▶️ Answer/Explanation
(a)(i) A
A is the correct answer because A contains single atoms of one element and the atoms are not joined to other atoms.
B is incorrect because B contains molecules of an element (diatomic molecules).
C is incorrect because C contains molecules of a compound (atoms of different elements bonded together).
D is incorrect because D contains an element but the atoms are joined together to form a giant covalent structure (e.g., diamond).
(a)(ii) C
C is the correct answer because C contains atoms of two different elements chemically bonded together, which is the definition of a compound.
A is incorrect because A is an element (single, uncombined atoms).
B is incorrect because B is an element (diatomic molecules of the same element).
D is incorrect because D is an element (a giant covalent structure of the same element).
(a)(iii) B
B is correct because B shows two atoms of the same element chemically bonded together, e.g., \(H_2\).
A is incorrect because A contains single, uncombined atoms.
C is incorrect because C contains molecules of a compound, e.g., \(H_2O\).
D is incorrect because D contains a giant covalent structure that could have the formula for an element like C (graphite/diamond).
(b)
M1: It contains two different elements / two different types (or sizes) of atoms.
M2: The particles are not chemically joined / not chemically bonded together.
Explanation: A mixture consists of two or more elements or compounds not chemically bonded together. In the diagram, two distinct particle types (e.g., atoms of one element and molecules of another) are simply intermingled without bonds between them.
▶️ Answer/Explanation
(a)(i) Fluorine has the fewest number of electron shells / energy levels.
Additional detail: In Group 7, as you go down the group, atoms gain an extra electron shell (energy level). Fluorine, being at the top of the group, has the fewest shells (two), making its atomic radius the smallest. IGNORE references to protons, neutrons, atomic number, and mass number as the question is specifically about atomic size.
(a)(ii) C
Reasoning: Iodine (I) is in period 5 (so has 5 occupied electron shells) and group 7 (so has 7 electrons in its outer shell).
• A is incorrect; this describes an element like arsenic (period 4, group 15).
• B is incorrect; this describes an element like selenium (period 4, group 16).
• D is incorrect; this has the number of shells and outer electrons reversed for iodine.
(b)(i) Example description:
• Does not need heating.
• Reacts very quickly / violently / explosively.
Marking guidance: Must imply a quicker reaction than chlorine. The answer should include two points: one about no heating required and one about an extremely vigorous reaction.
(b)(ii) As the atoms get bigger / atomic size increases, the reactivity decreases.
Alternative wording: Reactivity decreases down the group as the atomic radius increases.
Explanation: This trend occurs because larger atoms have their outermost electrons farther from the nucleus and are more shielded. This makes it harder for them to gain an electron (which is what defines halogen reactivity) as the nuclear attraction on the incoming electron is weaker.
▶️ Answer/Explanation
(a)(i) Oxygen (or air)
Explanation: Rusting is an oxidation reaction. For iron to rust, it requires both water and oxygen from the air.
(a)(ii) Hydrated iron(III) oxide
Explanation: Rust is a hydrated form of iron(III) oxide, often represented as \( \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \).
(b)(i)
• Plastic acts as a barrier / coating (1).
• Therefore it stops oxygen / water (from air) getting to the iron surface (1).
Explanation: Method A is a barrier method. The plastic physically separates the iron from the two reactants needed for rusting (oxygen and water), preventing the electrochemical reaction from occurring.
(b)(ii) Galvanising
Explanation: Coating iron or steel with a layer of zinc is known as galvanising.
(b)(iii)
• Zinc is more reactive than iron / is higher in the reactivity series (1).
• Zinc reacts / oxidises / corrodes in preference to iron (1).
Explanation: This is sacrificial protection. When the zinc coating is damaged, an electrochemical cell is formed where zinc, being more reactive, acts as the anode and loses electrons (oxidises) more readily than iron. The iron becomes the cathode and is protected from oxidation. The zinc sacrificially corrodes to protect the iron.
▶️ Answer/Explanation
(a)(i) 14
Explanation: Mass number = Number of protons + Number of neutrons. For F: \(7 + 7 = 14\).
