ESAT Chemistry Mock Module 3 Worked Solutions

Correct Answer: D. Hexene decolourises it from orange to colourless; hexane leaves it orange

Fastest Approach (🚀):
Alkene has $\text{C}=\text{C}$ $\Rightarrow$ addition $\Rightarrow$ decolourised.
Alkane saturated $\Rightarrow$ no reaction $\Rightarrow$ stays orange.

Matches Option D.

Step-by-Step Breakdown:

1. What makes an alkene reactive here

Hexene contains a $\text{C}=\text{C}$ double bond. One of the two bonds can open, allowing a bromine molecule to add across it:
$\text{C}_6\text{H}_{12}+\text{Br}_2\rightarrow\text{C}_6\text{H}_{12}\text{Br}_2$

The orange bromine is consumed, so the solution turns colourless.

2. Why the alkane does not react

Hexane is saturated — every bond is a single bond and every carbon already carries its full complement of hydrogen. There is nothing for bromine to add to, so the solution stays orange.

Alkanes do react with bromine, but only by substitution and only in ultraviolet light. Under the conditions of this test, nothing happens.

3. Reading the result

The colour change is a removal of colour: orange to colourless. It is not a precipitate, so 'milky white' describes a different test entirely — that is limewater with carbon dioxide, and Option E imports it.

4. Why this is the standard test

It is quick, needs no heating, and distinguishes saturated from unsaturated in seconds. The same test shows that vegetable oils are unsaturated while animal fats are largely saturated.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Both decolourise the bromine water — Alkane Reacts
  Alkanes do not react with bromine water in the absence of ultraviolet light.
• B. Neither decolourises the bromine water — Reaction Denied
  The alkene does react.
• C. Hexane decolourises it; hexene leaves it orange — Reversed
  The two compounds swapped.
• E. Both turn the bromine water milky white — Wrong Observation
  Milky white describes the limewater test for carbon dioxide.

Common Mistake (⚠️):
Saying the solution turns white or milky. Bromine water is decolourised — the orange disappears and nothing is precipitated.

Takeaway (📌):
Bromine water decolourises with alkenes by addition across the double bond, and is unchanged by alkanes.

Correct Answer: C. Zinc is more reactive than copper, because it loses electrons more readily to form ions

Fastest Approach (🚀):
Zinc displaces copper; copper cannot displace zinc.
Zinc more reactive.

Matches Option C.

Step-by-Step Breakdown:

1. Read the successful reaction

$\text{Zn}+\text{CuSO}_4\rightarrow\text{ZnSO}_4+\text{Cu}$

Zinc has taken the sulfate from copper, so zinc has formed ions while copper has been forced out of solution as the metal.

2. Read the failed reaction

Copper added to zinc sulfate does nothing. Copper cannot displace zinc, which confirms the order rather than merely failing to contradict it.

3. What reactivity means here

A metal's reactivity is its tendency to lose electrons and form positive ions. Zinc does this more readily than copper:
$\text{Zn}\rightarrow\text{Zn}^{2+}+2\text{e}^{-}$
$\text{Cu}^{2+}+2\text{e}^{-}\rightarrow\text{Cu}$

Zinc is oxidised, copper reduced — a redox reaction, and the reactivity series is essentially a ranking of how readily each metal is oxidised.

4. What is observed

The blue solution fades as $\text{Cu}^{2+}$ is removed, and a brown deposit of copper appears on the zinc. The mixture also warms, since the reaction is exothermic.

5. Why a third metal is not needed

Two experiments, one positive and one negative, settle the order of two metals completely. A third would be needed only to place a third metal, so Option E asks for more than the question requires.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Copper is more reactive than zinc — Order Reversed
  Reversed — copper failed to displace zinc.
• B. The two metals are equally reactive — Contradicts Data
  Equal reactivity would mean neither reaction occurs.
• D. Zinc is more reactive because it is denser — Wrong Cause
  Density has no bearing on reactivity.
• E. Nothing can be concluded without testing a third metal — Over-cautious
  Two experiments settle the order of two metals.

Common Mistake (⚠️):
Reading the reaction the wrong way round. The metal added as an element is the more reactive one when a reaction occurs — it is the one displacing.

Takeaway (📌):
Reactivity is the tendency to form positive ions. A more reactive metal displaces a less reactive one from solution, and is oxidised in doing so.

Correct Answer: A. Carbon dioxide; oxygen relights a glowing splint

Fastest Approach (🚀):
Milky limewater $=$ carbon dioxide.
Relights a glowing splint $=$ oxygen.

Matches Option A.

Step-by-Step Breakdown:

1. The five tests worth knowing

Hydrogen — a lighted splint gives a squeaky pop.
Oxygen — a glowing splint relights.
Carbon dioxide — limewater turns milky.
Chlorine — damp blue litmus turns red, then is bleached white.
Ammonia — damp red litmus turns blue, and it has a sharp smell.

2. Apply them

Milky limewater identifies carbon dioxide. The precipitate is calcium carbonate, formed as the gas reacts with the calcium hydroxide solution.

Oxygen is identified by relighting a glowing splint.

3. The distinction that is always tested

A lighted splint for hydrogen — it burns explosively, hence the pop.
A glowing splint for oxygen — it relights, because oxygen supports combustion but does not itself burn.

Confusing lighted with glowing swaps the two gases, and Options A and C are both built on it.

4. Why oxygen does not burn

Combustion is reaction with oxygen. Oxygen is the oxidising agent, not the fuel, so it supports burning without burning itself — which is exactly what the relighting splint demonstrates.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Hydrogen; oxygen is identified by a squeaky pop — Tests Swapped
  Hydrogen does not affect limewater, and the pop is hydrogen's test.
• C. Carbon dioxide; oxygen is identified by a squeaky pop — Second Test Wrong
  Right first gas, but the pop identifies hydrogen.
• D. Oxygen; carbon dioxide relights a glowing splint — Gases Swapped
  The two gases swapped.
• E. Chlorine; oxygen turns damp litmus paper white — Wrong Gas
  Chlorine does not turn limewater milky.

Common Mistake (⚠️):
Swapping the splint tests. Hydrogen needs a lighted splint and pops; oxygen needs a glowing one and relights it.

Takeaway (📌):
Limewater milky is carbon dioxide; glowing splint relights is oxygen; squeaky pop is hydrogen; bleached litmus is chlorine.

Correct Answer: D. $78\%$ nitrogen, $21\%$ oxygen, about $1\%$ argon, with carbon dioxide a trace

Fastest Approach (🚀):
Nitrogen $78$, oxygen $21$, argon $1$.

Matches Option D.

Step-by-Step Breakdown:

1. The composition

Nitrogen, about $78\%$
Oxygen, about $21\%$
Argon, about $0.9\%$
Carbon dioxide, about $0.04\%$
plus water vapour, which varies — hence the specification of dry air.

2. Why nitrogen dominates

Nitrogen exists as $\text{N}_2$ with a very strong triple bond, so it is chemically inert under ordinary conditions and accumulates. Oxygen, being reactive, is continuously consumed by respiration and combustion and replenished by photosynthesis.

3. Why carbon dioxide's share is misleading

At about $0.04\%$ it is a trace, yet it drives the greenhouse effect. Abundance and importance are quite separate — which is exactly the point the atmospheric chemistry section is making.

