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1. Calculate the number of moles of \(\text{NaOH}\): \(n(\text{NaOH}) = \text{concentration} \times \text{volume} = 0.150\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 3.75 \times 10^{-3}\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation: \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Thus, \(n(\text{H}_2\text{SO}_4) = \frac{3.75 \times 10^{-3}\text{ mol}}{2} = 1.875 \times 10^{-3}\text{ mol}\). 3. Calculate the concentration of \(\text{H}_2\text{SO}_4\): \(\text{Concentration} = \frac{\text{moles}}{\text{volume}} = \frac{1.875 \times 10^{-3}\text{ mol}}{0.01875\text{ dm}^3} = 0.100\text{ mol/dm}^3\).

1 mark for the correct option A. [Method: Calculate moles of NaOH = 0.00375 mol; calculate moles of sulfuric acid = 0.001875 mol; divide by volume in dm³ to find concentration = 0.100 mol/dm³]

The pipette must be rinsed with water to clean it and then with the solution it will measure (sodium hydroxide) so that any remaining water drops do not dilute the solution. Similarly, the burette must be rinsed with water and then with the acid it will contain to prevent dilution.

1 mark for the correct option A.

The ion has a \(3-\text{ charge}\), meaning it has 3 more electrons than protons. Since it has 18 electrons, the number of protons is \(18 - 3 = 15\). The nucleon number (protons + neutrons) is 31. Therefore, the number of neutrons is \(31 - 15 = 16\).

1 mark for the correct option A. [Method: Protons = Electrons - charge = 18 - 3 = 15; Neutrons = Mass number - Protons = 31 - 15 = 16]

Across Period 3 from left to right (sodium to argon), the character of elements changes from metallic (e.g., sodium, magnesium) to non-metallic (e.g., sulfur, chlorine). Correspondingly, their oxides change from basic (e.g., sodium oxide is a basic metal oxide) to acidic (e.g., sulfur dioxide is an acidic non-metal oxide).

1 mark for the correct option B.

1. The reaction with sodium hydroxide forming a white precipitate that dissolves in excess indicates the presence of \(\text{Al}^{3+}\) or \(\text{Zn}^{2+}\) ions (calcium forms a white precipitate that does not dissolve, and magnesium also forms an insoluble precipitate). 2. No precipitate with barium nitrate indicates the absence of sulfate ions, ruling out zinc sulfate. 3. A white precipitate with silver nitrate indicates the presence of chloride ions. Therefore, the compound is aluminium chloride, \(\text{AlCl}_3\).

1 mark for the correct option A. [Method: NaOH test identifies amphoteric hydroxide of Al3+ or Zn2+; silver nitrate test identifies Cl-; absence of reaction with barium nitrate rules out sulfate]

1. The gas that turns limewater cloudy is carbon dioxide, which shows that solid \(Z\) is a carbonate. 2. The reaction of the metal ion with excess aqueous ammonia to form a dark blue solution is characteristic of copper(II) ions, \(\text{Cu}^{2+}\). Therefore, the green solid \(Z\) is copper(II) carbonate.

1 mark for the correct option A. [Method: Limewater test identifies carbonate; excess ammonia test identifies copper(II)]

The polymer has a continuous carbon-carbon backbone with no heteroatoms (such as oxygen or nitrogen) in the chain, which indicates it is an addition polymer. The repeating unit is \(-\text{CH}_2-\text{CH(Cl)}-\), which comes from the monomer chloroethene, \(\text{CH}_2=\text{CHCl}\).

1 mark for the correct option B.

1. Since only \(W\) and \(Y\) react with dilute acid, they are more reactive than \(X\) and \(Z\). 2. Since the oxide of \(Y\) is reduced by carbon but the oxide of \(W\) is not, \(W\) is more reactive than \(Y\) (i.e., \(W > Y\)). 3. Since \(Z\) displaces \(X\) from solution, \(Z\) is more reactive than \(X\) (i.e., \(Z > X\)). Combining these facts gives the order of reactivity: \(W > Y > Z > X\).

1 mark for the correct option A.

First, write the balanced chemical equation: \(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). Calculate the moles of \(\text{NaOH}\): \(n(\text{NaOH}) = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}\). Using the stoichiometry of the reaction, the moles of \(\text{H}_2\text{SO}_4\) required are: \(n(\text{H}_2\text{SO}_4) = 0.00250 / 2 = 0.00125\text{ mol}\). Finally, find the concentration of \(\text{H}_2\text{SO}_4\): \(C = 0.00125\text{ mol} / 0.0200\text{ dm}^3 = 0.0625\text{ mol/dm}^3\).

1 mark for the correct option B. Award 0 marks for incorrect options.

A volumetric pipette is designed to measure and deliver a highly accurate fixed volume (such as \(25.0\text{ cm}^3\)) of solution. A burette is designed to deliver variable volumes of liquid during titration.

1 mark for the correct option A.

An element with 15 protons is phosphorus (P). The mass number (nucleon number) is the sum of protons and neutrons: \(15 + 16 = 31\). The charge of the ion is determined by the difference between protons and electrons: \(15 - 18 = -3\). Therefore, the representation is \(^{31}_{15}\text{P}^{3-}\).

1 mark for the correct option B.

The evolution of a gas that turns limewater cloudy indicates the presence of carbonate ions (\(\text{CO}_3^{2-}\)). The formation of a light blue precipitate with aqueous sodium hydroxide, which is insoluble in excess, is the characteristic test for copper(II) ions (\(\text{Cu}^{2+}\)).

1 mark for the correct option A.

During addition polymerization, the double bond of propene opens up to link monomers together. The repeating unit is derived from the monomer by replacing the double bond with single bonds extending outside parentheses: \(-[\text{CH}_2-\text{CH}(\text{CH}_3)]-\).

1 mark for the correct option C.

Metal X is the most reactive because it is the only one that can react with cold water. Metal Y is more reactive than Z because it reacts with dilute hydrochloric acid while Z does not react with dilute acid. This gives the order of reactivity: X > Y > Z.

1 mark for the correct option A.

Methyl orange is yellow in alkaline solutions (such as aqueous sodium hydroxide). As acid is added, the solution is neutralized. The end-point occurs when the indicator changes to an orange color just before turning red with excess acid.

1 mark for the correct option A.

The standard test for the ammonium ion is to add aqueous sodium hydroxide and warm the mixture gently. This reaction liberates ammonia gas, \(\text{NH}_3\), which can be identified because it is alkaline and turns damp red litmus paper blue.

1 mark for the correct option B.

The balanced chemical equation for the reaction is: \(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). First, calculate the number of moles of sulfuric acid reacted: \(\text{moles of H}_2\text{SO}_4 = 0.100\text{ mol/dm}^3 \times (20.0 / 1000)\text{ dm}^3 = 0.00200\text{ mol}\). According to the reaction stoichiometry, 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{NaOH}\). Thus, the moles of \(\text{NaOH}\) present = \(2 \times 0.00200\text{ mol} = 0.00400\text{ mol}\). Finally, calculate the concentration of \(\text{NaOH}\): \(\text{concentration} = 0.00400\text{ mol} / 0.0250\text{ dm}^3 = 0.160\text{ mol/dm}^3\).

1 mark for the correct option B. (Award 1 mark for correct calculation showing 0.160 mol/dm³; reject 0.080 mol/dm³ which fails to use the 1:2 reaction ratio).

Hydrochloric acid is a strong acid and aqueous ammonia is a weak base. The titration of a strong acid with a weak base results in an acidic salt solution at the endpoint, which has a pH of approximately 5. Methyl orange is the suitable indicator for this titration as its pH transition range matches the acidic endpoint. Since the acid is added to the alkali in the flask, the starting mixture is alkaline (yellow) and changes to orange at the endpoint.

1 mark for the correct option A. Reject options suggesting phenolphthalein or thymolphthalein as they change color in the alkaline pH range which is unsuitable for a strong acid-weak base titration.

The ion has a 2- charge (\(X^{2-}\)) and contains 18 electrons. This means it has 2 more electrons than protons. Therefore, the number of protons is \(18 - 2 = 16\). The mass number (nucleon number) is the sum of protons and neutrons. Therefore, the number of neutrons is \(34 - 16 = 18\).

