulate in landfills or pollute oceans.
(ii) Traditional synthetic addition polymers like poly(ethene) have very stable, non-polar carbon-carbon single bonds in their backbones, which microorganisms cannot digest.
(iii) Lactic acid (2-hydroxypropanoic acid) contains both a carboxylic acid group (\(-\text{COOH}\)) and an alcohol group (\(-\text{OH}\)) on the same molecule. Thus, it undergoes self-condensation to form a polyester containing ester linkages.
(iv) The formula of 2-hydroxypropanoic acid is \(\text{CH}_3\text{—CH(OH)—COOH}\).

(a)(i) [1 mark] Addition polymerisation.
(ii) [2 marks] 1 mark for correct C=C double bond with correct substituents; 1 mark for showing fully displayed methyl and methyl ester group details.
(iii) [1 mark] Correct repeat unit showing single C-C bond with extension bonds.
(b)(i) [2 marks] 1 mark for molecules joining/reacting together; 1 mark for the release of a small molecule / water.
(ii) [3 marks] 1 mark for drawing correct ester link (\(\text{—O—CO—}\)); 1 mark for correct alternating block sequence; 1 mark for drawing two complete repeat units with open bonds at the ends.
(c)(i) [1 mark] Biodegrades/decomposes naturally / reduces landfill footprint.
(ii) [1 mark] Chemically inert / strong C-C covalent bonds.
(iii) [1 mark] Ester linkage.
(iv) [1 mark] Correct structural/displayed formula of 2-hydroxypropanoic acid.

Detailed walk-through:
(a) The reaction between a metal and an acid produces hydrogen gas. The standard chemical test for hydrogen is to introduce a lighted splint into the mouth of the test tube, which produces a characteristic squeaky pop sound.
(b)(i) Doubling the concentration means there are twice as many hydrogen ions in the same volume of solution. This increases the frequency of collisions between the hydrogen ions and the zinc surface, thereby increasing the rate of reaction.
(ii) Decreasing the temperature decreases the average kinetic energy of the reactant particles. This means they move slower and collide less frequently. Furthermore, a significantly smaller proportion of the colliding particles will possess the required activation energy, which decreases the frequency of successful collisions.
(c)(i) The initial rate is highest at the start, so the curve is steepest at \(t=0\). As reactants are consumed, the rate decreases, and the curve becomes less steep, eventually becoming flat (horizontal) when all the zinc has reacted.
(ii) Zinc powder has a much larger surface area than granules. This increases the rate of reaction, so curve 'P' starts much steeper. Since the mass of zinc is the same and the acid is in excess, the same number of moles of hydrogen gas are produced, so both curves level off at the exact same final volume.
(iii) Larger surface area allows more reactant particles to be in contact at any one time, increasing collision frequency. The overall yield (final volume) is determined solely by the moles of limiting reactant (zinc), which is kept constant.

(a) [2 marks] 1 mark for identifying hydrogen; 1 mark for lighted splint test with squeaky pop.
(b)(i) [3 marks] 1 mark for stating the rate of reaction increases; 1 mark for stating there are more particles per unit volume; 1 mark for stating that collision frequency increases.
(ii) [3 marks] 1 mark for stating the rate of reaction decreases; 1 mark for stating particles have less kinetic energy (or fewer particles have energy \(\ge\) activation energy); 1 mark for stating the frequency of successful collisions decreases.
(c)(i) [1 mark] Correct curve starting at origin and flattening out.
(ii) [2 marks] 1 mark for curve 'P' having a steeper initial gradient; 1 mark for curve 'P' levelling off at the exact same volume plateau.
(iii) [2 marks] 1 mark for stating powder has a larger surface area leading to more frequent collisions; 1 mark for stating that the same mass of zinc produces the same amount (volume) of hydrogen gas.

