2024 Edexcel IGCSE Chemistry 模擬試題連答案詳解

First, find the number of protons in a neutral atom of X: \(\text{Number of protons} = \text{Mass number} - \text{Number of neutrons} = 70 - 39 = 31\). A neutral atom of X has 31 protons and 31 electrons. The \(X^{2+}\) ion is formed by losing two electrons, so the number of electrons is \(31 - 2 = 29\). Therefore, the ion has 31 protons and 29 electrons.

1 mark for the correct option (A).

1. Assume a 100 g sample of the oxide. This contains 70.0 g of Fe and 30.0 g of O (since 100% - 70.0% = 30.0%). 2. Convert mass to moles: Moles of Fe = \(70.0 / 56 = 1.25\) mol; Moles of O = \(30.0 / 16 = 1.875\) mol. 3. Divide by the smallest value to find the simplest mole ratio: Ratio of Fe = \(1.25 / 1.25 = 1\); Ratio of O = \(1.875 / 1.25 = 1.5\). 4. Convert to whole numbers by multiplying both by 2: Fe = 2, O = 3. The empirical formula is \(Fe_2O_3\).

1 mark for the correct option (B).

Potassium reacts vigorously with water to form potassium hydroxide (an alkaline solution) and hydrogen gas. The chemical equation is \(2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)\). Potassium hydroxide is strongly alkaline, which causes universal indicator to turn blue/purple. Therefore, statement C is incorrect because an alkaline (basic) solution is formed, not an acidic one.

1 mark for identifying the incorrect statement (C).

An ester linkage is formed by the reaction between a carboxyl group (\(-COOH\)) and a hydroxyl group (\(-OH\)) with the elimination of a water molecule. To form a long-chain polyester, both monomers must have these functional groups at both ends. Therefore, a dicarboxylic acid (two \(-COOH\) groups) and a diol (two \(-OH\) groups) are required.

1 mark for the correct option (B).

To calculate the relative atomic mass: \(\text{Relative atomic mass} = \frac{(69 \times 60.1) + (71 \times 39.9)}{100} = \frac{4146.9 + 2832.9}{100} = \frac{6979.8}{100} = 69.798\). Rounding to 1 decimal place gives 69.8.

M1: Correct substitution of values into formula: \(((69 \times 60.1) + (71 \times 39.9)) / 100\) (1.1 marks)
M2: Final answer of 69.8 (1.1 marks). Reject 69.798 (not given to 1 decimal place).

Metals consist of a giant lattice of positive metal ions arranged in regular layers surrounded by a sea of delocalised electrons. When a force is applied, these layers of ions can slide over each other. The delocalised electrons are free to move with the layers, maintaining the metallic bonding throughout the structure.

M1: Mention that the structure has layers of positive ions / atoms that can slide over each other (1.1 marks)
M2: Mention that the delocalised electrons (or metallic bonds) hold the structure together as the layers slide / non-directional bonding prevents shattering (1.1 marks)

To test for ammonium ions, add sodium hydroxide solution to the sample and warm the mixture gently. This reaction releases ammonia gas. Test the gas with damp red litmus paper, which will turn blue because ammonia is alkaline.

M1: Add sodium hydroxide (solution) and heat / warm (1.1 marks)
M2: Gas produced turns damp red litmus paper blue (1.1 marks). Reject blue litmus paper. Accept 'damp universal indicator paper turns blue/purple'.

When lithium is added to water, it reacts less vigorously than sodium or potassium. The observations are that it floats on the water, moves slowly across the surface, bubbles / fizzes (due to the production of hydrogen gas), and gradually decreases in size until it disappears.

M1 & M2: Any two from the following (1.1 marks each):
- Floats on water
- Moves on the surface
- Effervescence / bubbles / fizzing
- Decreases in size / disappears / dissolves
- Heat is released. Reject 'burns with red flame' (unless ignition is specified, which does not happen spontaneously with lithium).

First, calculate the number of moles of each element in 100 g of the compound:
Moles of \(\text{S} = \frac{25.2}{32} = 0.7875\text{ mol}\)
Moles of \(\text{F} = \frac{74.8}{19} = 3.9368\text{ mol}\)

Next, divide both by the smallest number of moles (0.7875):
Ratio of \(\text{S} = \frac{0.7875}{0.7875} = 1\)
Ratio of \(\text{F} = \frac{3.9368}{0.7875} \approx 5\)

The empirical formula is \(\text{SF}_5\).