(a)(ii) 2.5
Explanation: The atomic number (number of protons) is 7, which is nitrogen. Its electronic configuration is 2.5 (two electrons in the first shell, five in the second).
(a)(iii)
M1: They have the same number of protons / same atomic number.
M2: They have different numbers of neutrons.
Explanation: Isotopes are atoms of the same element (same proton number) with different numbers of neutrons, hence different mass numbers. F and G both have 7 protons but 7 and 8 neutrons respectively.
(a)(iv)
M1: (It has) three more electrons than protons.
M2: Electrons have a negative charge and protons have a positive charge / The overall charge is negative because the negative charges outweigh the positive charges.
Explanation: Species H has 7 protons (+) and 10 electrons (−). The net charge is \( (7+) + (10-) = 3- \), making it a negative ion (anion).
(b) 12.01
Explanation and Calculation:
The relative atomic mass \(A_r\) is the weighted mean mass of the isotopes. \[ A_r = \frac{(98.930 \times 12) + (1.070 \times 13)}{100} \] Calculate step-by-step:
\(98.930 \times 12 = 1187.16\)
\(1.070 \times 13 = 13.91\)
Sum: \(1187.16 + 13.91 = 1201.07\)
Divide by 100: \(1201.07 \div 100 = 12.0107\)
Rounded to two decimal places: 12.01
Mark Scheme Note: The answer must be given to two decimal places. Award 1 mark if the sum (1201.07) is found but not divided by 100, provided the final answer is to 2 d.p.
▶️ Answer/Explanation
(a)
Any one from:
• add more limewater (to cover the glass tube on the left)
• the glass tube on the left should be longer / should be submerged in the limewater.
Explanation: For gas to bubble through the limewater, the delivery tube from syringe A must be below the liquid surface.
(b)(i)
Volume of carbon dioxide removed = initial volume – final volume = \(76 \, \text{cm}^3 – 66 \, \text{cm}^3 = 10 \, \text{cm}^3\).
Percentage by volume of \( \text{CO}_2 = \frac{10}{76} \times 100 \approx 13.2\% \) (to 3 significant figures).
(b)(ii)
The limewater turns from colourless to cloudy/milky.
Explanation: Carbon dioxide reacts with calcium hydroxide in limewater to form insoluble calcium carbonate, which appears as a white precipitate.
(b)(iii)
• The percentage/amount of carbon dioxide in the air is too small (approximately 0.04%).
• Therefore, the change in volume in the syringes would be too small to measure accurately with this apparatus (less than 1 cm³ change on a 76 cm³ sample).
• The gas syringes are not precise/accurate enough to measure such a small volume change.
(c)(i)
Copper(II) oxide (CuO) / copper oxide.
Explanation: Copper reacts with oxygen when heated to form copper(II) oxide, a black solid.
(c)(ii)
The powder has a greater surface area (than larger pieces of copper).
Explanation: A greater surface area allows more copper atoms to be exposed to the oxygen gas at once, so the reaction occurs more quickly, ensuring all oxygen is removed efficiently in the experiment.
(c)(iii)
• Argon has a full outer shell of electrons (eight electrons in its outer shell).
• Therefore, it does not readily lose, gain, or share electrons / it has a stable electron configuration.
• This makes it extremely unreactive as it has no tendency to form chemical bonds.
Explanation: This is a characteristic of all noble gases (Group 0), which are chemically inert due to their stable electronic structures.
▶️ Answer/Explanation
(a)(i) Y
Reason: Y contains an oxygen atom in an –OH group (butan-1-ol), so it is not a hydrocarbon (compounds containing only hydrogen and carbon).
(a)(ii) V
Reason: V is methylpropane, a saturated alkane. Its molecular formula is \(\text{C}_4\text{H}_{10}\), which simplifies to the empirical formula \(\text{CH}_{2.5}\). However, the question states “empirical formula \(\text{CH}_2\)”, which is the empirical formula for alkenes (e.g., butene). There seems to be a discrepancy. Based on the mark scheme, the intended saturated hydrocarbon with empirical formula \(\text{CH}_2\) is likely a cycloalkane, but among U–Y, V is a saturated alkane. The provided answer in the mark scheme is V, so it is accepted here.
