Chemistry 5070 / 21 — October/November 2024

(ii)

an atom that forms an ion that gives a lilac colour in a flame test

______

(iii)

an atom of a monatomic gas

______

(iv)

an atom of an element that is used in the treatment of the domestic water supply to remove tastes and odours

______

DifficultyMedium

Worked solution

Answer

None of the given configurations (D, E, F, G, H) represent chlorine.

Note: The mark scheme provided in the source material appears to contain an error for this part. Chlorine (configuration 2, 8, 7) is used to remove tastes and odours and kill bacteria in water treatment. None of the atoms D-H match chlorine. If forced to choose based on a potential typo in the question's diagram bank, no valid answer exists among D-H. However, following the strict mark scheme provided which lists 'E' for this part, we note this is chemically incorrect as E is helium.

Correct Chemistry Answer: Chlorine ( $Cl_2$ ) is used. Its configuration is 2, 8, 7.

Final answer

Chlorine (not present in bank; mark scheme error noted)

Detailed explanation

Walkthrough

Chlorine is the substance used in the domestic water supply to kill bacteria and remove tastes and odours. Chlorine is in Group VII and has the electronic configuration 2, 8, 7. None of the atoms D (2, 4), E (2), F (2, 6), G (2, 8, 2), or H (2, 8, 8, 1) represent chlorine.

The provided mark scheme lists 'E' (helium) for this part, which is chemically incorrect. Helium is an inert noble gas and is not used in water treatment. This is likely an error in the source mark scheme or a mismatch in the question bank. In a real exam, you would write 'chlorine' or note the discrepancy.

Key Takeaways

Chlorine is added to water supplies for disinfection.

Common Mistakes

Assuming the answer must be one of the given letters even if it is chemically wrong.
Forgetting that chlorine is a Group VII element (7 outer electrons).

Things to Be Careful About

Always check if the required element is actually present in the bank. If not, state the correct chemical fact. Do not force an incorrect answer to fit a flawed mark scheme in a real exam context, though for this exercise, the mark scheme's 'E' is noted as erroneous.

Techniques used

recall substances used in water purificationidentify chlorine from its properties

(v)

an atom that forms a stable ion by losing two electrons.

______

(b)

Deduce the number of protons and neutrons in the chromium atom shown.

^{53}_{24}\text{Cr}

number of protons = ______
number of neutrons = ______

Q215MMediumChemical ReactionsChemical EnergeticsMetalsThe Periodic TableStoichiometry

(a)

Iron is extracted in a blast furnace by the reduction of iron(III) oxide with carbon monoxide.

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

2M

(i)

This reaction is a redox reaction.

State the meaning of the term redox reaction.

______

(ii)

Explain how carbon monoxide acts as a reducing agent in this reaction.

______

(b)

Calcium carbonate is added to the blast furnace. The calcium carbonate undergoes thermal decomposition.

\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2

The thermal decomposition of calcium carbonate is endothermic.

Complete the reaction pathway diagram in Fig. 2.1 to show:

the reactant and products
a labelled arrow for the activation energy, $E_a$
a labelled arrow for the enthalpy change, $\Delta H$ .

Fig. 2.1

DifficultyMedium

Worked solution

Answer

The diagram shows an endothermic reaction pathway:

Reactants ( $\text{CaCO}_3$ ) are on the left at a lower energy level.
Products ( $\text{CaO} + \text{CO}_2$ ) are on the right at a higher energy level.
A curve rises from reactants to a peak (activation energy hump) and falls to products.
An upward arrow from the reactant level to the peak is labelled $E_a$ (activation energy).
An upward arrow from the reactant level to the product level is labelled $\Delta H$ (enthalpy change).

Final answer

See diagram description: endothermic profile with reactants below products, $E_a$ arrow from reactants to peak, $\Delta H$ arrow from reactants to products.

Detailed explanation

Walkthrough

The reaction $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ is stated to be endothermic. This means the products have more energy than the reactants.

Energy Levels: Draw a horizontal line for reactants on the left. Draw a higher horizontal line for products on the right.
Pathway: Draw a curve starting at the reactant level, rising to a peak (the transition state), and then falling to the product level. The peak must be higher than the product level.
Activation Energy ( $E_a$ ): Draw an almost vertical upward arrow from the reactant energy line to the top of the peak. Label it $E_a$ or 'activation energy'. This represents the energy needed to start the reaction.
Enthalpy Change ( $\Delta H$ ): Draw an upward arrow from the reactant energy line to the product energy line. Label it $\Delta H$ or 'enthalpy change'. Since it is endothermic, this arrow points up, indicating a positive $\Delta H$ .

Key Takeaways

Endothermic reactions have products at a higher energy level than reactants.
$E_a$ is always measured from the reactant level to the top of the hump.
$\Delta H$ is measured from the reactant level to the product level.

Common Mistakes

Drawing the product level below the reactant level (that would be exothermic).
Drawing the $\Delta H$ arrow from the bottom of the graph (zero energy) instead of from the reactant level.
Forgetting to label the arrows.
Drawing the $E_a$ arrow from the product level.

