An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V1) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
卷二 (選擇題 Extended)
Answer all 40 multiple-choice questions on the separate answer sheet in soft pencil.
80 題目 · 80 分
題目 1 · 選擇題
1 分
The table shows information about four different atoms, W, X, Y and Z.
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. Looking at the table, atoms W and X both have 6 protons (which means they are carbon atoms) but have different numbers of neutrons (W has 6, while X has 8). Therefore, W and X are isotopes of the same element.
評分準則
Award 1 mark for the correct answer A.
題目 2 · 選擇題
1 分
Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes.
Which row identifies the product formed at each electrode?
During the electrolysis of aqueous copper(II) sulfate using inert electrodes: - At the positive electrode (anode), hydroxide ions ($$\text{OH}^-$$) from water are discharged in preference to sulfate ions ($$\text{SO}_4^{2-}$$) to form oxygen gas ($$\text{O}_2$$). - At the negative electrode (cathode), copper(II) ions ($$\text{Cu}^{2+}$$) are discharged in preference to hydrogen ions ($$\text{H}^+$$) because copper is less reactive than hydrogen. This produces copper metal ($$\text{Cu}$$).
評分準則
Award 1 mark for the correct answer B.
題目 3 · 選擇題
1 分
Which polymer is formed by condensation polymerisation?
A.poly(chloroethene)
B.poly(ethene)
C.polyester
D.poly(propene)
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解題
Addition polymerisation involves the linking of monomers containing double bonds (such as ethene, propene, and chloroethene) without forming any other products. Poly(chloroethene), poly(ethene), and poly(propene) are all addition polymers. Polyester (such as terylene) is formed by a condensation reaction between a dicarboxylic acid and a diol, releasing small molecules (such as water), and is therefore a condensation polymer.
評分準則
Award 1 mark for the correct answer C.
題目 4 · 選擇題
1 分
Four metal oxides are heated strongly with carbon powder.
Which of these oxides can be reduced by heating with carbon?
A.1 and 2 only
B.3 and 4 only
C.1, 2 and 4 only
D.1, 2, 3 and 4
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解題
A metal oxide can be reduced by carbon if the metal is less reactive than carbon. In the reactivity series, carbon is more reactive than zinc, iron, and copper, but less reactive than magnesium. Therefore, copper(II) oxide, iron(III) oxide, and zinc oxide (1, 2, and 4) are reduced by heating with carbon, whereas magnesium oxide (3) is not.
評分準則
Award 1 mark for the correct answer C.
題目 5 · 選擇題
1 分
An unknown salt solution is tested using aqueous reagents.
- The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. - The addition of aqueous ammonia produces a green precipitate that is insoluble in excess.
Which cation is present in the solution?
A.chromium(III)
B.iron(II)
C.iron(III)
D.copper(II)
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解題
According to the qualitative analysis notes: - Chromium(III) ($$\text{Cr}^{3+}$$) produces a green precipitate that dissolves in excess aqueous sodium hydroxide to form a green solution, and is insoluble in excess aqueous ammonia. - Iron(II) ($$\text{Fe}^{2+}$$) produces a green precipitate that is insoluble in both excess aqueous sodium hydroxide and excess aqueous ammonia. Therefore, the cation present is iron(II).
評分準則
Award 1 mark for the correct answer B.
題目 6 · 選擇題
1 分
A student titrated dilute hydrochloric acid from a burette into a conical flask containing aqueous sodium hydroxide and a few drops of thymolphthalein indicator.
Which colour change is observed at the end-point of this titration?
A.blue to colourless
B.colourless to blue
C.pink to colourless
D.colourless to pink
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解題
Thymolphthalein is an indicator that is blue in alkaline solutions (pH above 10) and colourless in acidic or neutral solutions. Since the acid is added from the burette to the alkali in the conical flask, the solution initially starts as alkaline (blue) and changes to neutral/acidic (colourless) at the end-point. Thus, the observed colour change is blue to colourless.
評分準則
Award 1 mark for the correct answer A.
題目 7 · 選擇題
1 分
Which process is endothermic?
A.the combustion of methane
B.the decomposition of calcium carbonate
C.the reaction of anhydrous copper(II) sulfate with water
D.the neutralisation of an acid by an alkali
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解題
An endothermic process is one that absorbs thermal energy from the surroundings. Thermal decomposition of calcium carbonate requires continuous heating to break down into calcium oxide and carbon dioxide, making it endothermic. Combustion of methane, the reaction of anhydrous copper(II) sulfate with water, and the neutralisation of an acid by an alkali are all exothermic processes that release thermal energy to the surroundings.
評分準則
Award 1 mark for the correct answer B.
題目 8 · 選擇題
1 分
Which statement explains why copper is used to make electrical wiring?
A.It is a transition element with a high melting point.
B.It is ductile and has delocalised electrons that are free to move.
C.It is resistant to corrosion and does not react with water.
D.It is strong and has a low density.
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解題
Copper is used for electrical wiring because of its physical properties: it is ductile, meaning it can be drawn out into long thin wires, and it is an excellent conductor of electricity because it has a metallic lattice structure with delocalised valence electrons that are free to move and carry charge. While it is a transition metal with a high melting point and is relatively unreactive, these properties do not directly explain why it is used as wiring.
評分準則
Award 1 mark for the correct answer B.
題目 9 · single-select
1 分
An experiment is set up to compare the rates of diffusion of four different gases: carbon dioxide, \(CO_2\), hydrogen chloride, \(HCl\), ammonia, \(NH_3\), and methane, \(CH_4\). Which gas will diffuse the furthest distance in a given time at room temperature?
A.carbon dioxide
B.hydrogen chloride
C.ammonia
D.methane
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解題
The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (or simply, lighter gases diffuse faster). Lighter gases have smaller relative molecular masses (\(M_r\)). The relative molecular masses of the gases are: \(CO_2 = 44\), \(HCl = 36.5\), \(NH_3 = 17\), and \(CH_4 = 16\). Methane has the smallest \(M_r\) and therefore diffuses the fastest and furthest.
評分準則
1 mark for identifying methane as having the smallest relative molecular mass and thus diffusing the furthest.
題目 10 · single-select
1 分
An atom of element X is represented by having 15 protons, 16 neutrons, and 15 electrons. Which row identifies the nucleon number of X and the group number of X in the Periodic Table?
A.nucleon number = 31, group number = V
B.nucleon number = 31, group number = VI
C.nucleon number = 15, group number = V
D.nucleon number = 15, group number = VI
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解題
The nucleon number of an atom is the total number of protons and neutrons in its nucleus: \(15 + 16 = 31\). The group number of an element is determined by the number of valence (outer shell) electrons. With 15 electrons, the electronic configuration of X is 2, 8, 5, which means it has 5 valence electrons and belongs to Group V.
評分準則
1 mark for correctly identifying both the nucleon number as 31 and the group number as V.
題目 11 · single-select
1 分
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which statement correctly describes the reaction that occurs at one of the electrodes?
A.Lead ions gain electrons at the anode to form lead metal.
B.Lead ions gain electrons at the cathode to form lead metal.
C.Bromide ions gain electrons at the anode to form bromine gas.
D.Bromide ions lose electrons at the cathode to form bromine gas.
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解題
During the electrolysis of molten lead(II) bromide, positive lead(II) ions (\(Pb^{2+}\)) are attracted to the negative electrode (cathode), where they gain electrons (reduction) to form lead metal: \(Pb^{2+} + 2e^- \rightarrow Pb\). Negative bromide ions (\(Br^-\)) migrate to the positive electrode (anode), where they lose electrons (oxidation) to form bromine gas: \(2Br^- \rightarrow Br_2 + 2e^-\).
評分準則
1 mark for selecting the statement that lead ions gain electrons at the cathode to form lead metal.
題目 12 · single-select
1 分
Four oxides are listed below: 1. Aluminium oxide, \(Al_2O_3\) 2. Carbon dioxide, \(CO_2\) 3. Copper(II) oxide, \(CuO\) 4. Sulfur dioxide, \(SO_2\). Which of these oxides will react with both aqueous sodium hydroxide and dilute hydrochloric acid?
A.1 only
B.1 and 3
C.2 and 4
D.3 only
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解題
An oxide that reacts with both acids and bases is classified as an amphoteric oxide. Aluminium oxide (\(Al_2O_3\)) is amphoteric and can react with both dilute hydrochloric acid and aqueous sodium hydroxide. Carbon dioxide and sulfur dioxide are acidic oxides (react only with bases), and copper(II) oxide is a basic oxide (reacts only with acids).
評分準則
1 mark for identifying only aluminium oxide (1) as the oxide that reacts with both aqueous sodium hydroxide and dilute hydrochloric acid.
題目 13 · single-select
1 分
Equal masses of four different metals are added to separate, identical volumes of dilute sulfuric acid. The observations are recorded: Metal 1 dissolves rapidly with vigorous bubbling, and the solution remains colorless; Metal 2 shows no reaction and remains unchanged; Metal 3 dissolves slowly with gentle bubbling, and the solution remains colorless; Metal 4 reacts extremely violently, posing a dangerous explosion risk. Which order shows the metals arranged from least reactive to most reactive?
The reactivity of a metal with dilute acid is indicated by the rate of hydrogen gas production (bubbling). Metal 2 is the least reactive as it does not react at all. Metal 3 is next, reacting slowly. Metal 1 reacts rapidly, and Metal 4 is the most reactive, reacting extremely violently. Therefore, the correct order from least to most reactive is 2, then 3, then 1, then 4.
評分準則
1 mark for the correct arrangement of the metals from least to most reactive.
題目 14 · single-select
1 分
An addition polymer has the structure represented by \(-[-CH_2-CH(CH_3)-CH_2-CH(CH_3)-]-\). Which monomer is used to produce this polymer?
A.ethane
B.ethene
C.propane
D.propene
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解題
The polymer shown is polypropene. It is formed by the addition polymerisation of propene monomer, \(CH_2=CH-CH_3\). In addition polymerisation, the double bond of the alkene monomer breaks to form a long single-bonded carbon chain.
評分準則
1 mark for identifying propene as the monomer of polypropene.
題目 15 · single-select
1 分
In an acid-base titration, a student uses a pipette to transfer a specific volume of sodium hydroxide solution into a conical flask, then adds a few drops of methyl orange indicator. Which piece of apparatus should be used to add dilute hydrochloric acid to the conical flask, and what is the color change of the indicator at the end-point?
A.apparatus = burette, color change = yellow to orange
B.apparatus = burette, color change = orange to yellow
C.apparatus = measuring cylinder, color change = yellow to pink
D.apparatus = measuring cylinder, color change = pink to yellow
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解題
Dilute hydrochloric acid is added using a burette to allow precise control over the volume added. Methyl orange is yellow in alkaline solutions (such as aqueous sodium hydroxide). At the end-point, as the solution is neutralised, the color changes from yellow to orange.
評分準則
1 mark for selecting the correct apparatus (burette) and the correct color change (yellow to orange).
題目 16 · single-select
1 分
A colorless solution is tested with the following results: 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. Which aqueous ions are present in the solution?
A.copper(II) and chloride
B.iron(II) and sulfate
C.iron(III) and sulfate
D.chromium(III) and chloride
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解題
The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(Fe^{2+}\) (whereas chromium(III) forms a green precipitate that dissolves in excess to give a green solution). The formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate indicates the presence of sulfate ions, \(SO_4^{2-}\). Thus, the ions present are iron(II) and sulfate.
評分準則
1 mark for correctly identifying the cation as iron(II) and the anion as sulfate.
題目 17 · 選擇題
1 分
In a gaseous diffusion experiment, four gases are released at the same temperature and pressure. Which gas will diffuse the fastest?
A.Argon, \(\text{Ar}\)
B.Carbon dioxide, \(\text{CO}_2\)
C.Fluorine, \(\text{F}_2\)
D.Nitrogen, \(\text{N}_2\)
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解題
The rate of diffusion of a gas is inversely proportional to its relative molecular mass. This means that lighter gas molecules diffuse faster. The relative molecular masses of the given gases are: Ar = 40, \(\text{CO}_2\) = 44, \(\text{F}_2\) = 38, and \(\text{N}_2\) = 28. Since nitrogen has the lowest relative molecular mass, it will diffuse the fastest.
評分準則
[1 mark] - Correct choice D is selected.
題目 18 · 選擇題
1 分
Chlorine has two naturally occurring isotopes: \({}^{35}\text{Cl}\) and \({}^{37}\text{Cl}\). Which statement about these isotopes is correct?
A.\({}^{35}\text{Cl}\) and \({}^{37}\text{Cl}\) have different numbers of electrons in their outer shells.
B.\({}^{37}\text{Cl}\) has two more neutrons than \({}^{35}\text{Cl}\).
C.The chemical properties of \({}^{35}\text{Cl}\) and \({}^{37}\text{Cl}\) are different because their nucleon numbers are different.
D.\({}^{35}\text{Cl}\) and \({}^{37}\text{Cl}\) have different atomic numbers.
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解題
Isotopes are atoms of the same element with the same atomic number (number of protons) but different nucleon numbers (number of neutrons). The atomic number of chlorine is 17. Therefore, \({}^{35}\text{Cl}\) has 18 neutrons (35 - 17) and \({}^{37}\text{Cl}\) has 20 neutrons (37 - 17). Thus, \({}^{37}\text{Cl}\) has two more neutrons than \({}^{35}\text{Cl}\). Both isotopes have identical chemical properties because they have the same electronic configuration.
