Cambridge IGCSE · Thinka 原創模擬試題

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

Thinka Nov 2024 (V1) Cambridge International A Level-Style Mock — Chemistry (0620)

280 分300 分鐘2024
分析模擬試題
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.

Paper 11 選擇題 (Core)

There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
An atom of sodium has a nucleon number of 23 and a proton number of 11. What is the composition of a sodium ion, \(\text{Na}^+\)?
  1. A.11 protons, 12 neutrons, 10 electrons
  2. B.11 protons, 12 neutrons, 11 electrons
  3. C.11 protons, 12 neutrons, 12 electrons
  4. D.12 protons, 11 neutrons, 10 electronsHyvoksytyt: A; hylätyt muut valinnat (B, C, D).
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解題

The proton number of sodium is 11, so it has 11 protons. The number of neutrons is found by subtracting the proton number from the nucleon number: 23 - 11 = 12 neutrons. A sodium ion, \(\text{Na}^+\), is formed when a sodium atom loses one electron, so it has 11 - 1 = 10 electrons.

評分準則

1 mark for correct option A.
題目 2 · multiple_choice
1 分
Which pieces of apparatus are used to carry out an acid-base titration? 1. To measure exactly 25.0 cm\(^3\) of sodium hydroxide solution into a conical flask. 2. To add dilute hydrochloric acid until the indicator changes colour.
  1. A.1: pipette, 2: burette
  2. B.1: measuring cylinder, 2: pipette
  3. C.1: burette, 2: measuring cylinder
  4. D.1: beaker, 2: burette
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解題

A pipette is used to measure and transfer a highly accurate, fixed volume of liquid (such as 25.0 cm\(^3\)). A burette is used to add variable volumes of a solution with high accuracy until an endpoint is reached.

評分準則

1 mark for correct option A.
題目 3 · multiple_choice
1 分
Molten lead(II) bromide is electrolysed using carbon electrodes. Which statement about this electrolysis is correct?
  1. A.Lead is formed at the negative electrode (cathode).
  2. B.Bromine gas is formed at the negative electrode (cathode).
  3. C.The mass of the carbon anodes decreases.
  4. D.Lead ions gain electrons at the positive electrode (anode).
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解題

During the electrolysis of molten lead(II) bromide, positive lead ions (\(\text{Pb}^{2+}\)) are attracted to the negative electrode (cathode) where they gain electrons to form lead metal. Negative bromide ions (\(\text{Br}^-\)) are attracted to the positive electrode (anode) where they lose electrons to form bromine gas. The carbon electrodes are inert and do not change in mass.

評分準則

1 mark for correct option A.
題目 4 · multiple_choice
1 分
A student prepares a sample of hydrated copper(II) sulfate crystals by reacting excess copper(II) oxide with dilute sulfuric acid. Which process is used to remove the unreacted copper(II) oxide from the reaction mixture?
  1. A.Condensation
  2. B.Crystallisation
  3. C.Distillation
  4. D.Filtration
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解題

Excess copper(II) oxide is an insoluble solid base. After the reaction is complete, the unreacted copper(II) oxide remains suspended in the copper(II) sulfate solution. Filtration is used to separate the insoluble solid residue from the soluble salt solution (filtrate).

評分準則

1 mark for correct option D.
題目 5 · multiple_choice
1 分
Ethene molecules undergo addition polymerisation to form poly(ethene). Which statement about this process is correct?
  1. A.The polymer has a double carbon-carbon bond in its repeating unit.
  2. B.The monomer is a saturated hydrocarbon.
  3. C.There is no change in the empirical formula when ethene forms poly(ethene).
  4. D.Water is produced as a side product during this reaction.
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解題

Addition polymerisation involves monomer molecules (which are unsaturated hydrocarbons like ethene) joining together without losing any atoms. Therefore, the empirical formula of poly(ethene), \(\text{CH}_2\), is the same as that of ethene, \(\text{C}_2\text{H}_4\) (which simplifies to \(\text{CH}_2\)). No small molecules like water are lost, and the double bonds are converted to single bonds.

評分準則

1 mark for correct option C.
題目 6 · multiple_choice
1 分
A chromatogram is run using a safe food dye. The solvent front travels 8.0 cm from the baseline. The dye spot travels 6.0 cm from the baseline. What is the \(R_f\) value of this food dye?
  1. A.0.25
  2. B.0.75
  3. C.1.33
  4. D.4.80
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解題

The retardation factor (\(R_f\)) is calculated by dividing the distance moved by the substance by the distance moved by the solvent front: \(R_f = \frac{\text{distance travelled by dye}}{\text{distance travelled by solvent}} = \frac{6.0 \text{ cm}}{8.0 \text{ cm}} = 0.75\).

評分準則

1 mark for correct option B.
題目 7 · multiple_choice
1 分
How many shared pairs of electrons and how many non-bonding outer-shell electrons are there in one molecule of water, \(\text{H}_2\text{O}\)?
  1. A.2 shared pairs and 2 non-bonding outer-shell electrons
  2. B.2 shared pairs and 4 non-bonding outer-shell electrons
  3. C.4 shared pairs and 2 non-bonding outer-shell electrons
  4. D.4 shared pairs and 4 non-bonding outer-shell electrons
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解題

A water molecule has two single O-H covalent bonds, meaning there are 2 shared pairs of electrons. Oxygen is in Group VI, so it has 6 electrons in its outer shell. Since 2 of these electrons are shared in the covalent bonds, 4 non-bonding outer-shell electrons remain on the oxygen atom. Hydrogen has no non-bonding electrons.

評分準則

1 mark for correct option B.
題目 8 · multiple_choice
1 分
Which gas is a major greenhouse gas released by the decomposition of vegetation and waste in landfill sites?
  1. A.Carbon monoxide
  2. B.Methane
  3. C.Nitrogen dioxide
  4. D.Sulfur dioxide
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解題

Methane is a powerful greenhouse gas that is released during the anaerobic decomposition of organic waste in landfills and by the digestive processes of livestock.

評分準則

1 mark for correct option B.
題目 9 · 選擇題
1 分
A student performs a paper chromatography experiment to identify the food dyes present in a green sweet.

She places a spot of the green dye mixture on the baseline of a chromatogram, along with spots of three pure reference dyes: Yellow 4, Yellow 5, and Blue 1.

After running the chromatogram in a suitable solvent, the green spot separates into two spots. One spot travels the same distance as Yellow 5. The other spot travels the same distance as Blue 1.

Which dyes are present in the green sweet?
  1. A.Yellow 4 only
  2. B.Yellow 5 and Blue 1
  3. C.Yellow 4 and Blue 1
  4. D.Yellow 4 and Yellow 5
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解題

Chromatography separates mixtures into their components based on solubility. Since the spot from the green sweet separated into two spots that matched the heights (and therefore the \(R_f\) values) of Yellow 5 and Blue 1, the green sweet contains Yellow 5 and Blue 1. It does not contain Yellow 4 because no spot corresponding to the height of Yellow 4 was detected from the green sweet.

評分準則

1 mark for the correct option B.
題目 10 · 選擇題
1 分
An atom of an element has a nucleon number of 31 and contains 16 neutrons in its nucleus.

Which row in the table correctly identifies the group and period of this element in the Periodic Table?

$$\begin{array}{|c|c|c|} \hline & \text{Group} & \text{Period} \\ \hline \text{A} & \text{V} & 3 \\ \hline \text{B} & \text{V} & 4 \\ \hline \text{C} & \text{VI} & 3 \\ \hline \text{D} & \text{VI} & 4 \\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
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解題

1. Calculate the atomic number (number of protons): \(31 - 16 = 15\).
2. Write down the electronic configuration for atomic number 15: \(2, 8, 5\).
3. Identify the group from the number of outer shell electrons: 5 outer electrons means Group V.
4. Identify the period from the number of electron shells: 3 shells occupied means Period 3.
Therefore, the element is in Group V, Period 3, which corresponds to row A.

評分準則

1 mark for the correct option A.
題目 11 · 選擇題
1 分
During the electrolysis of molten lead(II) bromide, \(\text{PbBr}_2\), using inert carbon electrodes, several observations and processes occur.

Which statements are correct?

1. Lead ions move towards the negative electrode (cathode) where they lose electrons.
2. Bromide ions move towards the positive electrode (anode) where they gain electrons.
3. A grey liquid metal forms at the negative electrode.
4. A brown gas is evolved at the positive electrode.
  1. A.1 and 2
  2. B.1 and 3
  3. C.2 and 4
  4. D.3 and 4
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解題

Let's analyse each statement:
- Statement 1 is incorrect: Lead ions (\(\text{Pb}^{2+}\)) move to the negative electrode (cathode) but they gain electrons (reduction) to form lead atoms: \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\).
- Statement 2 is incorrect: Bromide ions (\(\text{Br}^-\)) move to the positive electrode (anode) where they lose electrons (oxidation) to form bromine molecules: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\).
- Statement 3 is correct: Metallic lead forms as a grey liquid at the cathode under the high temperatures used.
- Statement 4 is correct: Bromine vapor forms as a brown gas at the anode.

Therefore, only statements 3 and 4 are correct.

評分準則

1 mark for the correct option D.
題目 12 · 選擇題
1 分
Copper(II) sulfate is a soluble salt. It can be prepared by reacting insoluble copper(II) oxide with dilute sulfuric acid.

Which sequence of steps describes the correct method to obtain pure, dry crystals of hydrated copper(II) sulfate from these starting materials?
  1. A.add excess copper(II) oxide to dilute sulfuric acid $\rightarrow$ filter $\rightarrow$ heat the filtrate to dryness
  2. B.add excess copper(II) oxide to dilute sulfuric acid $\rightarrow$ filter $\rightarrow$ heat the filtrate to the point of crystallisation $\rightarrow$ cool $\rightarrow$ filter and dry the crystals
  3. C.add excess dilute sulfuric acid to copper(II) oxide $\rightarrow$ filter $\rightarrow$ evaporate all the water $\rightarrow$ wash the residue with water
  4. D.mix equal volumes of dilute sulfuric acid and copper(II) oxide $\rightarrow$ evaporate the mixture to dryness $\rightarrow$ wash with acid
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解題

To prepare a soluble salt from an acid and an insoluble base:
1. Add excess insoluble base (copper(II) oxide) to the acid (dilute sulfuric acid) to ensure all the acid is neutralised.
2. Filter the mixture to remove the unreacted excess copper(II) oxide.
3. Heat the filtrate (copper(II) sulfate solution) until the crystallization point is reached.
4. Allow the solution to cool so that hydrated crystals can form.
5. Filter to collect the crystals, wash with a small volume of cold distilled water to remove soluble impurities, and dry them.

評分準則

1 mark for the correct option B.
題目 13 · 選擇題
1 分
A section of a synthetic polymer chain is shown below:

$$-\text{CH}_2-\text{CH(Cl)}-\text{CH}_2-\text{CH(Cl)}-\text{CH}_2-\text{CH(Cl)}-$$

Which monomer is used to produce this polymer?
  1. A.$\text{CH}_2=\text{CH}_2$
  2. B.$\text{CH}_2=\text{CHCl}$
  3. C.$\text{CHCl}=\text{CHCl}$
  4. D.$\text{CH}_3-\text{CH}_2\text{Cl}$
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解題

The polymer shown is an addition polymer, poly(chloroethene) (PVC). Its repeating unit is \(-\text{CH}_2-\text{CH(Cl)}-\). To identify the monomer, we replace the single carbon-carbon bond of the repeating unit with a double bond, which gives \(\text{CH}_2=\text{CHCl}\).

評分準則

1 mark for the correct option B.
題目 14 · 選擇題
1 分
An aqueous solution of an unknown salt $X$ is tested.

- The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide.
- The addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.

What is the chemical name of salt $X$?
  1. A.copper(II) chloride
  2. B.copper(II) sulfate
  3. C.iron(II) chloride
  4. D.iron(II) sulfate
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解題

1. The reaction with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. This confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\).
2. The reaction with barium nitrate in the presence of dilute nitric acid produces a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).

Therefore, salt \(X\) is iron(II) sulfate.

評分準則

1 mark for the correct option D.
題目 15 · 選擇題
1 分
Which row correctly describes the trends in colour, physical state at room temperature, and reactivity of the Group VII elements (halogens) as you go down the group?

$$\begin{array}{|c|c|c|c|} \hline & \text{Colour} & \text{Physical state at room temp.} & \text{Reactivity} \\ \hline \text{A} & \text{gets lighter} & \text{gas to solid} & \text{increases} \\ \hline \text{B} & \text{gets lighter} & \text{solid to gas} & \text{decreases} \\ \hline \text{C} & \text{gets darker} & \text{gas to solid} & \text{decreases} \\ \hline \text{D} & \text{gets darker} & \text{solid to gas} & \text{increases} \\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
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解題

As you go down Group VII (the halogens):
- **Colour**: gets darker (fluorine is pale yellow, chlorine is pale green, bromine is red-brown, iodine is grey-black).
- **Physical state at room temperature**: changes from gas to liquid to solid (i.e., gas to solid).
- **Reactivity**: decreases because the outer shell is further from the nucleus, making it harder to attract an incoming electron.

This matches row C.

評分準則

1 mark for the correct option C.
題目 16 · 選擇題
1 分
Which statement about alkenes is correct?
  1. A.Alkenes are saturated hydrocarbons containing only single carbon-carbon bonds.
  2. B.Alkenes react with aqueous bromine, causing the colour of the solution to change from colourless to orange.
  3. C.Alkenes are formed when long-chain alkanes are broken down by cracking.
  4. D.Alkenes have the general molecular formula $\text{C}_n\text{H}_{2n+2}$.
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解題

- Option A is incorrect because alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.
- Option B is incorrect because alkenes decolourise aqueous bromine (changing the colour from orange-brown to colourless).
- Option C is correct because the cracking of long-chain alkanes produces shorter-chain alkanes and alkenes.
- Option D is incorrect because the general formula for alkenes is \(\text{C}_n\text{H}_{2n}\).

評分準則

1 mark for the correct option C.
題目 17 · 選擇題
1 分
Which separation technique is used to obtain pure water from seawater?
  1. A.filtration
  2. B.simple distillation
  3. C.fractional distillation
  4. D.paper chromatography
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解題

Simple distillation is used to separate a volatile solvent (such as water) from non-volatile solutes (such as dissolved salts in seawater). The water evaporates, is condensed in the condenser, and is collected as pure water.

評分準則

1 mark for correct answer B. Correctly identifies simple distillation as the separation technique.
題目 18 · 選擇題
1 分
An atom of carbon-12 and an atom of carbon-14 are isotopes of carbon. Which statement about these two atoms is correct?
  1. A.They have a different number of protons.
  2. B.They have a different number of electron shells.
  3. C.They have the same number of neutrons.
  4. D.They have the same chemical properties because they have the same number of outer shell electrons.
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解題

Isotopes of the same element have the same number of protons and electrons, and therefore the same electronic configuration. Since chemical properties depend on the arrangement of outer shell electrons, isotopes have identical chemical properties.