1 mark for the correct option A. Reject options with incorrect proton and neutron combinations.

In Period 3, sodium, magnesium, and aluminium are giant metallic structures. Silicon is a giant covalent structure. Phosphorus, sulfur, chlorine, and argon are simple molecular/atomic structures. Therefore, the structural trend changes from giant metallic, to giant covalent, to simple molecular.

1 mark for the correct option A. Reject incorrect structural, bonding, or electrical trends across Period 3.

The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). Note that chromium(III) ions also form a green precipitate but it is soluble in excess NaOH to give a green solution. The formation of a white precipitate with aqueous barium nitrate in acidic conditions indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the compound is iron(II) sulfate.

1 mark for the correct option C. Reject options indicating other cations or anions.

Heating a compound with aqueous sodium hydroxide to produce a gas that turns damp red litmus paper blue (ammonia gas) is the characteristic test for ammonium ions, \(\text{NH}_4^+\). Adding dilute hydrochloric acid to produce a gas that turns limewater cloudy (carbon dioxide) is the characteristic test for carbonate ions, \(\text{CO}_3^{2-}\). Therefore, compound \(Z\) is ammonium carbonate.

1 mark for the correct option A. Reject options with incorrect cation/anion pairings.

The reaction between a dicarboxylic acid and a diol forms a polyester. The linkage formed in a polyester is an ester linkage (\(-\text{COO}-\)). In condensation polymerization of these monomers, a molecule of water (\(\text{H}_2\text{O}\)) is eliminated for every ester linkage formed.

1 mark for the correct option C. Reject options suggesting amide linkages or other eliminated molecules.

Metal \(W\) reacts with dilute hydrochloric acid to produce hydrogen, indicating it is more reactive than hydrogen. Metal \(X\) does not react, indicating it is less reactive than hydrogen. Thus, \(W\) is more reactive than \(X\) (\(W > X\)). Since metal \(Y\) can reduce the oxide of metal \(W\), metal \(Y\) is more reactive than metal \(W\) (\(Y > W\)). Therefore, the order of reactivity from most to least reactive is \(Y \rightarrow W \rightarrow X\).

1 mark for the correct option C. Reject all other order sequences.

First, calculate the number of moles of sulfuric acid used in the titration: \(\text{moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3 = 0.100 \text{ mol/dm}^3 \times \frac{15.0}{1000} \text{ dm}^3 = 0.00150 \text{ mol}\). According to the balanced equation, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Therefore, \(\text{moles of } \text{NaOH} = 2 \times 0.00150 \text{ mol} = 0.00300 \text{ mol}\). Finally, find the concentration of the sodium hydroxide solution: \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00300 \text{ mol}}{0.0250 \text{ dm}^3} = 0.120 \text{ mol/dm}^3\).

1 mark for the correct option C.
- Award 1 mark for calculating the correct concentration of 0.120 mol/dm^3.
- Incorrect option A uses 1:2 ratio incorrectly.
- Incorrect option B assumes a 1:1 reaction ratio.

For the given ion \(^{34}_{16}X^{2-}\): The atomic number (bottom number) is 16, which represents the number of protons. The mass number (top number) is 34. The number of neutrons is found by subtracting the atomic number from the mass number: \(34 - 16 = 18\) neutrons. Since the ion has a \(2-\)\ charge, it has 2 more electrons than protons: \(16 + 2 = 18\) electrons. Therefore, there are 16 protons, 18 neutrons, and 18 electrons.

1 mark for the correct option B.
- Award 1 mark for identifying the correct combination of protons (16), neutrons (18), and electrons (18).

The formation of a cream precipitate with acidified silver nitrate indicates the presence of bromide ions (\(\text{Br}^-\)). The reaction of the solution with dropwise aqueous sodium hydroxide producing a green precipitate that is insoluble in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. Therefore, the salt is iron(II) bromide.

1 mark for the correct option B.
- Award 1 mark for identifying iron(II) bromide based on correct anion and cation test results.

Polyesters are formed via condensation polymerisation when dicarboxylic acid monomers react with diol monomers. Each ester linkage is created with the elimination of a water molecule. A dicarboxylic acid and a diamine would react to form a polyamide (such as nylon), not a polyester.

1 mark for the correct option B.
- Award 1 mark for identifying that a dicarboxylic acid and a diol react to form a polyester.

A more reactive metal will displace a less reactive metal from its aqueous salt solution. Since \(X\) displaces all three other metals, it is the most reactive. \(Y\) displaces two metals (\(W\) and \(Z\)), so it is second. \(Z\) displaces only one metal (\(W\)), so it is third. \(W\) cannot displace any other metal, making it the least reactive. Therefore, the order of reactivity is \(X \rightarrow Y \rightarrow Z \rightarrow W\).

1 mark for the correct option A.
- Award 1 mark for ordering the metals correctly from most to least reactive.

Methyl orange is an indicator that is yellow in alkaline solutions (such as aqueous sodium hydroxide) and red in highly acidic solutions. During a titration where acid is added to an alkali, the mixture starts as yellow. At the exact neutralisation end-point, the indicator turns to its intermediate orange colour. Thus, the colour change observed is yellow to orange.

1 mark for the correct option A.
- Award 1 mark for identifying the correct end-point colour change of methyl orange when acid is added to alkali.

Isotopes are defined as atoms of the same element that contain the same number of protons (and thus the same atomic number and electronic structure) but different numbers of neutrons, resulting in different mass numbers. Because they have the same electronic configuration, they have identical chemical properties.

1 mark for the correct option C.
- Award 1 mark for selecting the correct definition/statement regarding isotopes.

By identifying the repeating unit in the polymer chain, we see it is \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\). This repeating unit contains three carbon atoms in total. The corresponding monomer must have a double bond between the carbon atoms of the main chain: \(\text{CH}_2=\text{CH}-\text{CH}_3\). This hydrocarbon with three carbons and one carbon-carbon double bond is propene. Its polymer is poly(propene).

1 mark for the correct option D.
- Award 1 mark for correctly identifying propene as the monomer from the given addition polymer structure.

1. Moles of \( \text{NaOH} \) used = \( \frac{20.0}{1000} \times 0.100 = 0.00200 \text{ mol} \). 2. From the balanced equation, 2 moles of \( \text{NaOH} \) react with 1 mole of \( \text{H}_2\text{SO}_4 \). So, moles of \( \text{H}_2\text{SO}_4 \) = \( \frac{0.00200}{2} = 0.00100 \text{ mol} \). 3. Concentration of \( \text{H}_2\text{SO}_4 \) = \( \frac{0.00100}{25.0 / 1000} = 0.0400 \text{ mol/dm}^3 \).

Award 1 mark for the correct calculation of concentration: 0.040 mol/dm³.

Rinsing the conical flask with the reactant solution before starting adds unmeasured moles of that reactant (due to remaining droplets), leading to an inaccurate titration volume. Distilled water does not add any extra moles of the reactant.

Award 1 mark for identifying the conical flask as the correct glassware.

Thymolphthalein is blue in basic solutions and colorless in acidic or neutral solutions. Since the conical flask initially contains sodium hydroxide (an alkali), the solution is blue. At the end-point, when neutralized by the hydrochloric acid, the solution turns colorless.

Award 1 mark for the correct color change from blue to colorless.

An ion with a 2- charge has 2 more electrons than protons. Since it has 18 electrons, the number of protons is \( 18 - 2 = 16 \). The number of neutrons is found by subtracting the protons from the mass number: \( 34 - 16 = 18 \).

Award 1 mark for identifying the correct combination: 16 protons and 18 neutrons.

The period number indicates the number of occupied electron shells (Period 3 means 3 shells). The group number indicates the number of valency (outer-shell) electrons (Group VI means 6 valency electrons). Y is sulfur, which forms stable \( \text{Y}^{2-} \) ions and is a non-metal.

Award 1 mark for the statement that the element has 3 occupied electron shells and 6 valency electrons.

