An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 22: 選擇題
Answer all 40 multiple-choice questions. For each question, there are four possible answers, A, B, C, and D. You may use a calculator.
36 题目 · 36 分
题目 1 · 選擇題
1 分
Four metals, \(W\), \(X\), \(Y\) and \(Z\), are investigated using displacement reactions.
\begin{itemize} \item Metal \(W\) displaces metal \(X\) and metal \(Y\) from aqueous solutions of their nitrates. \item Metal \(X\) displaces metal \(Y\) from its nitrate solution. \item Metal \(Z\) does not react with aqueous solutions of the nitrates of \(W\), \(X\) or \(Y\). \end{itemize}
What is the order of reactivity of the four metals, starting with the most reactive?
A.\(W \rightarrow X \rightarrow Y \rightarrow Z\)
B.\(Z \rightarrow Y \rightarrow X \rightarrow W\)
C.\(W \rightarrow Y \rightarrow X \rightarrow Z\)
D.\(X \rightarrow W \rightarrow Z \rightarrow Y\)
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解题
A more reactive metal displaces a less reactive metal from its compound.
1. \(W\) displaces \(X\) and \(Y\), so \(W\) is more reactive than both \(X\) and \(Y\). 2. \(X\) displaces \(Y\), so \(X\) is more reactive than \(Y\) (order so far: \(W > X > Y\)). 3. \(Z\) does not displace \(W\), \(X\), or \(Y\), so \(Z\) is the least reactive of all four.
Therefore, the order of reactivity from most to least reactive is \(W \rightarrow X \rightarrow Y \rightarrow Z\).
评分标准
A is correct [1 mark]. B reverses the order entirely. C has \(Y\) more reactive than \(X\), which contradicts the fact that \(X\) displaces \(Y\). D has \(X\) as the most reactive metal, contradicting \(W\) displacing \(X\).
题目 2 · 選擇題
1 分
Three unlabelled metals, \(\text{P}\), \(\text{Q}\), and \(\text{R}\), were tested as follows:
• Only metal \(\text{P}\) reacts with cold water to produce a gas. • Metal \(\text{Q}\) reacts with dilute hydrochloric acid, but metal \(\text{R}\) does not. • When heated with copper(II) oxide, metal \(\text{Q}\) reduces it to copper, but metal \(\text{R}\) does not.
What is the order of reactivity of the metals, from most reactive to least reactive?
Metal \(\text{P}\) is the only metal reactive enough to react with cold water, making it the most reactive. Metal \(\text{Q}\) reacts with dilute hydrochloric acid and reduces copper(II) oxide, whereas metal \(\text{R}\) does neither. Therefore, \(\text{Q}\) is more reactive than \(\text{R}\). The decreasing order of reactivity is \(\text{P} \rightarrow \text{Q} \rightarrow \text{R}\).
评分标准
A is correct [1]. B, C, D are incorrect based on the observations that P is the most reactive (reacts with cold water) and Q is more reactive than R (reacts with acid and reduces CuO).
题目 3 · 選擇題
1 分
In a titration, \(20.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), is completely neutralised by \(25.0\text{ cm}^3\) of dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
What is the concentration of the dilute sulfuric acid?
A.\(0.0400\text{ mol/dm}^3\)
B.\(0.0800\text{ mol/dm}^3\)
C.\(0.125\text{ mol/dm}^3\)
D.\(0.160\text{ mol/dm}^3\)
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解题
First, calculate the moles of \(\text{NaOH}\): \[\text{moles of NaOH} = \frac{20.0}{1000} \times 0.100 = 0.00200\text{ mol}\] Using the stoichiometric ratio \(2\text{NaOH} : 1\text{H}_2\text{SO}_4\): \[\text{moles of H}_2\text{SO}_4 = \frac{0.00200}{2} = 0.00100\text{ mol}\] Now, calculate the concentration of \(\text{H}_2\text{SO}_4\): \[\text{concentration} = \frac{0.00100}{\frac{25.0}{1000}} = 0.0400\text{ mol/dm}^3\]
评分标准
A is correct [1]. B uses the incorrect 1:1 mole ratio (0.0800 mol/dm³). C and D represent inverted or unscaled volume calculation errors.
题目 4 · 選擇題
1 分
An oxide of phosphorus contains \(56.4\%\) phosphorus by mass.
[\(A_\text{r}: \text{P}, 31; \text{O}, 16\)]
What is the empirical formula of this oxide?
A.\(\text{PO}\)
B.\(\text{PO}_2\)
C.\(\text{P}_2\text{O}_3\)
D.\(\text{P}_2\text{O}_5\)
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解题
Percentage of phosphorus \(= 56.4\%\), so percentage of oxygen \(= 100 - 56.4 = 43.6\%\). Moles of \(\text{P} = \frac{56.4}{31} = 1.819\text{ mol}\) Moles of \(\text{O} = \frac{43.6}{16} = 2.725\text{ mol}\) Divide by the smallest number of moles: \(\text{P}: \frac{1.819}{1.819} = 1.00\) \(\text{O}: \frac{2.725}{1.819} = 1.50\) Multiply by 2 to obtain whole numbers: \(\text{P}_2\text{O}_3\).
评分标准
C is correct [1]. A, B, and D correspond to incorrect rounding or inversion of molar masses.
题目 5 · 選擇題
1 分
Tests are carried out on an aqueous solution of compound \(\text{X}\).
• Adding aqueous sodium hydroxide produces a green precipitate that does not dissolve in excess sodium hydroxide. • Adding dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is compound \(\text{X}\)?
A.chromium(III) sulfate
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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解题
The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Chromium(III) forms a green precipitate that dissolves in excess \(\text{NaOH}\)). The formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, compound \(\text{X}\) is iron(II) sulfate.
评分标准
C is correct [1]. A is incorrect because chromium(III) hydroxide is soluble in excess NaOH. B is incorrect because barium nitrate tests for sulfate, not chloride. D is incorrect because iron(III) forms a red-brown precipitate with NaOH.
题目 6 · 選擇題
1 分
Which statement correctly compares the manufacture of ethanol by the catalytic hydration of ethene with its manufacture by fermentation?
A.Fermentation is a continuous process whereas the hydration of ethene is a batch process.
B.Fermentation produces a purer product directly than the hydration of ethene.
C.Hydration of ethene uses a renewable raw material.
D.Hydration of ethene is a much faster process than fermentation.
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解题
Catalytic hydration of ethene is a continuous and fast industrial reaction operating at high temperature and pressure, whereas fermentation is a slow batch process. Fermentation uses renewable resources (sugars), whereas ethene is obtained from crude oil (non-renewable). Hydration produces pure ethanol directly, whereas fermentation produces an impure aqueous mixture requiring fractional distillation.
评分标准
D is correct [1]. A is incorrect because fermentation is a batch process and hydration is continuous. B is incorrect because fermentation yields an impure product. C is incorrect because ethene is derived from non-renewable crude oil.
题目 7 · 選擇題
1 分
Pieces of four different metals, W, X, Y and Z, are placed separately into aqueous solutions of their nitrates.
What is the order of reactivity of the metals, from most reactive to least reactive?
A.Y $\rightarrow$ W $\rightarrow$ Z $\rightarrow$ X
B.X $\rightarrow$ Z $\rightarrow$ W $\rightarrow$ Y
C.Y $\rightarrow$ Z $\rightarrow$ W $\rightarrow$ X
D.W $\rightarrow$ Y $\rightarrow$ Z $\rightarrow$ X
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解题
A more reactive metal displaces a less reactive metal from its nitrate solution. - Metal Y displaces W, X, and Z, so Y is the most reactive. - Metal W displaces X and Z, so W is more reactive than X and Z. - Metal Z displaces only X, so Z is more reactive than X. - Metal X displaces none of the metals, so X is the least reactive.
Therefore, the order of reactivity from most reactive to least reactive is: Y $\rightarrow$ W $\rightarrow$ Z $\rightarrow$ X.
评分标准
A is correct [1]. B reverses the order (least to most reactive). C places Z above W incorrectly. D places W as the most reactive metal.
题目 8 · 選擇題
1 分
A student carries out a titration to find the concentration of an aqueous solution of sodium hydroxide, $\text{NaOH}$.
A $25.0\text{ cm}^3$ portion of $\text{NaOH(aq)}$ requires exactly $20.0\text{ cm}^3$ of $0.0500\text{ mol/dm}^3$ dilute sulfuric acid, $\text{H}_2\text{SO}_4\text{(aq)}$, for complete neutralisation.
2. Use the stoichiometric ratio from the equation ($2\text{NaOH} : 1\text{H}_2\text{SO}_4$): $$\text{moles of } \text{NaOH} = 2 \times 0.00100\text{ mol} = 0.00200\text{ mol}$$
3. Calculate the concentration of $\text{NaOH}$: $$\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00200\text{ mol}}{\frac{25.0}{1000}\text{ dm}^3} = 0.0800\text{ mol/dm}^3$$
评分标准
C is correct [1]. A results from dividing by 2 instead of multiplying by 2 for the mole ratio ($0.0200\text{ mol/dm}^3$). B results from omitting the 2:1 mole ratio entirely ($0.0400\text{ mol/dm}^3$). D results from an extra factor of 2 ($0.160\text{ mol/dm}^3$).
题目 9 · 選擇題
1 分
A sample of an oxide of iron contains $1.12\text{ g}$ of iron and $0.48\text{ g}$ of oxygen.
[$A_r$: $\text{Fe} = 56$, $\text{O} = 16$]
What is the empirical formula of this iron oxide?
