JAMB UTME - Chemistry - 2025

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The pH of a salt solution depends on the strength of the acid and base from which the salt was formed. A salt formed from a strong acid and a strong base produces a neutral solution with a pH of 7, because neither ion undergoes hydrolysis in water.

Sodium chloride (NaCl) is formed from hydrochloric acid (HCl, a strong acid) and sodium hydroxide (NaOH, a strong base). When dissolved in water, the Na⁺ and Cl^- ions do not react with water, so the solution remains neutral at pH 7.

The other salts behave differently:

• Ammonium chloride (NH₄Cl) is formed from a weak base (NH₃) and a strong acid (HCl). The ammonium ion hydrolyses in water to release H⁺ ions, making the solution acidic (pH below 7).
• Ammonium tetraoxosulphate(VI) ((NH₄)₂SO₄) is also formed from a weak base and a strong acid, producing an acidic solution.
• Sodium trioxocarbonate(IV) (Na₂CO₃) is formed from a strong base (NaOH) and a weak acid (H₂CO₃). The carbonate ion hydrolyses in water to produce OH^- ions, making the solution basic (pH above 7).

Exam tip: To predict the pH of a salt solution, identify the parent acid and base. Strong acid + strong base gives pH 7; strong acid + weak base gives pH below 7; weak acid + strong base gives pH above 7.

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Tetraoxosulphate(VI) acid is the IUPAC name for sulphuric acid, \(\text{H}_2\text{SO}_4\). Its large-scale industrial manufacture uses the Contact process.

The Contact process involves three main stages:

1. Sulphur is burned in air to produce sulphur(IV) oxide: \(\text{S} + \text{O}_2 \rightarrow \text{SO}_2\)
2. Sulphur(IV) oxide is oxidised to sulphur(VI) oxide using a vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) catalyst at about 450 \(^\circ\text{C}\): \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\)
3. The \(\text{SO}_3\) is absorbed in concentrated \(\text{H}_2\text{SO}_4\) to form oleum (\(\text{H}_2\text{S}_2\text{O}_7\)), which is then carefully diluted with water to give concentrated sulphuric acid.

The other named processes serve different purposes. The Haber process manufactures ammonia from nitrogen and hydrogen. The Frasch process is used for mining sulphur deposits underground using superheated water. The Bosch process (or water-gas shift reaction) produces hydrogen from carbon monoxide and steam. None of these produces sulphuric acid.

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Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions, the equilibrium shifts in the direction that tends to counteract that change. For pressure changes, the key factor is the difference in the total number of moles of gas on each side of the equation.

If there is no change in volume during a gaseous reaction, this means the total number of moles of gaseous products equals the total number of moles of gaseous reactants. In such a case, increasing or decreasing the pressure gives the system no direction in which to shift, because neither the forward nor the backward reaction would reduce the total number of gas molecules.

Therefore, a change in pressure will have no effect on the equilibrium position when the reaction involves equal moles of gas on both sides.

For example, in the reaction:

\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]

there are 2 moles of gas on each side, so pressure changes do not shift the equilibrium.

Pressure only affects equilibrium when there is an unequal number of moles of gas on either side. In those cases, increasing pressure favours the side with fewer moles of gas, and decreasing pressure favours the side with more moles.

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Permalloy is an alloy composed of iron (Fe) and nickel (Ni), typically in a ratio of approximately 20% iron and 80% nickel, though the exact composition can vary.

Permalloy is valued for its exceptionally high magnetic permeability, meaning it is very easily magnetised even by weak magnetic fields. This property makes it useful in:

• Transformer cores
• Magnetic shielding
• Recording heads in data storage devices
• Sensitive magnetic instruments

The other metals listed form different alloys with iron:

• Iron + tin = tin plate (tin-coated iron, used for food cans)
• Iron + chromium (plus nickel and carbon) = stainless steel
• Iron + copper is not a standard named alloy

The defining feature of permalloy is that it is a nickel-iron alloy.

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Solubility is defined as the maximum amount of a solute that can dissolve in a given quantity of solvent at a particular temperature to form a saturated solution. Its importance lies directly in the fact that it determines the amount of solute that can dissolve in a given amount of solvent.

This knowledge is practically essential in:

• Preparing solutions of known concentration in the laboratory
• Predicting whether a precipitate will form when solutions are mixed
• Understanding purification techniques such as recrystallisation
• Pharmaceutical formulation (ensuring a drug dissolves sufficiently for absorption)

The other options describe colligative properties - effects that arise after a solute has been dissolved:

• Lowering the vapour pressure and raising the boiling point are consequences of having dissolved particles in solution, not descriptions of what solubility itself tells us.
• Knowing chemical reactions of a gas with a liquid relates to chemical reactivity, not solubility as a concept.

Solubility is fundamentally about how much dissolves, and that is its primary importance.