評分準則
[1 mark] - Correct choice B is selected.
題目 19 · 選擇題
1 分
Aqueous copper(II) sulfate is electrolysed using inert carbon electrodes. What are the products formed at the positive electrode (anode) and negative electrode (cathode)?
A.anode: copper; cathode: oxygen
B.anode: oxygen; cathode: copper
C.anode: hydrogen; cathode: copper
D.anode: oxygen; cathode: hydrogen
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解題
At the positive anode (+), hydroxide ions (\(\text{OH}^-\)) from the water are discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)) to produce oxygen gas (\(\text{O}_2\)). At the negative cathode (-), copper ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions (\(\text{H}^+\)) because copper is lower than hydrogen in the reactivity series, forming copper metal.
評分準則
[1 mark] - Correct choice B is selected.
題目 20 · 選擇題
1 分
In the extraction of iron, the following reaction occurs in the blast furnace: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). Which statement correctly identifies the substance that is oxidised and the reducing agent?
A.\(\text{Fe}_2\text{O}_3\) is oxidised and \(\text{CO}\) is the reducing agent.
B.\(\text{CO}\) is oxidised and \(\text{Fe}_2\text{O}_3\) is the reducing agent.
C.\(\text{CO}\) is oxidised and \(\text{CO}\) is the reducing agent.
D.\(\text{Fe}_2\text{O}_3\) is reduced and \(\text{CO}_2\) is the reducing agent.
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解題
Carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)), which means it is oxidised. The substance that is oxidised acts as the reducing agent because it reduces the other reactant (iron(III) oxide). Therefore, carbon monoxide is both the substance oxidised and the reducing agent.
評分準則
[1 mark] - Correct choice C is selected.
題目 21 · 選擇題
1 分
The reaction for the Haber process is: \(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\) (where the forward reaction is exothermic). Which set of conditions will produce the highest yield of ammonia at equilibrium?
A.high temperature and high pressure
B.high temperature and low pressure
C.low temperature and high pressure
D.low temperature and low pressure
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解題
To maximise the yield of ammonia (the forward reaction): 1. Temperature: Since the forward reaction is exothermic, lowering the temperature shifts the equilibrium to the right to produce more heat, increasing the yield. 2. Pressure: There are 4 moles of gas on the reactant side and 2 moles of gas on the product side. Increasing the pressure shifts the equilibrium to the side with fewer gas molecules (the right), increasing the yield. Therefore, low temperature and high pressure will produce the highest yield.
評分準則
[1 mark] - Correct choice C is selected.
題目 22 · 選擇題
1 分
Metal X reacts vigorously with cold water. Metal Y reacts with steam but does not react with cold water. Metal Z does not react with dilute hydrochloric acid. What is the order of reactivity of these three metals, starting with the most reactive?
A.X \(\rightarrow\) Y \(\rightarrow\) Z
B.Y \(\rightarrow\) X \(\rightarrow\) Z
C.Z \(\rightarrow\) Y \(\rightarrow\) X
D.X \(\rightarrow\) Z \(\rightarrow\) Y
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解題
Metal X is the most reactive because it reacts with cold water. Metal Y is moderately reactive as it requires steam to react. Metal Z is the least reactive because it cannot even displace hydrogen from dilute acid. Therefore, the reactivity order from most reactive to least reactive is X \(\rightarrow\) Y \(\rightarrow\) Z.
評分準則
[1 mark] - Correct choice A is selected.
題目 23 · 選擇題
1 分
Terylene is a synthetic polyester. Which pair of monomers is used in a condensation polymerisation reaction to produce Terylene?
A.a dicarboxylic acid and a diamine
B.a dicarboxylic acid and a diol
C.a diamine and a diol
D.a dicarboxylic acid and an alkene
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解題
Polyesters are formed by condensation polymerisation when a dicarboxylic acid monomer reacts with a diol monomer. The ester linkages are formed with the elimination of water molecules.
評分準則
[1 mark] - Correct choice B is selected.
題目 24 · 選擇題
1 分
An aqueous solution of an unknown salt S is tested. The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. The addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of S?
A.chromium(III) chloride
B.iron(II) sulfate
C.iron(III) sulfate
D.iron(II) chloride
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解題
The green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\) (chromium(III) ions also form a green precipitate, but it is soluble in excess sodium hydroxide). The white precipitate with acidified barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, salt S is iron(II) sulfate.
評分準則
[1 mark] - Correct choice B is selected.
題目 25 · 選擇題
1 分
Under the same conditions of temperature and pressure, which gas diffuses the slowest?
A.carbon monoxide, \(CO\)
B.chlorine, \(Cl_2\)
C.methane, \(CH_4\)
D.neon, \(Ne\)
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解題
The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (\(M_r\)). Therefore, the gas with the highest \(M_r\) diffuses the slowest. Let's calculate the \(M_r\) for each gas: - Carbon monoxide (\(CO\)): \(12 + 16 = 28\) - Chlorine (\(Cl_2\)): \(35.5 \times 2 = 71\) - Methane (\(CH_4\)): \(12 + 4 = 16\) - Neon (\(Ne\)): \(20\)
Chlorine has the largest relative molecular mass and will therefore diffuse the slowest.
評分準則
1 mark: B is selected as the correct answer.
題目 26 · 選擇題
1 分
How many protons, neutrons and electrons are in the phosphide ion, \(P^{3-}\), formed from an atom of phosphorus with nucleon number 31?
A.15 protons, 16 neutrons, 18 electrons
B.15 protons, 16 neutrons, 12 electrons
C.18 protons, 13 neutrons, 15 electrons
D.15 protons, 31 neutrons, 18 electrons
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解題
Phosphorus (\(P\)) has atomic number 15, which means a neutral atom has 15 protons and 15 electrons. - The number of neutrons is calculated by subtracting the atomic number from the nucleon number: \(31 - 15 = 16\) neutrons. - The phosphide ion has a charge of \(3-\), meaning it has gained 3 electrons: \(15 + 3 = 18\) electrons.
Therefore, the phosphide ion contains 15 protons, 16 neutrons, and 18 electrons.
評分準則
1 mark: A is selected as the correct answer.
題目 27 · 選擇題
1 分
A dilute aqueous solution of sodium sulfate is electrolysed using inert electrodes.
Which row identifies the product formed at each electrode?
During the electrolysis of dilute aqueous sodium sulfate (\(Na_2SO_4\)): - At the positive electrode (anode), hydroxide ions (\(OH^-\)) from water are discharged in preference to sulfate ions (\(SO_4^{2-}\)), producing oxygen gas: \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\). - At the negative electrode (cathode), hydrogen ions (\(H^+\)) from water are discharged in preference to sodium ions (\(Na^+\)), producing hydrogen gas: \(2H^+ + 2e^- \rightarrow H_2\).
Thus, oxygen is produced at the anode and hydrogen at the cathode.
評分準則
1 mark: A is selected as the correct answer.
題目 28 · 選擇題
1 分
Which statement about exothermic and endothermic reactions is correct?
A.In an endothermic reaction, more energy is released when bonds are formed than is absorbed when bonds are broken.
B.In an exothermic reaction, the temperature of the surroundings decreases.
C.In an endothermic reaction, the products have more energy than the reactants.
D.In an exothermic reaction, the enthalpy change (\(\Delta H\)) is positive.
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解題
Let's evaluate each statement: - Option A is incorrect: in an endothermic reaction, more energy is released when bonds are formed than is absorbed when bonds are broken. - Option B is incorrect: in an exothermic reaction, the temperature of the surroundings decreases. - Option C is correct: since energy is taken in from the surroundings in an endothermic reaction, the total chemical energy of the products is greater than that of the reactants. - Option D is incorrect: exothermic reactions have a negative enthalpy change (\(\Delta H < 0\)).
評分準則
1 mark: C is selected as the correct answer.
題目 29 · 選擇題
1 分
The table shows whether displacement reactions occur when metals P, Q, R and S are added to aqueous solutions of their salts.
| Metal added | Salt of P | Salt of Q | Salt of R | Salt of S | | :---: | :---: | :---: | :---: | :---: | | P | - | no reaction | reaction | no reaction | | Q | reaction | - | reaction | reaction | | R | no reaction | no reaction | - | no reaction | | S | reaction | no reaction | reaction | - |
What is the order of reactivity of the metals, from most reactive to least reactive?
A.Q \(\rightarrow\) S \(\rightarrow\) P \(\rightarrow\) R
B.Q \(\rightarrow\) P \(\rightarrow\) S \(\rightarrow\) R
C.R \(\rightarrow\) P \(\rightarrow\) S \(\rightarrow\) Q
D.R \(\rightarrow\) S \(\rightarrow\) P \(\rightarrow\) Q
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解題
To find the order of reactivity, we analyze how many displacement reactions each metal can perform: - Metal Q displaces P, R, and S (3 reactions). Thus, Q is the most reactive. - Metal S displaces P and R (2 reactions). Thus, S is the second most reactive. - Metal P only displaces R (1 reaction). Thus, P is third. - Metal R does not displace any of the other metals (0 reactions). Thus, R is the least reactive.
Reflecting this, the order of reactivity from most to least reactive is: Q \(\rightarrow\) S \(\rightarrow\) P \(\rightarrow\) R.
評分準則
1 mark: A is selected as the correct answer.
題目 30 · 選擇題
1 分
In a catalytic converter, nitrogen monoxide and carbon monoxide react to form less harmful gases:
\(2NO + 2CO \rightarrow N_2 + 2CO_2\)
Which statement about this reaction is correct?
A.Carbon monoxide is reduced to carbon dioxide.
B.Nitrogen monoxide is oxidised to nitrogen gas.
C.Carbon monoxide acts as an oxidising agent.
D.Nitrogen monoxide is reduced because it loses oxygen.
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解題
Let's analyze the reaction in terms of redox definitions: - Carbon monoxide (\(CO\)) gains oxygen to form carbon dioxide (\(CO_2\)). Since gain of oxygen is oxidation, \(CO\) is oxidised (acting as a reducing agent). - Nitrogen monoxide (\(NO\)) loses oxygen to form nitrogen gas (\(N_2\)). Since loss of oxygen is reduction, \(NO\) is reduced (acting as an oxidising agent).
Therefore, statement D is correct.
評分準則
1 mark: D is selected as the correct answer.
題目 31 · 選擇題
1 分
Which statement describes a chemical property of aqueous ethanoic acid?
A.It reacts with copper metal to produce copper(II) ethanoate and hydrogen gas.
B.It reacts with sodium carbonate to produce carbon dioxide gas.
C.It turns universal indicator paper dark blue.
D.It reacts with sodium hydroxide in a highly endothermic reaction.
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解題
Let's evaluate each option: - Option A is incorrect because copper is below hydrogen in the reactivity series and does not react with dilute acids to produce hydrogen gas. - Option B is correct because ethanoic acid is an acid and reacts with carbonates to produce a salt, water, and carbon dioxide gas. - Option C is incorrect because ethanoic acid is weak acid, so it turns universal indicator yellow/orange, not blue (which indicates an alkali). - Option D is incorrect because neutralisation reactions are exothermic (release thermal energy), not endothermic.
評分準則
1 mark: B is selected as the correct answer.
題目 32 · 選擇題
1 分
A portion of an addition polymer structure is shown:
\([-CH_2-CH(Cl)-CH_2-CH(Cl)-CH_2-CH(Cl)-]\)
What is the monomer used to make this polymer?
A.chloroethane
B.chloroethene
C.dichloroethane
D.dichloroethene
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解題
The given polymer structure is poly(chloroethene), which has the repeating unit \([-CH_2-CH(Cl)-]\). Addition polymers are made from unsaturated monomers (containing a carbon-carbon double bond). To find the monomer, we replace the single bond in the repeating unit with a double bond: \(CH_2=CHCl\), which is chloroethene.
Saturated compounds like chloroethane cannot undergo addition polymerisation.
評分準則
1 mark: B is selected as the correct answer.
題目 33 · MCQ
1 分
Two gas jars, one containing sulfur dioxide, \(SO_2\), and the other containing carbon dioxide, \(CO_2\), are opened at the same time in a closed room at constant temperature. Which statement about their diffusion is correct?
A.Sulfur dioxide diffuses faster because it has a larger relative molecular mass.
B.Carbon dioxide diffuses faster because it has a smaller relative molecular mass.
C.Both gases diffuse at the same rate because they are at the same temperature.
D.Neither gas diffuses because both are denser than air.
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解題
The rate of diffusion of a gas depends on its relative molecular mass (\(M_r\)). The relative molecular mass of carbon dioxide (\(CO_2\)) is \(12 + 2(16) = 44\), while that of sulfur dioxide (\(SO_2\)) is \(32 + 2(16) = 64\). Since carbon dioxide has a smaller relative molecular mass, it diffuses faster than sulfur dioxide.
評分準則
1 mark for selecting the correct option (B).
題目 34 · MCQ
1 分
An ion of an isotope of chlorine is represented as \(^{37}_{17}\text{Cl}^{-}\). How many protons, neutrons, and electrons are in this ion?