Test 1 indicates the presence of iron(II) ions, \( \text{Fe}^{2+} \), because they form a green precipitate with sodium hydroxide that is insoluble in excess (unlike chromium(III) which dissolves in excess). Test 2 indicates the presence of sulfate ions, \( \text{SO}_4^{2-} \), because they form a white precipitate of barium sulfate with acidified barium nitrate. Thus, the salt is iron(II) sulfate.

Award 1 mark for identifying the salt as iron(II) sulfate.

Poly(ethene) is made by addition polymerisation, which forms no byproduct. Nylon is a polyamide, made by condensation polymerisation, which eliminates water molecules as a byproduct.

Award 1 mark for identifying the correct row (Row A).

X is the most reactive because it displaces all other metals (W, Y, Z) from their oxides. Z is the next most reactive because it displaces W and Y. W is next because it displaces only Y. Y is the least reactive because it cannot displace any other metal. The reactivity order is X > Z > W > Y.

Award 1 mark for identifying the correct reactivity order: X to Z to W to Y.

First, calculate the number of moles of NaOH used: \(n(\text{NaOH}) = 0.0250 \text{ dm}^3 \times 0.100 \text{ mol/dm}^3 = 0.00250 \text{ mol}\). According to the stoichiometric ratio, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Thus, \(n(\text{H}_2\text{SO}_4) = 0.00250 \text{ mol} / 2 = 0.00125 \text{ mol}\). Finally, calculate the volume of sulfuric acid: \(V = n / c = 0.00125 \text{ mol} / 0.0500 \text{ mol/dm}^3 = 0.0250 \text{ dm}^3 = 25.0 \text{ cm}^3\).

[1] B is the correct answer. Method: 1 mark for applying mole ratio and calculating the correct volume of 25.0 cm³.

The ion \(\text{X}^{3-}\) has 18 electrons in total (2 + 8 + 8). Since the ion has a 3- charge, it has gained 3 electrons compared to the neutral atom. Therefore, the neutral atom has \(18 - 3 = 15\) electrons. Since the proton number equals the number of electrons in a neutral atom, the proton number is 15 (which is phosphorus). An element with 15 electrons has an electronic configuration of 2,8,5. Because it has 5 electrons in its outer shell, it belongs to Group V (or Group 15).

[1] A is the correct answer. 1 mark for identifying the correct Group (V) and proton number (15).

Iron(II) ions (\(\text{Fe}^{2+}\)) react with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide, which is insoluble in excess sodium hydroxide. Chromium(III) (\(\text{Cr}^{3+}\)) also forms a green precipitate, but it is soluble in excess sodium hydroxide to give a green solution.

[1] B is the correct answer. 1 mark for identifying the cation as Iron(II) based on insolubility in excess NaOH.

A polyamide (such as nylon) is formed by condensation polymerisation of a dicarboxylic acid and a diamine. This reaction eliminates water molecules and forms amide linkages, -CONH-, between the monomer units.

[1] B is the correct answer. 1 mark for recalling that reacting dicarboxylic acids with diamines forms amide linkages.

Since Metal W displaces both X and Y, W is more reactive than both X and Y. Metal Y displaces X, meaning Y is more reactive than X (Y > X). Metal Z cannot displace any other metal, so Z is the least reactive of all four. Putting these observations together gives the reactivity order: W > Y > X > Z.

[1] A is the correct answer. 1 mark for correctly determining the relative reactivity order of all four metals.

A volumetric pipette is calibrated to measure and deliver a single fixed volume (e.g., 25.0 cm³) with extremely high accuracy. A burette is designed to deliver variable, controlled volumes accurately, which is essential for finding the precise equivalence point (end-point) of the titration.

[1] A is the correct answer. 1 mark for identifying the correct functions and precision requirements of pipettes and burettes.

The repeating unit of the addition polymer is -CH(CH3)-CH2-. To find the monomer, we locate the two carbon atoms in the main chain of the repeating unit and add a double bond between them: CH(CH3)=CH2. This is propene, which contains 3 carbon atoms and one carbon-carbon double bond.

[1] B is the correct answer. 1 mark for identifying the monomer of the polymer as propene.

Effervescence with acid producing a gas that turns limewater cloudy indicates the presence of carbonate ions (\(\text{CO}_3^{2-}\)). The addition of aqueous silver nitrate to the acidified solution produces a white precipitate, indicating the presence of chloride ions (\(\text{Cl}^-\)), because silver chloride (\(\text{AgCl}\)) is a white, insoluble solid.

[1] A is the correct answer. 1 mark for correctly identifying carbonate (from gas/limewater test) and chloride (from silver nitrate test).

First, calculate the volume of dilute sulfuric acid used from the burette: \( 21.20\text{ cm}^3 - 5.20\text{ cm}^3 = 16.00\text{ cm}^3 \). Next, calculate the number of moles of sodium hydroxide used: \( \text{moles of NaOH} = \frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol} \). From the balanced equation, \( 2 \) moles of \( \text{NaOH} \) react with \( 1 \) mole of \( \text{H}_2\text{SO}_4 \). Therefore, the number of moles of sulfuric acid reacted is: \( \text{moles of H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol} \). Finally, calculate the concentration of the sulfuric acid: \( \text{concentration} = \frac{0.00125\text{ mol}}{16.00 / 1000\text{ dm}^3} = 0.078125\text{ mol/dm}^3 \approx 0.0781\text{ mol/dm}^3 \).

1 mark for the correct option B. Award marks for identifying: (1) Volume of acid = 16.00 cm3, (2) Moles of NaOH = 0.00250 mol, (3) Moles of H2SO4 = 0.00125 mol, (4) Concentration = 0.0781 mol/dm3.

The conical flask is the vessel in which the reaction takes place and is used to hold the transferred reactant from the pipette. It must be rinsed with distilled water only. Any remaining water on the walls of the conical flask does not change the amount of moles of reactant delivered to it by the pipette. If it were rinsed with the solution it will contain, extra reactant would remain on the walls, introducing systematic error. The pipette and burette must be rinsed with their respective solutions to prevent dilution.

1 mark for B. Conical flask is rinsed with distilled water only to prevent adding extra moles of analyte or reagent. Rinsing pipette and burette with water would dilute the solutions used, so they must be rinsed with the solution they will contain.

The number of protons (atomic number) is 16, which corresponds to the element sulfur (\( \text{S} \)). The mass number (nucleon number) is the sum of protons and neutrons: \( 16 + 18 = 34 \). Since the particle has 18 electrons and 16 protons, there is a surplus of 2 electrons, giving it a charge of \( 2- \). Therefore, the correct chemical symbol is \( ^{34}_{16}\text{S}^{2-} \).

1 mark for A. Deduce: (1) Protons = 16 gives sulfur, S. (2) Mass number = 34. (3) 18 electrons vs 16 protons gives a charge of 2-.

Across Period 3: Sodium, magnesium, and aluminium are metals with metallic bonding. Silicon is a metalloid with a giant covalent structure. Phosphorus, sulfur, chlorine, and argon are non-metals with simple molecular/atomic structures. Thus, the bonding changes from metallic to giant covalent to simple molecular. The number of outer-shell electrons increases, elements change from metals to non-metals, and oxides change from basic to acidic.

1 mark for C. Reject A (metals to non-metals), reject B (outer-shell electrons increase), reject D (basic to acidic oxides).

The flame test produces a lilac flame, which is the characteristic test result for potassium ions (\( \text{K}^+ \)). Acidifying the solution and adding aqueous silver nitrate yields a cream precipitate, which indicates the presence of bromide ions (\( \text{Br}^- \)). Thus, the salt is potassium bromide (\( \text{KBr} \)).

1 mark for B. Lilac flame indicates potassium. Cream precipitate with acidified silver nitrate indicates bromide.

Carbonate ions react with dilute acids to produce carbon dioxide gas. The identity of this gas is confirmed by bubbling it through limewater, which turns cloudy/milky. Option A describes the test for ammonia gas. Option B describes the test for oxygen gas. Option D describes the test for sulfate ions, although carbonates would also form a precipitate, it would dissolve in the nitric acid with effervescence, whereas sulfate's precipitate remains insoluble.

