Chemistry 5070/21 — May/June 2024

Walkthrough

The Haber process combines nitrogen and hydrogen to make ammonia. The reaction is slow without a catalyst, so iron is used to speed it up. Iron is a transition metal and is a common catalyst. From the list, iron is the only substance that fits.

Key Takeaways

The Haber process uses an iron catalyst to increase the rate of reaction without being used up.

Common Mistakes

Choosing vanadium(V) oxide – that is the catalyst for the Contact process (making sulfuric acid), not the Haber process.

Things to Be Careful About

Remember the specific catalyst for each industrial process: iron for Haber, vanadium(V) oxide for Contact.

Walkthrough

A giant covalent structure is a network of atoms held together by covalent bonds in a giant lattice. Diamond and graphite are forms of carbon with giant covalent structures. Silicon(IV) oxide (silica) also has a giant covalent structure, similar to diamond. From the list, both carbon and silicon(IV) oxide qualify. The mark scheme allows either.

Key Takeaways

Giant covalent structures include diamond, graphite, and silicon(IV) oxide. They have high melting points and are hard (except graphite is soft).

Common Mistakes

Choosing glucose or methanol – these are simple molecular substances with low melting points.

Things to Be Careful About

Carbon exists in several forms; the question does not specify which, so carbon is acceptable. Silicon(IV) oxide is also correct.

Walkthrough

Hydrated copper(II) sulfate contains water of crystallisation. When heated, the water is driven off, leaving anhydrous copper(II) sulfate. The blue hydrated salt turns white when anhydrous. This is a classic test for water: adding water to white anhydrous copper(II) sulfate turns it blue.

Key Takeaways

Hydrated copper(II) sulfate is blue; anhydrous copper(II) sulfate is white. Heating removes water of crystallisation.

Common Mistakes

Confusing hydrated and anhydrous forms – the question specifies 'hydrated', so the blue one.

Things to Be Careful About

The colour change is blue to white on heating, and white to blue on adding water.

Walkthrough

During treatment of domestic water, carbon (often activated carbon) is used to remove tastes and odours. It adsorbs impurities onto its surface. From the list, carbon is the correct substance.

Key Takeaways

Activated carbon is used in water treatment to remove tastes and odours by adsorption.

Common Mistakes

Choosing chlorine – chlorine is used to kill bacteria, not to remove tastes and odours.

Things to Be Careful About

Carbon removes tastes and odours; chlorine disinfects. The question specifically asks for tastes and odours.

Walkthrough

The empirical formula is the simplest whole-number ratio of atoms in a compound. Glucose has the molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$. Dividing all subscripts by 6 gives $\text{CH}_2\text{O}$. Methanoic acid is $\text{CH}_2\text{O}_2$, which is already simplest and does not match. Methanol is $\text{CH}_4\text{O}$, which also does not match. So glucose is the only one with empirical formula $\text{CH}_2\text{O}$.

Key Takeaways

The empirical formula is the simplest ratio; the molecular formula is a multiple of it. Glucose's empirical formula is $\text{CH}_2\text{O}$.

Common Mistakes

Choosing methanoic acid – its formula is $\text{CH}_2\text{O}_2$, not $\text{CH}_2\text{O}$.

Things to Be Careful About

Check the ratio of C:H:O in the molecular formula before deciding.

Walkthrough

The question states that calcium carbide ($\text{CaC}_2$) reacts with water ($\text{H}_2\text{O}$) to form ethyne ($\text{C}_2\text{H}_2$) and calcium hydroxide. We need to write the formula for calcium hydroxide and balance the equation.

1. Identify formulae: Calcium is in Group II, so it forms $\text{Ca}^{2+}$. Hydroxide is $\text{OH}^-$. To balance charges, calcium hydroxide is $\text{Ca(OH)}_2$. Water is $\text{H}_2\text{O}$. Ethyne is given as $\text{C}_2\text{H}_2$. Calcium carbide is given as $\text{CaC}_2$.
2. Write unbalanced equation: $\text{CaC}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{C}_2\text{H}_2$.
3. Balance: There are 2 hydrogens and 1 oxygen on the left (in water) but 2 hydroxide groups (2 O, 2 H) and 2 H in ethyne on the right. Total H on right = 2 (from hydroxide) + 2 (from ethyne) = 4. Total O on right = 2. So we need 2 water molecules on the left: $2\text{H}_2\text{O}$.
4. Check balance:
   • Ca: 1 left, 1 right.
   • C: 2 left, 2 right.
   • H: 4 left ($2 \times 2$), 4 right (2 in hydroxide + 2 in ethyne).
   • O: 2 left, 2 right.