A.17 protons, 20 neutrons, 18 electrons
B.17 protons, 20 neutrons, 16 electrons
C.17 protons, 37 neutrons, 18 electrons
D.18 protons, 20 neutrons, 17 electrons
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解題
The atomic number is 17, so there are 17 protons. The nucleon number (mass number) is 37, so the number of neutrons is \(37 - 17 = 20\). Since it is a negatively charged ion with a 1- charge, it has gained one electron, giving it \(17 + 1 = 18\) electrons.
評分準則
1 mark for selecting the correct option (A).
題目 35 · MCQ
1 分
A strip of metal X is placed in an aqueous solution of metal Y sulfate. A pink-brown solid is deposited on the surface of metal X. Which row correctly identifies metals X and Y?
A.X = copper, Y = zinc
B.X = iron, Y = copper
C.X = silver, Y = copper
D.X = zinc, Y = magnesium
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解題
The pink-brown solid deposited is copper metal. This indicates that metal Y is copper (\(Cu\)). For copper to be displaced from its solution, metal X must be more reactive than copper. Iron (\(Fe\)) is more reactive than copper and can displace it, whereas silver (\(Ag\)) is less reactive and cannot.
評分準則
1 mark for selecting the correct option (B).
題目 36 · MCQ
1 分
A chromatogram of a food dye is run using an organic solvent. The solvent front travels \(10.0\text{ cm}\) from the baseline. A yellow spot travels \(4.5\text{ cm}\) from the baseline. What is the \(R_f\) value of the yellow spot and what does it indicate about its solubility?
A.\(R_f = 0.45\); it is more soluble in the solvent than a spot with \(R_f = 0.80\).
B.\(R_f = 0.45\); it is less soluble in the solvent than a spot with \(R_f = 0.80\).
C.\(R_f = 2.22\); it is more soluble in the solvent than a spot with \(R_f = 0.80\).
D.\(R_f = 2.22\); it is less soluble in the solvent than a spot with \(R_f = 0.80\).
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解題
The \(R_f\) value is calculated as \(\text{distance moved by spot} / \text{distance moved by solvent} = 4.5\text{ cm} / 10.0\text{ cm} = 0.45\). A lower \(R_f\) value (0.45 compared to 0.80) means the substance travels less with the mobile phase, indicating it is less soluble in the organic solvent.
評分準則
1 mark for selecting the correct option (B).
題目 37 · MCQ
1 分
Ethanol is manufactured industrially by the catalytic hydration of ethene. Which reaction conditions are required for this process?
The industrial hydration of ethene to produce ethanol requires a temperature of \(300^\circ\text{C}\), a high pressure of \(60\text{ atm}\) (equivalent to \(6000\text{ kPa}\)), and a phosphoric acid catalyst.
評分準則
1 mark for selecting the correct option (A).
題目 38 · MCQ
1 分
Which piece of apparatus should be used to measure exactly \(25.0\text{ cm}^3\) of aqueous sodium hydroxide into a conical flask, and which indicator is suitable for titrating it with dilute hydrochloric acid?
A.measuring cylinder; methyl orange
B.volumetric pipette; methyl orange
C.measuring cylinder; universal indicator
D.volumetric pipette; universal indicator
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解題
A volumetric pipette is used to deliver an extremely accurate, fixed volume of a solution (e.g., \(25.0\text{ cm}^3\)). Methyl orange is a suitable indicator for a strong acid-strong base titration because it has a sharp, distinct color change at the end-point. Universal indicator is not suitable because its color changes are too gradual.
評分準則
1 mark for selecting the correct option (B).
題目 39 · MCQ
1 分
Which statement about condensation polymers is correct?
A.They are formed from monomers containing at least one carbon-carbon double bond.
B.Their formation produces a polymer as the only product.
C.Polyesters are formed by the reaction of a dicarboxylic acid and a diol.
D.Terylene is an addition polymer used to make clothing fibers.
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解題
Condensation polyesters are formed by the reaction between a dicarboxylic acid and a diol, which results in the elimination of small water molecules. Addition polymers are the ones formed from monomers with carbon-carbon double bonds, and they produce no side products.
評分準則
1 mark for selecting the correct option (C).
題目 40 · MCQ
1 分
An unknown white solid salt X is tested: - Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. - Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of salt X?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(III) sulfate
D.iron(II) chloride
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解題
The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (\(Fe^{2+}\)). (Note that chromium(III) also forms a green precipitate but it dissolves in excess NaOH to give a green solution). The formation of a white precipitate with acidified barium nitrate confirms the presence of sulfate ions (\(SO_4^{2-}\)). Therefore, the salt is iron(II) sulfate.
評分準則
1 mark for selecting the correct option (B).
題目 41 · 選擇題
1 分
The atomic number of a phosphorus atom is 15 and its mass number is 31. What is the number of protons, neutrons and electrons in a phosphide ion, \(P^{3-}\)?
A.protons: 15, neutrons: 16, electrons: 18
B.protons: 15, neutrons: 16, electrons: 12
C.protons: 15, neutrons: 31, electrons: 18
D.protons: 18, neutrons: 16, electrons: 15
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解題
The atomic number (15) gives the number of protons (15). In a neutral phosphorus atom, there are also 15 electrons, but the phosphide ion \(P^{3-}\) has gained 3 electrons to form a stable octet, resulting in 18 electrons (15 + 3). The mass number (31) is the sum of protons and neutrons, so the number of neutrons is 31 - 15 = 16. Therefore, the phosphide ion contains 15 protons, 16 neutrons, and 18 electrons.
評分準則
[1 mark] awarded for selecting option A. - Reject option B (incorrect electron count representing a positive ion). - Reject option C (neutron count incorrectly matched to mass number). - Reject option D (protons and electrons reversed).
題目 42 · 選擇題
1 分
A student aims to prepare pure, dry crystals of magnesium sulfate, \(MgSO_4\). Which combination of reactants and method of preparation should be selected?
A.reactants: magnesium nitrate and dilute sulfuric acid; method: precipitation followed by filtration
B.reactants: magnesium metal and dilute nitric acid; method: titration followed by evaporation
C.reactants: magnesium oxide and dilute sulfuric acid; method: adding excess solid, filtering, heating filtrate to saturation and cooling
D.reactants: magnesium carbonate and dilute hydrochloric acid; method: titration followed by crystallization
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解題
Magnesium sulfate is a soluble salt. To prepare a soluble salt from an insoluble reactant (like magnesium oxide), the general method is adding the solid reactant to the dilute acid (sulfuric acid) in excess to ensure all acid has reacted. The excess solid is filtered off to obtain a pure solution of the salt. The filtrate is then heated to its saturation point, and left to cool slowly so that crystals form. Therefore, option C is correct.
評分準則
[1 mark] awarded for selecting option C. - Reject option A: Magnesium nitrate and sulfuric acid do not produce pure magnesium sulfate directly in a standard preparation pathway, and titration is used for soluble reactants. - Reject option B: Dilute nitric acid would produce magnesium nitrate, not magnesium sulfate. - Reject option D: Hydrochloric acid would produce magnesium chloride, and titration is reserved for soluble bases/acids.
題目 43 · 選擇題
1 分
The equation for the reaction between iron(II) ions and chlorine gas is shown.
A.Iron(II) ions are reduced because they gain electrons.
B.Iron(II) ions are oxidised because they lose electrons.
C.Chlorine gas is oxidised because it gains electrons.
D.Chlorine gas is reduced because it loses electrons.
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解題
In this reaction, the iron(II) ions (\(Fe^{2+}\)) lose electrons to form iron(III) ions (\(Fe^{3+}\)), which means they are oxidised (Oxidation is Loss of electrons). The chlorine molecules (\(Cl_2\)) gain electrons to form chloride ions (\(Cl^-\)), which means chlorine is reduced (Reduction is Gain of electrons). Thus, option B is correct.
評分準則
[1 mark] awarded for selecting option B. - Reject option A: Iron(II) ions lose electrons and are oxidised, not reduced. - Reject option C: Chlorine gains electrons, so it is reduced, not oxidised. - Reject option D: Chlorine is reduced because it gains, not loses, electrons.
題目 44 · 選擇題
1 分
The reaction between calcium carbonate and dilute hydrochloric acid is investigated. Which change increases the rate of reaction solely by increasing the collision frequency of the particles without changing the proportion of particles with energy greater than or equal to the activation energy?
A.increasing the concentration of the acid
B.increasing the temperature of the acid
C.adding a catalyst to the mixture
D.using larger pieces of calcium carbonate
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解題
Increasing the concentration of the acid increases the number of reacting particles per unit volume, which directly increases the frequency of collisions between the reactant particles. However, it does not alter the kinetic energy of the particles, so the proportion of collisions that have energy greater than or equal to the activation energy remains unchanged. Therefore, option A is correct. Increasing temperature (B) and adding a catalyst (C) both increase the proportion of particles/collisions with sufficient energy.
評分準則
[1 mark] awarded for selecting option A. - Reject option B: Increasing temperature increases the collision frequency AND the proportion of particles with energy \(\ge E_a\). - Reject option C: Adding a catalyst decreases the activation energy, changing the proportion of particles with enough energy. - Reject option D: Using larger pieces decreases the surface area, which decreases the collision frequency and rate.
題目 45 · 選擇題
1 分
The structure of a synthetic polymer is shown.
\([-CO-C_6H_4-CO-NH-C_6H_{12}-NH-]_n\)
Which row correctly identifies the type of polymer and the functional group linkage present?
A.type of polymer: polyamide; linkage: amide
B.type of polymer: polyamide; linkage: ester
C.type of polymer: polyester; linkage: amide
D.type of polymer: polyester; linkage: ester
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解題
The polymer contains the amide linkage \(-CONH-\). A polymer built from monomers linked together by amide linkages is classified as a polyamide (e.g., Nylon). Therefore, option A is correct.
評分準則
[1 mark] awarded for selecting option A. - Reject option B: The amide linkage is not an ester linkage. - Reject options C and D: Polyesters contain the ester linkage \(-COO-\), not the amide linkage.
題目 46 · 選擇題
1 分
Equal masses of four different metals, J, K, L and M, are added separately to equal volumes of aqueous copper(II) sulfate of the same concentration. The temperature rise during each reaction is measured.
- Metal J: \(0.0\ ^\circ\text{C}\) - Metal K: \(14.5\ ^\circ\text{C}\) - Metal L: \(8.2\ ^\circ\text{C}\) - Metal M: \(21.3\ ^\circ\text{C}\)
What is the correct order of reactivity of the metals, starting with the most reactive?
A.M \(\rightarrow\) K \(\rightarrow\) L \(\rightarrow\) J
B.J \(\rightarrow\) L \(\rightarrow\) K \(\rightarrow\) M
C.M \(\rightarrow\) L \(\rightarrow\) K \(\rightarrow\) J
D.J \(\rightarrow\) K \(\rightarrow\) L \(\rightarrow\) M
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解題
A displacement reaction is exothermic, and the larger the difference in reactivity between the displacing metal and copper, the larger the temperature rise. Thus, the metal with the highest temperature rise (M) is the most reactive, followed by K, then L. Metal J did not react (no temperature rise), meaning it is the least reactive of the four (and less reactive than copper). The order from most reactive to least reactive is M \(\rightarrow\) K \(\rightarrow\) L \(\rightarrow\) J.
評分準則
[1 mark] awarded for selecting option A. - Reject option B: This represents the reverse order (least to most reactive). - Reject options C and D: These contain incorrect relative orderings of K and L.
題目 47 · 選擇題
1 分
A student titrates aqueous sodium hydroxide in a conical flask with dilute hydrochloric acid from a burette, using methyl orange as the indicator. What is the colour change observed in the conical flask at the end-point of this titration?
A.yellow to orange
B.orange to yellow
C.red to yellow
D.yellow to colourless
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解題
Methyl orange is yellow in alkaline solutions (such as aqueous sodium hydroxide in the conical flask). As the acid is added from the burette, the solution reaches neutralisation at the end-point where the indicator changes colour from yellow to orange. (If excess acid is added, it turns red). Therefore, the correct end-point colour change is yellow to orange.
評分準則
[1 mark] awarded for selecting option A. - Reject option B: This represents a titration of acid with alkali (from yellow/orange back to alkaline). - Reject option C: Red to yellow is incorrect for adding acid to alkali. - Reject option D: Methyl orange does not become colourless.
題目 48 · 選擇題
1 分
Propene gas is bubbled through a solution of bromine in an organic solvent. Which statement about this reaction is correct?
A.The reaction is a substitution reaction and the solution remains orange-brown.
B.The reaction is an addition reaction and the orange-brown color is decolourised.
C.The reaction is a substitution reaction and 1,2-dibromopropane is formed.
D.The reaction is an addition reaction and the solution turns green.
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解題
Propene is an alkene containing a carbon-carbon double bond, which undergoes addition reactions with halogens. Bromine reacts via an addition reaction to form 1,2-dibromopropane. During this reaction, the characteristic orange-brown colour of bromine is decolourised (becomes colourless). Thus, option B is correct.
評分準則
[1 mark] awarded for selecting option B. - Reject option A: Alkenes undergo addition, not substitution, and the solution is decolourised. - Reject option C: The reaction is an addition reaction, not substitution. - Reject option D: The solution does not turn green; it becomes colourless.
題目 49 · MCQ
1 分
Which gas will diffuse the fastest at room temperature and pressure?
