JAMB UTME - Chemistry - 2025

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.

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.

Carbohydrates follow the general formula \(\text{C}_x(\text{H}_2\text{O})_y\). To find the value of x for fructose, you need to know the molecular formula of fructose.

Fructose has the molecular formula C₆H₁₂O₆. Rewriting this in the carbohydrate general formula:

\[\text{C}_6\text{H}_{12}\text{O}_6 = \text{C}_6(\text{H}_2\text{O})_6\]

Comparing with \(\text{C}_x(\text{H}_2\text{O})_y\), the value of x = 6 (and y = 6 as well).

Fructose is a monosaccharide (simple sugar) that is an isomer of glucose. Both have the same molecular formula C₆H₁₂O₆, but they differ in structural arrangement: glucose is an aldose (contains an aldehyde group) while fructose is a ketose (contains a ketone group).

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 \(\Delta S > 0\).

The reaction is:

\[\text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s)\]

On the reactant side, there are 2 moles of gas (1 mole of NH₃ + 1 mole of HCl). On the product side, NH₄Cl is a solid, so there are effectively 0 moles of gas.

According to Le Chatelier's principle, when the pressure of a gaseous system at equilibrium is increased, the equilibrium shifts towards the side with fewer moles of gas to reduce the pressure.

Since the product side has fewer gaseous moles than the reactant side, increasing the pressure will shift the equilibrium to the right, favouring the product (NH₄Cl).

Note that changing pressure shifts the position of equilibrium but does not change the equilibrium constant (K). The equilibrium constant is only affected by changes in temperature, not pressure or concentration.

Exam tip: When applying Le Chatelier's principle to pressure changes, count only the moles of gaseous species on each side. Solids and liquids are not affected by pressure changes.

A covalent bond is formed when two atoms share a pair of electrons. When the two bonded atoms have different electronegativities, they do not attract the shared electron pair equally. The more electronegative atom pulls the electron pair closer to itself, creating a partial negative charge on that atom and a partial positive charge on the other. This results in a polar covalent bond.

Therefore, a polar covalent bond is characterised by the unequal sharing of an electron pair between atoms.

By contrast, when two atoms of the same element (or very similar electronegativity) bond, the electrons are shared equally, producing a non-polar covalent bond (e.g., H₂, Cl₂, O₂).

The options mentioning "lone pair" are incorrect in this context. A lone pair is an unshared pair of electrons that belongs entirely to one atom and is not involved in bonding between the two atoms. Covalent bonds (whether polar or non-polar) involve bonding pairs, not lone pairs.

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).

Dilute nitric acid (HNO\(_3\)) is an oxidising acid, which means it usually oxidises the metal and is itself reduced to nitrogen oxides (such as NO or NO\(_2\)) rather than producing hydrogen gas. This is different from non-oxidising acids like dilute HCl or dilute H\(_2\)SO\(_4\), which readily liberate H\(_2\) with reactive metals.

However, magnesium (Mg) is an exception. Because magnesium is extremely reactive (high up in the electrochemical series), it reacts so vigorously with very dilute HNO\(_3\) that the reaction proceeds faster than the acid can act as an oxidising agent. The result is that hydrogen gas is liberated:

\[\text{Mg} + 2\text{HNO}_3\text{(very dilute)} \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2\uparrow\]

Copper (Cu) is below hydrogen in the activity series and cannot displace hydrogen from any acid under normal conditions. Zinc (Zn) reacts with dilute HNO\(_3\) but produces NO gas rather than H\(_2\), because it is not reactive enough to overcome the oxidising nature of the acid. Calcium (Ca) is very reactive but reacts explosively with water itself and, in practice with dilute HNO\(_3\), produces nitrogen oxides or ammonia rather than clean H\(_2\) liberation; the standard examination answer for this question is magnesium.

The boiling point of a substance depends on the strength of its intermolecular forces and, to a lesser extent, its molecular mass. The key intermolecular forces in order of strength are: hydrogen bonding > dipole-dipole > van der Waals (London dispersion).

Consider the four compounds:

• CH₃CH₂OH (ethanol, M_r = 46): contains an -OH group, so it forms hydrogen bonds. Boiling point approximately 78 °C.
• CH₃CH₂CH₂Cl (1-chloropropane, M_r = 78.5): polar C-Cl bond gives dipole-dipole interactions but no hydrogen bonding. Boiling point approximately 47 °C.
• CH₃CH₂CH₂CH₂CH₂CH₃ (hexane, M_r = 86): non-polar, only van der Waals forces. Boiling point approximately 69 °C.
• CH₃CH₂CH₂OH (propan-1-ol, M_r = 60): contains an -OH group, so it forms hydrogen bonds, and has a higher molecular mass than ethanol. Boiling point approximately 97 °C.

Propan-1-ol (CH₃CH₂CH₂OH) has the highest boiling point. It combines hydrogen bonding (the strongest intermolecular force among these molecules) with a greater molecular mass than ethanol, giving it stronger overall intermolecular attractions.

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.

Gold is one of the least reactive metals, sitting at the very bottom of the reactivity series. Because of this extremely low reactivity, gold does not combine with other elements under natural conditions. It therefore occurs in the earth's crust as the free (uncombined) metal, often found as nuggets or flakes in alluvial deposits.

Since gold already exists in its elemental form, there is no need for chemical extraction from an ore. Metals higher in the reactivity series (such as iron, aluminium, or sodium) form stable compounds with oxygen, sulphur, or other elements and must be chemically reduced to obtain the pure metal. Gold does not form such compounds naturally, so it is simply recovered by physical methods like panning or washing.