評分準則

1 mark for correct answer D. Explains the connection between isotopes, outer shell electrons, and chemical properties.
題目 19 · 選擇題
1 分
A student investigates the reaction between dilute hydrochloric acid and excess calcium carbonate chips. Which change would decrease the initial rate of the reaction?
  1. A.using a larger volume of the same dilute hydrochloric acid
  2. B.using the same mass of calcium carbonate as a fine powder
  3. C.lowering the temperature of the acid
  4. D.increasing the concentration of the acid
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解題

Lowering the temperature reduces the average kinetic energy of the reacting particles. As a result, there are fewer collisions per second and a smaller proportion of collisions have energy greater than or equal to the activation energy, decreasing the reaction rate.

評分準則

1 mark for correct answer C. Identifies that lowering temperature decreases the reaction rate.
題目 20 · 選擇題
1 分
Which statement about the Group I elements lithium, sodium and potassium is correct?
  1. A.Sodium is harder than lithium.
  2. B.Potassium reacts more vigorously with water than sodium.
  3. C.Their melting points increase down the group.
  4. D.They form acidic oxides when burned in oxygen.
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解題

Reactivity of Group I alkali metals increases down the group. Therefore, potassium reacts more vigorously with water than sodium, which in turn reacts more vigorously than lithium.

評分準則

1 mark for correct answer B. Correctly identifies the trend in reactivity down Group I.
題目 21 · 選擇題
1 分
How many shared pairs of electrons are there in a molecule of nitrogen, \(\text{N}_2\)?
  1. A.1
  2. B.2
  3. C.3
  4. D.6
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解題

Each nitrogen atom has 5 electrons in its outer shell. To achieve a stable octet, two nitrogen atoms share 3 electrons each, forming a triple covalent bond. This triple bond consists of 3 shared pairs of electrons.

評分準則

1 mark for correct answer C. Correctly determines the number of shared pairs of electrons in a triple covalent nitrogen-nitrogen bond.
題目 22 · 選擇題
1 分
A student prepares a pure sample of hydrated copper(II) sulfate crystals by adding excess insoluble copper(II) oxide to warm dilute sulfuric acid. Which sequence of steps should they perform next to obtain the crystals?
  1. A.filter the mixture \(\rightarrow\) evaporate the filtrate to dryness
  2. B.filter the mixture \(\rightarrow\) heat the filtrate to crystallisation point \(\rightarrow\) leave to cool and crystallise \(\rightarrow\) filter and dry the crystals
  3. C.heat the mixture to crystallisation point \(\rightarrow\) filter the mixture \(\rightarrow\) dry the residue
  4. D.crystallise the mixture directly \(\rightarrow\) filter to remove excess copper(II) oxide \(\rightarrow\) wash with water
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解題

The unreacted insoluble copper(II) oxide must first be removed by filtration. The filtrate containing dissolved copper(II) sulfate is then heated to evaporate water until the crystallisation point is reached. The hot solution is allowed to cool and crystallise. Finally, the crystals are filtered from the remaining solution and dried using filter paper.

評分準則

1 mark for correct answer B. Identifies the correct order of experimental steps to prepare and isolate crystals from an insoluble reactant.
題目 23 · 選擇題
1 分
Which statement correctly classifies the oxides of sulfur, calcium, and aluminium?
  1. A.Sulfur dioxide is acidic, calcium oxide is basic, and aluminium oxide is amphoteric.
  2. B.Sulfur dioxide is basic, calcium oxide is acidic, and aluminium oxide is amphoteric.
  3. C.Sulfur dioxide is amphoteric, calcium oxide is basic, and aluminium oxide is acidic.
  4. D.Sulfur dioxide is acidic, calcium oxide is amphoteric, and aluminium oxide is basic.
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解題

Sulfur dioxide is a non-metal oxide, so it is an acidic oxide. Calcium oxide is a metal oxide, which reacts with acids to form salts, making it a basic oxide. Aluminium oxide reacts with both acids and bases, classifying it as an amphoteric oxide.

評分準則

1 mark for correct answer A. Correctly groups sulfur dioxide as acidic, calcium oxide as basic, and aluminium oxide as amphoteric.
題目 24 · 選擇題
1 分
Which compound immediately decolorizes aqueous bromine?
  1. A.ethane
  2. B.ethanol
  3. C.ethene
  4. D.ethyl ethanoate
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解題

Ethene is an alkene (an unsaturated hydrocarbon with a carbon-carbon double bond). It reacts rapidly with aqueous bromine in an addition reaction, causing the orange-brown colour of the bromine water to disappear (decolorize).

評分準則

1 mark for correct answer C. Identifies ethene as the unsaturated hydrocarbon that decolorizes bromine water.
題目 25 · 選擇題
1 分
How many protons, neutrons, and electrons are present in one sodium ion, \(^{23}_{11}\text{Na}^+\)?
  1. A.11 protons, 12 neutrons, 10 electrons
  2. B.11 protons, 12 neutrons, 11 electrons
  3. C.11 protons, 12 neutrons, 12 electrons
  4. D.12 protons, 11 neutrons, 10 electrons
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解題

The atomic number is 11, which represents the number of protons (11). The mass number is 23, so the number of neutrons is \(23 - 11 = 12\). The sodium ion has a \(+1\) charge, meaning it has lost 1 electron, leaving it with \(11 - 1 = 10\) electrons.

評分準則

Award 1 mark for selecting correct option A. Correctly identifies 11 protons, 12 neutrons, and 10 electrons.
題目 26 · 選擇題
1 分
Which piece of apparatus is used to accurately measure out a fixed volume of \(25.0\text{ cm}^3\) of sodium hydroxide solution into a conical flask, and which is used to add the dilute hydrochloric acid gradually until the end-point is reached?
  1. A.Pipette for sodium hydroxide; burette for dilute hydrochloric acid
  2. B.Burette for sodium hydroxide; pipette for dilute hydrochloric acid
  3. C.Measuring cylinder for sodium hydroxide; pipette for dilute hydrochloric acid
  4. D.Pipette for sodium hydroxide; measuring cylinder for dilute hydrochloric acid
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解題

A volumetric pipette is specifically designed to measure and deliver an extremely accurate fixed volume (such as \(25.0\text{ cm}^3\)) of liquid. A burette is designed to deliver variable volumes of solution dropwise to find the precise end-point of a titration.

評分準則

Award 1 mark for selecting option A, matching the correct apparatus to their respective volumetric tasks.
題目 27 · 選擇題
1 分
Molten lead(II) bromide is electrolyzed using inert carbon electrodes. What are the products formed at the anode and cathode?
  1. A.Anode: lead; Cathode: bromine
  2. B.Anode: bromine; Cathode: lead
  3. C.Anode: oxygen; Cathode: lead
  4. D.Anode: hydrogen; Cathode: bromine
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解題

Lead(II) bromide consists of positive lead ions, \(\text{Pb}^{2+}\), and negative bromide ions, \(\text{Br}^-\). During electrolysis, positive lead ions move to the cathode (negative electrode) where they gain electrons to form lead metal. Negative bromide ions move to the anode (positive electrode) where they lose electrons to form bromine gas.

評分準則

Award 1 mark for selecting option B, which correctly identifies bromine at the anode and lead at the cathode.
題目 28 · 選擇題
1 分
A student prepares dry crystals of copper(II) sulfate by reacting excess solid copper(II) oxide with dilute sulfuric acid. Which sequence of steps should the student perform after the reaction is complete to obtain pure, dry crystals?
  1. A.filter \(\rightarrow\) evaporate to dryness \(\rightarrow\) wash with water
  2. B.filter \(\rightarrow\) heat to crystallizing point \(\rightarrow\) cool \(\rightarrow\) filter off crystals and dry them
  3. C.evaporate to dryness \(\rightarrow\) filter \(\rightarrow\) wash with ethanol
  4. D.cool \(\rightarrow\) filter \(\rightarrow\) evaporate to dryness \(\rightarrow\) wash with acid
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解題

First, the excess insoluble copper(II) oxide must be filtered off to obtain a pure copper(II) sulfate solution. The solution is then heated to the point of crystallization (evaporating some water to make it saturated), cooled slowly so crystals can form, and finally, the crystals are filtered out of the remaining solution and dried.

評分準則

Award 1 mark for option B, describing the correct experimental sequence of filtration, crystallization, cooling, and drying.
題目 29 · 選擇題
1 分
Which statement correctly describes the formation and structure of poly(ethene)?
  1. A.It is formed by a condensation reaction where water is produced, and the polymer contains double bonds.
  2. B.It is formed by an addition reaction, and the polymer contains only single carbon-to-carbon bonds.
  3. C.It is formed by an addition reaction, and the polymer contains double carbon-to-carbon bonds.
  4. D.It is formed by a condensation reaction, and the polymer contains only single carbon-to-carbon bonds.
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解題

Poly(ethene) is formed via addition polymerization. During this reaction, many ethene monomer molecules join together without the loss of any small molecules, and the double bonds in ethene open up to form a polymer chain containing only single carbon-to-carbon bonds.

評分準則

Award 1 mark for option B, identifying addition polymerization and the presence of only single carbon-to-carbon bonds.
題目 30 · 選擇題
1 分
An aqueous solution of an unknown salt is tested. Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. Which cation is present in the solution?
  1. A.\(\text{Fe}^{2+}\)
  2. B.\(\text{Fe}^{3+}\)
  3. C.\(\text{Cu}^{2+}\)
  4. D.\(\text{Cr}^{3+}\)
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解題

Iron(II) ions (\(\text{Fe}^{2+}\)) react with hydroxide ions to form a green precipitate of iron(II) hydroxide, which does not dissolve when excess sodium hydroxide is added. In contrast, chromium(III) ions also form a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution.

評分準則

Award 1 mark for option A, identifying \(\text{Fe}^{2+}\) as the cation forming a green precipitate insoluble in excess sodium hydroxide.
題目 31 · 選擇題
1 分
A student sets up a paper chromatography experiment to separate the dyes in a green food coloring. Which experimental setup is correct?
  1. A.The start line is drawn in ink, and the solvent level is above the start line.
  2. B.The start line is drawn in pencil, and the solvent level is above the start line.
  3. C.The start line is drawn in ink, and the solvent level is below the start line.
  4. D.The start line is drawn in pencil, and the solvent level is below the start line.
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解題

The start line must be drawn in pencil because graphite is insoluble in the chromatography solvents and will not run. The solvent level in the beaker must be below the start line so that the spots of food coloring do not dissolve directly into the solvent pool at the bottom.

評分準則

Award 1 mark for option D, indicating the pencil line and the solvent level below the start line.
題目 32 · 選擇題
1 分
Which air pollutant is correctly matched with its main source of release into the atmosphere?
  1. A.Carbon monoxide — complete combustion of hydrocarbon fuels
  2. B.Oxides of nitrogen — car engines where nitrogen and oxygen react at high temperatures
  3. C.Sulfur dioxide — decomposition of vegetation in modern landfills
  4. D.Methane — combustion of coal in power stations
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解題

Oxides of nitrogen are formed inside car engines due to the high temperature, which provides enough energy for nitrogen and oxygen from the air to react together. Carbon monoxide comes from incomplete combustion, sulfur dioxide from burning sulfur-containing fossil fuels, and methane mainly from agricultural activities and decomposing waste.

評分準則

Award 1 mark for option B, identifying the high-temperature reaction in car engines as the source of oxides of nitrogen.
題目 33 · 選擇題
1 分
A student performs chromatography on a sample of black ink. The start line is drawn 1.5 cm from the bottom of the chromatography paper. After running the chromatogram, the solvent front is 6.0 cm above the start line. A red spot in the ink has travelled 4.5 cm from the start line. What is the \(R_f\) value of this red spot?
  1. A.0.25
  2. B.0.60
  3. C.0.75
  4. D.1.33
查看答案詳解

解題

The value of \(R_f\) is calculated using the formula: \(R_f = \frac{\text{distance travelled by the spot}}{\text{distance travelled by the solvent front}}\). Substituting the given values: \(R_f = \frac{4.5\text{ cm}}{6.0\text{ cm}} = 0.75\). The distance from the bottom of the paper (1.5 cm) is not used in the calculation.

評分準則

Award 1 mark for the correct calculation of \(R_f\) (4.5 divided by 6.0) leading to 0.75.
題目 34 · 選擇題
1 分
An atom of phosphorus has the symbol \(^{31}_{15}\text{P}\). Which statement correctly describes the subatomic particles in this atom?
  1. A.It has 15 protons, 15 neutrons and 16 electrons.
  2. B.It has 15 protons, 16 neutrons and 15 electrons.
  3. C.It has 16 protons, 15 neutrons and 15 electrons.
  4. D.31 protons, 15 neutrons and 15 electrons.
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解題

The atomic number (bottom number, 15) represents the number of protons, which is also equal to the number of electrons in a neutral atom (15). The mass number (top number, 31) represents the total number of protons and neutrons. The number of neutrons is found by subtracting the atomic number from the mass number: \(31 - 15 = 16\).

評分準則

Award 1 mark for the correct identification of the number of protons (15), neutrons (16), and electrons (15).
題目 35 · 選擇題
1 分
Which pair of aqueous solutions, when mixed together, will produce a precipitate of the insoluble salt silver chloride?
  1. A.silver nitrate and sodium chloride
  2. B.silver carbonate and hydrochloric acid
  3. C.silver nitrate and copper
  4. D.silver oxide and sodium chloride
查看答案詳解

解題

To prepare an insoluble salt by precipitation, two soluble salt solutions must react together. Silver nitrate (\(\text{AgNO}_3\)) and sodium chloride (\(\text{NaCl}\)) are both soluble in water. Mixing them produces a precipitate of silver chloride (\(\text{AgCl}\)) and soluble sodium nitrate (\(\text{NaNO}_3\)).

評分準則

Award 1 mark for choosing the two soluble starting reactants that successfully yield a precipitate of silver chloride.
題目 36 · 選擇題
1 分
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at the anode and the cathode?
  1. A.Anode: bromine; Cathode: lead
  2. B.Anode: hydrogen; Cathode: lead
  3. C.Anode: lead; Cathode: bromine
  4. D.Anode: oxygen; Cathode: hydrogen
查看答案詳解

解題

During the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)), the positive lead ions (\(\text{Pb}^{2+}\)) migrate to the negative electrode (cathode) where they gain electrons to form lead metal. The negative bromide ions (\(\text{Br}^-\)) migrate to the positive electrode (anode) where they lose electrons to form bromine gas.