A.Carbon dioxide, \(CO_2\)
B.Carbon monoxide, \(CO\)
C.Nitrogen dioxide, \(NO_2\)
D.Sulfur dioxide, \(SO_2\)
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解題
The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (\(M_r\)). This means the gas with the smallest \(M_r\) will diffuse the fastest.
Since carbon monoxide has the lowest relative molecular mass (\(28\)), it will diffuse the fastest.
評分準則
1 mark for the correct option B.
題目 50 · MCQ
1 分
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Which row correctly identifies the product formed at each electrode?
A.Positive electrode (anode): chlorine gas; Negative electrode (cathode): hydrogen gas
B.Positive electrode (anode): oxygen gas; Negative electrode (cathode): hydrogen gas
C.Positive electrode (anode): chlorine gas; Negative electrode (cathode): sodium metal
D.Positive electrode (anode): oxygen gas; Negative electrode (cathode): sodium metal
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解題
During the electrolysis of concentrated aqueous sodium chloride: - At the positive electrode (anode), chloride ions (\(Cl^-\)) are discharged in preference to hydroxide ions (\(OH^-\)) because they are in high concentration, producing chlorine gas (\(Cl_2\)). - At the negative electrode (cathode), hydrogen ions (\(H^+\)) are discharged in preference to sodium ions (\(Na^+\)) because hydrogen is less reactive than sodium, producing hydrogen gas (\(H_2\)).
評分準則
1 mark for the correct option A.
題目 51 · MCQ
1 分
Three metals, \(X\), \(Y\) and \(Z\), were added separately to aqueous solutions of their nitrates. The observations are recorded in the table below:
\(\begin{array}{|c|c|c|c|} \hline \text{Metal} & \text{Solution of } X(NO_3)_2 & \text{Solution of } Y(NO_3)_2 & \text{Solution of } Z(NO_3)_2 \\ \hline X & - & \text{no reaction} & \text{displacement reaction} \\ \hline Y & \text{displacement reaction} & - & \text{displacement reaction} \\ \hline Z & \text{no reaction} & \text{no reaction} & - \\ \hline \end{array}\)
What is the correct order of reactivity of the metals, from least reactive to most reactive?
A.\(Z \rightarrow X \rightarrow Y\)
B.\(Z \rightarrow Y \rightarrow X\)
C.\(Y \rightarrow X \rightarrow Z\)
D.\(X \rightarrow Y \rightarrow Z\)
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解題
A more reactive metal displaces a less reactive metal from its salt solution. - Metal \(Y\) displaces both \(X\) and \(Z\), so \(Y\) is the most reactive. - Metal \(X\) displaces \(Z\) but cannot displace \(Y\), so \(X\) is more reactive than \(Z\) but less reactive than \(Y\). - Metal \(Z\) cannot displace either \(X\) or \(Y\), so \(Z\) is the least reactive.
Therefore, the order of reactivity from least reactive to most reactive is \(Z \rightarrow X \rightarrow Y\).
評分準則
1 mark for the correct option A.
題目 52 · MCQ
1 分
Which statement about ethanoic acid is correct?
A.It turns damp blue litmus paper red.
B.It reacts with copper metal to produce hydrogen gas.
C.It acts as a strong acid that fully ionises in aqueous solution.
D.It reacts with sodium hydroxide to produce sodium ethanoate and hydrogen gas.
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解題
Ethanoic acid is a weak carboxylic acid. - Like all acids, it turns damp blue litmus paper red. Thus, option A is correct. - Copper is below hydrogen in the reactivity series, so it does not react with dilute acids to produce hydrogen gas, making option B incorrect. - It only partially ionises in water, so it is a weak acid, making option C incorrect. - It reacts with sodium hydroxide (an alkali) in a neutralisation reaction to produce sodium ethanoate and water (not hydrogen), making option D incorrect.
評分準則
1 mark for the correct option A.
題目 53 · MCQ
1 分
What is the total number of atoms in \(0.5\text{ moles}\) of water molecules, \(H_2O\)?
(Assume the Avogadro constant is \(6.0 \times 10^{23}\text{ mol}^{-1}\))
A.\(3.0 \times 10^{23}\)
B.\(9.0 \times 10^{23}\)
C.\(1.8 \times 10^{24}\)
D.\(6.0 \times 10^{23}\)
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解題
1. Calculate the number of water molecules in \(0.5\text{ moles}\): \(\text{Number of molecules} = 0.5 \times 6.0 \times 10^{23} = 3.0 \times 10^{23}\text{ molecules}\)
2. Each molecule of water (\(H_2O\)) contains 3 atoms (2 hydrogen atoms and 1 oxygen atom).
3. Calculate the total number of atoms: \(\text{Total atoms} = 3 \times 3.0 \times 10^{23} = 9.0 \times 10^{23}\text{ atoms}\)
評分準則
1 mark for the correct option B.
題目 54 · MCQ
1 分
Which row correctly pairs the polymer with its type of polymerisation and the link present in its structure?
A.Polymer: nylon; Type of polymerisation: condensation; Link: amide
B.Polymer: Terylene; Type of polymerisation: addition; Link: ester
C.Polymer: poly(ethene); Type of polymerisation: condensation; Link: carbon–carbon single bond
D.Polymer: protein; Type of polymerisation: addition; Link: amide
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解題
- Nylon is a polyamide formed by condensation polymerisation and contains amide links. This makes option A correct. - Terylene is a polyester formed by condensation polymerisation, not addition, making option B incorrect. - Poly(ethene) is formed by addition polymerisation, not condensation, making option C incorrect. - Proteins are polyamides formed by condensation polymerisation, not addition, making option D incorrect.
評分準則
1 mark for the correct option A.
題目 55 · MCQ
1 分
An aqueous solution of salt \(S\) 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 silver nitrate produces a cream precipitate.
What is the identity of salt \(S\)?
A.chromium(III) bromide
B.iron(II) bromide
C.iron(II) chloride
D.iron(III) bromide
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解題
- The reaction with aqueous sodium hydroxide producing a green precipitate insoluble in excess indicates the presence of iron(II) ions (\(Fe^{2+}\)). Note that chromium(III) also produces a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution. - The reaction with dilute nitric acid and aqueous silver nitrate producing a cream precipitate indicates the presence of bromide ions (\(Br^-\)).
Therefore, salt \(S\) is iron(II) bromide.
評分準則
1 mark for the correct option B.
題目 56 · MCQ
1 分
The following reversible reaction is at equilibrium in a closed container:
Which changes in temperature and pressure will both shift the position of equilibrium to the right to increase the yield of methanol, \(CH_3OH\)?
A.decrease temperature, decrease pressure
B.decrease temperature, increase pressure
C.increase temperature, decrease pressure
D.increase temperature, increase pressure
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解題
- Effect of temperature: The forward reaction is exothermic (\(\Delta H = -91\text{ kJ/mol}\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the direction of the exothermic reaction (to the right) to release heat. Therefore, temperature should be decreased. - Effect of pressure: There are 3 moles of gas on the left-hand side (\(1\text{ CO} + 2\text{ H}_2\)) and only 1 mole of gas on the right-hand side (\(1\text{ CH}_3\text{OH}\)). Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas to reduce the pressure. Therefore, pressure should be increased.
評分準則
1 mark for the correct option B.
題目 57 · MCQ
1 分
An ion of an isotope of sulfur is represented as \(^{34}_{16}\text{S}^{2-}\).
Which row shows the number of protons, neutrons and electrons in this ion?
A.protons: 16 | neutrons: 18 | electrons: 18
B.protons: 16 | neutrons: 18 | electrons: 14
C.protons: 18 | neutrons: 16 | electrons: 18
D.protons: 16 | neutrons: 34 | electrons: 18
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解題
The atomic number is 16, which represents the number of protons. The nucleon number is 34. The number of neutrons is the nucleon number minus the atomic number: \(34 - 16 = 18\). Since the ion has a charge of \(2-\), it has gained two extra electrons, so the number of electrons is \(16 + 2 = 18\).
評分準則
1 mark for the correct option A.
題目 58 · MCQ
1 分
Phosphorus is in Group V and Period 3 of the Periodic Table.
Which row shows the charge and the electronic configuration of a phosphide ion?
A phosphorus atom has the atomic number 15, with an electronic configuration of 2,8,5. Being in Group V, it gains 3 electrons to achieve a stable outer shell of 8. This forms a phosphide ion with an electronic configuration of 2,8,8 and a charge of \(-3\).
評分準則
1 mark for the correct option A.
題目 59 · MCQ
1 分
A student neutralises \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide with \(0.050\text{ mol/dm}^3\) dilute sulfuric acid.
What volume of the dilute sulfuric acid is needed to react completely with the sodium hydroxide?
A.\(12.5\text{ cm}^3\)
B.\(25.0\text{ cm}^3\)
C.\(50.0\text{ cm}^3\)
D.\(100.0\text{ cm}^3\)
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解題
1. Moles of \(\text{NaOH} = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.0025\text{ mol}\). 2. According to the balanced equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). \(\text{Moles of H}_2\text{SO}_4 = 0.0025 / 2 = 0.00125\text{ mol}\). 3. \(\text{Volume of H}_2\text{SO}_4 = \text{moles} / \text{concentration} = 0.00125\text{ mol} / 0.050\text{ mol/dm}^3 = 0.0250\text{ dm}^3 = 25.0\text{ cm}^3\).
評分準則
1 mark for the correct option B.
題目 60 · MCQ
1 分
An aqueous solution of salt X produces a green precipitate when aqueous sodium hydroxide is added. The precipitate is insoluble in excess sodium hydroxide and slowly turns brown at the surface on standing.
Which ion is present in salt X?
A.\(\text{Cr}^{3+}\)
B.\(\text{Fe}^{2+}\)
C.\(\text{Fe}^{3+}\)
D.\(\text{Cu}^{2+}\)
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解題
The formation of a green precipitate with sodium hydroxide that remains insoluble in excess and slowly oxidises to turn brown at the surface on standing is the characteristic test for iron(II) ions, \(\text{Fe}^{2+}\). Chromium(III) also produces a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution.
The repeating unit of the addition polymer is \(\text{–CH(CH}_3\text{)–CH}_2\text{–}\). Restoring the double bond between the two carbon atoms in the chain gives the monomer \(\text{CH}_3\text{CH=CH}_2\), which is propene.
評分準則
1 mark for the correct option A.
題目 62 · MCQ
1 分
Three metals, X, Y, and Z, are reacted with oxides of other metals or dilute hydrochloric acid.
- Metal X reduces the oxide of Y. - Metal Y reacts with dilute hydrochloric acid, but metal Z does not react with dilute hydrochloric acid.
Which list shows the metals in order of decreasing reactivity (most reactive first)?
A.X, Y, Z
B.Z, Y, X
C.Y, X, Z
D.X, Z, Y
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解題
Metal X reduces the oxide of Y, meaning X is more reactive than Y (X > Y). Metal Y reacts with dilute acid, whereas Z does not, which shows Y is more reactive than hydrogen and Z is less reactive than hydrogen (Y > Z). Thus, the order of decreasing reactivity is X, Y, Z.
評分準則
1 mark for the correct option A.
題目 63 · MCQ
1 分
Which row correctly identifies the products formed at each inert electrode when concentrated aqueous sodium chloride is electrolysed?
A.positive electrode: chlorine gas | negative electrode: hydrogen gas
B.positive electrode: oxygen gas | negative electrode: sodium metal
C.positive electrode: chlorine gas | negative electrode: sodium metal
D.positive electrode: oxygen gas | negative electrode: hydrogen gas
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解題
During the electrolysis of concentrated aqueous sodium chloride (brine): - Chloride ions (\(\text{Cl}^-\)) are discharged at the positive electrode (anode) to produce chlorine gas. - Hydrogen ions (\(\text{H}^+\)) are preferentially discharged at the negative electrode (cathode) because hydrogen is less reactive than sodium, producing hydrogen gas.
評分準則
1 mark for the correct option A.
題目 64 · MCQ
1 分
A reaction pathway diagram shows that the energy level of the products is lower than the energy level of the reactants.
Which statement about this reaction is correct?
A.The reaction is exothermic and the enthalpy change, \(\Delta H\), is negative.
B.The reaction is endothermic and the enthalpy change, \(\Delta H\), is positive.
C.The reaction is exothermic and the enthalpy change, \(\Delta H\), is positive.
D.The reaction is endothermic and the enthalpy change, \(\Delta H\), is negative.
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解題
If the energy level of the products is lower than that of the reactants, energy is released to the surroundings. Therefore, the reaction is exothermic. The enthalpy change, \(\Delta H\), is negative because \(H_{\text{products}} - H_{\text{reactants}} < 0\).
評分準則
1 mark for the correct option A.
題目 65 · 選擇題
1 分
Ammonia gas, \(\text{NH}_3\), and hydrogen bromide gas, \(\text{HBr}\), are released at opposite ends of a long horizontal glass tube at the same time.
A white solid of ammonium bromide forms where the two gases meet.
Which statement explains why the white solid forms closer to the hydrogen bromide end of the tube?
A.The relative molecular mass of \(\text{HBr}\) is greater than that of \(\text{NH}_3\), so \(\text{HBr}\) molecules diffuse more slowly.