Detailed walk-through:
(a) To find the solubility at \(50^\circ\text{C}\):
1. Create a saturated solution by adding excess potassium chlorate to distilled water and heating to a temperature slightly above \(50^\circ\text{C}\) (e.g., \(60^\circ\text{C}\)) with stirring until no more dissolves.
2. Allow the mixture to cool to exactly \(50^\circ\text{C}\) and maintain this temperature.
3. Filter the mixture quickly to remove the undissolved solid, yielding a saturated solution at \(50^\circ\text{C}\).
4. Record the mass of a clean, dry evaporating basin.
5. Transfer some of the saturated filtrate to the basin and record the mass of the basin + solution. Subtract to find the mass of the saturated solution.
6. Heat the basin to evaporate all the water. To prevent loss of solid through spitting, use a water bath towards the end.
7. Dry the solid completely by heating to constant mass (repeating heating and weighing until mass is stable). Record the final mass of basin + dry solute. Subtract to find the mass of the solute.
8. Calculate mass of water: \(\text{mass of water} = \text{mass of solution} - \text{mass of solute}\).
9. Scale to 100 g of water: \(\text{Solubility} = (\text{mass of solute} / \text{mass of water}) \times 100\).

(b) Sources of error:
- Splattering/spitting of solute during heating can reduce measured solute mass. Prevent by heating gently over a water bath.
- Damp solute containing residual water can increase measured solute mass. Prevent by heating to constant mass.
- Crystallisation of solute in a cold funnel during filtration. Prevent by pre-heating the funnel.

(c) To prove a solution is saturated at \(50^\circ\text{C}\):
Add a weighed amount of solid \(\text{KClO}_3\) to the solution. Stir thoroughly while keeping the temperature at exactly \(50^\circ\text{C}\). Filter the remaining solid, dry it completely, and reweigh. If the mass of the solid is unchanged, the solution could not hold any more solute, confirming it was saturated.

(a) [6 marks]
- 1 mark: Prepare a saturated solution at/above \(50^\circ\text{C}\) by dissolving excess solute.
- 1 mark: Keep/cool the mixture to exactly \(50^\circ\text{C}\) and filter off the excess solid.
- 1 mark: Weigh the empty evaporating basin, and weigh it again with a portion of the saturated filtrate.
- 1 mark: Evaporate the water completely to dryness.
- 1 mark: Heat to constant mass / dry the remaining solid and reweigh.
- 1 mark: Calculate the mass of water and scale to g of solute per 100 g of water.
(b) [3 marks] Any three correct pairs of error and corresponding mitigation (1 mark each):
- Solute spitting during heating -> use a water bath / heat slowly.
- Crystallisation during filtration -> use a pre-heated funnel.
- Solid not completely dry -> heat to constant mass.
(c) [4 marks]
- 1 mark: Add a known mass of solid potassium chlorate.
- 1 mark: Maintain temperature at exactly \(50^\circ\text{C}\) and stir.
- 1 mark: Filter and dry the recovered solid.
- 1 mark: Weigh the dry solid; if the mass is unchanged, the solution is saturated.

Detailed walk-through:
(a)(i) Zinc sulfide (\(\text{ZnS}\)) reacts with oxygen (\(\text{O}_2\)) during roasting to produce zinc oxide (\(\text{ZnO}\)) and sulfur dioxide (\(\text{SO}_2\)). The balanced equation is:
\(2\text{ZnS(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{ZnO(s)} + 2\text{SO}_2\text{(g)}\).
(ii) Sulfur dioxide is an acidic gas that dissolves in rainwater to form acid rain. This damages buildings, kills trees, and lowers the pH of freshwater bodies, which is toxic to aquatic life.
(b)(i) Carbon (coke) is added to the furnace. It is oxidized to carbon dioxide, which further reacts with carbon to produce carbon monoxide (the reducing agent).
(ii) Carbon monoxide reduces zinc oxide: \(\text{ZnO(s)} + \text{CO(g)} \rightarrow \text{Zn(g)} + \text{CO}_2\text{(g)}\).
(iii) Since the reduction temperature (\(1000^\circ\text{C}\)) exceeds zinc's boiling point (\(907^\circ\text{C}\)), zinc is formed as a gas/vapour. The gaseous zinc leaves the blast furnace and is separated from other gases by condensing it back to liquid zinc in a condenser.
(c)(i) Pure copper has a regular lattice of identical atoms arranged in layers that slide over each other easily. In brass, zinc atoms of a different size are introduced, disrupting this regular arrangement. Consequently, the layers cannot slide easily, making the alloy harder.