1 mark for C. Carbonates are tested by adding acid and testing the resulting gas (carbon dioxide) with limewater, yielding a cloudy/milky appearance.

The repeating unit has an ester linkage (\( -\text{O-CO}- \)) in the polymer backbone. This shows it is a polyester. It is a synthetic polymer formed by condensation polymerisation of a diol (\( \text{HO-CH}_2\text{-CH}_2\text{-OH} \)) and a dicarboxylic acid (\( \text{HOOC-C}_6\text{H}_4\text{-COOH} \)) with the elimination of water molecules.

1 mark for C. Reject A (it is a condensation polymer), reject B (polyamide has amide links), reject D (it is synthetic, not natural).

To determine the reactivity: (1) Metal \( X \) displaces three other metals, so \( X \) is the most reactive. (2) Metal \( Z \) reacts with two metals (\( W \) and \( Y \)), so \( Z \) is more reactive than \( W \) and \( Y \). (3) Metal \( W \) only reacts with \( Y \), so \( W \) is more reactive than \( Y \). (4) Metal \( Y \) shows no reaction at all, meaning it cannot displace any other metal and is the least reactive. Thus, the order from most reactive to least reactive is: \( X \rightarrow Z \rightarrow W \rightarrow Y \).

1 mark for A. Count the number of displacement reactions for each metal: X = 3, Z = 2, W = 1, Y = 0. The correct sequence is therefore X -> Z -> W -> Y.

First, calculate the number of moles of sodium hydroxide: moles of \text{NaOH} = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}. From the balanced chemical equation, 2 moles of \text{NaOH} react with 1 mole of \text{H}_2\text{SO}_4. Therefore, moles of \text{H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol}. Now, calculate the concentration of sulfuric acid: concentration = \frac{\text{moles}}{\text{volume}} = \frac{0.00125\text{ mol}}{0.0200\text{ dm}^3} = 0.0625\text{ mol/dm}^3.

1 mark for the correct calculation leading to B.

Rinsing with distilled water ensures that any previous impurities or chemical residues are washed out. However, distilled water left on the inside walls would dilute the acid being added. Rinsing next with a small volume of the acid coats the inner surfaces, ensuring that the final filled acid remains at its exact concentration.

1 mark for identifying the correct preparation steps.

The stable ion X^{3-} has 2 + 8 + 8 = 18 electrons. Since it has a 3- charge, the neutral atom must have 3 fewer electrons than the ion, which is 18 - 3 = 15 electrons. In a neutral atom, the number of protons equals the number of electrons, so the atomic number of X is 15. The electronic configuration of the neutral atom is 2, 8, 5. With 5 outer-shell electrons, it belongs to Group V.

1 mark for the correct number of protons and group.

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Because they have the same proton number, they have identical electronic configurations. Since chemical properties are determined by the outer-shell electrons, isotopes have identical chemical properties.

1 mark for the correct comparison of isotopes.

The formation of a light blue precipitate with aqueous sodium hydroxide, which is insoluble in excess, indicates the presence of copper(II) ions, \text{Cu}^{2+}. The formation of a yellow precipitate with aqueous silver nitrate in the presence of dilute nitric acid indicates the presence of iodide ions, \text{I}^-. Therefore, the salt is copper(II) iodide.

1 mark for identifying the correct cation and anion.

The polymer shown is poly(chloroethene), which is an addition polymer. Addition polymers are formed from unsaturated monomers containing carbon-carbon double bonds. The monomer is chloroethene, \text{CH}_2=\text{CHCl}. It is non-biodegradable and does not contain ester or amide linkages.

1 mark for identifying the correct monomer and polymer type.

Since metal W displaces all three other metals, it is the most reactive. Since metal Y displaces both X and Z but not W, it is less reactive than W but more reactive than X and Z. Since metal X displaces Z but not Y, X is more reactive than Z. Thus, the order of reactivity from most to least reactive is W > Y > X > Z.

1 mark for the correct reactivity order sequence.

First, find the moles of \text{HCl} used: moles = 0.0500\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00125\text{ mol}. According to the chemical equation, 1 mole of \text{Ca(OH)}_2 reacts with 2 moles of \text{HCl}. Thus, moles of \text{Ca(OH)}_2 = \frac{0.00125}{2} = 0.000625\text{ mol}. The volume of the calcium hydroxide solution is 20.0\text{ cm}^3 = 0.0200\text{ dm}^3. Therefore, its concentration = \frac{0.000625\text{ mol}}{0.0200\text{ dm}^3} = 0.03125\text{ mol/dm}^3, which rounds to 0.0313\text{ mol/dm}^3.

1 mark for the correct calculation of concentration.

First, calculate the moles of \(\text{NaOH}\) reacted: \(\text{moles} = \text{concentration} \times \text{volume in dm}^3 = 0.150 \times (25.0/1000) = 0.00375\text{ mol}\). According to the equation, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Therefore, \(\text{moles of H}_2\text{SO}_4 = 0.00375 / 2 = 0.001875\text{ mol}\). Finally, calculate the concentration of \(\text{H}_2\text{SO}_4\): \(\text{concentration} = \text{moles} / \text{volume in dm}^3 = 0.001875 / (18.75/1000) = 0.100\text{ mol/dm}^3\).

Award 1 mark for the correct option B. Reject all other options.

A volumetric pipette is designed to measure and deliver exactly \(25.0\text{ cm}^3\) of a liquid. Methyl orange indicator is yellow in alkaline solutions and turns orange at the end-point when titrated with an acid (becoming red in excess acid).

Award 1 mark for the correct option B. Reject all other options.

The ion has a \(2-\)_charge, meaning it has 2 more electrons than protons. Therefore, the number of protons (atomic number) = \(18 - 2 = 16\). The nucleon (mass) number is the sum of protons and neutrons = \(16 + 18 = 34\).

Award 1 mark for the correct option A. Reject all other options.

For Group VII (the halogens), melting and boiling points increase down the group because the molecules become larger and the intermolecular forces become stronger. Reactivity of Group I increases down the group, and they react with water to form alkaline solutions.

Award 1 mark for the correct option B. Reject all other options.

Iron(II) ions, \(\text{Fe}^{2+}\), react with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide which is insoluble in excess. Sulfate ions, \(\text{SO}_4^{2-}\), react with barium ions to form an insoluble white precipitate of barium sulfate. Therefore, substance Y is iron(II) sulfate.

Award 1 mark for the correct option B. Reject all other options.

Ammonia is the only common alkaline gas, which turns damp red litmus paper blue. It reacts with hydrogen chloride gas to form dense white solid fumes of ammonium chloride.

Award 1 mark for the correct option D. Reject all other options.

Terylene is a polyester formed by condensation polymerisation between a dicarboxylic acid monomer and a diol monomer, with the elimination of water molecules.

Award 1 mark for the correct option B. Reject all other options.

Metal Y is the most reactive because it reacts with cold water. Metal X is moderately reactive because it reacts with steam but not cold water. Metal Z is the least reactive because it is below hydrogen in the reactivity series and does not react with dilute acid. Thus, the order is Y > X > Z.

Award 1 mark for the correct option B. Reject all other options.

1. Write the balanced chemical equation: \(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). 2. Calculate the moles of \(\text{NaOH}\) used: \(\text{moles} = 0.0250\text{ dm}^3 \times 0.150\text{ mol/dm}^3 = 0.00375\text{ mol}\). 3. Determine the moles of \(\text{H}_2\text{SO}_4\) reacted using the 2:1 mole ratio: \(\text{moles of H}_2\text{SO}_4 = 0.00375 \div 2 = 0.001875\text{ mol}\). 4. Calculate the concentration of the acid: \(\text{concentration} = 0.001875\text{ mol} \div 0.01875\text{ dm}^3 = 0.100\text{ mol/dm}^3\).

1 mark for the correct calculation leading to option B.

Swirling the flask continuously during the addition of acid ensures proper mixing, and adding the acid dropwise near the endpoint prevents overshooting the neutralisation point. Rinsing the conical flask with the sodium carbonate solution would introduce additional unmeasured reactant moles, while too much indicator can skew the titration result because indicators are themselves weak acids or bases.