B.The relative molecular mass of \(\text{NH}_3\) is greater than that of \(\text{HBr}\), so \(\text{NH}_3\) molecules diffuse more slowly.
C.The relative molecular mass of \(\text{HBr}\) is smaller than that of \(\text{NH}_3\), so \(\text{HBr}\) molecules diffuse more quickly.
D.Both gases have the same relative molecular mass and diffuse at the same rate, but gravity acts more on \(\text{HBr}\).
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解題
According to the kinetic theory of gases, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (or simply, lighter gas molecules diffuse faster than heavier gas molecules). - The relative molecular mass of ammonia, \(M_r(\text{NH}_3) = 14 + (3 \times 1) = 17\). - The relative molecular mass of hydrogen bromide, \(M_r(\text{HBr}) = 1 + 80 = 81\).
Since ammonia molecules are much lighter than hydrogen bromide molecules, ammonia diffuses faster and travels a greater distance in the same time. Hence, the white solid forms closer to the end where hydrogen bromide was released.
評分準則
1 mark for selecting the correct option (A).
題目 66 · 選擇題
1 分
An atom of element \(X\) has the electronic configuration 2,8,6. An atom of element \(Y\) has the electronic configuration 2,8,2.
Which row shows the formula of the compound formed when element \(X\) reacts with element \(Y\) and the type of chemical bonding present?
A.Formula: \(Y_3X\), Bonding: covalent
B.Formula: \(YX\), Bonding: ionic
C.Formula: \(Y_2X_6\), Bonding: covalent
D.Formula: \(YX_2\), Bonding: ionic
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解題
- Element \(X\) (electronic configuration 2,8,6) has 6 valency electrons and needs to gain 2 electrons to achieve a stable octet, forming an anion \(X^{2-}\). - Element \(Y\) (electronic configuration 2,8,2) is a metal with 2 valency electrons and will lose these 2 electrons to achieve stability, forming a cation \(Y^{2+}\). - The combination of \(Y^{2+}\) and \(X^{2-}\) results in a compound with a 1:1 ratio, having the formula \(YX\). - Because the bond is formed by the transfer of electrons from a metal to a non-metal, the bonding is ionic.
評分準則
1 mark for selecting the correct option (B).
題目 67 · 選擇題
1 分
Aqueous copper(II) chloride is electrolysed using inert carbon (graphite) electrodes.
Which row correctly identifies the products formed at each electrode?
A.Anode (+): oxygen, Cathode (-): copper
B.Anode (+): chlorine, Cathode (-): hydrogen
C.Anode (+): chlorine, Cathode (-): copper
D.Anode (+): oxygen, Cathode (-): hydrogen
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解題
Aqueous copper(II) chloride contains the following ions: \(\text{Cu}^{2+}\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). - At the cathode (negative electrode), \(\text{Cu}^{2+}\) is lower in the reactivity series than \(\text{H}^+\), so copper ions are preferentially discharged to produce copper metal. - At the anode (positive electrode), halide ions (\(\text{Cl}^-\)) are present in a concentrated solution and are discharged in preference to hydroxide ions (\(\text{OH}^-\)), producing chlorine gas.
Therefore, the products are chlorine at the anode and copper at the cathode.
評分準則
1 mark for selecting the correct option (C).
題目 68 · 選擇題
1 分
Four different metals, \(P\), \(Q\), \(R\), and \(S\), have the following chemical properties: - Metal \(P\) reacts with steam but does not react with cold water. - Metal \(Q\) is found native (uncombined) in the Earth's crust. - The oxide of metal \(R\) can be reduced when heated with carbon powder. - Metal \(S\) reacts violently with cold water.
What is the order of reactivity of the four metals, from least reactive to most reactive?
A.\(Q \rightarrow R \rightarrow P \rightarrow S\)
B.\(Q \rightarrow P \rightarrow R \rightarrow S\)
C.\(S \rightarrow P \rightarrow R \rightarrow Q\)
D.\(R \rightarrow Q \rightarrow P \rightarrow S\)
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解題
We can deduce the reactivity of each metal from the given properties: 1. Metal \(S\) reacts violently with cold water, which indicates it is highly reactive (e.g., sodium or potassium). 2. Metal \(P\) reacts with steam but not cold water, which indicates moderate reactivity (e.g., magnesium, zinc, or iron). 3. Metal \(R\) is less reactive than carbon because its oxide is reduced by carbon, but it is more reactive than native metals. 4. Metal \(Q\) is found native (uncombined), which indicates it is extremely unreactive (e.g., gold or platinum).
Arranging from least to most reactive gives: \(Q \rightarrow R \rightarrow P \rightarrow S\).
評分準則
1 mark for selecting the correct option (A).
題目 69 · 選擇題
1 分
Which of the following polymers is formed through a condensation polymerisation reaction?
A.poly(ethene)
B.nylon
C.poly(propene)
D.poly(chloroethene)
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解題
Condensation polymerisation involves the reaction of monomer units with the elimination of a small molecule, such as water or hydrogen chloride. - Nylon is a polyamide formed by condensation polymerisation from diamines and dicarboxylic acids. - Poly(ethene), poly(propene), and poly(chloroethene) are all addition polymers formed by linking unsaturated monomer molecules containing \(\text{C=C}\) double bonds without forming any by-products.
評分準則
1 mark for selecting the correct option (B).
題目 70 · 選擇題
1 分
A series of qualitative analysis tests is performed on an unknown aqueous salt solution: 1. Dilute nitric acid is added followed by aqueous silver nitrate, resulting in a cream-coloured precipitate. 2. Aqueous sodium hydroxide is added dropwise, producing a green precipitate that is insoluble in excess.
Which two ions are present in the unknown salt solution?
A.\(\text{Fe}^{2+}\) and \(\text{Cl}^-\)
B.\(\text{Fe}^{2+}\) and \(\text{Br}^-\)
C.\(\text{Cu}^{2+}\) and \(\text{Br}^-\)
D.\(\text{Fe}^{3+}\) and \(\text{I}^-\)
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解題
- Test 1: The formation of a cream precipitate with acidified silver nitrate confirms the presence of bromide ions, \(\text{Br}^-\). (Chloride gives a white precipitate, and iodide gives a yellow precipitate). - Test 2: The formation of a green precipitate with aqueous sodium hydroxide that remains insoluble in excess is characteristic of iron(II) ions, \(\text{Fe}^{2+}\).
Therefore, the solution contains \(\text{Fe}^{2+}\) and \(\text{Br}^-\).
評分準則
1 mark for selecting the correct option (B).
題目 71 · 選擇題
1 分
Which of the following chemical processes is endothermic?
A.the combustion of methane
B.the thermal decomposition of calcium carbonate
C.the reaction between sodium metal and water
D.the neutralisation of hydrochloric acid with sodium hydroxide
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解題
An endothermic process is one that absorbs thermal energy from the surroundings. - Thermal decomposition of calcium carbonate requires a continuous supply of heat energy to break the strong ionic bonds within \(\text{CaCO}_3\), making it endothermic. - The combustion of methane, the reaction of sodium with water, and the neutralisation of an acid with an alkali are all highly exothermic processes that release heat to the surroundings.
評分準則
1 mark for selecting the correct option (B).
題目 72 · 選擇題
1 分
In an acid-base titration, a student finds that \(25.0\text{ cm}^3\) of potassium hydroxide solution, \(\text{KOH}\), is exactly neutralised by \(20.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) sulfuric acid, \(\text{H}_2\text{SO}_4\).
The chemical equation for this reaction is: \[2\text{KOH}(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\]
What is the concentration of the potassium hydroxide solution?
A.\(0.040\text{ mol/dm}^3\)
B.\(0.080\text{ mol/dm}^3\)
C.\(0.160\text{ mol/dm}^3\)
D.\(0.320\text{ mol/dm}^3\)
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解題
1. Calculate the number of moles of sulfuric acid used: \[\text{moles of }\text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.0020\text{ mol}\]
2. Determine the moles of KOH needed from the stoichiometric ratio in the balanced equation (\(2\text{KOH} : 1\text{H}_2\text{SO}_4\)): \[\text{moles of }\text{KOH} = 2 \times 0.0020\text{ mol} = 0.0040\text{ mol}\]
3. Calculate the concentration of the KOH solution: \[\text{concentration of }\text{KOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.0040\text{ mol}}{0.0250\text{ dm}^3} = 0.160\text{ mol/dm}^3\]
評分準則
1 mark for selecting the correct option (C).
題目 73 · 選擇題
1 分
A student carries out a titration by adding aqueous sodium hydroxide from a burette to a conical flask containing dilute sulfuric acid and a few drops of methyl orange. What is the colour change of the indicator at the end-point?
A.red to orange
B.orange to red
C.yellow to red
D.yellow to orange
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解題
Methyl orange is red in acidic solutions. As aqueous sodium hydroxide is added and neutralisation is achieved, the indicator turns orange at the exact end-point (neutral point). In excess alkali, it turns yellow. Therefore, the colour change at the end-point is from red to orange.
評分準則
1 mark for the correct option A.
題目 74 · 選擇題
1 分
The nucleon number of a sulfur atom is 32 and its atomic number is 16. How many protons, neutrons and electrons are in a sulfide ion, \(\text{S}^{2-}\)?
A.protons = 16, neutrons = 16, electrons = 18
B.protons = 16, neutrons = 16, electrons = 14
C.protons = 18, neutrons = 16, electrons = 16
D.protons = 16, neutrons = 18, electrons = 18
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解題
The atomic number is 16, which means a sulfur atom has 16 protons. The nucleon number is 32, so the number of neutrons is \(32 - 16 = 16\). A sulfide ion, \(\text{S}^{2-}\), is formed by gaining 2 electrons, so it has \(16 + 2 = 18\) electrons.
評分準則
1 mark for the correct option A.
題目 75 · 選擇題
1 分
An aqueous solution of a salt is tested. Adding dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. Adding aqueous ammonia dropwise produces a light blue precipitate which dissolves in excess ammonia to give a dark blue solution. What is the identity of the salt?
A.copper(II) bromide
B.copper(II) chloride
C.iron(II) bromide
D.zinc bromide
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解題
A cream precipitate with acidified silver nitrate confirms the presence of bromide ions (\(\text{Br}^-\)). A light blue precipitate with aqueous ammonia that dissolves in excess to form a dark blue solution confirms the presence of copper(II) ions (\(\text{Cu}^{2+}\)). Therefore, the salt is copper(II) bromide.
評分準則
1 mark for the correct option A.
題目 76 · 選擇題
1 分
Which statement about condensation polymerisation is correct?
A.A small molecule, such as water, is eliminated during the reaction.
B.Only monomers with carbon-carbon double bonds can be used.
C.The polymer is the only product formed.
D.Poly(ethene) is made by condensation polymerisation.
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解題
In condensation polymerisation, monomer molecules with two functional groups react to form a polymer chain while eliminating a small molecule, such as water or hydrogen chloride. Addition polymerisation does not produce any side product, and it requires monomers with carbon-carbon double bonds.
評分準則
1 mark for the correct option A.
題目 77 · 選擇題
1 分
Metal \(X\) reacts with cold water to produce hydrogen gas. Metal \(Y\) does not react with cold water but reacts with steam. Metal \(Z\) does not react with steam but reacts with dilute hydrochloric acid. What is the correct order of reactivity of these metals, from least reactive to most reactive?
A.Y → Z → X
B.Z → Y → X
C.X → Y → Z
D.Z → X → Y
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解題
Metal \(X\) is highly reactive as it reacts with cold water (e.g., sodium or calcium). Metal \(Y\) is moderately reactive as it requires steam to react (e.g., magnesium or zinc). Metal \(Z\) is less reactive because it does not react with water or steam but still reacts with acid (e.g., lead or tin). Thus, the order of reactivity from least to most reactive is \(Z \rightarrow Y \rightarrow X\).
評分準則
1 mark for the correct option B.
題目 78 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Which row correctly identifies the products formed at each electrode?
At the positive electrode (anode), chloride ions are oxidised to form chlorine gas. At the negative electrode (cathode), hydrogen ions from water are reduced in preference to sodium ions, forming hydrogen gas. Therefore, the products are chlorine at the positive electrode and hydrogen at the negative electrode.
評分準則
1 mark for the correct option A.
題目 79 · 選擇題
1 分
In an endothermic reaction, which statement about the energy changes is correct?
A.The energy absorbed in bond breaking is greater than the energy released in bond making.
B.The energy level of the reactants is higher than the energy level of the products.
C.The temperature of the surroundings increases during the reaction.
D.The enthalpy change, \(\Delta H\), is negative.
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解題
An endothermic reaction absorbs heat energy from the surroundings because the energy required to break the bonds in the reactants is greater than the energy released when new bonds in the products are made. This causes a decrease in the temperature of the surroundings, the energy level of the products is higher than that of the reactants, and the enthalpy change (\(\Delta H\)) is positive.
評分準則
1 mark for the correct option A.
題目 80 · 選擇題
1 分
Ethanol can be manufactured by the catalytic hydration of ethene. Which row shows the correct conditions for this industrial process?