(a)(iii) W
Reason: W is but-1-ene, an alkene. Alkenes undergo addition reactions with bromine (decolorizing bromine water) without requiring UV light. The question mentions “reacts with bromine in the presence of ultraviolet radiation”, which is characteristic of alkanes (substitution). The provided answer is W, but note: alkenes react with bromine readily at room temperature; UV light is needed for alkanes. There may be a misprint. Based on the mark scheme, W is the answer given.
(a)(iv) X
Reason: The repeat unit \(-\text{CH}(\text{CH}_3)-\text{CH}_2-\) comes from the polymerisation of but-2-ene (compound X). The \(\text{CH}_3\) side group indicates the monomer had a methyl group on one of the carbons in the double bond.
(a)(v) Displayed formula of but-1-ene or methylpropene (2-methylprop-1-ene).
Example for but-1-ene:
\(\displaystyle \text{H}-\text{C}=\text{C}-\text{C}-\text{C}-\text{H}\) (with H atoms arranged correctly: \(\text{H}_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3\))
(a)(vi) Any two from:
• Same general formula.
• Similar chemical properties.
• Gradual trend in physical properties (e.g., boiling point increases with chain length).
• Each consecutive member differs by a \(\text{CH}_2\) group.
(b)(i)
Step 1: Assume 100 g of compound Z.
Mass of C = 38.7 g, H = 9.7 g, O = 51.6 g.
Step 2: Find moles.
Moles of C = \( \frac{38.7}{12} = 3.225 \)
Moles of H = \( \frac{9.7}{1} = 9.7 \)
Moles of O = \( \frac{51.6}{16} = 3.225 \)
Step 3: Divide by smallest (3.225).
C : H : O = \( \frac{3.225}{3.225} : \frac{9.7}{3.225} : \frac{3.225}{3.225} \approx 1 : 3 : 1 \)
Empirical formula = \(\text{CH}_3\text{O}\).
(b)(ii)
Empirical formula mass of \(\text{CH}_3\text{O}\) = 12 + (3×1) + 16 = 31.
\( \frac{M_r}{ \text{Empirical mass}} = \frac{62}{31} = 2 \)
Molecular formula = \( (\text{CH}_3\text{O})_2 = \text{C}_2\text{H}_6\text{O}_2 \).
▶️ Answer/Explanation
(a)(i) \( \text{N}_2 + \text{O}_2 \rightarrow 2\text{NO} \)
Correct balancing is required, but the mark scheme allows for fractions and multiples.
(a)(ii) The reaction has a high activation energy (\(E_a\)), so a high temperature is required to give the reactant molecules enough energy to overcome this barrier and react.
(a)(iii) Oxides of nitrogen contribute to acid rain, which can damage buildings, harm aquatic life, and damage forests. They can also cause respiratory problems in humans and contribute to smog formation.
(b)(i) A catalyst provides an alternative reaction pathway with a lower activation energy. More colliding particles now have energy equal to or greater than this lower activation energy, resulting in a higher frequency of successful collisions and an increased rate of reaction. The catalyst remains chemically unchanged at the end.
(b)(ii) Increasing the pressure of a gas increases the concentration of gas particles (more particles in a given volume). This brings the particles closer together, leading to more frequent collisions per unit time, and therefore a higher rate of reaction.
(c)(i)
The diagram should show a central nitrogen atom bonded to three hydrogen atoms. Each N–H bond should be represented by a shared pair of electrons (e.g., a pair of dots or a line). The nitrogen atom should also have one lone pair (non-bonding pair) of electrons.
M1: One pair of electrons between the N and each H.
M2: Two non-bonding electrons on the N.
(c)(ii) A covalent bond involves the strong electrostatic attraction between the positively charged nuclei of the bonded atoms and the shared pair(s) of negatively charged electrons.
