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

1. Sand is insoluble in water, so filtration (Y) is used to separate the solid sand from the soluble aqueous sodium chloride. 2. Nitrogen and oxygen in liquid air have different boiling points, so fractional distillation (X) is used to separate them. 3. Hexane and water are immiscible liquids that form two separate layers, so a separating funnel (Z) is used. 4. To obtain copper(II) sulfate crystals from its solution, the solution is heated to saturation and then allowed to cool during crystallization (W).

Award 1 mark for identifying the correct matching sequence (Option A).

1. Hematite is iron(III) oxide, which is the ore that acts as the primary source of iron (1-Q). 2. Coke is carbon, which burns to release heat and reacts to form carbon monoxide, the main reducing agent (2-S). 3. Limestone is calcium carbonate, which thermal decomposes to calcium oxide and reacts with silicon(IV) oxide to form liquid slag (3-P). 4. Air is blown into the bottom of the furnace to provide the oxygen required for the combustion of coke (4-R).

Award 1 mark for identifying the correct matching sequence (Option A).

I. Terylene is a polyester formed by condensation polymerization of a dicarboxylic acid and a diol, resulting in ester linkages (\(-COO-\)), so it matches with Z. II. Nylon is a polyamide formed by condensation polymerization of a dicarboxylic acid and a diamine, resulting in amide linkages (\(-CONH-\)), so it matches with Y. III. Poly(ethene) is an addition polymer formed from ethene monomers, resulting in a saturated hydrocarbon chain with only \(C-C\) single bonds, so it matches with X.

Award 1 mark for identifying the correct matching sequence (Option A).

Ethanol and water are miscible liquids with different boiling points (ethanol boils at \(78^\circ\text{C}\) and water at \(100^\circ\text{C}\)). Therefore, fractional distillation is the correct method used to separate them based on their differing boiling points.

[1 mark] Fractional distillation. Reject: simple distillation or distillation alone.

In the blast furnace, carbon dioxide reacts with hot coke (carbon) to form carbon monoxide, which is the reducing agent that reduces iron(III) oxide to iron. The chemical equation for this process is \(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\).

[1 mark] \(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\) (or alternative correct representations like \(\text{C} + \text{CO}_2 \rightarrow 2\text{CO}\)).

Condensation polymerization occurs when monomers containing two different functional groups react together to form a long polymer chain, releasing a small molecule (usually water or hydrogen chloride) as a by-product at each linkage point.

[1 mark] Condensation polymerization. Reject: addition polymerization.

Petroleum is a complex mixture of miscible hydrocarbons with different boiling points. Since their boiling points are relatively close and they are completely miscible, they are separated into different fractions based on their boiling points using fractional distillation.

Award 1 mark for 'Fractional distillation'. Reject: 'distillation' or 'simple distillation'.

The atomic number (proton number) of chlorine is 17. The 1- charge indicates that the chloride ion has gained one extra electron. Therefore, the total number of electrons is 17 + 1 = 18.

1 mark for the correct number of electrons (18).

The ion has a 3+ charge, which means it has lost 3 electrons compared to its neutral atom. Since the ion has 10 electrons, the neutral atom must have 10 + 3 = 13 electrons. In a neutral atom, the number of protons equals the number of electrons, so the number of protons is 13.

1 mark for the correct number of protons (13).

The nucleon number (mass number) of this iron isotope is 56, and the proton number (atomic number) is 26. The number of neutrons is calculated by subtracting the proton number from the nucleon number: 56 - 26 = 30.

1 mark for the correct number of neutrons (30).

The mass number (nucleon number) of the potassium ion is 39, which represents the total number of protons and neutrons in the nucleus.
The atomic number of potassium is 19, which is the number of protons.
The number of neutrons, \( X \), is calculated as:
\( X = \text{mass number} - \text{atomic number} = 39 - 19 = 20 \).

1 mark for 20.