評分準則

Award 1 mark for correctly identifying that bromine is formed at the anode and lead is formed at the cathode.
題目 37 · 選擇題
1 分
A student carries out an acid-base titration. Which pieces of apparatus are most suitable for measuring exactly \(25.0\text{ cm}^3\) of the alkali into a conical flask and for adding the acid during the titration?
  1. A.To measure alkali: beaker; To add acid: measuring cylinder
  2. B.To measure alkali: volumetric pipette; To add acid: burette
  3. C.To measure alkali: burette; To add acid: volumetric pipette
  4. D.To measure alkali: measuring cylinder; To add acid: beaker
查看答案詳解

解題

A volumetric pipette is designed to measure and deliver a highly accurate, fixed volume of liquid (such as \(25.0\text{ cm}^3\)). A burette is designed to measure and deliver variable, precise volumes of liquid during a reaction, making it suitable for adding the acid in a titration.

評分準則

Award 1 mark for selecting the volumetric pipette for measuring the alkali and the burette for adding the acid.
題目 38 · 選擇題
1 分
Which statement about the polymer poly(ethene) is correct?
  1. A.It is formed by a condensation reaction with the elimination of water.
  2. B.It is unsaturated because it contains double covalent bonds.
  3. C.It is formed from ethene monomers that have the molecular formula \(\text{C}_2\text{H}_4\).
  4. D.It has a different percentage by mass of carbon than its monomer.
查看答案詳解

解題

Poly(ethene) is an addition polymer formed when ethene monomers (\(\text{C}_2\text{H}_4\)) join together. Since it is formed by addition polymerisation, no small molecules like water are eliminated, and the repeating unit has the same empirical formula and percentage composition as the monomer, but with only single covalent bonds (saturated).

評分準則

Award 1 mark for identifying that poly(ethene) is formed from ethene monomers with the molecular formula \(\text{C}_2\text{H}_4\).
題目 39 · 選擇題
1 分
A solid sample is heated with aqueous sodium hydroxide. A gas is evolved that turns damp red litmus paper blue. Which ion is present in the solid sample?
  1. A.ammonium (\(\text{NH}_4^+\))
  2. B.carbonate (\(\text{CO}_3^{2-}\))
  3. C.chloride (\(\text{Cl}^-\))
  4. D.copper(II) (\(\text{Cu}^{2+}\))
查看答案詳解

解題

Heating an ammonium salt with aqueous sodium hydroxide releases ammonia gas (\(\text{NH}_3\)). Ammonia is an alkaline gas, which is identified because it turns damp red litmus paper blue. Therefore, the solid contains the ammonium ion (\(\text{NH}_4^+\)).

評分準則

Award 1 mark for identifying the ammonium ion based on the evolution of ammonia gas.
題目 40 · 選擇題
1 分
Which statement describes a characteristic of members of the same homologous series?
  1. A.They have the same physical properties such as boiling point and density.
  2. B.They have the same general formula and similar chemical properties.
  3. C.They have different general formulas but the same molecular mass.
  4. D.They have the same molecular formula but different arrangements of atoms.
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解題

Members of the same homologous series share the same general formula, have similar chemical properties (because they contain the same functional group), and display a gradual trend in physical properties (such as boiling point increasing as molecular size increases).

評分準則

Award 1 mark for selecting the correct characteristic of a homologous series (same general formula and similar chemical properties).

Paper 21 選擇題 (Extended)

There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct.
32 題目 · 32 分
題目 1 · 選擇題
1 分
An electrolysis cell contains concentrated aqueous sodium chloride using inert carbon electrodes. What are the products at the electrodes and the change in pH of the electrolyte during electrolysis?
  1. A.Cathode: hydrogen; Anode: oxygen; pH: remains unchanged
  2. B.Cathode: sodium; Anode: chlorine; pH: decreases
  3. C.Cathode: hydrogen; Anode: chlorine; pH: increases
  4. D.Cathode: chlorine; Anode: hydrogen; pH: decreases
查看答案詳解

解題

During the electrolysis of concentrated aqueous sodium chloride:
1. At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged preferentially over hydroxide ions (\(\text{OH}^-\)) because of their high concentration, forming chlorine gas (\(\text{Cl}_2\)).
2. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged preferentially over sodium ions (\(\text{Na}^+\)) because hydrogen is less reactive than sodium, forming hydrogen gas (\(\text{H}_2\)).
3. The remaining ions in solution are sodium (\(\text{Na}^+\)) and hydroxide (\(\text{OH}^-\)), which form sodium hydroxide (\(\text{NaOH}\)), an alkaline solution. Therefore, the pH of the electrolyte increases.

評分準則

1 mark for the correct option. Award 1 mark for identifying hydrogen at the cathode, chlorine at the anode, and an increase in pH. Reject all other combinations.
題目 2 · 選擇題
1 分
A student runs a chromatogram of a food dye mixture. The solvent front travels 8.0 cm from the baseline. Spot X travels 3.2 cm from the baseline. What is the \(R_f\) value of spot X, and is spot X more or less soluble in the mobile phase than a spot with an \(R_f\) value of 0.75?
  1. A.\(R_f = 0.40\); spot X is more soluble in the mobile phase
  2. B.\(R_f = 0.40\); spot X is less soluble in the mobile phase
  3. C.\(R_f = 2.50\); spot X is more soluble in the mobile phase
  4. D.\(R_f = 2.50\); spot X is less soluble in the mobile phase
查看答案詳解

解題

The retention factor is calculated as: \(R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}} = \frac{3.2\text{ cm}}{8.0\text{ cm}} = 0.40\).
A higher \(R_f\) value means the substance is more soluble in the mobile phase (solvent) and travels further. Since spot X has an \(R_f\) value of 0.40, which is lower than 0.75, spot X is less soluble in the mobile phase.

評分準則

1 mark for the correct option. Award 1 mark for the correct calculation of \(R_f = 0.40\) and the correct deduction that it is less soluble.
題目 3 · 選擇題
1 分
A student prepares a pure, dry sample of zinc sulfate crystals by reacting excess zinc powder with dilute sulfuric acid. Which sequence of experimental steps is correct to obtain the crystals?
  1. A.Filter the reaction mixture \(\rightarrow\) heat the filtrate to dryness \(\rightarrow\) wash the remaining dry powder with water
  2. B.Filter the reaction mixture \(\rightarrow\) heat the filtrate to the point of crystallisation \(\rightarrow\) allow to cool \(\rightarrow\) filter and dry the crystals with filter paper
  3. C.Heat the reaction mixture to dryness \(\rightarrow\) filter to remove excess zinc \(\rightarrow\) dry the remaining crystals in an oven
  4. D.Heat the reaction mixture to the point of crystallisation \(\rightarrow\) filter to remove excess zinc \(\rightarrow\) wash the filtrate with cold water
查看答案詳解

解題

The correct procedure for preparing a soluble salt from an acid and an insoluble metal/base is:
1. React excess solid (zinc) with the acid to ensure all the acid is fully neutralized.
2. Filter the mixture to remove the unreacted excess zinc powder.
3. Heat the filtrate to evaporate water until the point of crystallisation (saturation) is reached.
4. Allow the saturated solution to cool so that crystals can form.
5. Filter the crystals from the remaining liquid and dry them between sheets of filter paper. Heating to dryness (as in Option A and C) would result in anhydrous powder rather than hydrated crystals.

評分準則

1 mark for the correct option. Reject options that involve evaporating to dryness or filtering at incorrect stages.
題目 4 · 選擇題
1 分
A \(25.0\text{ cm}^3\) sample of aqueous sodium hydroxide of unknown concentration is titrated against \(0.0500\text{ mol/dm}^3\) sulfuric acid. Exactly \(20.0\text{ cm}^3\) of sulfuric acid is required for complete neutralization. \$$\text{2NaOH(aq)} + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O(l)}\$$ What is the concentration of the sodium hydroxide solution?
  1. A.0.0200 mol/dm³
  2. B.0.0400 mol/dm³
  3. C.0.0800 mol/dm³
  4. D.0.160 mol/dm³
查看答案詳解

解題

Step 1: Calculate moles of sulfuric acid: \(\text{moles} = \text{concentration} \times \text{volume in dm}^3 = 0.0500\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.00100\text{ mol}\).
Step 2: Use the stoichiometry of the balanced chemical equation. \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\).
Therefore, \(\text{moles of NaOH} = 2 \times 0.00100\text{ mol} = 0.00200\text{ mol}\).
Step 3: Calculate the concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00200\text{ mol}}{25.0 / 1000\text{ dm}^3} = 0.0800\text{ mol/dm}^3\).

評分準則

1 mark for the correct option. Accept \(0.0800\text{ mol/dm}^3\). Method marks are incorporated within the single mark.
題目 5 · 選擇題
1 分
An ion of element X is represented by \(^{37}_{17}\text{X}^-\). Which row correctly describes the number of protons, neutrons, and electrons in this ion?
  1. A.Protons: 17; Neutrons: 20; Electrons: 16
  2. B.Protons: 17; Neutrons: 20; Electrons: 18
  3. C.Protons: 17; Neutrons: 37; Electrons: 18
  4. D.Protons: 18; Neutrons: 20; Electrons: 17
查看答案詳解

解題

For the ion \(^{37}_{17}\text{X}^-\):
1. The atomic number is 17, which means there are 17 protons.
2. The mass number is 37. The number of neutrons is calculated as: \(\text{mass number} - \text{atomic number} = 37 - 17 = 20\).
3. The ion has a \(1-\$$ charge, meaning it has gained 1 electron compared to a neutral atom. The number of electrons is: \)17 + 1 = 18\).

評分準則

1 mark for the correct option. Reject any options with incorrect particle counts.
題目 6 · 選擇題
1 分
Which statement about polymers is correct?
  1. A.Terylene is a polyamide formed by condensation polymerisation.
  2. B.Nylon is a polyester formed by the elimination of water molecules.
  3. C.Addition polymerisation of ethene produces a polymer containing carbon-carbon double bonds.
  4. D.Hydrolysis of proteins produces amino acids.
查看答案詳解

解題

Let's analyze the statements:
- Option A is incorrect because Terylene is a polyester, not a polyamide.
- Option B is incorrect because Nylon is a polyamide, not a polyester.
- Option C is incorrect because addition polymerisation of ethene results in poly(ethene), which contains only carbon-carbon single bonds.
- Option D is correct because proteins are polyamides (condensation polymers) that can be hydrolysed back into their constituent monomer amino acids.

評分準則

1 mark for the correct option. Reject options with incorrect definitions of polymers or reaction products.
題目 7 · 選擇題
1 分
A solid salt Y is dissolved in distilled water to make an aqueous solution.
- Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess.
- Adding dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the chemical formula of salt Y?
  1. A.\(\text{Fe}_2(\text{SO}_4)_3\)
  2. B.\(\text{FeSO}_4\)
  3. C.\(\text{Cr}_2(\text{SO}_4)_3\)
  4. D.\(\text{FeCl}_2\)
查看答案詳解

解題

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

評分準則

1 mark for the correct option. Reject other salts that do not match both qualitative test results.
題目 8 · 選擇題
1 分
When but-2-ene is reacted with steam in the presence of an acid catalyst, an addition reaction occurs. What is the name and molecular formula of the product formed?
  1. A.butan-1-ol, \(\text{C}_4\text{H}_8\text{O}\)
  2. B.butan-2-ol, \(\text{C}_4\text{H}_{10}\text{O}\)
  3. C.butane, \(\text{C}_4\text{H}_{10}\)
  4. D.butan-2-ol, \(\text{C}_4\text{H}_8\text{O}\)
查看答案詳解

解題

But-2-ene is an alkene with the structure \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\).
Reacting it with steam (water, \(\text{H}_2\text{O}\)) in the presence of an acid catalyst results in hydration (an addition reaction).
The hydrogen atom (\(\text{H}\)) adds to one of the double-bonded carbons, and the hydroxyl group (\(-\text{OH}\)) adds to the other double-bonded carbon.
This forms butan-2-ol, \(\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH}_3\).
The molecular formula of butan-2-ol (an alcohol with four carbons) is \(\text{C}_4\text{H}_{10}\text{O}\).

評分準則

1 mark for the correct option. Reject options with incorrect chemical structures or incorrect molecular formulas.
題目 9 · 選擇題
1 分
An ion \( X^{2-} \) has a nucleon number of 34 and contains 18 electrons. What is the composition of the nucleus of an atom of isotope \( X-36 \)?
  1. A.16 protons and 18 neutrons
  2. B.16 protons and 20 neutrons
  3. C.18 protons and 18 neutrons
  4. D.18 protons and 20 neutrons Gold-36 is not the element, X is Sulfur (S-36).
查看答案詳解

解題

1. The ion has a 2- charge and contains 18 electrons, which means the neutral atom of element X has \( 18 - 2 = 16 \) electrons, and therefore 16 protons. 2. An isotope of element X (which has 16 protons) with a nucleon number of 36 (isotope \( X-36 \)) contains \( 36 - 16 = 20 \) neutrons. 3. The nucleus of an atom of this isotope consists of 16 protons and 20 neutrons.

評分準則

1 mark for the correct option B. Award 1 mark for identifying the correct number of protons (16) and neutrons (20). Reject other options.
題目 10 · 選擇題
1 分
A student titrates \( 25.0\text{ cm}^3 \) of \( 0.0500\text{ mol/dm}^3 \) sulfuric acid (\( \text{H}_2\text{SO}_4 \)) with sodium hydroxide solution (\( \text{NaOH} \)). The volume of \( \text{NaOH} \) required for complete neutralisation is \( 20.0\text{ cm}^3 \). What is the concentration of the sodium hydroxide solution?
  1. A.0.0313 mol/dm\(^3\)
  2. B.0.0625 mol/dm\(^3\)
  3. C.0.125 mol/dm\(^3\)
  4. D.0.250 mol/dm\(^3\)
查看答案詳解

解題

1. Write the balanced equation: \( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \). 2. Calculate the moles of sulfuric acid: \( n = 0.0250\text{ dm}^3 \times 0.0500\text{ mol/dm}^3 = 0.00125\text{ mol} \). 3. Use the stoichiometric ratio (1:2) to find the moles of sodium hydroxide: \( n(\text{NaOH}) = 2 \times 0.00125\text{ mol} = 0.00250\text{ mol} \). 4. Calculate the concentration of sodium hydroxide: \( \text{concentration} = \frac{0.00250\text{ mol}}{0.0200\text{ dm}^3} = 0.125\text{ mol/dm}^3 \).