1 mark for identifying the correct practical procedure for high accuracy (option C).

Since the ion \(X^{3+}\) has 18 electrons, the neutral atom of \(X\) must have \(18 + 3 = 21\) electrons, meaning its proton (atomic) number is 21. The nucleon number (mass number) is the sum of protons and neutrons: \(21 + 24 = 45\). Thus, statement A is correct.

1 mark for identifying the correct nucleon number based on atomic structure calculations (option A).

As you go down Group VII, the molecular size increases, resulting in stronger intermolecular forces of attraction (van der Waals forces). More thermal energy is required to overcome these forces, which causes the boiling points to increase down the group. Colors become darker, and reactivity decreases down the group.

1 mark for identifying the correct trend of boiling points down Group VII (option B).

The formation of a green precipitate with sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(Fe^{2+}\) (chromium(III) forms a green precipitate that is soluble in excess sodium hydroxide). The formation of a white precipitate with silver nitrate confirms the presence of chloride ions, \(Cl^-\). Therefore, the salt is iron(II) chloride.

1 mark for deducing the correct ionic identity of the salt from the qualitative analysis results (option B).

The polymer shown is an addition polymer. To find the monomer, identify the repeating unit, which is \(-CH_2-CH(CH_3)-\). Restoring the double bond between the two carbon atoms in the backbone yields \(CH_2=CH(CH_3)\), which is propene (or \(CH_3CH=CH_2\)).

1 mark for identifying the correct monomer propene from the polymer repeating unit (option B).

Nylon is a polyamide made by condensation polymerisation, containing amide linkages (\(-CONH-\)). Terylene is a polyester also made by condensation polymerisation, containing ester linkages (\(-COO-\)). Therefore, option C is correct.

1 mark for correctly matching the linkages and polymerisation class for both synthetic polymers (option C).

Since the oxide of Y cannot be reduced by carbon and metal Y reacts violently with acid, Y is the most reactive metal. X is of intermediate reactivity because it reacts slowly with acid and its oxide is reduced by carbon. Z is the least reactive because it does not react with acid and its oxide is reduced by carbon. The order of reactivity is therefore Y > X > Z.

1 mark for correctly ordering the three metals by decreasing reactivity (option B).

(a) A volumetric pipette is used to measure accurate fixed volumes.
(b) Reading at eye level prevents parallax error, and reading at the bottom of the meniscus ensures accuracy.
(c) Phenolphthalein is pink in alkaline solutions and turns colourless at the neutral end-point when acid is added.
(d)(i) Sulfuric acid is dibasic: \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\).
(ii) \text{Moles of } NaOH = \frac{18.40}{1000} \times 0.150 = 0.00276\text{ mol}.
(iii) Moles of \(H_2SO_4 = \frac{0.00276}{2} = 0.00138\text{ mol}\). Concentration of \(H_2SO_4 = \frac{0.00138}{0.0250} = 0.0552\text{ mol/dm}^3\).

(a) [1] Pipette (accept volumetric pipette, reject graduated pipette / measuring cylinder).
(b) [2] One mark for reading at the bottom of the meniscus; one mark for reading at eye level (or flat surface).
(c) [2] Pink [1] to colourless [1]. (Reject clear for colourless).
(d)(i) [2] Correct formula of reactants and products [1], correct balancing [1].
(ii) [1.3] \(0.00276\text{ mol}\) [1] (allow 1 mark for correct working \frac{18.40 \times 0.150}{1000}); correct sig figs/rounding [0.3].
(iii) [4] Moles of acid = moles of alkali / 2 = 0.00138\text{ mol} [1]; divided by volume in dm3 (0.0250) [1]; concentration = \(0.0552\text{ mol/dm}^3\) [1]; correct units [1].

(a)(i) A weak acid partially ionises to produce hydrogen ions in water.
(ii) A strong base completely dissociates to produce hydroxide ions in water.
(b) Universal indicator shows a spectrum of colours and changes gradually, whereas a titration requires a single, sharp colour change at the end-point.
(c)(i) Ethanoic acid reacts with potassium hydroxide to form potassium ethanoate and water: \(CH_3COOH + KOH \rightarrow CH_3COOK + H_2O\).
(ii) \text{Moles of } CH_3COOH = \frac{24.60}{1000} \times 0.100 = 0.00246\text{ mol}\). Mole ratio is 1:1, so \text{moles of } KOH = 0.00246\text{ mol}\). \text{Concentration of } KOH \text{ in mol/dm}^3 = \frac{0.00246}{0.0200} = 0.123\text{ mol/dm}^3\). \text{Concentration in g/dm}^3 = 0.123 \times 56 = 6.888\text{ g/dm}^3\ (which is 6.89 to 3 s.f.).

(a)(i) [2] Acid partially [1] ionises/dissociates in aqueous solution [1].
(ii) [2] Base fully [1] ionises/dissociates in aqueous solution [1].
(b) [2] No sharp/sudden colour change [1]; has a continuous range of colours/gradual change [1].
(c)(i) [2] Correct products and reactants [1], correct balanced equation [1].
(ii) [4.3] Moles of acid = \(0.00246\text{ mol}\) [1]; moles of alkali = \(0.00246\text{ mol}\) [1]; conc in mol/dm3 = \(0.123\) [1]; conc in g/dm3 = \(6.89\) [1] (allow 6.888); correct final answer to 3 sig figs with units [0.3].

(a) Proton number is the number of protons in the nucleus of an atom.
(b) Chlorine-37 (Cl, Z=17) with a -1 charge has 17 protons, 37 - 17 = 20 neutrons, and 17 + 1 = 18 electrons. Potassium-39 (K, Z=19) with a +1 charge has 19 protons, 39 - 19 = 20 neutrons, and 19 - 1 = 18 electrons.
(c) Chemical properties depend on the number and arrangement of valence electrons, which are identical for isotopes.
(d)(i) Elements are ordered by increasing atomic/proton number.
(ii) Vertical columns (groups) contain elements with the same number of outer shell electrons; horizontal rows (periods) contain elements with the same number of electron shells.
(e) A calcium atom has configuration 2, 8, 8, 2. The Ca2+ ion loses 2 outer electrons, leaving 2, 8, 8.

(a) [1] Number of protons in the nucleus.
(b) [4.3] Chlorine ion: 17 protons [0.5], 20 neutrons [0.5], 18 electrons [1]. Potassium ion: 19 protons [0.5], 20 neutrons [0.5], 18 electrons [1]. Clarity of response [0.3].
(c) [2] Same number of outer shell/valence electrons [1], chemical reactions depend on outer shell electrons [1].
(d)(i) [1] Increasing proton/atomic number.
(ii) [2] Groups: same number of outer shell electrons [1]; Periods: same number of electron shells [1].
(e) [2] Correct electronic configuration 2, 8, 8 [2] (allow 1 mark for showing correct shell capacity but wrong outer shell).

(a)(i) In Group I, reactivity increases down the group. This is because atomic radius increases, shielding increases, so the electrostatic attraction between the positive nucleus and negative outer electron decreases, making the outer electron easier to lose.
(ii) Down Group VII, molecular size increases, meaning intermolecular forces (Van der Waals) become stronger, so melting point increases. Density also increases.
(b) Noble gases have a stable octet (or duet for helium) outer shell, so they do not need to gain, lose, or share electrons.
(c)(i) Phosphorus has 15 electrons, which has electronic configuration 2, 8, 5.
(ii) 5 valence electrons means Group V. 3 occupied shells means Period 3.
(iii) Magnesium forms \(Mg^{2+}\) ions. Phosphorus forms \(P^{3-}\) ions. To balance the charges, the formula is \(Mg_3P_2\).

