▶️ Answer/Explanation
(a) B (filtration)
Explanation: Filtration is the correct physical separation method for separating an insoluble solid (sand) from a liquid (salt solution). The sand particles are too large to pass through the filter paper, while the salt solution (filtrate) passes through.
• A is not correct because crystallisation is used to obtain a solid solute from its solution, not to separate a solid from a liquid mixture.
• C is not correct because fractional distillation is used to separate miscible liquids with different boiling points.
• D is not correct because simple distillation is used to separate a pure solvent from a solution (as in part (b)), not to separate a solid from a liquid.
(b)(i)
X: thermometer
Y: (Liebig) condenser
Z: beaker
Details: This setup is for simple distillation. The thermometer (X) monitors the boiling point of the vapor. The Liebig condenser (Y) cools and condenses the water vapor back into liquid water. The beaker (Z) collects the pure water distillate.
(b)(ii) salt (sodium chloride solid)
Explanation: When the separation (simple distillation) is complete, all the water has been vaporized, condensed, and collected. The non-volatile solute, sodium chloride (salt), remains in the flask as a solid residue because its boiling point is much higher than that of water.
▶️ Answer/Explanation
(a) A (substance S)
• A is the correct answer because S only contains one dye as it produces only one spot.
• B is not correct because T does not only contain one dye as it produces two spots.
• C is not correct because U does not only contain one dye as it produces three spots.
• D is not correct because V does not only contain one dye as it produces two spots.
(b)
• Substance V contains blue (B) and red (R) dyes.
• This is because V has spots at the same height (i.e., they have traveled the same distance from the baseline) as those from the pure blue (B) and red (R) reference dyes.
(c) (i)
• Measurement of distance moved by spot Y: ~6.0 cm (acceptable range: 5.7–6.1 cm).
• Measurement of distance moved by solvent front: ~9.0 cm (acceptable range: 8.7–9.1 cm).
• Using the formula:
\( R_f = \frac{\text{distance moved by spot Y}}{\text{distance moved by solvent}} = \frac{6.0}{9.0} = 0.67 \) (or \( 0.\overline{6} \))
• Final answer (1–4 significant figures acceptable): \( R_f = 0.67 \)
(c) (ii)
• The spot from the yellow food dye (Y) does not move as far up the paper as the spot from the red food dye (R).
• In paper chromatography, a substance that is more soluble in the solvent (mobile phase) will travel further. Since Y travels a shorter distance than R, it is less soluble in the solvent used.
▶️ Answer/Explanation
(a)(i) magnesium
Explanation: Magnesium burns in air with a bright white flame, a characteristic test for Group 2 metals.
(a)(ii) Silver, because it is the least reactive (of the metals listed).
Explanation: Unreactive metals, such as silver, gold, and platinum, are often found native (uncombined) in the Earth’s crust because they do not readily react with other elements to form compounds. In the reactivity series, silver is below copper and is very unreactive.
(b) Explanation including the following points:
- M1: Method 1 / heating the metal oxide/lead(II) oxide with carbon.
- M2: Because lead is less reactive than iron (and iron is obtained from iron oxide by carbon extraction). OR Carbon is more reactive than lead.
- M3: Equation: \( 2\text{PbO} + \text{C} \rightarrow 2\text{Pb} + \text{CO}_2 \) OR \( \text{PbO} + \text{C} \rightarrow \text{Pb} + \text{CO} \) OR \( \text{PbO} + \text{CO} \rightarrow \text{Pb} + \text{CO}_2 \).
Explanation: The method of extraction depends on the metal’s position in the reactivity series. Metals less reactive than carbon (like lead, tin, iron) can be extracted by reduction with carbon. Metals more reactive than carbon (like aluminium) require electrolysis.
(c) Explanation containing the following points:
- Pure metal: Particles/ions/atoms are the same size and in a regular arrangement/lattice, so layers can easily slide over each other.
- Alloy:
- M2: Diagram of alloy structure showing at least three layers with at least one different-sized particle/atom/ion (see mark scheme image).
- M3: Having different-sized particles disrupts/breaks up the regular arrangement/lattice.
- M4: Therefore, it is harder for the layers to slide over each other, making the alloy harder.
Explanation: In a pure metal, the uniform layers of atoms can slip past each other relatively easily under stress, making it malleable. In an alloy, atoms of different sizes distort the regular lattice, preventing easy sliding and increasing hardness and strength.
▶️ Answer/Explanation
(a) \(CH_3OH\) (structural formula of methanol).
(b)(i) Fermentation.