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The corrosion of iron is specifically called rusting. Rusting occurs when iron reacts with oxygen and water (moisture) over time to form hydrated iron(III) oxide, commonly known as rust:

\[4\text{Fe} + 3\text{O}_2 + 6\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3\]

The iron(III) hydroxide gradually dehydrates to form the familiar reddish-brown rust (Fe₂O₃ . xH₂O). Both oxygen and water must be present for rusting to occur; iron does not rust in dry air or in air-free water.

The other options are different processes entirely: burning (combustion) is a rapid reaction with oxygen involving flame and heat; galvanizing is a method of preventing corrosion by coating iron with a layer of zinc; alloying is mixing metals together to form an alloy (such as stainless steel), which is also a corrosion-prevention strategy, not a corrosion process.

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A reducing agent is a substance that causes another substance to be reduced by donating electrons to it. In the process of donating electrons, the reducing agent itself is oxidized (its oxidation state increases).

In the reaction:

\[\text{Mg}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{MgCl}_{2(aq)} + \text{H}_{2(g)}\]

Magnesium changes from an oxidation state of 0 (as the free element) to +2 (in MgCl₂). It loses two electrons:

\[\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\]

These electrons are gained by hydrogen ions (H⁺), which are reduced from +1 to 0 (as H₂ gas). Magnesium is the reducing agent because it supplies the electrons that reduce the hydrogen ions.

The defining characteristic of a reducing agent is that it is oxidized. Being a metal or being a solid are physical properties that do not explain why magnesium acts as a reducing agent in this reaction.

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Colloids are classified based on the physical states of the dispersed phase (the substance spread throughout) and the dispersion medium (the substance in which it is spread). The main types include:

A commercial deodorant spray works by dispersing tiny liquid droplets (the fragrance and active ingredients) into the air (a gas). This makes it an aerosol - a colloid in which a liquid is dispersed in a gas.

It is not a foam (gas in liquid/solid), not an emulsion (liquid in liquid), and not a sol (solid in liquid).

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The empirical formula is the simplest whole-number ratio of atoms in a compound. The empirical formula mass (sometimes called empirical mass) is the molar mass corresponding to that simplest formula.

The molecular formula given is C₆H₁₂O₆. To find the empirical formula, divide all subscripts by their greatest common factor:

\[\text{GCF of } 6, 12, 6 = 6\]

\[\text{Empirical formula} = \text{C}_1\text{H}_2\text{O}_1 = \text{CH}_2\text{O}\]

Now calculate the empirical formula mass using the given atomic masses (H = 1, C = 12, O = 16):

\[\text{Empirical mass} = 12 + 2(1) + 16 = 30\]

As a check, the molecular mass of C₆H₁₂O₆ is 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180. Dividing by the empirical mass: 180 / 30 = 6, confirming that the molecular formula is exactly 6 times the empirical formula.

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In the electrolysis of concentrated sodium chloride solution (brine) using inert graphite electrodes, the products depend on the concentration of the solution and the electrode positions.

At the anode (positive electrode), negatively charged ions migrate and are discharged. In concentrated NaCl solution, both chloride ions (Cl^-) and hydroxide ions (OH^-) from water are present. However, because the chloride ion concentration is very high, chloride ions are preferentially discharged at the anode:

\[2\text{Cl}^{-}(aq) \rightarrow \text{Cl}_2(g) + 2e^{-}\]

This produces chlorine gas, which can be identified by its greenish-yellow colour and its ability to bleach damp litmus paper.

At the cathode, hydrogen gas is produced from the reduction of water (since Na⁺ ions are too electropositive to be discharged). Oxygen gas would be the anode product only in the electrolysis of dilute sodium chloride or dilute sulphuric acid, where hydroxide ions are discharged instead of chloride ions. Water vapour and hydrogen gas are not anode products in this process.

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The number of molecules in a sample is found by multiplying the number of moles by Avogadro's number. The relationship is:

\[N = n \times N_A\]

where \(N\) is the number of molecules, \(n\) is the number of moles, and \(N_A\) is Avogadro's number (\(6.02 \times 10^{23}\)).

Substituting the given values:

\[N = 4.0 \times 6.02 \times 10^{23}\]

\[N = 24.08 \times 10^{23}\]

\[N = 2.408 \times 10^{24}\]

Rounding to three significant figures gives \(2.41 \times 10^{24}\) molecules.

A common error is dividing instead of multiplying. Dividing 6.02 by 4 would give \(1.51 \times 10^{23}\), which is the number of molecules in 0.25 moles, not 4.0 moles. When you have more than one mole, the number of molecules must be greater than Avogadro's number, so any answer smaller than \(6.02 \times 10^{23}\) can be immediately ruled out.

