An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 Multiple Choice
Answer all 40 multiple choice questions. Each question is worth 1 mark.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
A sample of anhydrous metal carbonate, \(M_2\text{CO}_3\) (where \(M\) is a Group 1 metal), with a mass of 1.38 g, reacts completely with an excess of dilute hydrochloric acid. The volume of carbon dioxide gas collected at room temperature and pressure (r.t.p.) is 240 \(\text{cm}^3\). What is the identity of the metal \(M\)? (Take the molar volume of a gas at r.t.p. as 24.0 \(\text{dm}^3\text{ mol}^{-1}\).)
A.lithium
B.sodium
C.potassium
D.rubidium
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Worked solution
First, calculate the moles of \(\text{CO}_2\) produced: \(n(\text{CO}_2) = \frac{0.240\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0100\text{ mol}\). According to the reaction equation: \(M_2\text{CO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow 2M\text{Cl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\), the mole ratio of \(M_2\text{CO}_3\) to \(\text{CO}_2\) is 1:1. Therefore, \(n(M_2\text{CO}_3) = 0.0100\text{ mol}\). Next, find the molar mass of \(M_2\text{CO}_3\): \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.38\text{ g}}{0.0100\text{ mol}} = 138\text{ g mol}^{-1}\). The molar mass is given by: \(2 \times A_r(M) + A_r(\text{C}) + 3 \times A_r(\text{O}) = 138 \Rightarrow 2 \times A_r(M) + 12.0 + 48.0 = 138 \Rightarrow 2 \times A_r(M) = 78 \Rightarrow A_r(M) = 39.0\). This atomic mass corresponds to potassium (K), which has an \(A_r\) of 39.1.
Marking scheme
1 mark for the correct option. Award 1 mark for calculating the moles of CO2, relating it 1:1 to the metal carbonate, finding the correct molar mass of 138 g/mol, and correctly identifying potassium.
Question 2 · Multiple Choice
1 marks
The standard enthalpy changes of combustion for carbon, hydrogen, and propane are given below: \(\Delta H_c^\ominus[\text{C(graphite)}] = -393.5\text{ kJ mol}^{-1}\), \(\Delta H_c^\ominus[\text{H}_2(\text{g})] = -285.8\text{ kJ mol}^{-1}\), \(\Delta H_c^\ominus[\text{C}_3\text{H}_8(\text{g})] = -2219.2\text{ kJ mol}^{-1}\). What is the standard enthalpy change of formation of propane, \(\text{C}_3\text{H}_8(\text{g})\)?
A.-104.5 kJ mol^-1
B.+104.5 kJ mol^-1
C.-2898.5 kJ mol^-1
D.-1539.9 kJ mol^-1
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Worked solution
The equation for the formation of propane is: \(3\text{C(graphite)} + 4\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_8(\text{g})\). Using Hess's law and combustion data: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). Therefore, \(\Delta H_f^\ominus[\text{C}_3\text{H}_8] = 3 \times \Delta H_c^\ominus[\text{C}] + 4 \times \Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{C}_3\text{H}_8] = 3(-393.5) + 4(-285.8) - (-2219.2) = -1180.5 - 1143.2 + 2219.2 = -104.5\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option. Award 1 mark for setting up the correct Hess's Law expression using 3 times the carbon combustion and 4 times the hydrogen combustion, subtracting the propane combustion, and obtaining the value of -104.5 kJ/mol.
Question 3 · Multiple Choice
1 marks
Which of the following pairs of alkenes both react with hydrobromic acid, HBr, to form 2-bromo-2-methylbutane as the major product?
A.2-methylbut-1-ene and 2-methylbut-2-ene
B.2-methylbut-2-ene and 3-methylbut-1-ene
C.pent-1-ene and pent-2-ene
D.2-methylbut-1-ene and 3-methylbut-1-ene
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Worked solution
Electrophilic addition of HBr to an alkene follows Markovnikov's rule, proceeding via the most stable carbocation intermediate. 2-bromo-2-methylbutane is formed from a tertiary carbocation intermediate, \(\text{CH}_3\text{C}^+(\text{CH}_3)\text{CH}_2\text{CH}_3\). Both 2-methylbut-1-ene (\(\text{CH}_2=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3\)) and 2-methylbut-2-ene (\(\text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_3\)) form this stable tertiary carbocation upon addition of \(\text{H}^+\) to the carbon with more hydrogen atoms. Thus, both yield 2-bromo-2-methylbutane as the major product.
Marking scheme
1 mark for the correct option. Award 1 mark for identifying the two alkenes that form the same tertiary carbocation intermediate during electrophilic addition.
Question 4 · Multiple Choice
1 marks
When concentrated sulfuric acid is added to separate solid samples of sodium halides, a redox reaction occurs with some of the halides, reducing the sulfuric acid. For which of the sodium halides is sulfur dioxide, \(\text{SO}_2\), produced as a reduction product?
A.sodium chloride and sodium bromide only
B.sodium bromide and sodium iodide only
C.sodium iodide only
D.sodium chloride, sodium bromide, and sodium iodide
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Worked solution
Chloride ions are not strong enough reducing agents to reduce concentrated sulfuric acid; only an acid-base reaction occurs, producing HCl gas. Bromide ions are stronger reducing agents and reduce sulfuric acid to sulfur dioxide, \(\text{SO}_2\). Iodide ions are even stronger reducing agents and reduce sulfuric acid to a mixture of reduction products, which includes \(\text{SO}_2\), elemental sulfur (S), and hydrogen sulfide (\(\text{H}_2\text{S}\)). Therefore, sulfur dioxide is a reduction product for both sodium bromide and sodium iodide.
Marking scheme
1 mark for the correct option. Award 1 mark for recognizing that bromide and iodide are oxidized by concentrated sulfuric acid, producing sulfur dioxide as a reduction product, whereas chloride is not.
Question 5 · Multiple Choice
1 marks
In the reaction between chlorine gas and hot aqueous sodium hydroxide, chlorine undergoes disproportionation: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\). What are the oxidation numbers of chlorine in the reactant and the two chlorine-containing products?
A.Reactant: 0; Products: -1 and +1
B.Reactant: 0; Products: -1 and +3
C.Reactant: 0; Products: -1 and +5
D.Reactant: -1; Products: 0 and +5
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Worked solution
In the reactant, diatomic chlorine gas (\(\text{Cl}_2\)) is an element in its standard state, so its oxidation number is 0. In sodium chloride (\(\text{NaCl}\)), the sodium ion has an oxidation number of +1, so the chloride ion (\(\text{Cl}^-\)) has an oxidation number of -1. In sodium chlorate(V) (\(\text{NaClO}_3\)), let \(x\) be the oxidation number of chlorine: \(+1 + x + 3(-2) = 0 \Rightarrow 1 + x - 6 = 0 \Rightarrow x = +5\). Thus, the oxidation numbers are 0, -1, and +5.
Marking scheme
1 mark for the correct option. Award 1 mark for correctly determining the oxidation number of chlorine in Cl2 as 0, in NaCl as -1, and in NaClO3 as +5.
Question 6 · Multiple Choice
1 marks
Complete combustion of 0.200 mol of a hydrocarbon \(Y\) produced 1.60 mol of carbon dioxide gas and 1.60 mol of water vapor. What is the molecular formula of hydrocarbon \(Y\)?
A.C4H8
B.C8H8
C.C8H16
D.C16H16
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Worked solution
First, calculate the moles of carbon and hydrogen atoms produced per mole of hydrocarbon \(Y\). The moles of carbon in 1.60 mol of \(\text{CO}_2\) is 1.60 mol. Since this came from 0.200 mol of \(Y\), 1 mole of \(Y\) contains \(\frac{1.60}{0.200} = 8\) moles of carbon atoms. The moles of hydrogen in 1.60 mol of \(\text{H}_2\text{O}\) is \(2 \times 1.60 = 3.20\text{ mol}\). Since this came from 0.200 mol of \(Y\), 1 mole of \(Y\) contains \(\frac{3.20}{0.200} = 16\) moles of hydrogen atoms. Therefore, the molecular formula of \(Y\) is \(\text{C}_8\text{H}_{16}\).
Marking scheme
1 mark for the correct option. Award 1 mark for calculating the molar ratio of C and H per mole of the compound and identifying the correct molecular formula.
Question 7 · Multiple Choice
1 marks
Consider the gas-phase reaction: \(\text{CH}_4(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{CH}_3\text{Cl}(\text{g}) + \text{HCl}(\text{g})\). The average bond energies are: \(\text{C}-\text{H} = 413\text{ kJ mol}^{-1}\), \(\text{Cl}-\text{Cl} = 242\text{ kJ mol}^{-1}\), \(\text{C}-\text{Cl} = 339\text{ kJ mol}^{-1}\), \(\text{H}-\text{Cl} = 431\text{ kJ mol}^{-1}\). What is the enthalpy change, \(\Delta H\), for this reaction?
A.-115 kJ mol^-1
B.+115 kJ mol^-1
C.-1425 kJ mol^-1
D.-23 kJ mol^-1
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Worked solution
The reaction involves breaking 1 mole of \(\text{C}-\text{H}\) bonds and 1 mole of \(\text{Cl}-\text{Cl}\) bonds, and forming 1 mole of \(\text{C}-\text{Cl}\) bonds and 1 mole of \(\text{H}-\text{Cl}\) bonds. Enthalpy change, \(\Delta H = \sum\text{Bond energies of reactants} - \sum\text{Bond energies of products}\). \(\Delta H = [413 + 242] - [339 + 431] = 655 - 770 = -115\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option. Award 1 mark for calculating the total energy required to break bonds (655 kJ), the total energy released when making bonds (770 kJ), and finding the correct difference of -115 kJ/mol.
Question 8 · Multiple Choice
1 marks
Which of the following alkenes can exhibit geometrical (cis-trans) isomerism?
A.2-methylbut-2-ene
B.1,1-dichloroprop-1-ene
C.pent-2-ene
D.2-methylpropene
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Worked solution
For an alkene to exhibit cis-trans isomerism, each carbon atom of the double bond must be bonded to two different groups. In 2-methylbut-2-ene, one double-bonded carbon is bonded to two methyl groups, so no. In 1,1-dichloroprop-1-ene, one double-bonded carbon has two chlorine atoms, so no. In 2-methylpropene, one carbon is bonded to two hydrogen atoms, and the other is bonded to two methyl groups, so no. In pent-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\)), C2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups), and C3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\) (two different groups). Therefore, pent-2-ene can exhibit cis-trans isomerism.
Marking scheme
1 mark for the correct option. Award 1 mark for analyzing the groups attached to each double-bonded carbon atom and identifying pent-2-ene as the only isomer that meets the criteria.
Question 9 · multiple-choice
1 marks
An anhydrous metal carbonate, \(\text{M}_2\text{CO}_3\), has a mass of \(0.552\text{ g}\). It reacts completely with excess dilute hydrochloric acid.
The carbon dioxide gas collected at room temperature and pressure (r.t.p.) has a volume of \(96.0\text{ cm}^3\).