(a)(i) [3] Reactivity increases [1]; outer electron is further from nucleus / more shells / more shielding [1]; weaker attraction to nucleus / easier to lose electron [1].
(ii) [2] Density increases [1]; melting/boiling point increases [1].
(b) [2] Full outer shell of electrons [1]; stable electronic configuration / no tendency to lose/gain/share electrons [1].
(c)(i) [1] Phosphorus (accept P).
(ii) [2] Group V / 15 [1]; Period 3 [1].
(iii) [2.3] \(Mg_3P_2\) [2] (allow 1 mark for correct ions \(Mg^{2+}\) and \(P^{3-}\)), correct formatting [0.3].

(a)(i) Sodium hydroxide test: a light blue precipitate of copper(II) hydroxide is formed, which does not dissolve in excess. Aqueous ammonia test: a light blue precipitate is formed, which dissolves in excess ammonia to form a deep blue solution.
(ii) Sulfate test: acidify with dilute hydrochloric acid or nitric acid, then add barium chloride or barium nitrate solution. A white precipitate of barium sulfate forms.
(b)(i) Heating hydrated copper(II) sulfate removes water of crystallisation, leaving anhydrous copper(II) sulfate.
(ii) The liquid is water. Anhydrous cobalt(II) chloride paper turns from blue to pink in the presence of water, or white anhydrous copper(II) sulfate turns blue.
(c) Test for ammonium ion: add aqueous sodium hydroxide and heat gently. Ammonia gas is evolved, which is alkaline and turns damp red litmus paper blue.

(a)(i) [4] NaOH test: light blue precipitate [1]; insoluble in excess [1]. Ammonia test: light blue precipitate [1]; dissolves in excess to give deep/dark blue solution [1].
(ii) [3] Acidify with dilute nitric/hydrochloric acid [1]; add barium nitrate/chloride solution [1]; white precipitate [1] (reject acidify with sulfuric acid).
(b)(i) [1] Anhydrous copper(II) sulfate.
(ii) [2] Add to anhydrous cobalt(II) chloride [1]; turns from blue to pink [1] (OR anhydrous copper(II) sulfate [1] turns from white to blue [1]).
(c) [2.3] Add aqueous sodium hydroxide and warm [1]; gas evolved turns damp red litmus paper blue [1]; identification of ammonia gas [0.3].

(a)(i) Monomer: A small molecule that joins together with other monomers to form a polymer.
(ii) Polymer: A large, long-chain molecule made of many repeating units joined by covalent bonds.
(b)(i) The monomer of poly(ethene) is ethene: a carbon-carbon double bond with four hydrogen atoms attached: \(H_2C=CH_2\).
(ii) The polymer contains single carbon-carbon bonds. Three repeating units mean a chain of 6 carbons, each with 2 hydrogens: \(-(CH_2-CH_2)_3-\).
(c)(i) Nylon is made by condensation polymerisation, releasing water or hydrogen chloride.
(ii) The amide linkage consists of a carbonyl group bonded to an amine group: \(-C(=O)-N(-H)-\).
(iii) Proteins have the same amide (peptide) linkage.
(d) Non-biodegradability means they persist in landfills, taking up valuable land space and causing physical pollution.
(e) Modern polymers can be made biodegradable (containing linkages like esters that can be hydrolysed by microbes) or photodegradable (broken down by UV light).

(a)(i) [1] Simple molecule / starting material that can be linked to form a polymer.
(ii) [1] Large molecule / macromolecule made of many repeating monomer units.
(b)(i) [1] Correct drawing of ethene showing C=C double bond and 4 C-H single bonds.
(ii) [3] 6 carbon atoms in a single-bonded chain [1]; correct hydrogens on all carbons [1]; open-ended continuation bonds at both ends [1].
(c)(i) [1] Condensation polymerisation.
(ii) [2] Correct amide linkage showing \(-C(=O)-N(-H)-\) with all bonds [2] (allow 1 mark if C=O and N-H are shown but continuation bonds are incorrect).
(iii) [1] Protein / polypeptide.
(d) [1.3] Non-biodegradable [1]; stays in landfills for a very long time / fills up landfill space [0.3].
(e) [2] Made biodegradable [1]; made photodegradable / recyclable [1].

(a) Decreasing reactivity order: Sodium > Calcium > Magnesium > Iron > Copper.
(b)(i) Magnesium is more reactive than copper, so it displaces copper from copper(II) sulfate solution. Observations: the blue colour of the solution fades (as \(Cu^{2+}\) ions are removed), a red-brown solid (copper) is deposited, and the mixture gets warm (exothermic reaction).
(ii) Ionic equation: \(Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)\).
(iii) Copper is lower in the reactivity series than iron, so it is less reactive and cannot displace iron ions from solution.
(c)(i) Equation: \(2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(l)\).
(ii) Iron(III) oxide is the oxidising agent. It loses oxygen to become iron (or the iron(III) ion, \(Fe^{3+}\), gains three electrons to form iron atoms: \(Fe^{3+} + 3e^- \rightarrow Fe\)).

(a) [2] Correct order: Sodium, Calcium, Magnesium, Iron, Copper [2] (allow 1 mark if one metal is misplaced).
(b)(i) [2] Blue solution fades / becomes colourless [1]; red-brown solid forms [1] (accept container gets warm).
(ii) [2] Correct ionic species and products [1]; correct state symbols [1].
(iii) [2] Copper is less reactive than iron [1]; cannot displace iron from its compound [1].
(c)(i) [2] Correct reactants and products [1]; correct balancing [1].
(ii) [2.3] Iron(III) oxide / \(Fe_2O_3\) (or \(Fe^{3+}\)) [1]; because it loses oxygen / iron ion gains electrons (is reduced) [1]; clarity of explanation [0.3].

(a) Methyl orange is a suitable indicator for strong acid-weak base titrations. It is yellow in alkali, and turns red/pink at the end-point.
(b)(i) \text{Moles of } HCl = \frac{25.00}{1000} \times 0.100 = 0.0025\text{ mol}.
(ii) From the equation, 2 moles of \(HCl\) react with 1 mole of \(Na_2CO_3\). Therefore, \text{moles of } Na_2CO_3 = \frac{0.0025}{2} = 0.00125\text{ mol}.
(iii) The sample titrated was \(25.0\text{ cm}^3\). The total volume of the solution is \(250\text{ cm}^3\), which is 10 times larger. Total moles of \(Na_2CO_3 = 0.00125 \times 10 = 0.0125\text{ mol}.
(iv) The relative formula mass, \)M_r\), is \frac{\text{mass}}{\text{moles}} = \frac{3.575\text{ g}}{0.0125\text{ mol}} = 286.
(v) The \(M_r\) of anhydrous \(Na_2CO_3 = 2 \times 23 + 12 + 3 \times 16 = 106.
Mass of \)xH_2O = 286 - 106 = 180.
Since the \(M_r\) of water is 18, \(18x = 180 \Rightarrow x = 10.
So the formula is \)Na_2CO_3 \cdot 10H_2O\).

(a) [3] Methyl orange [1]; red (or pink) in acid [1]; yellow in alkali [1]. (Accept phenolphthalein: colourless in acid [1]; pink in alkali [1]).
(b)(i) [1.3] \(0.0025\text{ mol}\) [1]; calculation shown correctly [0.3].
(ii) [2] Moles of acid / 2 [1]; \(0.00125\text{ mol}\) [1] (consequential on (b)(i)).
(iii) [2] Moles in sample \times 10 [1]; \(0.0125\text{ mol}\) [1] (consequential on (b)(ii)).
(iv) [2] Mass / total moles [1]; 286 [1] (consequential on (b)(iii)).
(v) [2] Calculation of \(M_r\) of anhydrous salt (106) and subtraction from 286 [1]; \(x = 10\) [1].

(a) Methyl orange is a suitable indicator for a strong acid-weak base titration. The initial color in alkaline sodium carbonate is yellow, and the end-point is reached when it turns to a stable orange or pink-orange color.

(b)(i) Moles of \(\text{HCl} = \text{concentration} \times \text{volume in dm}^3 = 0.150 \times (22.40 / 1000) = 0.00336\text{ mol}\).

(ii) From the equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\). Moles of \(\text{Na}_2\text{CO}_3 = 0.00336 / 2 = 0.00168\text{ mol}\).