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Baking powder is a mixture of sodium hydrogen carbonate (baking soda) and a solid acid. The acid is needed because sodium hydrogen carbonate only releases carbon dioxide gas, the substance that makes dough rise, when it reacts with an acid:

\[ \text{NaHCO}_3 + \text{acid} \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{salt} \]

Tartaric acid, \( \text{C}_4\text{H}_6\text{O}_6 \), is a naturally occurring organic acid obtained mainly from grapes, and it is one of the classic solid acids used to formulate baking powder because it is stable when dry but dissolves and reacts quickly once water is added to the dough or batter. The same acid, because it has a pleasant sharp taste and is safe to consume in small amounts, is also added to soft drinks to give them their tart, refreshing flavour and to help balance the sweetness of the sugar in the drink.

The other acids listed do not fit both uses. Fatty acids are found in oils and fats and are not used as leavening or flavouring agents in drinks. Boric acid is toxic in the concentrations relevant to food and is not permitted as a food additive. Citric acid is common in fruit-flavoured drinks but is not the acid traditionally paired with sodium hydrogen carbonate in the classic baking-soda and soft-drink formulation being tested here.

When a question links a food acid to two different uses at once, check that the acid is both chemically suited to the reaction involved (reacting with a base to release gas) and safe and pleasant enough to be consumed directly, since not every food-grade acid satisfies both conditions.

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Pinking (also called knocking or engine knock) is the premature, uncontrolled auto-ignition of the fuel-air mixture inside an engine cylinder. It produces a metallic rattling noise and can seriously damage the engine over time.

Before petrol is sold, it is evaluated for its octane rating. The octane number measures how resistant the fuel is to pinking. A higher octane number means the petrol is less likely to auto-ignite prematurely and will burn smoothly during the power stroke.

The evaluation of petrol from crude oil before sale is therefore carried out to prevent pinking. Fuels with low octane ratings are either blended with higher-octane components or treated (for example, by adding anti-knock agents) before they are released for sale.

The other options describe petroleum refining processes, not problems to be prevented:

• Reforming is a refinery process that converts low-octane hydrocarbons into higher-octane ones (e.g. converting straight-chain alkanes into cyclic or branched structures).
• Pyrolysis is thermal decomposition of organic materials at high temperatures in the absence of oxygen.
• Cracking is the breaking down of large hydrocarbon molecules into smaller, more useful ones.

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Anti-freeze is a substance added to the water in a vehicle's radiator to lower its freezing point, preventing the coolant from solidifying in cold weather and potentially cracking the engine block.

Among the options given, propanetriol (glycerol or propane-1,2,3-triol, HOCH\(_2\)CH(OH)CH\(_2\)OH) has historically been used as an anti-freeze. Glycerol is highly soluble in water in all proportions, and when dissolved, it significantly depresses the freezing point of the mixture. This is a colligative property: the dissolved solute particles disrupt the formation of ice crystals, requiring a lower temperature to freeze.

The other options are unsuitable:

• Propanol (CH\(_3\)CH\(_2\)CH\(_2\)OH) is a flammable alcohol not used in radiator systems.
• Propanone (acetone, CH\(_3\)COCH\(_3\)) is a volatile ketone and a solvent, not an anti-freeze.
• Propanal (CH\(_3\)CH\(_2\)CHO) is an aldehyde with no anti-freeze application.

In modern practice, ethylene glycol (ethane-1,2-diol) is more commonly used, but among the four choices provided, propanetriol is the correct answer.

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Chlorine has an atomic number of 17, meaning a neutral chlorine atom has 17 electrons. These electrons fill the available subshells in order of increasing energy:

1. \(1s\) holds 2 electrons (total: 2)
2. \(2s\) holds 2 electrons (total: 4)
3. \(2p\) holds 6 electrons (total: 10)
4. \(3s\) holds 2 electrons (total: 12)
5. \(3p\) holds the remaining 5 electrons (total: 17)

The electron configuration is therefore: \(1s^2\,2s^2\,2p^6\,3s^2\,3p^5\)

Examining the other options:

• \(1s^2\,2s^2\,2p^6\,3s^2\,3p^4\) gives only 16 electrons (this is sulphur).
• \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\) gives 18 electrons (this is argon).
• \(1s^2\,2s^2\,2p^5\,3s^2\,3p^6\) gives 17 electrons but violates the aufbau principle - the \(2p\) subshell must fill completely (6 electrons) before the \(3s\) subshell begins.

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A blowing agent is a substance used to produce a cellular structure (foam) in materials such as plastics, rubber, and insulation. The term "blowing" refers specifically to the process of expanding a material by generating gas bubbles within it during manufacture.

Chlorofluorocarbons (CFCs) were widely used as blowing agents in the foam industry because they vaporise at low temperatures, creating uniform gas pockets that give foam products their lightweight, insulating structure. Expanded polystyrene, polyurethane foam, and similar products were traditionally manufactured using CFCs as the blowing agent.

While CFCs also have applications as refrigerants and aerosol propellants, those uses are distinct from the role of a blowing agent. A refrigerant absorbs and releases heat during phase changes in a cooling cycle, and a propellant provides pressure to expel contents from a container. Neither of these functions involves creating foam. The tyre industry does not use CFCs as blowing agents.

The key word in the question is "blowing agent," which points specifically to foam production.