The equation is balanced.

Key Takeaways

• Know the formula for common ions like hydroxide ($\text{OH}^-$) and combine them with metal ions (e.g., $\text{Ca}^{2+} + 2\text{OH}^- \rightarrow \text{Ca(OH)}_2$).
• Balancing equations involves checking atom counts for each element on both sides.

Common Mistakes

• Writing the formula for calcium hydroxide as $\text{CaOH}$ or $\text{CaO}_2\text{H}_2$ instead of $\text{Ca(OH)}_2$.
• Forgetting to balance the water molecules (using 1 instead of 2).
• Writing state symbols where not required (though often good practice, the mark scheme here focuses on formulae and balance).

Things to Be Careful About

• Ensure the formula for calcium hydroxide includes brackets around the hydroxide group: $\text{Ca(OH)}_2$, not $\text{CaO}_2\text{H}_2$.
• The mark scheme awards one mark for the correct formula of calcium hydroxide as a product and one for the balanced equation. Both are needed for full marks.

Walkthrough

The problem states calcium carbide is ionic and gives the formula $\text{CaC}_2$.

1. Identify cation charge: Calcium is in Group II of the Periodic Table. It loses 2 electrons to achieve a stable octet, forming the $\text{Ca}^{2+}$ ion.
2. Determine anion charge: The formula is $\text{CaC}_2$. Since the compound is electrically neutral, the total positive charge must equal the total negative charge. We have one $\text{Ca}^{2+}$ ($+2$ charge). Therefore, the $\text{C}_2$ part must have a charge of $-2$.
3. Write ion formula: The ion is the carbide ion, consisting of two carbon atoms with a $2-$ charge: $\text{C}_2^{2-}$.

Key Takeaways

• Ionic compounds are neutral; total positive charge = total negative charge.
• Group II metals form $2+$ ions.
• Some anions are polyatomic (like $\text{C}_2^{2-}$ or $\text{O}_2^{2-}$).

Common Mistakes

• Assuming the carbide ion is just $\text{C}^{2-}$. The formula $\text{CaC}_2$ shows two carbon atoms per calcium atom, so the ion must contain two carbons.
• Forgetting the charge on the ion.

Things to Be Careful About

• The question asks for the formula of the carbide ion, not the element. Include the charge: $\text{C}_2^{2-}$.

Walkthrough

This is a three-step mole calculation. First convert the volume of solution from $\text{cm}^3$ to $\text{dm}^3$ because concentration is in $\text{mol / dm}^3$. Then multiply concentration by volume to find the amount of ammonium nitrite. The equation shows a 1:1 mole ratio between $\text{NH}_4\text{NO}_2$ and $\text{N}_2$, so the amount of nitrogen is the same. Finally multiply this amount by the molar gas volume at r.t.p., $24\,\text{dm}^3\text{/mol}$, to get the volume of gas. Round to two significant figures at the end.

Key Takeaways

• Amount (mol) = concentration ($\text{mol / dm}^3$) x volume ($\text{dm}^3$).
• At r.t.p., one mole of any gas occupies $24\,\text{dm}^3$.
• The balanced equation gives the mole ratio needed to convert amount of reactant to amount of product.

Common Mistakes

• Forgetting to convert $25.0\,\text{cm}^3$ to $0.0250\,\text{dm}^3$ before using the concentration.
• Using the wrong mole ratio (the equation is 1:1 here).
• Rounding to two significant figures too early; keep the full value until the final step.
• Omitting the unit $\text{dm}^3$ in the final answer.

Things to Be Careful About

• The answer must be given to two significant figures, so $0.0798$ becomes $0.080$.
• Use the molar gas volume at r.t.p. ($24\,\text{dm}^3\text{/mol}$), not s.t.p. ($22.4\,\text{dm}^3\text{/mol}$).
• The mark scheme awards one mark for the amount of ammonium nitrite, one for the volume calculation, and one for the final rounded answer.

Walkthrough

Increasing the temperature gives the particles more kinetic energy, so they move faster. This has two effects: collisions happen more often, and a greater proportion of collisions have energy at least equal to the activation energy. The mark scheme accepts either 'more successful collisions' or 'more collisions with energy equal to or greater than the activation energy' as the second point.

Key Takeaways

• Temperature affects rate through particle kinetic energy and collision frequency.
• Only collisions with energy equal to or greater than the activation energy lead to reaction.

Common Mistakes

• Saying only 'particles move faster' without linking it to successful collisions.
• Saying 'more collisions' without mentioning energy/activation energy; the mark scheme wants the successful/effective collision idea.