A.$\text{FeO}$
B.$\text{Fe}_2\text{O}_3$
C.$\text{Fe}_3\text{O}_4$
D.$\text{Fe}_3\text{O}_2$
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解题
1. Find moles of each element: $$\text{moles of Fe} = \frac{1.12\text{ g}}{56\text{ g/mol}} = 0.020\text{ mol}$$ $$\text{moles of O} = \frac{0.48\text{ g}}{16\text{ g/mol}} = 0.030\text{ mol}$$
Therefore, the empirical formula is $\text{Fe}_2\text{O}_3$.
评分标准
B is correct [1]. A corresponds to a 1:1 mole ratio ($\text{FeO}$). C corresponds to $\text{Fe}_3\text{O}_4$. D corresponds to inverted mole ratio subscripts ($\text{Fe}_3\text{O}_2$).
题目 10 · 選擇題
1 分
An unknown solid X is dissolved in water to form a colourless solution. Two chemical tests are performed on portions of this solution:
1. Aqueous sodium hydroxide is added dropwise until in excess: a green precipitate forms which does not dissolve in excess sodium hydroxide. 2. Dilute nitric acid is added, followed by aqueous barium nitrate: a white precipitate forms.
What is the identity of solid X?
A.chromium(III) sulfate
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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解题
Test 1: Iron(II) ions ($\text{Fe}^{2+}$) react with aqueous sodium hydroxide to form a green precipitate of $\text{Fe(OH)}_2$, which is insoluble in excess. Test 2: Sulfate ions ($\text{SO}_4^{2-}$) react with aqueous barium nitrate in the presence of dilute nitric acid to form an insoluble white precipitate of barium sulfate ($\text{BaSO}_4$).
Combining these deductions identifies X as iron(II) sulfate.
评分标准
C is correct [1]. A chromium(III) gives a green precipitate that dissolves in excess $\text{NaOH(aq)}$. B iron(II) chloride would give no precipitate with aqueous barium nitrate. D iron(III) sulfate gives a red-brown precipitate with aqueous sodium hydroxide.
题目 11 · 選擇題
1 分
Ethanol is manufactured industrially either by the fermentation of aqueous glucose or by the catalytic hydration of ethene.
Which statement correctly compares these two processes?
A.Fermentation uses a phosphoric(V) acid catalyst, whereas hydration uses yeast as a catalyst.
B.Fermentation is a continuous process, whereas hydration is a batch process.
D.Fermentation produces a purer yield of ethanol directly than hydration.
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解题
Fermentation uses glucose obtained from crops/plants, which is a renewable raw material. In contrast, catalytic hydration uses ethene obtained from the cracking of crude oil fractions, which is a non-renewable (fossil) resource.
Evaluating the other options: - A is incorrect: yeast is used in fermentation, while phosphoric(V) acid is the catalyst for the hydration of ethene. - B is incorrect: fermentation is a batch process, whereas hydration of ethene is a continuous process. - D is incorrect: hydration of ethene produces a much purer product directly than fermentation, which produces dilute ethanol requiring fractional distillation.
评分标准
C is correct [1]. A inverts the catalysts used for the two processes. B inverts batch and continuous process descriptions. D incorrectly claims fermentation produces a purer product.
题目 12 · 選擇題
1 分
Four unlabelled metals, \(\text{W}\), \(\text{X}\), \(\text{Y}\) and \(\text{Z}\), are placed into separate aqueous solutions containing the nitrates of each metal. The results are summarized:
- \(\text{W}\) displaces \(\text{X}\) and \(\text{Z}\), but does not displace \(\text{Y}\). - \(\text{X}\) does not displace any of the other three metals. - \(\text{Y}\) displaces \(\text{W}\), \(\text{X}\) and \(\text{Z}\). - \(\text{Z}\) displaces \(\text{X}\) only.
What is the order of reactivity of the four metals, starting with the most reactive?
A more reactive metal displaces a less reactive metal from its aqueous salt solution. - \(\text{Y}\) displaces all three other metals (\(\text{W}\), \(\text{X}\), and \(\text{Z}\)), so \(\text{Y}\) is the most reactive. - \(\text{W}\) displaces \(\text{X}\) and \(\text{Z}\), so it is more reactive than \(\text{X}\) and \(\text{Z}\), but less reactive than \(\text{Y}\). - \(\text{Z}\) displaces \(\text{X}\) only, meaning \(\text{Z}\) is more reactive than \(\text{X}\) but less reactive than \(\text{W}\). - \(\text{X}\) displaces none of the metals, so it is the least reactive.
Therefore, the order of decreasing reactivity is \(\text{Y} > \text{W} > \text{Z} > \text{X}\).
评分标准
A [1 mark] - Deduce that Y is the most reactive, followed by W, Z, and then X as the least reactive.
题目 13 · 選擇題
1 分
A student carries out a titration using \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), and dilute hydrochloric acid, \(\text{HCl}\).
Exactly \(20.0\text{ cm}^3\) of the dilute hydrochloric acid is required to reach the end-point.
What is the concentration of the dilute hydrochloric acid?
A.\(0.080\text{ mol/dm}^3\)
B.\(0.125\text{ mol/dm}^3\)
C.\(0.250\text{ mol/dm}^3\)
D.\(0.500\text{ mol/dm}^3\)
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解题
1. Calculate the amount in moles of \(\text{NaOH}\): \[n(\text{NaOH}) = c \times V = 0.100\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00250\text{ mol}\]
2. From the balanced equation, \(1\text{ mol}\) of \(\text{NaOH}\) reacts with \(1\text{ mol}\) of \(\text{HCl}\): \[n(\text{HCl}) = 0.00250\text{ mol}\]
3. Calculate the concentration of \(\text{HCl}\): \[c(\text{HCl}) = \frac{n}{V} = \frac{0.00250\text{ mol}}{\frac{20.0}{1000}\text{ dm}^3} = \frac{0.00250}{0.0200} = 0.125\text{ mol/dm}^3\]
评分标准
B [1 mark] - Correct calculation of amount of NaOH and conversion to HCl concentration: 0.125 mol/dm³.
题目 14 · 選擇題
1 分
An oxide of phosphorus contains \(43.7\%\) phosphorus by mass.
The relative molecular mass, \(M_\text{r}\), of the oxide is \(284\).
D [1 mark] - Calculate empirical formula P₂O₅ and scale by Mr 284/142 = 2 to give P₄O₁₀.
题目 15 · 選擇題
1 分
An unknown solid, \(\text{Q}\), is dissolved in water to make an aqueous solution.
Two chemical tests are carried out on separate portions of the solution: 1. Aqueous sodium hydroxide is added dropwise until in excess: a green precipitate forms which is insoluble in excess. 2. Dilute nitric acid followed by aqueous barium nitrate is added: a white precipitate forms.
What is the identity of solid \(\text{Q}\)?
A.copper(II) sulfate
B.iron(II) sulfate
C.iron(III) sulfate
D.iron(II) chloride
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解题
- The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). - The formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, solid \(\text{Q}\) is iron(II) sulfate.
评分标准
B [1 mark] - Identify Fe²⁺ from green precipitate with NaOH(aq) and SO₄²⁻ from white precipitate with Ba(NO₃)₂(aq).
题目 16 · 選擇題
1 分
Which row correctly identifies the catalyst, temperature, and pressure used in the industrial manufacture of ethanol by the catalytic addition of steam to ethene?
The manufacture of ethanol by the hydration of ethene involves reacting ethene with steam under the following essential conditions: - Catalyst: concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) - Temperature: approximately \(300\ ^\circ\text{C}\) - Pressure: approximately \(60\text{ atm}\) (or \(6000\text{ kPa}\))
Therefore, row A is correct.
评分标准
A [1 mark] - Correct identification of phosphoric acid catalyst, 300 °C, and 60 atm.
题目 17 · 選擇題
1 分
Three metals, \(\text{J}\), \(\text{K}\), and \(\text{L}\), were added separately to aqueous solutions of their nitrates. The observations are recorded:
- \(\text{J}\) displaces \(\text{K}\) from aqueous \(\text{KNO}_3\) - \(\text{J}\) does not displace \(\text{L}\) from aqueous \(\text{L(NO}_3)_2\) - \(\text{K}\) does not displace \(\text{L}\) from aqueous \(\text{L(NO}_3)_2\)
Which statement about the metals is correct?
A.Metal \(\text{L}\) is the most reactive of the three metals.
B.Metal \(\text{L}\) is the least reactive of the three metals.
C.Metal \(\text{K}\) will displace metal \(\text{J}\) from its aqueous nitrate.
D.Metal \(\text{J}\) is more reactive than metal \(\text{L}\).
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解题
From the displacement observations: - \(\text{J}\) displaces \(\text{K}\), so \(\text{J}\) is more reactive than \(\text{K}\) (\(\text{J} > \text{K}\)). - \(\text{J}\) does not displace \(\text{L}\), so \(\text{L}\) is more reactive than \(\text{J}\) (\(\text{L} > \text{J}\)). - Combining these gives the reactivity order: \(\text{L} > \text{J} > \text{K}\).
Therefore, \(\text{L}\) is the most reactive of the three metals.
评分标准
A is correct [1 mark]. B is incorrect because \(\text{K}\) is the least reactive, not \(\text{L}\). C is incorrect because \(\text{K}\) is less reactive than \(\text{J}\), so it cannot displace \(\text{J}\). D is incorrect because \(\text{L}\) is more reactive than \(\text{J}\).
题目 18 · 選擇題
1 分
A student titrates \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), with dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
\(20.0\text{ cm}^3\) of the sulfuric acid is required for complete neutralisation.