4. How the gases are separated

By fractional distillation of liquid air: cool air until it liquefies, then warm it slowly. Nitrogen boils off first at $-196\ ^{\circ}\text{C}$, argon next, and oxygen last at $-183\ ^{\circ}\text{C}$. It is a physical process, because air is a mixture.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $78\%$ oxygen, $21\%$ nitrogen, $1\%$ argon — Swapped
  Nitrogen and oxygen swapped.
• B. $21\%$ nitrogen, $78\%$ carbon dioxide, $1\%$ oxygen — Proportions Wrong
  Carbon dioxide is a trace gas, not the majority.
• C. $50\%$ nitrogen, $50\%$ oxygen — Proportions Wrong
  Far too much oxygen.
• E. $78\%$ nitrogen, $21\%$ carbon dioxide, $1\%$ oxygen — Swapped
  Carbon dioxide and oxygen swapped.

Common Mistake (⚠️):
Swapping nitrogen and oxygen. Oxygen is the reactive one we depend on, which makes it feel like the majority — but it is only about a fifth.

Takeaway (📌):
Dry air: $78\%$ nitrogen, $21\%$ oxygen, $1\%$ argon, $0.04\%$ carbon dioxide. Separated by fractional distillation of liquid air.

Correct Answer: B. $25\ \text{g}$

Fastest Approach (🚀):
$0.25\times 100 = 25\ \text{g}$.

Matches Option B.

Step-by-Step Breakdown:

1. Rearrange the relationship

$n = \dfrac{m}{M_r}$, so $m = n\times M_r$.

2. Substitute

$m = 0.25\times 100 = 25\ \text{g}$

3. Check the direction

One mole weighs $100\ \text{g}$, so a quarter of a mole weighs a quarter of that. Any answer above $100\ \text{g}$ has divided where it should have multiplied, which disposes of Option A immediately.

4. The triangle

mass at the top, moles and $M_r$ below. Cover the quantity you want and the arrangement of the other two reads off. It is the same structure as speed, distance and time, and it removes any need to remember which way round the formula goes.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $400\ \text{g}$ — Inverted
  Dividing rather than multiplying.
• C. $100\ \text{g}$ — Quantity Ignored
  The mass of one mole, ignoring the quantity.
• D. $0.0025\ \text{g}$ — Wrong Direction
  Dividing by $M_r$ as though converting to moles.
• E. $4\ \text{g}$ — Arithmetic Error
  Using $M_r = 16$, or another arithmetic slip.

Common Mistake (⚠️):
Dividing $M_r$ by the number of moles, giving $400\ \text{g}$. Fewer moles must mean less mass.

Takeaway (📌):
$m = n\times M_r$ going one way, $n = \dfrac{m}{M_r}$ the other. Sanity-check against the mass of one mole.

Correct Answer: D. Close together but able to move past one another; fixed volume, no fixed shape

Fastest Approach (🚀):
Close together $\Rightarrow$ fixed volume.
Free to move past each other $\Rightarrow$ no fixed shape.

Matches Option D.

Step-by-Step Breakdown:

1. The three states

Solid — particles touching, in a regular arrangement, vibrating about fixed positions. Fixed shape and fixed volume.

Liquid — particles still touching, but irregularly arranged and able to slide past one another. Fixed volume, takes the shape of its container.

Gas — particles far apart, moving rapidly and randomly. Neither fixed shape nor fixed volume; it fills whatever it is put in.

2. Which fact explains which property

Spacing explains volume. Particles that touch cannot be pushed much closer, so solids and liquids are nearly incompressible and hold their volume. Gas particles have space between them, so a gas compresses readily.

Freedom of movement explains shape. Fixed positions give a fixed shape; the ability to move past neighbours lets liquids and gases flow.

3. Why liquids sit between the two

A liquid has the spacing of a solid and the mobility of a gas — which is exactly why it has one fixed property and one that is not.

4. Why Option D is impossible

Particles that are far apart and in fixed positions describes nothing. If they were fixed in place with gaps between them, the substance would have both a fixed volume and a fixed shape, which no state does with that spacing.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Far apart, moving rapidly and randomly; no fixed shape or volume — Wrong State
  Describes a gas.
• B. Close together, vibrating about fixed positions; fixed shape and volume — Wrong State
  Describes a solid.
• C. Far apart and in fixed positions; fixed volume, no fixed shape — Internally Inconsistent
  Far apart and fixed in place describes no real state.
• E. Close together and in fixed positions; no fixed volume — Internally Inconsistent
  Fixed positions with no fixed volume is contradictory.

Common Mistake (⚠️):
Saying liquid particles are far apart. They are as close as in a solid — which is why liquids barely compress — but they are free to move.

Takeaway (📌):
Spacing sets the volume, mobility sets the shape. A liquid has solid-like spacing and gas-like mobility.

Correct Answer: B. There are more acid particles in the same volume, so collisions with the metal happen more often

Fastest Approach (🚀):
More particles per unit volume $\Rightarrow$ more collisions per second.

Matches Option B.

Step-by-Step Breakdown:

1. What concentration changes

Doubling the concentration puts twice as many acid particles into the same volume. Twice as many are therefore adjacent to the metal surface at any moment, so collisions occur about twice as often.

Rate depends on the frequency of successful collisions, and doubling the frequency roughly doubles the rate.

2. What concentration does not change

The speed of the particles, which depends only on temperature. So the energy of each collision is unchanged, and the proportion of collisions that succeed is unchanged too.

That is the whole distinction between this factor and temperature, which changes both.

3. Why the other options belong elsewhere

Faster particles and more energetic collisions describe temperature.
Lowering the activation energy describes a catalyst.
Increasing surface area describes dividing the solid more finely, which is a separate factor and is not what changing the acid's concentration does.

4. Why 'roughly'

The relationship between concentration and rate is not always exactly proportional — it depends on the reaction's mechanism. At this level the qualitative statement is what is required: more concentrated means faster.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. The acid particles move faster, so they collide with more energy — Wrong Factor
  Describes the effect of raising the temperature.
• C. The activation energy is lowered by the extra concentration — Wrong Factor
  Describes a catalyst.
• D. The metal's surface area increases — Wrong Factor
  Describes dividing the solid more finely.
• E. The reaction becomes exothermic at higher concentrations — Irrelevant
  The energy change of a reaction does not depend on concentration.

Common Mistake (⚠️):
Explaining a concentration change with the mechanism for a temperature change. Concentration alters how often particles meet, not how hard they hit.

Takeaway (📌):
Concentration raises collision frequency alone. Temperature raises frequency and energy. A catalyst lowers the activation energy.

Correct Answer: D. Each carbon in graphite forms three covalent bonds, leaving one delocalised electron per atom; in diamond all four outer electrons are bonded

Fastest Approach (🚀):
Graphite: $3$ bonds, $1$ spare electron $\Rightarrow$ conducts.
Diamond: $4$ bonds, none spare $\Rightarrow$ insulator.

Matches Option D.

Step-by-Step Breakdown:

1. Diamond

Each carbon forms four covalent bonds, to four other carbons, in a rigid three-dimensional tetrahedral network. Every outer electron is committed to a bond, so none is free to move and diamond does not conduct.

That same network is why diamond is the hardest natural substance and melts extraordinarily high — breaking it needs covalent bonds broken throughout.