(iii) Concentration of \(\text{Na}_2\text{CO}_3 = \text{moles} / \text{volume in dm}^3 = 0.00168 / 0.0250 = 0.0672\text{ mol/dm}^3\).

(iv) Molar mass of \(\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g/mol}\).
Concentration in \(\text{g/dm}^3 = 0.0672 \times 106.0 = 7.12\text{ g/dm}^3\).

(c) A volumetric pipette has a much smaller percentage error and is designed to deliver an extremely precise, fixed volume (e.g., \(25.0\text{ cm}^3\)), whereas a measuring cylinder is less accurate and has a larger tolerance.

Part (a):
- Methyl orange (1)
- Color change: yellow (1) to orange / pink-orange / red (1) [OR Phenolphthalein (1), pink (1) to colorless (1)]

Part (b):
- (i) \(0.00336\text{ mol}\) (2) [1 mark for formula/working \(0.150 \times 0.0224\)]
- (ii) \(0.00168\text{ mol}\) (2) [1 mark for dividing answer (i) by 2]
- (iii) \(0.0672\text{ mol/dm}^3\) (2) [1 mark for formula/working \(\text{moles} / 0.025\)]
- (iv) \(7.12\text{ g/dm}^3\) (2) [1 mark for finding \(M_r = 106\), 1 mark for multiplying conc by 106]

Part (c):
- Pipette is more accurate / precise / has smaller percentage error than a measuring cylinder (1.3)

(a)(i) Nylon is a polyamide formed by the condensation reaction between a diamine and a dicarboxylic acid. Therefore, the reacting functional groups are the amine group (\(-\text{NH}_2\)) and the carboxylic acid group (\(-\text{COOH}\)) (or acid chloride group, \(-\text{COCl}\)).

(ii) The structure should show the amide link (\(-\text{NH}-\text{CO}-\)) joining the two blocks: \(\text{[-NH-}\Box\text{-NH-CO-}\Box\text{-CO-]_n}\) with open/continuation bonds at each end to indicate a polymer chain.

(iii) In the condensation reaction between a diamine and a dicarboxylic acid, water (\(\text{H}_2\text{O}\)) is eliminated.

(b)(i) In addition polymerisation, the monomer is unsaturated (contains a \(\text{C}=\text{C}\) double bond) and forms only one product (the polymer). In condensation polymerisation, monomers have two functional groups and form both the polymer and a small molecule (like water).

(ii) Poly(ethene) is made of repeating \(-\text{CH}_2-\text{CH}_2-\) units. A section with four carbon atoms is: \(-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\) with continuation bonds at each end.

(c) Non-biodegradable plastics persist in landfills for hundreds of years, taking up space and posing ingestion/entanglement hazards to wildlife. Incineration/burning of plastics releases toxic fumes (such as carbon monoxide or highly acidic hydrogen chloride gas from PVC) and carbon dioxide, which is a major greenhouse gas contributing to global warming.

Part (a):
- (i) Amine / amino (1) and carboxylic acid / carboxyl (1) (accept dicarboxylic acid / acyl chloride)
- (ii) Structure showing amide link \(-\text{NH}-\text{CO}-\) correctly linked (1); two correct block shapes representing monomer chains (1); continuation bonds at both ends (1)
- (iii) Water / \(\text{H}_2\text{O}\) (1) (accept hydrogen chloride / \(\text{HCl}\))

Part (b):
- (i) Addition polymerisation produces only one product / polymer (1); condensation polymerisation produces a polymer and a small molecule (e.g. water) (1) [OR addition requires carbon-carbon double bonds, condensation requires functional groups (1+1)]
- (ii) Four carbon atoms in a single-bonded chain (1); correct number of hydrogen atoms (8) and continuation bonds at both ends (1)

Part (c):
- (i) Landfills take up space because plastics do not decompose / ingestion/harm to wildlife (1); burning produces toxic gases / carbon dioxide (greenhouse gas) (1.3 for both detailed explanation points)

(a)(i) A cation is a positively charged ion. Particle A has 11 positive protons and 10 negative electrons, resulting in a net charge of \(1+\).

(ii) Halogens belong to Group VII and have 7 outer-shell electrons. Particle C is a neutral atom (17 protons, 17 electrons) with electronic configuration 2,8,7, which has 7 valence electrons, identifying it as a halogen.

(iii) Isotopes are particles of the same element with the same number of protons but different numbers of neutrons. Particles B and C both have 17 protons (chlorine), but B has 18 neutrons while C has 20 neutrons. Thus, B is an isotope of C (in ionic form).

(b)(i) Particle D has 12 electrons. Its electronic configuration is 2,8,2.

(ii) Group 2: because it has 2 electrons in its outer shell. Period 3: because it has 3 electron shells in total.

(c) Particle A has 11 protons and 10 electrons, so it is a cation with a charge of \(1+\) (\(\text{A}^+\)). Particle B has 17 protons and 18 electrons, so it is an anion with a charge of \(1-\space\) (\(\text{B}^-\)). The charges are equal and opposite, so they combine in a 1:1 ratio to form the neutral ionic compound \(\text{AB}\) (or \(\text{NaCl}\)).

Part (a):
- (i) Particle A (1) and explanation: has more protons than electrons / has 11 protons and 10 electrons / net positive charge (1)
- (ii) Particle C (1) and explanation: has 17 protons/electrons, electronic configuration 2,8,7 / has 7 outer shell electrons (1)
- (iii) Particle B (1) and explanation: same number of protons / both have 17 protons (1) but different number of neutrons / B has 18 and C has 20 (0.3)

Part (b):
- (i) 2,8,2 (1)
- (ii) Group 2 (1); Period 3 (1); Explanation: group number equals number of outer-shell electrons (2) and period number equals number of occupied electron shells (3) (1 mark for both explanations combined)

Part (c):
- Formula: \(\text{AB}\) (or \(\text{NaCl}\)) (1)
- Explanation: \(\text{A}\) forms a \(1+\) ion and \(\text{B}\) forms a \(1-\space\) ion, so they combine in a 1:1 ratio to be electrically neutral (1)

(a)(i) Aqueous sodium hydroxide reacts with iron(II) ions to form a green precipitate of iron(II) hydroxide, \(\text{Fe(OH)}_2\).

(ii) On standing, the green precipitate turns brown because it is oxidized by atmospheric oxygen to iron(III) hydroxide, \(\text{Fe(OH)}_3\).

(iii) The ionic equation is: \(\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}\).

(b)(i) Adding aqueous barium chloride to a solution containing sulfate ions produces an insoluble white precipitate of barium sulfate.

(ii) This confirms the presence of the sulfate ion, \(\text{SO}_4^{2-}\).

(iii) Dilute hydrochloric acid is added to acidify the solution and react with any carbonate ions (\(\text{CO}_3^{2-}\)) or sulfite ions (\(\text{SO}_3^{2-}\)) present. If not removed, these ions would react with barium ions to form white precipitates of barium carbonate or barium sulfite, giving a false positive.

(c) To test for ammonium ions (\(\text{NH}_4^+\)), add aqueous sodium hydroxide to the solution and warm gently. Ammonia gas is released, which can be identified because it turns damp red litmus paper blue.

Part (a):
- (i) Green (1) precipitate / solid (1)
- (ii) Turns brown (1); due to oxidation by oxygen in the air / oxidation of Fe(II) to Fe(III) (1)
- (iii) \(\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}\) [1 mark for correct formulas, 1 mark for balancing, 0.3 mark for state symbols]

Part (b):
- (i) White precipitate (1)
- (ii) Sulfate / \(\text{SO}_4^{2-}\) (1)
- (iii) To remove/react with carbonate / sulfite ions (1) which would also form a white precipitate with barium chloride (1)

Part (c):
- Add aqueous sodium hydroxide and warm (1)
- Gas turns damp red litmus paper blue (1)

(a) Magnesium is the most reactive because it displaces zinc, lead, and copper. Zinc is next because it displaces lead and copper, but not magnesium. Lead is next because it only displaces copper. Copper is the least reactive as it cannot displace any of the other metals. Order: Magnesium > Zinc > Lead > Copper.