(a)(i) [2 marks] 1 mark for correct formulas of reactants and products; 1 mark for correct balancing.
(ii) [2 marks] 1 mark for identifying acid rain; 1 mark for explaining that sulfur dioxide dissolves in water to form an acid that damages ecosystems or structures.
(b)(i) [1 mark] Coke / carbon.
(ii) [2 marks] 1 mark for correct formulas of reactants and products; 1 mark for correct balancing.
(iii) [3 marks] 1 mark for identifying the state as a gas/vapour; 1 mark for linking this to the temperature being above zinc's boiling point; 1 mark for explaining that it is separated by cooling and condensation.
(c)(i) [3 marks] 1 mark for stating that zinc and copper atoms have different sizes; 1 mark for stating that the regular layered arrangement is disrupted; 1 mark for explaining that the layers cannot slide over each other easily.

Detailed walk-through:
(a) An acid is defined as a proton donor. A weak acid only partially dissociates (or ionises) in water to release hydrogen ions, leaving most of the acid as molecules in equilibrium.
(b)(i) Because hydrochloric acid dissociates fully, it has a high concentration of hydrogen ions, corresponding to a very low pH (\(\approx 1\)). Ethanoic acid, being a weak acid, has a lower concentration of hydrogen ions, giving a higher pH (\(\approx 3\)).
(ii) Magnesium reacts with acids to produce hydrogen gas. The rate of this reaction is directly proportional to the concentration of hydrogen ions. The strong acid reacts rapidly with vigorous bubbling, whereas the weak acid reacts much more slowly with gentle bubbling.
(c)(i) Nitric acid reacts with copper(II) carbonate to form copper(II) nitrate, carbon dioxide, and water:
\(2\text{HNO}_3\text{(aq)} + \text{CuCO}_3\text{(s)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\).
(ii) Sulfuric acid reacts with iron(III) oxide to form iron(III) sulfate and water:
\(3\text{H}_2\text{SO}_4\text{(aq)} + \text{Fe}_2\text{O}_3\text{(s)} \rightarrow \text{Fe}_2(\text{SO}_4)_3\text{(aq)} + 3\text{H}_2\text{O(l)}\).

(a) [2 marks] 1 mark for proton donor; 1 mark for partial dissociation/ionisation in water.
(b)(i) [2 marks] 1 mark for stating hydrochloric acid has a lower pH (accept 1-2); 1 mark for stating ethanoic acid has a higher pH (accept 3-5).
(ii) [3 marks] 1 mark for observing rapid bubbling/dissolving with hydrochloric acid and slow bubbling/dissolving with ethanoic acid; 1 mark for stating hydrochloric acid has a higher concentration of hydrogen ions; 1 mark for linking ion concentration to collision frequency and rate of reaction.
(c)(i) [3 marks] 1 mark for correct formulas of all reactants and products; 1 mark for correct balancing; 1 mark for correct state symbols.
(ii) [3 marks] 1 mark for correct formulas of all reactants and products; 1 mark for correct balancing; 1 mark for correct state symbols.

Detailed walk-through:
(a) Isotopes are atoms of the same element that have the identical proton number (atomic number) but different nucleon numbers (mass numbers) due to different numbers of neutrons.
(b) Silicon has atomic number 14. Therefore, any neutral silicon atom has 14 protons and 14 electrons.
- Silicon-28 has a mass number (nucleon number) of 28. Neutrons = 28 - 14 = 14.
- Silicon-30 has a mass number of 30. Neutrons = 30 - 14 = 16.
(c) Calculate the weighted average mass:
\(A_r = \frac{(28 \times 92.23) + (29 \times 4.68) + (30 \times 3.09)}{100}\)
\(28 \times 92.23 = 2582.44\)
\(29 \times 4.68 = 135.72\)
\(30 \times 3.09 = 92.70\)
\(\text{Sum} = 2582.44 + 135.72 + 92.70 = 2810.86\)
\(A_r = \frac{2810.86}{100} = 28.1086\)
Rounding to two decimal places yields \(28.11\).
(d)(i) Medical use: e.g., Cobalt-60 for radiotherapy to kill cancer cells, or tracer isotopes in medical imaging.
(ii) Industrial use: e.g., Beta radiation emitters for measuring paper or plastic sheet thickness, or tracing leaks in underground pipes.
(iii) Chemical reactions only involve valence electrons. Since isotopes have the same electronic configuration, they undergo the exact same chemical reactions.