A.temperature: 300 °C, pressure: 60 atm, catalyst: phosphoric acid
B.temperature: 35 °C, pressure: 1 atm, catalyst: yeast
C.temperature: 300 °C, pressure: 1 atm, catalyst: nickel
D.temperature: 35 °C, pressure: 60 atm, catalyst: phosphoric acid
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解題
The industrial hydration of ethene to make ethanol requires a temperature of \(300\ ^\circ\text{C}\), a pressure of \(60\text{ atm}\) (or \(6000\text{ kPa}\)), and an acid catalyst (phosphoric acid). Yeast is used in the fermentation of sugar at lower temperatures (around \(35\ ^\circ\text{C}\)) and atmospheric pressure.
Answer all structured theoretical and quantitative questions in the spaces provided.
13 題目 · 158.39999999999998 分
題目 1 · structured
12.3 分
Lithium has two stable isotopes: lithium-6 and lithium-7.
(a) Define the term isotopes. [2]
(b) Describe the atomic structure of a neutral lithium-7 atom in terms of the number of protons, neutrons, and electrons. [3]
(c) State why both isotopes of lithium have identical chemical properties. [1.3]
(d) A sample of naturally occurring lithium consists of 7.5% lithium-6 and 92.5% lithium-7. Calculate the relative atomic mass of this sample of lithium. Give your answer to two decimal places. [3]
(e) Explain, in terms of its electronic configuration, why lithium is placed in Group I of the Periodic Table. [3]
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解題
(a) Isotopes are defined as atoms of the same element containing the same number of protons but different numbers of neutrons.
(b) From the Periodic Table, lithium has an atomic number of 3, so it has 3 protons. In a neutral atom, the number of electrons equals the number of protons (3 electrons). The mass number is 7, so the number of neutrons is \(7 - 3 = 4\) neutrons.
(c) Chemical properties are determined by the number and arrangement of electrons in the outer shell. Since isotopes of lithium have the same atomic number, they have the same electronic configuration, hence identical chemical properties.
(e) Lithium has an atomic number of 3, meaning its electronic configuration is 2, 1. Because it has exactly one electron in its outer/valence shell, it belongs to Group I.
評分準則
(a) - Same number of protons / atomic number [1] - Different number of neutrons / nucleon number [1]
(d) - Correct formula setup: \(((6 \times 7.5) + (7 \times 92.5)) / 100\) [1] - Intermediate evaluation: \(692.5 / 100\) [1] - Final answer: 6.93 (must be to 2 decimal places) [1]
(e) - Electronic configuration of Li is 2,1 [1] - Group number is determined by outer shell electrons [1] - Since there is 1 outer shell electron, it is in Group I [1]
題目 2 · structured
12.3 分
A student carries out a titration to determine the concentration of a sample of aqueous sodium hydroxide, \(\text{NaOH}\). They titrate \(25.0\text{ cm}^3\) of the \(\text{NaOH}\)(aq) with dilute nitric acid, \(\text{HNO}_3\)(aq), of concentration \(0.100\text{ mol/dm}^3\), using methyl orange as the indicator.
(a) Name the piece of apparatus used to measure exactly \(25.0\text{ cm}^3\) of aqueous sodium hydroxide. [1]
(b) State the colour change of the methyl orange indicator at the end-point. [2]
(c) The student performs the titration three times and obtains the following titres of nitric acid: \(21.4\text{ cm}^3\), \(21.2\text{ cm}^3\), and \(21.3\text{ cm}^3\). Calculate the average titre. [1.3]
(d) Write the balanced chemical equation for the reaction, including state symbols. [3]
(e) Calculate the concentration of the aqueous sodium hydroxide in \(\text{mol/dm}^3\). Show your working. [5]
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解題
(a) A volumetric pipette (or bulb pipette) is used to measure accurate, fixed volumes of liquids like \(25.0\text{ cm}^3\).
(b) Methyl orange is yellow in alkaline solutions (aqueous sodium hydroxide) and turns red in acidic solutions. At the neutral end-point, the transition color is orange (or peach).
(c) Average titre = \(\frac{21.4 + 21.2 + 21.3}{3} = 21.3\text{ cm}^3\).
(d) The chemical equation is: \(\text{HNO}_3\text{(aq)} + \text{NaOH(aq)} \rightarrow \text{NaNO}_3\text{(aq)} + \text{H}_2\text{O(l)}\)
(e) Steps to calculate concentration: 1. Moles of \(\text{HNO}_3\) used = \(\text{concentration} \times \text{volume (dm}^3\text{)} = 0.100 \times \frac{21.3}{1000} = 0.00213\text{ mol}\). 2. According to the balanced equation, \(1\text{ mol}\) of \(\text{HNO}_3\) reacts with \(1\text{ mol}\) of \(\text{NaOH}\). Therefore, moles of \(\text{NaOH}\) in \(25.0\text{ cm}^3\) = \(0.00213\text{ mol}\). 3. Concentration of \(\text{NaOH}\) = \(\frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00213}{25.0 / 1000} = 0.0852\text{ mol/dm}^3\).
(c) - Correct average titre: 21.3 \(\text{cm}^3\) [1.3]
(d) - Correct formulae and balancing: \(\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}\) [2] - State symbols correct: (aq) for reactants and sodium nitrate, (l) for water [1]
(e) - Moles of \(\text{HNO}_3\): \(0.100 \times (21.3 / 1000) = 0.00213\text{ mol}\) [1.5] - Reacting mole ratio is 1:1, so moles of \(\text{NaOH}\) = \(0.00213\text{ mol}\) [1.5] - Concentration of \(\text{NaOH}\): \(0.00213 / 0.025 = 0.0852\text{ mol/dm}^3\) [2]
題目 3 · structured
12.3 分
Dilute sulfuric acid, \(\text{H}_2\text{SO}_4\), reacts with aqueous potassium hydroxide, \(\text{KOH}\), in a neutralization reaction.
(a) Define the term neutralization. [2]
(b) Write the chemical equation for the reaction between dilute sulfuric acid and aqueous potassium hydroxide. [2]
(c) Calculate the number of moles of \(\text{KOH}\) present in \(20.0\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) \(\text{KOH}\). [2.3]
(d) Determine the number of moles of \(\text{H}_2\text{SO}_4\) required to fully react with this amount of \(\text{KOH}\). [2]
(e) Calculate the concentration of the sulfuric acid in \(\text{mol/dm}^3\) if \(15.0\text{ cm}^3\) of the acid was required to reach the end-point. [4]
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解題
(a) Neutralization is a chemical reaction in which an acid reacts with a base or alkali to produce a salt and water only.
(b) Dilute sulfuric acid (diprotic acid) reacts with potassium hydroxide to form potassium sulfate and water: \(\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}\)
(c) Moles of \(\text{KOH}\) = \(\text{concentration} \times \text{volume in dm}^3 = 0.150 \times \frac{20.0}{1000} = 0.00300\text{ mol}\).
(d) Based on the balanced chemical equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ moles}\) of \(\text{KOH}\). Thus, moles of \(\text{H}_2\text{SO}_4 = \frac{0.00300}{2} = 0.00150\text{ mol}\).
(e) Concentration of \(\text{H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00150}{15.0 / 1000} = 0.100\text{ mol/dm}^3\).
評分準則
(a) - Reaction of acid with a base/alkali [1] - Producing a salt and water [1]
Alkenes such as propene, \(\text{C}_3\text{H}_6\), undergo addition polymerization to form poly(propene).
(a) Explain what is meant by the term addition polymerization. [2]
(b) Draw the displayed formula of propene, showing all atoms and all bonds. [2]
(c) Describe the structure of poly(propene) by drawing a diagram of its repeat unit. [3]
(d) Terylene is a condensation polymer. State two differences between addition polymerization and condensation polymerization. [4]
(e) State one environmental issue associated with the disposal of non-biodegradable addition polymers in landfills. [1.3]
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解題
(a) Addition polymerization involves monomers with carbon-carbon double bonds joining together to form a long-chain polymer, where no other products are formed.
(b) Propene displayed formula contains three carbon atoms, one carbon-carbon double bond, and single bonds to hydrogens: ``` H H H | | | H - C = C - C - H | H ```
(c) The repeat unit of poly(propene) opens the C=C double bond, leaving single bonds on either side extending out of brackets: ``` H H | | - C - C - | | H CH3 ```
(d) Differences: 1. Addition polymerization produces only the polymer product, whereas condensation polymerization produces the polymer plus a small molecule byproduct (such as water or hydrogen chloride). 2. Addition polymerization uses monomers with C=C double bonds (unsaturated), while condensation polymerization uses monomers with two different functional groups (e.g., dicarboxylic acids and diols).
(e) Because they are non-biodegradable, they remain intact for hundreds of years, filling up landfills and presenting long-term waste disposal challenges.
評分準則
(a) - Monomers joining to form a long chain molecule/polymer [1] - No other product is formed / single product [1]
(b) - Double bond C=C present with correct valencies [1] - Complete structure showing all atoms and bonds including -CH3 group [1]
(c) - Single C-C bond in the backbone with continuation bonds extending outwards [1] - Correct groups attached (H, H, H, and -CH3) [1] - Brackets and 'n' correctly positioned to represent a repeat unit [1]
(d) - Addition: only one product / Condensation: polymer + small molecule (water) [2] - Addition: monomers have C=C double bonds / Condensation: monomers have functional groups (e.g., -OH, -COOH) [2]
(e) - Polymers are non-biodegradable / do not decompose [1] - Takes up landfill space / sight pollution [0.3]
題目 5 · structured
12.3 分
The relative reactivity of four metals (A, B, C, and D) was investigated by placing each metal separately into aqueous solutions of the other metal nitrates. The results of the displacement reactions are summarized below: - Metal A: displaces B and D from their nitrate solutions, but does not react with C. - Metal B: does not displace any of the other three metals. - Metal C: displaces A, B, and D from their nitrate solutions. - Metal D: displaces B, but does not react with A or C.
(a) Deduce the order of reactivity of the four metals, from least reactive to most reactive. [3]
(b) Identify which of the metals could be copper if the other three metals are zinc, magnesium, and iron. Explain your choice. [3.3]
(c) Write a word equation and an ionic equation for the displacement reaction between zinc and aqueous copper(II) sulfate. [4]
(d) State one physical observation that would be made during the reaction between zinc and aqueous copper(II) sulfate. [2]
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解題
(a) We analyze the reactions: - C is the most reactive because it displaces all other metals (A, B, D). - A is next because it displaces B and D, but not C. - D is less reactive than A (since D cannot displace A), but more reactive than B (since D displaces B). - B is the least reactive because it cannot displace any other metal. Therefore, the order of reactivity (least to most reactive) is: B < D < A < C.
(b) In the reactivity series of the given metals (zinc, magnesium, iron, and copper), the order of reactivity from least to most is: copper < iron < zinc < magnesium. Since Metal B is the least reactive, it corresponds to copper.
(d) Observations: - The blue color of the aqueous copper(II) sulfate solution fades or turns colorless. - A red-brown solid (copper metal) deposits on the zinc.
評分準則
(a) - Identifies B as least reactive and C as most reactive [1] - Places D and A in correct relative order (D < A) [1] - Correct final sequence: B < D < A < C [1]
(b) - Identifies Metal B [1] - States that copper is the least reactive among the listed metals [1] - Connects this to the fact that B does not displace any other metal [1.3]
(c) - Word equation correct [1.5] - Ionic equation correct with state symbols: \(\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}\)[2.5]
(d) - Any one of: - Blue solution fades / becomes colorless [1] - Red-brown/brown solid deposits on the zinc [1] - Zinc dissolves/gets smaller [1]
題目 6 · structured
12.3 分
A green crystalline solid, Compound X, was analyzed by a series of qualitative tests.
(a) When dilute hydrochloric acid was added to solid X, a gas was evolved that turned limewater cloudy. (i) Identify the gas. [1] (ii) Identify the anion present in Compound X. [1.3]
(b) Solid X was dissolved in distilled water to form Solution X. (i) Adding aqueous sodium hydroxide to Solution X produced a green precipitate that was insoluble in excess sodium hydroxide. Identify the cation present in X. [2] (ii) Describe the chemical test and positive result used to confirm the presence of chloride ions in a solution. [4]
(c) Describe how to perform a flame test and state the flame color produced by calcium ions. [4]
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解題
(a) (i) The gas that turns limewater cloudy is carbon dioxide, \(\text{CO}_2\). (ii) Carbon dioxide gas is produced when carbonates react with acid. Therefore, the anion is carbonate, \(\text{CO}_3^{2-}\).
(b) (i) Iron(II) ions, \(\text{Fe}^{2+}\), react with sodium hydroxide to form a dirty-green precipitate of iron(II) hydroxide, which does not dissolve in excess sodium hydroxide. (ii) To test for chloride ions, acidify the test solution with dilute nitric acid (to remove carbonate impurities) and then add aqueous silver nitrate. A white precipitate of silver chloride confirms chloride ions.
(c) A flame test is performed by dipping a clean platinum or nichrome wire into concentrated hydrochloric acid, then dipping it into the solid sample, and placing the wire into a non-luminous (blue) Bunsen burner flame. Calcium ions produce an orange-red (or brick-red) flame.
(b) - (i) Iron(II) / \(\text{Fe}^{2+}\) [2] (Reject: iron / iron(III)) - (ii) Add dilute nitric acid [1], add aqueous silver nitrate [1], white precipitate forms [2]
(c) - Use a clean platinum/nichrome wire [1] - Dip in concentrated hydrochloric acid and solid [1] - Place in non-luminous / blue Bunsen flame [1] - Flame color: orange-red / brick-red [1]
題目 7 · structured
12.3 分
Soluble and insoluble salts are prepared using different experimental methods.