A halide ion is a negatively charged ion formed from a Group VII halogen element (fluorine, chlorine, bromine, or iodine).
- \( \text{W} \) has 8 protons (oxygen) and 10 electrons, which represents an oxide ion, \( \text{O}^{2-} \).
- \( \text{X} \) has 9 protons (fluorine) and 9 electrons, which represents a neutral fluorine atom, \( \text{F} \).
- \( \text{Y} \) has 12 protons (magnesium) and 10 electrons, which represents a magnesium ion, \( \text{Mg}^{2+} \).
- \( \text{Z} \) has 17 protons (chlorine) and 18 electrons, which represents a chloride ion, \( \text{Cl}^- \).

Therefore, particle \( \text{Z} \) represents a halide ion.

1 mark for Z.

1. Identify the difference in charges: Magnesium ions (\(Mg^{2+}\)) and oxide ions (\(O^{2-}\)) have higher charges (+2 and -2) than sodium ions (\(Na^+\)) and chloride ions (\(Cl^-\)) (+1 and -1). 2. Relate to attraction: The electrostatic forces of attraction between oppositely charged ions in magnesium oxide are much stronger, requiring significantly more thermal energy to break.

M1: For stating that the ions in \(MgO\) have higher charges (\(Mg^{2+}\) and \(O^{2-}\)) than the ions in \(NaCl\) (\(Na^+\) and \(Cl^-\)) [1] M2: For stating that the electrostatic attraction between oppositely charged ions is stronger in \(MgO\) (or requires more energy to overcome) [1] [Reject: reference to covalent bonds or intermolecular forces]

1. Silicon(IV) oxide has a giant covalent structure where all valence electrons are shared in localized bonds, meaning there are no mobile ions or delocalised electrons to conduct electricity. 2. Molten sodium chloride has an ionic structure where the lattice has broken down, allowing the ions (\(Na^+\) and \(Cl^-\)) to move freely and carry the electric current.

M1: States that \(SiO_2\) has no mobile ions or delocalised electrons / free-moving electrons [1] M2: States that molten \(NaCl\) has mobile / free-moving ions [1]

1. Identify the molecular structure: Tetrachloromethane has a simple molecular structure with weak intermolecular forces (forces between molecules). 2. Explain what happens during boiling: When \(CCl_4\) boils, only these weak intermolecular forces are broken, not the strong covalent bonds inside the molecules. This requires very little thermal energy.

M1: Identifies that \(CCl_4\) consists of simple molecules / has weak intermolecular forces [1] M2: States that these weak intermolecular forces (forces between molecules) require little energy to overcome OR that covalent bonds are not broken [1] [Reject: Reference to breaking covalent bonds]

Silicon(IV) oxide, \(\text{SiO}_2\), has a macromolecular (giant covalent) lattice. Melting it requires breaking many strong covalent bonds throughout the giant structure, which requires a very large amount of thermal energy. In contrast, carbon dioxide, \(\text{CO}_2\), consists of simple molecules. Changing its state only requires overcoming the weak intermolecular forces (forces of attraction between the molecules), not breaking the strong covalent bonds within the molecules. Thus, very little thermal energy is needed.

Award 1 mark for each point:
- M1: Identify that silicon(IV) oxide has a giant covalent (macromolecular) structure with strong covalent bonds (that must be broken/require a lot of energy to break). [1]
- M2: Identify that carbon dioxide has a simple molecular structure with weak intermolecular forces / weak forces between molecules (that require little energy to overcome). [1]

Note: Reject any statement suggesting that covalent bonds within carbon dioxide molecules are broken during state changes.

Under heat, anhydrous copper(II) nitrate decomposes to form solid copper(II) oxide, nitrogen dioxide gas, and oxygen gas. The balanced chemical equation is: \( 2\text{Cu(NO}_3)_2 \rightarrow 2\text{CuO} + 4\text{NO}_2 + \text{O}_2 \).

1 mark for correct formulae of products: \( \text{CuO} \), \( \text{NO}_2 \), and \( \text{O}_2 \). 1 mark for correct balancing of the entire equation.