2. Graphite

Each carbon forms only three covalent bonds, producing flat hexagonal layers. Carbon has four outer electrons, so one per atom is left over and becomes delocalised between the layers.

Those delocalised electrons are free to move along the layers, so graphite conducts — which is why it is used for electrodes and in electrolysis cells.

3. The other property that follows

The layers are held to each other only by weak intermolecular forces, so they slide over one another easily. That is why graphite is soft and slippery, and why it works as a lubricant and marks paper.

One structural difference — three bonds against four — explains conductivity, hardness and lubrication together.

4. Why the other options fail

Graphite contains no ions; the mobile charges are electrons. Its covalent bonds are not weak — it is the forces between layers that are. And it is not a metal, though the delocalised electrons make it behave like one electrically.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Graphite contains ions that are free to move — Wrong Carrier
  Graphite contains no ions; the charge carriers are electrons.
• B. Diamond is harder, and hard substances cannot conduct — Wrong Cause
  Hardness does not determine conductivity.
• C. Graphite has weaker covalent bonds, which break to release electrons — Bond Strength
  The covalent bonds within a layer are strong.
• E. Graphite is a metal and diamond is a non-metal — Classification Wrong
  Graphite is a non-metal that happens to conduct.

Common Mistake (⚠️):
Saying graphite's covalent bonds are weak. The bonds within a layer are strong; the weak forces are the ones holding one layer to the next, and that explains softness rather than conduction.

Takeaway (📌):
Graphite: three bonds, one delocalised electron per atom, conducts and is soft. Diamond: four bonds, none free, insulating and hard.

Correct Answer: E. Copper at the cathode, oxygen at the anode

Fastest Approach (🚀):
Cathode: $\text{Cu}$ below $\text{H}$ $\Rightarrow$ copper.
Anode: no halide $\Rightarrow$ oxygen.

Matches Option E.

Step-by-Step Breakdown:

1. List the ions

From the salt: $\text{Cu}^{2+}$ and $\text{SO}_4^{2-}$.
From the water: $\text{H}^{+}$ and $\text{OH}^{-}$.

2. The cathode

$\text{Cu}^{2+}$ and $\text{H}^{+}$ compete. The less reactive one is discharged, and copper sits below hydrogen in the reactivity series — so copper is deposited as a pink-brown layer on the electrode:
$\text{Cu}^{2+}+2\text{e}^{-}\rightarrow\text{Cu}$

This is the opposite of the sodium chloride case, where the metal is more reactive than hydrogen and hydrogen wins.

3. The anode

$\text{SO}_4^{2-}$ and $\text{OH}^{-}$ compete. Sulfate is a stable polyatomic ion and is not discharged; with no halide present, oxygen is released from the hydroxide ions:
$4\text{OH}^{-}\rightarrow\text{O}_2+2\text{H}_2\text{O}+4\text{e}^{-}$

4. What is observed

The blue colour fades as $\text{Cu}^{2+}$ is removed from solution, copper plates the cathode, and bubbles of oxygen appear at the anode. The solution becomes acidic, since $\text{OH}^{-}$ is consumed and $\text{H}^{+}$ is left behind.

5. Why the electrodes are specified as inert

With copper electrodes instead, the anode dissolves to replace the copper removed at the cathode — the purification process for copper, and a different answer to the same-looking question.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Hydrogen at the cathode, oxygen at the anode — Reactivity Reversed
  Copper is below hydrogen, so copper is deposited.
• B. Copper at the cathode, sulfur at the anode — Wrong Anode Product
  Sulfate is not discharged; oxygen comes from the hydroxide.
• C. Copper at the cathode, hydrogen at the anode — Electrodes Swapped
  Hydrogen is a cathode product, never an anode one.
• D. Hydrogen at the cathode, sulfur dioxide at the anode — Both Wrong
  Both predictions wrong for this solution.

Common Mistake (⚠️):
Predicting hydrogen at the cathode by applying the sodium chloride result. The competition is decided by where the metal sits relative to hydrogen, and copper is below it.

Takeaway (📌):
Cathode: the metal if it is below hydrogen, otherwise hydrogen. Anode: the halogen if a halide is present, otherwise oxygen.

Correct Answer: E. $+6$

Fastest Approach (🚀):
$2x-14 = -2 \implies 2x = 12 \implies x = +6$.

Matches Option E.

Step-by-Step Breakdown:

1. Set the known states

Seven oxygens at $-2$ each:
$7\times(-2) = -14$

2. Sum to the ion's charge

The oxidation states in an ion total the charge on the ion, which is $-2$. With $x$ for each chromium and two of them:
$2x+(-14) = -2$

3. Solve

$2x = 12$
$x = +6$

4. The step that is missed

The equation gives $2x = 12$, and stopping there gives $+12$ — the combined state of both chromium atoms, not the state of one. An oxidation state is always quoted per atom, so the final division by two is compulsory. Option A is that error.

5. Why $+6$ makes sense

Chromium's highest oxidation state is $+6$, and dichromate is a strong oxidising agent for the same reason permanganate is: the metal is at the top of its range and has a long way to fall. Reduced, it goes to $\text{Cr}^{3+}$ — the familiar orange-to-green colour change.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $+12$ — Not Divided
  The combined total for both atoms, not the state per atom.
• B. $+3$ — Wrong Species
  The state of chromium after reduction, $\text{Cr}^{3+}$.
• C. $+7$ — Wrong Ion
  The state of manganese in permanganate, not chromium here.
• D. $-6$ — Sign Error
  Sign error in solving.

Common Mistake (⚠️):
Answering $+12$ by forgetting there are two chromium atoms. Oxidation state is per atom, so divide by the number of them.

Takeaway (📌):
Assign the known states, sum to the ion's charge, then divide by the number of atoms of the unknown element.

Correct Answer: E. The double bond opens and the monomers join end to end, with no other product formed

Fastest Approach (🚀):
Double bond opens, monomers join, nothing else formed.

Matches Option E.

Step-by-Step Breakdown:

1. What happens

Each ethene molecule's double bond opens, leaving a bond available at each end. Thousands of these units link into a long chain:
$n\,\text{C}_2\text{H}_4\rightarrow(\text{C}_2\text{H}_4)_n$

2. Why it is called addition polymerisation

The monomers simply add to one another. No atoms are lost, so the repeating unit has exactly the same atoms as the monomer, and the polymer's empirical formula matches the monomer's. Losing a small molecule such as water would make it condensation polymerisation, which needs monomers with two functional groups — so Option A describes the wrong mechanism.

3. Why the polymer is saturated

The double bonds are consumed in forming the chain. The product is a long saturated hydrocarbon, which is why poly(ethene) is unreactive, does not decolourise bromine water, and is not biodegradable. Option D has this backwards.

4. Why Option C is impossible

Joining thousands of monomers gives a molecule thousands of times heavier. Poly(ethene) has an $M_r$ in the tens of thousands.

5. Why Option E is wrong

Addition polymerisation requires an unsaturated monomer — a $\text{C}=\text{C}$ bond is exactly what opens. Saturated monomers cannot polymerise this way, which is why ethane does not.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Water is lost as each pair of monomers joins — Wrong Mechanism
  Describes condensation polymerisation.
• B. The reaction requires the monomers to be saturated — Reversed
  The monomer must be unsaturated for addition to occur.
• C. The polymer has a lower relative molecular mass than the monomer — Impossible
  The polymer is thousands of times heavier.
• D. The polymer contains one double bond for every monomer unit — Reversed
  The double bonds are used up forming the chain.