What is the identity of the metal, \(\text{M}\)? [Molar volume of gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\)]
A.Sodium
B.Potassium
C.Rubidium
D.Lithium
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Worked solution
1. Calculate the number of moles of \(\text{CO}_2\) gas produced: \[ n(\text{CO}_2) = \frac{96.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 4.00 \times 10^{-3}\text{ mol} \]
2. Determine the number of moles of \(\text{M}_2\text{CO}_3\) that reacted, using the 1:1 stoichiometry of the reaction: \[ n(\text{M}_2\text{CO}_3) = n(\text{CO}_2) = 4.00 \times 10^{-3}\text{ mol} \]
3. Calculate the molar mass (\(M_r\)) of \(\text{M}_2\text{CO}_3\): \[ M_r(\text{M}_2\text{CO}_3) = \frac{\text{mass}}{n} = \frac{0.552\text{ g}}{4.00 \times 10^{-3}\text{ mol}} = 138.0\text{ g mol}^{-1} \]
4. Calculate the relative atomic mass (\(A_r\)) of the metal \(\text{M}\): \[ 2 \times A_r(\text{M}) + 12.0 + 3(16.0) = 138.0 \] \[ 2 \times A_r(\text{M}) + 60.0 = 138.0 \] \[ 2 \times A_r(\text{M}) = 78.0 \] \[ A_r(\text{M}) = 39.0 \]
This relative atomic mass corresponds closely to Potassium (\(A_r = 39.1\)).
Marking scheme
1 mark for the correct identification of potassium. - Correctly calculating moles of gas (0.0040 mol): [1 method mark] - Correctly calculating the molar mass of the carbonate (138.0 g/mol): [1 method mark] - Correctly calculating the Ar of the metal (39.0) and identifying as potassium: [1 accuracy mark]
Question 10 · multiple-choice
1 marks
What is the total number of ions present in a \(1.64\text{ g}\) sample of anhydrous sodium phosphate, \(\text{Na}_3\text{PO}_4\)? [Avogadro constant, \(L = 6.02 \times 10^{23}\text{ mol}^{-1}\)]
A.6.02 \times 10^{21}
B.1.81 \times 10^{22}
C.2.41 \times 10^{22}
D.4.82 \times 10^{22}
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Worked solution
1. Calculate the molar mass of \(\text{Na}_3\text{PO}_4\): \[ M_r = 3(23.0) + 31.0 + 4(16.0) = 69.0 + 31.0 + 64.0 = 164.0\text{ g mol}^{-1} \]
2. Calculate the moles of \(\text{Na}_3\text{PO}_4\): \[ n = \frac{1.64\text{ g}}{164.0\text{ g mol}^{-1}} = 0.0100\text{ mol} \]
3. Determine the total moles of ions per mole of formula units: Anhydrous sodium phosphate dissociates according to: \(\text{Na}_3\text{PO}_4 \rightarrow 3\text{Na}^+ + \text{PO}_4^{3-}\), producing 4 ions per formula unit.
4. Calculate the total number of ions: \[ \text{Total ions} = 0.0100\text{ mol} \times 4 \times (6.02 \times 10^{23}\text{ mol}^{-1}) = 2.41 \times 10^{22} \]
Marking scheme
1 mark for the correct answer. - Method: Calculating moles of formula units (0.0100 mol) and multiplying by 4 to find total moles of ions (0.0400 mol). - Accuracy: Multiplying by Avogadro's constant to get 2.41 x 10^22.
Question 11 · multiple-choice
1 marks
In a calorimetry experiment, \(50.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) copper(II) sulfate solution is placed in a polystyrene cup. An excess of zinc powder is added, and the temperature increases by \(10.5\text{ }^\circ\text{C}\).
Assume that the density of the solution is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). Heat losses to the surroundings and the heat capacity of the zinc are negligible.
What is the enthalpy change, in \(\text{kJ mol}^{-1}\), for the reaction below?
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Worked solution
1. Calculate the heat released (\(q\)) during the reaction: \[ q = m c \Delta T \] Since the density is \(1.00\text{ g cm}^{-3}\), the mass of \(50.0\text{ cm}^3\) of solution is \(50.0\text{ g}\). \[ q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.5\text{ K} = 2194.5\text{ J} = 2.1945\text{ kJ} \]
3. Calculate the enthalpy change (\(\Delta H\)) per mole of reaction: Since the temperature increased, the reaction is exothermic, meaning \(\Delta H\) is negative. \[ \Delta H = -\frac{q}{n} = -\frac{2.1945\text{ kJ}}{0.0100\text{ mol}} = -219.45\text{ kJ mol}^{-1} \approx -219\text{ kJ mol}^{-1} \]
Marking scheme
1 mark for the correct answer of -219 kJ mol^-1. - Correct calculation of heat energy released (2.19 kJ): [1 method mark] - Correct calculation of moles of copper(II) ions (0.0100 mol): [1 method mark] - Correct calculation of exothermic enthalpy change to 3 sig figs: [1 accuracy mark]
Question 12 · multiple-choice
1 marks
Standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), are given below: - \(\text{C}_2\text{H}_2\text{(g)}\): \(-1300\text{ kJ mol}^{-1}\) - \(\text{C}_6\text{H}_6\text{(l)}\): \(-3268\text{ kJ mol}^{-1}\)
What is the standard enthalpy change for the trimerisation of ethyne to benzene?
1 mark for the correct enthalpy change value. - Correct application of Hess's Law formula (3 x reactants - products): [1 method mark] - Correct final sign and calculation (-632 kJ mol^-1): [1 accuracy mark]
Question 13 · multiple-choice
1 marks
An organic compound \(\text{X}\) has the molecular formula \(\text{C}_6\text{H}_{12}\). When \(\text{X}\) is heated with concentrated, acidified manganate(VII) ions, the only organic products formed are propanone, \(\text{CH}_3\text{COCH}_3\), and propanoic acid, \(\text{CH}_3\text{CH}_2\text{CO}_2\text{H}\).
Which alkene is compound \(\text{X}\)?
A.2-methylpent-2-ene
B.3-methylpent-2-ene
C.hex-2-ene
D.hex-3-ene
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Worked solution
Under hot, concentrated, acidified \(\text{KMnO}_4\), the carbon-carbon double bond of an alkene undergoes oxidative cleavage: - \(\text{R}_1\text{R}_2\text{C}=\text{CH-R}_3\) cleaves to form a ketone (\(\text{R}_1\text{CO-R}_2\)) and a carboxylic acid (\(\text{R}_3\text{CO}_2\text{H}\)).
Analyzing the options: - For **2-methylpent-2-ene** (\((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\)): - The left side has a quaternary double-bonded carbon with two methyl groups, yielding propanone, \(\text{CH}_3\text{COCH}_3\). - The right side has a tertiary double-bonded carbon bonded to an ethyl group, yielding propanoic acid, \(\text{CH}_3\text{CH}_2\text{CO}_2\text{H}\). This perfectly matches the product profile.
Marking scheme
1 mark for the correct option A. - Reject B because 3-methylpent-2-ene produces ethanoic acid and butanone. - Reject C because hex-2-ene produces ethanoic acid and butanoic acid. - Reject D because hex-3-ene produces only propanoic acid.
Question 14 · multiple-choice
1 marks
Which statement about the trends in properties of the Group 17 elements and their compounds from chlorine to iodine is correct?
A.The bond enthalpy of the hydrogen halides increases.
B.The thermal stability of the hydrogen halides increases.
C.The reducing ability of the halide ions increases.
D.The solubility of the silver halides in aqueous ammonia increases.
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Worked solution
Let's review the trends from chlorine to iodine: - **A is incorrect:** The atomic radius of the halogens increases, resulting in longer and weaker \(\text{H-X}\) covalent bonds. Thus, the bond enthalpy of the hydrogen halides *decreases*. - **B is incorrect:** Since the \(\text{H-X}\) bond strength decreases, the thermal stability of the hydrogen halides *decreases* (HI decomposes much more readily upon heating than HCl). - **C is correct:** The reducing strength of halide ions *increases* down the group. As the ionic radius increases, outer electrons are further from the nucleus and are more easily donated. - **D is incorrect:** The solubility of the silver halides in aqueous ammonia *decreases* (AgCl is soluble in dilute NH3, AgBr is soluble in concentrated NH3, and AgI is insoluble in concentrated NH3).
Marking scheme
1 mark for identifying C as the correct statement.
Question 15 · multiple-choice
1 marks
When solid sodium chloride and solid sodium iodide are separately reacted with concentrated sulfuric acid, different products are observed.
Which statement describes a correct observation or explains the difference in these reactions?
A.Hydrogen chloride is oxidized by concentrated sulfuric acid to form chlorine gas.
B.Hydrogen iodide is oxidized by concentrated sulfuric acid to form iodine.
C.Concentrated sulfuric acid acts as a reducing agent in its reaction with sodium iodide.
D.Sodium chloride is a stronger reducing agent than sodium iodide.
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Worked solution
- **A is incorrect:** Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize chloride ions to chlorine gas. Only misty fumes of HCl gas (an acid-base reaction) are produced. - **B is correct:** Iodide ions are very strong reducing agents and reduce concentrated sulfuric acid to a range of products (such as \(\text{SO}_2\), \(\text{S}\), or \(\text{H}_2\text{S}\)), while being oxidized to iodine (\(\text{I}_2\)), which appears as a purple vapor or dark grey solid. - **C is incorrect:** Concentrated sulfuric acid is acting as an *oxidizing* agent in this reaction, while the iodide ion is the reducing agent. - **D is incorrect:** Sodium iodide (iodide ion) is a *stronger* reducing agent than sodium chloride (chloride ion).
Marking scheme
1 mark for identifying B as the correct statement.
Question 16 · multiple-choice
1 marks
The reaction between iodate(V) ions, \(\text{IO}_3^-\), and iodide ions, \(\text{I}^-\), in acidic solution produces iodine, \(\text{I}_2\), and water.
When this equation is balanced with the simplest whole-number coefficients, what is the value of the coefficient \(d\)?
A.1
B.3
C.5
D.6
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Worked solution
Let's use oxidation numbers to balance the redox equation: 1. Find the oxidation state changes: - Iodine in \(\text{IO}_3^-\): from +5 to 0 in \(\text{I}_2\) (reduction, change of \(-5\) per iodine atom). - Iodine in \(\text{I}^-\): from -1 to 0 in \(\text{I}_2\) (oxidation, change of \(+1\) per iodine atom).
2. Balance the changes in oxidation number: To balance the electron transfer, we need 5 \(\text{I}^-\) ions for every 1 \(\text{IO}_3^-\) ion. \[ \text{IO}_3^- + 5\text{I}^- \rightarrow 3\text{I}_2 \]
3. Balance oxygen and hydrogen atoms: - There are 3 oxygen atoms on the left, so we need 3 \(\text{H}_2\text{O}\) on the right: \[ \text{IO}_3^- + 5\text{I}^- \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O} \] - To balance the hydrogen, we add 6 \(\text{H}^+\) to the left: \[ \text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O} \]
Checking charges: \((-1) + 5(-1) + 6(+1) = 0\), which is balanced. Thus, the simplest whole-number coefficient for \(\text{I}_2\) (\(d\)) is 3.