The mark scheme accepts either description: attraction between nuclei and shared electrons, or attraction between shared electrons and nuclei.
(c)(iii) Ammonia has a simple molecular structure. The covalent bonds within each molecule are strong. However, the forces between the molecules (intermolecular forces) are weak. These weak intermolecular forces require only a small amount of energy to overcome when changing from liquid to gas, resulting in a low boiling point.
Mention of breaking covalent bonds would be incorrect and score zero.
▶️ Answer/Explanation
(a) A method to produce dry, hydrated crystals:
- React the reactants: Add barium carbonate powder to dilute hydrochloric acid in a beaker, one spatula at a time, until the barium carbonate is in excess (i.e., until some solid remains undissolved and effervescence/fizzing from \( \text{CO}_2 \) production stops).
- Remove excess reactant: Filter the mixture to remove the excess, unreacted barium carbonate solid. The filtrate is an aqueous solution of barium chloride.
- Concentrate the solution: Gently heat the filtrate in an evaporating basin to evaporate some water until the solution is saturated. This can be identified when crystals start to form on the surface or on a glass rod dipped into the solution.
- Crystallise: Allow the hot, saturated solution to cool. Crystals of hydrated barium chloride (\( \text{BaCl}_2 \cdot 2\text{H}_2\text{O} \)) will form as the solubility decreases with cooling.
- Separate the crystals: Filter the mixture to collect the crystals, separating them from the remaining solution (the mother liquor).
- Dry the crystals: Leave the crystals in a warm place to dry, or dry them between filter papers. Avoid oven drying at high temperatures, as this may drive off the water of crystallisation.
Note: Evaporating the solution to dryness would yield an anhydrous powder, not hydrated crystals, so the cooling/crystallisation step is essential.
(b) Test for sulfate ions in a solution also containing carbonate ions:
- First, add dilute acid (e.g., nitric acid or hydrochloric acid) to the test solution. This will react with the carbonate ions (\( \text{CO}_3^{2-} \)) to produce carbon dioxide gas (effervescence), removing them from the solution. This step is crucial because barium carbonate (\( \text{BaCO}_3 \)) is also a white precipitate and would give a false positive if the carbonate ions were not removed first.
- Then, add a few drops of barium chloride solution to the acidified test solution. The formation of a white precipitate confirms the presence of sulfate ions, due to the formation of insoluble barium sulfate (\( \text{BaSO}_4 \)).
Ionic equations:
For carbonate removal: \( \text{CO}_3^{2-} (\text{aq}) + 2\text{H}^+ (\text{aq}) \rightarrow \text{CO}_2 (\text{g}) + \text{H}_2\text{O} (\text{l}) \)
For sulfate test: \( \text{Ba}^{2+} (\text{aq}) + \text{SO}_4^{2-} (\text{aq}) \rightarrow \text{BaSO}_4 (\text{s}) \)
▶️ Answer/Explanation
(a)(i) & (ii)
The points should be plotted correctly on the provided grid (Mass on x-axis, Temperature on y-axis). A smooth curve of best fit should be drawn, passing through or near most points, but ignoring the anomalous result (which, from the mark scheme, appears to be the first data point at 0.0g, 17.0°C, or possibly the point at 2.0g, 12.9°C which shows a smaller decrease than expected). The curve will show a steep initial decrease in temperature that gradually levels off.
(a)(iii)
Polystyrene is a good thermal insulator / is a poor conductor of heat. This reduces heat transfer (gain or loss) between the reaction mixture and the surroundings. Therefore, the temperature change measured is closer to the true value for the reaction itself, making the results more accurate. A glass beaker would conduct heat more easily, leading to greater heat exchange with the environment and a less accurate measurement.
(a)(iv)
A possible reason is that the student recorded the initial temperature of the acid before adding the first portion of sodium carbonate, but the temperature might still have been decreasing from the initial setup/measuring process. Alternatively, the student may have forgotten to stir the mixture thoroughly after adding a portion, leading to a localised temperature measurement that wasn’t representative of the whole mixture.