Common Mistake (⚠️):
Confusing addition with condensation polymerisation. Addition loses nothing; condensation eliminates a small molecule, usually water, at each link.

Takeaway (📌):
Addition polymerisation: unsaturated monomer, double bond opens, monomers join, no other product, saturated polymer.

Correct Answer: B. Low density, good electrical conductivity, and resistance to corrosion from its oxide layer

Fastest Approach (🚀):
Aircraft: light. Cables: conducting and light.
Both: corrosion-resistant.

Matches Option B.

Step-by-Step Breakdown:

1. Aircraft bodies

Low density is decisive — every kilogram costs fuel for the life of the aircraft. Aluminium alloys give useful strength at roughly a third the density of steel.

2. Overhead cables

Good electrical conductivity is needed, and copper conducts better. But aluminium is far lighter, so an aluminium cable of equal conductance is lighter than a copper one — which matters when it must hang between pylons under its own weight. It is also cheaper.

3. Corrosion resistance, for both

Aluminium is in fact a reactive metal. It reacts immediately with oxygen to form a thin, tough, impermeable layer of aluminium oxide that seals the surface and stops further attack. The metal behaves as though unreactive because the oxide protects it.

That is why Option E is subtly wrong: reactivity does not prevent corrosion. The reactivity causes the oxide layer, and the layer prevents further corrosion.

4. Why the other options fail

Aluminium has a low density, not high; a moderate melting point; and it is not magnetic. Iron is the magnetic one, and it is also the metal that rusts progressively — because iron oxide flakes away rather than sealing the surface, exposing fresh metal underneath.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. High density and high reactivity — Property Wrong
  Aluminium has a low density.
• C. Very high melting point and magnetic behaviour — Property Wrong
  Aluminium is not magnetic and has a moderate melting point.
• D. Low melting point and high density — Property Wrong
  Aluminium's density is low, not high.
• E. High reactivity, which prevents corrosion — Causation Confused
  Reactivity causes the oxide layer; the layer prevents corrosion.

Common Mistake (⚠️):
Calling aluminium unreactive. It is reactive, and the protective oxide layer that forms instantly is what makes it behave otherwise.

Takeaway (📌):
Uses follow properties. Aluminium: low density, good conductor, protected by its own oxide layer despite being reactive.

Correct Answer: A. Sulfate, $\text{SO}_4^{2-}$

Fastest Approach (🚀):
Barium chloride $+$ acid, white precipitate $\Rightarrow$ sulfate.

Matches Option A.

Step-by-Step Breakdown:

1. The anion tests

Carbonate — add dilute acid; effervescence, and the gas turns limewater milky.
Sulfate — add dilute hydrochloric acid then barium chloride; a white precipitate of barium sulfate.
Halides — add dilute nitric acid then silver nitrate:
chloride, white precipitate
bromide, cream precipitate
iodide, yellow precipitate

2. Apply it

Barium chloride giving a white precipitate identifies sulfate:
$\text{Ba}^{2+}+\text{SO}_4^{2-}\rightarrow\text{BaSO}_4(\text{s})$

3. Why the acid is added first

Barium carbonate is also a white insoluble solid, so a carbonate would give a false positive. Adding hydrochloric acid first destroys any carbonate present — it fizzes away as carbon dioxide — so any precipitate that then forms must be the sulfate.

That is why the test is specified as acidified barium chloride, and it is the detail examiners look for.

4. Why halides give nothing here

Barium chloride, bromide and iodide are all soluble, so no precipitate forms. Halides need silver nitrate instead, and their three precipitate colours distinguish them from one another.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Carbonate, $\text{CO}_3^{2-}$ — Eliminated By Acid
  Acidifying first destroys any carbonate present.
• C. Chloride, $\text{Cl}^{-}$ — Wrong Reagent
  Chloride needs silver nitrate; barium chloride is soluble.
• D. Iodide, $\text{I}^{-}$ — Wrong Reagent
  Iodide gives a yellow precipitate with silver nitrate.
• E. Bromide, $\text{Br}^{-}$ — Wrong Reagent
  Bromide gives a cream precipitate with silver nitrate.

Common Mistake (⚠️):
Omitting the acid, or not knowing why it is there. Without it a carbonate gives a white precipitate too and the test cannot distinguish them.

Takeaway (📌):
Acidified barium chloride tests for sulfate; acidified silver nitrate tests for halides, by precipitate colour; dilute acid tests for carbonate.

Correct Answer: D. Carbon monoxide — the main cause of acid rain

Fastest Approach (🚀):
Acid rain: $\text{SO}_2$ and $\text{NO}_x$.
Carbon monoxide is toxic, not acidic.

Matches Option D.

Step-by-Step Breakdown:

1. The pollutants and their origins

Sulfur dioxide — from sulfur impurities in fossil fuels. Dissolves in rainwater to form acidic solutions.

Nitrogen oxides — formed when nitrogen and oxygen from the air combine in the high temperatures of an engine. Also acidic in water, and involved in photochemical smog.

Carbon monoxide — from incomplete combustion where oxygen is limited. Colourless and odourless.

Carbon dioxide — from complete combustion of any carbon-based fuel.

2. Their effects

Acid rain: from $\text{SO}_2$ and $\text{NO}_x$. It damages trees, acidifies lakes, and erodes limestone buildings and statues.

Toxicity: carbon monoxide binds to haemoglobin far more strongly than oxygen does, so the blood cannot carry oxygen. It is deadly and gives no warning, being odourless.

Global warming: carbon dioxide and methane trap outgoing infrared radiation.

3. Spot the error

Carbon monoxide does not cause acid rain. It contains no sulfur or nitrogen and forms no acid in water. Its danger is entirely a matter of toxicity.

4. Why the distinction matters

Catalytic converters address carbon monoxide and nitrogen oxides; removing sulfur from fuel and scrubbing flue gases addresses sulfur dioxide. Different problems need different remedies, so attributing the wrong effect to a pollutant points at the wrong solution.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Sulfur dioxide — acid rain — Correct Pairing
  Correct — sulfur dioxide is a principal cause of acid rain.
• B. Carbon monoxide — toxic, binding to haemoglobin in place of oxygen — Correct Pairing
  Correct — carbon monoxide binds to haemoglobin.
• C. Carbon dioxide — a greenhouse gas contributing to global warming — Correct Pairing
  Correct — carbon dioxide is a greenhouse gas.
• E. Nitrogen oxides — acid rain and photochemical smog — Correct Pairing
  Correct — nitrogen oxides cause both.

Common Mistake (⚠️):
Grouping all pollutants under one effect. Each has a distinct origin and a distinct consequence, and they are controlled by different means.

Takeaway (📌):
Acid rain from $\text{SO}_2$ and $\text{NO}_x$; toxicity from CO; global warming from $\text{CO}_2$ and methane.

Correct Answer: C. $36\ \text{g}$

Fastest Approach (🚀):
$n(\text{H}_2) = \dfrac{4}{2} = 2$.
$2:2$ ratio, so $n(\text{H}_2\text{O}) = 2$ and $m = 36\ \text{g}$.