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Brine is a concentrated solution of sodium chloride (NaCl) in water. During electrolysis of brine, the ions present are Na⁺, Cl^-, H⁺ (from water), and OH^- (from water).

At the cathode (negative electrode), reduction takes place. The two cations competing for discharge are Na⁺ and H⁺. Because hydrogen ions are much easier to reduce than sodium ions (sodium has a very negative standard electrode potential), H⁺ ions are preferentially discharged:

\[2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\]

The gas produced at the cathode is therefore hydrogen.

At the anode (positive electrode), chloride ions are oxidised to produce chlorine gas. Sodium hydroxide remains in solution. Steam is not produced during electrolysis, and oxygen would only appear at the anode if a dilute solution were used instead of concentrated brine.

Exam tip: In electrolysis of brine, remember the three products: hydrogen at the cathode, chlorine at the anode, and sodium hydroxide in solution.

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Aluminium (Al) is the metal widely used as a packaging material. It is used to make drink cans, food containers, and aluminium foil for wrapping food.

Aluminium is ideal for packaging because of several key properties:

• Low density - it is lightweight, making it easy to transport and handle.
• Malleability - it can be rolled into very thin sheets (foil) or pressed into various container shapes.
• Corrosion resistance - it forms a tough, protective oxide layer (\(\text{Al}_2\text{O}_3\)) on its surface that prevents further reaction with air or moisture, keeping the contents safe.
• Non-toxicity - it does not contaminate food or beverages.
• Recyclability - aluminium can be recycled repeatedly without losing its properties.

The other metals are unsuitable for packaging:

• Copper (Cu) is too heavy and expensive for everyday packaging, and it can impart a metallic taste to food.
• Iron (Fe) rusts readily when exposed to moisture and air, which would contaminate and spoil packaged goods.
• Sodium (Na) is an extremely reactive alkali metal that reacts violently with water and catches fire in air, making it entirely impractical for packaging.

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The equation Na\(_2\)X \(\rightleftharpoons\) 2Na\(^+\) + X\(^{2-}\) shows the compound Na\(_2\)X dissociating into its constituent ions: sodium ions (Na\(^+\)) and an anion X\(^{2-}\).

This dissociation into oppositely charged ions is the hallmark of an ionic bond. In ionic bonding, one or more electrons are transferred from a metal atom (here, sodium) to a non-metal atom (here, X). Sodium loses one electron to form Na\(^+\), while X gains two electrons to form X\(^{2-}\). Two sodium atoms are needed to supply the two electrons that X requires.

The other bond types do not fit:

• A covalent bond involves the sharing of electron pairs between atoms, not transfer and complete dissociation into ions.
• A dative (coordinate) bond is a special covalent bond where both shared electrons come from one atom.
• A metallic bond involves a "sea" of delocalised electrons shared among metal atoms in a lattice - it does not produce discrete anions like X\(^{2-}\).

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A lone pair is a pair of valence electrons on an atom that is not shared in a bond. To find which molecule has the highest number of lone pairs, draw the Lewis structure of each molecule and count all lone pairs on every atom.

CH₄: Carbon has four bonding pairs (one to each hydrogen) and no lone pairs. Each hydrogen also has no lone pairs. Total lone pairs: 0.

NH₃: Nitrogen has three bonding pairs (one to each hydrogen) and one lone pair. Total lone pairs: 1.

H₂O: Oxygen has two bonding pairs (one to each hydrogen) and two lone pairs. Total lone pairs: 2.

CO₂: Carbon forms two double bonds (one to each oxygen) and has no lone pairs. Each oxygen in a double bond with carbon retains two lone pairs. Total lone pairs: 2 + 2 = 4.

CO₂ has the highest total number of lone pairs (four), making it the correct answer.

Exam tip: When counting lone pairs, remember to include those on every atom in the molecule, not just the central atom.

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The spontaneity of a reaction is determined by the Gibbs free energy change, given by:

\[\Delta G = \Delta H - T\Delta S\]

A reaction is spontaneous when \(\Delta G\) is negative.

In this question:

• \(\Delta H\) is positive and small (slightly endothermic).
• \(\Delta S\) is positive and large (significant increase in disorder).

Substituting into the equation:

\[\Delta G = (\text{small positive}) - T \times (\text{large positive})\]

Since \(T\) (absolute temperature in Kelvin) is always positive, the term \(T\Delta S\) will be a large positive number. Subtracting this large positive value from a small positive \(\Delta H\) gives:

\[\Delta G = \text{small positive} - \text{large positive} = \text{negative}\]

A negative \(\Delta G\) means the reaction is spontaneous.

Exam tip: When \(\Delta H\) is positive but \(\Delta S\) is also positive and large, the entropy term dominates, and the reaction is spontaneous, especially at higher temperatures. This is called an entropy-driven reaction.

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Iron is extracted from its ore in a blast furnace. The iron that comes directly out of the blast furnace is called pig iron. It contains about 3-4% carbon along with smaller amounts of impurities such as silicon, manganese, phosphorus, and sulphur.