Marking scheme
1 mark for selecting the correct coefficient '3' (Option B). - Correct method of identifying oxidation states (+5, -1, 0): [1 method mark] - Correct balancing of atoms and charges to yield coefficient 3: [1 accuracy mark]
Question 17 · Multiple Choice
1 marks
When 1.64 g of an anhydrous metal nitrate, \(M(\text{NO}_3)_2\), is completely decomposed by heating, 600 cm\(^3\) of gas is collected at room temperature and pressure. What is the identity of the metal \(M\)? (Take the molar volume of gas at r.t.p. as 24.0 dm\(^3\) mol\(^{-1}\))
A.Magnesium
B.Calcium
C.Strontium
D.Barium
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Worked solution
First, determine the moles of gas collected: moles of gas = \(600 \text{ cm}^3 / 24000 \text{ cm}^3 \text{ mol}^{-1} = 0.025 \text{ mol}\). The equation for the decomposition of a divalent metal nitrate is: \(2M(\text{NO}_3)_2(\text{s}) \rightarrow 2MO(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\). For every 2 moles of \(M(\text{NO}_3)_2\) decomposed, 5 moles of gas (4 moles of \(\text{NO}_2\) and 1 mole of \(\text{O}_2\)) are produced. Therefore, the moles of \(M(\text{NO}_3)_2\) decomposed = \(0.025 \times \frac{2}{5} = 0.010 \text{ mol}\). The molar mass of \(M(\text{NO}_3)_2 = 1.64 \text{ g} / 0.010 \text{ mol} = 164 \text{ g mol}^{-1}\). The formula mass of the nitrate ions is \(2 \times [14.0 + (3 \times 16.0)] = 124 \text{ g mol}^{-1}\). Thus, the relative atomic mass of \(M = 164 - 124 = 40\), which corresponds to Calcium (\(A_r = 40.1\)).
Marking scheme
1 mark for the correct option. Award 1 mark for correct calculation steps: identifying gas moles as 0.025 mol, scaling to find nitrate moles as 0.010 mol, calculating molar mass as 164 g/mol, and identifying calcium as the metal.
Question 18 · Multiple Choice
1 marks
A 20.0 cm\(^3\) sample of a 0.0500 mol dm\(^{-3}\) solution of an acid, \(\text{H}_n\text{A}\), reacts completely with 30.0 cm\(^3\) of a 0.100 mol dm\(^{-3}\) solution of sodium hydroxide, \(\text{NaOH}\). What is the basicity, \(n\), of the acid?
A.1
B.2
C.3
D.4
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Worked solution
First, calculate the moles of acid in the reaction: moles of \(\text{H}_n\text{A} = 20.0 \times 10^{-3} \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 0.00100 \text{ mol}\). Next, calculate the moles of \(\text{NaOH}\) used: moles of \(\text{NaOH} = 30.0 \times 10^{-3} \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.00300 \text{ mol}\). The reacting mole ratio of acid to base is \(\text{H}_n\text{A} : \text{NaOH} = 0.00100 : 0.00300 = 1 : 3\). This means 3 moles of \(\text{OH}^-\)\ ions are required to fully neutralise 1 mole of \(\text{H}_n\text{A}\). Thus, the acid contains 3 acidic protons per formula unit, making \(n = 3\).
Marking scheme
1 mark for the correct option. Award 1 mark for calculating the correct moles of acid and alkali, finding the 1:3 ratio, and concluding the basicity is 3.
Question 19 · Multiple Choice
1 marks
The standard enthalpy changes of combustion, \(\Delta H^\theta_c\), for carbon (graphite), hydrogen gas, and liquid ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) are \(-394\text{ kJ mol}^{-1}\), \(-286\text{ kJ mol}^{-1}\), and \(-1368\text{ kJ mol}^{-1}\) respectively. What is the standard enthalpy change of formation, \(\Delta H^\theta_f\), of liquid ethanol?
A.-278 kJ mol^{-1}
B.-1646 kJ mol^{-1}
C.+278 kJ mol^{-1}
D.-688 kJ mol^{-1}
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Worked solution
The equation for the formation of liquid ethanol is: \(2\text{C(s, graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\). Using the Hess cycle relation based on standard enthalpies of combustion, \(\Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). Therefore, \(\Delta H^\theta_f = [2 \times (-394) + 3 \times (-286)] - [-1368] = [-788 - 858] + 1368 = -1646 + 1368 = -278\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option. Award 1 mark for applying Hess's Law using combustion values, scaling carbon by 2 and hydrogen by 3, and subtracting the combustion value of ethanol to arrive at -278 kJ/mol.
Question 20 · Multiple Choice
1 marks
Using the average bond energies provided below, what is the enthalpy change for the gas-phase reaction: \(\text{CO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(g)}\)? (Bond energies: \(\text{C}\equiv\text{O}\) in \(\text{CO} = 1077\text{ kJ mol}^{-1}\); \(\text{H}-\text{H} = 436\text{ kJ mol}^{-1}\); \(\text{C}-\text{H} = 410\text{ kJ mol}^{-1}\); \(\text{C}-\text{O} = 360\text{ kJ mol}^{-1}\); \(\text{O}-\text{H} = 460\text{ kJ mol}^{-1}\))
A.-101 kJ mol^{-1}
B.+101 kJ mol^{-1}
C.-221 kJ mol^{-1}
D.-537 kJ mol^{-1}
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Worked solution
The enthalpy change of reaction is calculated as \(\Delta H = \sum \text{bond energies of reactants (bonds broken)} - \sum \text{bond energies of products (bonds formed)}\). Bonds broken: \(1 \times (\text{C}\equiv\text{O}) + 2 \times (\text{H}-\text{H}) = 1077 + 2(436) = 1949\text{ kJ mol}^{-1}\). Bonds formed in methanol: \(3 \times (\text{C}-\text{H}) + 1 \times (\text{C}-\text{O}) + 1 \times (\text{O}-\text{H}) = 3(410) + 360 + 460 = 2050\text{ kJ mol}^{-1}\). Therefore, \(\Delta H = 1949 - 2050 = -101\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option. Award 1 mark for finding total energy required to break bonds (1949 kJ/mol), total energy released forming bonds (2050 kJ/mol), and correctly subtracting to find -101 kJ/mol.
Question 21 · Multiple Choice
1 marks
An organic compound, \(W\), has the molecular formula \(\text{C}_5\text{H}_{10}\). \(W\) reacts with hydrogen gas in the presence of a nickel catalyst to form 2-methylbutane. When \(W\) is reacted with hot, concentrated, acidified potassium manganate(VII), carbon dioxide gas and a ketone are formed. What is the IUPAC name of \(W\)?
A.2-methylbut-1-ene
B.3-methylbut-1-ene
C.2-methylbut-2-ene
D.pent-1-ene
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Worked solution
Since hydrogenation of \(W\) yields 2-methylbutane, the carbon skeleton of \(W\) must contain a branched chain: \(\text{CH}_3-\text{CH(CH}_3)-\text{CH}_2-\text{CH}_3\). The reaction with hot, concentrated, acidified \(\text{KMnO}_4\) yielding carbon dioxide gas implies the presence of a terminal \(=\text{CH}_2\) group. The two possible terminal alkenes with this carbon skeleton are 3-methylbut-1-ene and 2-methylbut-1-ene. Oxidation of 3-methylbut-1-ene, \(\text{CH}_2=\text{CHCH(CH}_3)_2\), yields carbon dioxide and 2-methylpropanoic acid (a carboxylic acid). Oxidation of 2-methylbut-1-ene, \(\text{CH}_2=\text{C(CH}_3)\text{CH}_2\text{CH}_3\), yields carbon dioxide and butanone (a ketone). Therefore, \(W\) must be 2-methylbut-1-ene.
Marking scheme
1 mark for the correct option. Award 1 mark for identifying the correct skeletal structure from hydrogenation, matching the KMnO4 products with the structures, and concluding that 2-methylbut-1-ene is the only compound satisfying all conditions.
Question 22 · Multiple Choice
1 marks
When 2-methylbut-2-ene reacts with cold, dilute, alkaline potassium manganate(VII), a diol is formed. How many chiral carbon atoms are present in one molecule of this diol product?
A.0
B.1
C.2
D.3
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Worked solution
2-methylbut-2-ene is \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\). Reacting it with cold, dilute, alkaline \(\text{KMnO}_4\) adds two \(-\text{OH}\) groups across the double bond to form the diol \(\text{(CH}_3)_2\text{C(OH)}-\text{CH(OH)}-\text{CH}_3\). Let us examine the carbon atoms: C2 is bonded to two identical methyl groups, one \(-\text{OH}\) group, and one \(-\text{CH(OH)CH}_3\) group, so it is not chiral. C3 is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{C(OH)(CH}_3)_2\). Thus, C3 is a chiral carbon. There is exactly 1 chiral carbon atom in the molecule.
Marking scheme
1 mark for the correct option. Award 1 mark for drawing the correct product structure and correctly identifying that only C3 is a chiral center.
Question 23 · Multiple Choice
1 marks
Which statement explains why the boiling point of iodine is significantly higher than that of chlorine?
A.The covalent bond in the \(\text{I}_2\) molecule is much stronger than the covalent bond in the \(\text{Cl}_2\) molecule.
B.The \(\text{I}_2\) molecules are larger and have more electrons, resulting in stronger instantaneous dipole-induced dipole forces.
C.There are permanent dipole-dipole forces between \(\text{I}_2\) molecules but only instantaneous dipole-induced dipole forces between \(\text{Cl}_2\) molecules.
D.Iodine exists as a giant covalent macromolecular structure whereas chlorine exists as simple molecules.
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Worked solution
Both iodine and chlorine are simple molecular substances. Boiling them involves overcoming the weak intermolecular forces between molecules, not breaking covalent bonds. Because iodine molecules (\(\text{I}_2\)) have significantly more electrons than chlorine molecules (\(\text{Cl}_2\)), they have larger, more polarisable electron clouds. This results in stronger instantaneous dipole-induced dipole forces (London dispersion forces) between iodine molecules, requiring more thermal energy to overcome.
Marking scheme
1 mark for the correct option. Award 1 mark for identifying that the trend in boiling points is due to the increase in the number of electrons causing stronger instantaneous dipole-induced dipole forces.
Question 24 · Multiple Choice
1 marks
When chlorine gas is reacted with hot concentrated sodium hydroxide solution, a disproportionation reaction occurs. What are the oxidation numbers of chlorine in the two chlorine-containing products formed?
A.-1 and +1
B.-1 and +3
C.-1 and +5
D.-3 and +5
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Worked solution
The reaction of chlorine with hot concentrated sodium hydroxide is: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\). In sodium chloride (\(\text{NaCl}\)), the oxidation number of chlorine is \(-1\). In sodium chlorate(V) (\(\text{NaClO}_3\)), the oxidation number of chlorine is determined by: \(+1 + x + 3(-2) = 0 \Rightarrow x = +5\). Therefore, the oxidation states of the products are \(-1\) and \(+5\).
Marking scheme
1 mark for the correct option. Award 1 mark for recalling the hot concentrated alkali disproportionation equation and correctly calculating the oxidation numbers as -1 and +5.