評分準則

1 mark for the correct option C. Award 1 mark for correct molar ratio step and concentration calculation.
題目 11 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. What are the products formed at each electrode and how does the pH of the electrolyte change?
  1. A.Anode: chlorine; Cathode: hydrogen; pH: decreases
  2. B.Anode: chlorine; Cathode: hydrogen; pH: increases
  3. C.Anode: oxygen; Cathode: sodium; pH: increases
  4. D.Anode: oxygen; Cathode: hydrogen; pH: remains unchanged
查看答案詳解

解題

During the electrolysis of concentrated aqueous sodium chloride: 1. At the anode (+), chloride ions (\( \text{Cl}^- \)) are discharged in preference to hydroxide ions because of their high concentration, producing chlorine gas. 2. At the cathode (-), hydrogen ions (\( \text{H}^+ \)) are discharged in preference to sodium ions because hydrogen is lower in the reactivity series, producing hydrogen gas. 3. Sodium ions (\( \text{Na}^+ \)) and hydroxide ions (\( \text{OH}^- \)) remain in the solution, forming sodium hydroxide, which is alkaline, so the pH increases.

評分準則

1 mark for the correct option B. Reject options with incorrect products or incorrect pH trend.
題目 12 · 選擇題
1 分
Which pair of reagents and method is most suitable for preparing a pure, dry sample of the insoluble salt lead(II) sulfate?
  1. A.Mixing aqueous lead(II) nitrate and dilute sulfuric acid, filtering, washing the residue with distilled water, and drying.
  2. B.Reacting solid lead(II) oxide with dilute sulfuric acid, filtering, and evaporating the filtrate to dryness.
  3. C.Mixing aqueous lead(II) nitrate and aqueous sodium sulfate, and evaporating the filtrate to dryness.
  4. D.Reacting lead metal with dilute sulfuric acid, filtering, and drying the residue.
查看答案詳解

解題

Lead(II) sulfate is an insoluble salt. The standard preparation method for insoluble salts is precipitation. This involves mixing two soluble salt solutions (such as lead(II) nitrate and dilute sulfuric acid), filtering the mixture to collect the precipitate, washing the residue with distilled water to remove soluble impurities, and then drying it.

評分準則

1 mark for the correct option A. Reject methods involving reacting metals directly with acids where insoluble salt layers prevent complete reaction, or evaporative crystallization for insoluble salts.
題目 13 · 選擇題
1 分
A synthetic polymer is formed by the reaction of a dicarboxylic acid and a diamine. Which row correctly identifies the type of polymerisation and the linkage formed?
  1. A.Type of polymerisation: addition; Linkage formed: amide
  2. B.Type of polymerisation: addition; Linkage formed: ester
  3. C.Type of polymerisation: condensation; Linkage formed: ester
  4. D.Type of polymerisation: condensation; Linkage formed: amide
查看答案詳解

解題

The reaction of a dicarboxylic acid and a diamine involves the elimination of small water molecules, which is a condensation polymerisation. The functional group formed by the reaction between the carboxylic acid group (\(-\text{COOH}\)) and the amine group (\(-\text{NH}_2\)) is an amide linkage (\(-\text{CO}-\text{NH}-\)). This polymer is a polyamide (such as nylon).

評分準則

1 mark for the correct option D. Reject addition polymerisation or ester linkage combinations.
題目 14 · 選擇題
1 分
A student is given an unknown green crystalline solid, \( Y \). When aqueous sodium hydroxide is added to a solution of \( Y \), a green precipitate is formed which is insoluble in excess. When dilute nitric acid followed by aqueous barium nitrate is added to a solution of \( Y \), a white precipitate is formed. What is the identity of \( Y \)?
  1. A.Chromium(III) sulfate
  2. B.Iron(II) chloride
  3. C.Iron(II) sulfate
  4. D.Iron(III) sulfate
查看答案詳解

解題

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

評分準則

1 mark for the correct option C. Reject chromium(III) because its hydroxide is soluble in excess NaOH. Reject chloride because it would give a white precipitate with silver nitrate, not barium nitrate.
題目 15 · 選擇題
1 分
A chromatogram of a food dye was produced using a suitable solvent. The solvent front moved \( 8.0\text{ cm} \) from the baseline. A spot in the dye moved \( 5.2\text{ cm} \) from the baseline. What is the \( R_f \) value of this spot, and which type of agent would be needed to view the spots if the substances were colourless?
  1. A.R\(_f\) = 0.65; locating agent
  2. B.R\(_f\) = 0.65; oxidising agent
  3. C.R\(_f\) = 1.54; locating agent
  4. D.R\(_f\) = 1.54; indicator
查看答案詳解

解題

1. The \( R_f \) value is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front: \( R_f = 5.2 / 8.0 = 0.65 \). 2. If the substances on the chromatogram were colourless, a locating agent is sprayed onto the chromatogram to make the spots visible.

評分準則

1 mark for the correct option A. Reject options with incorrect math (e.g., dividing 8.0 by 5.2) or incorrect chemical agents.
題目 16 · 選擇題
1 分
What is the volume occupied by \( 3.55\text{ g} \) of chlorine gas, \( \text{Cl}_2 \), at room temperature and pressure (r.t.p.)? [Molar volume of a gas at r.t.p. is \( 24\text{ dm}^3 \)]
  1. A.0.05 dm\(^3\)
  2. B.1.20 dm\(^3\)
  3. C.2.40 dm\(^3\)
  4. D.85.2 dm\(^3\)
查看答案詳解

解題

1. Calculate the relative molecular mass of chlorine gas, \( \text{Cl}_2 \): \( M_r = 2 \times 35.5 = 71 \). 2. Calculate the number of moles of chlorine gas: \( \text{moles} = \frac{3.55\text{ g}}{71\text{ g/mol}} = 0.05\text{ mol} \). 3. Calculate the volume of chlorine gas at r.t.p.: \( \text{volume} = 0.05\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.20\text{ dm}^3 \).

評分準則

1 mark for the correct option B. Award 1 mark for correct steps: determining Mr of Cl2 (71), mole calculation (0.05 mol), and correct gas volume multiplication.
題目 17 · multiple_choice
1 分
During the electrolysis of concentrated aqueous sodium chloride using inert electrodes, which products are formed at the electrodes, and what change occurs in the pH of the remaining electrolyte?
  1. A.Anode product: chlorine; Cathode product: hydrogen; pH of electrolyte: increases
  2. B.Anode product: oxygen; Cathode product: sodium; pH of electrolyte: decreases
  3. C.Anode product: chlorine; Cathode product: sodium; pH of electrolyte: remains unchanged
  4. D.Anode product: oxygen; Cathode product: hydrogen; pH of electrolyte: increases
查看答案詳解

解題

At the anode (positive electrode), halide ions are selectively discharged in a concentrated solution, producing chlorine gas, \(\text{Cl}_2(\text{g})\). At the cathode (negative electrode), hydrogen ions are discharged preferentially over sodium ions because hydrogen is lower in the reactivity series, producing hydrogen gas, \(\text{H}_2(\text{g})\). As \(\text{H}^+\) and \(\text{Cl}^-\) ions are removed, \(\text{Na}^+\) and \(\text{OH}^-\) ions remain in the solution, forming sodium hydroxide which is alkaline. Thus, the pH of the remaining electrolyte increases.

評分準則

Award 1 mark for selecting the correct option (A). [Method/Accuracy: 1 mark for identifying correct anode and cathode products along with the pH change]
題目 18 · multiple_choice
1 分
The repeating unit of a synthetic polyester is shown below: \(-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CO}-\). Which monomers react to form this polymer?
  1. A.\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\) and \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\dots\text{COOH}\)
  2. B.\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{COOH}\) and \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\dots\text{COOH}\)
  3. C.\(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) and \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\dots\text{OH}\)
  4. D.\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\dots\text{OH}\) and \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\)
查看答案詳解

解題

A polyester is formed by the condensation reaction between a diol (containing two \(-\text{OH}\) groups) and a dicarboxylic acid (containing two \(-\text{COOH}\) groups). Breaking the ester link (\(-\text{O}-\text{CO}-\)) in the polymer repeating unit reveals that the diol part has two carbon atoms, giving \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\). The dicarboxylic acid part has five carbon atoms in total (three in the central chain and two in the carbonyl groups), giving \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\dots\text{COOH}\). This corresponds to option A.

評分準則

Award 1 mark for the correct identification of the diol and dicarboxylic acid monomers (A).
題目 19 · multiple_choice
1 分
In a titration, \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), is neutralized by \(20.0\text{ cm}^3\) of dilute sulfuric acid, \(\text{H}_2\text{SO}_4\), according to the equation: \(2\text{NaOH}(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\). What is the concentration of the sulfuric acid?
  1. A.\(0.0400\text{ mol/dm}^3\)
  2. B.\(0.0625\text{ mol/dm}^3\)
  3. C.\(0.125\text{ mol/dm}^3\)
  4. D.\(0.250\text{ mol/dm}^3\)
查看答案詳解

解題

Step 1: Calculate moles of \(\text{NaOH} = \text{volume in dm}^3 \times \text{concentration} = 0.0250 \times 0.100 = 0.00250\text{ mol}\). Step 2: From the balanced equation, \(2\text{ moles of NaOH}\) react with \(1\text{ mole of H}_2\text{SO}_4\). Thus, moles of \(\text{H}_2\text{SO}_4 = 0.00250 / 2 = 0.00125\text{ mol}\). Step 3: Calculate the concentration of \(\text{H}_2\text{SO}_4 = \text{moles} / \text{volume in dm}^3 = 0.00125 / 0.0200 = 0.0625\text{ mol/dm}^3\).

評分準則

Award 1 mark for the correct calculation of concentration (B).
題目 20 · multiple_choice
1 分
A student wishes to prepare a pure, dry sample of the insoluble salt, barium sulfate (\(\text{BaSO}_4\)). Which pair of aqueous solutions should be mixed together, and what is the correct sequence of steps to obtain the pure, dry product from the mixture?
  1. A.Solutions: \(\text{BaCl}_2(\text{aq})\) and \(\text{Na}_2\text{SO}_4(\text{aq})\); Sequence: filter \(\rightarrow\) wash residue with distilled water \(\rightarrow\) dry residue
  2. B.Solutions: \(\text{BaCO}_3(\text{s})\) and \(\text{H}_2\text{SO}_4(\text{aq})\); Sequence: filter \(\rightarrow\) evaporate filtrate \(\rightarrow\) crystallise
  3. C.Solutions: \(\text{BaCl}_2(\text{aq})\) and \(\text{Na}_2\text{SO}_4(\text{aq})\); Sequence: evaporate mixture to dryness \(\rightarrow\) wash with distilled water \(\rightarrow\) dry residue
  4. D.Solutions: \(\text{Ba(OH)}_2(\text{aq})\) and \(\text{H}_2\text{SO}_4(\text{aq})\); Sequence: titration \(\rightarrow\) evaporate filtrate \(\rightarrow\) crystallise
查看答案詳解

解題

To prepare an insoluble salt, we use a precipitation reaction by mixing two soluble salts. Barium chloride and sodium sulfate are both soluble in water. When mixed, they react to form insoluble barium sulfate and soluble sodium chloride. The insoluble product is separated by filtration, washed with distilled water to remove soluble impurities (such as unreacted ions and sodium chloride), and then dried in an oven or with filter paper.

評分準則

Award 1 mark for selecting the correct reactants and separation process (A).
題目 21 · multiple_choice
1 分
A series of halogen displacement experiments is carried out using aqueous solutions of halogens and halide salts. Which row correctly identifies the observations for each reaction mixture?
  1. A.\(\text{Cl}_2 + \text{KBr}\): solution turns orange; \(\text{Br}_2 + \text{KI}\): solution turns brown; \(\text{I}_2 + \text{KCl}\): no reaction
  2. B.\(\text{Cl}_2 + \text{KBr}\): no reaction; \(\text{Br}_2 + \text{KI}\): solution turns brown; \(\text{I}_2 + \text{KCl}\): solution turns pale green
  3. C.\(\text{Cl}_2 + \text{KBr}\): solution turns orange; \(\text{Br}_2 + \text{KI}\): no reaction; \(\text{I}_2 + \text{KCl}\): solution turns pale green
  4. D.\(\text{Cl}_2 + \text{KBr}\): solution turns brown; \(\text{Br}_2 + \text{KI}\): no reaction; \(\text{I}_2 + \text{KCl}\): no reaction
查看答案詳解

解題

1. Chlorine is more reactive than bromine, so it displaces bromide ions: \(\text{Cl}_2(\text{aq}) + 2\text{KBr}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{Br}_2(\text{aq})\). The solution turns orange. 2. Bromine is more reactive than iodine, so it displaces iodide ions: \(\text{Br}_2(\text{aq}) + 2\text{KI}(\text{aq}) \rightarrow 2\text{KBr}(\text{aq}) + \text{I}_2(\text{aq})\). The solution turns brown. 3. Iodine is less reactive than chlorine, so no reaction occurs between \(\text{I}_2(\text{aq})\) and \(\text{KCl}(\text{aq})\).

評分準則

Award 1 mark for the correct combination of observations in all three reactions (A).
題目 22 · multiple_choice
1 分
An unknown solid X is dissolved in water to make a solution. The solution is tested, and the following observations are made: 1) Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. 2) Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of solid X?
  1. A.Iron(II) chloride
  2. B.Iron(II) sulfate
  3. C.Iron(III) sulfate
  4. D.Chromium(III) sulfate
查看答案詳解

解題

1. A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also gives a green precipitate, but it dissolves in excess sodium hydroxide to give a green solution). 2. The formation of a white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Thus, the solid is iron(II) sulfate.

評分準則

Award 1 mark for identifying the correct salt (B).
題目 23 · multiple_choice
1 分
An ion of element Y has a charge of \(2-\). This ion has the electronic configuration \(2, 8, 8\). What is the proton number and the group number of element Y in the Periodic Table?
  1. A.Proton number: 16; Group number: Group VI
  2. B.Proton number: 18; Group number: Group VIII / Group 0
  3. C.Proton number: 20; Group number: Group II
  4. D.Proton number: 16; Group number: Group VIII / Group 0
查看答案詳解

解題

The ion \(\text{Y}^{2-}\) has 18 electrons (electronic configuration \(2, 8, 8\)). Because it has a charge of \(2-\), it must have 2 more electrons than protons. Therefore, the number of protons (atomic number) is \(18 - 2 = 16\). Element Y (neutral atom) has 16 electrons, meaning its electronic configuration is \(2, 8, 6\). Because it has 6 valency electrons, it belongs to Group VI of the Periodic Table.

評分準則

Award 1 mark for identifying both the proton number and group number correctly (A).
題目 24 · multiple_choice
1 分
A chromatogram of a sample of food coloring was obtained using water as the solvent. The solvent front moved a distance of \(8.0\text{ cm}\) from the baseline. A spot containing a single yellow dye moved a distance of \(5.2\text{ cm}\) from the baseline. What is the \(R_f\) value of this yellow dye, and is it more soluble in water than a dye with an \(R_f\) value of \(0.35\)?
  1. A.\(R_f\) value: \(0.65\); Yes, it is more soluble in water
  2. B.\(R_f\) value: \(0.65\); No, it is less soluble in water
  3. C.\(R_f\) value: \(1.54\); Yes, it is more soluble in water
  4. D.\(R_f\) value: \(1.54\); No, it is less soluble in water
查看答案詳解

解題

The \(R_f\) value is calculated by dividing the distance moved by the substance by the distance moved by the solvent front: \(R_f = 5.2 / 8.0 = 0.65\). In a chromatogram, a substance with a higher \(R_f\) value travels further because it has a greater solubility in the mobile phase (water) than a substance with a lower \(R_f\) value. Therefore, it is more soluble in water than a dye with an \(R_f\) of \(0.35\).