Things to Be Careful About

The arrows for $E_a$ and $\Delta H$ must be clearly labelled. The $E_a$ arrow goes to the peak of the curve, not just partway up. The $\Delta H$ arrow is the vertical difference between reactant and product levels.

Techniques used

draw endothermic energy profilelabel activation energy barrierlabel enthalpy change

(c)

Describe how slag is formed in the blast furnace.

Include a symbol equation in your answer.

______

(d)

Iron is a transition element.

Transition elements have coloured compounds.

State two other physical properties that are typical of transition elements and not of Group I metals.

______
______

(e)

The equation shows the decomposition of iron pentacarbonyl, $\text{Fe(CO)}_5$ , in a closed container.

\text{Fe(CO)}_5(\text{l}) \rightleftharpoons \text{Fe}(\text{s}) + 5\text{CO}(\text{g})

4M

(i)

Predict and explain what happens to the position of equilibrium when the pressure is decreased. The temperature remains the same.

prediction = ______
explanation = ______

(ii)

This reaction can be used to produce pure iron.

Describe and explain, by referring to the equation, how a sample of pure iron that is free from $\text{Fe(CO)}_5(\text{l})$ is produced from the equilibrium mixture.

______

(f)

Iron reacts with hot concentrated sulfuric acid.

The products are iron(III) sulfate, sulfur dioxide and a liquid that turns anhydrous copper(II) sulfate blue.

Construct the symbol equation for this reaction.

______

DifficultyMedium

Worked solution

Answer

The liquid that turns anhydrous copper(II) sulfate blue is water ( $\text{H}_2\text{O}$ ).
Reactants: $\text{Fe}$ and $\text{H}_2\text{SO}_4$ .
Products: $\text{Fe}_2(\text{SO}_4)_3$ , $\text{SO}_2$ , $\text{H}_2\text{O}$ .

Balanced equation:

2\text{Fe} + 6\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{SO}_2 + 6\text{H}_2\text{O}

Final answer

2Fe + 6H2SO4 -> Fe2(SO4)3 + 3SO2 + 6H2O

Detailed explanation

Walkthrough

Identify products: The question states products are iron(III) sulfate ( $\text{Fe}_2(\text{SO}_4)_3$ ), sulfur dioxide ( $\text{SO}_2$ ), and a liquid that turns anhydrous copper(II) sulfate blue. Anhydrous copper(II) sulfate is white and turns blue in the presence of water. So the liquid is water ( $\text{H}_2\text{O}$ ).
Write unbalanced equation: $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{SO}_2 + \text{H}_2\text{O}$ .
Balance:
- Balance Iron: Need 2 Fe on left. $2\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{SO}_2 + \text{H}_2\text{O}$ .
- Balance Sulfate/Sulfur: Right side has 3 sulfate groups in $\text{Fe}_2(\text{SO}_4)_3$ plus 1 S in $\text{SO}_2$ . Total S = 4? Wait. Let's look at the oxidation. Fe goes from 0 to +3 (loses 3e- per Fe, so 6e- total). S in $\text{H}_2\text{SO}_4$ is +6. S in $\text{SO}_2$ is +4 (gains 2e-). To balance electrons: 6e- lost by Fe, so need 3 S atoms to gain 2e- each. So 3 $\text{SO}_2$ produced.
- This means 3 $\text{H}_2\text{SO}_4$ are reduced to $\text{SO}_2$ . The other $\text{H}_2\text{SO}_4$ provides the sulfate ions for $\text{Fe}_2(\text{SO}_4)_3$ . $\text{Fe}_2(\text{SO}_4)_3$ needs 3 sulfate ions, so 3 more $\text{H}_2\text{SO}_4$ .
- Total $\text{H}_2\text{SO}_4$ = 3 (reduced) + 3 (acidic) = 6.
- Equation: $2\text{Fe} + 6\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{SO}_2 + \text{H}_2\text{O}$ .
- Balance Hydrogen: Left has 12 H. Right needs 6 $\text{H}_2\text{O}$ .
- Final: $2\text{Fe} + 6\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{SO}_2 + 6\text{H}_2\text{O}$ .
- Check Oxygen: Left = 24. Right = 12 (sulfate) + 6 (SO2) + 6 (water) = 24. Balanced.

Key Takeaways

Anhydrous copper(II) sulfate test identifies water.
Concentrated sulfuric acid acts as an oxidising agent with metals, producing $\text{SO}_2$ instead of $\text{H}_2$ .
Balancing redox equations involving concentrated acids requires careful counting of sulfate ions vs reduced sulfur.

Common Mistakes

Producing $\text{H}_2$ gas instead of $\text{SO}_2$ (this happens with dilute acids, not hot concentrated).
Writing iron(II) sulfate ( $\text{FeSO}_4$ ) instead of iron(III) sulfate ( $\text{Fe}_2(\text{SO}_4)_3$ ). The question states iron(III).
Incorrect balancing, e.g., $\text{Fe} + 2\text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}$ (this is for iron(II) and wrong stoichiometry for iron(III)).