2. Use the stoichiometric ratio from the equation (\(2\text{ NaOH} : 1\text{ H}_2\text{SO}_4\)): \[ n(\text{H}_2\text{SO}_4) = \frac{0.00250}{2} = 0.00125\text{ mol} \]
B is correct [1 mark]. A (\(0.125\text{ mol/dm}^3\)) is incorrect (fails to divide by the mole ratio 2). C (\(0.0313\text{ mol/dm}^3\)) is incorrect (divided by 4 instead of 2). D (\(0.0800\text{ mol/dm}^3\)) is incorrect (inverted titration volume ratio).
题目 19 · 選擇題
1 分
A \(4.80\text{ g}\) sample of an iron oxide contains \(3.36\text{ g}\) of iron and \(1.44\text{ g}\) of oxygen.
\([A_r\text{: Fe, } 56\text{; O, } 16]\)
What is the empirical formula of this iron oxide?
A.\(\text{FeO}\)
B.\(\text{Fe}_3\text{O}_4}\)
C.\(\text{Fe}_2\text{O}_3}\)
D.\(\text{FeO}_2\)
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解题
1. Moles of \(\text{Fe} = \frac{3.36}{56} = 0.060\text{ mol}\) 2. Moles of \(\text{O} = \frac{1.44}{16} = 0.090\text{ mol}\) 3. Divide by the smallest number of moles: \(\text{Fe}: \frac{0.060}{0.060} = 1.0\) \(\text{O}: \frac{0.090}{0.060} = 1.5\) 4. Multiply by 2 to obtain whole numbers: \(\text{Fe}_2\text{O}_3\).
评分标准
C is correct [1 mark]. A is incorrect (assumes 1:1 molar ratio). B is incorrect (assumes 3:4 ratio). D is incorrect (assumes 1:2 ratio).
题目 20 · 選擇題
1 分
An aqueous solution of salt \(\text{X}\) is tested as follows:
- Addition of aqueous sodium hydroxide produces a green precipitate that does not dissolve in excess sodium hydroxide. - Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of salt \(\text{X}\)?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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解题
1. The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). 2. The formation of a white precipitate with aqueous barium nitrate in the presence of dilute nitric acid confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, salt \(\text{X}\) is iron(II) sulfate.
评分标准
B is correct [1 mark]. A is incorrect because chromium(III) forms a green precipitate that is soluble in excess \(\text{NaOH}\). C is incorrect because chloride gives a precipitate with silver nitrate, not barium nitrate. D is incorrect because iron(III) forms a red-brown precipitate.
题目 21 · 選擇題
1 分
Which statement about the manufacture of ethanol by fermentation is correct?
A.It requires a temperature of \(300^\circ\text{C}\) and a phosphoric acid catalyst.
B.It is a continuous process that produces pure ethanol directly.
C.It produces ethanol and water as the only products.
D.It uses renewable raw materials and produces a dilute aqueous solution of ethanol.
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解题
Fermentation converts glucose from renewable plant sources (such as sugarcane or maize) into ethanol and carbon dioxide using yeast at around \(30\text{--}35^\circ\text{C}\) under anaerobic conditions. It produces a dilute aqueous solution of ethanol which requires fractional distillation to concentrate.
Option A describes the hydration of ethene (catalytic addition of steam). Option B is incorrect because fermentation is a batch process, not continuous. Option C is incorrect because carbon dioxide is also produced.
评分标准
D is correct [1 mark]. A is incorrect (describes industrial hydration of ethene). B is incorrect (fermentation is a batch process). C is incorrect (carbon dioxide is also a product of fermentation).
题目 22 · 選擇題
1 分
Three metals, W, X, and Y, are investigated.
• Metal W displaces metal X from an aqueous solution of its nitrate. • Metal X reduces the oxide of metal Y when heated, but does not react with the oxide of metal W. • Carbon reduces the oxide of metal X when heated, but does not reduce the oxide of metal W.
Which list shows the metals in order of decreasing reactivity?
A.W, X, Y
B.W, Y, X
C.Y, X, W
D.X, W, Y
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解题
Metal W displaces metal X from aqueous solution, so W is more reactive than X (\(\text{W} > \text{X}\)). Metal X reduces the oxide of metal Y, meaning X displaces Y from its oxide, so X is more reactive than Y (\(\text{X} > \text{Y}\)). Therefore, the order of decreasing reactivity (most reactive to least reactive) is W, X, Y.
评分标准
A is correct [1 mark]. B is incorrect because X is more reactive than Y. C is the reverse order (increasing reactivity). D incorrectly places X as more reactive than W.
题目 23 · 選擇題
1 分
In a titration, \( 25.0\text{ cm}^3 \) of \( 0.200\text{ mol/dm}^3 \) aqueous sodium hydroxide, \( \text{NaOH} \), is neutralised by \( 20.0\text{ cm}^3 \) of dilute sulfuric acid, \( \text{H}_2\text{SO}_4 \).
Step 2: Use the stoichiometric ratio from the equation (\(2\text{NaOH} : 1\text{H}_2\text{SO}_4\)): \(\text{moles of }\text{H}_2\text{SO}_4 = \frac{0.00500}{2} = 0.00250\text{ mol}\).
Step 3: Calculate the concentration of \(\text{H}_2\text{SO}_4\): \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00250\text{ mol}}{0.0200\text{ dm}^3} = 0.125\text{ mol/dm}^3\).
评分标准
B is correct [1 mark]. A results from dividing by 2 twice. C results from omitting the 2:1 mole ratio. D results from multiplying by 2 instead of dividing.
题目 24 · 選擇題
1 分
A compound contains \( 40.0\% \) carbon, \( 6.7\% \) hydrogen, and \( 53.3\% \) oxygen by mass. Its relative molecular mass, \( M_r \), is 60. [ \( A_r \): C, 12; H, 1; O, 16 ]
Step 2: Divide by the smallest value (3.33): • C = 1, H = 2, O = 1 \(\rightarrow\) Empirical formula = \( \text{CH}_2\text{O} \).
Step 3: Empirical formula mass = \( 12 + (2 \times 1) + 16 = 30 \).
Step 4: Scale factor = \( \frac{M_r}{\text{empirical mass}} = \frac{60}{30} = 2 \). Therefore, the molecular formula is \( (\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2 \).
评分标准
C is correct [1 mark]. A is the empirical formula, not the molecular formula. B has an incorrect \( M_r \) (44). D has an incorrect \( M_r \) (60, but wrong composition).
题目 25 · 選擇題
1 分
A sample of a solid salt is dissolved in distilled water to form solution X. Two tests are carried out on separate portions of solution X.
• Addition of aqueous sodium hydroxide produces a green precipitate that does not dissolve in excess. • Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
Which compound is present in solution X?
A.chromium(III) sulfate
B.copper(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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解题
The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \( \text{Fe}^{2+} \). The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \( \text{SO}_4^{2-} \). Therefore, the compound is iron(II) sulfate.
评分标准
C is correct [1 mark]. A is incorrect because chromium(III) forms a green precipitate that dissolves in excess aqueous sodium hydroxide. B is incorrect because copper(II) forms a light blue precipitate. D is incorrect because iron(III) forms a red-brown precipitate.
题目 26 · 選擇題
1 分
Ethanol can be manufactured by the fermentation of aqueous glucose or by the catalytic addition of steam to ethene (hydration).
Which statement comparing these two processes is correct?
A.Fermentation uses a renewable resource, whereas hydration uses a non-renewable resource.
B.Fermentation is a continuous process, whereas hydration is a batch process.
C.Fermentation requires a higher temperature than hydration.
D.Hydration produces an impure product requiring fractional distillation, whereas fermentation produces pure ethanol.
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解题
Fermentation uses glucose derived from plant matter (crops), which is a renewable raw material. Hydration uses ethene obtained from the cracking of petroleum fractions, which is a non-renewable fossil fuel source. Therefore, statement A is correct. Statement B is incorrect because fermentation is a batch process while hydration is continuous. Statement C is incorrect because fermentation requires \( 25\text{--}35^\circ\text{C} \) whereas hydration requires around \( 300^\circ\text{C} \). Statement D is incorrect because fermentation produces an impure dilute mixture requiring fractional distillation, whereas hydration yields pure ethanol.
评分标准
A is correct [1 mark]. B is incorrect (fermentation is batch; hydration is continuous). C is incorrect (hydration requires higher temperature, \( \approx 300^\circ\text{C} \)). D is incorrect (fermentation requires fractional distillation to obtain pure ethanol).
题目 27 · 選擇題
1 分
Four unknown metals, W, X, Y, and Z, are investigated. The experimental findings are listed.
- Metal W displaces metal Y from an aqueous solution of its nitrate, but does not react with an aqueous solution of the nitrate of metal X. - The oxide of metal Z is reduced by heating with carbon, but the oxide of metal X cannot be reduced by heating with carbon. - Metal Z reacts with dilute hydrochloric acid to produce hydrogen gas, but metal Y does not react with dilute hydrochloric acid.
What is the correct order of reactivity of the four metals, from most reactive to least reactive?
A.X > W > Y > Z
B.X > W > Z > Y
C.W > X > Z > Y
D.Y > Z > W > X
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解题
1. W displaces Y but not X, so X is more reactive than W, and W is more reactive than Y (X > W > Y). 2. The oxide of X cannot be reduced by carbon, but the oxide of Z can, so X is more reactive than Z (and carbon). 3. Metal Z reacts with dilute acid, whereas metal Y does not react with dilute acid, meaning Z is more reactive than Y (Z > Y). 4. Combining the observations: X is the most reactive, followed by W, then Z, and Y is the least reactive (X > W > Z > Y).