Pig iron is brittle due to its high carbon content and is not suitable for most engineering applications in its raw form. It must be further processed to produce more useful forms of iron and steel:

• Cast iron is produced by remelting pig iron and allowing it to cool in moulds. It still has a relatively high carbon content (2-4%) but has improved properties for casting.
• Wrought iron is nearly pure iron (less than 0.1% carbon) produced by removing most of the carbon and impurities from pig iron. It is soft, malleable, and resistant to corrosion.
• Carbon steel is produced by carefully controlling the carbon content of iron (0.1-1.5%) to achieve specific mechanical properties.

The iron that comes directly from the blast furnace, before any further refining, is pig iron.

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Ethyne (commonly known as acetylene, C₂H₂) is the organic gas used in the oxy-acetylene torch for welding and cutting metals. When ethyne burns in pure oxygen, it produces an extremely hot flame reaching temperatures above 3,000 °C - hot enough to melt steel and other metals.

The combustion reaction is:

\[2\text{C}_2\text{H}_2(g) + 5\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 2\text{H}_2\text{O}(g)\]

Ethyne produces such a high temperature because it is an unsaturated hydrocarbon with a carbon-carbon triple bond, which stores a large amount of energy. This makes it far superior to other hydrocarbons for metalwork.

Methane, propene, and butane can all burn, but their flames do not reach the extreme temperatures required to cut through metals. Ethyne's unique suitability for this industrial application is a frequently tested fact in organic chemistry.

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Tristearin is a fat (triglyceride) formed from glycerol and three molecules of stearic acid. Alkaline hydrolysis of a triglyceride is the reaction of the fat with a strong alkali such as sodium hydroxide. This reaction is known as saponification and produces soap (the sodium salt of the fatty acid) and glycerol as a by-product:

\[\text{(C}_{17}\text{H}_{35}\text{COO)}_3\text{C}_3\text{H}_5 + 3\text{NaOH} \rightarrow 3\text{C}_{17}\text{H}_{35}\text{COONa} + \text{C}_3\text{H}_5\text{(OH)}_3\]

The by-product is glycerol, also known as propane-1,2,3-triol. Its structural formula shows three hydroxyl (-OH) groups, one on each of the three carbon atoms. An alcohol with three -OH groups is classified as a trihydric alkanol.

A dihydric alkanol has two -OH groups (e.g. ethane-1,2-diol). A secondary alkanol has the -OH group on a carbon bonded to two other carbon atoms. A tertiary alkanol has the -OH on a carbon bonded to three other carbons. Glycerol has two primary -OH groups and one secondary -OH group, but its defining classification is that it is trihydric, since it carries three hydroxyl groups in total.

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Potassium trioxochlorate(V) (KClO₃, potassium chlorate) can decompose on heating to produce oxygen gas. However, it is not recommended for modern laboratory preparation of oxygen because it forms explosive mixtures with carbonaceous materials.

KClO₃ is a powerful oxidising agent. When it comes into contact with organic (carbon-containing) substances such as rubber tubing, cork stoppers, paper, or wood, it can react violently and cause explosions. Even small amounts of organic impurities can trigger a dangerous, rapid exothermic decomposition.

The safer modern alternative for laboratory preparation of oxygen is heating hydrogen peroxide (H₂O₂) with manganese(IV) oxide (MnO₂) as a catalyst, which does not carry the same explosion risk.

While KClO₃ is certainly hazardous, the specific reason it is not recommended is the formation of explosive mixtures with carbonaceous materials, not simply a general statement of being hazardous or explosive on its own.

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A reaction is feasible (spontaneous) when the Gibbs free energy change is negative: \(\Delta G < 0\). The relationship between Gibbs free energy, enthalpy, and entropy is:

\[\Delta G = \Delta H - T\Delta S\]

The question states that \(\Delta G^\circ\) is negative. To determine which combination of \(\Delta H\) and \(\Delta S\) guarantees this, consider each option:

• If \(\Delta H\) is positive and \(\Delta S\) is positive: \(\Delta G = (+) - T(+)\). This is negative only at sufficiently high temperatures, so feasibility is not guaranteed at all temperatures.
• If \(\Delta H = 0\) and \(\Delta S\) is positive: \(\Delta G = 0 - T(+) = -T\Delta S\), which is always negative at any temperature above 0 K. This guarantees a negative \(\Delta G\).
• If \(\Delta H\) is positive and \(\Delta S\) is negative: \(\Delta G = (+) - T(-) = (+) + T|\Delta S|\), which is always positive. The reaction would never be feasible.
• If \(\Delta S = 0\): \(\Delta G = \Delta H\), and feasibility depends entirely on the sign of \(\Delta H\), which is not specified.

The only option that ensures \(\Delta G\) is negative under all conditions is \(\Delta H = 0\) and \(\Delta S\) is positive, because the \(-T\Delta S\) term is always negative when \(\Delta S > 0\).