Things to Be Careful About

The mark scheme awards 1 mark for correct formulae and 1 mark for correct balancing. Ensure $\text{Fe}_2(\text{SO}_4)_3$ is written correctly with the subscript 2 and 3. The balancing is the tricky part; using the electron transfer method helps.

Techniques used

identify products from descriptionbalance redox equation with correct formulae

Q36MMedium-EasyElectrochemistry

Fig. 3.1 shows the apparatus used for the electrolysis of concentrated aqueous sodium chloride using graphite electrodes.

Fig. 3.1

(a)

Label the cathode on Fig. 3.1.

(b)

Explain why concentrated aqueous sodium chloride conducts electricity.

______

(c)

2M

(i)

Name the product formed at the cathode.

______

(ii)

Chlorine is formed at the anode.

Construct the ionic half-equation for the reaction at the anode.

______

(d)

Graphite is suitable as an electrode because it conducts electricity.

State one other property of graphite that makes it suitable for use as an electrode.

______

(e)

State the product formed at the cathode when molten sodium chloride is electrolysed.

______

Q414MMediumOrganic ChemistryChemical Energetics

This question is about alkanes and alkenes.

(a)

But-1-ene belongs to the alkene homologous series.

Members of the same homologous series differ from one member to the next by a $-\text{CH}_2-$ group and have similar chemical properties.

State two other characteristics of a homologous series.

______
______

(b)

Fig. 4.1 shows the displayed formula of but-1-ene.

Fig. 4.1

3M

(i)

Explain how Fig. 4.1 shows that but-1-ene is an unsaturated compound.

______

(ii)

Give the structural formula of but-1-ene.

______

(iii)

Draw the displayed formula of an isomer of but-1-ene.

(c)

Undecane, $\text{C}_{11}\text{H}_{24}$ , is present in the kerosene/paraffin fraction from the distillation of petroleum.

3M

(i)

Give one use of the kerosene/paraffin fraction.

______

(ii)

When undecane is cracked, shorter hydrocarbon molecules are formed.

Construct the symbol equation for a reaction in which undecane is cracked and the only products are butane, propene and ethene.

______

(d)

Propane reacts with chlorine to form chloropropane and one other product, X.

\text{C}_3\text{H}_8 + \text{Cl}_2 \rightarrow \text{C}_3\text{H}_7\text{Cl} + \text{X}

2M

(i)

Name product X.

______

(ii)

State the essential condition for this reaction.

______

(e)

Ethene reacts with bromine at room temperature.

Fig. 4.2 shows the displayed formulae of the reactants and product.

Fig. 4.2

4M

(i)

Calculate the enthalpy change of this reaction in $\text{kJ / mol}$ .

Use the bond energies in Table 4.1.

Table 4.1

type of bond	$\text{C=C}$	$\text{C–H}$	$\text{Br–Br}$	$\text{C–C}$	$\text{C–Br}$
bond energy in $\text{kJ / mol}$	612	413	193	347	290

enthalpy change = ______ $\text{kJ / mol}$

DifficultyMedium

Worked solution

Working

Bonds broken (endothermic):
1 $\times$ C=C: $612 \text{ kJ / mol}$
1 $\times$ Br–Br: $193 \text{ kJ / mol}$
Total energy to break bonds = $612 + 193 = 805 \text{ kJ / mol}$

Bonds formed (exothermic):
1 $\times$ C–C: $347 \text{ kJ / mol}$
2 $\times$ C–Br: $2 \times 290 = 580 \text{ kJ / mol}$
Total energy released forming bonds = $347 + 580 = 927 \text{ kJ / mol}$

Enthalpy change = energy to break bonds $-$ energy released forming bonds
Enthalpy change = $805 - 927 = -122 \text{ kJ / mol}$

Answer

$-122$

Final answer

-122

Detailed explanation

Walkthrough

The enthalpy change of a reaction can be estimated from bond energies using: $\Delta H = \sum(\text{bond energies broken}) - \sum(\text{bond energies formed})$ . Bond breaking is endothermic (positive), bond making is exothermic (negative).

From Fig. 4.2, in ethene ( $\text{C}_2\text{H}_4$ ) and bromine ( $\text{Br}_2$ ), the bonds broken are one C=C double bond and one Br–Br single bond. Energy absorbed = $612 + 193 = 805 \text{ kJ / mol}$ .

In 1,2-dibromoethane ( $\text{C}_2\text{H}_4\text{Br}_2$ ), the bonds formed are one C–C single bond and two C–Br single bonds (the four C–H bonds remain unchanged and cancel out). Energy released = $347 + 2(290) = 347 + 580 = 927 \text{ kJ / mol}$ .

$\Delta H = 805 - 927 = -122 \text{ kJ / mol}$ . The negative sign indicates the reaction is exothermic.

Key Takeaways

Bond energy calculations use $\Delta H = \text{bonds broken} - \text{bonds formed}$ . Only count bonds that actually change; unchanged bonds (like C–H here) can be ignored on both sides. Bond breaking is always positive (energy in), bond forming is always negative (energy out).

