Chemistry 0620 | Paper 2 Mock 01 | Multiple Choice (Extended)

The student is correct. In an exothermic reaction the chemicals give out energy to the surroundings, so they must end up with less energy than they started with. On the diagram that shows as the product level lying below the reactant level, which is exactly what this profile shows.

Expressed as a calculation, \(\Delta H = E_{\text{products}} - E_{\text{reactants}}\), and a lower product energy makes \(\Delta H\) negative, which is the sign convention for exothermic.

The claim that the products are always higher describes an endothermic reaction instead, and equal levels would mean no net energy change at all, which is not a real reaction outcome. Temperature does not decide the direction either; it affects the rate. Fix the rule now: exothermic means products below reactants, endothermic means products above.

A homologous series is a family of compounds with the same general formula, whose members differ by CH₂ and share the same functional group. Reactions occur where a bond is polar or can break easily, and that is exactly what the functional group provides: -OH in alcohols, C=C in alkenes, -COOH in carboxylic acids. Because every member carries the same reactive site, all members undergo the same reactions.

The hydrocarbon chain is comparatively unreactive, so counting hydrogen atoms or comparing relative molecular masses predicts physical trends such as boiling point and viscosity, not chemical behaviour. Colour is irrelevant, since most of these compounds are colourless.

The name says it: a hydrocarbon is a compound containing only hydrogen and carbon, with no other element present. Methane \(\text{CH}_4\), octane \(\text{C}_8\text{H}_{18}\) and the alkanes generally, formula \(\text{C}_n\text{H}_{2n+2}\), are all hydrocarbons, and crude oil is a mixture of thousands of them.

Something made only of carbon atoms is an element such as graphite or diamond, not a compound and not a hydrocarbon. Adding oxygen to carbon and hydrogen makes a different family altogether, for example ethanol \(\text{C}_2\text{H}_5\text{OH}\), which is a compound of three elements and therefore not a hydrocarbon. A mixture of water and carbon is not a compound at all, since its components are not chemically joined.

The word "only" is what earns the mark; do not omit it.

Two horizontal lines appear on a finished chromatogram. The lower one is the pencil baseline where the sample was applied; the upper one is the solvent front, the highest level the solvent reached before the paper was taken out and that level was marked. On this diagram the lower line carries the label P and the upper line carries the label Q, so Q marks the solvent front.

The remaining label sits on a circle part-way up the paper, which is a separated spot rather than a level. The front must be pencilled in immediately, before the solvent evaporates and its position is lost, because both distances used in the R_f calculation are measured between these marks.

A correct row needs a property that is both true of ethanol and the actual reason for the use.

Ethanol is a flammable liquid that burns in air to release energy, \( \mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O} \), so the row linking use as a fuel with burning to release energy is the correct pairing. The other fuel row contradicts itself by claiming ethanol does not burn. The solvent row fails on the property rather than the use, because ethanol mixes with water in all proportions, so calling it insoluble is false; that miscibility is part of why it is such a useful solvent. Ethanol is not a strong acid either, and it is sulfuric acid that car batteries contain.

The mass needed is 32 g. Mass = moles × A_r = 0.5 × 64 = 32 g, which makes sense because half a mole must weigh half of the molar mass.

Choosing 64 g weighs out a full mole and ignores the word half. Choosing 128 g multiplies by 2 instead of by 0.5, a very common slip when the fraction is written as a word rather than a decimal. Choosing 16 g quarters the molar mass, treating half a mole as half of a half. Writing the numbers into mass = n × A_r before calculating stops all three errors, and gives about 3.01 × 10²³ copper atoms.

The two polymers come from different alkenes, so their repeat units differ. Poly(ethene) is made from CH₂=CH₂ and repeats -CH₂-CH₂-, while poly(propene) is made from CH₂=CHCH₃ and repeats -CH₂-CH(CH₃)-, so every repeat unit of poly(propene) carries a methyl side group hanging off the backbone. Containing the same two elements does not make two compounds identical, because structure decides identity.

Neither polymer keeps a double bond, since the C=C is used up in joining the units, and neither is made by condensation, because no small molecule is lost. Poly(ethene) contains no chlorine; that is poly(chloroethene).