評分準則

Award 1 mark for the correct \(R_f\) value and solubility comparison (A).
題目 25 · multiple_choice
1 分
Concentrated aqueous sodium chloride is electrolysed using carbon electrodes. Which row correctly describes the product at each electrode and the change in the pH of the remaining electrolyte?
  1. A.Cathode: Hydrogen; Anode: Chlorine; pH of solution: Increases
  2. B.Cathode: Sodium; Anode: Chlorine; pH of solution: Decreases
  3. C.Cathode: Hydrogen; Anode: Oxygen; pH of solution: No change
  4. D.Cathode: Sodium; Anode: Oxygen; pH of solution: Increases
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解題

At the cathode, hydrogen ions are reduced to form hydrogen gas, which is preferred over sodium ions. At the anode, chloride ions are oxidised to chlorine gas because the solution is concentrated. The removal of hydrogen and chloride ions leaves behind sodium and hydroxide ions in solution, raising the pH. Therefore, the correct row is A.

評分準則

1 mark for identifying that hydrogen is produced at the cathode, chlorine at the anode, and the pH increases.
題目 26 · multiple_choice
1 分
Which pair of reactants is most suitable for preparing a pure, dry sample of the insoluble salt lead(II) sulfate?
  1. A.Aqueous lead(II) nitrate and dilute sulfuric acid
  2. B.Solid lead(II) oxide and dilute sulfuric acid
  3. C.Solid lead(II) carbonate and dilute sulfuric acid
  4. D.Aqueous lead(II) nitrate and solid barium sulfate
查看答案詳解

解題

Lead(II) sulfate is an insoluble salt. The best method to prepare an insoluble salt is by precipitation, which requires two soluble starting materials. Aqueous lead(II) nitrate is soluble, and dilute sulfuric acid is soluble. Solid reactants like lead(II) oxide or carbonate will become coated in the insoluble lead(II) sulfate, preventing the reaction from completing. Therefore, A is the correct choice.

評分準則

1 mark for selecting the two soluble starting reagents suitable for precipitation.
題目 27 · multiple_choice
1 分
A polyester is formed by condensation polymerisation. Which monomer pair could react together to form a polyester?
  1. A.A dicarboxylic acid and a diol
  2. B.A dicarboxylic acid and a diamine
  3. C.A dialcohol and a diamine
  4. D.Two different alkenes
查看答案詳解

解題

A polyester is formed via condensation polymerisation of a dicarboxylic acid containing two carboxylic acid groups and a diol containing two alcohol groups, forming ester linkages and releasing water. Therefore, A is correct.

評分準則

1 mark for identifying a dicarboxylic acid and a diol as the monomers for a polyester.
題目 28 · multiple_choice
1 分
An unknown solution X is tested. The addition of aqueous sodium hydroxide to X produces a green precipitate that is insoluble in excess sodium hydroxide. The addition of dilute nitric acid followed by aqueous barium nitrate to X produces a white precipitate. What is the identity of solution X?
  1. A.Iron(II) sulfate
  2. B.Iron(III) sulfate
  3. C.Chromium(III) sulfate
  4. D.Iron(II) chloride
查看答案詳解

解題

The test with aqueous sodium hydroxide yields a green precipitate insoluble in excess, confirming the presence of iron(II) ions (chromium(III) forms a green precipitate but dissolves in excess to a green solution). The test with barium nitrate and nitric acid yields a white precipitate, confirming the presence of sulfate ions. Thus, the solution is iron(II) sulfate. A is correct.

評分準則

1 mark for identifying iron(II) sulfate based on the cation and anion confirmation tests.
題目 29 · multiple_choice
1 分
A chromatography experiment is carried out on a sample of food colouring. The solvent front travels 8.0 cm from the baseline. A spot of yellow dye in the food colouring travels 5.2 cm from the baseline. What is the Rf value of this yellow dye?
  1. A.0.65
  2. B.1.54
  3. C.5.20
  4. D.0.80
查看答案詳解

解題

The retention factor (Rf value) is the ratio of the distance travelled by the substance to the distance travelled by the solvent front: Rf = 5.2 cm / 8.0 cm = 0.65. Therefore, A is correct.

評分準則

1 mark for the correct calculation of the Rf value as 0.65.
題目 30 · multiple_choice
1 分
An element X has the proton number 15. Which statement about element X is correct?
  1. A.It is a non-metal in Group V and Period 3 of the Periodic Table.
  2. B.It is a metal in Group III and Period 5 of the Periodic Table.
  3. C.It is a non-metal in Group III and Period 5 of the Periodic Table.
  4. D.It is a metal in Group V and Period 3 of the Periodic Table.
查看答案詳解

解題

With a proton number of 15, the element is phosphorus. Its electronic configuration is 2, 8, 5. The 5 outer electrons place it in Group V, and the 3 occupied electron shells place it in Period 3. Since it is phosphorus, it is a non-metal. Thus, A is correct.

評分準則

1 mark for identifying the correct periodic table position and non-metallic character.
題目 31 · multiple_choice
1 分
A student performs an acid-base titration to determine the concentration of a solution of hydrochloric acid. Which piece of apparatus should be rinsed only with distilled water, and not with the solution it is to contain, before the titration is carried out?
  1. A.The conical flask
  2. B.The burette
  3. C.The pipette
  4. D.Both the burette and the pipette
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解題

Rinsing the conical flask with the solution it is to contain would introduce an unknown extra number of moles of that solution, spoiling the accuracy. Distilled water is acceptable because it does not alter the moles of reagent added by the pipette. In contrast, the pipette and burette must be rinsed with their respective solutions to prevent dilution. Therefore, A is correct.

評分準則

1 mark for identifying the conical flask as the vessel that should only be rinsed with distilled water.
題目 32 · multiple_choice
1 分
Which molecular formula represents an organic compound belonging to the homologous series of carboxylic acids?
  1. A.C3H6O2
  2. B.C3H8O
  3. C.C3H6O
  4. D.C3H4O2
查看答案詳解

解題

Carboxylic acids belong to the homologous series with the general formula CnH2nO2. For a compound with 3 carbon atoms, the molecular formula is C3H6O2, which represents propanoic acid. Thus, A is correct.

評分準則

1 mark for identifying the correct molecular formula representing a carboxylic acid.

Paper 31 Theory (Core)

Answer all questions. Write your answers in the spaces provided on the question paper. You may use a calculator.
8 題目 · 80 分
題目 1 · Structured
10 分
This question is about atomic structure and the Periodic Table.

(a) Complete the descriptions to show the relative charge and relative mass of a proton, neutron and electron.
- Relative charge of a neutron
- Relative mass of a proton
- Relative charge of an electron [3]

(b) An atom of chlorine is represented as \(^{37}_{17}\text{Cl}\).
(i) State the number of protons, neutrons and electrons in this atom of chlorine. [3]
(ii) Deduce the electronic configuration of this chlorine atom. [1]

(c) Explain why argon is very unreactive. [1]

(d) Define the term 'isotopes'. [2]
查看答案詳解

解題

(a) Relative charge of a neutron is 0. Relative mass of a proton is 1. Relative charge of an electron is -1.
(b) (i) Protons = 17, Neutrons = 20 (since 37 - 17 = 20), Electrons = 17.
(ii) The electronic configuration of a chlorine atom (atomic number 17) is 2,8,7.
(c) Argon is a noble gas in Group VIII with a full outer shell of 8 electrons, which is a stable configuration and does not readily gain, lose, or share electrons.
(d) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

評分準則

(a) [3 marks]
- Relative charge of neutron: 0 [1]
- Relative mass of proton: 1 [1]
- Relative charge of electron: -1 (or 1-) [1]

(b) [4 marks]
- (i) Protons: 17 [1]
- (i) Neutrons: 20 [1]
- (i) Electrons: 17 [1]
- (ii) 2,8,7 [1]

(c) [1 mark]
- Full outer shell of electrons / stable electron arrangement [1]

(d) [2 marks]
- Atoms of the same element / same number of protons / same atomic number [1]
- with different numbers of neutrons / different nucleon numbers [1]
題目 2 · Structured
10 分
A student uses paper chromatography to separate and identify food colorings in a sweet.

(a) Describe how the student would set up and carry out this paper chromatography experiment. [4]

(b) State why the start line on the chromatography paper must be drawn in pencil and not ink. [1]

(c) Explain why the level of the solvent in the beaker must be below the start line. [1]

(d) After running the chromatogram, three distinct spots are observed from the sweet's food coloring.
(i) Explain how this observation shows that the food coloring is a mixture. [1]
(ii) One spot travelled 4.5 cm from the baseline, while the solvent front travelled 9.0 cm. Calculate the \(R_f\) value of this spot. Show your working. [2]
(iii) State what is meant by the term 'solvent'. [1]
查看答案詳解

解題

(a) Draw a baseline in pencil near the bottom of the chromatography paper. Place a spot of the food coloring on the baseline. Suspend the paper vertically in a beaker containing a suitable solvent (such as water or ethanol), ensuring the solvent level is below the baseline. Cover the beaker and allow the solvent to run up the paper until near the top, then remove the paper and mark the solvent front.
(b) Ink is soluble in the solvent and would run/spread up the paper, interfering with the results; pencil (graphite) is insoluble and will not run.
(c) If the solvent is above the start line, the food coloring spots will dissolve/wash directly into the solvent reservoir instead of moving up the paper.
(d) (i) It has separated into multiple spots, indicating more than one substance is present.
(ii) \(R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent}} = \frac{4.5}{9.0} = 0.50\).
(iii) A solvent is a liquid substance in which a solute dissolves to form a solution.

評分準則

(a) [4 marks]
- Draw a pencil baseline near bottom of paper AND place spot of coloring on it [1]
- Place paper in a beaker with a solvent [1]
- Ensure solvent level is below the baseline [1]
- Allow solvent to travel up the paper / cover beaker to prevent evaporation [1]

(b) [1 mark]
- Pencil is insoluble / does not run / does not contaminate chromatogram [1]
- Reject: 'pencil does not react'

(c) [1 mark]
- Otherwise the spots would dissolve in the solvent / wash off the paper [1]

(d) [4 marks]
- (i) Separates into more than one spot / has multiple components [1]
- (ii) Working: 4.5 / 9.0 [1]
- (ii) Correct value: 0.5 (or 0.50) [1]
- (iii) Liquid in which solute dissolves [1]
題目 3 · Structured
10 分
This question is about the electrolysis of molten lead(II) bromide using inert carbon electrodes.

(a) Define the term 'electrolysis'. [2]

(b) State the name of:
(i) the positive electrode [1]
(ii) the negative electrode [1]

(c) Describe what is observed during this electrolysis at:
(i) the positive electrode [1]
(ii) the negative electrode [1]

(d) Write a word equation for the reaction that occurs during this electrolysis. [1]

(e) Explain why solid lead(II) bromide does not conduct electricity but molten lead(II) bromide does. [2]

(f) Suggest a suitable material for the inert electrodes. [1]
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解題

(a) Electrolysis is the breakdown of an ionic compound, molten or in aqueous solution, by the passage of electricity.
(b) (i) Anode
(ii) Cathode
(c) (i) Red-brown or orange-brown gas/fumes are evolved (bromine gas).
(ii) A grey liquid or silvery bead of metal is formed at the bottom (lead metal).
(d) lead(II) bromide \(\rightarrow\) lead + bromine
(e) In solid lead(II) bromide, the ions are in fixed positions in a giant ionic lattice and cannot move. When molten, the lattice breaks down and the ions are free to move and carry the charge.
(f) Graphite (or carbon or platinum).

評分準則

(a) [2 marks]
- Breakdown of an ionic compound / substance [1]
- By the passage of electricity [1]

(b) [2 marks]
- (i) Anode [1]
- (ii) Cathode [1]

(c) [2 marks]
- (i) Brown/red-brown fumes or gas [1]
- (ii) Silvery / grey liquid / beads [1]

(d) [1 mark]
- lead(II) bromide \(\rightarrow\) lead + bromine [1]
- Accept: lead bromide \(\rightarrow\) lead + bromine

(e) [2 marks]
- Ions in solid are in fixed positions / cannot move [1]
- Ions in molten are free to move / mobile [1]
- Reject: electrons are free to move

(f) [1 mark]
- Graphite / carbon / platinum [1]
題目 4 · Structured
10 分
A student prepares crystals of copper(II) sulfate by reacting excess copper(II) oxide with dilute sulfuric acid.

(a) State the name of this type of reaction. [1]

(b) Explain why copper(II) oxide is added in excess. [1]

(c) Describe the method to obtain pure, dry crystals of copper(II) sulfate from the reaction mixture containing excess copper(II) oxide. [5]

(d) Write the chemical equation for the reaction between copper(II) oxide and dilute sulfuric acid. [2]

(e) State the color of copper(II) sulfate crystals. [1]
查看答案詳解

解題

(a) Neutralization (or acid-base reaction).
(b) To ensure that all of the sulfuric acid is completely reacted and used up.
(c) First, filter the mixture to remove the unreacted excess copper(II) oxide. Heat the filtrate (copper(II) sulfate solution) in an evaporating basin until the crystallization point is reached (when crystals start to form on a glass rod). Leave the hot, saturated solution to cool and crystallize slowly. Filter off the crystals from the remaining solution. Pat dry the crystals with filter paper or leave them in a warm oven to dry.
(d) \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\)
(e) Blue.

評分準則

(a) [1 mark]
- Neutralisation / acid-base [1]

(b) [1 mark]
- To make sure all the acid reacts / is used up [1]

(c) [5 marks]
- Filter (to remove excess copper(II) oxide) [1]
- Heat / evaporate the filtrate / solution [1]
- to the point of crystallization / to obtain a saturated solution [1]
- Leave to cool (to form crystals) [1]
- Filter off crystals AND dry with filter paper / in a warm oven [1]

(d) [2 marks]
- \(\text{CuO} + \text{H}_2\text{SO}_4\) reactants [1]
- \(\text{CuSO}_4 + \text{H}_2\text{O}\) products [1]

(e) [1 mark]
- Blue [1]
題目 5 · Structured
10 分
This question is about polymers.