(a) [2 marks] 1 mark for same number of protons / same element; 1 mark for different number of neutrons.
(b) [4 marks]
- 4 marks if all 8 values are correct.
- 3 marks if 6-7 values are correct.
- 2 marks if 4-5 values are correct.
- 1 mark if 2-3 values are correct.
(c) [4 marks] 1 mark for setting up the correct weighted average expression; 1 mark for calculating intermediate terms correctly; 1 mark for dividing by 100; 1 mark for final answer as 28.11 (must be 2 decimal places).
(d)(i) [1 mark] Cancer treatment / tracer / sterilising medical equipment.
(ii) [1 mark] Paper/foil thickness gauging / detecting pipe leaks.
(iii) [1 mark] They have the same number of outer-shell electrons / electronic configuration.

(a) Gas syringe
(b) Volume produced during the second minute = 38 - 24 = 14 cm³.
Time interval = 60 s.
Average rate = 14 / 60 = 0.23 cm³/s (or 7/30 cm³/s).
(c)(i) The rate increases / reaction is faster.
(ii) Powder has a greater surface area, leading to more frequent collisions between reactant particles.
(d) One of the reactants (hydrochloric acid) has been completely used up.
(e) The curve for powder would have a steeper gradient (rise more rapidly) at the start and would level off earlier (at the same final volume of gas).
(f) Temperature of the acid, concentration of the acid, volume of the acid, mass of calcium carbonate (any two).

(a) [1 mark] Gas syringe. Accept: syringe. Reject: delivery tube.
(b) [1 mark] Correct subtraction of volume (38 - 24 = 14 cm³) and dividing by 60 s.
[1 mark] Correct final value with unit: 0.23 cm³/s (accept 0.233 or 7/30). Deduct 1 mark if unit is missing or incorrect (e.g., cm³/min).
(c)(i) [1 mark] Rate increases / reaction is faster.
(ii) [1 mark] Larger/greater surface area (of powder) leading to more frequent collisions / higher frequency of collisions (per unit time).
(d) [1 mark] The limiting reactant / hydrochloric acid is completely used up.
(e) [1 mark] Curve is steeper / has a larger gradient initially.
[1 mark] Levels off at the same volume but at an earlier time.
(f) [2 marks] Any two from:
- Concentration of acid
- Volume of acid
- Mass of calcium carbonate
- Temperature of the mixture
(1 mark for each correct variable)

(a) Green.
(b)(i) Green precipitate formed.
(ii) Precipitate remains insoluble / does not dissolve.
(c) Green precipitate formed, which is insoluble in excess aqueous ammonia.
(d) Add dilute hydrochloric acid (or dilute nitric acid) to the solution, then add aqueous barium chloride (or barium nitrate). A white precipitate is formed.
(e) Add aqueous sodium hydroxide and warm the mixture. Test the gas evolved with damp red litmus paper, which will turn blue.

(a) [1 mark] Green (accept pale green/grey-green; reject blue-green or blue).
(b)(i) [1 mark] Green precipitate.
(ii) [1 mark] Insoluble in excess / precipitate remains.
(c) [1 mark] Green precipitate.
[1 mark] Insoluble in excess / precipitate remains.
(d) [1 mark] Add dilute hydrochloric acid OR dilute nitric acid.
[1 mark] Add barium chloride solution OR barium nitrate solution.
[1 mark] White precipitate (consequent on adding a barium salt).
Note: If sulfuric acid is added, 0 marks for reagents.
(e) [1 mark] Add aqueous sodium hydroxide and heat/warm.
[1 mark] Gas / ammonia produced turns damp red litmus paper blue.