评分标准
B is correct [1 mark]. A incorrect: places Y as more reactive than Z. C incorrect: places W ahead of X. D incorrect: inverted order of reactivity.
题目 28 · 選擇題
1 分
A student titrates \(25.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), with dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
The equation for the neutralisation reaction is shown.
Exactly \(20.0\text{ cm}^3\) of the dilute sulfuric acid is required to neutralise the sodium hydroxide completely.
What is the concentration of the dilute sulfuric acid in \(\text{mol/dm}^3\)?
A.0.0625
B.0.125
C.0.0313
D.0.250
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解题
Step 1: Calculate the amount in moles of \(\text{NaOH}\) used: \[\text{moles of NaOH} = 0.100\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00250\text{ mol}\]
Step 2: Use the stoichiometric ratio from the balanced equation (\(2\text{ NaOH} : 1\text{ H}_2\text{SO}_4\)) to find moles of \(\text{H}_2\text{SO}_4\): \[\text{moles of H}_2\text{SO}_4 = \frac{0.00250}{2} = 0.00125\text{ mol}\]
Step 3: Calculate the concentration of the sulfuric acid: \[\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00125}{0.0200\text{ dm}^3} = 0.0625\text{ mol/dm}^3\]
评分标准
A is correct [1 mark]. B incorrect: omits the 1:2 mole ratio (gives 0.125 mol/dm³). C incorrect: divides by 2 twice (gives 0.0313 mol/dm³). D incorrect: multiplies by 2 instead of dividing by 2 (gives 0.250 mol/dm³).
题目 29 · 選擇題
1 分
A compound contains only phosphorus and chlorine. Analysis shows that the compound contains \(22.55\%\) phosphorus by mass.
Step 1: Determine percentage of chlorine by mass: \[\%\text{ Cl} = 100\% - 22.55\% = 77.45\%\]
Step 2: Calculate moles of each element in a \(100\text{ g}\) sample: \[\text{Moles of P} = \frac{22.55}{31.0} = 0.7274\text{ mol}\] \[\text{Moles of Cl} = \frac{77.45}{35.5} = 2.1817\text{ mol}\]
Step 3: Divide by the smallest mole value to find the simplest whole number ratio: \[\text{Ratio of P} = \frac{0.7274}{0.7274} = 1\] \[\text{Ratio of Cl} = \frac{2.1817}{0.7274} = 3.00\]
Therefore, the empirical formula is \(\text{PCl}_3\).
评分标准
C is correct [1 mark]. A incorrect: assumes a 1:1 mole ratio. B incorrect: incorrect division or inversion of atomic masses. D incorrect: corresponds to phosphorus pentachloride (14.87% P).
题目 30 · 選擇題
1 分
A student carries out two tests on separate portions of an aqueous solution of salt Q.
- Test 1: Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. - Test 2: Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of salt Q?
A.iron(II) chloride
B.iron(II) sulfate
C.iron(III) chloride
D.iron(III) sulfate
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解题
Test 1: The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). Test 2: The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, salt Q is iron(II) sulfate.
评分标准
B is correct [1 mark]. A incorrect: chloride gives a white precipitate with silver nitrate, not barium nitrate. C incorrect: iron(III) gives a red-brown precipitate with sodium hydroxide. D incorrect: iron(III) gives a red-brown precipitate.
题目 31 · 選擇題
1 分
Ethanol is manufactured industrially by two methods:
- Method 1: Fermentation of aqueous glucose - Method 2: Catalytic addition of steam to ethene
Which statement correctly compares Method 1 and Method 2?
A.Method 1 uses a renewable raw material and operates at a lower temperature than Method 2.
B.Method 1 produces pure ethanol directly without the need for fractional distillation.
C.Method 1 is a continuous process whereas Method 2 is a batch process.
D.Method 2 uses a catalyst of yeast at \(300\text{ }^\circ\text{C}\).
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解题
Method 1 (fermentation) uses a renewable raw material (glucose from plants/sugar cane) and operates at a relatively low temperature (around \(30\text{ }^\circ\text{C}\) to \(37\text{ }^\circ\text{C}\)). Method 2 (hydration of ethene) uses a non-renewable raw material (crude oil) and operates at a higher temperature (around \(300\text{ }^\circ\text{C}\)). Hence, statement A is correct.
评分标准
A is correct [1 mark]. B incorrect: fermentation produces dilute ethanol which requires fractional distillation. C incorrect: fermentation is a batch process; hydration of ethene is continuous. D incorrect: yeast is the catalyst for fermentation, not hydration at 300 °C (which uses phosphoric acid).
题目 32 · 選擇題
1 分
The results of four experiments involving four metals, \(W\), \(X\), \(Y\) and \(Z\), are shown.
1. Metal \(W\) displaces metal \(X\) from an aqueous solution of \(X\text{SO}_4\). 2. Metal \(Z\) reacts vigorously with cold water, but metal \(W\) only reacts with steam. 3. Heating the carbonate of \(Y\) produces carbon dioxide, but heating the carbonate of \(Z\) produces no reaction. 4. Metal \(Y\) displaces metal \(X\) from an aqueous solution of \(X\text{SO}_4\), but does not displace metal \(W\) from \(W\text{SO}_4\).
What is the order of reactivity of the four metals, from most reactive to least reactive?
A.\(Z \rightarrow W \rightarrow Y \rightarrow X\)
B.\(X \rightarrow Y \rightarrow W \rightarrow Z\)
C.\(Z \rightarrow Y \rightarrow W \rightarrow X\)
D.\(W \rightarrow Z \rightarrow Y \rightarrow X\)
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解题
From experiment 1, \(W\) is more reactive than \(X\) (\(W > X\)). From experiment 2, \(Z\) reacts with cold water while \(W\) reacts only with steam, so \(Z > W\). From experiment 4, \(Y\) displaces \(X\) but cannot displace \(W\), so \(W > Y > X\). Combining these gives the overall reactivity order: \(Z > W > Y > X\).
评分标准
A is correct [1]. B is incorrect (reverses the entire reactivity series). C is incorrect (places \(Y\) above \(W\)). D is incorrect (places \(W\) above \(Z\)).
题目 33 · 選擇題
1 分
A student carries out a titration to determine the concentration of a solution of dilute hydrochloric acid using \(25.0\text{ cm}^3\) of aqueous sodium hydroxide.
Which row identifies the piece of apparatus that should be used to measure the \(25.0\text{ cm}^3\) of aqueous sodium hydroxide and the colour change observed at the end-point when methyl orange is used as the indicator?
A.apparatus: volumetric pipette ; colour change: yellow to orange
B.apparatus: volumetric pipette ; colour change: red to yellow
C.apparatus: measuring cylinder ; colour change: yellow to orange
D.apparatus: measuring cylinder ; colour change: red to yellow
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解题
A volumetric pipette is designed to measure a single fixed, accurate volume such as \(25.0\text{ cm}^3\). When methyl orange indicator is added to an alkali (aqueous sodium hydroxide in the conical flask), it is yellow. As hydrochloric acid is titrated into the flask, the indicator turns orange/red at the end-point.
评分标准
A is correct [1]. B is incorrect because the colour change from alkali to acid is yellow to orange/red, not red to yellow. C and D are incorrect because a measuring cylinder is not sufficiently accurate for volumetric analysis in titrations.
题目 34 · 選擇題
1 分
A compound contains \(40.0\%\) carbon, \(6.7\%\) hydrogen and \(53.3\%\) oxygen by mass.
The relative molecular mass, \(M_{\text{r}}\), of the compound is 60.
1. Calculate moles of each element in \(100\text{ g}\): - \(n(\text{C}) = \frac{40.0}{12} = 3.33\text{ mol}\) - \(n(\text{H}) = \frac{6.7}{1} = 6.70\text{ mol}\) - \(n(\text{O}) = \frac{53.3}{16} = 3.33\text{ mol}\)
2. Divide by the smallest value (\(3.33\)): - \(\text{C} : \text{H} : \text{O} = 1 : 2 : 1\) Empirical formula = \(\text{CH}_2\text{O}\) (empirical formula mass = \(12 + 2(1) + 16 = 30\)).
3. Determine the molecular formula: - Ratio = \(\frac{M_{\text{r}}}{\text{empirical mass}} = \frac{60}{30} = 2\) - Molecular formula = \((\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2\).
评分标准
B is correct [1]. A is incorrect (empirical formula, \(M_{\text{r}} = 30\)). C is incorrect (formula with \(M_{\text{r}} = 60\) but incorrect elemental mass percentages: \(60.0\%\) C, \(13.3\%\) H, \(26.7\%\) O). D is incorrect (\(M_{\text{r}} = 62\)).
题目 35 · 選擇題
1 分
A sample of solid \(\text{T}\) is dissolved in distilled water to form a colourless solution. The solution is divided into two portions.
- To the first portion, aqueous sodium hydroxide is added dropwise and then in excess. A green precipitate forms which does not dissolve in excess sodium hydroxide. - To the second portion, dilute nitric acid is added followed by aqueous barium nitrate. A white precipitate forms.
What is the identity of solid \(\text{T}\)?
A.chromium(III) sulfate
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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解题
A green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: \(\text{Cr}^{3+}\) forms a green precipitate that dissolves in excess \(\text{NaOH(aq)}\)).
A white precipitate formed upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, solid \(\text{T}\) is iron(II) sulfate (\(\text{FeSO}_4\)).