(a)(v)
The results show that after adding about 4.5 g of sodium carbonate, the temperature stops decreasing (remains constant at 11.0 °C for 4.5g and 5.0g). This indicates that no further reaction is taking place because the limiting reactant (the hydrochloric acid) has been completely used up. Adding more sodium carbonate no longer causes an endothermic temperature change.
(a)(vi)
The reaction is endothermic. This is shown by the decrease in temperature of the reaction mixture as sodium carbonate is added. An endothermic reaction takes in thermal energy from its surroundings, causing the temperature to fall. The results clearly demonstrate this cooling effect.
(b)(i)
The cotton wool plug allows the carbon dioxide gas to escape (so the mass loss can be measured) while preventing acid or solution from splashing out of the flask, which could affect the mass reading or be a safety hazard.
(b)(ii)
Calculation:
1. \( M_r(\text{Na}_2\text{CO}_3) = (2 \times 23) + 12 + (3 \times 16) = 106 \)
2. Moles of \(\text{Na}_2\text{CO}_3 = \frac{2.12}{106} = 0.02 \text{ mol}\)
3. From the equation, 1 mole of \(\text{Na}_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\).
4. \( M_r(\text{CO}_2) = 12 + (2 \times 16) = 44 \)
5. Mass of \(\text{CO}_2 = 0.02 \times 44 = 0.88 \text{ g}\)
Answer: 0.88 g
(b)(iii)
The actual mass produced might be less than the theoretical maximum (0.88 g) because:
1. The sodium carbonate may not be pure.
2. Some carbon dioxide gas produced may dissolve in the aqueous solution/acid before escaping.
3. The reaction may not have gone to completion.
4. Some gas may have escaped before the cotton wool was placed or was not all collected.
▶️ Answer/Explanation
(a)(i) fractional distillation
This method separates liquids with different boiling points. Ethanol boils at ~78°C and water at 100°C.
(a)(ii) evaporation (or boiling)
At point A (in the flask), liquid ethanol is heated and changes to a vapour/gas.
(a)(iii) condensation
At point B (in the condenser), the ethanol vapour cools and changes back to a liquid.
(b)(i) Moles of ethanol calculation:
Step 1: Calculate mass of ethanol.
\( \text{mass} = \text{volume} \times \text{density} = 15.5 \, \text{cm}^3 \times 0.79 \, \text{g/cm}^3 = 12.245 \, \text{g} \)
Step 2: Calculate moles using \( M_r = 46 \).
\( \text{moles} = \frac{\text{mass}}{M_r} = \frac{12.245}{46} = 0.266 \, \text{mol} \) (or 0.267 mol)
Alternative method: Moles per cm³ = \(0.79 / 46 = 0.0172 \, \text{mol/cm}^3\), then \(0.0172 \times 15.5 = 0.267 \, \text{mol}\).
(b)(ii) Number of molecules:
Molecules = moles × Avogadro’s constant
\( = 0.266 \, \text{mol} \times 6.00 \times 10^{23} \, \text{molecules/mol} = 1.60 \times 10^{23} \, \text{molecules} \)
Note: Using 0.267 mol gives \(1.60 \times 10^{23}\) (same to 3 s.f.).
(c)(i) Chemical test for water:
Add anhydrous copper(II) sulfate (white solid). If water is present, it turns blue.
Alternative: Add anhydrous cobalt chloride paper; it turns from blue to pink.
(c)(ii) Physical test for pure water:
Measure the boiling point. Pure water boils at exactly \(100 \, ^\circ \text{C}\) at standard atmospheric pressure.
Alternative: Measure the freezing point; pure water freezes at \(0 \, ^\circ \text{C}\).
(d)(i) Heat energy change (\(Q\)):
Step 1: Calculate temperature change, \(\Delta T\).
\( \Delta T = 70.5 \, ^\circ \text{C} – 21.0 \, ^\circ \text{C} = 49.5 \, ^\circ \text{C} \)
Step 2: Use \( Q = mc\Delta T \).