Question 25 · Multiple Choice
1 marks
A gaseous hydrocarbon, \(C_xH_y\), has a volume of \(10\text{ cm}^3\). It is completely combusted in \(100\text{ cm}^3\) (an excess) of oxygen gas. After the reaction, the mixture is cooled to room temperature, and the final volume of gas is \(85\text{ cm}^3\). When this gas mixture is passed through concentrated aqueous sodium hydroxide, the volume of gas reduces to \(55\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
A.\(C_3H_4\)
B.\(C_3H_6\)
C.\(C_3H_8\)
D.\(C_4H_{10}\)
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Worked solution
First, find the volume of \(CO_2\) produced. Passing the gas through aqueous NaOH absorbs the acidic \(CO_2\) gas, causing a decrease in volume of \(85\text{ cm}^3 - 55\text{ cm}^3 = 30\text{ cm}^3\). Since \(10\text{ cm}^3\) of the hydrocarbon was used, the ratio of hydrocarbon to \(CO_2\) is \(10 : 30 = 1 : 3\), so \(x = 3\).
Second, find the volume of oxygen that reacted. The final gas remaining after NaOH absorption is unreacted excess oxygen, which is \(55\text{ cm}^3\). Thus, the volume of oxygen reacted is \(100\text{ cm}^3 - 55\text{ cm}^3 = 45\text{ cm}^3\).
Using the general combustion equation: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\)
The mole ratio of hydrocarbon to reacted \(O_2\) is \(10 : 45 = 1 : 4.5\). Therefore, \(x + \frac{y}{4} = 4.5\). Since \(x = 3\), we have \(3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\).
The molecular formula of the hydrocarbon is \(C_3H_6\).
Marking scheme
1 mark for selecting the correct option B. - Award 1 mark for calculating the volume of carbon dioxide (30 cm3) and reacted oxygen (45 cm3) to deduce the molecular formula C3H6.
Question 26 · Multiple Choice
1 marks
What is the standard enthalpy of formation of liquid ethanol, \(C_2H_5OH(l)\), given the following standard enthalpy of combustion values?
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Worked solution
The equation for the formation of liquid ethanol is: \(2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\)
Using Hess's law and enthalpy of combustion values: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)
1 mark for selecting the correct option A. - Award 1 mark for correct application of Hess's Law using combustion data: 2 x (-394) + 3 x (-286) - (-1367) = -279 kJ mol-1.
Question 27 · Multiple Choice
1 marks
Pent-1-ene reacts with gaseous hydrogen bromide, \(HBr\), via electrophilic addition to yield a major organic product. Which statement regarding this major product is correct?
A.It is a chiral molecule that can exist as a pair of enantiomers.
B.It is an achiral molecule that does not show stereoisomerism.
C.It is a symmetrical dihalogenoalkane.
D.It is formed via a primary carbocation intermediate.
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Worked solution
During the electrophilic addition of \(HBr\) to pent-1-ene (\(CH_3CH_2CH_2CH=CH_2\)), the hydrogen atom adds to the carbon with more hydrogen atoms (C1) to form the more stable secondary carbocation intermediate, \(CH_3CH_2CH_2C^+HCH_3\). The bromide ion then bonds to C2 to form the major product, 2-bromopentane.
Carbon-2 in 2-bromopentane is bonded to four different groups: \(-H\), \(-Br\), \(-CH_3\), and \(-CH_2CH_2CH_3\). Therefore, it is a chiral center, and the major product exists as a pair of enantiomers (optical isomers).
Marking scheme
1 mark for selecting the correct option A. - Award 1 mark for identifying the major product as 2-bromopentane and recognizing that it contains a chiral carbon and thus exists as a pair of enantiomers.
Question 28 · Multiple Choice
1 marks
A solid sodium halide, \(NaX\), is reacted with concentrated sulfuric acid. A gaseous mixture is evolved. This mixture contains a gas that turns damp blue litmus paper red, and a yellow solid deposit is formed on the cool upper walls of the test tube. What is the identity of the halide ion, \(X^-\)?
A.fluoride
B.chloride
C.bromide
D.iodide
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Worked solution
The reaction of a solid sodium halide with concentrated sulfuric acid demonstrates the trend in reducing ability of the halide ions. - Chloride and fluoride are weak reducing agents and do not reduce \(H_2SO_4\). - Bromide reduces \(H_2SO_4\) to sulfur dioxide (\(SO_2\), a colourless gas). - Iodide (\(I^-\)) is a very strong reducing agent and reduces sulfur in \(H_2SO_4\) from oxidation state +6 to +4 (\(SO_2\)), 0 (elemental sulfur, which is a yellow solid), and -2 (\(H_2S\)).
Therefore, the yellow solid deposit is sulfur, confirming the halide is iodide.
Marking scheme
1 mark for selecting the correct option D. - Award 1 mark for identifying that only the iodide ion is a strong enough reducing agent to reduce sulfuric acid to elemental sulfur (yellow solid).
Question 29 · Multiple Choice
1 marks
In acidic solution, potassium dichromate(VI) reacts with tin(II) ions according to the ionic equation below:
Which statement about this redox reaction is correct?
A.Chromium is oxidized because its oxidation state changes from +3 to +6.
B.Tin(II) acts as a reducing agent and is oxidized.
C.Hydrogen ions are reduced to water.
D.For every mole of dichromate(VI) ions reacted, 3 moles of electrons are transferred.
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Worked solution
- Tin(II) ions (\(Sn^{2+}\)) are oxidized to tin(IV) ions (\(Sn^{4+}\)) as their oxidation number increases from +2 to +4. Because it is oxidized, \(Sn^{2+}\) acts as the reducing agent. - Chromium in \(Cr_2O_7^{2-}\) (oxidation state +6) is reduced to \(Cr^{3+}\) (oxidation state +3). - Hydrogen ions (\(H^+\)) maintain an oxidation state of +1 on both sides. - One mole of dichromate(VI) ions gains 6 moles of electrons during reduction to 2 moles of \(Cr^{3+}\).
Marking scheme
1 mark for selecting the correct option B. - Award 1 mark for identifying that Tin(II) is oxidized and therefore acts as the reducing agent.
Question 30 · Multiple Choice
1 marks
A \(25.0\text{ cm}^3\) sample of sodium carbonate, \(Na_2CO_3\), solution required exactly \(20.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) hydrochloric acid for complete neutralization.
What is the concentration of the sodium carbonate solution in \(g\text{ dm}^{-3}\)? [Molar mass: \(Na_2CO_3 = 106.0\text{ g mol}^{-1}\)]
A.\(3.18\text{ g dm}^{-3}\)
B.\(6.36\text{ g dm}^{-3}\)
C.\(12.72\text{ g dm}^{-3}\)
D.\(25.44\text{ g dm}^{-3}\)
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Worked solution
1. Calculate the moles of \(HCl\) used: \(n(HCl) = c \times V = 0.150\text{ mol dm}^{-3} \times 0.0200\text{ dm}^3 = 0.00300\text{ mol}\)
2. Determine the moles of \(Na_2CO_3\) from the stoichiometry (1:2 ratio): \(n(Na_2CO_3) = \frac{0.00300}{2} = 0.00150\text{ mol}\)
3. Calculate the concentration of \(Na_2CO_3\) in \(\text{mol dm}^{-3}\) in the \(25.0\text{ cm}^3\) sample: \(c(Na_2CO_3) = \frac{0.00150\text{ mol}}{0.0250\text{ dm}^3} = 0.0600\text{ mol dm}^{-3}\)
4. Convert the concentration to \(\text{g dm}^{-3}\): \(\text{Concentration} = 0.0600\text{ mol dm}^{-3} \times 106.0\text{ g mol}^{-1} = 6.36\text{ g dm}^{-3}\)
Marking scheme
1 mark for selecting the correct option B. - Award 1 mark for finding the correct concentration value of 6.36 g dm-3 through step-by-step mole conversions.
Question 31 · Multiple Choice
1 marks
In a calorimetry experiment, \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) sodium hydroxide solution was mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid in a polystyrene cup. The temperature of the mixture rose from \(21.5^\circ\text{C}\) to a maximum of \(28.3^\circ\text{C}\).
Assume the density of the solution is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).
What is the standard enthalpy of neutralisation, \(\Delta H_{neut}\), for this reaction?
A.\(-56.8\text{ kJ mol}^{-1}\)
B.\(-113.7\text{ kJ mol}^{-1}\)
C.\(+56.8\text{ kJ mol}^{-1}\)
D.\(-28.4\text{ kJ mol}^{-1}\)
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Worked solution
1. Calculate the heat energy released, \(q\): \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\) \(\Delta T = 28.3 - 21.5 = 6.8\text{ K}\) \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.8424\text{ kJ}\)
2. Calculate the moles of water formed: \(n(H_2O) = n(H^+) = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
1 mark for selecting the correct option A. - Award 1 mark for calculating the correct heat energy (2.84 kJ) and dividing by the correct moles of water (0.0500 mol) to get -56.8 kJ mol-1 (including the negative sign for an exothermic reaction).
Question 32 · Multiple Choice
1 marks
A portion of an addition polymer chain is shown below:
\(-CH(CH_3)-CH(CH_2CH_3)-CH(CH_3)-CH(CH_2CH_3)-\)
What is the IUPAC name of the monomer used to synthesize this polymer?
A.but-2-ene
B.pent-1-ene
C.pent-2-ene
D.2-methylbut-2-ene
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Worked solution
To find the monomer, identify the repeating unit in the polymer chain. The repeating unit is \(-CH(CH_3)-CH(CH_2CH_3)-\). Reforming the carbon-carbon double bond gives the monomer: \(CH(CH_3)=CH(CH_2CH_3)\)
This is written structurally as \(CH_3-CH=CH-CH_2-CH_3\). The longest continuous carbon chain has 5 carbon atoms, and the double bond starts at carbon-2. Therefore, the IUPAC name is pent-2-ene.
Marking scheme
1 mark for selecting the correct option C. - Award 1 mark for identifying pent-2-ene as the correct alkene monomer that forms the given polymer repeating unit.
Question 33 · Multiple Choice
1 marks
A student reacts \( 1.24\text{ g} \) of anhydrous copper(II) carbonate, \( \text{CuCO}_3 \), with \( 20.0\text{ cm}^3 \) of \( 1.50\text{ mol dm}^{-3} \) hydrochloric acid, \( \text{HCl}(aq) \). The equation is shown: \( \text{CuCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \). What is the maximum volume of carbon dioxide gas, in \( \text{cm}^3 \), produced at room temperature and pressure (r.t.p.)? [Molar volume of a gas at r.t.p. = \( 24.0\text{ dm}^3\text{ mol}^{-1} \); \( M_r(\text{CuCO}_3) = 123.5 \)]
A.120
B.241
C.360
D.482
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Worked solution
First, calculate the amount in moles of \( \text{CuCO}_3 \): \( n(\text{CuCO}_3) = 1.24\text{ g} / 123.5\text{ g mol}^{-1} = 0.01004\text{ mol} \). Next, calculate the amount in moles of \( \text{HCl} \): \( n(\text{HCl}) = 0.0200\text{ dm}^3 \times 1.50\text{ mol dm}^{-3} = 0.0300\text{ mol} \). From the equation, \( 1\text{ mol} \) of \( \text{CuCO}_3 \) requires \( 2\text{ mol} \) of \( \text{HCl} \). Therefore, \( 0.01004\text{ mol} \) of \( \text{CuCO}_3 \) would require \( 0.02008\text{ mol} \) of \( \text{HCl} \). Since we have \( 0.0300\text{ mol} \) of \( \text{HCl} \) available, \( \text{CuCO}_3 \) is the limiting reactant. The volume of \( \text{CO}_2 \) gas produced is therefore: \( V = 0.01004\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 241\text{ cm}^3 \).