Matches Option C.

Step-by-Step Breakdown:

1. Mass to moles

$n(\text{H}_2) = \dfrac{4.0}{2} = 2.0\ \text{mol}$

2. Apply the equation's ratio

The equation shows $2\text{H}_2$ giving $2\text{H}_2\text{O}$, so hydrogen and water are in a $1:1$ mole ratio:
$n(\text{H}_2\text{O}) = 2.0\ \text{mol}$

Note that oxygen is in a $1:2$ ratio with the hydrogen, so only $1.0\ \text{mol}$ of $\text{O}_2$ is consumed — worth reading off, because a variant of this question asks for exactly that.

3. Moles to mass

$m = 2.0\times 18 = 36\ \text{g}$

4. Check mass is conserved

$1.0\ \text{mol}$ of oxygen weighs $32\ \text{g}$, and $4+32 = 36\ \text{g}$ ✓ — the mass of water produced, exactly.

That check uses the reactant the question never mentioned, and it confirms the whole calculation in one line.

5. Why $18\ \text{g}$ is offered

That is the mass of one mole of water, and it is what you get by finding the ratio and forgetting that two moles of hydrogen were present.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $18\ \text{g}$ — Quantity Ignored
  The mass of one mole of water.
• B. $72\ \text{g}$ — Ratio Error
  Doubling the correct answer, as though the ratio were 1:2.
• D. $4\ \text{g}$ — Ratio On Masses
  Applying the mole ratio to the masses directly.
• E. $9\ \text{g}$ — Ratio Error
  Halving rather than doubling somewhere in the chain.

Common Mistake (⚠️):
Applying the $2:2$ ratio to the grams, giving $4\ \text{g}$ of water from $4\ \text{g}$ of hydrogen. Equal mole ratios do not mean equal masses.

Takeaway (📌):
Mass to moles, ratio from the equation, moles to mass. Conservation of mass across the whole equation is a free check.

Correct Answer: C. It stays constant at $0\ ^{\circ}\text{C}$, because the energy supplied is used to overcome the forces holding the particles in fixed positions

Fastest Approach (🚀):
Energy goes into breaking the arrangement, not into kinetic energy.
Temperature constant at $0\ ^{\circ}\text{C}$.

Matches Option C.

Step-by-Step Breakdown:

1. The shape of the heating curve

Sloped from $-10\ ^{\circ}\text{C}$ to $0\ ^{\circ}\text{C}$ — the ice warms.
Flat at $0\ ^{\circ}\text{C}$ — the ice melts.
Sloped from $0\ ^{\circ}\text{C}$ to $20\ ^{\circ}\text{C}$ — the water warms.

2. What temperature measures

The average kinetic energy of the particles. Temperature rises only when the particles are being made to move faster.

3. Where the energy goes during melting

It is used to overcome the forces holding particles in their fixed lattice positions, not to increase their speed. Since the average kinetic energy is unchanged, so is the temperature — even though energy is being supplied throughout.

4. Why energy is still being absorbed

Option E gets the observation right and the reason wrong. Energy is absorbed continuously during melting — that is the latent heat of fusion. It simply does not show up as a temperature change.

5. Why the two sloped sections differ in gradient

Ice and water have different specific heat capacities, so a given energy input raises their temperatures by different amounts. Water's is higher, so its section of the curve is shallower.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. It rises steadily, because energy is being supplied continuously — Plateau Ignored
  Ignores the plateau at the melting point.
• B. It falls to $0\ ^{\circ}\text{C}$, because melting is endothermic and cools the ice — Direction Wrong
  Endothermic means it absorbs energy; it does not cool the ice below its melting point.
• D. It rises quickly to $0\ ^{\circ}\text{C}$ and then continues rising at the same rate — Plateau Ignored
  Omits the flat section entirely.
• E. It stays constant because no energy is being absorbed during melting — Reason Wrong
  Right observation, wrong reason — energy is absorbed throughout.

Common Mistake (⚠️):
Assuming steady heating means a steadily rising temperature. During a change of state the energy goes into rearranging particles, and the temperature holds still.

Takeaway (📌):
Sloped sections warm the substance; flat sections change its state. Energy is absorbed throughout, but only the sloped parts raise the temperature.

Correct Answer: C. $-818\ \text{kJ mol}^{-1}$

Fastest Approach (🚀):
Broken: $1652+996 = 2648$.
Formed: $1610+1856 = 3466$.
$2648-3466 = -818\ \text{kJ mol}^{-1}$.

Matches Option C.

Step-by-Step Breakdown:

1. Count the bonds broken

$\text{CH}_4$ has four $\text{C}-\text{H}$ bonds: $4\times 413 = 1652$
$2\text{O}_2$ has two $\text{O}=\text{O}$ bonds: $2\times 498 = 996$

total broken $= 2648\ \text{kJ mol}^{-1}$

2. Count the bonds formed

$\text{CO}_2$ has two $\text{C}=\text{O}$ bonds: $2\times 805 = 1610$
$2\text{H}_2\text{O}$ has two $\text{O}-\text{H}$ bonds each, so four in total: $4\times 464 = 1856$

total formed $= 3466\ \text{kJ mol}^{-1}$

3. Subtract

$\Delta H = 2648-3466 = -818\ \text{kJ mol}^{-1}$

4. Check the sign

Combustion releases energy, so $\Delta H$ must be negative. A positive answer means the subtraction was reversed — Option B.

5. The two counting traps

Two bonds per water molecule, and two water molecules, giving four $\text{O}-\text{H}$ bonds. Counting two gives $3466-928 = 2538$ formed and the wrong answer.

Two $\text{C}=\text{O}$ bonds in one carbon dioxide molecule, since the structure is $\text{O}=\text{C}=\text{O}$. Counting one halves that contribution.

Multiplying the coefficient by the bonds per molecule is the step that has to be written down rather than done mentally.

6. Why the value is approximate

The accepted enthalpy of combustion of methane is about $-890\ \text{kJ mol}^{-1}$. Bond energies are averages across many compounds, so a calculation from them lands close but not exact — and the water here is formed as a gas rather than a liquid.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-2648\ \text{kJ mol}^{-1}$ — Incomplete
  The bonds broken alone, not the difference.
• B. $+818\ \text{kJ mol}^{-1}$ — Sign Reversed
  The subtraction reversed; combustion is exothermic.
• D. $-1570\ \text{kJ mol}^{-1}$ — Coefficient Dropped
  Counting two $\text{O}-\text{H}$ bonds rather than four.
• E. $-424\ \text{kJ mol}^{-1}$ — Coefficient Dropped
  Halving the correct answer, or counting one water molecule.

Common Mistake (⚠️):
Counting two $\text{O}-\text{H}$ bonds instead of four. The coefficient in the equation multiplies the bonds inside each molecule, and both factors must be applied.

Takeaway (📌):
$\Delta H = \text{broken}-\text{formed}$, with each bond energy multiplied by the coefficient and by the bonds per molecule.

Correct Answer: E. $\text{C}_2\text{H}_5\text{OH}$; carbon dioxide and water

Fastest Approach (🚀):
$n = 2$: $\text{C}_2\text{H}_5\text{OH}$.
Complete combustion $\Rightarrow \text{CO}_2+\text{H}_2\text{O}$.