Row 4 is correct: carbon monoxide comes from the incomplete combustion of fuels, when there is too little oxygen to oxidise all the carbon to CO₂.

The other rows have real sources attached to the wrong gas. Lightning in clouds provides enough energy to make oxides of nitrogen, not carbon monoxide. Burning sulfur impurities in coal gives sulfur dioxide, not methane; methane comes from decay, landfill sites and livestock. The reaction of N₂ with O₂ in hot engines makes oxides of nitrogen, not sulfur dioxide. Learning each pollutant with its own origin stops these scrambled pairings catching you out.

Neutralisation removes the acidic hydrogen from the carboxyl group and leaves a negatively charged ion: \[ \mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O} \] The ethanoate ion is therefore \( \mathrm{CH_3COO^-} \), carrying a single negative charge on the site where the hydrogen used to be.

The hydroxide ion comes from the alkali, not the acid, and is used up in forming water. The carbonate ion belongs to a different family and carries a 2- charge. \( \mathrm{CH_3OH} \) is methanol, a neutral molecule with no charge and no carboxyl group. To write any carboxylate ion, delete the acidic hydrogen and add one negative charge.

Inside these lamps the gas is at a high temperature and surrounds hot metal parts, so it must not attack them or take part in any reaction. Krypton and xenon can be used because they are unreactive, so they do not damage the hot lamp.

Each has a full outer shell of eight electrons, so there is no tendency to gain, lose or share electrons even at high temperature. They also let the metal parts run hotter and brighter without oxidising, since no oxygen is present.

These gases do not burn, so they cannot be flammable, and they are certainly not acids that could etch glass. They are not chosen as conductors either: the current passes through the gas only once a high voltage has ionised it.

Hydrated copper(II) sulfate is blue because of the water of crystallisation locked into its lattice. Heating drives that water off as steam:

\[\mathrm{CuSO_4\cdot 5H_2O\ (blue)} \rightleftharpoons \mathrm{CuSO_4\ (white)} + 5\mathrm{H_2O}\]

So during heating the solid changes from blue to white, and the vapour labelled in the diagram condenses to colourless water at the cooler upper part of the tube. The row recording a blue to white colour change with water as the liquid formed is the one that matches.

White to blue describes the reverse change, seen when water is dripped onto the cooled anhydrous solid, which also releases heat. No oil is produced, since nothing organic is present, and there is certainly a visible change, so the row reporting nothing happening is ruled out by the observation.

Non-metals form covalent oxides that dissolve in water to release hydrogen ions, making the solution acidic. The question states that P₄O₁₀ dissolves to give phosphoric acid, H₃PO₄, so its solution has a pH below 7 and the oxide is acidic. It also reacts with alkalis to form salts, in this case phosphates.

The evidence given, a non-metal element and an acid in water, rules out the alternatives directly: basic oxides come from metals and give alkaline solutions, neutral oxides such as carbon monoxide give no reaction and pH 7, and amphoteric oxides such as aluminium oxide react with both acids and alkalis. Use the element type as the first clue and the water test to confirm.

The strip is displacing silver, so metal X must be above silver in the reactivity series and must give a coloured solution when its own ions form.

The colour change is the decisive clue. The solution starts colourless and turns pale blue, and blue in aqueous solution means copper(II) ions. So copper atoms are dissolving as \( \text{Cu}^{2+} \) while silver ions are being deposited as shiny crystals:

\[ \text{Cu} + 2\text{AgNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{Ag} \]

Silver itself cannot be metal X, since a metal cannot displace its own ions and nothing would change. Gold and platinum both sit below silver, so no reaction would occur and the solution would stay colourless. Use the solution colour to name the metal that has dissolved.

A salt is named after the acid it is made from, because the negative ion of the salt comes from the acid. Hydrochloric acid gives chlorides, sulfuric acid gives sulfates and nitric acid gives nitrates. Checking the table against that rule, only row 2 is right, since sulfuric acid does form a sulfate. For example \( \mathrm{H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)} \).