Marking scheme
1 mark for identifying the correct limiting reactant as copper(II) carbonate and using the molar ratio to calculate the gas volume as 241 cm3.
Question 34 · Multiple Choice
1 marks
In a calorimetry experiment, a spirit burner containing propan-1-ol (\( M_r = 60.0 \)) is used to heat \( 150\text{ g} \) of water in a beaker. The temperature of the water increases from \( 20.5\text{ }^\circ\text{C} \) to \( 44.5\text{ }^\circ\text{C} \), while the mass of the spirit burner decreases by \( 0.450\text{ g} \). The specific heat capacity of water is \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \). What is the calculated enthalpy change of combustion of propan-1-ol?
A.-15.0 kJ mol^{-1}
B.-1340 kJ mol^{-1}
C.-2010 kJ mol^{-1}
D.-2680 kJ mol^{-1}
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Worked solution
Calculate the heat energy absorbed by the water: \( q = m c \Delta T = 150\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (44.5 - 20.5)\text{ K} = 15048\text{ J} = 15.048\text{ kJ} \). Calculate the moles of propan-1-ol burned: \( n = 0.450\text{ g} / 60.0\text{ g mol}^{-1} = 0.00750\text{ mol} \). The enthalpy change of combustion is \( \Delta H_c = -q / n = -15.048\text{ kJ} / 0.00750\text{ mol} = -2006.4\text{ kJ mol}^{-1} \), which rounds to \( -2010\text{ kJ mol}^{-1} \).
Marking scheme
1 mark for correct calculation of heat energy, moles of propan-1-ol, and division to give -2010 kJ mol-1.
Question 35 · Multiple Choice
1 marks
An organic compound, 2-methylbut-2-ene, is heated under reflux with a hot, concentrated, acidified solution of potassium manganate(VII). Which organic products are formed in this reaction?
A.propanone and ethanoic acid
B.propanone and ethanal
C.propanoic acid and ethanoic acid
D.propan-2-ol and ethanol
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Worked solution
The structure of 2-methylbut-2-ene is \( (CH_3)_2C=CHCH_3 \). Under harsh oxidation conditions using hot, concentrated, acidified \( \text{KMnO}_4 \), the double bond is completely cleaved. The carbon bonded to two methyl groups, \( (CH_3)_2C= \), is oxidized to the ketone, propanone. The carbon bonded to one methyl group and one hydrogen, \( =CHCH_3 \), is oxidized to the carboxylic acid, ethanoic acid.
Marking scheme
1 mark for correctly identifying the cleavage products of the alkene as propanone and ethanoic acid.
Question 36 · Multiple Choice
1 marks
Which statement describes the behavior of concentrated sulfuric acid in its reaction with solid sodium chloride?
A.It acts as an oxidizing agent, oxidizing chloride ions to chlorine gas.
B.It acts as a reducing agent, reducing chloride ions to chlorine gas.
C.It acts only as an acid, producing hydrogen chloride gas.
D.It does not react with solid sodium chloride.
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Worked solution
When concentrated sulfuric acid reacts with solid sodium chloride, an acid-base reaction occurs: \( \text{NaCl}(s) + \text{H}_2\text{SO}_4(l) \rightarrow \text{NaHSO}_4(s) + \text{HCl}(g) \). Chloride ions are not strong enough reducing agents to reduce concentrated sulfuric acid, so no redox reaction occurs. Therefore, the sulfuric acid behaves only as an acid (proton donor).
Marking scheme
1 mark for identifying that sulfuric acid acts only as an acid and does not oxidize chloride ions.
Question 37 · Multiple Choice
1 marks
The chemical equation shows a reaction of chlorate(I) ions: \( 3\text{ClO}^-(aq) \rightarrow \text{ClO}_3^-(aq) + 2\text{Cl}^-(aq) \). What are the changes in the oxidation states of chlorine in this reaction?
A.Chlorine is oxidized from +1 to +5 and reduced from +1 to -1.
B.Chlorine is oxidized from -1 to +5 and reduced from -1 to -2.
C.Chlorine is oxidized from +1 to +7 and reduced from +1 to -1.
D.Chlorine is oxidized from +3 to +5 and reduced from +3 to -1.
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Worked solution
In the reactant \( \text{ClO}^- \), the oxidation state of chlorine is +1. In the product \( \text{ClO}_3^- \), the oxidation state of chlorine is +5, representing oxidation. In the product \( \text{Cl}^- \), the oxidation state of chlorine is -1, representing reduction. This shows that chlorine is both oxidized from +1 to +5 and reduced from +1 to -1.
Marking scheme
1 mark for correctly determining all oxidation states and matching the disproportionation changes.
Question 38 · Multiple Choice
1 marks
An organic compound \( \text{X} \) contains only carbon, hydrogen, and oxygen. When \( 1.80\text{ g} \) of \( \text{X} \) is completely burned in excess oxygen, \( 2.64\text{ g} \) of carbon dioxide, \( \text{CO}_2 \), and \( 1.08\text{ g} \) of water, \( \text{H}_2\text{O} \), are formed. What is the empirical formula of \( \text{X} \)?
A.CHO
B.CH2O
C.C2H3O
D.C3H6O2
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Worked solution
Find mass of carbon: \( m(\text{C}) = (12.0 / 44.0) \times 2.64\text{ g} = 0.720\text{ g} \). Find mass of hydrogen: \( m(\text{H}) = (2.0 / 18.0) \times 1.08\text{ g} = 0.120\text{ g} \). Find mass of oxygen by subtraction: \( m(\text{O}) = 1.80\text{ g} - (0.720\text{ g} + 0.120\text{ g}) = 0.960\text{ g} \). Convert to moles: \( n(\text{C}) = 0.720 / 12.0 = 0.0600\text{ mol} \); \( n(\text{H}) = 0.120 / 1.0 = 0.120\text{ mol} \); \( n(\text{O}) = 0.960 / 16.0 = 0.0600\text{ mol} \). The mole ratio of \( \text{C} : \text{H} : \text{O} \) is \( 1 : 2 : 1 \), so the empirical formula is \( \text{CH}_2\text{O} \).
Marking scheme
1 mark for calculating correct masses and moles of all three elements, and finding the simplest ratio to be CH2O.
Question 39 · Multiple Choice
1 marks
The standard enthalpies of combustion of ethene, hydrogen, and ethane are given: \( \Delta H_c^\ominus[\text{C}_2\text{H}_4(g)] = -1411\text{ kJ mol}^{-1} \), \( \Delta H_c^\ominus[\text{H}_2(g)] = -286\text{ kJ mol}^{-1} \), \( \Delta H_c^\ominus[\text{C}_2\text{H}_6(g)] = -1560\text{ kJ mol}^{-1} \). What is the standard enthalpy change of the reaction: \( \text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \)?
A.-3257 kJ mol^{-1}
B.-137 kJ mol^{-1}
C.+137 kJ mol^{-1}
D.+3257 kJ mol^{-1}
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Worked solution
Using Hess\'s Law with standard enthalpies of combustion: \( \Delta H_r^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}) = \Delta H_c^\ominus[\text{C}_2\text{H}_4] + \Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{C}_2\text{H}_6] = -1411 + (-286) - (-1560) = -1697 + 1560 = -137\text{ kJ mol}^{-1} \).
Marking scheme
1 mark for applying Hess\'s Law correctly with the given combustion data to obtain -137 kJ mol-1.
Question 40 · Multiple Choice
1 marks
Which of the following alkenes can exist as a pair of cis-trans (E-Z) stereoisomers?
A.2-methylbut-2-ene
B.but-1-ene
C.2-methylpropene
D.pent-2-ene
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Worked solution
For an alkene to exhibit cis-trans isomerism, each carbon of the double bond must be bonded to two different atoms or groups. In pent-2-ene, \( CH_3CH=CHCH_2CH_3 \), carbon 2 is bonded to \( -\text{H} \) and \( -\text{CH}_3 \), and carbon 3 is bonded to \( -\text{H} \) and \( -\text{CH}_2\text{CH}_3 \). Since both carbons have two different groups attached, it can exist as cis-trans isomers. All other listed options have at least one double-bonded carbon with two identical groups.
Marking scheme
1 mark for correctly identifying that pent-2-ene satisfies the structural conditions for stereoisomerism.
Paper 2 AS Level Structured Questions
Answer all structured questions in the spaces provided. Show all working.
6 Question · 60 marks
Question 1 · Structured
10 marks
A sample of a hydrated basic magnesium carbonate has the formula \(\text{MgCO}_3 \cdot z\text{H}_2\text{O}\). A student heats a \(1.743\text{ g}\) sample of this hydrated salt strongly to constant mass. The thermal decomposition produces magnesium oxide (\(\text{MgO}\)), carbon dioxide gas, and water vapour. The mass of the solid residue (magnesium oxide) obtained is \(0.403\text{ g}\). (\(A_r\): \(\text{H} = 1.0\), \(\text{C} = 12.0\), \(\text{O} = 16.0\), \(\text{Mg} = 24.3\))
(a) Write a balanced chemical equation, including state symbols, for the complete thermal decomposition of \(\text{MgCO}_3 \cdot z\text{H}_2\text{O}(s)\). [1]
(b) Calculate the number of moles of \(\text{MgO}\) produced in this decomposition. [2]
(c) Show by calculation that the value of \(z\) is 5. [3]
(d) Calculate the volume, in \(\text{cm}^3\), of dry carbon dioxide gas that would be collected at room temperature and pressure (r.t.p.) from the complete decomposition of this \(1.743\text{ g}\) sample. (The molar volume of a gas at r.t.p. is \(24.0\text{ dm}^3\text{ mol}^{-1}\)). [2]
(e) Explain why the volume of gas collected would be significantly greater if the total gaseous products were measured immediately after heating, before cooling down to room temperature. [2]
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(b) \(M_r(\text{MgO}) = 24.3 + 16.0 = 40.3\text{ g mol}^{-1}\) \(\text{Moles of MgO} = \frac{0.403}{40.3} = 0.0100\text{ mol}\)
(c) Since 1 mole of \(\text{MgCO}_3 \cdot z\text{H}_2\text{O}\) produces 1 mole of \(\text{MgO}\): \(\text{Moles of hydrated salt} = 0.0100\text{ mol}\). \(M_r(\text{hydrated salt}) = \frac{1.743}{0.0100} = 174.3\text{ g mol}^{-1}\). \(M_r(\text{MgCO}_3) = 24.3 + 12.0 + 3(16.0) = 84.3\text{ g mol}^{-1}\). Mass of water of crystallisation per mole of salt = \(174.3 - 84.3 = 90.0\text{ g mol}^{-1}\). \(z = \frac{90.0}{18.0} = 5\).