Matches Option E.

Step-by-Step Breakdown:

1. Apply the general formula

Alcohols are $\text{C}_n\text{H}_{2n+1}\text{OH}$. With $n = 2$:
$2(2)+1 = 5$

so ethanol is $\text{C}_2\text{H}_5\text{OH}$.

2. Complete combustion

$\text{C}_2\text{H}_5\text{OH}+3\text{O}_2\rightarrow 2\text{CO}_2+3\text{H}_2\text{O}$

Any compound of carbon, hydrogen and oxygen burning completely in plenty of air gives carbon dioxide and water — nothing else.

3. Incomplete combustion

With a limited supply of air, carbon monoxide and carbon (soot) form instead. Carbon monoxide is colourless, odourless and toxic, binding to haemoglobin in place of oxygen — which is why the word 'completely' in the question matters.

4. The other reaction of alcohols

Oxidation, by an oxidising agent or by bacteria in air, converts ethanol to ethanoic acid — how wine turns to vinegar. That is a different reaction from combustion, which is what Option E confuses it with.

5. Why the functional group matters

The $-\text{OH}$ group determines the chemistry. Every alcohol burns, oxidises to a carboxylic acid, and reacts with sodium — regardless of chain length, which affects only the physical properties.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{C}_2\text{H}_6$; carbon dioxide and water — Wrong Compound
  $\text{C}_2\text{H}_6$ is ethane, not ethanol.
• B. $\text{C}_2\text{H}_5\text{OH}$; ethanoic acid only — Wrong Reaction
  Oxidation to ethanoic acid is a different reaction from combustion.
• C. $\text{C}_2\text{H}_5\text{OH}$; carbon monoxide and hydrogen — Incomplete Combustion
  Carbon monoxide comes from incomplete combustion.
• D. $\text{C}_2\text{H}_4\text{OH}$; carbon dioxide and water — Formula Error
  One hydrogen short of the alcohol formula.

Common Mistake (⚠️):
Writing ethanol as $\text{C}_2\text{H}_6$, which is ethane. The $-\text{OH}$ group is what makes it an alcohol and it must appear in the formula.

Takeaway (📌):
Alcohols are $\text{C}_n\text{H}_{2n+1}\text{OH}$. Complete combustion of any of them gives carbon dioxide and water.

Correct Answer: A. Aluminium is more reactive than carbon, so carbon cannot reduce its oxide; iron is less reactive, so carbon can

Fastest Approach (🚀):
Aluminium above carbon $\Rightarrow$ carbon cannot reduce it.
Iron below carbon $\Rightarrow$ carbon reduction works.

Matches Option A.

Step-by-Step Breakdown:

1. Extraction is reduction

A metal ore is usually the metal's oxide, so extracting the metal means removing the oxygen — reduction. Something must take the oxygen, and carbon is the cheap industrial choice.

2. Where carbon sits

Carbon is placed in the reactivity series between zinc and iron. It can take oxygen from the oxide of any metal below it, because those metals hold their oxygen less strongly.

Iron is below carbon, so in the blast furnace carbon and carbon monoxide reduce iron(III) oxide to iron.

Aluminium is above carbon. It holds its oxygen more strongly than carbon does, so heating aluminium oxide with carbon achieves nothing.

3. What is used instead

Electrolysis of molten aluminium oxide, dissolved in cryolite to lower the melting point. Electrical energy does what carbon cannot.

4. Why the method matters economically

Electrolysis is far more energy-intensive than carbon reduction, so metals above carbon are correspondingly expensive to produce, and recycling them saves a large fraction of that energy. That is why aluminium recycling is worth far more per tonne than steel recycling.

5. The three-way rule

Above carbon: electrolysis.
Below carbon: reduction with carbon.
Below hydrogen and very unreactive, such as gold: found native, requiring no reduction at all.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Aluminium oxide has a higher melting point than iron oxide — Wrong Cause
  Melting point is not what determines whether carbon can reduce the oxide.
• C. Aluminium is less reactive than carbon, so a stronger method is needed — Order Reversed
  Aluminium is more reactive than carbon, not less.
• D. Aluminium ore contains no oxygen to remove — Factually Wrong
  Bauxite is an aluminium oxide ore.
• E. Carbon reacts with aluminium to form a carbide, which is why electrolysis is cheaper — Economics Reversed
  Electrolysis is more expensive, not cheaper.

Common Mistake (⚠️):
Getting the direction backwards and calling aluminium less reactive than carbon. Its higher reactivity is precisely why carbon cannot reduce it.

Takeaway (📌):
Extraction is reduction. Carbon reduces the oxides of metals below it; metals above carbon need electrolysis.

Correct Answer: B. $\text{Cu}^{2+}$; potassium gives a lilac flame

Fastest Approach (🚀):
Blue precipitate $\Rightarrow \text{Cu}^{2+}$.
Potassium $\Rightarrow$ lilac.

Matches Option B.

Step-by-Step Breakdown:

1. Precipitates with sodium hydroxide

$\text{Cu}^{2+}$ — blue precipitate.
$\text{Fe}^{2+}$ — green precipitate, darkening on standing as it oxidises.
$\text{Fe}^{3+}$ — orange-brown precipitate.
$\text{Al}^{3+}$, $\text{Ca}^{2+}$, $\text{Mg}^{2+}$ — white precipitates, distinguished because the aluminium one redissolves in excess sodium hydroxide.

Blue therefore identifies copper(II).

2. Flame colours

Lithium — crimson red
Sodium — yellow
Potassium — lilac
Calcium — orange-red
Copper — blue-green

Potassium gives lilac, which is Option B's second half.

3. Why sodium is the awkward one

Sodium's intense yellow masks other colours, so the smallest contamination swamps a lilac potassium flame. That is why the wire is cleaned with acid between tests, and why a potassium flame is often viewed through blue glass to filter the yellow out.

4. Why two different methods

Group 1 and 2 metals form white, indistinguishable precipitates, so flame colour is the practical test for them. Transition metals give distinctly coloured precipitates, so sodium hydroxide identifies them directly. Each method covers what the other cannot.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\text{Fe}^{2+}$; potassium gives a yellow flame — Both Wrong
  $\text{Fe}^{2+}$ gives green, and yellow is sodium.
• C. $\text{Fe}^{3+}$; potassium gives a crimson flame — Both Wrong
  $\text{Fe}^{3+}$ gives orange-brown, and crimson is lithium.
• D. $\text{Cu}^{2+}$; potassium gives a yellow flame — Flame Wrong
  Right cation, but yellow is sodium's flame.
• E. $\text{Ca}^{2+}$; potassium gives a green flame — Precipitate Wrong
  $\text{Ca}^{2+}$ gives a white precipitate.

Common Mistake (⚠️):
Attributing yellow to potassium. Yellow is sodium; potassium is lilac, and sodium contamination is exactly what makes the lilac hard to see.

Takeaway (📌):
Sodium hydroxide: blue is copper(II), green is iron(II), orange-brown is iron(III), white with redissolving is aluminium. Flames: sodium yellow, potassium lilac, lithium crimson.

Correct Answer: B. Chlorine kills microorganisms; fluoride helps to prevent tooth decay

Fastest Approach (🚀):
Chlorine: kills microbes. Fluoride: teeth.

Matches Option B.