评分标准
C is correct [1]. A is incorrect because chromium(III) hydroxide dissolves in excess aqueous sodium hydroxide to form a green solution. B is incorrect because chloride ions do not form a white precipitate with barium nitrate. D is incorrect because iron(III) ions produce a red-brown precipitate with aqueous sodium hydroxide.
题目 36 · 選擇題
1 分
Which statement comparing the production and reactions of ethanol is correct?
A.Fermentation of aqueous glucose requires a phosphoric acid catalyst at \(300\,^\circ\text{C}\).
B.Hydration of ethene with steam produces pure ethanol in a continuous process.
C.Complete combustion of ethanol produces carbon monoxide and water.
D.Oxidation of ethanol by acidified aqueous potassium manganate(VII) produces ethyl ethanoate.
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解题
The manufacture of ethanol by the catalytic addition of steam to ethene (hydration) is a continuous process that produces relatively pure ethanol.
- Statement A is incorrect because fermentation uses yeast (enzymes) at roughly \(30\text{--}35\,^\circ\text{C}\), whereas phosphoric acid at \(300\,^\circ\text{C}\) is used for ethene hydration. - Statement C is incorrect because complete combustion produces carbon dioxide and water, not carbon monoxide. - Statement D is incorrect because oxidation of ethanol with acidified aqueous potassium manganate(VII) forms ethanoic acid, not ethyl ethanoate.
评分标准
B is correct [1]. A is incorrect (confuses conditions of fermentation with hydration). C is incorrect (incomplete combustion produces carbon monoxide; complete combustion yields carbon dioxide). D is incorrect (oxidation produces ethanoic acid, while esterification produces ethyl ethanoate).
Answer all questions. Write your answers in the spaces provided. Show all your working and use appropriate units.
6 题目 · 79.98 分
题目 1 · structured
13.33 分
This question is about metals and the reactivity series.
(a) A student investigates the displacement reactions of four metals, \(\text{W}\), \(\text{X}\), \(\text{Y}\) and \(\text{Z}\), by adding each metal to aqueous solutions of metal nitrates. The results are shown: - Metal \(\text{W}\) displaces \(\text{X}^{2+}\), \(\text{Y}^{2+}\) and \(\text{Z}^{2+}\). - Metal \(\text{X}\) displaces \(\text{Z}^{2+}\) only. - Metal \(\text{Y}\) displaces \(\text{X}^{2+}\) and \(\text{Z}^{2+}\). - Metal \(\text{Z}\) does not displace any metal from its nitrate solution.
Deduce the order of reactivity of the four metals, from most reactive to least reactive.
(b) Zinc reacts with aqueous copper(II) sulfate in a displacement reaction. (i) Write the ionic equation, including state symbols, for the reaction between zinc and aqueous copper(II) sulfate. (ii) Explain in terms of electron transfer why zinc acts as a reducing agent in this reaction.
(c) When powdered metal \(\text{X}\) is heated strongly with iron(III) oxide, a vigorous reaction takes place to produce liquid iron and the oxide of \(\text{X}\), \(\text{X}_2\text{O}_3\). (i) Name this type of redox reaction. (ii) Identify the oxidising agent in this reaction and explain your choice. (iii) Deduce the relative position of metal \(\text{X}\) compared to iron in the reactivity series.
(d) Explain why aluminium appears unreactive and resists corrosion even though it is placed high in the reactivity series.
(e) Steel hulls of ships can be protected from rusting by attaching blocks of zinc. (i) Name this method of rust prevention. (ii) Explain how the zinc blocks protect the steel hull from rusting.
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解题
(a) Metal W displaces all other metals so it is the most reactive. Metal Y displaces X and Z, so Y > X. Metal X displaces only Z, so X > Z. Metal Z displaces none, so it is the least reactive. Order: W > Y > X > Z.
(b)(i) Zinc atoms react with copper(II) ions: \(\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}\). (ii) Zinc loses electrons (each Zn atom loses 2 electrons to become \(\text{Zn}^{2+}\)). A reducing agent donates/loses electrons.
(c)(i) Displacement / thermite reaction. (ii) Iron(III) oxide / \(\text{Fe}_2\text{O}_3\) is the oxidising agent because it loses oxygen to \(\text{X}\) (or \(\text{Fe}^{3+}\) gains electrons to form \(\text{Fe}\)). (iii) Metal \(\text{X}\) is more reactive than iron because it displaces iron from its oxide.
(d) Aluminium reacts rapidly with oxygen in air to form a thin, non-porous, protective oxide layer (\(\text{Al}_2\text{O}_3\)) on its surface that prevents underlying metal from contacting oxygen/water.
(e)(i) Sacrificial protection. (ii) Zinc is higher in the reactivity series than iron, so zinc oxidises/loses electrons preferentially instead of the iron hull.
评分标准
(a) [2 marks] M1: W most reactive and Z least reactive [1] M2: Correct full sequence: W > Y > X > Z / W, Y, X, Z [1]
(b) [3 marks] (i) M1: Correct species: \(\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\) [1] M2: Correct state symbols: (s), (aq), (aq), (s) [1] (ii) M3: Zinc loses electrons / electron donor [1]
(c) [3 marks] (i) M1: Displacement / thermite / redox [1] (ii) M2: Iron(III) oxide / \(\text{Fe}_2\text{O}_3\) / \(\text{Fe}^{3+}\) AND loses oxygen / gains electrons [1] (iii) M3: Metal X is higher / above iron / more reactive than iron [1]
(d) [2 marks] M1: Forms a protective / impervious layer of aluminium oxide [1] M2: Prevents oxygen/water reaching the metal underneath [1]
(e) [2 marks] (i) M1: Sacrificial protection / sacrificial anode [1] (ii) M2: Zinc is more reactive than iron / zinc oxidises/corrodes/loses electrons instead of iron [1]
题目 2 · structured
13.33 分
A student carries out a titration to determine the concentration of a solution of ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\).
(a) Name the piece of apparatus used to: (i) accurately measure \(25.0\text{ cm}^3\) of \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\) into a conical flask (ii) deliver the variable volume of standard potassium hydroxide solution, \(\text{KOH(aq)}\).
(b) The student adds a few drops of phenolphthalein indicator to the ethanedioic acid in the conical flask and titrates with \(\text{KOH(aq)}\). State the colour change observed at the end-point.
(c) The equation for the neutralisation reaction is: \[\text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{KOH(aq)} \rightarrow \text{K}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{H}_2\text{O(l)}\] In the titration, \(25.0\text{ cm}^3\) of \(0.0800\text{ mol/dm}^3\) \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\) requires \(20.0\text{ cm}^3\) of \(\text{KOH(aq)}\) for complete neutralisation. (i) Calculate the number of moles of \(\text{H}_2\text{C}_2\text{O}_4\) in \(25.0\text{ cm}^3\) of solution. (ii) Deduce the number of moles of \(\text{KOH}\) that reacted. (iii) Calculate the concentration of the \(\text{KOH(aq)}\) in \(\text{mol/dm}^3\). (iv) Calculate the concentration of the \(\text{KOH(aq)}\) in \(\text{g/dm}^3\). [\(A_\text{r}\): \(\text{K} = 39\), \(\text{O} = 16\), \(\text{H} = 1\)]
(d) Describe how pure, dry crystals of potassium ethanedioate, \(\text{K}_2\text{C}_2\text{O}_4\), can be prepared from the solutions of ethanedioic acid and potassium hydroxide without using an indicator in the final preparation.
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解题
(a)(i) Volumetric pipette / bulb pipette. (ii) Burette.
(b) Acidic solution is colourless with phenolphthalein; at neutralisation / alkaline end-point it turns pink.
(c)(i) \(\text{Moles of } \text{H}_2\text{C}_2\text{O}_4 = \text{concentration} \times \text{volume} = 0.0800 \times \frac{25.0}{1000} = 0.00200\text{ mol} = 2.00 \times 10^{-3}\text{ mol}\). (ii) From the mole ratio \(1 : 2\): \(\text{Moles of KOH} = 2 \times 0.00200 = 0.00400\text{ mol}\). (iii) \(\text{Concentration of KOH} = \frac{\text{moles}}{\text{volume (dm}^3)} = \frac{0.00400}{0.0200} = 0.200\text{ mol/dm}^3\). (iv) \(M_\text{r}(\text{KOH}) = 39 + 16 + 1 = 56\text{ g/mol}\). \(\text{Concentration in g/dm}^3 = 0.200 \times 56 = 11.2\text{ g/dm}^3\).
(d) Method for preparing pure dry salt: 1. Mix \(25.0\text{ cm}^3\) of ethanedioic acid and \(20.0\text{ cm}^3\) of \(\text{KOH}\) without adding indicator. 2. Heat the neutral solution to evaporate water until saturation / crystallisation point is reached. 3. Leave the hot saturated solution to cool to form crystals. 4. Filter the mixture to collect the crystals, wash with a small volume of cold distilled water, and dry between filter papers / in a warm oven.
(d) [4 marks] MP1: Mix the exact reacting volumes of acid and alkali (or \(25.0\text{ cm}^3\) acid + \(20.0\text{ cm}^3\) alkali) without indicator [1] MP2: Heat / evaporate until crystallisation point / saturation point [1] MP3: Cool to crystallise [1] MP4: Filter, rinse with cold distilled water, and dry with filter paper / desiccator / warm oven (not strong heating) [1]
题目 3 · structured
13.33 分
This question is about chemical calculations and stoichiometry.