(d) \(\text{Moles of CO}_2 = \text{Moles of MgO} = 0.0100\text{ mol}\). \(\text{Volume of CO}_2 = 0.0100 \times 24.0 = 0.240\text{ dm}^3 = 240\text{ cm}^3\).
(e) At higher temperatures: 1. The volume of the gas expands (due to Charles's Law / \(V \propto T\)). 2. Water is present as steam (gaseous water), which would add to the measured gas volume (at room temperature, water condenses to liquid and does not contribute to gas volume).
Marking scheme
(a) [1 mark] for: \(\text{MgCO}_3\cdot z\text{H}_2\text{O}(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g) + z\text{H}_2\text{O}(g)\). (Allow state symbols only if completely correct). (b) [1 mark] for correct molar mass of \(\text{MgO} = 40.3\text{ g mol}^{-1}\). [1 mark] for correct calculation of moles = \(0.0100\text{ mol}\). (c) [1 mark] for setting moles of hydrated salt = \(0.0100\text{ mol}\) and calculating \(M_r = 174.3\). [1 mark] for \(M_r(\text{MgCO}_3) = 84.3\). [1 mark] for showing that \(z = 5\) (must show clear calculation steps). (d) [1 mark] for identifying moles of \(\text{CO}_2 = 0.0100\text{ mol}\). [1 mark] for converting to \(\text{cm}^3\): \(0.0100 \times 24000 = 240\text{ cm}^3\) (or \(0.240\text{ dm}^3\)). (e) [1 mark] for stating that gas expands at higher temperature. [1 mark] for stating that water is present as steam (gas) at high temperature, which adds to the total volume (whereas it is a liquid at r.t.p.).
Question 2 · Structured
10 marks
A sample of impure ammonium sulfate, \((\text{NH}_4)_2\text{SO}_4\), with a mass of \(1.50\text{ g}\) was heated with \(50.0\text{ cm}^3\) of \(0.500\text{ mol dm}^{-3}\) sodium hydroxide solution, \(\text{NaOH}\)(aq). The ammonia gas evolved was expelled completely by boiling the mixture. The excess sodium hydroxide remaining in the reaction mixture required \(25.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\)(aq), for complete neutralisation. (\(A_r\): \(\text{H} = 1.0\), \(\text{N} = 14.0\), \(\text{O} = 16.0\), \(\text{S} = 32.1\))
(a) Write ionic equations, including state symbols, for: (i) the reaction between ammonium ions and hydroxide ions. [1] (ii) the reaction between hydrogen ions and hydroxide ions during the titration. [1]
(b) Calculate: (i) the number of moles of sodium hydroxide initially added to the sample. [1] (ii) the number of moles of sodium hydroxide that remained in excess. [1] (iii) the number of moles of ammonium sulfate present in the impure sample. [2] (iv) the percentage purity by mass of the ammonium sulfate in the sample. Give your answer to three significant figures. [3]
(c) Suggest why the mixture must be heated and boiled before the excess sodium hydroxide is titrated. [1]
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(b) (i) \(\text{Moles of NaOH initially added} = \frac{50.0}{1000} \times 0.500 = 0.0250\text{ mol}\). (ii) \(\text{Moles of HCl used} = \frac{25.0}{1000} \times 0.200 = 0.00500\text{ mol}\). Since the reaction ratio of \(\text{NaOH}\) and \(\text{HCl}\) is 1:1, \(\text{moles of excess NaOH} = 0.00500\text{ mol}\). (iii) \(\text{Moles of NaOH reacted with ammonium sulfate} = 0.0250 - 0.00500 = 0.0200\text{ mol}\). From the ionic equation in (a)(i), 1 mole of \(\text{NH}_4^+\) reacts with 1 mole of \(\text{OH}^-\). Thus, \(\text{moles of NH}_4^+ \text{ reacted} = 0.0200\text{ mol}\). Since 1 mole of \((\text{NH}_4)_2\text{SO}_4\) contains 2 moles of \(\text{NH}_4^+\): \(\text{Moles of }(\text{NH}_4)_2\text{SO}_4 = \frac{0.0200}{2} = 0.0100\text{ mol}\). (iv) \(M_r\text{ of }(\text{NH}_4)_2\text{SO}_4 = 2(14.0 + 4.0) + 32.1 + 4(16.0) = 132.1\text{ g mol}^{-1}\). \(\text{Mass of }(\text{NH}_4)_2\text{SO}_4 = 0.0100 \times 132.1 = 1.321\text{ g}\). \(\text{Percentage purity} = \frac{1.321}{1.50} \times 100\% = 88.067\% = 88.1\%\).
(c) To completely expel all the dissolved ammonia gas, because dissolved ammonia is alkaline and would react with the hydrochloric acid during titration, leading to an artificially high titer of acid.
Marking scheme
(a)(i) [1 mark] for correct equation with state symbols: \(\text{NH}_4^+(aq) + \text{OH}^-(aq) \rightarrow \text{NH}_3(g) + \text{H}_2\text{O}(l)\). (a)(ii) [1 mark] for correct equation with state symbols: \(\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)\). (b)(i) [1 mark] for \(0.0250\text{ mol}\). (b)(ii) [1 mark] for \(0.00500\text{ mol}\). (b)(iii) [1 mark] for calculating moles of \(\text{NaOH}\) reacted = \(0.0200\text{ mol}\). [1 mark] for dividing by 2 to get \(0.0100\text{ mol}\) of \((\text{NH}_4)_2\text{SO}_4\). (b)(iv) [1 mark] for \(M_r = 132.1\). [1 mark] for mass of pure salt = \(1.321\text{ g}\) (allow ECF from (b)(iii)). [1 mark] for percentage purity = \(88.1\%\) (must be 3 significant figures). (c) [1 mark] for stating that dissolved ammonia is basic/alkaline and must be expelled to prevent it from reacting with the acid during titration.
Question 3 · Structured
10 marks
Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), is a liquid fuel that can be burned completely in oxygen.
(a) Define the term *standard enthalpy change of combustion*, \(\Delta H_c^\bottom\). [2]
(b) Write a balanced chemical equation, including state symbols, for the complete combustion of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\). [2]
(c) Use the standard enthalpy changes of combustion given in the table below to calculate the standard enthalpy change of formation of propan-1-ol, \(\Delta H_f^\bottom[\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)]\).
Draw a labelled Hess's Law cycle and show your working. [4]
(d) Suggest two reasons why the experimental value for the enthalpy change of combustion of propan-1-ol, obtained using a simple spirit burner and copper calorimeter in a school laboratory, is significantly less exothermic than the data book value. [2]
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Worked solution
(a) The enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (298 K, 100 kPa).
(c) Formation equation: \(3\text{C}(s) + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\) (\(\Delta H_f^\bottom\)) Both reactants and products can be burned to form \(3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\). By Hess's Law: \(\Delta H_f^\bottom + \Delta H_c^\bottom[\text{propan-1-ol}] = 3\Delta H_c^\bottom[\text{C}(s)] + 4\Delta H_c^\bottom[\text{H}_2(g)]\) \(\Delta H_f^\bottom + (-2021.0) = 3(-393.5) + 4(-285.8)\) \(\Delta H_f^\bottom - 2021.0 = -1180.5 - 1143.2 = -2323.7\) \(\Delta H_f^\bottom = -2323.7 + 2021.0 = -302.7\text{ kJ mol}^{-1}\).
(d) Any two of: 1. Heat loss to the surrounding air / calorimeter. 2. Incomplete combustion of propan-1-ol (forming carbon/carbon monoxide instead of carbon dioxide). 3. Loss of fuel by evaporation from the wick of the spirit burner.
Marking scheme
(a) [1 mark] for: enthalpy change when 1 mole of substance is burned completely in oxygen. [1 mark] for: under standard conditions (298 K and 100 kPa / 1 bar). (b) [1 mark] for correct formulas of reactants and products: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), \(\text{O}_2\), \(\text{CO}_2\), \(\text{H}_2\text{O}\). [1 mark] for correct balancing and state symbols: \((l)\) for propan-1-ol and water, \((g)\) for oxygen and carbon dioxide. (c) [1 mark] for drawing a correct Hess's Law cycle with arrows showing combustion pathways of both sides to \(3\text{CO}_2 + 4\text{H}_2\text{O}\). [1 mark] for \(\Sigma \Delta H_c^\bottom(\text{reactants}) = 3(-393.5) + 4(-285.8) = -2323.7\text{ kJ mol}^{-1}\). [1 mark] for correct algebraic setup: \(\Delta H_f^\bottom = -2323.7 - (-2021.0)\). [1 mark] for final value with correct sign and units: \(-302.7\text{ kJ mol}^{-1}\). (d) [2 marks] (1 mark for each valid reason, max 2): - Heat lost to surroundings. - Incomplete combustion. - Evaporation of fuel.
Question 4 · Structured
10 marks
Alkenes undergo addition reactions due to the presence of a carbon-carbon double bond, \(\text{C}=\text{C}\).
(a) (i) Explain why 2-methylbut-2-ene, \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\), does not exhibit cis-trans isomerism, while but-2-ene, \(\text{CH}_3\text{CH}=\text{CHCH}_3\), does. [2] (ii) Draw the structural formula for the trans-isomer of but-2-ene, showing its spatial arrangement clearly. [2]
(b) 2-methylbut-2-ene reacts with hydrogen bromide, \(\text{HBr}\), at room temperature to produce two isomeric bromoalkanes. (i) Name the major product formed in this reaction. [1] (ii) Explain, in terms of carbocation stability, why one of the isomeric bromoalkanes is formed as the major product. [2] (iii) Write a complete mechanism for the reaction of 2-methylbut-2-ene with \(\text{HBr}\) to form the major product. Include curly arrows, relevant dipoles, charges, and lone pairs. [3]
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Worked solution
(a) (i) In 2-methylbut-2-ene, one of the doubly-bonded carbon atoms (C2) is attached to two identical groups (two methyl groups, \(\text{-CH}_3\)). For cis-trans isomerism to occur, both carbon atoms of the double bond must be attached to two different groups. In but-2-ene, each double-bonded carbon has one \(\text{-H}\) and one \(\text{-CH}_3\) group attached, so it can form cis-trans isomers. (ii) Structural formula of trans-but-2-ene: H3C H \ / C = C / \ H CH3
(b) (i) 2-bromo-2-methylbutane (ii) The reaction proceeds via carbocation intermediates. The addition of \(\text{H}^+\) to 2-methylbut-2-ene can form either a tertiary carbocation, \(\text{(CH}_3)_2\text{C}^+\text{-CH}_2\text{CH}_3\), or a secondary carbocation, \(\text{(CH}_3)_2\text{CH-C}^+\text{HCH}_3\). A tertiary carbocation is more stable than a secondary carbocation because of the positive inductive effect of three electron-releasing alkyl groups, which disperses the positive charge on the carbon atom more effectively. (iii) Mechanism: Step 1: Electrophilic addition. The \(\pi\)-bond electrons of the double bond attack the \(\text{H}^{\delta+}\) of the polar \(\text{H-Br}\) molecule. A curly arrow goes from the double bond to the \(\text{H}\) of \(\text{H-Br}\). Another curly arrow goes from the \(\text{H-Br}\) bond to the \(\text{Br}\). Step 2: Formation of the tertiary carbocation intermediate: \(\text{(CH}_3)_2\text{C}^+\text{-CH}_2\text{CH}_3\) and a bromide ion, \(\text{:Br}^-\). Step 3: Nucleophilic attack. A curly arrow goes from the lone pair of the \(\text{:Br}^-\) ion to the positively charged carbon atom of the carbocation to form 2-bromo-2-methylbutane.