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The equation shows ethanol (C₂H₅OH) reacting with ethanoic acid (CH₃COOH) to form ethyl ethanoate (CH₃COOC₂H₅) and water (H₂O).

This is an esterification reaction. Esterification is the reaction between a carboxylic acid and an alcohol to produce an ester and water. It is typically catalysed by a concentrated strong acid such as tetraoxosulphate(VI) acid (H₂SO₄), and the reaction is reversible, indicated by the equilibrium sign (⇌).

The general equation is:

\[\text{Carboxylic acid} + \text{Alcohol} \xrightleftharpoons{\text{H}_2\text{SO}_4} \text{Ester} + \text{Water}\]

The other options do not apply here:

• Condensation is a broader term for reactions where two molecules combine with the loss of a small molecule (often water). While esterification is technically a type of condensation, the specific and more accurate name for this reaction between an acid and an alcohol is esterification.
• Fermentation is the enzymatic conversion of sugars (such as glucose) to ethanol and carbon(IV) oxide.
• Hydrolysis is the reverse of esterification, where an ester is broken down by water (or alkali) into the parent acid and alcohol.

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A physical change is a change in which no new substance is formed. The original substance can be recovered by simple physical methods (such as evaporation, filtration, or condensation), and no chemical bonds are broken or formed between different types of atoms.

Dissolving sodium chloride in water is a physical change. When NaCl dissolves, the ionic lattice breaks apart and Na⁺ and Cl^- ions become surrounded by water molecules (hydration). However, no new chemical substance is created. The sodium chloride can be fully recovered by evaporating the water. The process is reversible.

The other options all involve chemical changes:

• Addition of water to calcium oxide: CaO + H₂O → Ca(OH)₂. A new substance (calcium hydroxide) is formed, and considerable heat is released.
• Corrosion of iron: Iron reacts with oxygen and moisture to form iron(III) oxide (rust, Fe₂O₃). New substances are produced irreversibly.
• Burning of charcoal: C + O₂ → CO₂. Combustion produces a completely different substance (carbon dioxide) and is irreversible.

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Ammonium trioxonitrate(V) is the IUPAC name for ammonium nitrate, NH₄NO₃. When heated gently (thermal decomposition), it breaks down as follows:

\[\text{NH}_4\text{NO}_3 \xrightarrow{\text{heat}} \text{N}_2\text{O} + 2\text{H}_2\text{O}\]

The products are dinitrogen monoxide (N₂O, also known as nitrous oxide or laughing gas) and water (H₂O).

To verify, check that the equation is balanced:

• N: 2 on each side (1 from NH₄⁺ and 1 from NO₃^- on the left; 2 in N₂O on the right).
• H: 4 on each side (4 in NH₄⁺ on the left; 4 in 2H₂O on the right).
• O: 3 on each side (3 in NO₃^- on the left; 1 in N₂O + 2 in 2H₂O on the right).

The other options are incorrect: NO₂ and H₂O would not balance correctly with the given reactant; N₂O and O₂ would leave hydrogen unaccounted for; and NO₃ is not a stable molecular product of thermal decomposition.

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Solubility is defined as the mass of solute that dissolves in 100 g of solvent (water) to form a saturated solution at a given temperature.

From the question, the mass of the saturated solution is 12.5 g and the mass of anhydrous salt recovered after evaporation is 2 g. The mass of water in the solution is therefore:

\[\text{Mass of water} = 12.5 - 2 = 10.5 \text{ g}\]

Solubility is calculated as:

\[\text{Solubility} = \frac{\text{Mass of solute}}{\text{Mass of solvent}} \times 100\]

\[\text{Solubility} = \frac{2}{10.5} \times 100 = 19.05 \text{ g per 100 g of water}\]

The calculated value of 19.05 g/100 g water is closest to 19.05, which rounds to approximately 19 g/100 g. Among the available options, 19.05 does not match any value exactly. However, if the question intends the mass of water to be taken as 10 g (a common simplification in some exam settings where the saturated solution mass is approximated), the calculation becomes:

\[\text{Solubility} = \frac{2}{10} \times 100 = 20.0 \text{ g per 100 g of water}\]

The intended answer is therefore 19.05 g/100 g water by strict calculation, but the closest provided value is 20.0 g per 100 g of water.

Exam tip: Always identify the mass of solute and the mass of solvent separately from the total solution mass before applying the solubility formula.

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This question tests the ability to read and interpret a solubility-temperature graph. The graph plots solubility (y-axis) against temperature in °C (x-axis) for four substances: X, Y, Z, and Q.

Examining each curve on the graph:

• X shows a steep, curved increase in solubility with temperature. Its solubility rises slowly at lower temperatures but accelerates sharply at higher temperatures, giving a concave-upward curve.
• Y follows a straight line from the origin, rising at a constant rate as temperature increases. This is a steady, linear increase in solubility.
• Z appears as a nearly horizontal line at about 45 units of solubility, meaning its solubility barely changes regardless of temperature. Z is essentially temperature-independent.
• Q rises to a peak around 40 to 50 °C and then decreases, meaning its solubility first increases and then drops with further heating.