(a) Poly(ethene) is an addition polymer made from the monomer ethene.
(i) State the formula of ethene and describe its structural feature that allows it to polymerize. [1]
(ii) Describe the structural differences between the monomer ethene and a section of the polymer poly(ethene). [2]
(iii) State the type of chemical bond holding the atoms together in a poly(ethene) chain. [1]

(b) Plastics made from synthetic polymers pose environmental concerns.
(i) Explain why non-biodegradable plastics persist in landfill sites for a long time. [1]
(ii) Give one disadvantage of disposing of plastics by burning them. [1]

(c) Nylon is a synthetic polyamide polymer.
(i) Name two natural macromolecules (polymers) found in living organisms. [2]
(ii) State the type of linkage present in nylon. [1]
(iii) State the name of the type of polymerization reaction used to produce nylon. [1]
查看答案詳解

解題

(a) (i) Ethene is \(\text{C}_2\text{H}_4\). It contains a carbon-carbon double bond (\(\text{C}=\text{C}\)) which can break open to link monomers together.
(ii) Ethene has a double bond between carbon atoms and is unsaturated, whereas poly(ethene) contains only single carbon-carbon bonds and is saturated.
(iii) Covalent bond.
(b) (i) They are non-biodegradable, meaning they cannot be broken down or decomposed by microbes/bacteria.
(ii) Burning plastics releases toxic gases / greenhouse gases (such as carbon dioxide).
(c) (i) Proteins and starch (or cellulose/DNA).
(ii) Amide linkage (or peptide linkage).
(iii) Condensation polymerization.

評分準則

(a) [4 marks]
- (i) \(\text{C}_2\text{H}_4\) / contains a carbon-carbon double bond / \(\text{C}=\text{C}\) [1]
- (ii) Monomer has double bonds / is unsaturated [1]
- (ii) Polymer has only single bonds / is saturated [1]
- (iii) Covalent [1]

(b) [2 marks]
- (i) Microorganisms / decomposers / bacteria cannot break them down [1]
- (ii) Releases toxic gases / greenhouse gases / carbon dioxide [1]
- Reject: 'causes pollution' (too vague)

(c) [4 marks]
- (i) Any two from: protein / starch / cellulose / DNA / glycogen [2]
- (ii) Amide / amide linkage [1]
- (iii) Condensation [1]
題目 6 · Structured
10 分
A student carries out an acid-base titration to determine the volume of dilute hydrochloric acid required to neutralize 25.0 cm³ of sodium hydroxide solution.

(a) Name the piece of apparatus used to measure:
(i) exactly 25.0 cm³ of sodium hydroxide solution into the conical flask. [1]
(ii) the volume of dilute hydrochloric acid added. [1]

(b) Methyl orange indicator is added to the sodium hydroxide solution before starting the titration.
State the color of methyl orange:
(i) in the sodium hydroxide solution. [1]
(ii) at the end-point of the titration. [1]

(c) The student performs three titrations and records the volumes of dilute hydrochloric acid added:
- Titration 1: 23.9 cm³
- Titration 2: 22.4 cm³
- Titration 3: 22.6 cm³
(i) Identify which titration result is anomalous and should be ignored. [1]
(ii) Calculate the average volume of dilute hydrochloric acid required using the concordant results. Show your working and state the unit. [2]

(d) Write a chemical equation for the reaction between hydrochloric acid (\(\text{HCl}\)) and sodium hydroxide (\(\text{NaOH}\)). [2]

(e) State the chemical name of the salt produced in this reaction. [1]
查看答案詳解

解題

(a) (i) Pipette
(ii) Burette
(b) (i) Yellow
(ii) Orange (or peach)
(c) (i) Titration 1 is anomalous because it is much larger than the other two results.
(ii) The concordant results are 22.4 cm³ and 22.6 cm³.
Average = \(\frac{22.4 + 22.6}{2} = 22.5\text{ cm}^3\).
(d) \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)
(e) Sodium chloride

評分準則

(a) [2 marks]
- (i) Volumetric pipette / pipette [1]
- (ii) Burette [1]

(b) [2 marks]
- (i) Yellow [1]
- (ii) Orange / peach [1]

(c) [3 marks]
- (i) Titration 1 [1]
- (ii) Working: \(\frac{22.4 + 22.6}{2}\) [1]
- (ii) Answer: 22.5 AND unit: \(\text{cm}^3\) [1]

(d) [2 marks]
- Reactants: \(\text{HCl} + \text{NaOH}\) [1]
- Products: \(\text{NaCl} + \text{H}_2\text{O}\) [1]

(e) [1 mark]
- Sodium chloride [1]
題目 7 · Structured
10 分
A student is given a mixture of two salts, Salt A and Salt B. Tests are carried out to identify the ions present.

(a) Salt A is a green solid. The student dissolves it in water to make a green solution.
(i) When aqueous sodium hydroxide is added to the solution of Salt A, a green precipitate is formed that remains insoluble in excess sodium hydroxide. Identify the cation present in Salt A. [1]
(ii) When dilute nitric acid followed by aqueous barium nitrate is added to a solution of Salt A, a dense white precipitate is formed. Identify the anion present in Salt A. [1]
(iii) Deduce the chemical formula of Salt A. [1]

(b) Salt B is a white solid. When dilute hydrochloric acid is added to solid Salt B, effervescence is observed and a colorless gas is produced.
(i) This gas is bubbled through limewater, which turns cloudy. Name the gas. [1]
(ii) Identify the anion present in Salt B. [1]

(c) Describe a chemical test to identify each of the following gases, including the positive result:
(i) Chlorine gas. [2]
(ii) Ammonia gas. [2]
(iii) Oxygen gas. [1]
查看答案詳解

解題

(a) (i) Iron(II) ion (\(\text{Fe}^{2+}\))
(ii) Sulfate ion (\(\text{SO}_4^{2-}\))
(iii) \(\text{FeSO}_4\)
(b) (i) Carbon dioxide (\(\text{CO}_2\))
(ii) Carbonate ion (\(\text{CO}_3^{2-}\))
(c) (i) Test: Use damp blue litmus paper. Result: The paper is bleached white.
(ii) Test: Use damp red litmus paper. Result: The paper turns blue.
(iii) Test: Use a glowing splint. Result: The splint relights.

評分準則

(a) [3 marks]
- (i) Iron(II) / \(\text{Fe}^{2+}\) [1]
- (ii) Sulfate / \(\text{SO}_4^{2-}\) [1]
- (iii) \(\text{FeSO}_4\) [1]

(b) [2 marks]
- (i) Carbon dioxide / \(\text{CO}_2\) [1]
- (ii) Carbonate / \(\text{CO}_3^{2-}\) [1]

(c) [5 marks]
- (i) Test: damp blue litmus paper [1]
- (i) Result: bleached / white [1]
- (ii) Test: damp red litmus paper [1]
- (ii) Result: turns blue [1]
- (iii) Test: glowing splint AND Result: relights [1]
題目 8 · Structured
10 分
Ethanol, \(\text{C}_2\text{H}_5\text{OH}\), belongs to the homologous series of alcohols.

(a) (i) State two characteristics of a homologous series. [2]
(ii) Draw the displayed structure of ethanol, showing all atoms and all bonds. [1]

(b) Ethanol can be manufactured by the fermentation of glucose.
(i) State two reaction conditions required for fermentation to occur efficiently. [2]
(ii) State the name of the substance in yeast that acts as a catalyst in this reaction. [1]

(c) Ethanol is used as a fuel.
(i) Write a word equation for the complete combustion of ethanol. [1]
(ii) Explain why the combustion of ethanol is described as an exothermic reaction, in terms of temperature change and energy transfer. [2]

(d) State the name of the alcohol that contains only one carbon atom. [1]
查看答案詳解

解題

(a) (i) Any two of: same general formula, consecutive members differ by a \(\text{CH}_2\) group, similar chemical properties, show a trend in physical properties, same functional group.
(ii) Displayed structure: two carbon atoms linked by a single bond. The first carbon has three hydrogen atoms attached. The second carbon has two hydrogen atoms and one oxygen atom attached. The oxygen atom has a hydrogen atom attached (\(\text{H}_3\text{C}-\text{CH}_2-\text{O}-\text{H}\), with all bonds explicitly drawn).
(b) (i) Yeast, warm temperature (e.g. 30–40 °C), and anaerobic conditions (absence of oxygen).
(ii) Enzymes.
(c) (i) ethanol + oxygen \(\rightarrow\) carbon dioxide + water
(ii) It is exothermic because it transfers thermal energy to the surroundings, causing the temperature of the surroundings to increase.
(d) Methanol.

評分準則

(a) [3 marks]
- (i) Any two from: same general formula / same functional group / similar chemical properties / trend in physical properties / consecutive members differ by \(\text{CH}_2\) [2]
- (ii) Correct displayed structure showing all atoms and bonds (must show C-C, C-H, C-O, and O-H bonds) [1]

(b) [3 marks]
- (i) Any two from: presence of yeast / temperature between 20-40 °C / anaerobic conditions / absence of oxygen / water present [2]
- (ii) Enzyme / zymase [1]

(c) [3 marks]
- (i) ethanol + oxygen \(\rightarrow\) carbon dioxide + water [1]
- (ii) Exothermic because it releases heat/energy [1]
- (ii) to the surroundings / raises temperature of surroundings [1]

(d) [1 mark]
- Methanol [1]

Paper 41 Theory (Extended)

Answer all questions. Write your answers in the spaces provided on the question paper. You should show all your working and use appropriate units.
6 題目 · 79.98 分
題目 1 · structured
13.33 分
Element X has an atomic number of 15. Element Y has an atomic number of 17.

(a) (i) Deduce the electronic configuration of an atom of X. [1]
(ii) State the group number and period number of X in the Periodic Table. [2]

(b) Draw a dot-and-cross diagram to show the bonding in a molecule of the compound formed between X and Y, which has the formula \(XY_3\). Show outer shell electrons only. [4]

(c) X has two isotopes, \(^{31}\text{X}\) and \(^{33}\text{X}\).
(i) Define the term isotopes. [2]
(ii) Explain why these two isotopes have identical chemical properties. [1]

(d) Explain, in terms of structure and bonding, why element Y (which exists as diatomic molecules \(Y_2\)) has a very low boiling point. [3.33]
查看答案詳解

解題

(a) (i) Atomic number 15 corresponds to phosphorus, which has the electronic configuration 2,8,5.
(ii) Group: V (or 15) because it has 5 outer electrons. Period: 3 because it has 3 electron shells occupied.

(b) In \(XY_3\) (which represents phosphorus trichloride, \(\text{PCl}_3\)):
- The central X atom shares one pair of electrons with each of the three Y atoms (3 single covalent bonds).
- The central X atom has one lone pair of electrons (2 non-bonding electrons) in its outer shell to complete its octet of 8 electrons.
- Each Y atom has three lone pairs of electrons (6 non-bonding electrons) and shares one pair with X to complete its octet.

(c) (i) Isotopes are atoms of the same element with the same number of protons (or atomic number) but different numbers of neutrons (or nucleon number).
(ii) Chemical properties are determined by the number and arrangement of outer shell electrons. Since both isotopes have the same electronic configuration, they react in the same way.

(d) Element Y exists as simple diatomic molecules (\(Y_2\)). It has a simple molecular structure with weak intermolecular forces (or van der Waals' forces) between the molecules. Very little thermal energy is required to overcome these weak intermolecular forces, resulting in a very low boiling point.

評分準則

(a) (i) 2,8,5 [1]
(ii) Group V / 15 [1] and Period 3 [1]
(b) Three shared pairs of electrons between X and three Y atoms [1]
One lone pair on central X atom [1]
Three lone pairs (6 non-bonding electrons) on each Y atom [2] (1 mark if only some Y atoms are fully correct)
(c) (i) Atoms of the same element with the same number of protons / atomic number [1] but different numbers of neutrons / mass number [1]
(ii) They have the exact same electronic configuration / same number of outer shell electrons [1]
(d) Simple molecular structure / molecular lattice [1]
Weak intermolecular forces / forces between molecules [1]
Little energy required to break/overcome these forces [1.33] (do NOT accept breaking of covalent bonds)
題目 2 · structured
13.33 分
A student carries out a titration to find the concentration of a solution of sulfuric acid, \(\text{H}_2\text{SO}_4\).
The student titrates \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) sodium hydroxide, \(\text{NaOH}\), with the sulfuric acid.
The equation for the reaction is:
\(\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)

(a) Name a suitable indicator for this titration and state the colour change observed in the conical flask if the acid is added from the burette to the alkali. [3]

(b) The average titre of sulfuric acid is \(18.75\text{ cm}^3\).
Calculate:
(i) the number of moles of \(\text{NaOH}\) in \(25.0\text{ cm}^3\) of the solution. [1]
(ii) the number of moles of \(\text{H}_2\text{SO}_4\) that reacted with this amount of \(\text{NaOH}\). [1]
(iii) the concentration of the sulfuric acid in \(\text{mol/dm}^3\). Give your answer to 3 significant figures. [2]
(iv) the concentration of the sulfuric acid in \(\text{g/dm}^3\). [2]

(c) Describe the practical steps required to obtain pure, dry crystals of sodium sulfate from the titrated mixture. [4.33]
查看答案詳解

解題

(a) Suitable indicators include:
- Methyl orange: starts yellow in alkali (NaOH) and turns pink/orange at the end-point.
- Phenolphthalein: starts pink/red in alkali and turns colourless at the end-point.

(b) (i) \(\text{moles of NaOH} = \text{volume in dm}^3 \times \text{concentration} = \frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol}\)
(ii) From the balanced equation, \(1\text{ mole of H}_2\text{SO}_4\) reacts with \(2\text{ moles of NaOH}\).
\(\text{moles of H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol}\)
(iii) \(\text{Concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00125}{18.75 / 1000} = 0.066667 \approx 0.0667\text{ mol/dm}^3\) (to 3 significant figures).
(iv) \(M_r(\text{H}_2\text{SO}_4) = (2 \times 1.0) + 32.1 + (4 \times 16.0) = 98.1\text{ g/mol}\) (or \(98\text{ g/mol}\)).
\(\text{Concentration in g/dm}^3 = 0.066667\text{ mol/dm}^3 \times 98.1\text{ g/mol} = 6.54\text{ g/dm}^3\) (or \(0.0667 \times 98 = 6.53\text{ g/dm}^3\)).

(c) To prepare pure, dry crystals:
1. Repeat the titration using the exact same volumes of acid (\(18.75\text{ cm}^3\)) and alkali (\(25.0\text{ cm}^3\)) but without adding any indicator to prevent contamination.
2. Transfer the resulting neutral sodium sulfate solution to an evaporating basin.
3. Heat the solution to evaporate the water until the crystallization point is reached (indicated by crystals starting to form on a cold glass rod).
4. Allow the hot saturated solution to cool down slowly to room temperature so that large crystals of sodium sulfate form.
5. Filter the mixture to separate the crystals from the remaining mother liquor.
6. Gently dry the crystals between sheets of filter paper or in a warm oven (do not overheat to avoid dehydration).