Common Mistakes

Forgetting that two C–Br bonds are formed (writing $347 + 290$ instead of $347 + 2 \times 290$ ). Reversing the formula to get $927 - 805 = +122$ . Forgetting the negative sign.

Things to Be Careful About

The unit is kJ / mol. The answer must include the correct sign; $-122$ is exothermic. Bond energies are average values, so the calculation gives an estimate, but the mark scheme expects the exact arithmetic result from the given table.

Techniques used

sum bond energies for bonds brokensum bond energies for bonds formedcalculate enthalpy change as broken minus formed

(ii)

Describe the colour change when a sample of excess ethene is added to a few drops of aqueous bromine.

from ______ to ______

Q514MMediumStoichiometryAcids, Bases and SaltsChemical ReactionsExperimental Techniques and Chemical Analysis

A student adds large pieces of copper(II) carbonate to dilute hydrochloric acid. The copper(II) carbonate is in excess.

(a)

Complete the equation by adding state symbols for the products.

\text{CuCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CuCl}_2(\_\_\_\_\_\_) + \text{H}_2\text{O}(\_\_\_\_\_\_) + \text{CO}_2(\_\_\_\_\_\_)

(b)

Fig. 5.1 shows how the mass of the reaction mixture changes with time as the reaction proceeds.

Fig. 5.1

4M

(i)

In another experiment, powdered copper(II) carbonate is used instead of large pieces of copper(II) carbonate. All other conditions and the mass of copper(II) carbonate stay the same.

Draw a line on the grid in Fig. 5.1 to show how the mass of the reaction mixture changes with time.

(ii)

The initial experiment is repeated using large pieces of copper(II) carbonate and hydrochloric acid of a higher concentration.

All other conditions stay the same.

Describe and explain the difference in rate of reaction when hydrochloric acid of a higher concentration is used.

______

DifficultyMedium-Easy

Worked solution

Answer

The rate of reaction increases (the reaction is faster).

Explanation:
A higher concentration of hydrochloric acid means there are more hydrogen ions (or acid particles) per unit volume of solution. The particles are more crowded and the distance between them is smaller. This leads to a greater frequency of collisions per second between the acid particles and the copper(II) carbonate, increasing the rate of reaction.

Answer

Rate increases because higher concentration means more particles per unit volume, leading to more frequent collisions.

Final answer

Rate increases because higher concentration means more particles per unit volume, leading to more frequent collisions.

Detailed explanation

Walkthrough

Concentration is defined as the number of particles of solute per unit volume of solution. When the concentration of the acid is increased, there are more $\text{H}^+$ ions in the same volume. According to collision theory, for a reaction to occur, particles must collide with sufficient energy. With more particles packed into the same space, the average distance between them is reduced, so they collide more frequently. More collisions per second means more successful collisions per second, which directly increases the rate of reaction.

Key Takeaways

Rate is proportional to collision frequency.
Higher concentration → more particles per unit volume → shorter distances between particles → more frequent collisions.

Common Mistakes

Saying "there are more particles" without specifying "per unit volume" or "in the same volume".
Attributing the rate increase to particles having more energy (that is an explanation for increased temperature, not concentration).

Things to Be Careful About

The mark scheme explicitly does not award a mark for simply saying "because". You must state the effect (rate increases) and then provide the two linked reasons (more crowded/smaller distance, AND more collisions per second).

Techniques used

apply collision theory to concentration effects

(c)

Excess copper(II) carbonate is added to $22.0\,\text{cm}^3$ of $0.500\,\text{mol / dm}^3$ hydrochloric acid.

Calculate the volume of carbon dioxide released measured at room temperature and pressure.

Give your answer to two significant figures.

volume of carbon dioxide gas = ______ $\text{dm}^3$

DifficultyMedium-Easy

Worked solution

Working

First, calculate the moles of hydrochloric acid used. The volume must be converted from $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000.

\text{moles of HCl} = 0.500 \times \frac{22.0}{1000} = 0.011 \text{ mol}

From the balanced equation in part (a):

\text{CuCO}_3 + 2\text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} + \text{CO}_2

The mole ratio of $\text{HCl}$ to $\text{CO}_2$ is 2 : 1. Since copper(II) carbonate is in excess, hydrochloric acid is the limiting reactant.

\text{moles of CO}_2 = \frac{0.011}{2} = 0.0055 \text{ mol}

At room temperature and pressure (r.t.p.), 1 mole of any gas occupies $24\text{ dm}^3$ .

\text{volume of CO}_2 = 0.0055 \times 24 = 0.132 \text{ dm}^3

Rounding to two significant figures gives $0.13\text{ dm}^3$ .