Things to Be Careful About

• The question asks to 'describe and explain', so both the effect on rate and the reason are needed.
• Use the phrase 'successful collisions' or 'collisions with energy equal to or greater than the activation energy'.

Walkthrough

Decreasing the concentration means fewer ammonium nitrite particles in the same volume, so the particles are less crowded. As a result, collisions between reactant particles happen less frequently, so the rate decreases. The mark scheme separates these into two points: fewer particles per unit volume, and lower collision frequency.

Key Takeaways

• Concentration affects rate through the frequency of collisions.
• Lower concentration leads to fewer particles per unit volume, fewer collisions per second, and a lower rate.

Common Mistakes

• Saying 'fewer particles' without specifying 'per unit volume' or 'less crowded'.
• Confusing concentration with temperature; concentration does not change particle energy.

Things to Be Careful About

• The mark scheme wants two distinct ideas: crowding and collision frequency.
• Do not say particles move more slowly; decreasing concentration does not change their speed.

Walkthrough

A pH meter is one way to measure pH. Another standard method is to use universal indicator paper (or universal indicator solution): add a drop of the solution to the paper, then compare the colour produced with the colour chart supplied with the indicator. The colour corresponds to a pH value.

Key Takeaways

• pH can be measured with a pH meter or with an indicator and a colour chart.
• Universal indicator gives a colour that changes gradually across the pH range.

Common Mistakes

• Naming litmus paper, which only distinguishes acid/alkali and does not give a pH value.
• Saying 'use universal indicator' without mentioning comparing the colour to a pH chart; the chart is the part that gives the pH.

Things to Be Careful About

• The mark scheme awards one mark for universal indicator paper and one for matching the colour with a pH colour chart.
• 'Other way' means not a pH meter, so do not describe using a pH meter.

Walkthrough

Potassium iodide is ionic. In the solid, a giant lattice of K⁺ and I⁻ ions is held together by strong electrostatic forces. Melting means breaking these forces, which needs a large amount of energy, hence a high melting point.

Key Takeaways

• Ionic compounds have high melting points because of strong ionic bonds.
• The forces are electrostatic attractions between oppositely charged ions.

Common Mistakes

• Saying 'strong bonds between molecules' – ionic compounds do not have molecules.
• Omitting that the attraction is between positive and negative ions.

Things to Be Careful About

• Use 'electrostatic' not just 'attraction' to be precise.
• Mention that breaking these forces requires energy.

Walkthrough

Potassium is a Group I metal with one outer electron. Iodine is a Group VII non-metal that needs one electron to complete its octet. In the reaction, two potassium atoms each transfer one electron to one iodine molecule. The iodine molecule splits into two iodide ions, each having gained one electron. This electron transfer forms K⁺ and I⁻ ions.

Key Takeaways

• Metals lose electrons to form positive ions.
• Non-metals gain electrons to form negative ions.
• Electron transfer must be balanced.

Common Mistakes

• Saying 'iodine gains one electron' – must be per molecule (two electrons).
• Not specifying that each potassium atom loses one electron.

Things to Be Careful About

• Use 'lose' and 'gain' correctly.
• Mention the number of electrons per atom and per molecule.

Walkthrough

In concentrated aqueous potassium iodide, the ions present are K⁺, I⁻, H⁺ and OH⁻. At the anode, iodide ions are discharged in preference to hydroxide ions because iodide is higher in the discharge series (easier to oxidise). At the cathode, hydrogen ions are discharged in preference to potassium ions because hydrogen is lower in the reactivity series (easier to reduce). Thus iodine forms at the anode and hydrogen at the cathode.

Key Takeaways

• In aqueous electrolysis, the ion that is easier to discharge is selected.
• Halide ions (except fluoride) are discharged at the anode over hydroxide.
• Hydrogen is discharged at the cathode over metals above hydrogen in reactivity series.

Common Mistakes

• Predicting potassium at the cathode – incorrect because potassium is too reactive.
• Predicting oxygen at the anode – only if iodide is not present.

Things to Be Careful About

• State symbols: iodine is (aq) or (s) depending on concentration; hydrogen is (g).
• Remember the discharge series order.

Walkthrough

Acidified potassium manganate(VII) is a strong oxidising agent. It oxidises iodide ions to iodine. The manganate(VII) ion is reduced to Mn²⁺, which is almost colourless in aqueous solution. The iodine produced gives a brown colour. Therefore, the purple colour fades to colourless and a brown colour appears.