評分準則

(a) Methyl orange [1]; yellow [1] to pink / orange [1] (OR Phenolphthalein [1]; pink [1] to colourless [1])
(b) (i) \(0.00250\text{ mol}\) / \(2.5 \times 10^{-3}\text{ mol}\) [1]
(ii) \(0.00125\text{ mol}\) / \(1.25 \times 10^{-3}\text{ mol}\) (allow ecf = (b)(i) / 2) [1]
(iii) \(0.0667\text{ mol/dm}^3\) (allow ecf = (b)(ii) / 0.01875) [2] (1 mark for working, 1 mark for correct value to 3 sf)
(iv) \(6.53\text{ g/dm}^3\) or \(6.54\text{ g/dm}^3\) (allow ecf = (b)(iii) \(\times 98\)) [2] (1 mark for \(M_r = 98\), 1 mark for final value)
(c) Repeat titration with same volumes but without indicator [1.33]
Heat / evaporate solution to crystallization point / saturation [1]
Allow to cool (and crystallize) [1]
Filter crystals and dry between filter papers / in desiccator [1]
題目 3 · structured
13.33 分
A student investigates the electrolysis of aqueous copper(II) sulfate, \(\text{CuSO}_4(\text{aq})\), using two different setups:
Setup A: Inert carbon (graphite) electrodes.
Setup B: Copper electrodes.

(a) For Setup A:
(i) State the observations made at the anode and write an ionic half-equation for the reaction occurring there. [3]
(ii) State the observations made at the cathode and write an ionic half-equation for the reaction occurring there. [3]
(iii) Describe and explain the change in pH of the electrolyte during the electrolysis. [2.33]

(b) For Setup B:
(i) Describe how the mass of each electrode changes during the process and explain why, including any half-equations. [3]
(ii) State one industrial application of this specific type of electrolysis. [2]
查看答案詳解

解題

(a) (i) At the anode (positive electrode) in Setup A, hydroxide ions (\(\text{OH}^-\)) from water are discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)). This produces oxygen gas, so bubbles of a colourless gas are observed.
Half-equation: \(4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-\).

(ii) At the cathode (negative electrode) in Setup A, copper ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions (\(\text{H}^+\)) because copper is lower in the reactivity series. A pink-brown solid of copper metal is deposited on the electrode.
Half-equation: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\).

(iii) The pH of the electrolyte decreases (it becomes acidic). During electrolysis, \(\text{Cu}^{2+}\) and \(\text{OH}^-\) are removed from the solution. This leaves \(\text{H}^+\) and \(\text{SO}_4^{2-}\) ions in solution, which is equivalent to dilute sulfuric acid. The increasing concentration of \(\text{H}^+\) ions causes the pH to fall.

(b) (i) At the anode (copper), copper atoms are oxidized to copper ions and enter the solution: \(\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\), so the anode mass decreases.
At the cathode (copper), copper ions from the solution are reduced to copper atoms and deposit onto the electrode: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\), so the cathode mass increases.
The decrease in mass of the anode is equal to the increase in mass of the cathode.

(ii) Industrial applications include:
- Purification/refining of copper (making pure copper for electrical wiring).
- Electroplating objects with copper.

評分準則

(a) (i) Observation: bubbles of a colourless gas / effervescence [1]
Half-equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) [2] (allow 1 mark for unbalanced or correct species with wrong charges)
(ii) Observation: pink / brown / pink-brown solid deposits [1]
Half-equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) [2] (allow 1 mark for unbalanced or correct species with wrong charges)
(iii) pH decreases / becomes acidic [1]
Because \(\text{H}^+\) ions remain in solution while \(\text{OH}^-\) ions are discharged (leaving sulfuric acid) [1.33]
(b) (i) Anode mass decreases because copper atoms lose electrons to form ions / dissolve: \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\) [1.5]
Cathode mass increases because copper ions gain electrons to form copper metal / deposit: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) [1.5]
(b) (ii) Refining / purification of copper OR electroplating [2]
題目 4 · structured
13.33 分
A student plans to prepare pure, dry crystals of hydrated nickel(II) sulfate, \(\text{NiSO}_4 \cdot 6\text{H}_2\text{O}\), starting from insoluble nickel(II) carbonate, \(\text{NiCO}_3\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).

(a) Write a balanced chemical equation, including state symbols, for the reaction between nickel(II) carbonate and sulfuric acid. [3]

(b) Describe a detailed experimental procedure that the student should follow to prepare a pure, dry sample of hydrated nickel(II) sulfate crystals. Your description should explain:
- why excess nickel(II) carbonate is used [1.33]
- how the excess nickel(II) carbonate is removed [1]
- how the crystals are obtained from the solution [2]
- how the crystals are dried without losing their water of crystallization. [1]

(c) Calculate the maximum mass of hydrated nickel(II) sulfate, \(\text{NiSO}_4 \cdot 6\text{H}_2\text{O}\), that can be produced from \(5.95\text{ g}\) of nickel(II) carbonate.
[Relative atomic masses, \(M_r\): \(\text{Ni} = 59\), \(\text{C} = 12\), \(\text{O} = 16\), \(\text{S} = 32\), \(\text{H} = 1\)] [5]
查看答案詳解

解題

(a) The reaction between a metal carbonate and an acid yields a salt, carbon dioxide, and water:
\(\text{NiCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{NiSO}_4(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
State symbols are: (s) for solid nickel(II) carbonate, (aq) for aqueous sulfuric acid, (aq) for the dissolved nickel(II) sulfate salt, (g) for carbon dioxide gas, and (l) for liquid water.

(b) 1. Add excess nickel(II) carbonate to a measured volume of warm dilute sulfuric acid in a beaker and stir.
- Excess nickel(II) carbonate is used to ensure that all the sulfuric acid is completely neutralised and reacted.
2. Filter the mixture using a funnel and filter paper.
- This removes the excess unreacted solid nickel(II) carbonate as the residue, leaving a blue-green solution of nickel(II) sulfate as the filtrate.
3. Heat the filtrate in an evaporating basin to evaporate some of the water until the crystallization point is reached (test by seeing if crystals form on a cold glass rod).
4. Allow the hot saturated solution to cool down slowly. This allows hydrated nickel(II) sulfate crystals to form.
5. Filter off the crystals from the remaining liquid.
6. Dry the crystals gently by pressing them between pieces of filter paper or placing them in a desiccator. Do not heat them strongly with a Bunsen burner, as this would drive off the water of crystallization, leaving anhydrous nickel(II) sulfate.

(c) 1. Calculate the relative formula mass (\(M_r\)) of \(\text{NiCO}_3\):
\(M_r(\text{NiCO}_3) = 59 + 12 + (3 \times 16) = 119\).
2. Calculate the number of moles of \(\text{NiCO}_3\) reacted:
\(\text{moles of NiCO}_3 = \frac{5.95}{119} = 0.050\text{ mol}\).
3. From the chemical equation, the mole ratio of \(\text{NiCO}_3\) to \(\text{NiSO}_4 \cdot 6\text{H}_2\text{O}\) is \(1:1\).
\(\text{moles of NiSO}_4 \cdot 6\text{H}_2\text{O} = 0.050\text{ mol}\).
4. Calculate the relative formula mass (\(M_r\)) of \(\text{NiSO}_4 \cdot 6\text{H}_2\text{O}\):
\(M_r = 59 + 32 + (4 \times 16) + 6 \times (2 \times 1 + 16) = 59 + 32 + 64 + (6 \times 18) = 155 + 108 = 263\).
5. Calculate the mass of hydrated nickel(II) sulfate:
\(\text{mass} = \text{moles} \times M_r = 0.050\text{ mol} \times 263\text{ g/mol} = 13.15\text{ g}\) (or \(13.2\text{ g}\) to 3 sf).

評分準則

(a) Correct chemical formulae for reactants and products: \(\text{NiCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NiSO}_4 + \text{CO}_2 + \text{H}_2\text{O}\) [1]
Balanced equation [1]
State symbols: \(\text{NiCO}_3(\text{s})\), \(\text{H}_2\text{SO}_4(\text{aq})\), \(\text{NiSO}_4(\text{aq})\), \(\text{CO}_2(\text{g})\), \(\text{H}_2\text{O}(\text{l})\) [1]
(b) Excess used: to ensure all acid is completely reacted / neutralised [1.33]
Removal: filter to remove excess solid nickel(II) carbonate [1]
Crystals: heat to crystallization point / saturate AND leave to cool [2]
Drying: pat dry with filter paper / place in desiccator / do not heat strongly (which would lose water of crystallization) [1]
(c) \(M_r(\text{NiCO}_3) = 119\) [1]
Moles of \(\text{NiCO}_3 = 5.95 / 119 = 0.050\text{ mol}\) [1]
Mole ratio is \(1:1\) / moles of hydrated salt = \(0.050\text{ mol}\) [1]
\(M_r(\text{NiSO}_4 \cdot 6\text{H}_2\text{O}) = 263\) [1]
Mass = \(0.050 \times 263 = 13.15\text{ g}\) (accept \(13.2\text{ g}\)) [1]
題目 5 · structured
13.33 分
Polymers are long-chain molecules formed from many smaller units called monomers.

(a) Polymer A is an addition polymer with the following repeating unit:
\(-[\text{CH}(\text{C}_6\text{H}_5)-\text{CH}_2]-\)
(i) Draw the structure of the monomer used to make Polymer A. Show all atoms and bonds. [2]
(ii) State one major environmental problem associated with the disposal of addition polymers such as Polymer A. [1.33]

(b) Nylon is a synthetic polyamide formed by condensation polymerisation.
(i) Name the two types of functional groups that react together to form a polyamide. [2]
(ii) Draw the structural formula of nylon, showing two repeating units. Represent the carbon rings or chains with blocks as \(-[\text{HN}-\square-\text{NH}-\text{CO}-\square-\text{CO}]-\). Show the amide linkages clearly. [4]

(c) Terylene is a polyester.
(i) State the name of the linkage in Terylene and draw its structure showing all bonds. [2]
(ii) Compare the production of addition polymers and condensation polymers. In your answer, refer to the number of products formed. [2]
查看答案詳解

解題

(a) (i) The monomer of an addition polymer is an alkene. We replace the single bond between the carbon atoms of the backbone in the repeating unit with a double bond. The structure is phenylethene (styrene): \(\text{H}_2\text{C}=\text{CH}(\text{C}_6\text{H}_5)\). All bonds must be shown, particularly the double bond \(\text{C}=\text{C}\).
(ii) Addition polymers are inert and non-biodegradable, meaning they are not broken down by microbes, causing them to persist in landfills for hundreds of years. Burning them can also release toxic gases.

(b) (i) A polyamide is formed by the reaction of diamines (containing amine / amino groups, \(-\text{NH}_2\)) and dicarboxylic acids (containing carboxylic acid groups, \(-\text{COOH}\)).
(ii) Nylon has amide linkages (\(-\text{CO}-\text{NH}-\)). A structural formula showing two repeating units:
\(-[\text{HN}-\square-\text{NH}-\text{CO}-\square-\text{CO}-\text{HN}-\square-\text{NH}-\text{CO}-\square-\text{CO}]-\)
The amide linkage must show all bonds clearly: \(\text{-N(H)-C(=O)-}\).

(c) (i) Terylene is a polyester, so it contains ester linkages. Structure: \(\text{-O-C(=O)-}\).
(ii) Addition polymerisation involves the joining of unsaturated monomer molecules with no other products formed (only the polymer is produced). Condensation polymerisation involves the reaction of monomer molecules with two different functional groups, forming the polymer and releasing a small molecule (such as \(\text{H}_2\text{O}\) or \(\text{HCl}\)) as a side product (two products formed).

評分準則

(a) (i) Correct structure of monomer showing \(\text{C}=\text{C}\) double bond and all atoms (\(\text{C}_6\text{H}_5\)- group can be represented as a ring or text) [2]
(ii) Non-biodegradable / fill up landfills / produce toxic gases when incinerated [1.33]
(b) (i) Amine / amino group [1] and carboxylic acid group [1] (accept dicarboxylic acid / diamine)
(ii) Diagram showing two repeating units [2]
Amide linkages (\(-\text{CO}-\text{NH}-\) or \(\text{-NH-CO-}\)) shown correctly with all bonds [2]
(c) (i) Ester linkage [1] and its structure showing \(\text{-C(=O)-O-}\) bonds [1]
(ii) Addition polymerisation produces only one product (the polymer) [1]
Condensation polymerisation produces two products (the polymer and a small molecule like water / HCl) [1]
題目 6 · structured
13.33 分
A student is provided with a green crystalline solid, Compound Z, which contains two different cations and two different anions.

(a) The student dissolves a sample of Z in water to make an aqueous solution.
(i) To a portion of this solution, aqueous sodium hydroxide is added dropwise until in excess. A green precipitate is formed which is insoluble in excess sodium hydroxide.
Identify the metal cation present in Z. [1]
(ii) To another portion of the solution, the student adds dilute nitric acid followed by aqueous barium nitrate. A white precipitate is formed.
Identify one of the anions present in Z and write an ionic equation for this reaction. [3]
(iii) To a third portion of the solution, the student adds dilute nitric acid followed by aqueous silver nitrate. A cream precipitate is formed.
Identify the second anion present in Z and write an ionic equation for this reaction. [3]

(b) Solid Z is heated in a test-tube. A gas is evolved which turns damp red litmus paper blue.
(i) Identify this gas. [1]
(ii) State the name of the other cation that must have been present in Z to produce this gas. [1]
(iii) Describe a chemical test (other than litmus paper) to confirm the identity of the gas identified in (b)(i), including the observation. [2]

(c) Suggest a possible formula for anhydrous Compound Z that is consistent with all the above observations. [2.33]
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解題

(a) (i) A green precipitate with sodium hydroxide that is insoluble in excess indicates either Iron(II), \(\text{Fe}^{2+}\), or Nickel(II), \(\text{Ni}^{2+}\). (Chromium(III) also gives a green precipitate but it is soluble in excess sodium hydroxide). Iron(II) is most common.
(ii) The white precipitate formed with barium nitrate in the presence of nitric acid confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
The ionic equation is: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\).
(iii) The cream precipitate formed with silver nitrate in the presence of nitric acid confirms the presence of bromide ions, \(\text{Br}^-\).
The ionic equation is: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\).

(b) (i) A gas that turns damp red litmus paper blue is alkaline. This gas is ammonia, \(\text{NH}_3\).
(ii) To produce ammonia gas upon heating, the solid must contain ammonium ions, \(\text{NH}_4^+\).
(iii) Ammonia gas can be tested by holding a glass rod dipped in concentrated hydrochloric acid near the gas. Observation: White smoke (or dense white fumes) of ammonium chloride is formed.

