▶️ Answer/Explanation
(a)(i) Boron / B
Explanation: The element with atomic number 5 is boron. Its symbol is B. It is located in Group 13, Period 2 of the Periodic Table.
(a)(ii) Na / Mg / Al
Explanation: Any metallic element from Period 3 (Na, Mg, Al) is acceptable. Sodium (Na), Magnesium (Mg), and Aluminium (Al) are all metals found in Period 3.
(a)(iii) Silicon / Si
Explanation: The number of protons in an atom defines its atomic number. An atom with 14 protons has an atomic number of 14, which corresponds to the element silicon (Si).
(a)(iv) Nitrogen / N
Explanation: The electronic configuration 2.5 indicates an atom with 2 electrons in its first shell and 5 in its second, giving a total of 7 electrons (and therefore 7 protons, atomic number 7). This is nitrogen (N). Allow N\(_2\).
(a)(v) Aluminium oxide
Explanation: The element with atomic number 13 is aluminium (Al), a metal in Group 13. It reacts with oxygen (a non-metal in Group 16) to form an ionic compound, aluminium oxide. The formula is Al\(_2\)O\(_3\).
(b)(i) D (Group 0)
Explanation: Group 0 contains the noble gases (e.g., helium, neon, argon). These elements have full outer electron shells, making them very stable and chemically unreactive under standard conditions.
A is not correct because Group 2 elements (alkaline earth metals) are reactive, readily losing two electrons to form 2+ ions.
B is not correct because Group 5 elements (like nitrogen and phosphorus) can form various compounds and are not all unreactive.
C is not correct because Group 6 elements (like oxygen and sulfur) are reactive, especially oxygen which readily gains electrons.
(b)(ii) B (lithium)
Explanation: Reactivity in Group 1 (alkali metals) increases down the group. Lithium, being at the top of the group, has the smallest atomic radius and the highest ionization energy among the listed options, making it the least likely to lose its outer electron. Therefore, it is the least reactive.
A is not correct because caesium, being at the bottom of Group 1, is the most reactive alkali metal listed.
C is not correct because potassium is more reactive than lithium and sodium.
D is not correct because sodium is more reactive than lithium.
▶️ Answer/Explanation
(a)
Note: The other changes are incorrect for these processes. Diffusion involves movement of particles from high to low concentration, evaporation is liquid to gas at the surface, and freezing is liquid to solid.
(b)
A correct answer will contain three distinct marks, aligning with the mark scheme:
M1: Measure the melting point (or boiling point) of each solid. [1 mark]
M2: The solid that has a fixed/sharp melting point is the pure substance. [1 mark]
Reasoning: Pure substances have a definite, sharp melting point because their particles are arranged in a regular pattern and melt at a single, specific temperature (e.g., pure ice melts exactly at \(0^\circ \text{C}\)).
M3: The solid that melts over a range of temperatures is the mixture. [1 mark]
Reasoning: Mixtures contain different substances that interfere with each other’s structure, causing melting to occur over a broader temperature range. The presence of impurities lowers and broadens the melting point.
Alternative (if using boiling point):
M1: Measure the boiling point.
M2: The solid that boils at a fixed temperature is pure.
M3: The solid that boils over a range of temperatures is a mixture.
▶️ Answer/Explanation
(a)(i) Q
Explanation: Gold is a very unreactive metal. It does not react with water or dilute acids. In the table, only metal Q shows no reaction with both water and dilute hydrochloric acid.
(a)(ii) Any one of: might be explosive / dangerous / unsafe
Explanation: Metal R reacts with water very quickly to produce hydrogen gas. This indicates it is a very reactive metal (like potassium or sodium). Such metals often react violently or explosively with acids, so the experiment was likely omitted for safety reasons.
(a)(iii) R, S, P, Q
Explanation: The reactivity order is deduced from the vigour of reactions.
- R is most reactive: reacts very quickly with water.
- S is next: reacts quickly with water and very quickly with acid.
- P is less reactive: no reaction with water, slow reaction with acid.
- Q is least reactive: no reaction with either.