M1: Correctly dividing percentages by atomic masses to find moles: 0.7875 and 3.9368 (1.1 marks)
M2: Finding simplest whole number ratio 1:5 to give \(\text{SF}_5\) (1.1 marks)

1. Calculate \(M_r(\text{NaOH}) = 23 + 16 + 1 = 40\).
2. Calculate moles of \(\text{NaOH} = \frac{2.00}{40} = 0.050\text{ mol}\).
3. Convert volume to dm\(^3\): \(\frac{250}{1000} = 0.250\text{ dm}^3\).
4. Calculate concentration: \(\text{Concentration} = \frac{0.050}{0.250} = 0.20\text{ mol/dm}^3\).

M1: Calculate moles of \(\text{NaOH}\): \(\frac{2.00}{40} = 0.050\text{ mol}\) (1.1 marks)
M2: Divide moles by volume in dm\(^3\): \(\frac{0.050}{0.250} = 0.20\text{ mol/dm}^3\) (1.1 marks). Accept 0.2.

Fermentation of glucose requires specific conditions: presence of yeast (which contains the enzymes), anaerobic conditions (to prevent the oxidation of ethanol to ethanoic acid), and a warm temperature (around 30-40 °C, which is the optimum temperature for the yeast enzymes).

M1 & M2: Any two from the following (1.1 marks each):
- presence of yeast / zymase
- anaerobic conditions / absence of oxygen / absence of air
- temperature in the range of 25 °C to 40 °C

In industrial catalytic cracking, long-chain alkanes are passed over a heated catalyst of silica (silicon dioxide) or alumina (aluminium oxide) at a high temperature in the range of 600 °C to 700 °C to break them down into shorter-chain alkanes and alkenes.

M1: Catalyst: silica / silicon dioxide OR alumina / aluminium oxide OR zeolite (1.1 marks)
M2: Temperature: 600 °C - 700 °C (accept any temperature or range between 500 °C and 750 °C) (1.1 marks)

Step 1: Write down the formula for calculating relative atomic mass: \( A_r = \frac{\sum(\text{isotopic mass} \times \text{abundance})}{100} \). Step 2: Substitute the values into the equation: \( A_r = \frac{(24 \times 78.9) + (26 \times 21.1)}{100} \). Step 3: Calculate the numerator: \( 1893.6 + 548.6 = 2442.2 \). Step 4: Divide by 100 to find the relative atomic mass: \( 24.422 \). Step 5: Round the final value to 1 decimal place, which gives \( 24.4 \).

M1: for showing a correct calculation method: \( \frac{(24 \times 78.9) + (26 \times 21.1)}{100} \) [1 mark]. M2: for the final correct answer of 24.4 (must be to 1 decimal place) [1.2 marks].

Copper has a giant metallic lattice structure. There is a strong electrostatic force of attraction between the positive copper ions and the surrounding sea of delocalised electrons. Overcoming these extremely strong metallic bonds in order to melt the copper requires a very large amount of thermal energy.

M1: for stating copper has a giant metallic structure with strong electrostatic attraction between positive ions and delocalised electrons [1 mark]. M2: for explaining that a large amount of energy is required to overcome/break these metallic bonds [1.2 marks].

In the reactivity series, aluminium is positioned above carbon, making it more reactive. Because of this, carbon is not a strong enough reducing agent to displace aluminium from its oxide or remove the oxygen from aluminium oxide. Therefore, carbon reduction cannot be used, and aluminium must be extracted using electrolysis instead.

M1: for stating aluminium is more reactive than carbon (or carbon is less reactive than aluminium) [1 mark]. M2: for explaining that carbon cannot displace aluminium / reduce aluminium oxide / remove the oxygen from aluminium oxide [1.2 marks].

To test for ammonium ions, add aqueous sodium hydroxide solution to the test solution and warm the mixture gently. This reaction produces ammonia gas. Test the gas evolved by holding damp red litmus paper near the mouth of the tube. The paper will turn blue, confirming the presence of ammonia gas and therefore ammonium ions in the original solution.

M1: for adding sodium hydroxide solution and heating/warming [1 mark]. M2: for testing the gas with damp red litmus paper and observing that it turns blue [1.2 marks]. Reject blue litmus paper. Accept indicator paper turning blue.