(c) Based on the ions identified:
Cations: \(\text{Fe}^{2+}\) (or \(\text{Ni}^{2+}\)) and \(\text{NH}_4^+\).
Anions: \(\text{SO}_4^{2-}\) and \(\text{Br}^-\).
To make an electrically neutral compound, we can combine them. For example:
Charges: \(\text{NH}_4^+\) (\(+1\)) + \(\text{Fe}^{2+}\) (\(+2\)) = \(+3\).
\(\text{SO}_4^{2-}\) (\(-2\)) + \(\text{Br}^-\) (\(-1\)) = \(-3\).
Thus, a possible formula is \(\text{NH}_4\text{Fe(SO}_4)\text{Br}\) or \(\text{NH}_4\text{Ni(SO}_4)\text{Br}\). Alternatively, \(\text{(NH}_4)_2\text{Fe(SO}_4)\text{Br}_2\) is also electrically neutral and acceptable.

評分準則

(a) (i) \(\text{Fe}^{2+}\) / Iron(II) (accept \(\text{Ni}^{2+}\) / Nickel(II)) [1]
(ii) Sulfate / \(\text{SO}_4^{2-}\) [1]
Ionic equation: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\) [2] (1 mark for correct species, 1 mark for state symbols)
(iii) Bromide / \(\text{Br}^-\) [1]
Ionic equation: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) [2] (1 mark for correct species, 1 mark for state symbols)
(b) (i) Ammonia / \(\text{NH}_3\) [1]
(ii) Ammonium / \(\text{NH}_4^+\) [1]
(iii) Test: Glass rod dipped in concentrated hydrochloric acid [1]
Observation: Dense white smoke / fumes [1]
(c) \(\text{NH}_4\text{Fe(SO}_4)\text{Br}\) OR \(\text{(NH}_4)_2\text{Fe(SO}_4)\text{Br}_2\) (accept \(\text{Ni}\) in place of \(\text{Fe}\)) [2.33] (allow ecf from ions identified in (a) and (b))

Paper 61 Alternative to Practical

Answer all questions. Write your answers in the spaces provided on the question paper. Notes for use in qualitative analysis are provided.
4 題目 · 40 分
題目 1 · Practical & Planning
10 分

A student wants to investigate how the concentration of aqueous sodium thiosulfate affects its rate of reaction with dilute hydrochloric acid. When these two solutions are mixed, a precipitate of sulfur forms slowly, making the solution go cloudy and obscure a black cross drawn on a piece of paper placed under the reaction flask.

Plan an investigation to determine how the rate of this reaction changes with the concentration of sodium thiosulfate.

In your plan, you should include:

  • a list of the apparatus required
  • a detailed step-by-step method, including how the concentration of sodium thiosulfate will be varied
  • the variables that must be kept constant (controlled variables)
  • how you will use the experimental measurements to compare the rates of reaction.
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解題

To investigate the effect of concentration on the rate of reaction:

1. Apparatus: Conical flask, measuring cylinders (for measuring volumes of liquid), white paper marked with a black cross, stopwatch (for measuring time taken for reaction).

2. Method:
- Place the conical flask on the paper marked with a cross.
- Add a measured volume (e.g., 50 cm³) of sodium thiosulfate solution to the flask.
- Add a measured volume (e.g., 10 cm³) of dilute hydrochloric acid and start the stopwatch immediately.
- Swirl the flask once and look vertically down through the neck of the flask. Stop the stopwatch when the cross is completely obscured by the yellow precipitate of sulfur.
- Record the time taken (\(t\)).
- Repeat the experiment using different concentrations of sodium thiosulfate. This is achieved by mixing different ratios of sodium thiosulfate and distilled water (e.g., 40 cm³ thiosulfate + 10 cm³ water, 30 cm³ thiosulfate + 20 cm³ water, etc.) ensuring the total volume of the mixture remains constant.

3. Variables to control: Temperature of the reactants, total volume of the reaction mixture, and the concentration/volume of hydrochloric acid used.

4. Analyzing results: The rate is inversely proportional to the time taken (\(\text{Rate} \propto 1/t\)). Compare the times or calculate \(1/t\) for each concentration. A shorter time (higher \(1/t\)) indicates a faster rate of reaction.

評分準則

[1 mark] Apparatus: conical flask, measuring cylinder(s), stopwatch, paper with cross.
[4 marks] Method:
- Measure fixed volume of sodium thiosulfate and place in conical flask over the cross (1 mark).
- Add a fixed volume of acid and start the timer immediately (1 mark).
- Stop timer when cross is no longer visible and record time (1 mark).
- Repeat with different concentrations of sodium thiosulfate by diluting with water (1 mark).
[2 marks] Variables:
- Keep total volume of mixture constant (1 mark).
- Keep temperature constant OR keep concentration/volume of acid constant (1 mark).
[2 marks] Rate calculation:
- Rate is proportional to 1/time (or 1/t) (1 mark).
- Shorter time means higher rate (1 mark).
[1 mark] General clarity and logical sequence of the proposed method.
題目 2 · Practical & Planning
10 分

A student investigated the pigments present in three different blue felt-tip pens (A, B, and C) and a standard blue dye (D) using paper chromatography.

(a) State why the start line on the chromatography paper must be drawn in pencil and not ink. [1]

(b) State why the solvent level in the beaker must be below the start line. [1]

(c) During the experiment, the solvent front traveled a distance of 8.0 cm from the start line. One of the pigments in pen A moved a distance of 5.2 cm. Calculate the \(R_f\) value of this pigment. Show your working. [2]

(d) Suggest why a lid is placed on the beaker during the chromatography run. [1]

(e) If pen B produced only one spot on the chromatogram, what can you conclude about the composition of the ink in pen B? [1]

(f) In a different chromatography experiment, colorless amino acids are separated.
(i) State what must be sprayed on the chromatogram to make the spots visible. [1]
(ii) Describe how this substance is used safely. [1]
(iii) Explain how the student can identify the specific amino acids present once they are visible. [2]

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解題

(a) Pencil lead (graphite) is insoluble in chromatography solvents (such as water or ethanol), so it will not dissolve, run, or interfere with the chromatogram, whereas ink is soluble and would separate along with the samples.

(b) If the solvent level is above the start line, the spots of dyes will dissolve directly into the solvent in the bottom of the beaker instead of traveling up the paper.

(c) Calculation of \(R_f\) value:
\(R_f = \frac{\text{Distance traveled by substance}}{\text{Distance traveled by solvent front}}\)
\(R_f = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\)

(d) Placing a lid on the beaker prevents the solvent from evaporating and ensures that the atmosphere inside the beaker is saturated with solvent vapor, which helps the solvent travel up the paper evenly.

(e) It is a pure substance / single dye because it did not separate into multiple spots.

(f) (i) A locating agent (or ninhydrin) is sprayed on the paper.
(ii) Spraying should be done in a fume cupboard because the locating agent may be toxic/harmful/flammable, or safety goggles/gloves should be worn.
(iii) Calculate the \(R_f\) values of the spots and compare them to standard/known reference \(R_f\) values of amino acids (or run known samples of amino acids alongside on the same chromatogram and match their heights).

評分準則

[1 mark] (a) Pencil is insoluble (in the solvent) / graphite will not run or dissolve.
[1 mark] (b) To prevent the sample/spots from dissolving into the solvent pool.
[2 marks] (c) Calculation:
- Correct formula or division: 5.2 / 8.0 (1 mark)
- Correct answer: 0.65 (no units) (1 mark)
[1 mark] (d) To prevent evaporation of solvent / to saturate the atmosphere inside the beaker.
[1 mark] (e) It is a pure substance / contains only one dye.
[3 marks] (f):
- (i) Locating agent / ninhydrin (1 mark)
- (ii) Spray in a fume cupboard / wear eye protection/gloves (1 mark)
- (iii) Compare Rf values with known standards / run known amino acids alongside and compare heights (1 mark)
題目 3 · Practical & Planning
10 分

A student carried out a titration to determine the concentration of a solution of sodium hydroxide, \(\text{NaOH}\), using dilute hydrochloric acid, \(\text{HCl}\), with a concentration of \(0.100\text{ mol/dm}^3\).

(a) The student filled a burette with the \(0.100\text{ mol/dm}^3\) hydrochloric acid. The initial burette reading was \(1.2\text{ cm}^3\). After adding the acid to the sodium hydroxide solution in a conical flask until the end-point was reached, the final burette reading was \(27.0\text{ cm}^3\).
(i) State the volume of hydrochloric acid added during this titration. [1]
(ii) Suggest why the student should swirl the conical flask continuously during the addition of the acid. [1]
(iii) Explain why a white tile is placed under the conical flask. [1]

(b) Methyl orange indicator was used in the flask. State the color change of the indicator at the end-point:
from ___________ to ___________. [2]

(c) The student repeated the titration three more times. The volumes of acid added in these titrations were \(24.1\text{ cm}^3\), \(24.5\text{ cm}^3\), and \(24.3\text{ cm}^3\).
(i) State which of the four titrations (including the first one) is an anomaly and should be excluded from the average calculation. [1]
(ii) Calculate the average volume of hydrochloric acid used, using only the concordant results. Show your working. [2]

(d) Suggest how the student could obtain a pure sample of sodium chloride crystals from the neutralised mixture without any indicator present. [2]

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解題

(a) (i) Volume of acid added = Final reading - Initial reading = \(27.0\text{ cm}^3 - 1.2\text{ cm}^3 = 25.8\text{ cm}^3\).
(ii) Swirling ensures thorough mixing of the acid and alkali so that the reaction is complete and the end-point is detected accurately.
(iii) The white tile provides a neutral background, making the color change of the indicator much easier to see clearly.

(b) In alkaline solution (sodium hydroxide), methyl orange is yellow. At the neutralisation end-point, it changes to orange (or pink/red if excess acid is added). Thus, the color change is from yellow to orange (or pink/red).

(c) (i) The first titration volume (\(25.8\text{ cm}^3\)) is anomalous because it is not within \(\pm 0.2\text{ cm}^3\) of the other results (concordant results are \(24.1\), \(24.5\), and \(24.3\text{ cm}^3\)).
(ii) Average of concordant results:
\(\text{Average} = \frac{24.1 + 24.5 + 24.3}{3} = \frac{72.9}{3} = 24.3\text{ cm}^3\).

(d) To obtain pure sodium chloride crystals:
- Repeat the titration using the same volumes (\(25.0\text{ cm}^3\) of \(\text{NaOH}\) and \(24.3\text{ cm}^3\) of \(\text{HCl}\)) but without adding any indicator.
- Heat the resulting sodium chloride solution to evaporate most of the water until the crystallization point is reached.
- Allow the hot, saturated solution to cool slowly so crystals form, then filter and dry the crystals with filter paper.

評分準則

[1 mark] (a)(i) 25.8 cm³
[1 mark] (a)(ii) To ensure complete mixing / reactants react fully.
[1 mark] (a)(iii) To make the color change of the indicator easier to see clearly.
[2 marks] (b) Yellow (1 mark) to orange / pink / red (1 mark).
[1 mark] (c)(i) Titration 1 (or 25.8 cm³).
[2 marks] (c)(ii) Calculation:
- Correct values chosen: 24.1, 24.5, 24.3 (1 mark)
- Correct calculation: 24.3 cm³ (1 mark)
[2 marks] (d) Preparation:
- Repeat titration without indicator using 24.3 cm³ of acid (1 mark).
- Evaporate to crystallization point and allow to cool / crystallize (1 mark).
題目 4 · Practical & Planning
10 分

A student carried out tests on a green solid, Compound E, which is a soluble salt. Complete the table of observations and conclusions below.

Tests and observations:

(a) Test 1: A portion of solid E was heated in a dry test-tube. A colorless gas was given off which turned limewater cloudy.
(i) Name the gas given off. [1]
(ii) Identify the anion present in Compound E. [1]

(b) Test 2: Dilute nitric acid was added to an aqueous solution of E, followed by aqueous silver nitrate. No precipitate was formed.
What conclusion can be drawn about the presence of halide ions in E? [1]

(c) Test 3: Aqueous sodium hydroxide was added dropwise, then in excess, to a portion of the solution of E. A green precipitate was formed, which was insoluble in excess.
(i) Identify the cation present in E. [1]
(ii) Explain why the green precipitate does not dissolve in excess sodium hydroxide, referring to the chemical nature of the precipitate. [2]

(d) Test 4: Aqueous ammonia was added dropwise, then in excess, to another portion of the solution of E.
(i) Describe the observation you would expect to see. [2]
(ii) State the chemical formula of Compound E. [1]

(e) Describe how you would test a solution of E to show that sulfate ions are not present. State the reagents used and the observation for a negative result. [1]

查看答案詳解

解題

(a) (i) The gas that turns limewater cloudy is carbon dioxide (\(\text{CO}_2\)).
(ii) The presence of carbon dioxide indicates that the anion in Compound E is carbonate (\(\text{CO}_3^{2-}\)).

(b) The absence of a precipitate when silver nitrate is added means that halide ions (chloride, bromide, and iodide) are absent.

(c) (i) Iron(II) ions (\(\text{Fe}^{2+}\)) form a green precipitate with sodium hydroxide which is insoluble in excess. (Note: Chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to give a green solution, so E must contain \(\text{Fe}^{2+}\)).
(ii) The green precipitate is iron(II) hydroxide, \(\text{Fe(OH)}_2\). It is a basic oxide/hydroxide and is not amphoteric, so it does not react with or dissolve in excess sodium hydroxide (unlike amphoteric hydroxides like zinc or chromium hydroxide).

(d) (i) With aqueous ammonia, iron(II) ions form a green precipitate that is insoluble in excess ammonia.
(ii) Compound E contains \(\text{Fe}^{2+}\) and \(\text{CO}_3^{2-}\), so its formula is \(\text{FeCO}_3\).

(e) To test for sulfate ions, add dilute hydrochloric acid (or nitric acid) followed by aqueous barium chloride (or barium nitrate). If sulfate ions are absent, no white precipitate is formed (the solution remains clear).

評分準則

[1 mark] (a)(i) Carbon dioxide / CO2
[1 mark] (a)(ii) Carbonate / CO3^2-
[1 mark] (b) Halide ions (Cl-, Br-, I-) are absent.
[1 mark] (c)(i) Iron(II) / Fe^2+
[2 marks] (c)(ii) The precipitate is iron(II) hydroxide (1 mark) which is basic / not amphoteric / does not react with excess NaOH (1 mark).
[2 marks] (d)(i) Green precipitate (1 mark) which is insoluble in excess (1 mark).
[1 mark] (d)(ii) FeCO3
[1 mark] (e) Add dilute hydrochloric acid/nitric acid and aqueous barium chloride/barium nitrate, and observe no white precipitate.

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