The correct inference is that the solubility of Y increases steadily as temperature increases. The word "steadily" is key here: Y's straight-line graph means its solubility rises at a uniform, constant rate per degree of temperature increase. X also increases with temperature, but its increase is not steady; it accelerates (curves upward), so the rate of increase itself changes.

The claim that the solubility of X and Y is the same at all temperatures is incorrect because the two curves only intersect at a single point; at all other temperatures, their solubilities differ. The claim that the solubility of X, Y, and Z is temperature dependent is incorrect because Z is nearly flat, showing its solubility is essentially independent of temperature. The claim that the solubility of Z increases as temperature increases is directly contradicted by Z's horizontal line on the graph.

Exam tip: When a question uses the word "steadily," look for a straight-line relationship on the graph. A curve that bends upward or downward represents a changing rate of increase, not a steady one.

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Transition metals are widely used as catalysts in both industrial and biological processes. The key electronic property that makes them effective catalysts is the presence of partially filled d-orbitals.

Partially filled d-orbitals allow transition metals to:

• Exhibit variable oxidation states by losing different numbers of d-electrons, enabling them to participate in electron-transfer steps during a reaction.
• Form temporary bonds with reactant molecules by accepting or donating electron density through their available d-orbitals, which lowers the activation energy of the reaction.
• Provide a suitable surface or coordination environment for reactants to adsorb, orient, and react more readily.

A filled d-orbital would mean the metal has no vacant d-orbitals to accept electron density from reactants, greatly reducing its catalytic ability. Similarly, f-orbitals (whether filled or partially filled) are associated with the inner transition metals (lanthanides and actinides) and are not the primary reason for catalytic activity in the common transition metal catalysts such as iron, nickel, platinum, and palladium.

Exam tip: The two key properties of transition metals to remember are variable oxidation states and catalytic activity, both arising from partially filled d-orbitals.

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To name this compound using IUPAC nomenclature, first expand the condensed structural formula (CH\(_3\))\(_2\)CHCH(OH)CH\(_2\)C(CH\(_3\))\(_3\):

\[\text{CH}_3-\underset{|}{\overset{\text{CH}_3}{\text{CH}}}-\underset{|}{\overset{\text{OH}}{\text{CH}}}-\text{CH}_2-\underset{|}{\overset{\text{CH}_3}{\underset{|}{\overset{}{\text{C}}}}}(\text{CH}_3)_2\]

Step 1: Find the longest carbon chain containing the OH group.

Tracing through the backbone: CH\(_3\)-CH-CH(OH)-CH\(_2\)-C-CH\(_3\) gives 6 carbons, so the parent chain is hexane.

Step 2: Number to give the OH group the lowest locant.

Numbering from the end nearest the OH group:

• C1: CH\(_3\) (end near the isopropyl group)
• C2: CH (bearing one CH\(_3\) branch)
• C3: CH-OH
• C4: CH\(_2\)
• C5: C (bearing two CH\(_3\) branches)
• C6: CH\(_3\)

OH is on carbon 3. Numbering from the other end would place OH on carbon 4, which is higher, so this direction is correct.

Step 3: Identify substituents.

• Carbon 2: one methyl group
• Carbon 5: two methyl groups

There are three methyl substituents at positions 2, 5, and 5.

Step 4: Construct the name.

The IUPAC name is 2,5,5-trimethylhexan-3-ol.

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Copper is deposited from CuSO₄ solution by the reduction of Cu²⁺ ions:

\[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\]

This means each mole of copper requires 2 moles of electrons (2 faradays) to be deposited.

Step 1: Calculate the moles of copper to be deposited.

\[n_{\text{Cu}} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{64} = 0.0703125 \text{ mol}\]

Step 2: Calculate the total charge required.

Since 1 mole of Cu requires 2 faradays:

\[Q = n_{\text{Cu}} \times 2 \times F = 0.0703125 \times 2 \times 96500\]

\[Q = 0.140625 \times 96500 = 13570.3 \text{ C}\]

Step 3: Calculate the time using \(Q = It\).

\[t = \frac{Q}{I} = \frac{13570.3}{2.5} = 5428 \text{ sec}\]

The time required is 5428 seconds.

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A diagonal relationship in the periodic table refers to the similarity in properties between an element in Period 2 and the element diagonally below and to its right in Period 3. This occurs because moving one period down increases size and metallic character, while moving one group to the right decreases them, so the two effects partially cancel out.

The well-established diagonal pairs are:

• Li (Period 2, Group 1) and Mg (Period 3, Group 2)
• Be (Period 2, Group 2) and Al (Period 3, Group 13)
• B (Period 2, Group 13) and Si (Period 3, Group 14)

From the given options, B and Si is one of the classic diagonal pairs. They share similar properties such as forming covalent compounds, acting as semiconductors, and forming acidic oxides.