A pure compound consists of only one substance, meaning all its intermolecular forces or bonds are identical, so it melts at a sharp, single, fixed temperature. An impure mixture contains different substances that disrupt the regular structure, meaning different bonds break at different temperatures, causing the mixture to melt gradually over a wide range of temperatures.

M1: for stating that a pure compound melts at a sharp / fixed / single temperature [1 mark]. M2: for stating that an impure mixture melts over a range of temperatures [1.2 marks].

Potassium reacts extremely vigorously with water because it is highly reactive. The reaction is strongly exothermic, causing the potassium metal to melt into a spherical ball. It floats on the water and moves rapidly across the surface. The heat generated ignites the hydrogen gas produced, which burns with a characteristic lilac flame. Effervescence (bubbling) is also observed as hydrogen gas is evolved, and the potassium eventually disappears.

Award 1 mark for the first correct observation and 1.2 marks for the second correct observation up to a maximum of 2.2 marks. Correct observations include: potassium melts into a shiny ball / potassium floats / potassium moves rapidly on the surface / burns with a lilac flame / effervescence / bubbles of gas / potassium disappears. Do not accept 'gas is produced' without observation of bubbles or fizzing.

Step 1: Divide the percentage mass of each element by its relative atomic mass to find the molar amount. For sulfur: \( \text{moles of S} = \frac{29.6}{32} = 0.925 \text{ mol} \). For fluorine: \( \text{moles of F} = \frac{70.4}{19} = 3.705 \text{ mol} \). Step 2: Divide both mole values by the smaller number to determine the simplest whole-number ratio. For sulfur: \( \frac{0.925}{0.925} = 1 \). For fluorine: \( \frac{3.705}{0.925} \approx 4 \). Step 3: Write down the resulting empirical formula, which is \( \text{SF}_4 \).

M1: for calculating the correct molar amounts of sulfur (0.925) and fluorine (3.705) by dividing mass percents by atomic masses [1 mark]. M2: for dividing by the smaller value to obtain a 1:4 ratio and stating the correct empirical formula \( \text{SF}_4 \) [1.2 marks].

Step 1: Write down the balanced chemical equation: \( \text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \). This shows a 1:1 reaction ratio. Step 2: Calculate the amount of hydrochloric acid in moles: \( \text{moles of HCl} = \text{concentration} \times \text{volume} = 0.120 \text{ mol/dm}^3 \times \frac{25.0}{1000} \text{ dm}^3 = 0.00300 \text{ mol} \). Step 3: Use the stoichiometric ratio to find the moles of NaOH: \( \text{moles of NaOH} = 0.00300 \text{ mol} \). Step 4: Calculate the concentration of the sodium hydroxide solution: \( \text{concentration of NaOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00300}{\frac{20.0}{1000}} = 0.150 \text{ mol/dm}^3 \).

M1: for calculating the moles of hydrochloric acid: \( 0.00300 \text{ mol} \) [1 mark]. M2: for dividing the moles of sodium hydroxide by its volume in \( \text{dm}^3 \) to find the correct concentration of \( 0.150 \text{ mol/dm}^3 \) (must be to 3 significant figures) [1.2 marks].

To test for the presence of ammonium (\(\text{NH}_4^+\)) ions in a solid sample: 1. Add sodium hydroxide solution to the solid and warm the mixture gently to release ammonia gas. 2. Test the escaping gas with damp red litmus paper. Ammonia gas is alkaline and will dissolve in the moisture on the paper, turning the red litmus paper blue.

M1: Add sodium hydroxide (solution / aqueous) and warm/heat (1 mark)
M2: (Gas produced/ammonia) turns damp red litmus paper blue (1 mark)
(Reject: blue litmus paper / dry litmus paper)

To draw the repeat unit of poly(chloroethene): 1. Change the double bond in the chloroethene monomer (\(\text{CH}_2=\text{CHCl}\)) into a single carbon-carbon bond (\(\text{-C-C-}\)). 2. Keep the atoms attached to each carbon the same: one carbon has two hydrogen atoms, and the other carbon has one hydrogen atom and one chlorine atom. 3. Add open-ended continuation bonds extending outwards from each of the carbon atoms to represent the repeating polymer chain, usually enclosed in brackets with an 'n' subscript.