Marking scheme
(a)(i) [1 mark] for stating that C2 of 2-methylbut-2-ene has two identical methyl groups. [1 mark] for stating that cis-trans isomerism requires both double-bonded carbons to be attached to two different groups. (a)(ii) [1 mark] for correct arrangement around \(\text{C}=\text{C}\) showing planar \(120^\circ\) angles. [1 mark] for placing methyl groups on opposite sides of the double bond. (b)(i) [1 mark]: 2-bromo-2-methylbutane. (b)(ii) [1 mark] for stating that the tertiary carbocation is more stable than the secondary carbocation. [1 mark] for explaining that alkyl groups are electron-donating / have a positive inductive effect which disperses the positive charge. (b)(iii) [1 mark] for correct dipole \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) and curly arrow from \(\text{C}=\text{C}\) double bond to \(\text{H}\), and curly arrow from \(\text{H-Br}\) bond to \(\text{Br}\). [1 mark] for drawing the correct tertiary carbocation intermediate and \(\text{:Br}^-\) with a lone pair and negative charge. [1 mark] for curly arrow from the lone pair on the bromide ion to the \(\text{C}^+\) of the carbocation.
Question 5 · Structured
10 marks
This question is about Group 17 chemistry. Solid sodium halides react differently with concentrated sulfuric acid depending on the reducing power of the halide ions.
(a) When concentrated sulfuric acid is added to solid sodium chloride, a misty gas is evolved. (i) Identify the misty gas. [1] (ii) Write a balanced equation for this reaction. [1] (iii) Explain, in terms of oxidation numbers, why this reaction is not classified as a redox reaction. [2]
(b) When concentrated sulfuric acid is added to solid sodium iodide, a vigorous reaction occurs, producing a complex mixture of products. (i) A dark grey solid and a gas with a smell of rotten eggs are observed. Identify both of these products. [2] (ii) Name the yellow solid that can also be formed in this reaction. [1] (iii) Explain, in terms of the relative reducing abilities of halide ions, why sodium iodide reacts so differently from sodium chloride when treated with concentrated sulfuric acid. [3]
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Worked solution
(a) (i) Hydrogen chloride / \(\text{HCl}\) (ii) \(\text{NaCl}(s) + \text{H}_2\text{SO}_4(l) \rightarrow \text{NaHSO}_4(s) + \text{HCl}(g)\) (or \(\text{2NaCl}(s) + \text{H}_2\text{SO}_4(l) \rightarrow \text{Na}_2\text{SO}_4(s) + \text{2HCl}(g)\)) (iii) The oxidation numbers of all atoms remain unchanged: \(\text{Na}\) is \(+1\), \(\text{Cl}\) is \(-1\), \(\text{H}\) is \(+1\), \(\text{S}\) is \(+6\), and \(\text{O}\) is \(-2\) in both reactants and products. Since there is no change in oxidation numbers, it is not a redox reaction.
(b) (i) Dark grey solid: Iodine / \(\text{I}_2\). Gas with rotten egg smell: Hydrogen sulfide / \(\text{H}_2\text{S}\). (ii) Sulfur / \(\text{S}\) (iii) Iodide ions (\(\text{I}^-\)) are stronger reducing agents than chloride ions (\(\text{Cl}^-\)). The iodide ion is larger and has more electron shells (greater shielding) than the chloride ion. Therefore, the outermost electrons in iodide are less strongly attracted to the nucleus and are lost more easily. Because of this, iodide ions can reduce sulfuric acid (where sulfur is in the \(+6\) oxidation state) to sulfur (\(0\)) and hydrogen sulfide (\(-2\)), whereas chloride ions are too weak reducing agents to reduce sulfuric acid (acting only as a base).
Marking scheme
(a)(i) [1 mark] for Hydrogen chloride (accept \(\text{HCl}\)). (a)(ii) [1 mark] for \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\). (a)(iii) [1 mark] for stating that oxidation numbers do not change. [1 mark] for showing specific values: e.g., \(\text{Cl}\) remains \(-1\) and \(\text{S}\) remains \(+6\). (b)(i) [1 mark] for Iodine (accept \(\text{I}_2\)). [1 mark] for Hydrogen sulfide (accept \(\text{H}_2\text{S}\)). (b)(ii) [1 mark] for Sulfur (accept \(\text{S}\)). (b)(iii) [1 mark] for stating that iodide is a stronger reducing agent than chloride. [1 mark] for explanation: iodide ion is larger / has more shielding, so outermost electrons are lost more easily. [1 mark] for connecting to reaction: iodide reduces sulfur in \(\text{H}_2\text{SO}_4\) from \(+6\) to lower oxidation states (like \(0\) or \(-2\)), whereas chloride cannot reduce sulfuric acid.
Question 6 · Structured
10 marks
Acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), is a powerful oxidizing agent that can be used to titrate iron(II) ions, \(\text{Fe}^{2+}\), in solution.
(a) Determine the oxidation number of chromium in the dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\). [1]
(b) (i) Write the half-equation for the reduction of dichromate(VI) ions, \(\text{Cr}_2\text{O}_7^{2-}\), to chromium(III) ions, \(\text{Cr}^{3+}\), in acidic solution. [1] (ii) Write the half-equation for the oxidation of iron(II) ions, \(\text{Fe}^{2+}\), to iron(III) ions, \(\text{Fe}^{3+}\). [1] (iii) Combine these two half-equations to write the overall ionic equation for this reaction. [1]
(c) A student titrated a \(25.0\text{ cm}^3\) sample of an aqueous solution of iron(II) sulfate, \(\text{FeSO}_4\), with \(0.0200\text{ mol dm}^{-3}\) potassium dichromate(VI) solution, \(\text{K}_2\text{Cr}_2\text{O}_7\). The student found that exactly \(22.5\text{ cm}^3\) of the potassium dichromate(VI) solution was required to reach the end-point.
(i) Calculate the number of moles of \(\text{Cr}_2\text{O}_7^{2-}\) ions used in the titration. [1] (ii) Determine the number of moles of \(\text{Fe}^{2+}\) ions present in the \(25.0\text{ cm}^3\) sample. [1] (iii) Calculate the concentration, in \(\text{mol dm}^{-3}\), of the iron(II) sulfate solution. [1] (iv) Calculate the mass, in grams, of anhydrous iron(II) sulfate, \(\text{FeSO}_4\), needed to prepare \(250\text{ cm}^3\) of this iron(II) sulfate solution. Give your answer to three significant figures. (\(A_r\): \(\text{Fe} = 55.8\), \(\text{S} = 32.1\), \(\text{O} = 16.0\)) [3]
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(c) (i) \(\text{Moles of Cr}_2\text{O}_7^{2-} = \frac{22.5}{1000} \times 0.0200 = 4.50 \times 10^{-4}\text{ mol}\). (ii) From the balanced equation, 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\) reacts with 6 moles of \(\text{Fe}^{2+}\). \(\text{Moles of Fe}^{2+} = 6 \times 4.50 \times 10^{-4} = 2.70 \times 10^{-3}\text{ mol}\). (iii) \(\text{Concentration of FeSO}_4 = \frac{2.70 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.108\text{ mol dm}^{-3}\). (iv) \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + 4(16.0) = 151.9\text{ g mol}^{-1}\). \(\text{Moles of FeSO}_4 \text{ needed for } 250\text{ cm}^3 = 0.108 \times 0.250 = 0.0270\text{ mol}\). \(\text{Mass of anhydrous FeSO}_4 = 0.0270 \times 151.9 = 4.1013\text{ g} = 4.10\text{ g}\) (to 3 sig fig).
Marking scheme
(a) [1 mark] for \(+6\) (or 6, or VI). (b)(i) [1 mark]: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). (b)(ii) [1 mark]: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\). (b)(iii) [1 mark]: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\). (c)(i) [1 mark]: \(4.50 \times 10^{-4}\text{ mol}\). (c)(ii) [1 mark]: \(2.70 \times 10^{-3}\text{ mol}\) (allow ECF from (b)(iii) and (c)(i)). (c)(iii) [1 mark]: \(0.108\text{ mol dm}^{-3}\) (allow ECF). (c)(iv) [1 mark] for correct molar mass of \(\text{FeSO}_4 = 151.9\text{ g mol}^{-1}\). [1 mark] for calculating moles in \(250\text{ cm}^3 = 0.0270\text{ mol}\) (allow ECF). [1 mark] for final mass = \(4.10\text{ g}\) (allow ECF, must be to 3 significant figures).
Paper 3 Advanced Practical Skills
Perform the practical experiments as instructed and complete the questions.
3 Question · 39.99 marks
Question 1 · Practical
13.33 marks
Perform the titration to determine the concentration of hydrogen peroxide in FA 1.
**Introduction** Commercial hair bleach contains hydrogen peroxide, \(\text{H}_2\text{O}_2\). In this experiment, you will determine the exact concentration of \(\text{H}_2\text{O}_2\) in a diluted sample by titration with a standard solution of potassium manganate(VII), \(\text{KMnO}_4\).
**Procedure** 1. Pipette 25.0 cm\(^3\) of **FA 1** into a clean conical flask. 2. Use a measuring cylinder to add approximately 20 cm\(^3\) of **FA 3** into the same flask. 3. Fill the burette with the potassium manganate(VII) solution, **FA 2**. 4. Titrate **FA 1** with **FA 2** until a permanent pale pink colour is obtained. 5. Repeat the titration as necessary to obtain concordant results (within 0.10 cm\(^3\)).
**Tasks** 1. Record all your titration results in a single, clearly structured table showing initial and final burette readings, and the volume of **FA 2** added. 2. State the average volume of **FA 2** to be used in your calculations. 3. Calculate the number of moles of \(\text{MnO}_4^-\n\) ions in your average titre volume. 4. Using the equation below, calculate the number of moles of \(\text{H}_2\text{O}_2\) present in 25.0 cm\(^3\) of **FA 1**: \(2\text{MnO}_4^-(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{O}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})\) 5. Calculate the concentration, in mol dm\(^{-3}\), of \(\text{H}_2\text{O}_2\) in **FA 1**. 6. Calculate the concentration, in g dm\(^{-3}\), of \(\text{H}_2\text{O}_2\) in the original commercial hair bleach (before dilution). [\(M_r\) of \(\text{H}_2\text{O}_2 = 34.0\)] 7. Calculate the percentage uncertainty in the volume of **FA 1** delivered by the 25.0 cm\(^3\) pipette (error of pipette is \(\pm 0.06\text{ cm}^3\)). Explain why a student should not use a 50 cm\(^3\) measuring cylinder instead of a pipette for this step.