(a) A hydrocarbon, \(\text{Q}\), contains \(85.7\%\) carbon and \(14.3\%\) hydrogen by mass. [\(A_\text{r}\): \(\text{C} = 12\), \(\text{H} = 1\)] (i) Show by calculation that the empirical formula of \(\text{Q}\) is \(\text{CH}_2\). (ii) The relative molecular mass, \(M_\text{r}\), of hydrocarbon \(\text{Q}\) is 56. Deduce the molecular formula of \(\text{Q}\). (iii) Draw the displayed formula of an unbranched alkene with this molecular formula.
(b) A student heats a sample of hydrated magnesium sulfate, \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\), to remove all water of crystallisation. The results obtained are: - Mass of hydrated salt = \(4.92\text{ g} - Mass of anhydrous \)\text{MgSO}_4\) remaining = \(2.40\text{ g} [\)A_\text{r}\): \(\text{Mg} = 24\), \(\text{S} = 32\), \(\text{O} = 16\), \(\text{H} = 1\)] (i) Calculate the mass of water driven off during heating. (ii) Calculate the number of moles of anhydrous \(\text{MgSO}_4\) formed. (iii) Calculate the number of moles of \(\text{H}_2\text{O}\) lost. (iv) Determine the value of integer \(x\).
(c) Calcium carbonate reacts with excess dilute hydrochloric acid: \[\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\] Calculate the volume of carbon dioxide gas, in \(\text{dm}^3\) at room temperature and pressure (r.t.p.), produced when \(10.0\text{ g}\) of calcium carbonate reacts completely. [\(M_\text{r}(\text{CaCO}_3) = 100\); 1 mole of any gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
(d) Define the term *limiting reactant*.
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解题
(a)(i) \(\text{Moles of C} = \frac{85.7}{12} = 7.142\text{ mol}\) \(\text{Moles of H} = \frac{14.3}{1} = 14.30\text{ mol}\) Dividing by smallest: \(\text{C} = \frac{7.142}{7.142} = 1\), \(\text{H} = \frac{14.30}{7.142} = 2.00\). Empirical formula is \(\text{CH}_2\).
(ii) Empirical formula mass of \(\text{CH}_2 = 12 + (2 \times 1) = 14\). \(\text{Number of empirical units} = \frac{56}{14} = 4\). Molecular formula = \(\text{C}_4\text{H}_8\).
(iii) Displayed formula of but-1-ene: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\) showing all C-H and C=C/C-C single bonds, or but-2-ene: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\).
(c) \(\text{Moles of } \text{CaCO}_3 = \frac{10.0}{100} = 0.100\text{ mol}\). From equation: 1 mole \(\text{CaCO}_3\) gives 1 mole \(\text{CO}_2\). \(\text{Moles of } \text{CO}_2 = 0.100\text{ mol}\). \(\text{Volume of } \text{CO}_2 = 0.100 \times 24.0\text{ dm}^3 = 2.40\text{ dm}^3\).
(d) The reactant that is completely used up first, which stops the reaction / limits the amount of product formed.
评分标准
(a) [4 marks] (i) M1: Moles of \(\text{C} = 7.14\) and Moles of \(\text{H} = 14.3\) [1] M2: Simplest integer ratio \(1 : 2\) leading to \(\text{CH}_2\) [1] (ii) M3: \(\text{C}_4\text{H}_8\) [1] (iii) M4: Fully displayed formula of but-1-ene or but-2-ene showing every single/double bond and all H atoms [1]
(d) [1 mark] M1: Reactant that is used up first / not in excess / determines/limits the amount of product formed [1]
题目 4 · structured
13.33 分
This question concerns qualitative analysis and chemical identification tests.
(a) Solid \(\text{M}\) is a mixture of two ionic compounds, compound \(\text{A}\) and compound \(\text{B}\). A student carries out three tests on solid \(\text{M}\) and its aqueous solution: - **Test 1:** Aqueous sodium hydroxide is added dropwise to an aqueous solution of \(\text{M}\) until in excess. A green precipitate is formed which dissolves in excess aqueous sodium hydroxide to give a green solution. - **Test 2:** Dilute nitric acid followed by aqueous barium nitrate is added to another portion of the aqueous solution of \(\text{M}\). A white precipitate forms. - **Test 3:** Solid \(\text{M}\) is heated gently with aqueous sodium hydroxide and aluminium foil. A colourless, pungent gas is evolved which turns damp red litmus paper blue.
(i) Identify the cation present from the result of Test 1. (ii) Identify the anion identified by Test 2 and write the ionic equation, with state symbols, for the formation of the white precipitate. (iii) Name the pungent gas released in Test 3 and deduce the anion that produced this gas.
(b) Describe a chemical test to distinguish between aqueous sodium chloride, \(\text{NaCl(aq)}\), and aqueous sodium iodide, \(\text{NaI(aq)}\). Include the reagent used and the observations for each solution.
(c) Describe: (i) a chemical test to confirm the presence of water (ii) a physical test to confirm that a liquid sample is pure water.
(d) Sulfur dioxide, \(\text{SO}_2\), is an acidic, reducing gas. Describe a chemical test to confirm the presence of sulfur dioxide gas and state the observation.
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解题
(a)(i) The cation giving a green precipitate that dissolves in excess aqueous \(\text{NaOH}\) to form a green solution is the chromium(III) ion, \(\text{Cr}^{3+}\). (ii) Dilute nitric acid and barium nitrate test for sulfate ions, \(\text{SO}_4^{2-}\). The white precipitate is barium sulfate: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). (iii) Heating with \(\text{NaOH}\) and aluminium foil reduces nitrate ions (\(\text{NO}_3^-\)) to ammonia gas (\(\text{NH}_3\)), which turns damp red litmus blue.
(b) Reagent: Acidified aqueous silver nitrate (dilute nitric acid + \(\text{AgNO}_3\text{(aq)}\)). With \(\text{NaCl(aq)}\): white precipitate of \(\text{AgCl}\). With \(\text{NaI(aq)}\): yellow precipitate of \(\text{AgI}\).
(c)(i) Add anhydrous copper(II) sulfate; it turns from white to blue (or anhydrous cobalt(II) chloride paper turns from blue to pink). (ii) Measure the boiling point: pure water boils at exactly \(100^\circ\text{C}\) at 1 atm (or freezes at \(0^\circ\text{C}\)).
(d) Reagent: Acidified aqueous potassium manganate(VII), \(\text{KMnO}_4\). Observation: Colour changes from purple to colourless (decolourised).
(b) [3 marks] MP1: Add dilute nitric acid AND aqueous silver nitrate [1] MP2: Observation with chloride: white precipitate [1] MP3: Observation with iodide: yellow precipitate [1]
(c) [2 marks] (i) M1: Anhydrous copper(II) sulfate turns from white to blue OR anhydrous cobalt(II) chloride turns from blue to pink [1] (ii) M2: Boils at (exactly) \(100^\circ\text{C}\) / freezes at (exactly) \(0^\circ\text{C}\) [1]
Alcohols are a homologous series of organic compounds containing the \(-\text{OH}\) functional group.
(a) Ethanol can be manufactured industrially by two different methods: - **Method 1:** Fermentation of aqueous glucose - **Method 2:** Catalytic addition of steam to ethene
(i) State two essential conditions required for the fermentation of glucose in Method 1. (ii) Write the balanced chemical equation for the reaction in Method 2 and name the catalyst used. (iii) State one advantage of Method 1 compared to Method 2 and one advantage of Method 2 compared to Method 1.
(b) Propan-1-ol can be oxidised by heating with acidified potassium manganate(VII). (i) State the colour change observed during this oxidation reaction. (ii) Name the organic product formed and draw its displayed formula.
(c) Ethanol reacts with propanoic acid in the presence of concentrated sulfuric acid catalyst to form an ester, \(\text{E}\), and water. (i) State the name of ester \(\text{E}\). (ii) Draw the displayed formula of ester \(\text{E}\). (iii) State one use of esters in everyday products.
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解题
(a)(i) Essential conditions for fermentation: - Presence of yeast (source of zymase enzymes) - Anaerobic conditions / exclusion of air / oxygen - Temperature in range \(25^\circ\text{C}\) to \(35^\circ\text{C}\) - Aqueous solution (Any two)
(iii) Advantage of Method 1: Uses renewable plant resources / glucose (does not deplete crude oil) / operates at mild temperatures. Advantage of Method 2: Rapid/continuous process / produces pure ethanol without needing fractional distillation / 100% atom economy.
(b)(i) Acidified potassium manganate(VII) changes from purple to colourless. (ii) Oxidation of propan-1-ol yields propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)). Displayed formula shows three carbon atoms with single bonds between C-C, five C-H single bonds, one \(\text{C}=\text{O}\) double bond and one \(\text{C}-\text{O}-\text{H}\) group.
(c)(i) Ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) + Propanoic acid (\(\text{C}_2\text{H}_5\text{COOH}\)) forms ethyl propanoate. (ii) Displayed formula: \(\text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_3\) showing every bond explicitly. (iii) Perfumes, food flavourings, solvents, or plasticisers.