First, determine the moles of iron: \( n(\text{Fe}) = 11.2 / 56 = 0.20 \text{ mol} \). The balanced equation is \( 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \). The molar ratio of \( \text{Fe} \) to \( \text{Fe}_2\text{O}_3 \) is \( 2:1 \), so moles of \( \text{Fe}_2\text{O}_3 = 0.20 / 2 = 0.10 \text{ mol} \). The molar mass of \( \text{Fe}_2\text{O}_3 \) is \( (2 \times 56) + (3 \times 16) = 160 \text{ g/mol} \). Therefore, mass of \( \text{Fe}_2\text{O}_3 = 0.10 \times 160 = 16.0 \text{ g} \).

1 mark for calculating moles of iron as 0.20 mol or identifying the mole ratio of 2:1. 1 mark for calculating the correct mass of 16.0 g (allow 16 g).

First, calculate the moles of zinc: \( n(\text{Zn}) = 3.25 / 65 = 0.050 \text{ mol} \). Based on the 1:1 stoichiometric ratio in the equation, the moles of \( \text{H}_2 \) gas produced is also 0.050 mol. Volume of \( \text{H}_2 = 0.050 \times 24 = 1.2 \text{ dm}^3 \).

1 mark for calculating moles of Zn as 0.050 mol. 1 mark for multiplying moles by 24 to get 1.2 dm3 (allow 1.2).

From the balanced equation, \(2\text{ mol}\) of \(\text{Cu}(\text{NO}_3)_2\) produces \(4\text{ mol}\) of \(\text{NO}_2\).
This is a \(1:2\) mole ratio.

Therefore, moles of \(\text{NO}_2\) produced = \(0.05\text{ mol} \times 2 = 0.10\text{ mol}\).

Volume of \(\text{NO}_2\) produced = \(\text{moles} \times 24\text{ dm}^3/\text{mol}\)
Volume = \(0.10 \times 24 = 2.4\text{ dm}^3\) (or \(2400\text{ cm}^3\)).

M1: Deduce that the number of moles of \(\text{NO}_2\) is \(0.10\text{ mol}\) (or show multiplication of \(0.05\) by \(2\)) [1]
M2: Calculate the correct volume: \(2.4\text{ dm}^3\) (or \(2400\text{ cm}^3\)) [1]

The percentage yield is calculated using the formula:
\[\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]

\[\text{Percentage Yield} = \frac{11.12}{13.90} \times 100\%\]

\[\text{Percentage Yield} = 80.0\%\]

M1: Award [1] for showing correct working, i.e., \(\frac{11.12}{13.90} \times 100\) or equivalent.
M2: Award [1] for the correct final answer of \(80.0\%\) (or \(80\%\)).

1. Find the percentage of oxygen by mass:
\[100\% - 30.43\% = 69.57\%\text{ oxygen by mass}\]

2. Divide the percentages by the respective relative atomic masses to find the mole ratio:
\[\text{Moles of N} = \frac{30.43}{14} = 2.174\text{ mol}\]
\[\text{Moles of O} = \frac{69.57}{16} = 4.348\text{ mol}\]

3. Divide by the smallest value to find the simplest whole-number ratio:
\[\text{N} = \frac{2.174}{2.174} = 1\]
\[\text{O} = \frac{4.348}{2.174} = 2\]

Thus, the empirical formula is \(\text{NO}_2\).

M1: Show the division of mass percentages by relative atomic masses (e.g., \(\frac{30.43}{14}\) and \(\frac{69.57}{16}\)) to obtain a mole ratio of approx. \(2.17 : 4.35\) [1]
M2: State the correct empirical formula: \(\text{NO}_2\) [1]

1. As pressure increases from 100 kPa to 400 kPa at 250 degrees C, the percentage yield of G increases from 45% to 78%, which means the position of equilibrium shifts to the right.
2. As temperature increases from 250 degrees C to 450 degrees C at 100 kPa, the percentage yield of G decreases from 45% to 18%. A decrease in yield with rising temperature indicates the equilibrium has shifted to the left to absorb the heat, which means the forward reaction is exothermic.