The other rows swap the acids around: hydrochloric acid cannot make a nitrate or a sulfate, because it contains no nitrogen and no sulfur, and nitric acid cannot make a chloride, because it contains no chlorine. That is the check to apply. The elements in the salt must all be present in the reactants, since atoms are neither created nor destroyed in a chemical reaction.

Bubbles forming on the inside of the cup and the metal thinning show that the metal is dissolving in the acid and releasing hydrogen. Only a metal above hydrogen in the reactivity series can do this:

\[ \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]

Zinc sits above hydrogen, so it is steadily eaten away while hydrogen bubbles cling to the surface, which matches both observations.

Silver, gold and platinum all lie below hydrogen. None of them can displace hydrogen from a dilute acid, so a cup made from any of them would show no bubbles and would not lose thickness, which is precisely why they are used for coinage and jewellery. The gradual loss of metal is the key clue that the container itself is a reactant.

Intensely coloured compounds are a characteristic property of the transition elements, so the purple of potassium manganate(VII), KMnO₄, points to manganese, a transition metal. Its ion absorbs part of the visible spectrum and the light that reaches the eye appears purple, and manganese existing here in a high oxidation state also illustrates variable oxidation state.

Compare this with the alkali metals of Group I, whose compounds such as sodium chloride are white and dissolve to colourless solutions, and with the noble gases, which are unreactive and form almost no compounds at all.

The strong colour is put to practical use: manganate(VII) fades as it is reduced, so the colour change signals when a redox reaction is complete.

Electrode Y carries the + sign in the diagram, so Y is the anode. Negative ions are attracted to the positive electrode, and in molten lead(II) bromide the only negative ion is Br⁻.

At the anode the bromide ions lose electrons (oxidation):

2Br⁻ → Br₂ + 2e⁻

Bromine is a red-brown liquid that vaporises readily, so the coloured vapour at Y is brown bromine vapour, which matches the observation described.

Lead is the cathode product at X, not Y, because Pb²⁺ is positive and moves to the negative electrode. Oxygen and hydrogen cannot form because the compound is molten with no water present, so the only ions available are Pb²⁺ and Br⁻.

Group I reactivity increases down the group because the outer electron is further from the nucleus and better shielded, so it is lost more easily. Losing that electron is exactly what the reaction with water requires:

\[ 2\text{M} + 2\text{H}_2\text{O} \rightarrow 2\text{MOH} + \text{H}_2 \]

Of the three beakers, potassium is the lowest of these metals in Group I, so it loses its outer electron most readily and reacts most violently. It melts, skims the surface and the hydrogen ignites with a lilac flame. Sodium melts and fizzes rapidly but usually does not ignite, and lithium fizzes steadily without melting.

Do not assume the metal with the smallest atoms is the most reactive; for Group I the trend runs the other way.

Addition polymerisation is the one type of polymerisation with a single product. Each C=C opens and the units bond directly to one another, so every atom of every ethene molecule ends up in the chain and nothing is left over. The equation is n CH₂=CH₂ → -(CH₂-CH₂)_n-, giving poly(ethene) only.

Water is lost in condensation polymerisation, as in making nylon or a polyester, which needs monomers carrying functional groups at each end rather than a double bond; expecting water here is the usual confusion between the two types. No hydrogen is released, and ethane cannot form because no extra hydrogen is supplied.

Both tubes fizz, but the gases are different and the splint test tells them apart. Tube 2 contains a reactive metal and an acid, and a reactive metal displaces hydrogen from acid: \(\mathrm{Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)}\). Hydrogen is the gas that burns with a squeaky pop when a lighted splint is held over it, so hydrogen is the gas made in tube 2.

Tube 1 holds a carbonate and acid, which gives carbon dioxide. That gas puts a lighted splint out and turns limewater milky, so it can never pop. Oxygen relights a glowing splint instead of popping, and chlorine is not a product here.

Match each test to its gas: pop for hydrogen, relight for oxygen, limewater for carbon dioxide.