(a) Describe how a pure, dry sample of the insoluble salt lead(II) sulfate can be prepared from aqueous lead(II) nitrate and aqueous sodium sulfate. [5]
(b) Write the ionic equation, including state symbols, for the precipitation of lead(II) sulfate. [3]
(c) Explain why an acid-base titration is used to prepare sodium chloride from sodium hydroxide and hydrochloric acid, rather than simply adding excess sodium hydroxide to the acid. [3]
(d) State the general name given to the type of reaction that occurs when an acid reacts with an alkali. [1.3]
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解題
(a) Preparation steps: 1. Mix aqueous solutions of lead(II) nitrate and sodium sulfate in a beaker to precipitate lead(II) sulfate. 2. Filter the mixture using a funnel and filter paper to separate the solid precipitate (residue) from the filtrate. 3. Wash the precipitate (lead(II) sulfate) on the filter paper with distilled water to remove soluble sodium nitrate. 4. Dry the precipitate in a warm oven or pat dry gently with filter paper.
(b) The ionic equation showing the formation of the solid precipitate is: \(\text{Pb}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)}\)
(c) In this preparation, both reactants (sodium hydroxide and hydrochloric acid) are soluble, and the product (sodium chloride) is also highly soluble. Unlike reacting an acid with an insoluble base, any excess sodium hydroxide cannot be filtered off. Therefore, titration must be used to mix exactly equal chemical amounts of acid and alkali before crystallizing the pure salt.
(d) The reaction between an acid and an alkali is called neutralization.
評分準則
(a) - Mix / combine the two aqueous solutions [1] - Filter the mixture to obtain the precipitate / lead(II) sulfate [1] - Wash the residue / precipitate with distilled water [1] - Dry the solid on filter paper / in a warm oven [1] - Mention of 'filtrate' / 'residue' correctly used [1]
(b) - Reactants correctly identified as \(\text{Pb}^{2+}\) and \(\text{SO}_4^{2-}\) [1] - Product identified as \(\text{PbSO}_4\) [1] - State symbols correct: (aq) for reactants, (s) for product [1]
(c) - Both reactants and the product are soluble [1] - Excess reactant cannot be separated by filtration [1] - Titration ensures exact stoichiometric / neutral amounts of reactants are used [1]
(d) - Neutralization [1.3]
題目 8 · structured
12.3 分
Sodium (atomic number 11) and chlorine (atomic number 17) react together to form the ionic compound sodium chloride.
(a) State the electronic configurations of a sodium atom and a chlorine atom. [2]
(b) Explain, in terms of electron transfer, how sodium chloride is formed from sodium and chlorine atoms. [3.3]
(c) Describe the structure of solid sodium chloride and explain why it has a high melting point. [4]
(d) Explain why solid sodium chloride does not conduct electricity, but molten sodium chloride does. [3]
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解題
(a) Sodium has 11 electrons: 2 in the first shell, 8 in the second, and 1 in the outer shell (2,8,1). Chlorine has 17 electrons: 2 in the first shell, 8 in the second, and 7 in the outer shell (2,8,7).
(b) A sodium atom loses its single valence electron to form a positively charged sodium ion, \(\text{Na}^+\). A chlorine atom gains this electron to form a negatively charged chloride ion, \(\text{Cl}^-\). Both ions obtain stable noble gas configurations (full outer shells).
(c) Solid sodium chloride has a giant ionic lattice structure consisting of a regular, three-dimensional arrangement of alternating positive \(\text{Na}^+\) and negative \(\text{Cl}^-\) ions. Strong electrostatic attractions act in all directions between oppositely charged ions, which require a high temperature and significant thermal energy to overcome.
(d) In solid sodium chloride, the ions are held tightly in fixed positions in the lattice and cannot move to carry electric current. When molten, the lattice structure breaks down, and the ions become mobile, allowing them to migrate towards electrodes and conduct electricity.
評分準則
(a) - Sodium: 2,8,1 [1] - Chlorine: 2,8,7 [1]
(b) - Sodium loses 1 electron / transfer of 1 electron from sodium [1] - Chlorine gains 1 electron [1] - Formation of \(\text{Na}^+\) and \(\text{Cl}^-\)[1.3]
(c) - Giant ionic lattice [1] - Alternating positive and negative ions [1] - Strong electrostatic forces of attraction / ionic bonds [1] - Requires a large amount of energy to break [1]
(d) - In solid: ions are fixed / cannot move [1] - In molten: lattice breaks down / ions are free to move [1] - Mobile ions are able to carry electric charge [1]
題目 9 · structured
12 分
A student prepares a standard solution of sulfamic acid, \(\text{HSO}_3\text{NH}_2\), to determine the concentration of a sodium hydroxide solution.
(a) Describe how the student would prepare exactly \(250\text{ cm}^3\) of a standard solution of sulfamic acid from a known mass of solid sulfamic acid. [3]
(b) Name the pieces of apparatus used to: (i) measure exactly \(25.0\text{ cm}^3\) of the prepared sulfamic acid solution. [1] (ii) deliver the sodium hydroxide solution during the titration. [1]
(c) Methyl orange is used as the indicator. State the colour change observed at the end-point when sodium hydroxide is added from the burette to the sulfamic acid in the conical flask. [2]
(d) The average volume of sodium hydroxide required to neutralise \(25.0\text{ cm}^3\) of the \(0.100\text{ mol/dm}^3\) sulfamic acid was \(21.5\text{ cm}^3\). (i) Given that sulfamic acid is monoprotic, write the chemical equation for the neutralisation reaction with sodium hydroxide. [2] (ii) Calculate the concentration of the sodium hydroxide solution in \(\text{mol/dm}^3\). Give your answer to 3 significant figures. [3]
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解題
(a) 1. Dissolve the weighed mass of solid sulfamic acid in a beaker using a small volume of distilled water. 2. Transfer the solution quantitatively (including washings) into a \(250\text{ cm}^3\) volumetric flask. 3. Make up to the mark with distilled water until the bottom of the meniscus is on the line, and shake well to homogenise.
(b) (i) Volumetric pipette (or bulb pipette). (ii) Burette.
(c) The colour changes from red (acidic) to orange/yellow (at the end-point/alkaline transition).
(ii) Moles of sulfamic acid = \(\text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}\). Since the reaction is 1:1, moles of \(\text{NaOH} = 0.00250\text{ mol}\). Concentration of \(\text{NaOH} = \frac{\text{moles}}{\text{volume}} = \frac{0.00250\text{ mol}}{0.0215\text{ dm}^3} = 0.116\text{ mol/dm}^3\).
評分準則
(a) - Dissolve solid in beaker with distilled water [1] - Transfer with washings to a \(250\text{ cm}^3\) volumetric flask [1] - Fill to the mark with distilled water and invert/shake [1]
(b) (i) Volumetric pipette [1] (ii) Burette [1]
(c) - Red [1] - To orange/yellow [1]
(d) (i) - Correct reactants and products [1] - Balanced with state symbols [1] (ii) - Calculation of moles of acid \((0.00250\text{ mol})\) [1] - Equating moles of acid to moles of alkali \((1:1\text{ ratio})\) [1] - Final concentration calculation to 3 s.f. \((0.116\text{ mol/dm}^3)\) [1]
題目 10 · structured
12 分
An impure sample of limestone (mostly calcium carbonate, \(\text{CaCO}_3\)) of mass \(3.00\text{ g}\) was added to \(50.0\text{ cm}^3\) of \(1.00\text{ mol/dm}^3\) hydrochloric acid, \(\text{HCl}\), which was in excess.
(a) Write a balanced chemical equation, including state symbols, for the reaction between calcium carbonate and hydrochloric acid. [3]
(b) After the reaction was complete, the unreacted hydrochloric acid was titrated against sodium hydroxide. It required \(24.0\text{ cm}^3\) of \(1.00\text{ mol/dm}^3\) sodium hydroxide, \(\text{NaOH}\), for complete neutralisation.
Calculate: (i) the number of moles of hydrochloric acid originally added to the limestone. [1] (ii) the number of moles of sodium hydroxide used in the titration. [1] (iii) the number of moles of unreacted hydrochloric acid remaining in the flask. [1] (iv) the number of moles of hydrochloric acid that reacted with the calcium carbonate. [1] (v) the mass of calcium carbonate present in the \(3.00\text{ g}\) sample. [2] (vi) the percentage purity of the limestone sample. [3]
(b) (i) Moles of \(\text{HCl}\) original = \(0.0500\text{ dm}^3 \times 1.00\text{ mol/dm}^3 = 0.0500\text{ mol}\). (ii) Moles of \(\text{NaOH}\) used = \(0.0240\text{ dm}^3 \times 1.00\text{ mol/dm}^3 = 0.0240\text{ mol}\). (iii) Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\) has a 1:1 stoichiometry, unreacted \(\text{HCl} = 0.0240\text{ mol}\). (iv) Reacted \(\text{HCl} = 0.0500\text{ mol} - 0.0240\text{ mol} = 0.0260\text{ mol}\). (v) Stoichiometric ratio is \(1\text{ mol of CaCO}_3\) to \(2\text{ mol of HCl}\). Moles of \(\text{CaCO}_3 = \frac{0.0260}{2} = 0.0130\text{ mol}\). Mass of \(\text{CaCO}_3 = 0.0130\text{ mol} \times 100\text{ g/mol} = 1.30\text{ g}\). (vi) Percentage purity = \(\frac{1.30\text{ g}}{3.00\text{ g}} \times 100\% = 43.3\%\).
評分準則
(a) - Correct reactant and product formulas [1] - Correct balancing [1] - Correct state symbols [1]
(b) (i) \(0.0500\text{ mol}\) [1] (ii) \(0.0240\text{ mol}\) [1] (iii) \(0.0240\text{ mol}\) [1] (iv) \(0.0260\text{ mol}\) [1] (v) - Moles of \(\text{CaCO}_3 = 0.0130\text{ mol}\) [1] - Mass of \(\text{CaCO}_3 = 1.30\text{ g}\) (accept \(1.301\text{ g}\) if using \(M_r = 100.1\)) [1] (vi) - Dividing calculated mass by \(3.00\text{ g}\) [1] - Multiplying by 100 [1] - Correct final value \(43.3\%\) (accept \(43.3\%\) to \(43.4\%\)) [1]
題目 11 · structured
12 分
Chlorine is an element in Group VII of the Periodic Table.
(a) Define the term isotopes. [2]
(b) Complete the following table to show the number of subatomic particles in the given species: | Species | Protons | Neutrons | Electrons | | :--- | :---: | :---: | :---: | | \(^{35}\text{Cl}\) | | | | | \(^{37}\text{Cl}^-\)| | | | [3]
(c) Calculate the relative atomic mass of chlorine if the natural abundance of \(^{35}\text{Cl}\) is \(75.8\%\) and \(^{37}\text{Cl}\) is \(24.2\%\). Give your answer to 3 significant figures. [2]
(d) State and explain the trend in chemical reactivity of the halogens down Group VII. [3]
(e) Describe the change observed when chlorine gas is bubbled into aqueous potassium bromide. Write the ionic equation for this reaction. [2]
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解題
(a) Atoms of the same element with the same number of protons but different numbers of neutrons.
(d) Reactivity decreases down the group. This is because down the group, atomic radius increases and there is more shielding. Therefore, the attraction of the nucleus for an incoming electron weakens, making it harder to gain an electron.
Terylene is a synthetic polyester made by condensation polymerisation.
(a) Describe how condensation polymerisation differs from addition polymerisation. [1]
(b) Terylene is formed from ethane-1,2-diol, \(\text{HO-CH}_2\text{-CH}_2\text{-OH}\), and benzene-1,4-dicarboxylic acid, \(\text{HOOC-C}_6\text{H}_4\text{-COOH}\). (i) Draw the structure of one repeat unit of Terylene. Show all the atoms and bonds in the ester linkage explicitly. (You may represent the benzene ring as \(-\text{C}_6\text{H}_4-\)). [3] (ii) State the name of the small molecule eliminated during this reaction. [1]
(c) Nylon is a synthetic polyamide. (i) Identify the linkage group present in polyamides. [1] (ii) Draw the displayed structure of this polyamide linkage. [2]
(d) Biodegradable polyesters are being developed to reduce environmental pollution. (i) Explain what is meant by the term biodegradable. [1] (ii) State one environmental benefit of using biodegradable polyesters. [1] (iii) Name a natural polymer that contains the same type of linkage as Nylon. [2]
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解題
(a) Condensation polymerisation produces a small molecule (such as water) as a byproduct alongside the polymer, whereas addition polymerisation forms only the polymer.
(b) (i) Repeat unit structure: \(\text{[-O-CH}_2\text{-CH}_2\text{-O-CO-C}_6\text{H}_4\text{-CO-]_n}\) containing the explicit ester linkage \(\text{-O-C(=O)-}\). (ii) Water.
(c) (i) Amide linkage (or peptide linkage). (ii) Structure showing \(\text{-C(=O)-NH-}\) with all bonds shown explicitly.
(d) (i) Capable of being decomposed/broken down by microorganisms or biological activity. (ii) Reduces landfill waste / does not persist in the environment / decreases harm to marine life. (iii) Protein (or polypeptide).