Answer

0.13

Final answer

0.13

Detailed explanation

Walkthrough

Step 1: Find moles of the known reactant ( $\text{HCl}$ ). Use the formula $n = c \times V$ , ensuring volume is in $\text{dm}^3$ . $0.500 \times 0.0220 = 0.011\text{ mol}$ .
Step 2: Use the stoichiometric ratio from the balanced equation. 2 moles of $\text{HCl}$ produce 1 mole of $\text{CO}_2$ . So, $0.011 / 2 = 0.0055\text{ mol}$ of $\text{CO}_2$ .
Step 3: Convert moles of gas to volume at r.t.p. using the molar gas volume ( $24\text{ dm}^3\text{/mol}$ ). $0.0055 \times 24 = 0.132\text{ dm}^3$ .
Step 4: Round to the required significant figures (two). $0.132 \rightarrow 0.13$ .

Key Takeaways

Always convert volume in $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000 when using concentration in $\text{mol / dm}^3$ .
Identify the limiting reactant (here, $\text{HCl}$ , since the carbonate is in excess).
Molar gas volume at r.t.p. is $24\text{ dm}^3\text{/mol}$ (or $24000\text{ cm}^3\text{/mol}$ ).

Common Mistakes

Forgetting to divide the volume by 1000, leading to moles of $\text{HCl} = 11$ .
Using a 1:1 mole ratio instead of the correct 2:1 ratio from the balanced equation.
Forgetting to round to two significant figures at the end.

Things to Be Careful About

The question asks for the answer in $\text{dm}^3$ , not $\text{cm}^3$ . If you use $24000\text{ cm}^3\text{/mol}$ , you will get $132\text{ cm}^3$ , which must then be converted back to $0.13\text{ dm}^3$ .

Techniques used

calculate moles from concentration and volumeuse mole ratio from balanced equationcalculate gas volume at r.t.p.

(d)

3M

(i)

Describe the observations made when:

a few drops of aqueous ammonia are added to an aqueous solution containing copper(II) ions
______
excess aqueous ammonia is added to an aqueous solution containing copper(II) ions.
______

DifficultyEasy

Worked solution

Answer

A few drops of aqueous ammonia: a light blue precipitate forms.
Excess aqueous ammonia: the precipitate dissolves to form a dark blue (or deep blue) solution.

Answer

Light blue precipitate, soluble in excess to give a dark blue solution.

Final answer

Light blue precipitate, soluble in excess to give a dark blue solution

Detailed explanation

Walkthrough

Copper(II) ions ( $\text{Cu}^{2+}$ ) react with hydroxide ions provided by aqueous ammonia to form copper(II) hydroxide, which is insoluble and appears as a light blue precipitate.

\text{Cu}^{2+}(\text{aq}) + 2\text{NH}_3(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s}) + 2\text{NH}_4^+(\text{aq})

When excess aqueous ammonia is added, the precipitate dissolves because a soluble complex ion, tetraamminecopper(II), is formed. This gives a characteristic dark blue solution.

\text{Cu(OH)}_2(\text{s}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu(NH}_3\text{)}_4]^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})

Key Takeaways

Ammonia test for $\text{Cu}^{2+}$ : light blue ppt, soluble in excess to give dark blue solution.
This distinguishes copper(II) from other metal ions like $\text{Mg}^{2+}$ (white ppt, insoluble in excess) or $\text{Al}^{3+}$ (white ppt, soluble in excess to give colourless solution).

Common Mistakes

Saying "white precipitate" (that is for magnesium or aluminium).
Saying the precipitate is "insoluble in excess" (that is true for magnesium and aluminium hydroxides, but not copper).
Calling the final solution "blue" instead of "dark blue" or "deep blue".

Things to Be Careful About

Always describe both stages: the initial addition (few drops) and the addition of excess. Both observations are required for full marks.

Techniques used

identify cation from reaction with ammonia

(ii)

An ionic compound of copper has the formula $\text{Cu}_2\text{O}$ .

Deduce the oxidation number of copper in $\text{Cu}_2\text{O}$ .

______

(e)

Describe how to prepare crystals of ammonium chloride by reacting aqueous ammonia with dilute hydrochloric acid.

______

DifficultyMedium

Worked solution

Answer

Titrate aqueous ammonia with dilute hydrochloric acid (using an indicator like methyl orange or universal indicator) to find the exact volumes needed for neutralisation.
Repeat the titration without the indicator to obtain a neutral solution without contaminating it with indicator colour.
Mix the measured volumes of aqueous ammonia and hydrochloric acid together.
Evaporate the resulting filtrate (or solution) to the crystallisation point / heat until a saturated solution is formed.
Leave the solution to cool, then filter off the crystals and dry them between filter papers.

Answer

Titrate to find volumes, mix without indicator, evaporate to saturation, cool and filter/dry crystals.

Final answer

Titrate to find volumes, mix without indicator, evaporate to saturation, cool and filter/dry crystals.

Detailed explanation

Walkthrough

Ammonia is a soluble base (an alkali), and hydrochloric acid is a soluble acid. To prepare a soluble salt like ammonium chloride, we cannot simply add the solid base to the acid and filter off excess, because the base is already dissolved. Instead, we must use a titration to find the exact stoichiometric ratio.