[1 mark] Equilibrium shifts to the right when pressure increases.
[1 mark] Forward reaction is exothermic.
[1 mark] Explanation: Increasing temperature (at constant pressure) decreases the yield of G.

1. The reaction has 2 moles of gas on the reactant side (1 CH4 + 1 H2O) and 4 moles of gas on the product side (1 CO + 3 H2). According to Le Chatelier's principle, increasing pressure shifts the equilibrium to the side with fewer moles of gas to oppose the change, which is the reactant (left) side, thereby decreasing the percentage of CO.
2. At a constant pressure of 5 atm, increasing the temperature from 600 to 800 degrees C increases the percentage of CO from 32% to 68%. An increase in yield with higher temperature shows that the forward reaction is endothermic, as the equilibrium shifts to absorb heat.

[1 mark] States that there are fewer moles of gas on the left-hand side (2 moles) than the right-hand side (4 moles).
[1 mark] Explains that increasing pressure shifts equilibrium to the left / side with fewer gas moles.
[1 mark] Identifies reaction as endothermic because yield of CO increases as temperature increases.

1. In the presence of the catalyst, the time taken to reach equilibrium is significantly reduced (from 120 minutes to 5 minutes), which indicates that the catalyst increases the rate of reaction. However, the equilibrium yield remains unchanged at 96%, showing that a catalyst does not alter the position of equilibrium.
2. Comparing the reactions with the catalyst at 450 degrees C and 550 degrees C, the yield of SO3 decreases from 96% to 80% as temperature increases. According to Le Chatelier's principle, an increase in temperature shifts the equilibrium in the direction of the endothermic reaction. Since the yield of the forward product decreased, the backward reaction must be endothermic, meaning the forward reaction is exothermic.

[1 mark] Explains that catalyst increases rate (reaches equilibrium faster).
[1 mark] Explains that catalyst has no effect on the position of equilibrium / yield of product.
[1 mark] Identifies reaction as exothermic with reference to decreasing yield of SO3 (from 96% to 80%) as temperature increases.

First, add distilled water to the mixture and stir to dissolve the nickel(II) sulfate. Filter the mixture; the insoluble silicon(IV) oxide remains as residue on the filter paper, while the nickel(II) sulfate solution passes through as the filtrate. Heat the filtrate in an evaporating basin to evaporate some water until the crystallisation point is reached. Allow the hot saturated solution to cool slowly so that crystals of nickel(II) sulfate form. Finally, filter the crystals from the remaining liquid, wash them with a small volume of cold distilled water, and dry them using filter paper or in a warm oven.

[1 mark] Add distilled water to dissolve the nickel(II) sulfate and filter the mixture to remove the insoluble silicon(IV) oxide residue. [1 mark] Heat the filtrate (nickel(II) sulfate solution) until it is saturated / reaches crystallization point, then allow it to cool to form crystals. [0.5 mark] Filter the crystals and dry them with filter paper or in a warm oven.

The Rf value is calculated using the formula: Rf = distance travelled by substance / distance travelled by solvent. Here, Rf = 5.1 cm / 8.5 cm = 0.60. The baseline must be drawn in pencil because pencil lead (graphite) is insoluble in chromatography solvents. If ink were used, it would dissolve in the solvent and travel up the paper, separating into its own component dyes and interfering with the results of the chromatogram.

[1 mark] Correct calculation of Rf value: 5.1 / 8.5 = 0.60. [1 mark] Explanation that pencil/graphite is insoluble in the solvent and will not run. [0.5 mark] Explanation that ink is soluble and would run/separate, which would interfere with the chromatogram.

Since lead(II) sulfate is insoluble, it is prepared by precipitation. First, dissolve the solid lead(II) nitrate and solid sodium sulfate separately in distilled water. Mix the two solutions together to form a precipitate of lead(II) sulfate. Filter the mixture to separate the insoluble lead(II) sulfate residue from the soluble sodium nitrate filtrate. Wash the residue on the filter paper with distilled water to remove any remaining soluble impurities. Finally, dry the pure lead(II) sulfate precipitate in a warm oven or by pressing it between sheets of filter paper.