(a)(i) Sand
(ii) Sodium chloride solution (or aqueous sodium chloride / salt solution)
(b) Wash the residue (sand) with distilled water, then dry it in a warm oven (or leave it to dry in air/between filter papers).
(c) Heat the filtrate in an evaporating basin until a saturated solution is formed (or crystals begin to form on a glass rod). Leave the solution to cool so that crystals form. Filter the crystals from the remaining liquid and dry them on filter paper.
(d)(i) To monitor the temperature of the vapour to ensure it is at the boiling point of water (100 °C).
(ii) Cold water enters the condenser jacket at the bottom (lower inlet) and leaves at the top (upper outlet).

(a)(i) [1 mark] Sand / silicon dioxide.
(ii) [1 mark] Sodium chloride solution / aqueous sodium chloride. Reject: sodium chloride / salt / water alone.
(b) [1 mark] Wash the residue with distilled water.
[1 mark] Dry in a warm oven or leave to dry in air / between filter papers. Reject: heat strongly with a Bunsen burner.
(c) [1 mark] Heat/evaporate the filtrate to crystallization point / until crystals start to form.
[1 mark] Leave to cool (and crystallize).
[1 mark] Filter the crystals and dry them (with filter paper / in a warm oven).
(d)(i) [1 mark] To measure/monitor the temperature of the vapour (to show that pure water is distilling / boiling point is 100 °C).
(ii) [1 mark] Water enters at the bottom/lower inlet.
[1 mark] Water leaves at the top/upper outlet.

(a) Method:
1. Use a volumetric pipette to measure a fixed volume (e.g., 25.0 cm³) of Brand A vinegar into a clean conical flask.
2. Add a few drops of phenolphthalein indicator to the flask.
3. Fill a burette with the 0.10 mol/dm³ sodium hydroxide solution, ensuring the jet is filled, and record the initial reading.
4. Slowly run the sodium hydroxide solution from the burette into the conical flask, swirling constantly, until the solution just turns from colourless to permanent pink.
5. Record the final burette reading and calculate the volume of sodium hydroxide added (titre).
6. Repeat the procedure for Brand B and Brand C vinegars using the same volume of vinegar.
7. The brand of vinegar that requires the largest volume (titre) of sodium hydroxide solution to neutralise it is the most concentrated.

(b) Wear safety goggles / eye protection (as sodium hydroxide is corrosive/an irritant).
(c) Colourless to pink (accept pale pink).
(d) A pipette is more accurate / precise than a measuring cylinder (it has a smaller percentage error).
(e) The volume of sodium hydroxide solution needed would be larger, because the water remaining in the burette dilutes the sodium hydroxide solution, requiring more of it to neutralise the same amount of acid.

(a) [6 marks] 1 mark for each of the following points, up to a maximum of 6 marks:
- Pipette / volumetric pipette used to measure a fixed volume of vinegar (into a conical flask).
- Add phenolphthalein indicator (to the vinegar in the flask).
- Burette filled with sodium hydroxide solution.
- Run sodium hydroxide from the burette into the flask with swirling until a permanent colour change occurs / end-point is reached.
- Record initial and final burette readings (to find the volume added).
- Repeat the procedure with the other brands of vinegar (Brand B and Brand C), keeping the volume of vinegar constant.
- Conclusion: the vinegar requiring the largest volume of sodium hydroxide is the most concentrated.

(b) [1 mark] Wear safety goggles / gloves. (Accept: wash hands if spilled; reject: wear lab coat alone).
(c) [1 mark] Colourless to pink / pale pink. (Both colours in the correct order are required).
(d) [1 mark] More accurate / higher precision / smaller error. (Reject: easier to use).
(e) [1 mark] The volume needed is larger because the sodium hydroxide is diluted by the water in the burette.