Key Takeaways

• Manganate(VII) is purple; its reduced form is colourless.
• Iodine in aqueous solution is brown.
• This is a redox reaction.

Common Mistakes

• Saying 'purple to brown' – the manganate becomes colourless, not brown.
• Missing the brown colour from iodine.

Things to Be Careful About

• Acidified manganate(VII) is required for the reaction.
• The brown colour may be described as 'yellow-brown'.

Walkthrough

In the equation, each iodide ion loses one electron to become an iodine atom; two iodide ions lose two electrons overall to form one I₂ molecule. Loss of electrons is oxidation. Meanwhile, each chlorine atom in Cl₂ gains one electron to become Cl⁻; the chlorine molecule gains two electrons overall. Gain of electrons is reduction. Thus the reaction involves both oxidation and reduction.

Key Takeaways

• Oxidation = loss of electrons.
• Reduction = gain of electrons.
• Redox reactions involve simultaneous oxidation and reduction.

Common Mistakes

• Confusing which species is oxidised/reduced.
• Not stating the electron transfer explicitly.

Things to Be Careful About

• Use 'loses electrons' and 'gains electrons' precisely.
• Mention both species and the electron change.

Walkthrough

The system is at equilibrium, meaning the forward and backward reactions happen at the same rate. When we change the conditions, the position of equilibrium shifts to oppose the change — this is Le Chatelier's principle.

Increasing the temperature adds heat energy to the system. The forward reaction is endothermic, which means it absorbs heat energy. So the equilibrium shifts in the forward direction to use up the added heat. That moves the position of equilibrium to the right, toward the products (zinc oxide and sulfur dioxide).

The question says the pressure is kept constant, so we only need to think about the temperature change — no pressure effect to consider here.

Key Takeaways

• Le Chatelier's principle: a system at equilibrium opposes any change made to it.
• Increasing temperature favours the endothermic direction (the direction that absorbs heat).
• The forward reaction being endothermic is the clue that tells you which way the shift goes.

Common Mistakes

• Saying the equilibrium moves to the left — increasing temperature favours the endothermic direction, which here is the forward (right) direction.
• Confusing endothermic (absorbs heat) with exothermic (releases heat).
• Forgetting to give the reason (to absorb thermal energy) — the mark scheme awards a separate mark for it.

Things to Be Careful About

The mark scheme wants two points: (1) the position moves to the right / product side, and (2) the reason — to absorb thermal energy. Both are needed for the 2 marks. The phrase 'to absorb thermal energy' (or equivalent) is the second mark, so don't leave it out.

Walkthrough

Decreasing the pressure favours the side of the equilibrium with more moles of gas — the system shifts to increase the pressure again by producing more gas.

Count the moles of gas on each side. On the left, ZnSO$_3$ is a solid, so it contributes zero moles of gas. On the right, ZnO is a solid but SO$_2$ is a gas, so there is 1 mole of gas. The right-hand side therefore has more moles of gas.

So decreasing the pressure shifts the equilibrium to the right, toward the products.

Key Takeaways

• Decreasing pressure favours the side with more moles of gas.
• Only gaseous species count when considering pressure changes — solids and liquids are ignored.

Common Mistakes

• Counting the solids (ZnSO$_3$ and ZnO) as gas — they don't count for pressure effects.
• Saying the equilibrium moves to the left.
• Giving the direction without the reason; the mark scheme wants 'more moles of gas on the right-hand side'.

Things to Be Careful About

The two marks are for (1) the direction (to the right / product side) and (2) the reason (more moles of gas on the right-hand side). Both points must be stated. Note that even though there is only one mole of gas on the product side, the reactant side has none, so the product side has more gas.

Walkthrough

The balanced equation is:

The mole ratio of ZnSO$_3$ to ZnO is 1 : 1, so the number of moles of ZnO formed equals the number of moles of ZnSO$_3$ used.

First, find the relative formula masses:

Convert the mass of ZnSO$_3$ to moles:

Since the ratio is 1 : 1, 0.1759 mol of ZnO is formed. Convert to mass:

Key Takeaways

• Reacting-mass calculations: mass → moles (divide by $M_r$), use the mole ratio from the balanced equation, then moles → mass (multiply by $M_r$).
• The balanced equation tells you the mole ratio — here it's 1 : 1.

Common Mistakes

• Using the wrong $M_r$ — check each element's atomic mass (Zn = 65, S = 32, O = 16).
• Forgetting the 1 : 1 ratio.
• Rounding too early and getting a slightly different final answer (the mark scheme gives 14.2448 g; 14.2 g to 3 significant figures is correct).
• Omitting the unit 'g' in the final answer.