M1: Single C-C bond with open continuation bonds extending outwards from both carbon atoms (1 mark)
M2: Correct groups attached (two H atoms on one carbon, one H and one Cl on the other carbon) (1 mark)
(Accept: structure with or without brackets and 'n')

1. Calculate the moles of \(\text{NaNO}_3\) reacted: \(\text{Moles} = 4.25\text{ g} / 85\text{ g/mol} = 0.05\text{ mol}\). 2. Determine the moles of \(\text{O}_2\) produced using the 2:1 stoichiometric ratio: \(\text{Moles of O}_2 = 0.05\text{ mol} / 2 = 0.025\text{ mol}\). 3. Calculate the volume of \(\text{O}_2\) gas at r.t.p.: \(\text{Volume} = 0.025\text{ mol} \times 24\text{ dm}^3/\text{mol} = 0.6\text{ dm}^3\).

- M1: Calculate moles of \(\text{NaNO}_3\) as 0.05 mol (1). - M2: Calculate moles of \(\text{O}_2\) as 0.025 mol (1). - M3: Calculate volume of \(\text{O}_2\) as 0.6 \(\text{dm}^3\) (1). Accept 600 \(\text{cm}^3\) if the unit is explicitly stated by the candidate.

1. Calculate the moles of \(\text{H}_2\text{SO}_4\) used: \(\text{Moles} = 0.125\text{ mol/dm}^3 \times (20.0 / 1000)\text{ dm}^3 = 0.0025\text{ mol}\). 2. Determine the moles of \(\text{KOH}\) reacted using the 2:1 stoichiometric ratio: \(\text{Moles of KOH} = 2 \times 0.0025\text{ mol} = 0.0050\text{ mol}\). 3. Calculate the concentration of the \(\text{KOH}\) solution: \(\text{Concentration} = 0.0050\text{ mol} / 0.0250\text{ dm}^3 = 0.20\text{ mol/dm}^3\).

- M1: Calculate moles of \(\text{H}_2\text{SO}_4\) as 0.0025 mol (1). - M2: Calculate moles of \(\text{KOH}\) as 0.0050 mol (1). - M3: Calculate concentration of \(\text{KOH}\) as 0.20 \(\text{mol/dm}^3\) (1).

1. Energy needed to break reactant bonds: \(4 \times (\text{C}-\text{H}) + 1 \times (\text{Cl}-\text{Cl}) = 4(413) + 243 = 1895\text{ kJ/mol}\) (or considering only bonds broken: \(413 + 243 = 656\text{ kJ/mol}\)). 2. Energy released when product bonds are formed: \(3 \times (\text{C}-\text{H}) + 1 \times (\text{C}-\text{Cl}) + 1 \times (\text{H}-\text{Cl}) = 3(413) + 339 + 432 = 2010\text{ kJ/mol}\) (or considering only bonds formed: \(339 + 432 = 771\text{ kJ/mol}\)). 3. Calculate the enthalpy change: \(\Delta H = \text{Energy in} - \text{Energy out} = 1895 - 2010 = -115\text{ kJ/mol}\) (or \(656 - 771 = -115\text{ kJ/mol}\)).

- M1: Sum of reactant bond energies broken = 1895 kJ (or 656 kJ) (1). - M2: Sum of product bond energies formed = 2010 kJ (or 771 kJ) (1). - M3: Enthalpy change \(\Delta H\) with correct negative sign = -115 kJ/mol (1). Award 2 marks for a final value of +115.

1. Use the relative atomic mass formula: \(A_r = [(\text{mass of isotope 1} \times \text{abundance}) + (\text{mass of isotope 2} \times \text{abundance})] / 100\). 2. Substitute the values: \(A_r = [(69 \times 60.1) + (71 \times 39.9)] / 100\). 3. Calculate the numerator and final value: \(A_r = (4146.9 + 2832.9) / 100 = 6979.8 / 100 = 69.798\). 4. Round to 2 decimal places: \(69.80\).

- M1: Calculate total mass of 100 atoms: \((69 \times 60.1) + (71 \times 39.9)\) or \(4146.9 + 2832.9\) (1). - M2: Divide total mass by 100: \(6979.8 / 100\) (1). - M3: Provide final answer rounded to 2 decimal places: 69.80 (1). Award 2 marks for 69.8 or 69.798 (incorrect rounding or decimal places).