评分标准
(a) [6 marks] (i) Any two from [2]: - Yeast / enzymes [1] - Anaerobic / absence of oxygen / air [1] - Temperature between \(20^\circ\text{C}\) and \(40^\circ\text{C}\) (inclusive) [1] - Aqueous / in water [1] (ii) M3: \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\) [1] M4: Phosphoric acid / \(\text{H}_3\text{PO}_4\) [1] (iii) M5: Advantage of Method 1: uses renewable resources / less energy needed [1] M6: Advantage of Method 2: continuous / faster rate / purer product / higher yield [1]
(b) [3 marks] (i) M1: Purple to colourless / decolourised [1] (ii) M2: Propanoic acid [1] M3: Correct displayed formula showing all atoms and all bonds including \(\text{C}=\text{O}\) and \(\text{O}-\text{H}\) [1]
(c) [4 marks] (i) M1: Ethyl propanoate [1] (ii) M2: Correct ester linkage \(-\text{C}(=\text{O})-\text{O}-\) [1] M3: Fully correct displayed formula showing all atoms and bonds of \(\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3\) [1] (iii) M4: Perfumes / fragrances / food flavourings / solvents [1]
题目 6 · 結構題
13.33 分
A student carries out a titration experiment to determine the concentration of a sample of dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
(a) The student measures \(25.0\text{ cm}^3\) of \(0.120\text{ mol/dm}^3\) sodium hydroxide solution, \(\text{NaOH}\), into a conical flask using a volumetric pipette.
(i) Name the piece of apparatus used to add the dilute sulfuric acid slowly into the conical flask. [1]
(ii) State the colour change observed at the end-point when methyl orange indicator is added to the sodium hydroxide solution and titrated with dilute sulfuric acid.
Initial colour: ..................................................... Colour at end-point: ..................................................... [2]
(b) The chemical equation for the neutralisation reaction is:
The average volume of dilute sulfuric acid required to react completely with \(25.0\text{ cm}^3\) of the \(0.120\text{ mol/dm}^3\) \(\text{NaOH}\) solution is \(18.75\text{ cm}^3\).
(i) Calculate the amount, in moles, of \(\text{NaOH}\) present in \(25.0\text{ cm}^3\) of the \(0.120\text{ mol/dm}^3\) solution. [1]
(ii) Deduce the amount, in moles, of \(\text{H}_2\text{SO}_4\) that reacted with this amount of \(\text{NaOH}\). [1]
(iii) Calculate the concentration of the dilute sulfuric acid in \(\text{mol/dm}^3\). [2]
(iv) Calculate the concentration of the dilute sulfuric acid in \(\text{g/dm}^3\). [\(A_r\): \(\text{H}=1\), \(\text{O}=16\), \(\text{S}=32\)] [2]
(c) Write the ionic equation, including state symbols, for this neutralisation reaction. [2]
(d) Pure, dry crystals of hydrated sodium sulfate can be prepared from the solutions used in the titration.
Describe how you would prepare pure, dry crystals of sodium sulfate after determining the exact volumes required, without using an indicator. [2]
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解题
(a)(i) A burette is used to deliver accurately measured variable volumes of acid. (ii) Methyl orange is yellow in alkaline solution (\(\text{NaOH}\)) and turns orange / red at the end-point upon neutralisation.
(b)(i) \(\text{Moles of NaOH} = \text{concentration} \times \text{volume (in }\text{dm}^3\text{)} = 0.120 \times \frac{25.0}{1000} = 3.00 \times 10^{-3}\text{ mol}\) (or \(0.00300\text{ mol}\)). (ii) From the balanced equation, \(2\text{ mol}\) of \(\text{NaOH}\) reacts with \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\). \(\text{Moles of }\text{H}_2\text{SO}_4 = \frac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}\text{ mol}\) (or \(0.00150\text{ mol}\)). (iii) \(\text{Concentration of }\text{H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume (in }\text{dm}^3\text{)}} = \frac{1.50 \times 10^{-3}}{18.75 / 1000} = 0.0800\text{ mol/dm}^3\). (iv) \(M_r(\text{H}_2\text{SO}_4) = 2(1) + 32 + 4(16) = 98\text{ g/mol}\). \(\text{Concentration in g/dm}^3 = 0.0800\text{ mol/dm}^3 \times 98\text{ g/mol} = 7.84\text{ g/dm}^3\).
(c) The neutralisation reaction between a strong acid and a strong alkali involves hydrogen ions reacting with hydroxide ions to form water: \[\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\]
(d) 1. Combine \(25.0\text{ cm}^3\) of \(\text{NaOH}\) and \(18.75\text{ cm}^3\) of \(\text{H}_2\text{SO}_4\) without adding indicator. 2. Heat the mixture to evaporate water until the crystallisation point / saturation is reached. 3. Leave the solution to cool to form crystals, filter off the crystals, and dry them between sheets of filter paper.
评分标准
(a)(i) Burette [1] (a)(ii) Initial: yellow [1] End-point: orange / pink / red [1] (b)(i) \(3.00 \times 10^{-3}\) / \(0.00300\) (mol) [1] (b)(ii) \(1.50 \times 10^{-3}\) / \(0.00150\) (mol) [1] (allow ecf from (b)(i)) (b)(iii) \(\text{volume in dm}^3 = 0.01875\) OR \(\frac{1.50 \times 10^{-3}}{0.01875}\) [1] \(0.0800\) (\(\text{mol/dm}^3\)) [1] (allow ecf from (b)(ii)) (b)(iv) \(M_r(\text{H}_2\text{SO}_4) = 98\) [1] \(7.84\) (\(\text{g/dm}^3\)) [1] (allow ecf from (b)(iii)) (c) \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\) [1] Correct state symbols: \((\text{aq}) + (\text{aq}) \rightarrow (\text{l})\) [1] (dependent on correct formulae) (d) Any two from: - Mix exact reacting volumes without indicator [1] - Heat / evaporate to crystallisation point / until saturated [1] - Cool and filter (to obtain crystals) [1] - Dry crystals with / between filter paper(s) / in a desiccator / warm oven (R: direct heating to dryness) [1]
Paper 62: Alternative to Practical
Answer all questions. Write your answers in the spaces provided. Notes for use in qualitative analysis are provided.
4 题目 · 40 分
题目 1 · practical
10 分
A student investigated the temperature change when four different powdered metals, **A**, **B**, **C**, and **D**, were added separately to aqueous copper(II) sulfate.
In each experiment: - A measuring cylinder was used to place \(25.0\text{ cm}^3\) of aqueous copper(II) sulfate into a polystyrene cup. - The initial temperature of the solution was measured and recorded. - Exactly \(1.0\text{ g}\) of metal powder was added, the mixture was stirred continuously, and the highest temperature reached was recorded. - The polystyrene cup was rinsed and dried between experiments.
**(e)** State two variables that were kept constant to ensure the comparison was fair. [2]
**(f)** Predict the effect on the temperature rise if \(50.0\text{ cm}^3\) of the same aqueous copper(II) sulfate had been used instead of \(25.0\text{ cm}^3\) with \(1.0\text{ g}\) of metal **C**. Explain your answer. [1]
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解题
**(a)** Calculate each temperature change (highest temperature \(-\) initial temperature): - Metal **A**: \(38.5 - 21.5 = +17.0\text{ }^\circ\text{C}\) - Metal **C**: \(54.5 - 21.0 = +33.5\text{ }^\circ\text{C}\) - Metal **D**: \(31.0 - 22.5 = +8.5\text{ }^\circ\text{C}\)
**(b)** A polystyrene cup is a thermal insulator, which minimises heat loss to the surroundings compared to glass.
**(c)** The higher the temperature increase in a displacement reaction, the more reactive the metal: $$\mathbf{C} > \mathbf{A} > \mathbf{D} > \mathbf{B}$$
**(d)** Metal **B** gave a temperature change of \(0.0\text{ }^\circ\text{C}\), meaning no displacement reaction took place. Hence, metal **B** is copper (or a metal lower than copper in the reactivity series, such as silver).
**(e)** Controlled variables include: 1. Mass of metal powder (\(1.0\text{ g}\)) 2. Volume of copper(II) sulfate solution (\(25.0\text{ cm}^3\)) 3. Concentration of copper(II) sulfate solution 4. Surface area / state of subdivision of the metal powder
**(f)** The temperature rise would be approximately halved / smaller because doubling the volume of solution means the heat released by the reaction warms twice the mass of water.
评分标准
**(a) [3 marks]** - M1: Metal **A** = \(17.0\text{ }^\circ\text{C}\) [1] - M2: Metal **C** = \(33.5\text{ }^\circ\text{C}\) [1] - M3: Metal **D** = \(8.5\text{ }^\circ\text{C}\) [1]
**(b) [1 mark]** - M1: (Polystyrene is a) good thermal insulator / reduces heat loss (to the surroundings) / poor conductor of heat [1] *(R: prevents all heat loss)*
**(d) [2 marks]** - M1: Copper / \(\text{Cu}\) (or silver / gold / platinum) [1] - M2: No reaction occurred / metal is less reactive than copper / cannot displace copper(II) ions [1]
**(e) [2 marks]** - Any two from [1 mark each]: - Mass of metal powder / \(1.0\text{ g}\) - Volume of aqueous copper(II) sulfate / \(25.0\text{ cm}^3\) - Concentration of aqueous copper(II) sulfate - Particle size / surface area of metal / powdered form - Rate of stirring
**(f) [1 mark]** - M1: Temperature rise is halved / smaller / less AND because the same energy heats twice the volume / mass of liquid [1]
题目 2 · practical
10 分
A student determined the concentration of a sample of dilute hydrochloric acid, **solution X**, by titrating it with \(0.100\text{ mol/dm}^3\) aqueous sodium hydroxide, **solution Y**.
A \(25.0\text{ cm}^3\) portion of **solution Y** was transferred into a conical flask using a volumetric pipette. A few drops of methyl orange indicator were added. **Solution X** was placed in a burette and run into the conical flask until the end-point was reached.