(b)(i) galvanising / galvanisation
Explanation: This is the process of coating iron or steel with a layer of zinc to prevent rusting. The zinc acts as a protective barrier and can also provide sacrificial protection.
(b)(ii) Any one of: painting / oiling / greasing / sacrificial protection
Explanation: These are common barrier methods (paint, oil, grease) that exclude oxygen and water. Sacrificial protection involves attaching a more reactive metal (like magnesium or zinc) to the iron, which corrodes instead.
(c)(i) zinc displaces copper
Explanation: In the reaction, zinc atoms (Zn) remove oxygen from copper(II) oxide (CuO) to form zinc oxide (ZnO) and copper metal (Cu). A more reactive metal can displace a less reactive metal from its oxide.
(c)(ii)
- Oxidising agent: copper(II) oxide / CuO (1 mark)
- Explanation: because copper(II) oxide loses oxygen / gives oxygen to zinc / is reduced (1 mark)
Detailed explanation: Oxidation is the gain of oxygen; reduction is the loss of oxygen. In this reaction, zinc gains oxygen to form ZnO (zinc is oxidised). Copper(II) oxide loses oxygen to form Cu (copper(II) oxide is reduced). The substance that causes oxidation by being reduced itself is the oxidising agent. Therefore, CuO is the oxidising agent.
▶️ Answer/Explanation
(a)(i) Hydroxide ion / \(\text{OH}^-\)
Detail: Sodium hydroxide dissociates in water to produce \(\text{Na}^+\) and \(\text{OH}^-\) ions. The \(\text{OH}^-\) ions are responsible for the alkalinity (high pH).
(a)(ii) C (11)
A is not correct because 3 is acidic, not alkaline.
B is not correct because 6 is weakly acidic.
D is not correct because 14 is the pH of a very strong alkali like concentrated sodium hydroxide; ammonia solution is only weakly alkaline.
(b)(i) C (as a proton donor)
Detail: According to the Brønsted-Lowry theory, an acid is a proton (\(\text{H}^+\)) donor. Sulfuric acid donates protons to the ammonia (which acts as a base).
A and B are not correct because neutrons are not involved in acid-base reactions.
D is not correct because a proton acceptor is the definition of a base.
(b)(ii) C (pink | colourless)
Detail: Phenolphthalein is pink in alkaline solutions (like ammonia solution) and colourless in neutral or acidic solutions. Adding excess sulfuric acid neutralises the ammonia and then makes the solution acidic, causing the indicator to turn colourless.
A is not correct because phenolphthalein does not turn orange or red.
B is not correct for the same reason.
D is not correct because it reverses the colour change.
(c)(i)
M1: Ammonium ion is \(\text{NH}_4^+\) (charge 1+) and sulfate ion is \(\text{SO}_4^{2-}\) (charge 2-).
M2: To balance the charges, two ammonium ions (total charge 2+) are needed for every one sulfate ion (charge 2-), giving the formula \((\text{NH}_4)_2\text{SO}_4\).
(c)(ii)
Calculation:
\( M_r = 2 \times [14 + (4 \times 1)] + 32 + (4 \times 16) \)
\( = 2 \times [18] + 32 + 64 \)
\( = 36 + 32 + 64 = 132 \)
The relative formula mass is 132.
(c)(iii)
Method 1 (Using mass ratio):
M1: In 132 g of \((\text{NH}_4)_2\text{SO}_4\), the mass of nitrogen is \(2 \times 14 = 28\) g.
M2: Mass of nitrogen in 1 g of ammonium sulfate is \(28 \div 132\).
M3: Mass of nitrogen in 1000 g (1.0 kg) is \(1000 \times (28 \div 132) = 212.\overline{12}\) g ≈ 212 g (to 3 significant figures).
Method 2 (Using moles):
M1: Moles of ammonium sulfate = \(1000 \div 132 \approx 7.57576\) mol.
M2: Each mole of \((\text{NH}_4)_2\text{SO}_4\) contains 2 moles of N atoms.