Step-by-Step Breakdown:

1. Chlorine

Added to kill microorganisms — bacteria and viruses that cause diseases such as cholera and typhoid. It remains present in low concentration through the distribution network, so the water stays sterile all the way to the tap.

Chlorination is among the most significant public health measures ever adopted, and it is the reason waterborne epidemics are rare where it is practised.

2. Fluoride

Added to reduce tooth decay. Fluoride ions are incorporated into tooth enamel, making it more resistant to attack by the acids that bacteria produce from sugars.

It is a health measure, not a treatment of the water itself — the water is no cleaner for it.

3. What the other stages do

Neither additive removes solids or minerals. Those come earlier:
Screening and sedimentation remove large debris and settle out suspended matter.
Filtration through sand and gravel removes fine particles.
Only then is chlorine added.

So Option E attributes filtration's job to chlorine.

4. Why fluoridation is debated

The benefit is well established, but adding a substance to a public supply for a medical purpose is a decision about consent as much as chemistry. Some regions fluoridate and others do not, and the question is usually framed as ethical rather than scientific.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Chlorine improves the taste; fluoride removes dissolved minerals — Purpose Wrong
  Chlorine affects taste but is added to disinfect; fluoride removes nothing.
• C. Chlorine softens the water; fluoride kills bacteria — Both Wrong
  Chlorine does not soften water, and fluoride is not a disinfectant.
• D. Both are added to kill microorganisms — Second Purpose Wrong
  Fluoride is a dental measure, not a disinfectant.
• E. Chlorine removes suspended solids; fluoride neutralises acidity — Stage Confused
  Filtration removes suspended solids, not chlorine.

Common Mistake (⚠️):
Assuming both additives are there to disinfect. Only chlorine sterilises; fluoride is a dental measure with no effect on the water's safety.

Takeaway (📌):
Chlorine kills microorganisms and persists through the network; fluoride strengthens enamel against decay. Solids are removed earlier by filtration.

Correct Answer: A. $4.0\ \text{g dm}^{-3}$

Fastest Approach (🚀):
$0.100\times 40 = 4.0\ \text{g dm}^{-3}$.

Matches Option A.

Step-by-Step Breakdown:

1. Read what each unit means

$\text{mol dm}^{-3}$ — moles of solute in each cubic decimetre.
$\text{g dm}^{-3}$ — grams of solute in each cubic decimetre.

The volume is the same in both, so only the solute's quantity needs converting.

2. Convert moles to grams

$0.100\ \text{mol}\times 40\ \text{g mol}^{-1} = 4.0\ \text{g}$

So each cubic decimetre holds $4.0\ \text{g}$: the concentration is $4.0\ \text{g dm}^{-3}$.

3. Check the direction

$M_r$ is greater than $1$, so the number of grams must exceed the number of moles. Any answer smaller than $0.100$ has divided instead of multiplied, which rules out Options B and E.

4. Going the other way

$\text{concentration in mol dm}^{-3} = \dfrac{\text{concentration in g dm}^{-3}}{M_r}$

The same conversion, inverted. Deciding which is bigger first makes the direction impossible to get wrong.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0.0025\ \text{g dm}^{-3}$ — Inverted
  Dividing $0.1$ by $40$.
• C. $400\ \text{g dm}^{-3}$ — Inverted
  Dividing $40$ by $0.1$.
• D. $40\ \text{g dm}^{-3}$ — Quantity Ignored
  The mass of one mole, ignoring the concentration.
• E. $0.10\ \text{g dm}^{-3}$ — No Conversion
  Quoting the molar concentration with the wrong unit attached.

Common Mistake (⚠️):
Dividing by $M_r$ instead of multiplying. Grams are the smaller unit, so there are more of them per cubic decimetre than moles.

Takeaway (📌):
$\text{g dm}^{-3} = \text{mol dm}^{-3}\times M_r$. The volume is unchanged; only the solute's units convert.

Correct Answer: B. Sublimation; the reverse is deposition

Fastest Approach (🚀):
Solid $\to$ gas $=$ sublimation.
Gas $\to$ solid $=$ deposition.

Matches Option B.

Step-by-Step Breakdown:

1. Name the change

Solid straight to gas, with no liquid in between, is sublimation.

2. Name the reverse

Gas straight to solid is deposition. It is why frost forms directly on cold surfaces from water vapour, without dew forming first.

3. Why the other terms do not fit

Melting is solid to liquid, freezing liquid to solid, boiling and evaporation liquid to gas, condensation gas to liquid. Every one of them involves the liquid state, which this change skips.

Option E pairs the right forward term with a reverse that goes to the wrong state.

4. Why carbon dioxide behaves this way

At atmospheric pressure, solid carbon dioxide has no stable liquid phase — it needs about five times atmospheric pressure before a liquid can exist. So on a laboratory bench it passes straight to gas at about $-78\ ^{\circ}\text{C}$, which is why it is called dry ice: it cools without wetting anything.

5. Other examples

Iodine sublimes on gentle warming, giving a violet vapour that deposits as crystals on a cool surface — both directions visible in one experiment.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Evaporation; the reverse is condensation — Wrong States
  Evaporation begins from a liquid.
• C. Melting; the reverse is freezing — Wrong States
  Melting produces a liquid.
• D. Boiling; the reverse is condensation — Wrong States
  Boiling produces a gas from a liquid.
• E. Sublimation; the reverse is melting — Reverse Wrong
  Right forward term, but melting ends at a liquid.

Common Mistake (⚠️):
Calling it evaporation. Evaporation is from a liquid surface; sublimation starts from a solid and never passes through a liquid.

Takeaway (📌):
Sublimation is solid to gas; deposition is gas to solid. Every other change-of-state term involves the liquid phase.

Correct Answer: A. Effervescence; the salt is sodium ethanoate

Fastest Approach (🚀):
Acid $+$ carbonate $\Rightarrow$ fizzing $\text{CO}_2$.
Ethanoic acid $\Rightarrow$ ethanoate salt.

Matches Option A.

Step-by-Step Breakdown:

1. Recognise the reaction type

Ethanoic acid is a carboxylic acid, $\text{CH}_3\text{COOH}$. The $-\text{COOH}$ group releases $\text{H}^{+}$ in solution, so it behaves as an acid.

Acid plus carbonate gives salt, water and carbon dioxide:
$2\text{CH}_3\text{COOH}+\text{Na}_2\text{CO}_3\rightarrow 2\text{CH}_3\text{COONa}+\text{H}_2\text{O}+\text{CO}_2$

2. What is observed

Effervescence — bubbles of carbon dioxide. Passing it through limewater turns the limewater milky, which confirms the gas.

3. Naming the salt

Carboxylic acid names ending in -oic acid give salts ending in -oate. Ethanoic acid gives ethanoate, with the metal in front: sodium ethanoate.

Option C invents 'sodium ethanol', which mixes the alcohol name into a salt.

4. Why 'weak' does not mean 'unreactive'

Ethanoic acid is a weak acid — only partly ionised in solution — so it reacts more slowly and gives a higher pH than hydrochloric acid at the same concentration. But it undergoes every characteristic acid reaction, so Option A confuses weakness with inertness.