Chlorine is added to treated water as a disinfectant, so its main purpose is to kill bacteria and other microorganisms. It destroys the microbes that cause water-borne diseases such as cholera and typhoid, and because a small residual amount remains dissolved, the water is protected against recontamination as it travels through the pipes to homes.

Chlorine does not remove dissolved salts, since those stay in solution and would need distillation to be taken out. It does not remove large floating objects either, as those are caught by screening and settling long before this stage. Far from sweetening the water, too much chlorine gives an unpleasant taste and smell, which is why the dose is carefully controlled.

Each common colourless gas has its own test, so the confirming observation must be one that no other gas gives. Carbon dioxide reacts with the calcium hydroxide in limewater to form insoluble calcium carbonate, and that fine white solid is what makes the limewater look milky:

\[\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O}\]

So turning limewater milky is the observation that confirms carbon dioxide.

A squeaky pop with a lighted splint identifies hydrogen, relighting a glowing splint identifies oxygen, and turning damp red litmus blue identifies ammonia. Carbon dioxide in fact does the opposite of that last test, since its solution is weakly acidic and turns damp blue litmus red.

A metal displaces iron from iron(II) sulfate only if it is above iron in the series, because it must be able to give electrons to \( \text{Fe}^{2+} \) ions: \( \text{M} + \text{FeSO}_4 \rightarrow \text{MSO}_4 + \text{Fe} \).

Reading the ladder, calcium, magnesium and zinc all sit above the iron line, so each deposits a coating of iron. Copper is printed below iron, so it cannot reduce \( \text{Fe}^{2+} \) and no coating appears. That gives three metals producing a coating.

Counting four is the usual error, from assuming every metal added must do something. In practice calcium also attacks the water itself, so its coating is masked by fizzing, but by series position it still qualifies.

Choose the method whose separating principle matches the difference between the components. An insoluble solid keeps its particles in large clusters that cannot pass through the pores of filter paper, while the liquid flows through freely, so filtration is the correct method for this task, exactly as the first row of the table states.

Crystallisation is for recovering a solid that has dissolved, so it is the wrong tool when the solid never dissolved in the first place. Distillation is for obtaining the solvent from a solution using boiling point, which wastes energy and is unnecessary here. Chromatography separates dissolved substances by how far they travel on paper. The deciding question is always whether the solid is soluble or insoluble.

Count the atoms shown in the structure rather than recalling the formula.

Two carbon atoms are joined by a single bond. The left-hand carbon carries three hydrogens, one to its side and one above and below, and the right-hand carbon carries two hydrogens as well as a bond to the oxygen, which itself carries the last hydrogen. That gives \( \mathrm{CH_3CH_2OH} \), normally written \( \mathrm{C_2H_5OH} \). \( \mathrm{CH_3OH} \) is methanol and has only one carbon atom, so it cannot match a two-carbon chain. \( \mathrm{C_2H_6} \) is ethane and \( \mathrm{C_2H_4} \) is ethene; both ignore the oxygen drawn in the chain, and neither contains the \( \mathrm{-OH} \) group that makes a compound an alcohol.

Check the three rows of the table against what you know about ionic compounds. They are built from oppositely charged ions rather than molecules or free electrons, they have high melting points because the attractions in the giant lattice are strong, and they do not conduct as solids because the ions are fixed.

Column Y shows exactly that combination: ions, a high melting point and no conduction as a solid.

Column Z contains atoms and electrons and conducts as a solid, which describes a metal, where delocalised electrons move through a lattice of positive ions. Column X contains molecules with a low melting point, which describes a simple molecular substance held together by weak intermolecular forces.

A metal strip becomes coated with copper only if it is more reactive than copper, so it can push copper out of solution: \[\text{Cu}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Cu}(s)\]

The bigger the reactivity gap, the faster the electrons are handed over and the sooner the brown copper layer appears. Strip P was coated in only 5 s, against 40 s for Q and a full 3 minutes for R, so P is the most reactive of the three.

The misconception here is to read a long coating time as a "bigger" reaction because more time passed. Time is an inverse measure: a short time means a fast reaction. In displacement experiments, quickest change equals greatest reactivity.