[1 mark] Dissolve both solids in distilled water (separately) and mix the two solutions together to form the precipitate. [1 mark] Filter the mixture to collect the lead(II) sulfate residue. [0.5 mark] Wash the residue with distilled water and dry it (e.g., with filter paper or in a warm oven).

Fractional distillation separates miscible liquids based on their different boiling points. When the mixture of ethanol and water is heated, both liquids begin to evaporate. Since ethanol has a lower boiling point (\(78\ ^\circ\text{C}\)), its vapour rises up the fractionating column. The column has a temperature gradient (cooler at the top) and a high surface area. The water vapour (with a higher boiling point of \(100\ ^\circ\text{C}\)) condenses on the surfaces inside the column and runs back down into the flask, while ethanol vapour reaches the top, enters the condenser, and is liquefied to be collected first.

1 mark: State that separation is based on different boiling points (ethanol has the lower boiling point). 1 mark: Explain that ethanol vapour rises while water vapour condenses and falls back into the flask. 0.5 marks: Explain that the fractionating column provides a large surface area or temperature gradient to allow repeated condensation and evaporation for complete separation.

1. Calculate the number of moles of each element in \(100\text{ g}\) of the compound:
- Moles of \(\text{C} = 54.55 / 12 = 4.546\text{ mol}\)
- Moles of \(\text{H} = 9.09 / 1 = 9.09\text{ mol}\)
- Moles of \(\text{O} = 36.36 / 16 = 2.273\text{ mol}\)

2. Divide each value by the smallest number of moles (\(2.273\)) to find the simplest molar ratio:
- Ratio of \(\text{C} = 4.546 / 2.273 = 2\)
- Ratio of \(\text{H} = 9.09 / 2.273 = 4\)
- Ratio of \(\text{O} = 2.273 / 2.273 = 1\)

3. The simplest whole-number ratio of \(\text{C} : \text{H} : \text{O}\) is \(2 : 4 : 1\).

Therefore, the empirical formula of compound X is \(\text{C}_2\text{H}_4\text{O}\).

Award 1 mark for each of the following up to 4 marks:
- Dividing percentages by relative atomic masses for C, H, and O to find moles [1]
- Calculating correct mole values: \(4.55\) (or \(4.546\)) mol C, \(9.09\) mol H, and \(2.27\) (or \(2.273\)) mol O [1]
- Dividing by the smallest value to obtain the ratio of \(2 : 4 : 1\) [1]
- Correct empirical formula: \(\text{C}_2\text{H}_4\text{O}\) (allow ecf from incorrect ratio if working shown) [1]

1. Find the mass of water of crystallisation lost:
\(\text{Mass of water} = 5.88\text{ g} - 4.44\text{ g} = 1.44\text{ g}\)

2. Calculate the relative formula mass of anhydrous calcium chloride, \(\text{CaCl}_2\):
\(M_r(\text{CaCl}_2) = 40 + (2 \times 35.5) = 111\)

3. Calculate the number of moles of anhydrous \(\text{CaCl}_2\) and \(\text{H}_2\text{O}\):
- Moles of \(\text{CaCl}_2 = 4.44 / 111 = 0.04\text{ mol}\)
- Moles of \(\text{H}_2\text{O} = 1.44 / 18 = 0.08\text{ mol}\)

4. Determine the ratio of moles of water to moles of anhydrous salt:
\(x = 0.08 / 0.04 = 2\)

Therefore, the value of \(x\) is \(2\).