Step 1: Perform a titration of the aqueous ammonia with the dilute hydrochloric acid using an appropriate indicator to determine the exact volume of acid needed to neutralise a known volume of ammonia.
Step 2: Repeat the experiment without the indicator, using the volumes found in Step 1, to produce a pure, neutral solution of ammonium chloride without any indicator contamination.
Step 3: Evaporate some of the water from this solution by heating it until it reaches the crystallisation point (a saturated solution). This can be tested by dipping a cold glass rod into the solution and checking for crystal formation.
Step 4: Allow the saturated solution to cool slowly. As it cools, the solubility of ammonium chloride decreases, and crystals will form.
Step 5: Filter the crystals from the remaining solution (mother liquor) and press them between dry filter papers to absorb any remaining surface moisture.

Key Takeaways

Soluble salts from soluble bases and acids require titration to determine the correct proportions.
Crystallisation involves evaporating to saturation, cooling to form crystals, and filtering/drying.

Common Mistakes

Suggesting adding excess solid ammonia and filtering (ammonia is a gas/liquid in solution, not an insoluble solid like a metal oxide or carbonate).
Forgetting to mention evaporation or crystallisation; just saying "mix and heat" is not enough.
Not mentioning the removal of the indicator in the final preparation step.

Things to Be Careful About

The question asks for crystals, so the final steps must include cooling and filtering/drying.
"Evaporate to dryness" is wrong here; that would decompose the salt or leave impurities. You must evaporate to the crystallisation point and then cool.

Techniques used

plan preparation of a soluble salt from a soluble base and acid

Q610MMediumAtoms, Elements and CompoundsStoichiometryChemistry of the Environment

Fig. 6.1 shows the structures of zinc sulfide and disulfur dichloride.

Zinc sulfide has a structure similar to diamond.

Fig. 6.1

(a)

2M

(i)

Explain why zinc sulfide does not conduct electricity.

Use the information in Fig. 6.1.

______

(ii)

Predict one other physical property of zinc sulfide.

______

(b)

Explain why disulfur dichloride has a low melting point.

Use the information in Fig. 6.1.

______

(c)

Complete Fig. 6.2 to show the dot-and-cross diagram for the electronic configuration of disulfur dichloride.

Show only the outer shell electrons.

Fig. 6.2

DifficultyMedium

Worked solution

Answer

Each overlap area (Cl–S, S–S, S–Cl) contains one shared pair (2 electrons). Each sulfur atom has two lone pairs (4 electrons) in its non-overlapping region. Each chlorine atom has three lone pairs (6 electrons) in its non-overlapping region.

Working

Total outer shell electrons: $2 \times 7$ (from Cl) + $2 \times 6$ (from S) = $14 + 12 = 26$ electrons.
Bonding pairs: 3 pairs = 6 electrons.
Remaining lone pair electrons: $26 - 6 = 20$ electrons (10 pairs).
Distribution: 3 lone pairs on each Cl (6 electrons each) and 2 lone pairs on each S (4 electrons each). $6+6+4+4 = 20$ . Correct.

Final answer

One shared pair in each overlap; two lone pairs on each S; three lone pairs on each Cl.

Detailed explanation

Walkthrough

Disulfur dichloride has the structure Cl–S–S–Cl. We need to draw a dot-and-cross diagram showing only outer shell electrons.

Count valence electrons: Cl is in Group VII (7 outer electrons), S is in Group VI (6 outer electrons). Total = $2(7) + 2(6) = 26$ electrons.
Draw the skeleton: four overlapping circles in a chain Cl–S–S–Cl.
Place shared pairs: one pair (2 electrons) in each of the three overlap regions (Cl–S, S–S, S–Cl). This uses 6 electrons.
Distribute remaining electrons as lone pairs to complete octets:
- Each Cl needs 6 more electrons (3 lone pairs).
- Each S needs 4 more electrons (2 lone pairs).
- Total lone pair electrons: $2(6) + 2(4) = 20$ . Total electrons = $6 + 20 = 26$ . Matches.

Key Takeaways

Dot-and-cross diagrams for simple molecules require placing shared pairs in overlaps and lone pairs on the outer atoms to satisfy the octet rule.

Common Mistakes

Forgetting lone pairs on sulfur (it needs 2 lone pairs, not 1). Putting too many or too few electrons on chlorine (must be 3 lone pairs = 6 electrons). Using the same symbol (dots or crosses) for both atoms' electrons; usually, one atom's electrons are dots and the other's are crosses, but in the overlap, they must be mixed or clearly distinguished.

Things to Be Careful About

The question says 'show only the outer shell electrons'. Do not draw inner shells. Ensure the circles overlap correctly to represent the bonds. The mark scheme awards 1 mark for the whole diagram if correct.

Techniques used

draw dot-and-cross diagram for simple covalent moleculedistribute outer shell electrons to satisfy octet ruleshow shared pairs and lone pairs correctly

(d)

Disulfur dichloride reacts with water as shown.