\( Q = 100 \, \text{g} \times 4.2 \, \text{J/g/}^\circ \text{C} \times 49.5 \, ^\circ \text{C} = 20790 \, \text{J} \)
This can be rounded to \(20800 \, \text{J}\) or written as \(20.79 \, \text{kJ}\).
(d)(ii) Molar enthalpy change (\(\Delta H\)):
Step 1: Convert \(Q\) to kJ.
\( Q = 20790 \, \text{J} = 20.79 \, \text{kJ} \)
Step 2: Calculate \(\Delta H\) for 0.0200 mol of ethanol burned.
\( \Delta H = \frac{-20.79 \, \text{kJ}}{0.0200 \, \text{mol}} = -1039.5 \, \text{kJ/mol} \)
The sign is negative because combustion is an exothermic reaction (heat is released).
Final answer: \(\Delta H = -1040 \, \text{kJ/mol}\) (to 3 significant figures).
▶️ Answer/Explanation
(a)(i) Displacement (ALLOW redox)
A more reactive metal (manganese) displaces a less reactive metal ion (chromium) from its salt solution. This is a single displacement (redox) reaction.
(a)(ii) most reactive manganese
least reactive cadmium
From Table 1: Mn displaces Cr, so Mn > Cr. Cr displaces Cd, so Cr > Cd. Sn does not displace Cd, so Cd > Sn. Therefore, the order is Mn > Cr > Cd > Sn.
(a)(iii) (manganese) chromium cadmium tin
Any three metals that correctly fit into a standard reactivity series (e.g., potassium, sodium, calcium, magnesium, aluminium, zinc, iron, lead, copper, silver, gold) are acceptable here. The answer given is from the mark scheme for the specific series being constructed.
(b) Copper added to magnesium sulfate solution:
• M1: No colour change.
• M2: Copper is less reactive than magnesium / copper cannot displace magnesium.
Copper is below magnesium in the reactivity series, so no reaction occurs. The blue solution remains blue, and the brown copper metal remains brown.
Zinc added to iron sulfate solution:
• M3: Zinc turns (from light grey to) dark grey / becomes coated in a dark grey metal.
• M4: Solution turns (from green to) colourless.
• M5: Zinc is more reactive than iron / zinc displaces iron / zinc reduces iron ions.
Zinc is above iron in the reactivity series. Zinc atoms displace Fe\(^{2+}\) ions from solution, forming iron metal (dark grey coating) and colorless zinc sulfate solution.
(c)(i) Any two from:
• Concentration of dilute sulfuric acid (M1)
• Temperature (M2)
• Surface area of the metal / size of piece of metal (M3)
To make a fair comparison and determine a valid order of reactivity, these factors that affect the rate of reaction must be kept constant for each metal tested.
(c)(ii) Calcium sulfate forms a (solid) layer / coating around the calcium metal. (ALLOW calcium sulfate prevents the sulfuric acid coming into contact with calcium).
The insoluble product, calcium sulfate (\(CaSO_4\)), coats the remaining calcium metal, forming a barrier that prevents further contact between the acid and the metal, stopping the reaction.
(d) Show by calculation:
Method 1: Comparing moles
• M1: Moles of aluminium = \( \frac{1.00}{27} = 0.0370\) mol (3 s.f.)
• From the equation: \( 2Al + 3H_2SO_4 \)
Moles of \(H_2SO_4\) needed for all Al = \( 0.0370 \times \frac{3}{2} = 0.0556\) mol
• M2: Moles of \(H_2SO_4\) available = 0.0600 mol.
Since 0.0600 mol > 0.0556 mol, sulfuric acid is in excess.
Method 2: Comparing mass
• M1: Moles of \(H_2SO_4\) available = 0.0600 mol.
From the equation: Moles of Al needed for all acid = \( 0.0600 \times \frac{2}{3} = 0.0400\) mol.
• M2: Mass of Al needed = \( 0.0400 \times 27 = 1.08\) g.
Since only 1.00 g of Al is present (< 1.08 g), sulfuric acid is in excess.
Either method clearly shows that the amount of sulfuric acid present is more than required to completely react with the given aluminium.