M3: Mass of nitrogen = \(7.57576 \times 2 \times 14 = 212.\overline{12}\) g ≈ 212 g.
▶️ Answer/Explanation
(a) (i)
• M1: Alkanes
• M2: Because its molecular formula fits the general formula for alkanes, \(C_nH_{2n+2}\). For \(C_2H_6\), \(n=2\), so \(2n+2=6\), which matches.
(a) (ii)
• D
Explanation: Compound D is \(C_3H_8\). Its empirical formula (simplest whole number ratio) is also \(C_3H_8\), so they are the same.
(a) (iii)
• M1: Isomers are compounds with the same molecular formula but different structural/displayed formulae.
• M2: Butane (straight-chain):
CH3-CH2-CH2-CH3
• M3: Methylpropane (branched-chain):
CH3-CH(CH3)-CH3
Note: \(C_4H_{10}\) is butane. The two isomers are butane (unbranched) and methylpropane (branched, also called isobutane).
(b)
Mark scheme points:
• M1: Heat/vaporize the crude oil.
• M2: Pass the vapor into a fractionating column/tower.
• M3: Fractions/compounds separate due to their different boiling points (temperature gradient in the column).
• M4: Compound D (\(C_3H_8\), propane) is collected at the top of the column in the refinery gas fraction.
Award maximum 3 marks. If the description is of a lab process, max 3. If confused with cracking, only M1 can be awarded.
(c) (i)
• Addition (polymer)
Reject “additional”.
(c) (ii)
• Poly(propene) / Polypropene
Allow polypropylene.
(c) (iii)
• M1: Correct repeat unit:
-[-CH(CH3)-CH2-]- or displayed as:
• M2: Brackets around the repeat unit, ‘n’ subscript, and extension bonds (single bonds extending out of the brackets) shown correctly.
Ignore bond angles. Allow use of CH3. M2 is dependent on M1.
▶️ Answer/Explanation
(a) Sulfur (S)
The solid sulfur (S) formed is the yellow precipitate that obscures the cross.
(b) Any two from:
- Concentration of hydrochloric acid.
- Concentration of sodium thiosulfate solution.
- Height of the eye above the flask / viewing position.
- Same size/shape of conical flask.
- Same size/darkness of the cross.
To make it a fair test, all variables except temperature must be controlled.
(c) Any one from:
- The thiosulfate solution would cool down / not remain at the required (higher) temperature during the reaction.
- Larger (percentage) errors in timing because the times become very small (the reaction is very fast).
- Risk of some solution evaporating at higher temperatures, changing concentrations.
At higher temperatures, the reaction is extremely fast, making timing imprecise and difficult.
(d)
- All five points plotted correctly (e.g., (20, 0.0025), (30, 0.0053), (40, 0.012), (50, 0.023), (60, 0.042)).
- A smooth curve of best fit drawn through the points, showing a positive, non-linear relationship.
The graph shows that rate increases with temperature, initially slowly then more rapidly.
(e) (i) Example answer: 0.016 s\(^{-1}\) (accept 0.015–0.017 s\(^{-1}\)).
To obtain this, draw a vertical line from 45°C on the x-axis up to the curve, then a horizontal line to the y-axis to read the rate.
(e) (ii) Using \( \text{time} = \frac{1}{\text{rate}} \).
If rate = 0.016 s\(^{-1}\), then \( \text{time} = \frac{1}{0.016} = 62.5 \) s (accept answers from 58.8 s to 62.5 s based on a rate between 0.017 and 0.016).
(e) (iii) As temperature increases, the rate of reaction increases. Not linear/directly proportional; the increase is steeper at higher temperatures.
The relationship is positive and exponential-like, as shown by the curve.
(f) Explanation includes:
- Increasing temperature increases the (mean) kinetic energy of the particles.
- Particles move faster and collide more frequently.
- A greater proportion of collisions have energy equal to or greater than the activation energy (successful collisions).
- Therefore, the frequency of successful collisions per second increases, so the rate of reaction increases.
This is a direct application of collision theory, linking temperature to energy and collision success.