1. Calculate moles of \(\text{Zn}\) reacted: \(\text{Moles of Zn} = 3.25\text{ g} / 65\text{ g/mol} = 0.050\text{ mol}\). 2. Determine theoretical mass of \(\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}\) crystals: Molar mass of crystals \(M_r = 65 + 32 + (4 \times 16) + 7 \times (18) = 287\text{ g/mol}\). Theoretical mass \(= 0.050\text{ mol} \times 287\text{ g/mol} = 14.35\text{ g}\). 3. Calculate percentage yield: \(\text{Percentage yield} = (11.48\text{ g} / 14.35\text{ g}) \times 100 = 80\%\).

- M1: Calculate moles of zinc as 0.050 mol (1). - M2: Calculate theoretical mass of hydrated zinc sulfate crystals as 14.35 g (1). - M3: Calculate percentage yield as 80% (1).

**Test for ammonium ions (\(\text{NH}_4^+\)):**
1. Add a small volume of aqueous sodium hydroxide (\(\text{NaOH}\)) to the solid sample (or its solution) in a boiling tube.
2. Warm the mixture gently using a Bunsen burner.
3. Hold a piece of damp red litmus paper at the mouth of the tube.
4. **Observation:** The damp red litmus paper turns blue, indicating the presence of ammonia gas, which confirms the presence of ammonium ions.

**Test for sulfate ions (\(\text{SO}_4^{2-}\)):**
1. Dissolve the solid sample in deionised water to prepare an aqueous solution.
2. Add a few drops of dilute hydrochloric acid (\(\text{HCl}\)) to acidify the solution (to remove any carbonate impurities).
3. Add a few drops of barium chloride solution (\(\text{BaCl}_2\)).
4. **Observation:** A white precipitate of barium sulfate (\(\text{BaSO}_4\)) forms, confirming the presence of sulfate ions.

- **Ammonium Ion Test (1.5 marks total):**
- Add sodium hydroxide solution AND warm/heat (1 mark).
- Test the gas produced with damp red litmus paper, which turns blue (0.5 marks). (Reject: dry litmus paper, blue litmus paper).
- **Sulfate Ion Test (2 marks total):**
- Dissolve the sample and add dilute hydrochloric acid / dilute nitric acid (1 mark). (Reject: sulfuric acid).
- Add barium chloride solution / barium nitrate solution (1 mark).
- **Sulfate Ion Observation (1 mark total):**
- White precipitate / white ppt (1 mark).

**Procedure:**
1. Measure a known volume (e.g., \(50\text{ cm}^3\)) of copper(II) sulfate solution of a known concentration using a measuring cylinder or pipette.
2. Transfer the solution into a polystyrene cup, which is supported inside a glass beaker to provide stability and extra insulation.
3. Place a thermometer into the solution and record its initial temperature. To ensure thermal equilibrium, record the temperature at regular intervals (e.g., every minute) for 3 minutes before adding the zinc.
4. Add a known mass of zinc powder (ensuring that zinc is in excess so that all copper(II) sulfate reacts completely).
5. Immediately place a lid on the polystyrene cup to minimise heat loss to the surrounding air.
6. Stir the mixture continuously using the thermometer or a magnetic stirrer to ensure even heat distribution.
7. Record the temperature at regular intervals (e.g., every minute) and note the maximum temperature reached by the mixture.

**Minimising Heat Loss:**
- Use a polystyrene cup because polystyrene is a good thermal insulator.
- Use a plastic lid on the cup to prevent heat loss through convection and evaporation.

- **Volume and reaction vessel (1 mark):** Measure a known volume of copper(II) sulfate solution using a measuring cylinder/pipette AND place it in a polystyrene cup (1 mark).
- **Temperature measurement (1 mark):** Measure/record the initial temperature of the solution AND record the maximum temperature reached (1 mark).
- **Reactants and stirring (1 mark):** Add a known mass of zinc (in excess) AND stir the mixture (1 mark).
- **Minimising heat loss (1 mark):** Use a polystyrene cup (as an insulator) AND use a lid on the cup (1 mark).
- **Logical structure (0.5 marks):** Steps are presented in a coherent, logical chronological sequence (0.5 marks).