\text{S}_2\text{Cl}_2 + 3\text{H}_2\text{O} \rightarrow 2\text{HCl} + \text{H}_2\text{S} + \text{H}_2\text{SO}_3

$13.5\,\text{g}$ of disulfur dichloride is reacted with $8.00\,\text{g}$ of water.

Show by calculation that water is in excess.

DifficultyMedium

Worked solution

Working

Relative formula mass of $\text{S}_2\text{Cl}_2$ : $M_r = (2 \times 32) + (2 \times 35.5) = 64 + 71 = 135$

\text{moles of } \text{S}_2\text{Cl}_2 = \frac{13.5}{135} = 0.100 \text{ mol}

\text{moles of } \text{H}_2\text{O} = \frac{8.00}{18} = 0.444 \text{ mol}

From the equation, 1 mol $\text{S}_2\text{Cl}_2$ reacts with 3 mol $\text{H}_2\text{O}$ .

Moles of $\text{H}_2\text{O}$ required to react with 0.100 mol $\text{S}_2\text{Cl}_2$ :

0.100 \times 3 = 0.300 \text{ mol}

Since 0.444 mol $\text{H}_2\text{O}$ is available and only 0.300 mol is required, water is in excess.

(Alternatively: moles of $\text{S}_2\text{Cl}_2$ that would react with 0.444 mol $\text{H}_2\text{O} = 0.444 / 3 = 0.148$ mol. Since 0.100 mol < 0.148 mol, $\text{S}_2\text{Cl}_2$ is limiting and water is in excess.)

Final answer

Water is in excess (0.444 mol available > 0.300 mol required).

Detailed explanation

Walkthrough

The goal is to show water is in excess. We do this by calculating the moles of each reactant and comparing them using the stoichiometric ratio from the balanced equation: $\text{S}_2\text{Cl}_2 + 3\text{H}_2\text{O} \rightarrow \dots$

Calculate $M_r$ of $\text{S}_2\text{Cl}_2$ : $S=32, Cl=35.5 \Rightarrow 2(32) + 2(35.5) = 135$ .
Calculate moles of $\text{S}_2\text{Cl}_2$ : $13.5 / 135 = 0.100$ mol.
Calculate moles of $\text{H}_2\text{O}$ : $M_r = 18$ . Moles = $8.00 / 18 = 0.444$ mol.
Use the ratio 1:3. For 0.100 mol of $\text{S}_2\text{Cl}_2$ , we need $0.100 \times 3 = 0.300$ mol of water.
We have 0.444 mol of water, which is more than 0.300 mol. Therefore, water is in excess.

Key Takeaways

To identify the excess reactant, calculate moles of both reactants, use the balanced equation to find how much of one is needed to react with the other, and compare with the actual amount available.

Common Mistakes

Using atomic mass instead of molecular mass for $\text{S}_2\text{Cl}_2$ (e.g., using 32+35.5=67.5). Forgetting the 1:3 ratio and comparing moles directly (0.100 vs 0.444 looks like excess, but you must prove it via stoichiometry). Rounding too early (0.444 vs 0.44).

Things to Be Careful About

State symbols are not required for this calculation. Ensure $M_r$ values are correct to 1 decimal place as per standard O Level practice (S=32.1 is sometimes used, but 32 is standard unless specified; mark scheme uses 135 implying S=32, Cl=35.5). The mark scheme accepts either showing water needed is 0.300 or showing S2Cl2 that would react is 0.148.

Techniques used

calculate molar mass of S2Cl2calculate moles of reactants from massuse stoichiometric ratio to determine limiting reactant

(e)

Sulfur dioxide is an air pollutant.

3M

(i)

State one adverse effect of sulfur dioxide in the air.

______

(ii)

Describe two ways of reducing the emissions of sulfur dioxide in the air.

______
______

DifficultyMedium-Easy

Worked solution

Answer

Flue gas desulphurisation: react $\text{SO}_2$ with calcium oxide ( $\text{CaO}$ ) or calcium carbonate ( $\text{CaCO}_3$ ) in power station chimneys.
Use low-sulfur fuels (or switch to renewable energy sources such as solar, wind, hydroelectric; burn less fossil fuel).

Final answer

Flue gas desulphurisation (using CaO or CaCO3) and use of low-sulfur fuels or renewable energy sources.

Detailed explanation

Walkthrough

To reduce $\text{SO}_2$ emissions, we can either remove it from the exhaust gases (end-of-pipe) or prevent its formation at the source.

Flue gas desulphurisation: Calcium oxide (from limestone) or calcium carbonate is sprayed into the flue gases. They react with $\text{SO}_2$ to form calcium sulfite/sulfate (a solid 'gypsum'), removing the pollutant before it enters the atmosphere.
Source reduction: Use fuels with less sulfur (low-sulfur coal/oil). Switch to energy sources that do not produce $\text{SO}_2$ , such as renewables (solar, wind, nuclear, hydro). Reduce overall fossil fuel consumption.

Key Takeaways

$\text{SO}_2$ can be removed from emissions chemically (using alkaline materials like lime/limestone) or prevented by changing the fuel source.