Award 1 mark for each of the following up to 4 marks:
- Calculation of the mass of water lost = \(1.44\text{ g}\) [1]
- Calculation of moles of anhydrous \(\text{CaCl}_2 = 0.04\text{ mol}\) [1]
- Calculation of moles of \(\text{H}_2\text{O} = 0.08\text{ mol}\) [1]
- Deduction of the mole ratio of \(1 : 2\) resulting in \(x = 2\) [1]

1. Calculate the moles of \(\text{Mg}\):
\(\text{Moles of Mg} = 0.12 / 24 = 0.005\text{ mol}\)

2. Calculate the moles of \(\text{HCl}\):
\(\text{Moles of HCl} = (25.0 / 1000) \times 0.50 = 0.0125\text{ mol}\)

3. Determine the limiting reactant:
According to the equation, \(1\text{ mol}\) of \(\text{Mg}\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).
Therefore, \(0.005\text{ mol}\) of \(\text{Mg}\) requires \(2 \times 0.005 = 0.010\text{ mol}\) of \(\text{HCl}\).
Since we have \(0.0125\text{ mol}\) of \(\text{HCl}\), which is in excess, \(\text{Mg}\) is the limiting reactant.

4. Calculate the volume of \(\text{H}_2\) gas:
Since \(\text{Mg}\) is the limiting reactant, the moles of \(\text{H}_2\) produced is equal to the moles of \(\text{Mg}\) reacted:
\(\text{Moles of H}_2 = 0.005\text{ mol}\)
\(\text{Volume of H}_2 = 0.005 \times 24000\text{ cm}^3\text{/mol} = 120\text{ cm}^3\).

Therefore, the volume of hydrogen gas produced is \(120\text{ cm}^3\).

Award 1 mark for each of the following up to 4 marks:
- Correctly calculating moles of \(\text{Mg} = 0.005\text{ mol}\) and moles of \(\text{HCl} = 0.0125\text{ mol}\) [1]
- Showing that \(\text{Mg}\) is the limiting reactant by comparing the mole ratio (e.g., stating that \(0.005\text{ mol}\) of Mg requires \(0.010\text{ mol}\) of HCl, leaving HCl in excess) [1]
- Correctly relating the moles of \(\text{H}_2\) produced to the limiting reactant (\(0.005\text{ mol}\) of \(\text{H}_2\)) [1]
- Calculation of the volume of hydrogen gas = \(120\text{ cm}^3\) (unit not strictly required but must be correct value) [1]

In the blast furnace, iron(III) oxide is reduced by carbon monoxide gas to produce molten iron and carbon dioxide gas. The balanced chemical equation is: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).

M1: Correct formulae of reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\) [1] M2: Correctly balanced equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) [1]

Limestone (calcium carbonate) thermally decomposes to form calcium oxide (CaO). This basic oxide reacts with the acidic silicon dioxide (SiO2) impurity to produce calcium silicate (CaSiO3), which is removed as slag. The equation is: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

M1: Calcium silicate (accept slag) [1] M2: Correct balanced equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]

Aluminium is higher in the reactivity series than carbon. Therefore, carbon is not a strong enough reducing agent to remove oxygen from aluminium oxide. Aluminium must instead be extracted by electrolysis of its molten oxide.

M1: Identify that aluminium is more reactive than carbon (or carbon is less reactive than aluminium) [1] M2: State that carbon cannot reduce aluminium oxide (or cannot displace aluminium / aluminium has too strong an affinity for oxygen) [1]

In the blast furnace, carbon monoxide acts as the reducing agent. It reduces iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) to iron (\(\text{Fe}\)), and is itself oxidized to carbon dioxide (\(\text{CO}_2\)). The balanced chemical equation is: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).

1 mark: Correct formulas of reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\) (or equivalent).
1 mark: Correctly balanced equation: \(1, 3 \rightarrow 2, 3\).

Limestone decomposes upon heating, which is known specifically as thermal decomposition. The reaction produces solid calcium oxide (lime) and carbon dioxide gas: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).

1 mark: Thermal decomposition (Reject: 'decomposition' on its own; Accept: 'thermal breakdown').
1 mark: Correct balanced equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).