5. Why Option E is wrong

A glowing splint relights in oxygen. Carbon dioxide extinguishes it, and is tested with limewater instead.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. No reaction, because ethanoic acid is too weak — Weak vs Unreactive
  Weak means partly ionised, not unreactive.
• C. Effervescence; the salt is sodium ethanol — Naming Error
  'Sodium ethanol' is not a salt name.
• D. A white precipitate; the salt is sodium carbonate — Wrong Observation
  No precipitate forms; the products are soluble plus a gas.
• E. Effervescence, and the gas relights a glowing splint — Wrong Gas Test
  A relighting splint indicates oxygen, not carbon dioxide.

Common Mistake (⚠️):
Assuming a weak acid does not react. Weak means partially ionised, not unreactive — it still gives all three characteristic acid reactions.

Takeaway (📌):
Carboxylic acids behave as acids: fizzing with carbonates, salt plus water with bases, hydrogen with metals. The salt ends in -oate.

Correct Answer: C. They form stable ions in more than one oxidation state, and their compounds are usually coloured

Fastest Approach (🚀):
Variable oxidation states and coloured compounds.
Group 1: always $1+$, always white.

Matches Option C.

Step-by-Step Breakdown:

1. What both share

Conductivity, metallic lustre, malleability, and the formation of positive ions. All are general metallic properties, so Options A, B and D cannot distinguish the two.

2. What is distinctive about transition metals

Variable oxidation states — iron forms $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$; copper forms $\text{Cu}^{+}$ and $\text{Cu}^{2+}$; manganese ranges from $+2$ to $+7$.

Coloured compounds — copper(II) salts blue, iron(II) pale green, iron(III) orange-brown, manganate(VII) deep purple.

Catalytic activity — iron in the Haber process, nickel in hydrogenation, and the variable oxidation states are what make that possible.

Higher densities and melting points than Group 1.

3. What Group 1 metals do instead

They form only $1+$ ions, because losing the single outer electron gives a noble gas configuration and losing a second would break into a full shell. Their compounds are almost all white and dissolve to colourless solutions.

4. Why Option E is backwards

Group 1 metals are the ones reacting vigorously with cold water — sodium fizzing across the surface is the standard demonstration. Transition metals are far less reactive: iron rusts only slowly, and copper does not react with water at all.

That difference in reactivity is also why transition metals are the useful structural and electrical materials, while Group 1 metals must be stored under oil.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. They conduct electricity — Shared Property
  All metals conduct.
• B. They form ions with a positive charge — Shared Property
  All metals form positive ions.
• D. They have a shiny appearance when freshly cut — Shared Property
  Both are shiny when freshly cut.
• E. They react vigorously with cold water — Reversed
  Group 1 metals are the vigorous ones with water.

Common Mistake (⚠️):
Choosing a general metallic property such as conductivity. Both families are metals, so only the properties Group 1 lacks can distinguish them.

Takeaway (📌):
Transition metals: variable oxidation states, coloured compounds, catalytic, dense, less reactive. Group 1: always $1+$, white compounds, very reactive.

Correct Answer: B. The liquid contains water, but this does not prove it is pure water

Fastest Approach (🚀):
White $\to$ blue $\Rightarrow$ water present.
Purity needs a boiling or freezing point.

Matches Option B.

Step-by-Step Breakdown:

1. What the test detects

White anhydrous copper(II) sulfate reacts with water to form the blue hydrated salt:
$\text{CuSO}_4+5\text{H}_2\text{O}\rightarrow\text{CuSO}_4{\cdot}5\text{H}_2\text{O}$

The colour change is a positive test for the presence of water.

2. Why it does not prove purity

Sea water, orange juice and dilute acid would all turn it blue, because all contain water. The test says water is there; it says nothing about what else is.

3. How purity is established

By a physical property. Pure water boils at exactly $100\ ^{\circ}\text{C}$ and freezes at exactly $0\ ^{\circ}\text{C}$ at standard pressure. Dissolved solutes raise the boiling point and lower the freezing point, and they also make the substance melt or boil over a range rather than at a fixed point.

That range is the general test for purity: a pure substance has a sharp melting and boiling point, a mixture does not.

4. The reverse reaction

Heating the blue hydrated salt drives the water off and returns it to white. The change is reversible, which is why the same sample can be reused, and it is one of the standard demonstrations of a reversible reaction.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. The liquid is pure water — Purity Assumed
  Any aqueous solution gives this result.
• C. The liquid is an acid — Wrong Property
  The test is unrelated to acidity.
• D. The liquid contains copper ions — Source Confused
  The copper comes from the reagent, not the liquid.
• E. The liquid is anhydrous — Reversed
  Anhydrous means without water — the opposite of what the result shows.

Common Mistake (⚠️):
Concluding the liquid is pure water. Any aqueous solution passes this test; purity is a physical measurement, not a chemical one.

Takeaway (📌):
Anhydrous copper(II) sulfate white to blue detects water. Purity is shown by a sharp boiling or melting point, not by a chemical test.

Correct Answer: A. $\text{C}_2\text{H}_6$

Fastest Approach (🚀):
Add $\text{H}_2$ across the double bond: $\text{C}_2\text{H}_4\to\text{C}_2\text{H}_6$.

Matches Option A.

Step-by-Step Breakdown:

1. What addition does

The $\text{C}=\text{C}$ double bond opens, and one hydrogen atom attaches to each carbon:
$\text{C}_2\text{H}_4+\text{H}_2\rightarrow\text{C}_2\text{H}_6$

2. Check against the general formulae

Ethene is an alkene, $\text{C}_n\text{H}_{2n}$ with $n = 2$.
Ethane is an alkane, $\text{C}_n\text{H}_{2n+2}$ with $n = 2$.

The difference is exactly the two hydrogens added, which is a useful check that the product is right.

3. The conditions

A nickel catalyst at about $150\ ^{\circ}\text{C}$. The catalyst provides a lower-activation-energy pathway; without it the reaction is impractically slow.

4. Where it is used

Hydrogenation of unsaturated vegetable oils, converting some $\text{C}=\text{C}$ bonds to single bonds. That raises the melting point and turns a liquid oil into a solid or semi-solid fat — the process behind margarine.

5. The other addition reactions of alkenes

With bromine: $\text{C}_2\text{H}_4+\text{Br}_2\rightarrow\text{C}_2\text{H}_4\text{Br}_2$, decolourising bromine water — the test for unsaturation.
With steam: $\text{C}_2\text{H}_4+\text{H}_2\text{O}\rightarrow\text{C}_2\text{H}_5\text{OH}$, the industrial route to ethanol — which is Option C, and the right answer to a different question.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\text{C}_2\text{H}_2$ — Direction Reversed
  Two hydrogens fewer — this would be a removal, not an addition.
• C. $\text{C}_2\text{H}_5\text{OH}$ — Wrong Reagent
  The product of adding steam, not hydrogen.
• D. $\text{C}_2\text{H}_4$ unchanged — Reaction Denied
  Addition does occur under these conditions.
• E. $\text{C}_4\text{H}_8$ — Carbons Doubled
  The carbon count is unchanged in an addition to one molecule.

Common Mistake (⚠️):
Choosing ethanol. Adding water across the double bond gives ethanol; adding hydrogen gives ethane. Read which reagent is named.

Takeaway (📌):
Alkene plus hydrogen gives the alkane, two hydrogens heavier. Alkene plus bromine gives the dibromide; alkene plus steam gives the alcohol.