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Worked solution
1. **Average Titre**: Let the average titre of FA 2 be 24.20 cm\(^3\). 2. **Moles of \(\text{MnO}_4^-\)**: \(n(\text{MnO}_4^-) = C \times V = 0.0200 \text{ mol dm}^{-3} \times \frac{24.20}{1000} \text{ dm}^3 = 4.84 \times 10^{-4} \text{ mol}\) 3. **Moles of \(\text{H}_2\text{O}_2\) in 25.0 cm\(^3\)**: Using the stoichiometry ratio \(2\text{MnO}_4^- : 5\text{H}_2\text{O}_2\): \(n(\text{H}_2\text{O}_2) = 4.84 \times 10^{-4} \times \frac{5}{2} = 1.21 \times 10^{-3} \text{ mol}\) 4. **Concentration of FA 1**: \([\text{H}_2\text{O}_2]_{\text{FA1}} = \frac{1.21 \times 10^{-3} \text{ mol}}{0.0250 \text{ dm}^3} = 0.0484 \text{ mol dm}^{-3}\) 5. **Concentration in Original Bleach**: Since FA 1 is diluted 10-fold: \([\text{H}_2\text{O}_2]_{\text{original}} = 0.0484 \text{ mol dm}^{-3} \times 10 = 0.484 \text{ mol dm}^{-3}\) \(\text{Concentration in g dm}^{-3} = 0.484 \text{ mol dm}^{-3} \times 34.0 \text{ g mol}^{-1} = 16.46 \text{ g dm}^{-3}\) 6. **Percentage Uncertainty**: \(\% \text{ Uncertainty} = \frac{0.06}{25.0} \times 100\% = 0.24\%\) Using a measuring cylinder instead of a pipette would introduce much larger experimental error because measuring cylinders have a lower precision (typically \(\pm 0.5\text{ cm}^3\) or higher).
Marking scheme
1. **Table of results (2.0 marks)**: - All columns and rows labelled with correct units (1 mark). - All burette readings to 0.05 cm\(^3\) and concordant titres within 0.10 cm\(^3\) (1 mark). 2. **Accuracy (3.0 marks)**: - Award 3 marks if the student's average titre is within 0.20 cm\(^3\) of the supervisor's value. - Award 2 marks for within 0.30 cm\(^3\); 1 mark for within 0.50 cm\(^3\). 3. **Calculations (5.33 marks)**: - Correct calculation of moles of \(\text{MnO}_4^-\) (1.0 mark). - Correct mole ratio multiplier (5/2) applied to find moles of \(\text{H}_2\text{O}_2\) (1.33 marks). - Correct concentration of FA 1 in mol dm\(^{-3}\) (1.0 mark). - Correctly multiplying by dilution factor (x10) to find original molar concentration (1.0 mark). - Correct calculation of mass concentration in g dm\(^{-3}\) to 3 significant figures (1.0 mark). 4. **Uncertainty analysis (3.0 marks)**: - Correct percentage uncertainty calculation (0.24%) (1.0 mark). - Explaining that measuring cylinders have much higher uncertainty/error than volumetric pipettes (2.0 marks).
Question 2 · Practical
13.33 marks
Determine the enthalpy change of the displacement reaction between zinc and copper(II) ions.
**Introduction** In this experiment, you will investigate the heat energy released when zinc powder reacts with aqueous copper(II) sulfate.
**Procedure** 1. Place the polystyrene cup inside the 250 cm\(^3\) glass beaker for thermal insulation. 2. Pipette 25.0 cm\(^3\) of **FA 4** into the polystyrene cup. 3. Place the thermometer in the solution and record its temperature every minute for 3 minutes, stirring gently. 4. At exactly 4 minutes, add all of the **FA 5** (zinc powder) to the cup. Stir the mixture vigorously, but do not record the temperature at 4 minutes. 5. Continue stirring and record the temperature of the mixture every minute from 5 minutes to 10 minutes.
**Tasks** 1. Record all your temperature measurements in a clearly structured table. 2. Plot a graph of temperature (y-axis) against time (x-axis) on grid paper. Draw two straight lines of best fit: one through the points before adding zinc (minutes 0 to 3) and one through the cooling points (minutes 5 to 10). Extrapolate both lines to 4 minutes to determine the maximum corrected temperature rise, \(\Delta T\). 3. Calculate the heat energy, \(q\), in joules, released during this reaction. [Assume the density of the solution is 1.00 g cm\(^{-3}\) and its specific heat capacity is 4.18 J g\(^{-1}\) K\(^{-1}\)] 4. Calculate the number of moles of \(\text{Cu}^{2+}\) ions in 25.0 cm\(^3\) of **FA 4**. 5. Calculate the enthalpy change of displacement, \(\Delta H\), in kJ mol\(^{-1}\). Include the correct sign. 6. State the main source of heat loss in this experimental set-up and suggest a modification to the apparatus that would reduce this heat loss.
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Worked solution
1. **Corrected Temperature Rise**: From the extrapolated graph, the temperature at 4 min before reaction is \(21.5^\circ\text{C}\). The extrapolated maximum cooling temperature at 4 min is \(53.0^\circ\text{C}\). \(\Delta T = 53.0 - 21.5 = 31.5^\circ\text{C}\) 2. **Heat Energy (q)**: \(q = m \times c \times \Delta T = 25.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{ K}^{-1} \times 31.5 \text{ K} = 3291.75 \text{ J} = 3.29 \text{ kJ}\) 3. **Moles of \(\text{Cu}^{2+}\)**: \(n(\text{Cu}^{2+}) = 0.800 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.0200 \text{ mol}\) 4. **Enthalpy Change (\(\Delta H\))**: \(\Delta H = -\frac{q}{n} = -\frac{3.29175 \text{ kJ}}{0.0200 \text{ mol}} = -164.6 \text{ kJ mol}^{-1} \approx -165 \text{ kJ mol}^{-1}\) (3 sig figs) 5. **Sources and Improvements**: The main source of heat loss is from the top of the open polystyrene cup. This can be reduced by using a plastic lid with a hole for the thermometer.
Marking scheme
1. **Table of temperature results (2.0 marks)**: - All temperatures recorded to 0.5 °C (1 mark). - Time and temperature headers with correct units (1 mark). 2. **Graph plotting and extrapolation (3.33 marks)**: - Axes labelled correctly with units, and appropriate scale (1 mark). - Correct plotting of all points (1 mark). - Two linear lines of best fit showing extrapolation to 4 minutes to find \(\Delta T\) (1.33 marks). 3. **Heat energy calculation (2.0 marks)**: - Mass of solution correct (25.0 g) used in \(q = mc\Delta T\) (1 mark). - Correct calculated value of \(q\) to 3 significant figures (1 mark). 4. **Moles calculation (1.0 mark)**: - Correctly calculates \(0.0200\text{ mol}\) of \(\text{Cu}^{2+}\). 5. **Enthalpy calculation (2.0 marks)**: - Correct value of \(\Delta H\) with negative sign (1 mark). - Correct units of \(\text{kJ mol}^{-1}\) (1 mark). 6. **Error and improvement (3.0 marks)**: - Identifying heat loss to surroundings / evaporation from open top (1 mark). - Suggesting a plastic lid / wrapping the cup in cotton wool / double-cupping (2 marks).
Question 3 · Practical
13.33 marks
Identify the cation and anion in three unknown solutions: FA 6, FA 7, and FA 8.
**Introduction** You are provided with three solutions: **FA 6**, **FA 7**, and **FA 8**. Each solution contains a single Group 2 metal cation and a single halide anion.
**Procedure** Carry out the following tests in separate test-tubes and record your observations systematically.
* **Test 1**: To a 1 cm depth of each solution, add a few drops of aqueous sodium hydroxide, \(\text{NaOH}\), then add excess. * **Test 2**: To a 1 cm depth of each solution, add a few drops of dilute sulfuric acid, \(\text{H}_2\text{SO}_4\). * **Test 3**: To a 1 cm depth of each solution, add a few drops of aqueous silver nitrate, \(\text{AgNO}_3\), then add dilute aqueous ammonia, followed by concentrated aqueous ammonia.
**Tasks** 1. Present your observations for all tests in a single table. 2. Deduce the chemical formula of each salt in **FA 6**, **FA 7**, and **FA 8**, providing key observational evidence for each choice. 3. Write the ionic equations (including state symbols) for: (i) The reaction of the cation in **FA 6** with sulfate ions. (ii) The reaction of the anion in **FA 8** with silver ions. (iii) The reaction of the cation in **FA 7** with hydroxide ions.
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Worked solution
1. **Observations Table**: * **FA 6**: - Test 1: No precipitate with \(\text{NaOH}\). - Test 2: Dense white precipitate with \(\text{H}_2\text{SO}_4\). - Test 3: White precipitate with \(\text{AgNO}_3\), which completely dissolves in dilute aqueous ammonia. * **FA 7**: - Test 1: White precipitate with \(\text{NaOH}\), insoluble in excess. - Test 2: No precipitate with \(\text{H}_2\text{SO}_4\). - Test 3: Yellow precipitate with \(\text{AgNO}_3\), which remains insoluble in both dilute and concentrated aqueous ammonia. * **FA 8**: - Test 1: Faint white precipitate (or no precipitate initially, but white precipitate on standing) with \(\text{NaOH}\). - Test 2: White precipitate with \(\text{H}_2\text{SO}_4\). - Test 3: Cream precipitate with \(\text{AgNO}_3\), which is insoluble in dilute ammonia but dissolves in concentrated ammonia.
2. **Deductions**: * **FA 6**: Cation is \(\text{Ba}^{2+}\) (forms dense sulfate precipitate); Anion is \(\text{Cl}^-\n\) (forms white silver halide precipitate soluble in dilute \(\text{NH}_3\)). Formula: \(\text{BaCl}_2\). * **FA 7**: Cation is \(\text{Mg}^{2+}\) (forms insoluble hydroxide precipitate, no sulfate precipitate); Anion is \(\text{I}^-\n\) (forms yellow silver halide precipitate insoluble in ammonia). Formula: \(\text{MgI}_2\). * **FA 8**: Cation is \(\text{Ca}^{2+}\) (forms faint hydroxide precipitate, moderate white sulfate precipitate); Anion is \(\text{Br}^-\n\) (forms cream silver halide precipitate soluble only in conc. \(\text{NH}_3\)). Formula: \(\text{CaBr}_2\).
1. **Observations table (6.0 marks)**: - 2.0 marks for complete and accurate observations of FA 6 across all three tests. - 2.0 marks for complete and accurate observations of FA 7 across all three tests. - 2.0 marks for complete and accurate observations of FA 8 across all three tests. 2. **Identifications with evidence (3.0 marks)**: - 1.0 mark for identifying FA 6 as \(\text{BaCl}_2\) with correct supporting reasoning. - 1.0 mark for identifying FA 7 as \(\text{MgI}_2\) with correct supporting reasoning. - 1.0 mark for identifying FA 8 as \(\text{CaBr}_2\) with correct supporting reasoning. 3. **Ionic equations (4.33 marks)**: - 1.5 marks for \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\) with correct state symbols. - 1.5 marks for \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) with correct state symbols. - 1.33 marks for \(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s})\) with correct state symbols.
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