▶️ Answer/Explanation
(a)(i) Measuring cylinder (Allow: pipette, burette).
This apparatus is used to measure a fixed volume of liquid accurately.
(a)(ii) To ensure the temperature is the same throughout the solution / to ensure heat is evenly distributed.
Stirring ensures a uniform temperature for an accurate thermometer reading.
(a)(iii) Blue.
Aqueous copper(II) ions (\(\text{Cu}^{2+}(aq)\)) give a characteristic blue colour.
(b)
(c)(i)
The formula for heat energy is \(Q = mc\Delta T\).
Mass of water, \(m = 50 \, \text{cm}^3 = 50 \, \text{g}\) (since \(1 \, \text{cm}^3 = 1 \, \text{g}\)).
Specific heat capacity, \(c = 4.2 \, \text{J g}^{-1} °C^{-1}\).
Temperature change, \(\Delta T = 3.3 \, °C\).
Therefore: \[ Q = 50 \times 4.2 \times 3.3 = 693 \, \text{J} \]
693 J is approximately 700 J, as required.
(c)(ii)
Step 1 – Moles of \(\text{CuSO}_4\): \[ \text{Moles} = \frac{\text{mass}}{M_r} = \frac{1.70}{159.5} = 0.01066 \, \text{mol} \, (\text{approximately } 0.0107 \, \text{mol}) \]
Step 2 – Enthalpy change per mole (using \(Q = 693 \, \text{J}\)): \[ \Delta H = \frac{Q}{\text{moles}} = \frac{693}{0.01066} = 64990 \, \text{J/mol} \approx 65.0 \, \text{kJ/mol} \]
Step 3 – Apply sign: The temperature increased, so the reaction is exothermic. Therefore, \(\Delta H\) is negative.
Final answer: \(\mathbf{\Delta H = -65 \, \text{kJ/mol}}\) (Allow range: -64 to -66 kJ/mol).
(d)
• The temperature decreased from 23.8 °C to 22.7 °C.
• This means the dissolving process absorbed heat from the surroundings.
• Therefore, the process is endothermic.
Explanation: For anhydrous \(\text{CuSO}_4\), dissolving is exothermic due to energy released during hydration of ions. For hydrated \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), the energy required to break the ionic lattice and separate the ions is greater than the energy released when they are hydrated (as they are already partially hydrated), resulting in an overall endothermic process.
▶️ Answer/Explanation
(a)(i) B (decomposition)
A is not correct because when sodium hydrogencarbonate is heated, combustion does not take place.
C is not correct because when sodium hydrogencarbonate is heated, oxidation does not take place.
D is not correct because when sodium hydrogencarbonate is heated, reduction does not take place.
(a)(ii) Because carbon dioxide gas is produced/given off. The gas forms bubbles within the cake mixture, causing it to rise and increase in volume.
(b)(i) To ensure the reaction is complete / to obtain a constant mass / to ensure all the \( \text{NaHCO}_3 \) has decomposed. Heating again ensures no further mass loss occurs, confirming the reaction has finished.
(b)(ii)
Advantage: To stop any solid (e.g., \( \text{Na}_2\text{CO}_3 \) or \( \text{NaHCO}_3 \)) from spitting out or being lost during heating.
Disadvantage: The gases (\( \text{CO}_2 \) and \( \text{H}_2\text{O} \) vapour) produced cannot easily escape, which might affect the reaction or cause pressure build-up.
(c)(i)
Mass of \( \text{NaHCO}_3 \) = (mass of crucible + \( \text{NaHCO}_3 \)) – (mass of empty crucible)
= \( 29.75 \, \text{g} – 26.50 \, \text{g} \)
= \( 3.25 \, \text{g} \)
(c)(ii)
Step 1: Calculate moles of \( \text{NaHCO}_3 \).
\( n = \frac{\text{mass}}{M_r} = \frac{3.25}{84} = 0.03869 \, \text{mol} \)
Step 2: Use reaction stoichiometry. From the equation: \( 2\text{NaHCO}_3 \rightarrow 1\text{Na}_2\text{CO}_3 \).