Alkene isomers with the molecular formula \(\text{C}_4\text{H}_8\) can be straight-chain or branched-chain: (1) \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\) is named but-1-ene; (2) \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) is named but-2-ene; (3) \(\text{CH}_2=\text{C}(\text{CH}_3)_2\) is a branched isomer named 2-methylpropene.

Award 1 mark for each correct IUPAC name: but-1-ene (allow 1-butene) [1]; but-2-ene (allow 2-butene) [1]; 2-methylpropene (allow methylpropene / methyl propene) [1]. Reject: butane, cyclobutane, methylcyclopropane, or any alkane/cycloalkane names.

Polyesters are formed by condensation polymerisation of a diol and a dicarboxylic acid. Breaking the ester bonds in the repeating unit: (1) The diol part \(\text{–O–CH}_2\text{–CH}_2\text{–O–}\) comes from the monomer ethane-1,2-diol, \(\text{HO–CH}_2\text{–CH}_2\text{–OH}\). (2) The dicarboxylic acid part \(\text{–CO–(CH}_2)_4\text{–CO–}\) comes from the monomer hexanedioic acid, \(\text{HOOC–(CH}_2)_4\text{–COOH}\). (3) The linkage containing the \(\text{–O–CO–}\) group is an ester linkage.

(a) 1 mark for identifying the diol: ethane-1,2-diol (accept glycol / ethylene glycol) OR formula: \(\text{HOCH}_2\text{CH}_2\text{OH}\); 1 mark for identifying the dicarboxylic acid: hexanedioic acid (accept adipic acid) OR formula: \(\text{HOOC(CH}_2)_4\text{COOH}\). (b) 1 mark for: ester linkage / ester bond.

Oxidation of a primary alcohol with three carbon atoms (propan-1-ol) produces a carboxylic acid with three carbon atoms, which is propanoic acid. The formula for propanoic acid showing its carboxylic functional group is \(\text{CH}_3\text{CH}_2\text{COOH}\) or \(\text{C}_2\text{H}_5\text{COOH}\). The standard laboratory reagent for oxidising alcohols to carboxylic acids is acidified potassium manganate(VII) (or acidified potassium dichromate(VI)), and the mixture must be heated.

(a) 1 mark for: propanoic acid. (b) 1 mark for: \(\text{CH}_3\text{CH}_2\text{COOH}\) or \(\text{C}_2\text{H}_5\text{COOH}\). (c) 1 mark for: Acidified potassium manganate(VII) (or \(\text{KMnO}_4\)) and heat / warm / reflux (or acidified potassium dichromate(VI) and heat).

(a) The polymer contains ester linkages (\(-\text{O}-\text{C}(=\text{O})-\)), which are formed by the elimination of a small molecule (water). Therefore, it is formed via condensation polymerization.

(b) To determine the monomers of a polyester, we mentally split the ester linkages:
- The oxygen-containing part comes from the diol: \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\) (ethane-1,2-diol).
- The carbonyl-containing part comes from the dicarboxylic acid: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) (butanedioic acid).

M1: For identifying 'condensation' (polymerization) [1]

M2: For the correct structural/displayed formula of the diol: \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\) [1]

M3: For the correct structural/displayed formula of the dicarboxylic acid: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) [1]

Note: Award marks for equivalent displayed formulas showing all bonds explicitly.

(a) The polymer contains ester linkages (\(-\text{O}-\text{C}(=\text{O})-\)), which are formed by the elimination of a small molecule (water). Therefore, it is formed via condensation polymerization.

(b) To determine the monomers of a polyester, we mentally split the ester linkages:
- The oxygen-containing part comes from the diol: \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\) (ethane-1,2-diol).
- The carbonyl-containing part comes from the dicarboxylic acid: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) (butanedioic acid).

M1: For identifying 'condensation' (polymerization) [1]

M2: For the correct structural/displayed formula of the diol: \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\) [1]

M3: For the correct structural/displayed formula of the dicarboxylic acid: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) [1]

Note: Award marks for equivalent displayed formulas showing all bonds explicitly.