Things to Be Careful About

The mark scheme awards a mark for each of: correct $M_r$ values, correct moles (25.5/145 = 0.1759 mol), and correct final mass (14.2448 g ≈ 14.2 g). Show all three stages of working to be sure of all three marks. Answer to 3 significant figures, since 25.5 has 3 significant figures.

Walkthrough

Zinc oxide reacts with both an acid (dilute hydrochloric acid) and a base (aqueous sodium hydroxide). An oxide that reacts with both acids and bases is called amphoteric. Zinc oxide is a well-known amphoteric oxide.

Sulfur dioxide is the oxide of a non-metal (sulfur). Non-metal oxides are generally acidic — they react with bases (alkalis) such as sodium hydroxide, but not with acids. So sulfur dioxide reacts with aqueous sodium hydroxide but not with dilute hydrochloric acid.

Key Takeaways

• Amphoteric oxides react with both acids and bases — zinc oxide is an example.
• Acidic oxides (usually non-metal oxides) react with bases but not with acids — sulfur dioxide is an example.

Common Mistakes

• Calling zinc oxide simply 'basic' — it is amphoteric because it reacts with both acids and alkalis.
• Calling sulfur dioxide amphoteric — it is acidic, reacting only with bases.
• Not naming both oxides' behaviour; the mark scheme awards one mark for each.

Things to Be Careful About

The two marks are for 'zinc oxide is amphoteric' and 'sulfur dioxide is acidic'. Use the correct terms — 'amphoteric' and 'acidic' — as the mark scheme accepts these exact descriptors.

Walkthrough

Write the word equation first: zinc sulfite + nitric acid → zinc nitrate + water + sulfur dioxide.

Identify the formulae:

• Zinc sulfite: ZnSO$_3$
• Nitric acid: HNO$_3$
• Zinc nitrate: Zn(NO$_3$)$_2$ (zinc is Zn$^{2+}$, nitrate is NO$_3^-$, so two nitrates are needed)
• Water: H$_2$O
• Sulfur dioxide: SO$_2$

Now balance the equation. There are two nitrate groups on the right, so we need 2 HNO$_3$ on the left:

Check the balance: Zn 1:1, S 1:1, N 2:2, H 2:2, O: left 3 + 6 = 9, right 6 + 1 + 2 = 9. Balanced.

State symbols: zinc sulfite is a solid (s), nitric acid is aqueous (aq), zinc nitrate is aqueous (aq), water is a liquid (l), sulfur dioxide is a gas (g).

Key Takeaways

• To construct an equation from a description: identify each substance's formula, then balance.
• The nitrate ion is NO$_3^-$, so zinc nitrate is Zn(NO$_3$)$_2$.
• State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.

Common Mistakes

• Writing the wrong formula for zinc nitrate (e.g. ZnNO$_3$ instead of Zn(NO$_3$)$_2$).
• Forgetting state symbols — the mark scheme gives a mark for them.
• Leaving the equation unbalanced.

Things to Be Careful About

The mark scheme says the state-symbol mark is dependent on correct formulae — so if the formulae are wrong, the state symbols mark is lost too. Make sure the equation is balanced and every species has its correct state symbol. The colourless liquid is water; the aqueous zinc salt is zinc nitrate.

Walkthrough

The question asks for one adverse effect of increased global warming. The mark scheme accepts 'climate change' as a direct answer. This is a straightforward recall point.

Key Takeaways

Global warming leads to climate change, which includes rising sea levels, extreme weather, and habitat loss.

Common Mistakes

• Giving a cause of global warming instead of an effect (e.g., 'burning fossil fuels').
• Giving a vague answer like 'bad for the environment' without specifying an effect.

Things to Be Careful About

The answer must be an adverse effect, not a cause or a general statement. 'Climate change' is the safest concise answer.

Walkthrough

The mark scheme awards three separate marks: (1) carbon dioxide absorbs thermal energy from the Earth, (2) that energy is reflected or emitted back towards the Earth, and (3) this reduces or stops the loss of thermal energy into space. Each step is a distinct point; a candidate must write all three to score full marks.

Key Takeaways

Greenhouse gases trap heat by absorbing outgoing thermal radiation and re-emitting it back to the surface, preventing it from escaping to space.

Common Mistakes

• Only stating 'carbon dioxide traps heat' without the three-step mechanism.
• Confusing absorption with reflection of sunlight (the mechanism concerns thermal/infrared radiation from the Earth, not visible light).
• Missing the final point about reduced loss to space.