(b)(ii) An explanation that includes four of the following points:
• The process (fermentation/respiration) needs to be anaerobic.
• In air (aerobic conditions), ethanol is not produced.
• In air, carbon dioxide and water are produced instead.
• If the temperature is above \(40^\circ C\), the enzymes in yeast become denatured/lose their structure.
• This causes fermentation to slow down or stop.
Alternatively: In air, ethanol would be oxidised to ethanoic acid (vinegar).
(c)(i)
Completed table:
• Propanoic acid: Molecular formula = \(C_3H_6O_2\).
• Missing compound name: Butanoic acid (Molecular formula given as \(C_4H_8O_2\)).
(c)(ii) Carboxylic acids (or alkanoic acids).
(d)(i) Acts as a catalyst / to speed up the reaction.
(d)(ii) Displayed formula of ethyl propanoate (ester from propanoic acid and ethanol):
Key features: Ester linkage (\(-COO-\)) correctly shown, alkyl chains from acid and alcohol attached correctly.
(d)(iii) Example:
• Property: Distinctive/sweet/fruity smell / volatile.
• Use: Used in perfumes / flavourings / food additives / solvents (for paints/varnishes).
Alternatively: Used in the production of soaps (via saponification).
Additional Details:
• Homologous series (c)(ii): A family of compounds with the same functional group, similar chemical properties, and a general formula where each successive member differs by \(CH_2\). For carboxylic acids, the general formula is \(C_nH_{2n}O_2\).
• Esterification (d)(i): Concentrated sulfuric acid acts as a catalyst and a dehydrating agent, shifting the equilibrium towards ester formation.
• Fermentation conditions (b)(ii): The optimum temperature range is typically 30–40°C. Higher temperatures denature yeast enzymes (zymase), while oxygen presence leads to aerobic respiration or oxidation of ethanol.
▶️ Answer/Explanation
(a) Butadiene is unsaturated because it contains carbon–carbon double bonds (\(C=C\)). It is a hydrocarbon because it contains only the elements carbon and hydrogen.
(b)(i) from orange to colourless.
(Accept yellow → colourless. The bromine water is decolorized because butadiene undergoes an addition reaction with bromine.)
(b)(ii) \(\Delta H = -190 \ \text{kJ/mol}\)
Working:
Bonds broken: \(2 \times C=C\) \((2 \times 612)\), \(1 \times C-C\) \((348)\), \(6 \times C-H\) \((6 \times 412)\), \(2 \times Br-Br\) \((2 \times 193)\)
Energy in = \(1224 + 348 + 2472 + 386 = 4430 \ \text{kJ}\)
Bonds formed: \(3 \times C-C\) \((3 \times 348)\), \(6 \times C-H\) \((6 \times 412)\), \(4 \times C-Br\) \((4 \times 276)\)
Energy out = \(1044 + 2472 + 1104 = 4620 \ \text{kJ}\)
\(\Delta H = \text{Bonds broken} – \text{Bonds formed} = 4430 – 4620 = -190 \ \text{kJ/mol}\) (exothermic).
(c)(i) \[2C_4H_6 + 7O_2 \rightarrow 2C + 4CO + 2CO_2 + 6H_2O\]
(The equation shows incomplete combustion, producing solid carbon (soot), carbon monoxide, carbon dioxide and water.)
(c)(ii) Carbon monoxide (\(CO\)) is poisonous/toxic. It reduces the capacity of blood to carry oxygen because it binds to haemoglobin more strongly than oxygen, forming carboxyhaemoglobin.
OR
Carbon dioxide (\(CO_2\)) is a greenhouse gas. Its release contributes to global warming/climate change by trapping heat in the atmosphere.
(c)(iii) The energy profile diagram should show:
• A horizontal line for reactants \((2C_4H_6 + 11O_2)\) at a higher energy level.
• A curve rising to a peak (activation energy) then falling to products \((8CO_2 + 6H_2O)\) at a lower energy level.
• \(\Delta H\) labelled as a downward arrow from reactants to products with value \(-3446 \ \text{kJ/mol}\).
• Activation energy \((E_a)\) labelled as an upward arrow from reactants to the peak of the curve.
(The reaction is highly exothermic, so products are much lower in energy than reactants.)
▶️ Answer/Explanation
(a) (i)
Zinc would react with the sulfuric acid / solution X.
Zinc is more reactive than hydrogen. If used as an electrode in an acidic solution, it would participate in a displacement reaction, dissolving and producing hydrogen gas, rather than acting as an inert electrode for the electrolysis.