The titration was carried out three times. The burette readings are shown in the table:
mean titre: ........................................................ \(\text{cm}^3\)
**(d)** Before filling the burette with **solution X**, the student washed the burette with distilled water but did not rinse it with **solution X**. State and explain the effect this error would have on the calculated titre value. [2]
**(e)** State why a volumetric pipette is preferred over a measuring cylinder for measuring **solution Y**. [1]
**(f)** Describe how the student would ensure that the end-point is determined as accurately as possible during the titration. [1]
**(b)** Methyl orange in alkali (**solution Y**) is yellow. At the end-point when acid is titrated into base, it turns orange (or red/pink at slight acid excess).
**(c)** Concordant titres are within \(0.20\text{ cm}^3\) of each other: - Titrations 2 and 3 are concordant (\(22.20\text{ cm}^3\) and \(22.40\text{ cm}^3\)). - Mean titre = \(\frac{22.20 + 22.40}{2} = 22.30\text{ cm}^3\).
**(d)** Water left in the burette dilutes the hydrochloric acid (**solution X**). Therefore, a greater volume of the more dilute acid is required to react with the fixed amount of alkali, resulting in a larger titre.
**(e)** A volumetric pipette provides a much higher degree of accuracy and precision (lower percentage uncertainty) than a measuring cylinder.
**(f)** Near the end-point, adding the acid dropwise while swirling continuously against a white tile ensures the exact point of colour change is not overshot.
评分标准
**(a) [3 marks]** - M1: Titration 1 = \(23.40\text{ cm}^3\) [1] - M2: Titration 2 = \(22.20\text{ cm}^3\) [1] - M3: Titration 3 = \(22.40\text{ cm}^3\) [1] *(Deduct 1 mark if not recorded to 2 decimal places ending in 0 or 5)*
**(b) [1 mark]** - M1: yellow to orange / yellow to pink / yellow to red [1]
**(c) [2 marks]** - M1: Identifies titrations 2 and 3 (or values \(22.20\) and \(22.40\)) [1] - M2: Correct calculation of mean = \(22.30\text{ cm}^3\) (ecf from student's concordant results) [1]
**(d) [2 marks]** - M1: Titre value increases / larger volume needed [1] - M2: Water dilutes **solution X** / lowers acid concentration [1]
**(e) [1 mark]** - M1: (Volumetric pipette is) more accurate / more precise / smaller percentage uncertainty [1]
**(f) [1 mark]** - M1: Add dropwise / drop by drop (near end-point) OR swirl flask continuously (during addition) OR place on a white tile (to see colour change clearly) [1]
题目 3 · practical
10 分
A student analysed two unknown substances: **solid J** and **solution K**.
### Tests on solid J **Solid J** was hydrated iron(II) sulfate, \(\text{FeSO}_4\cdot 7\text{H}_2\text{O}\).
**(a)** A small portion of **solid J** was placed in a dry test-tube and heated gently. Describe two observations you would expect to make. [2]
**(b)** The remainder of **solid J** was dissolved in distilled water to make an aqueous solution. The solution was divided into two test-tubes.
(i) To the first test-tube, aqueous sodium hydroxide was added dropwise until in excess. State your observations. [2]
(ii) To the second test-tube, dilute nitric acid was added, followed by aqueous barium nitrate. State your observation. [1]
### Tests on solution K **Solution K** was an aqueous solution of an ionic compound containing one cation and one anion.
**(c)** Aqueous ammonia was added dropwise and then in excess to a portion of **solution K**. - *Observations:* A white precipitate formed, which dissolved in excess aqueous ammonia to form a colourless solution. Identify the cation present in **solution K**. [1]
**(d)** To a second portion of **solution K**, dilute nitric acid was added followed by aqueous silver nitrate. - *Observations:* A yellow precipitate formed. Identify the anion present in **solution K**. [1]
**(e)** A third portion of **solution K** was mixed with aqueous sodium hydroxide and aluminium foil, then warmed gently. A gas was evolved. (i) State the test and observation used to identify this gas. [2]
(ii) Identify the gas evolved. [1]
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解题
**(a)** Heating a hydrated salt causes water of crystallisation to evaporate and condense as colourless droplets on the cooler upper sides of the test-tube. The green hydrated solid turns into a white/off-white anhydrous powder or decomposes to form steam.
**(b)(i)** Adding aqueous \(\text{NaOH}\) to \(\text{Fe}^{2+}\) produces an insoluble green precipitate of iron(II) hydroxide, \(\text{Fe(OH)}_2\), which does not dissolve in excess \(\text{NaOH}\).
**(b)(ii)** Adding dilute nitric acid followed by barium nitrate tests for sulfate ions (\(\text{SO}_4^{2-}\)). A white precipitate of barium sulfate forms.
**(c)** A white precipitate that dissolves in excess aqueous ammonia to form a colourless solution confirms the presence of zinc ions (\(\text{Zn}^{2+}\)).
**(d)** A yellow precipitate insoluble in dilute nitric acid upon adding aqueous silver nitrate confirms the presence of iodide ions (\(\text{I}^-\)).
**(e)(i)** The test for ammonia gas is holding damp red litmus paper near the mouth of the tube; it turns blue.
**(e)(ii)** The gas produced by reduction with \(\text{Al}\) and \(\text{NaOH}\) is ammonia (\(\text{NH}_3\)).
评分标准
**(a) [2 marks]** - Any two from [1 mark each]: - Drops of liquid / condensation / water droplets on the cooler walls of the test-tube - Steam / vapour given off - Colour change from green to white / grey / pale brown
**(b)(i) [2 marks]** - M1: Green precipitate [1] - M2: Insoluble in excess / precipitate remains [1] *(Allow: turns brown at the top / on standing)*
A student investigates the combustion of three different liquid alcohols: methanol, ethanol, and propan-1-ol.
**(a)** Plan an experiment to determine and compare the energy released per gram when each of the three liquid alcohols is burned to heat a sample of water.
You are provided with: - spirit burners containing methanol, ethanol, and propan-1-ol - a metal calorimeter (can) and a clamp stand - a thermometer - a measuring cylinder - an electronic balance - a supply of cold tap water - wooden splints.
Your plan should include: - the apparatus used and how it is assembled - the measurements to be taken - how to ensure the test is fair - how the energy released per gram can be calculated and compared. [6]
**(b)** State two reasons why the experimental energy values obtained from this simple calorimeter method are much lower than the theoretical data book values. [2]
**(c)** Suggest one practical improvement to the apparatus that would reduce heat loss to the surroundings. [1]
**(d)** State one safety precaution, other than wearing eye protection, that should be taken when using liquid alcohols. [1]
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解题
**(a)** A comprehensive plan must include: 1. **Setup & Procedure**: Measure a known volume (e.g., \(100\text{ cm}^3\)) of water using the measuring cylinder and pour it into the metal can. Clamp the can above the spirit burner. 2. **Initial Measurements**: Record the initial mass of the spirit burner containing the first alcohol using the balance. Record the initial temperature of the water using the thermometer. 3. **Heating**: Light the wick using a wooden splint and heat the water (e.g., until the temperature rises by \(20\text{ }^\circ\text{C}\) or for 2 minutes), stirring gently. 4. **Final Measurements**: Extinguish the flame, record the maximum temperature reached by the water, and reweigh the spirit burner to determine the mass of alcohol consumed (\(\Delta m = m_1 - m_2\)). 5. **Fair Testing**: Repeat the exact procedure for ethanol and propan-1-ol using the same starting volume of water, identical metal can, same height between wick and bottom of can, and same initial water temperature. 6. **Processing / Comparison**: Calculate the energy released per gram using: $$\text{Energy per gram} = \frac{m_{\text{water}} \times c \times \Delta T}{\Delta m_{\text{alcohol}}}$$ Alternatively, compare the value of \(\frac{\Delta T}{\Delta m}\) for each alcohol.
**(b)** Reasons for discrepancy: - Significant heat loss to the surrounding air. - Incomplete combustion of alcohol resulting in soot / carbon monoxide. - Heat absorbed by the metal can and thermometer. - Evaporation of alcohol from the wick during reweighing.
**(c)** Add a draught shield around the apparatus to reduce heat loss to moving air, or place a lid on the can.
**(d)** Alcohols are highly flammable; keep stock containers away from open flames, or use in a well-ventilated area.
评分标准
**(a) [6 marks]** - Award 1 mark for each of the following points (max 6): - **MP1:** Measure a known volume / mass of water into the metal can (using a measuring cylinder) [1] - **MP2:** Measure and record initial temperature of water AND initial mass of spirit burner [1] - **MP3:** Light the burner, heat the water, and extinguish flame [1] - **MP4:** Measure and record the highest / final temperature of water AND final mass of spirit burner [1] - **MP5:** Fair test variable kept constant (e.g. same volume of water / same height or distance of flame to can / same temperature increase / use of draught shield) [1] - **MP6:** Repeat the experiment with the other two alcohols [1] - **MP7:** Process results by calculating mass of alcohol burned (\(m_1 - m_2\)) AND temperature rise (\(T_2 - T_1\)) to calculate energy per gram / compare \(\frac{\Delta T}{\text{mass burned}}\) [1]
**(b) [2 marks]** - Any two from [1 mark each]: - Heat lost to the surroundings / air / atmosphere - Incomplete combustion of the fuel / alcohol - Heat absorbed by the metal can / thermometer / stand - Evaporation of alcohol (while weighing)
**(c) [1 mark]** - M1: Use a draught shield / windscreen OR put a lid on the calorimeter / can OR insulate the sides of the can [1]
**(d) [1 mark]** - M1: Keep cap on spirit burner when not in use / keep stock bottles away from naked flames / tie back long hair / ensure room is well-ventilated [1]
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