Things to Be Careful About

Use the term 'thermal energy' (or 'infrared radiation') rather than 'heat' alone. The mark scheme specifically mentions 'from the Earth' and 'into space' — include these direction phrases.

Walkthrough

The mark scheme lists three acceptable strategies: reforestation/afforestation (increases carbon dioxide absorption), decreasing use of fossil fuels (reduces emissions), and using renewable energy or hydrogen fuel (reduces carbon dioxide emissions). Any one of these scores the mark.

Key Takeaways

Global warming can be reduced by either removing carbon dioxide from the atmosphere (e.g., planting trees) or reducing carbon dioxide emissions (e.g., using renewable energy).

Common Mistakes

• Giving a strategy that does not directly address carbon dioxide (e.g., 'recycling' without linking to emissions).
• Giving a vague answer like 'stop pollution' without specifying how.
• Confusing adaptation (e.g., building flood defences) with mitigation.

Things to Be Careful About

The strategy must be specifically about carbon dioxide. If using reforestation, mention that it absorbs carbon dioxide; if using renewable energy, mention it reduces carbon dioxide emissions.

Walkthrough

A gas is made up of particles moving randomly and freely. Pressure comes from particles colliding with the walls of the container. At fixed temperature the particles keep the same average kinetic energy, so increasing the pressure must reduce the space in which the particles move. This makes the gas occupy a smaller volume. In terms of kinetic particle theory, the particles are pushed closer together and the average distance between particles decreases. The first mark is for the volume decrease; the second is for the particle separation explanation.

Key Takeaways

• Gas particles are widely spaced and in rapid random motion.
• At constant temperature, increasing pressure decreases the volume of a gas.
• The explanation uses particle separation, not particle speed.

Common Mistakes

• Saying the particles move faster; temperature is unchanged.
• Only stating the volume decreases without explaining that the particles are pushed closer together.
• Using 'more collisions' alone when the question asks for a kinetic particle theory explanation.

Things to Be Careful About

• The pressure is increased at room temperature, so temperature must be treated as constant.
• The answer should describe both the observed change (volume decreases) and the particle reason (separation decreases).

Walkthrough

Iodine sublimes, so it changes directly from a solid to a gas. In the solid, particles are close together in a regular arrangement and can only vibrate about fixed positions. When heated, the particles gain energy, overcome the forces holding them in place, and break away from their fixed positions. They move freely and randomly, so the separation increases and the arrangement becomes random. The mark scheme awards one mark for each of separation, arrangement and motion.

Key Takeaways

• Sublimation is a change from solid directly to gas.
• In a solid particles vibrate in fixed positions; in a gas they move freely and randomly.
• Increasing separation, changing from ordered to random, and changing from vibrating to moving are the three key changes.

Common Mistakes

• Saying the solid changes to a liquid first; the question specifies no liquid is formed.
• Writing 'arrangement changes from random to ordered'.
• Confusing vibration in fixed positions with free movement.

Things to Be Careful About

• Give the change in each column separately.
• Use the exact idea pair: ordered to random, and vibrating to moving from one place to another.
• 'Particles move apart' is the separation point; do not lose it by only discussing arrangement.

Walkthrough

At the same temperature, gas particles have the same average kinetic energy. Lighter particles must therefore move faster than heavier particles. Chlorine molecules have a lower relative molecular mass than iodine molecules, so the chlorine particles move more quickly and diffuse more quickly. The mark is awarded for identifying that chlorine has the lower relative molecular mass.

Key Takeaways

• Rate of diffusion depends on the mass of the particles.
• Lighter particles diffuse faster than heavier particles at the same temperature.
• Chlorine gas has a lower relative molecular mass than iodine gas.

Common Mistakes

• Saying chlorine diffuses faster because it has a higher molecular mass.
• Giving a description of diffusion without mentioning relative molecular mass.
• Comparing atoms instead of molecules; both are diatomic molecules, and the mark scheme says 'molecular mass'.

Things to Be Careful About

• The temperature and pressure are the same, so the only difference affecting the rate is the mass of the particles.
• Do not write 'less dense' alone; the expected idea is lower relative molecular mass.

Walkthrough

The notation shows the mass number (126) above the proton number (53). The proton number tells us the number of protons, so there are 53 protons. Neutrons are found by subtracting the proton number from the mass number: 126 - 53 = 73. A neutral iodine atom has the same number of electrons as protons, but the iodide ion has a 1- charge, so it has one extra electron: 53 + 1 = 54.

Key Takeaways

• Mass number = protons + neutrons.
• Proton number = number of protons.
• A negative ion has more electrons than protons; a positive ion has fewer.
• For iodide, the charge is 1-, so add one electron to the neutral atom count.