評分準則
(a) Condensation polymerisation releases a small molecule (such as water), addition does not [1]
(b) (i) - Correct ester linkage shown with all bonds \(\text{-C(=O)-O-}\) [1] - Correct diol remainder \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\) [1] - Correct dicarboxylic acid remainder \(\text{-CO-C}_6\text{H}_4\text{-CO-}\) with continuation bonds [1] (ii) Water [1]
(d) (i) Broken down by bacteria/microbes/fungi [1] (ii) Does not accumulate in ecosystems / fills up landfills less [1] (iii) Protein / polypeptide [2] (accept nylon-6,6 counterparts like silk/wool for [1])
題目 13 · structured
12 分
A student is provided with a green crystalline solid, Solid X, and carries out tests to identify its composition.
(a) (i) The student dissolves a sample of Solid X in distilled water. Adding aqueous sodium hydroxide to this solution produces a green precipitate that is insoluble in excess. Identify the cation present in Solid X. [1] (ii) Adding dilute hydrochloric acid to Solid X produces a colourless gas that turns limewater cloudy. Identify the gas and the anion present in Solid X. [2] (iii) Write the ionic equation for the reaction of this gas with limewater (calcium hydroxide solution) to form a white precipitate. Include state symbols. [2]
(b) The student also studies the reactivity of three metals: copper, iron, and zinc. (i) Order the three metals from least reactive to most reactive. [1] (ii) Describe two observations when a piece of zinc is placed in an aqueous solution of copper(II) sulfate. [2] (iii) Write the balanced chemical equation, including state symbols, for the displacement reaction between zinc and aqueous copper(II) sulfate. [3] (iv) Explain, in terms of electron transfer, why this displacement reaction is a redox reaction. [1]
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解題
(a) (i) Iron(II) ion / \(\text{Fe}^{2+}\) (ii) Gas: Carbon dioxide (\(\text{CO}_2\)). Anion: Carbonate ion (\(\text{CO}_3^{2-}\)). (iii) \(\text{Ca}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)}\)
(b) (i) Copper, Iron, Zinc. (ii) Red-brown solid deposits on the zinc; blue solution fades / turns colourless. (iii) \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\) (iv) Zinc loses electrons (is oxidised) and copper ions gain electrons (are reduced).
(b) (i) Copper, Iron, Zinc [1] (ii) Any two from: - Red-brown solid formed [1] - Blue solution fades / decolourises [1] - Zinc dissolves / gets smaller [1] (iii) - Correct formulas for reactants and products [1] - Correct balancing [1] - Correct state symbols: \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\)[1] (iv) Zinc loses electrons and copper gains electrons (both oxidation and reduction occur) [1]
Paper 6 (Alternative to Practical)
Answer all practical, graphical, qualitative analysis, and experimental planning questions.
4 題目 · 40 分
題目 1 · practical
10 分
A student investigates the concentration of citric acid in a commercial brand of lime juice. The citric acid in a \(25.0\text{ cm}^3\) sample of the juice is titrated against standard aqueous sodium hydroxide of concentration \(0.050\text{ mol/dm}^3\) using thymolphthalein indicator.
(a) Suggest why a volumetric pipette is used to measure the volume of the lime juice rather than a measuring cylinder. [1]
(b) State the colour change of the thymolphthalein indicator at the end-point when titrated with aqueous sodium hydroxide. From .................... to .................... [2]
(c) The initial reading on the burette was \(1.2\text{ cm}^3\) and the final reading was \(21.8\text{ cm}^3\). Calculate the volume of aqueous sodium hydroxide added during this titration. [1]
(d) Explain why a white tile is placed under the conical flask during the titration. [1]
(e) The student repeated the titration three times and obtained the following titres: \(20.6\text{ cm}^3\), \(20.5\text{ cm}^3\), and \(20.7\text{ cm}^3\). (i) Calculate the average volume of aqueous sodium hydroxide used in these three titrations. [1] (ii) Explain why the student did not include a fourth titre of \(22.1\text{ cm}^3\) in the calculation of the average. [1]
(f) Citric acid is a weak organic acid. State the expected pH of the lime juice before any sodium hydroxide is added. [1]
(g) Describe how the student could obtain a sample of pure, dry sodium citrate crystals from the neutralised solution obtained in the titration. [2]
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解題
(a) A volumetric pipette is much more accurate and precise for measuring a fixed volume of \(25.0\text{ cm}^3\) compared to a measuring cylinder. (b) Thymolphthalein is colourless in acidic solutions (like lime juice) and turns blue in alkaline solutions (at the end-point of the titration). (c) Volume of NaOH added = \(21.8\text{ cm}^3 - 1.2\text{ cm}^3 = 20.6\text{ cm}^3\). (d) A white tile provides a neutral background, making it easier to clearly observe the subtle colour change of the indicator. (e) (i) Average titre = \(\frac{20.6 + 20.5 + 20.7}{3} = 20.6\text{ cm}^3\). (ii) The titre of \(22.1\text{ cm}^3\) is an anomaly and is not concordant (within \(0.2\text{ cm}^3\)) with the other concordant results. (f) Since citric acid is a weak acid, the pH of lime juice is typically between 3 and 6. (g) The student should heat the solution in an evaporating basin until it reaches its crystallisation point (or is saturated). Then, let it cool slowly to allow crystals to form. Finally, filter the crystals to separate them from the remaining liquid and dry them between sheets of filter paper.
評分準則
(a) 1 mark for: more accurate/precise (than a measuring cylinder). (b) 1 mark for: colourless; 1 mark for: blue. (c) 1 mark for: \(20.6\text{ cm}^3\). (d) 1 mark for: to see the colour change more clearly / easily. (e) (i) 1 mark for: \(20.6\text{ cm}^3\); (ii) 1 mark for: it is not concordant / it is anomalous / not within \(0.2\text{ cm}^3\) of the others. (f) 1 mark for: any value in the range of 3 to 6 (inclusive). (g) 1 mark for: heat/evaporate to crystallisation point / until saturated; 1 mark for: filter off crystals AND dry with filter paper / in a warm oven (reject: heating to dryness).
題目 2 · practical
10 分
A student investigates the thermal stability of different metal carbonates by heating them and measuring the time taken for the carbon dioxide gas produced to turn limewater milky.
(a) When a sample of green copper(II) carbonate is heated, it decomposes to form a black solid and a gas. State the colour change of the solid during the heating. From .................... to .................... [2]
(b) Describe the test used to identify the gas produced, including the positive observation. [2]
(c) Write a chemical equation for the thermal decomposition of copper(II) carbonate, \(\text{CuCO}_3\). [2]
(d) The student heats sodium carbonate strongly using a Bunsen burner, but no gas is produced. Explain this observation in terms of the reactivity of sodium. [1]
(e) The time taken for the limewater to turn milky when heating equal masses of two other carbonates was recorded: - Calcium carbonate: \(45\text{ s\)} - Iron(II) carbonate: \(15\text{ s\)} (i) Use these results to deduce which metal is more reactive, calcium or iron. Explain your answer in terms of thermal stability. [2] (ii) State one variable, other than the mass of the carbonate, that must be kept constant to ensure a fair comparison. [1]
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解題
(a) Copper(II) carbonate is a green solid. Upon heating, it decomposes into black copper(II) oxide solid, so the colour change is from green to black. (b) The gas produced is carbon dioxide. To test for it, bubble the gas through limewater; the limewater turns cloudy or milky. (c) The thermal decomposition of copper(II) carbonate is represented by: \(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\). (d) Sodium is a very reactive metal (Group I). Reactive metals form highly stable carbonates that do not decompose at the standard temperatures produced by a laboratory Bunsen burner. (e) (i) Calcium is more reactive than iron. More reactive metals form more stable carbonates, which require more energy and thus more time (45 seconds vs 15 seconds) to decompose. (ii) To ensure a fair test, the distance of the flame from the test-tube, the size of the Bunsen flame, or the volume of limewater used must be kept constant.
評分準則
(a) 1 mark for: green; 1 mark for: black. (b) 1 mark for: test with limewater; 1 mark for: turns milky / cloudy. (c) 1 mark for: correct formulae for reactants and products (\(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\)); 1 mark for: fully balanced equation. (d) 1 mark for: sodium is very reactive / sodium carbonate is very stable / does not decompose. (e) (i) 1 mark for: calcium (is more reactive); 1 mark for: calcium carbonate is more thermally stable / takes longer to decompose; (ii) 1 mark for: strength of heating / distance of flame / volume of limewater.
題目 3 · practical
10 分
A student is provided with a mixture of two salts, salt A and salt B. Salt A is iron(II) sulfate and salt B is ammonium chloride. The student dissolves the mixture in distilled water to form a solution, which is divided into three portions to carry out the following tests.
(a) (i) To the first portion, aqueous sodium hydroxide is added dropwise, then in excess. State the observations. - Dropwise: .................... - In excess: .................... [2] (ii) To the second portion, dilute hydrochloric acid is added followed by aqueous barium chloride. State the observation and name the ion identified by this test. [2] (iii) To the third portion, dilute nitric acid is added followed by aqueous silver nitrate. State the observation and name the precipitate formed. [2]
(b) A separate sample of the solid mixture is heated with aqueous sodium hydroxide. (i) Name the gas produced. [1] (ii) Describe how this gas can be identified, including the test and expected observation. [2] (iii) Explain why this test must be performed in a well-ventilated laboratory or fume cupboard. [1]
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解題
(a) (i) Iron(II) ions (\(\text{Fe}^{2+}\)) react with sodium hydroxide to form a green precipitate of iron(II) hydroxide, which is insoluble in excess sodium hydroxide. (ii) The sulfate ion (\(\text{SO}_4^{2-}\)) from iron(II) sulfate reacts with barium ions to form a white precipitate of barium sulfate. (iii) The chloride ion (\(\text{Cl}^-\)) from ammonium chloride reacts with silver ions to form a white precipitate of silver chloride (\(\text{AgCl}\)). (b) (i) Ammonium ions (\(\text{NH}_4^+\)) react with hydroxide ions on heating to produce ammonia gas (\(\text{NH}_3\)). (ii) Ammonia gas is alkaline and can be identified by holding damp red litmus paper near the mouth of the tube, which turns blue. (iii) Ammonia is a pungent, toxic, and irritating gas, so it must be handled safely in a fume cupboard or well-ventilated area.
評分準則
(a) (i) 1 mark for: green precipitate; 1 mark for: insoluble / remains in excess. (ii) 1 mark for: white precipitate; 1 mark for: sulfate / \(\text{SO}_4^{2-}\). (iii) 1 mark for: white precipitate; 1 mark for: silver chloride / \(\text{AgCl}\). (b) (i) 1 mark for: ammonia / \(\text{NH}_3\). (ii) 1 mark for: damp red litmus paper; 1 mark for: turns blue. (iii) 1 mark for: ammonia is toxic / irritating / poisonous.
題目 4 · practical
10 分
Many plants contain coloured pigments in their leaves or petals that can be used as natural indicators. Plan an investigation to extract the red pigment from red cabbage leaves and determine whether the pigment is a single coloured substance or a mixture of different coloured compounds.
You are provided with red cabbage leaves, sand, ethanol, distilled water and common laboratory apparatus.
Your plan should include: - how to extract the coloured pigment from the leaves - the technique used to separate the components of the pigment - a detailed description of how this separation technique is carried out - how the results will show whether the pigment is a single substance or a mixture. [10]
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解題
To carry out this investigation: 1. Extraction: Cut or tear the red cabbage leaves into small pieces and place them in a mortar. Add a small quantity of sand to act as an abrasive, and add a suitable solvent (such as ethanol or hot water). Use a pestle to grind the leaves thoroughly until a dark red liquid is formed. Filter the mixture using a funnel and filter paper to obtain a clear, concentrated pigment extract. 2. Separation Technique: Identify the technique as paper chromatography. 3. Chromatography Procedure: Draw a straight baseline on a strip of chromatography paper using a pencil (approx. \(1.5\text{ cm}\) from the bottom). Use a capillary tube or a fine dropper to place a small, concentrated spot of the red cabbage extract onto the pencil line. Allow the spot to dry. Suspend the paper in a beaker containing a small volume of solvent (ethanol or water), ensuring that the solvent level is below the pencil line. Cover the beaker with a lid to saturate the atmosphere. Allow the solvent to run up the paper until it is near the top, then remove the paper and mark the solvent front with a pencil. 4. Analyzing Results: Observe the developed chromatogram. If the cabbage pigment is a single substance, only one coloured spot will appear on the paper. If the pigment is a mixture, multiple spots of different colours will appear at different heights along the paper.
評分準則
Award marks as follows (up to a maximum of 10): - 1 mark: Cut/crush/grind red cabbage leaves. - 1 mark: Use mortar and pestle (with sand). - 1 mark: Add solvent (ethanol / hot water) to extract pigment. - 1 mark: Filter the mixture to get a clear liquid extract. - 1 mark: Name paper chromatography as the separation technique. - 1 mark: Draw a baseline on the paper using a pencil. - 1 mark: Spot the extract onto the pencil baseline. - 1 mark: Place the paper in a beaker with the solvent level below the baseline. - 1 mark: Let the solvent run up the paper. - 1 mark: Explain results: 1 spot = single substance, more than 1 spot = mixture.
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