Common Mistakes

Suggesting 'catalytic converters' — these are primarily for nitrogen oxides and carbon monoxide from cars, not typically for $\text{SO}_2$ from power stations (though they can help, it's not the primary 5070 answer). Saying 'plant more trees' — trees absorb $\text{CO}_2$ , not significantly $\text{SO}_2$ .

Things to Be Careful About

Two ways are required for 2 marks. Ensure they are distinct. 'Use less fossil fuels' and 'use renewable energy' are often accepted as separate points, but 'flue gas desulphurisation' and 'low sulfur fuels' are the most chemically precise pair. Naming the renewable energy (solar, wind) is better than just saying 'renewables'.

Techniques used

identify methods for flue gas desulfurisationsuggest fuel switching or renewable energy alternatives

Q714MMedium-HardOrganic ChemistryAcids, Bases and SaltsStates of Matter

(a)

Propanoic acid can be represented by the formula $\text{CH}_3\text{CH}_2\text{COOH}$ .

8M

(i)

Propanoic acid reacts with methanol, $\text{CH}_3\text{OH}$ , to produce an ester.

Name the ester formed and draw its displayed formula.

name = ______

displayed formula

(ii)

Propanoic acid is a weak acid.

\text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_3\text{O}^+

Explain how this equation shows that:

$\text{CH}_3\text{CH}_2\text{COOH}$ is an acid by referring to proton transfer
______
$\text{CH}_3\text{CH}_2\text{COOH}$ is a weak acid.
______

(iii)

Propanoic acid reacts with magnesium.

Name the two products of this reaction.

______
______

(iv)

Magnesium is a solid at room temperature.

Describe the motion and separation of the particles in a solid.

motion = ______
separation = ______

(b)

Fig. 7.1 shows the simplified structures of two molecules that combine to form a polyamide.

Fig. 7.1

6M

(i)

Complete the diagram in Fig. 7.2 to show the structure of two repeat units of this polyamide.

Show all of the atoms and all of the bonds in the linkages.

Fig. 7.2

DifficultyMedium-Hard

Worked solution

Answer

Final answer

Completed diagram showing two repeat units with amide linkages -C(=O)-NH- and -NH-C(=O)-, left end ending in -NH- and right end ending in -C(=O)-

Detailed explanation

Walkthrough

The monomers are a dicarboxylic acid (shaded box, with $-\text{COOH}$ ends) and a diamine (open box, with $-\text{NH}_2$ ends). In condensation polymerisation, the $-\text{OH}$ from the carboxylic acid and an $-\text{H}$ from the amine are eliminated as water, forming an amide linkage $-\text{C}(=\text{O})-\text{NH}-$ .

The sequence of boxes is: open (amine), shaded (acid), open (amine), shaded (acid). We need to draw two repeat units, meaning we connect all four boxes.

Between the shaded box (acid) and the next open box (amine): the $-\text{C}(=\text{O})-$ end of the acid connects to the $-\text{NH}-$ end of the amine. Draw $-\text{C}(=\text{O})-\text{NH}-$ .
Between the open box (amine) and the next shaded box (acid): the $-\text{NH}-$ end of the amine connects to the $-\text{C}(=\text{O})-$ end of the acid. Draw $-\text{NH}-\text{C}(=\text{O})-$ .
Left end of the first open box: must end with $-\text{NH}-$ (derived from $-\text{NH}_2$ losing one H).
Right end of the last shaded box: must end with $-\text{C}(=\text{O})-$ (derived from $-\text{COOH}$ losing OH).

All atoms and bonds in the linkages must be shown explicitly (displayed formula style).

Key Takeaways

Condensation polymerisation between a dicarboxylic acid and a diamine forms a polyamide.
The amide linkage is $-\text{C}(=\text{O})-\text{NH}-$ .
End groups of the repeat unit chain reflect the remaining functional groups after water elimination.

Common Mistakes

Drawing the amide linkage backwards (e.g., $-\text{N}-\text{C}(=\text{O})-$ without the H on N, or connecting C to C).
Forgetting the hydrogen atom on the nitrogen in the amide linkage.
Not showing all bonds explicitly in the linkages (must be displayed formula, not skeletal).
Incorrect end groups (e.g., leaving $-\text{OH}$ or $-\text{H}_2\text{N}$ at the ends of a repeat unit diagram when it should be $-\text{C}(=\text{O})-$ and $-\text{NH}-$ ).

Things to Be Careful About

The diagram asks for 'two repeat units'. Ensure there are four boxes connected.
'Show all of the atoms and all of the bonds in the linkages' means you must draw the $\text{C=O}$ double bond and the $\text{N-H}$ bond explicitly in the amide groups.
Direction matters: carbonyl carbon from the acid must bond to the nitrogen from the amine.

Techniques used

complete a polymer repeat unit diagram from monomer structuresdraw amide linkages with all atoms and bondsensure correct orientation and end groups for the polymer chain

(ii)

Polyamides are polymers.

State the meaning of the term polymer.

______

(iii)

Polyamides are condensation polymers.

State one difference between condensation polymerisation and addition polymerisation.

______