Both gases set off at the same moment from opposite ends and stopped travelling at the same moment, when they met and formed the white ring. Since the times are equal, the distances directly compare their speeds.

The measurements show 36 cm travelled by ammonia against 24 cm by hydrogen chloride, so ammonia diffused faster, as it travelled the greater distance. The reason is its lower relative molecular mass, 17 against 36.5, which gives its molecules a higher average speed at the same temperature.

Claiming hydrogen chloride went further misreads the table, since 24 cm is the smaller value. Equal rates would have produced two equal distances of 30 cm each. And both gases plainly moved, otherwise no ring could have formed at all. Equal time is the hidden condition that makes distance a fair measure of rate.

Two separate conditions must both be met, so use the series shown in the diagram twice.

To displace zinc from zinc sulfate solution, the metal must be above zinc in the list, which rules out iron and copper: they sit below zinc and so cannot push zinc ions out of solution. That leaves potassium, calcium and magnesium.

The second condition is no reaction with cold water. Potassium reacts vigorously with cold water and calcium reacts steadily with it, both giving hydrogen and a hydroxide, so they are eliminated. Magnesium reacts only extremely slowly with cold water, needing steam for a rapid reaction, so it satisfies both requirements:

\[\mathrm{Mg + ZnSO_4 \rightarrow MgSO_4 + Zn}\]

The usual mistake is choosing the most reactive metal available. More reactive is not automatically better when a second condition limits the choice.

The extraction method depends on where a metal stands relative to carbon. Zinc, iron and copper all sit below carbon, so heating their oxides with carbon reduces them cheaply, for example \[\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\]

Aluminium is the metal extracted by electrolysis. It lies above carbon in the reactivity series, so carbon cannot take its oxygen away. Molten aluminium oxide dissolved in cryolite is electrolysed instead, and aluminium ions gain electrons at the cathode: \[\text{Al}^{3+} + 3\text{e}^{-} \rightarrow \text{Al}\]

Students often assume the most expensive or the most common metal is electrolysed. The deciding factor is only reactivity: above carbon means electrolysis, below carbon means reduction with carbon.

Rust is hydrated iron(III) oxide, \( \mathrm{Fe_2O_3}\cdot x\mathrm{H_2O} \), and it is orange-brown. Because the iron(III) oxide holds water loosely, the solid is porous and flaky, so it lifts off and lets fresh iron underneath be attacked. That is why a whole fork can eventually corrode through rather than being sealed by its own coating.

Silvery-white is the appearance of the clean steel itself, not the corrosion product. Bright green is the colour associated with copper corroding on old roofs, and with iron(II) compounds in solution. Deep blue points to copper(II) compounds. Learn the classic colours for the iron oxidation states: iron(II) species are usually pale green and iron(III) species are orange or red-brown.

The fertiliser washed into the pond supplies nitrate and phosphate, the nutrients that had been limiting algal growth, so the algae multiply rapidly across the surface. The mat they form is thick and opaque, so it blocks sunlight from reaching the water beneath, and the plants below can no longer photosynthesise. Without photosynthesis they make no glucose, run out of energy and die.

This is the first stage of eutrophication. Later, bacteria decompose the dead material and use up dissolved oxygen, which harms fish as well.

The algae shade the plants rather than lighting them, and nothing they do adds oxygen to the deep water. In the exam, link the bloom to light first, then to oxygen.

Read the two definitions in the table against OIL RIG. The substance X loses electrons and its oxidation number rises, so X is oxidised. The substance Y gains electrons and its oxidation number falls, so Y is reduced.

Now name the roles by what each does to its partner. X supplies the electrons that reduce Y, so X is the reducing agent. Y takes the electrons that oxidise X, so Y is the oxidising agent. The correct completion is therefore X as the reducing agent and Y as the oxidising agent.

Swapping them is the classic error, caused by naming the agent after what happens to it rather than what it does to the other substance. The two remaining choices break the reaction altogether: a redox reaction always needs one of each, since electrons lost by one species must be gained by another.