Common Mistakes

• Putting 53 electrons because the ion is confused with the neutral atom.
• Calculating neutrons as mass number + proton number.
• Swapping the proton number and mass number.

Things to Be Careful About

• The top number in the notation is the mass number; the bottom left number is the proton number.
• A 1- ion gains one electron; do not subtract it.
• The numbers should be integers; there are no units needed in the table.

Walkthrough

Compound A is propan-1-ol. It reacts with ethanoic acid ($\text{CH}_3\text{COOH}$) in the presence of an acid catalyst. This is an esterification reaction. The alcohol provides the propyl group ($\text{CH}_3\text{CH}_2\text{CH}_2\text{–}$) and the acid provides the ethanoate group ($\text{CH}_3\text{COO–}$). The ester is propyl ethanoate.

Key Takeaways

Esterification: acid + alcohol $\rightarrow$ ester + water. The name is alkyl (from alcohol) alkanoate (from acid).

Common Mistakes

Writing the condensed formula instead of the displayed formula. Mixing up the alkyl and alkanoate parts (e.g., ethyl propanoate).

Things to Be Careful About

The displayed formula must show the C=O double bond and the C–O single bond in the ester linkage. All C–H bonds must be drawn.

Walkthrough

Compound C is identified in the diagram as propanoic acid. The structural formula shows the carbon chain and the functional group without drawing every bond.

Key Takeaways

Carboxylic acids have the functional group –COOH. Propanoic acid has a 3-carbon chain.

Common Mistakes

Writing $\text{CH}_3\text{CH}_2\text{CO}_2\text{H}$ (acceptable but less common in 5070) or missing the H in the COOH group.

Things to Be Careful About

The question asks for the structural formula, not the displayed formula. $\text{CH}_3\text{CH}_2\text{COOH}$ is sufficient.

Walkthrough

Propanoic acid reacts with magnesium carbonate ($\text{MgCO}_3$). Acid + carbonate $\rightarrow$ salt + water + carbon dioxide. The acid is propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$), so the anion is propanoate ($\text{CH}_3\text{CH}_2\text{COO}^-$). The metal is magnesium ($\text{Mg}^{2+}$). To balance charges, two propanoate ions are needed for one magnesium ion. The salt is magnesium propanoate.

Key Takeaways

Carboxylic acids react like mineral acids with carbonates and metals, forming carboxylate salts.

Common Mistakes

Writing the formula as $\text{MgCH}_3\text{CH}_2\text{COO}$ (missing the subscript 2). Naming it "magnesium propionate" (common name, not IUPAC).

Things to Be Careful About

Magnesium has a 2+ charge ($\text{Mg}^{2+}$). The propanoate ion has a 1- charge ($\text{CH}_3\text{CH}_2\text{COO}^-$). The formula must be $\text{Mg(C}_3\text{H}_5\text{O}_2)_2$ or $\text{Mg(CH}_3\text{CH}_2\text{COO)}_2$.

Walkthrough

Reaction with $\text{MgCO}_3$: acid + carbonate $\rightarrow$ salt + water + carbon dioxide. So gas E is carbon dioxide ($\text{CO}_2$).
Reaction with Mg: acid + metal $\rightarrow$ salt + hydrogen. So gas F is hydrogen ($\text{H}_2$).

Key Takeaways

Acids react with carbonates to release $\text{CO}_2$. Acids react with reactive metals (above H in reactivity series) to release $\text{H}_2$.

Common Mistakes

Confusing the gases. Thinking $\text{O}_2$ is produced from carbonates.

Things to Be Careful About

Name the gas, not just the formula, unless asked for formula. "Carbon dioxide" and "hydrogen".

Walkthrough

Propanoic acid (C) reacts with ethanol in the presence of an acid catalyst. This is esterification. The alcohol (ethanol) provides the alkyl group "ethyl". The acid (propanoic acid) provides the alkanoate group "propanoate". The ester is ethyl propanoate.

Key Takeaways

Ester naming: alkyl group from the alcohol, alkanoate group from the carboxylic acid.

Common Mistakes

Writing "propyl ethanoate" (this is ester B, formed from propanol and ethanoic acid). Swapping the names.

Things to Be Careful About

Ethanol has 2 carbons -> ethyl. Propanoic acid has 3 carbons -> propanoate.

Walkthrough

This question tests your knowledge of the two ways polymers are made.

Addition polymerisation uses monomers that contain a carbon–carbon double bond, such as alkenes. The double bond opens and the monomers join end to end. No other substance is released, so the only product is the polymer. Because no atoms are lost, the polymer has the same empirical formula as the monomer.