The other options are not diagonal pairs:

• Be and Si are two groups apart diagonally, not one.
• Na and Be are in adjacent periods but Na is below and to the left of Be, not diagonal in the correct direction.
• Li and Al are separated by two groups in the diagonal direction.

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The structure shown is R-C(=O)-NH-H, which contains a carbonyl group (C=O) directly bonded to a nitrogen atom bearing hydrogen atoms. This is the defining arrangement of the amide functional group (-CONH₂).

An alkanamide (also called an amide) has the general formula R-CONH₂, where R is an alkyl group. The key feature distinguishing it from the other options is the simultaneous presence of both the C=O and the N-H bonds on the same carbon.

An alkylamine (R-NH₂) has nitrogen bonded to an alkyl group but no carbonyl. An alkanone (R-CO-R') has a carbonyl flanked by two carbon groups with no nitrogen. An amino acid would require both an amine group (-NH₂) and a carboxyl group (-COOH) on the same molecule, which is not the case here.

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Reforming is a petroleum refinery process that rearranges the molecular structure of hydrocarbons to produce higher-octane fuels and aromatic compounds. The most common type is catalytic reforming, which converts naphthenes (cycloalkanes) and straight-chain alkanes into aromatic hydrocarbons such as benzene, toluene, and xylene, typically using a platinum-based catalyst at high temperature.

The conversion of cyclohexane to benzene is a classic example of reforming. In this reaction, cyclohexane (C₆H₁₂) undergoes catalytic dehydrogenation, losing three molecules of hydrogen to form benzene (C₆H₆):

\[ \text{C}_6\text{H}_{12} \xrightarrow{\text{Pt catalyst, heat}} \text{C}_6\text{H}_6 + 3\text{H}_2 \]

This aromatization reaction increases the octane rating of the fuel fraction and produces valuable aromatic feedstocks for the chemical industry.

The other options describe different processes:

• Converting hexene to hexane is hydrogenation (addition of hydrogen across a double bond).
• Converting hexane to carbon(IV) oxide and water vapour is combustion (burning in oxygen).
• Converting hexane to butane and ethane is cracking (breaking larger molecules into smaller ones).

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When an inflated balloon is placed in a freezer, the temperature of the gas inside the balloon decreases. According to the kinetic molecular theory of gases, the average kinetic energy of gas molecules is directly proportional to absolute temperature:

\[\text{Average KE} = \tfrac{3}{2}kT\]

As temperature drops, the gas molecules move more slowly. Slower-moving molecules collide with the walls of the balloon less frequently and with less force, so they exert less pressure on the balloon walls. Since the balloon is flexible, it contracts until the internal and external pressures balance, causing the balloon to shrink.

The other options are incorrect:

• The volume does not increase - it decreases. This option contradicts what is observed.
• The number of gas molecules does not change. The balloon is sealed; no molecules escape or enter.
• Gas molecules never completely stop moving. According to kinetic theory, molecular motion only ceases at absolute zero (0 K), which is not reached in a domestic freezer.

This behaviour illustrates Charles's Law: at constant pressure, the volume of a gas is directly proportional to its absolute temperature (\(V \propto T\)).

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Petroleum is a naturally occurring substance found in underground rock formations. Its composition varies from one source to another because petroleum is a mixture of many different hydrocarbons and other organic compounds, not a single pure substance.

A pure substance (element or compound) has a fixed, definite composition regardless of its source. A mixture, however, consists of two or more substances combined in no fixed ratio, so its composition can differ from sample to sample.

Petroleum contains alkanes, cycloalkanes, aromatic hydrocarbons, and other compounds in varying proportions depending on the geological conditions under which it formed. This variable composition is precisely what defines it as a mixture and is why it must be separated into useful fractions by fractional distillation.

While petroleum is indeed a hydrocarbon-containing substance, a liquid, and a natural resource, none of those properties explain why its composition varies. Only the fact that it is a mixture accounts for this variability.

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An alkanoic acid (carboxylic acid) has the general formula \(\text{C}_n\text{H}_{2n}\text{O}_2\). The task is to find the molecular formula whose empirical formula is CH\(_2\)O and whose molecular formula fits the alkanoic acid pattern.

First, calculate the empirical formula mass of CH\(_2\)O:

\[12 + 2(1) + 16 = 30\]

The molecular formula must be a whole-number multiple of the empirical formula: \((\text{CH}_2\text{O})_n\). To satisfy the alkanoic acid general formula \(\text{C}_n\text{H}_{2n}\text{O}_2\), we need 2 oxygen atoms, which requires \(n = 2\):

\[(\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2\]

This is ethanoic acid (acetic acid, CH\(_3\)COOH), a well-known alkanoic acid.

The relative molecular mass is:

\[2(12) + 4(1) + 2(16) = 24 + 4 + 32 = 60\]

The answer is 60.