(b)(i) Zinc reacts with copper(II) nitrate to displace copper: \(\text{Zn} + \text{Cu(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Cu}\).

(ii) Zinc (\(\text{Zn}\)) is the reducing agent. During the reaction, zinc atoms lose two electrons to form zinc ions (\(\text{Zn}^{2+}\)), so zinc is oxidized. It reduces copper(II) ions (\(\text{Cu}^{2+}\)) by donating these electrons to them, forming copper metal.

(c)(i) Magnesium reacts with steam to form magnesium oxide and hydrogen gas: \(\text{Mg(s)} + \text{H}_2\text{O(g)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)}\).

(ii) During the reaction, magnesium burns with a bright white flame/light, leaving behind a white solid (magnesium oxide).

(d) Carbon can reduce copper oxide because copper is less reactive than carbon. However, aluminium is more reactive than carbon (higher in the reactivity series). Aluminium forms extremely stable bonds with oxygen, and carbon is not a strong enough reducing agent to remove oxygen from aluminium oxide.

Part (a):
- Magnesium > Zinc > Lead > Copper (2) [1 mark for partially correct order or reversed order]

Part (b):
- (i) \(\text{Zn} + \text{Cu(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Cu}\) (2) [1 mark for correct reactants and products, 1 mark for balancing]
- (ii) Zinc / \(\text{Zn}\) (1); explanation: zinc loses electrons / oxidation is loss (1); zinc transfers electrons to copper ions / reduces \(\text{Cu}^{2+}\) to \(\text{Cu}\) (1)

Part (c):
- (i) \(\text{Mg} + \text{H}_2\text{O} \rightarrow \text{MgO} + \text{H}_2\) (2) [1 mark for correct products \(\text{MgO}\) and \(\text{H}_2\), 1 mark for correct equation]
- (ii) Bright white flame / white solid formed (1)

Part (d):
- Aluminium is more reactive than carbon (1)
- Carbon cannot displace aluminium / cannot reduce aluminium oxide / aluminium has a stronger affinity for oxygen than carbon (1.3)

a(i) When aqueous sodium hydroxide is added to a solution containing copper(II) ions, a light blue precipitate of copper(II) hydroxide, \(Cu(OH)_2\), forms. This precipitate does not dissolve in excess sodium hydroxide.
a(ii) When aqueous ammonia is added dropwise to copper(II) ions, a light blue precipitate of \(Cu(OH)_2\) forms. In excess ammonia, the precipitate dissolves to form a soluble, deep blue tetraamminecopper(II) complex solution.
a(iii) Barium ions react with sulfate ions in an acidic solution to form an insoluble white precipitate of barium sulfate, \(BaSO_4\).
b) Copper(II) ions impart a characteristic blue-green color to a Bunsen burner flame.
c) Magnesium is more reactive than copper. It displaces copper from the solution: \(Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)\). The blue color of the \(Cu^{2+}\) solution fades to colorless as \(Mg^{2+}\) forms, and copper metal is deposited as a red-brown solid.
d) Based on the positive tests for \(Cu^{2+}\) and \(SO_4^{2-}\), the cation is copper(II) and the anion is sulfate.

- a(i): Blue precipitate [1]; insoluble in excess NaOH [1].
- a(ii): Blue precipitate [1]; dissolves/soluble in excess to form a deep blue solution [1].
- a(iii): White precipitate [1].
- b: Blue-green flame (accept: green flame) [1].
- c: Red-brown / brown solid forms [1]; blue solution fades / becomes colorless [1] (accept: effervescence).
- d: Cation: copper(II) / \(Cu^{2+}\) [1]; Anion: sulfate / \(SO_4^{2-}\) [1].

a(i) 15 + 3 = 18 \(cm^3\).
a(ii) Halfway between 30 and 40 is 35 \(cm^3\).
a(iii) 40 + 6 = 46 \(cm^3\).
b(i) Powdering the reactant increases the surface area, which increases the rate of reaction.
b(ii) A larger surface area increases the frequency of collisions between reactant particles (zinc and hydrogen ions), leading to more successful collisions per unit time.
c) Pushing the plunger in fully resets the starting volume to 0 \(cm^3\) and purges any trapped air that would cause an overestimation of the produced gas.
d) The acid is the limiting reactant, so once all the acid is consumed, the reaction stops.
e) An alternative setup involves downward displacement of water: a delivery tube from the flask runs under water into an inverted measuring cylinder filled with water. The gas displaces the water, allowing its volume to be read.

- a(i): 18 \(cm^3\) [1]
- a(ii): 35 \(cm^3\) [1]
- a(iii): 46 \(cm^3\) [1]
- b(i): Rate increases / reaction is faster [1]
- b(ii): Larger surface area [1]; more collisions per second / more frequent successful collisions [1]
- c: To reset starting volume to zero / ensure no air is trapped in syringe [1]
- d: Acid is used up / reacted fully / acid is the limiting reactant [1]
- e: Inverted measuring cylinder in a trough of water [1]; delivery tube connecting flask to the cylinder [1]

a) Titre = Final reading - Initial reading.
Titration 1: 23.4 - 0.0 = 23.4 \(cm^3\).
Titration 2: 34.6 - 10.1 = 24.5 \(cm^3\).
Titration 3: 26.9 - 3.5 = 23.4 \(cm^3\).
b) Concordant titres are within 0.2 \(cm^3\) of each other. Titration 1 (23.4) and Titration 3 (23.4) are identical. Titration 2 (24.5) is an outlier (anomalous).
Average = (23.4 + 23.4) / 2 = 23.4 \(cm^3\).
c) Methyl orange in alkali (NaOH) is yellow. At the neutralisation end-point, as acid is added, it turns orange (or pink if slightly over-titrated).
d)(i) Pipettes have a very narrow neck which reduces error in reading the meniscus, making them much more precise than measuring cylinders.
d)(ii) Practical steps include reading the meniscus at eye level to avoid parallax error, aligning the bottom of the meniscus with the calibration mark, or not blowing out the small droplet remaining at the tip.
e) Swirling ensures that the acid mixes immediately and completely with the alkali, giving an accurate color change at the exact neutralization point.

- a: Titration 1 titre = 23.4 \(cm^3\) [1]; Titration 2 titre = 24.5 \(cm^3\) [1]; Titration 3 titre = 23.4 \(cm^3\) [1].
- b: Selects Titrations 1 and 3 [1]; Calculates average = 23.4 \(cm^3\) [1] (reject if Titration 2 is included).
- c: Yellow [1] to orange / pink [1].
- d(i): More accurate / precise [1].
- d(ii): Read meniscus at eye level / bottom of meniscus on the line / do not blow out the last drop [1].
- e: To ensure thorough mixing / complete reaction [1].

An elegant experimental design for testing tensile strength must specify steps to guarantee a fair test.
1. Preparation of samples: The dimensions of the strips must be identical because a wider or longer strip behaves differently under tension.
2. Setup: Suspending the strip vertically from a clamp and attaching weights to the bottom allows gravity to apply a uniform downward force.
3. Procedure: Adding masses systematically (e.g., 50g at a time) ensures that the exact breaking point can be determined.
4. Variables & reliability: Multiple trials (at least 3) for each polymer are needed to calculate an average and identify anomalies. The temperature should be kept constant.
5. Conclusion: Comparing the average mass supported before breaking directly indicates strength; the higher the mass, the stronger the material.

- Prepare samples: Cut strips of the same length and width from each bag [1].
- Fair testing detail: Use a ruler and scissors to ensure dimensions are identical [1].
- Setup: Clamp the top of the plastic strip to a stand [1].
- Force application: Suspend a weight hanger from the bottom of the strip [1].
- Mass addition: Add masses systematically/one by one [1] (accept: add water slowly to a bucket).
- Measurement: Record the mass at which the strip breaks / snaps [1].
- Comparability: Repeat the procedure for the other two polymers [1].
- Reliability: Repeat the experiment multiple times (at least 3) for each polymer and calculate an average [1].
- Controlled variables: Keep temperature constant / use strips of the same thickness / cut from same direction of the bag [1].
- Conclusion: The strongest polymer is the one that supports the greatest average mass before breaking [1].