A large number written in front of a formula multiplies the whole formula, while a subscript applies only to the atom it follows. In 3CO₂ the subscript 2 gives two oxygen atoms per molecule and the 3 in front means three such molecules, so the oxygen total is \(3 \times 2 = 6\) atoms. There are also \(3 \times 1 = 3\) carbon atoms.

Answering 3 counts the molecules rather than the oxygen atoms, and answering 2 ignores the multiplier and counts a single molecule. An answer of 5 comes from adding \(3 + 2\) instead of multiplying. This distinction between the large number in front and the small subscript is exactly what makes equation balancing work, because only the front number may be changed when balancing.

Each nail sits in the same dilute acid, so the only variable is the coating metal. Bubbles of hydrogen come from the metal displacing hydrogen from the acid, and the further above hydrogen the metal is, the faster that happens.

Place the three coatings in the reactivity series: zinc is the highest, tin comes below it but is still above hydrogen, and copper lies below hydrogen. So the nail coated with zinc reacts fastest and bubbles most vigorously.

\[\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2}\]

The tin coated nail bubbles slowly, and the copper coated nail gives no bubbles at all because copper cannot displace hydrogen from a dilute acid. The results cannot be equal, since the whole point of the comparison is that reactivity differs. This is also why tin plating protects steel only while the coating stays unbroken.

Two carbon reactions happen in sequence near the base of the furnace. First the coke burns in the hot air blast, C + O₂ → CO₂, which releases the heat that raises the temperature to about 1500 ℃. That step makes carbon dioxide, not the reducing agent.

The carbon dioxide then rises through the white-hot coke and is itself reduced: CO₂ + C → 2CO. This is the step that finally produces carbon monoxide, which goes on to reduce the ore, Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Combustion alone is the common wrong choice because it involves carbon and air. The equation with 2O₂ giving 2CO does not balance oxygen, and 2CO → C + CO₂ destroys the reducing agent.

A neutral oxide reacts with neither acids nor alkalis and gives a solution of pH 7. Of the substances listed, water is itself the oxide of hydrogen, H₂O, and fits exactly, since pure water has pH 7 and is neither acid nor base in this classification.

Sodium oxide is basic and dissolves to give sodium hydroxide at about pH 13. Sulfur dioxide is acidic and forms sulfurous acid. Aluminium oxide is amphoteric, reacting with both acids and alkalis; its near-neutral pH in water reflects only its very low solubility, not neutrality. Carbon monoxide and nitrogen monoxide are the other neutral oxides worth remembering.

Ethanol and water are miscible with boiling points of 78℃ and 100℃, which are close enough that simple distillation lets some water vapour through with the ethanol. Fractional distillation adds a tall packed column in which the vapour condenses and re-evaporates many times on the way up. Each of these cycles enriches the rising vapour in the lower boiling component, so fractional distillation gives the best separation.

A separating funnel only works for immiscible liquids that form two layers, and these two mix completely. Filtration separates by particle size and cannot act on two liquids, and crystallisation needs a dissolved solid. Reach for a fractionating column whenever two miscible liquids have similar boiling points.

When a reactive metal meets water it forms a soluble hydroxide and hydrogen, which is why the universal indicator turns alkaline: \[2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\]

The more reactive the metal, the more hydroxide is produced and the higher the pH. Reading the colours, metal 4 gives purple, the strongly alkaline end of the scale, so it produces the most hydroxide and is the most reactive.

Green means the solution stayed neutral, so that metal did not react with water at all. Blue and dark blue show weaker alkalinity, that is, slower or partial reaction. Learn the indicator sequence green (neutral), blue, dark blue, purple (strongly alkaline), and read pH as a measure of how much hydroxide has formed.

The wax melts on the copper rod, because copper is a metal and metals are good conductors of heat. The delocalised electrons in a metal are free to move, so they gain kinetic energy at the hot end and carry it rapidly along the rod, and the vibrating lattice of ions passes energy on as well. Energy therefore reaches the wax blob quickly and melts it.

Wood has no free electrons. Its energy can only be passed slowly from one vibrating molecule to the next, so it is an insulator and its wax stays solid. That is exactly why saucepans have metal bases and wooden or plastic handles.

Saying both rods behave the same ignores the whole point of the fair test set up here.