Condensation polymerisation uses monomers with two functional groups, such as a carboxylic acid group and an alcohol group. As the monomers join, a small molecule is produced and removed; this is usually water, but the mark scheme simply says a simple molecule. Because atoms are lost as the small molecule, the polymer does not have the same empirical formula as the monomers.

For full marks you only need one correct comment about each type.

Key Takeaways

• Addition polymerisation: monomers with C=C bond, no other product, same empirical formula.
• Condensation polymerisation: two functional groups, small molecule formed, empirical formula changes.
• The mark scheme accepts any one correct statement about each type.

Common Mistakes

• Giving two comments about addition but only one about condensation, so a mark is lost.
• Saying water is always formed in condensation polymerisation; the mark scheme accepts a simple molecule, with water as the common example.
• Stating that addition polymers have a different empirical formula; it is the same because no atoms are lost.
• Suggesting condensation uses a C=C bond; that is addition polymerisation.

Things to Be Careful About

• The command word is describe, so the answer should be in words, not equations.
• Make sure the no-other-product idea is clear for addition, and say simple molecule formed for condensation.
• You do not need both examples from the list; one valid point on each side scores full marks.

Walkthrough

PET stands for poly(ethylene terephthalate). It is made by condensation polymerisation between a dicarboxylic acid and a diol. The functional group that joins the repeat units is the same ester group found in esters. You only need to name the linkage: ester.

Key Takeaways

• PET is a polyester.
• The ester linkage is formed when a carboxylic acid group reacts with an alcohol group, eliminating water.
• Condensation polymers have characteristic linkages: PET has ester linkages; nylon has amide linkages.

Common Mistakes

• Writing amide or peptide — those are linkages in polyamides such as nylon, not PET.
• Writing the name of the monomer, ethylene terephthalate, instead of the linkage type.

Things to Be Careful About

• The question asks for the type of linkage, not a drawing or a monomer.
• The ester linkage contains the –COO– arrangement; remembering this stops you confusing it with an amide.

Walkthrough

The percentages are by mass, so we can assume 100 g of polymer. That turns each percentage directly into a mass in grams.

Amounts in moles:

$\text{C}$: $\frac{47.1}{12} = 3.93$ mol

$\text{H}$: $\frac{6.5}{1} = 6.5$ mol

$\text{Cl}$: $\frac{46.4}{35.5} = 1.31$ mol

Divide by the smallest amount (1.31 mol):

C: $\frac{3.93}{1.31} = 3.0$

H: $\frac{6.5}{1.31} = 5.0$

Cl: $\frac{1.31}{1.31} = 1.0$

So the simplest whole-number ratio is C : H : Cl = 3 : 5 : 1, giving the empirical formula $\text{C}_3\text{H}_5\text{Cl}$.

Key Takeaways

• Percentage composition by mass can be treated as grams when finding an empirical formula.
• Convert mass to moles by dividing by relative atomic mass.
• Divide all mole amounts by the smallest value to get the simplest whole-number ratio.

Common Mistakes

• Using 35 for chlorine instead of 35.5.
• Rounding 3.925 to 4 before dividing; divide first, then round the final ratio.
• Forgetting to divide by the smallest number, so the formula is not in simplest whole-number ratio.
• Writing the formula in the wrong order; here the empirical formula is written C then H then Cl.

Things to Be Careful About

• Use $A_r$ values carefully: C = 12, H = 1, Cl = 35.5.
• Carry enough decimal places through the working; the ratio comes out at almost exactly 3 : 5 : 1.
• The percentages sum to 100.0 here, so no extra element such as oxygen needs to be considered.

Walkthrough

This is a recall question where you must give two distinct environmental problems caused by plastics. The mark scheme allows any two from: landfill sites may fill up, plastics accumulate in oceans, and burning plastics produces toxic gases. The answer above chooses the first two; either alternative is also accepted.

Key Takeaways

• Plastics are non-biodegradable and take up space in landfill.
• They can accumulate in oceans and harm marine life.
• Burning plastics can release toxic gases.

Common Mistakes

• Giving only one environmental problem when two are asked.
• Giving vague answers such as plastic is bad for the environment — the marks are for specific problems.
• Mentioning the same idea twice in different words, such as fills landfill and takes up space in landfill; this may be treated as one point.

Things to Be Careful About

• The mark scheme says any two from, so list two separate, recognisable problems.
• Acceptable alternatives may be phrased differently from the mark scheme, but they must clearly describe an environmental challenge.