(a) (ii)
Bubbles of gas / fizzing / effervescence at both electrodes.
During the electrolysis of dilute sulfuric acid, oxygen gas is produced at the anode (positive) and hydrogen gas at the cathode (negative). Both appear as bubbles.
(a) (iii)
B (a burning splint gives a squeaky pop).
Hydrogen gas is highly flammable. The test involves collecting the gas and holding a burning splint to the mouth of the test tube. The hydrogen ignites with a characteristic ‘squeaky pop’ sound.
A is incorrect because a glowing splint relights is the test for oxygen.
C is incorrect because a burning splint going out is typical for non-flammable gases like carbon dioxide.
D is incorrect because limewater turning cloudy is the test for carbon dioxide.
(b)
Description:
1. Add barium chloride solution (or barium nitrate solution).
2. A white precipitate forms.
The white precipitate is barium sulfate \( \text{BaSO}_4 \), which is insoluble. The reaction is: \[ \ce{Ba^{2+}_{(aq)} + SO_4^{2-}_{(aq)} -> BaSO4_{(s)}} \] The test is usually done after acidifying the sample with dilute hydrochloric acid to rule out carbonate or sulfite ions, which also form white precipitates but dissolve in acid.
(c) (i)
(Graduated) pipette (and pipette filler).
A pipette is designed to deliver a precise, fixed volume of liquid. A burette could also be used but is more commonly associated with variable additions during the titration itself.
(c) (ii)
Calculation of moles of KOH:
Volume in dm³ = \( 25.0 / 1000 = 0.0250 \text{ dm}^3 \)
Amount (mol) = concentration × volume = \( 0.125 \times 0.0250 \)
\[ \text{amount} = 0.003125 \text{ mol} \quad \text{(or } 3.125 \times 10^{-3} \text{ mol)} \] Formula: \( n = c \times V \) (where \( V \) is in dm³).
(c) (iii)
Calculation of volume of \( \text{H}_2\text{SO}_4 \):
From the equation: \( \text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow … \)
Moles of \( \text{H}_2\text{SO}_4 \) needed = moles of KOH / 2 = \( 0.003125 / 2 = 0.0015625 \text{ mol} \)
Volume of \( \text{H}_2\text{SO}_4 \) (in dm³) = moles / concentration = \( 0.0015625 / 0.10 = 0.015625 \text{ dm}^3 \)
Volume in cm³ = \( 0.015625 \times 1000 = 15.625 \text{ cm}^3 \)
\[ \text{Final answer} = 15.6 \text{ cm}^3 \quad \text{(to 3 significant figures)} \] Steps: 1) Use stoichiometry from the balanced equation. 2) Use \( V = n / c \). 3) Convert dm³ to cm³ by multiplying by 1000. Accept answers from 15.6 cm³ to 16 cm³ depending on rounding.
▶️ Answer/Explanation
(a) Calculation of the volume of \(CO_2\) produced:
- Calculate moles of \(K_2CO_3\): \[ n(K_2CO_3) = \frac{\text{mass}}{\text{molar mass}} = \frac{6.9}{138} = 0.0500 \, \text{mol} \]
- From the balanced equation, 1 mole of \(K_2CO_3\) produces 1 mole of \(CO_2\). Therefore, moles of \(CO_2\) produced = 0.0500 mol.
- Calculate volume of \(CO_2\) at room temperature and pressure (rtp): \[ \text{Volume of } CO_2 = n \times \text{molar volume} = 0.0500 \times 24 = 1.2 \, \text{dm}^3 \]
- Convert \(\text{dm}^3\) to \(\text{cm}^3\): \[ 1.2 \, \text{dm}^3 = 1.2 \times 1000 = 1200 \, \text{cm}^3 \]
Final answer: 1200 cm³
(b)(i)
Prediction: The yield of \(CO\) increases / higher yield of CO.
Reason: The forward reaction is endothermic (\(\Delta H = +41 \, \text{kJ/mol}\)). Increasing the temperature favours the endothermic direction to absorb the added heat, shifting the equilibrium position to the right, producing more \(CO\).
(b)(ii)
Prediction: No effect (on the yield of \(CO\)).
Reason: There are equal numbers of gaseous moles on both sides of the equation (2 moles on each side). Changing the pressure has no effect on the equilibrium position if the number of gaseous moles is the same on both sides.
Note: The mark scheme indicates that references to Le Chatelier’s principle are not required, but the underlying reasoning regarding endothermic/exothermic nature and mole counts is essential.