Moles of \( \text{Na}_2\text{CO}_3 = \frac{0.03869}{2} = 0.019345 \, \text{mol} \).
Step 3: Calculate mass of \( \text{Na}_2\text{CO}_3 \).
Mass = moles \( \times M_r = 0.019345 \times 106 = 2.05 \, \text{g} \) (to 3 significant figures).
Alternative method using mass ratio:
\( 2 \times 84 = 168 \, \text{g} \) of \( \text{NaHCO}_3 \) produces \( 106 \, \text{g} \) of \( \text{Na}_2\text{CO}_3 \).
Therefore, \( 3.25 \, \text{g} \) produces \( \frac{106}{168} \times 3.25 = 2.05 \, \text{g} \).
(d)(i)
Percentage yield = \( \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \)
= \( \frac{4.2}{4.8} \times 100\% = 87.5\% \) (accept 88%).
(d)(ii) Any one from:
• The sodium hydrogencarbonate used was impure.
• Not all of the sodium hydrogencarbonate reacted/decomposed (the reaction may not have gone to completion).
• Some product may have been lost during handling (though “other than spillages” is specified, this could include sublimation or dust loss).
• The weighing measurements may have had errors.
▶️ Answer/Explanation
(a) red lead oxide → lead(II) oxide + oxygen
Explanation: Heating red lead oxide (Pb\(_3\)O\(_4\)) causes thermal decomposition. The yellow solid formed is lead(II) oxide (PbO). The gas that relights a glowing splint is oxygen, which supports combustion.
(b)
Method 1: Empirical Formula
• Assume 100 g of sample → 86.6 g Pb, 13.4 g O.
• Moles of Pb = \( \frac{86.6}{207} = 0.4184 \) mol.
• Moles of O = \( \frac{13.4}{16} = 0.8375 \) mol.
• Ratio Pb:O = \( \frac{0.4184}{0.4184} : \frac{0.8375}{0.4184} = 1 : 2 \).
• Empirical formula = PbO\(_2\), which matches the molecular formula of lead(IV) oxide.
Method 2: Percentage Verification
• \(M_r\) of PbO\(_2\) = \(207 + (2 \times 16) = 239\).
• % Pb = \( \frac{207}{239} \times 100 = 86.6\% \).
• % O = \( \frac{32}{239} \times 100 = 13.4\% \).
This matches the given data.
(c)(i)
\[Pb_3O_4(s) + 4HNO_3(aq) \rightarrow 2Pb(NO_3)_2(aq) + PbO_2(s) + 2H_2O(l)\]
Explanation: Red lead oxide (Pb\(_3\)O\(_4\)) is a mixed oxide containing Pb(II) and Pb(IV). With dilute nitric acid, the Pb(II) component forms soluble lead(II) nitrate, while the Pb(IV) component precipitates as insoluble lead(IV) oxide. The state symbols are essential: (s) for solids, (aq) for aqueous, (l) for liquid.
(c)(ii) A description that includes the following key steps:
1. Reaction: Warm/heat the dilute nitric acid. Add the solid red lead oxide (Pb\(_3\)O\(_4\)) in portions with stirring until no more reacts (excess solid ensures all acid is used up).
2. Filtration: Filter the mixture to remove the unreacted red lead oxide and the insoluble brown lead(IV) oxide (PbO\(_2\)) solid. Collect the filtrate, which contains aqueous lead(II) nitrate.
3. Crystallisation: Heat/evaporate the filtrate (lead(II) nitrate solution) gently until crystals start to form on a glass rod (or until a saturated solution is obtained). Do not boil dry.
4. Cooling: Allow the solution to cool slowly, enabling more crystals to form.
5. Isolation: Filter off the formed lead(II) nitrate crystals.
6. Drying: Dry the crystals between sheets of filter paper or in a warm oven/desiccator.
Key Points: The method uses an insoluble reactant (red lead oxide) to produce a soluble salt (lead(II) nitrate). Filtration removes the insoluble products. Crystallisation and drying yield pure, dry crystals.