The phrase free in nature specifically means the metal is found uncombined. While gold is indeed unreactive (inert), the reason it does not require extraction is that it occurs free in nature, which is the more precise and direct answer to the question. Being "inert" describes a property; being "free in nature" describes the consequence of that property that directly answers why extraction is unnecessary.

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.

Acidified potassium dichromate (K₂Cr₂O₇) is a strong oxidising agent. When it oxidises a substance, the orange dichromate ion (Cr₂O₇^2-, containing Cr⁶⁺) is reduced to the green chromium(III) ion (Cr³⁺). This colour change from orange to green is the characteristic indicator that an oxidation reaction has occurred.

Human breath can contain ethanol (an alkanol) if the person has consumed alcohol. Ethanol is readily oxidised by acidified dichromate:

\[\text{C}_2\text{H}_5\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}\]

and further:

\[\text{CH}_3\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{COOH}\]

This is the principle behind the breathalyser test used to detect alcohol (alkanol) in a person's breath. The colour change from orange to green confirms the presence of an alkanol (alcohol).

Alkanones (ketones) are not easily oxidised by acidified dichromate under mild conditions, so they would not produce this colour change. Alkanals (aldehydes) and alkanoic acids (carboxylic acids) are not typically present in human breath in significant amounts.

This is a Faraday's law of electrolysis problem. The relationship between mass deposited, current, and time is:

\[m = \frac{M \times I \times t}{n \times F}\]

where \(m\) = mass deposited (g), \(M\) = molar mass, \(I\) = current (A), \(t\) = time (s), \(n\) = number of electrons transferred per ion, and \(F\) = Faraday constant (96500 C/mol).

For aluminium: \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\), so \(n = 3\), \(M = 27\), \(m = 9\) g, \(I = 18\) A.

Rearranging for time:

\[t = \frac{m \times n \times F}{M \times I}\]

\[t = \frac{9 \times 3 \times 96500}{27 \times 18}\]

\[t = \frac{2\,605\,500}{486}\]

\[t = 5360.49 \text{ seconds}\]

Converting to minutes:

\[t = \frac{5360.49}{60} = 89.34 \text{ minutes}\]

The time required is 89.34 minutes.

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.

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.

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.

A metal is said to occur in the free state (or native state) in nature when it is found as the pure, uncombined element rather than in a compound (such as an ore). This happens only for metals that are very low in the reactivity series, meaning they are resistant to reaction with oxygen, water, and acids.

Silver (Ag) is one such metal. It is found as native silver in the earth's crust because of its very low chemical reactivity. Gold and platinum are other classic examples of metals found in the free state.

Zinc and iron are too reactive to exist as free metals in nature. They readily react with oxygen and moisture to form oxides and other compounds, so they are always found as ores (e.g., zinc as zinc blende ZnS, iron as haematite Fe₂O₃). While copper can occasionally be found native, silver is the stronger answer here because it is less reactive than copper and more commonly occurs in the uncombined state.

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.

Choosing the right indicator for a titration depends on the strength of the acid and the base being used, because different indicators change colour at different pH ranges.

Ethanedioic acid (oxalic acid, HOOC-COOH) is a weak diprotic acid, and potassium hydroxide (KOH) is a strong base.

In a weak acid vs. strong base titration, the equivalence point lies in the basic/alkaline region (pH roughly 8-10). This is because at the equivalence point, the solution contains the conjugate base of the weak acid (potassium ethanedioate/oxalate), which hydrolyses to make the solution slightly basic.

Phenolphthalein changes colour in the pH range of approximately 8.2 to 10.0 (colourless in acid, pink in alkali). This range coincides with the equivalence point of a weak acid-strong base titration, making it the most suitable indicator.

Methyl orange (pH 3.1-4.4) and methyl red (pH 4.4-6.2) change colour in acidic pH ranges and would signal the endpoint too early, before the true equivalence point is reached. Methyl purple is less commonly used and is not the standard choice for this type of titration.

The cell notation (also called line notation) for an electrochemical cell follows the convention:

Anode | Anode ion || Cathode ion | Cathode

where the single vertical line (|) represents a phase boundary, and the double vertical line (||) represents the salt bridge separating the two half-cells.

For the reaction Mg + Pb²⁺ → Mg²⁺ + Pb:

• Anode (oxidation): Mg → Mg²⁺ + 2e^-. Magnesium is oxidized, so it is the anode.
• Cathode (reduction): Pb²⁺ + 2e^- → Pb. Lead ions are reduced, so lead is the cathode.

Applying the convention:

Mg | Mg²⁺ || Pb²⁺ | Pb

Using the notation in the options (where I = | and II = ||), this is written as Mg|Mg²⁺||Pb²⁺|Pb.

The anode always appears on the left and the cathode on the right. Within each half-cell, the metal (solid phase) is written adjacent to the outer edge, and the ion (aqueous phase) is written adjacent to the salt bridge.

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.

The melting and boiling points of a substance depend on the strength of the forces that must be overcome to change its state. For simple covalent molecules, there are two types of forces to consider:

1. Intramolecular forces (covalent bonds within the molecule): These are strong but are not broken during melting or boiling.
2. Intermolecular forces (forces between molecules): These are the forces that must be overcome during phase changes. In simple covalent molecules, these are typically weak van der Waals forces or dipole-dipole interactions.

Because the intermolecular forces are weak, relatively little energy is needed to separate the molecules from one another. This is why simple covalent substances such as water, methane, and carbon dioxide have low melting and boiling points compared to ionic or metallic substances.

The other options do not explain the low melting and boiling points:

• Having definite shapes describes molecular geometry but does not determine the strength of forces between molecules.
• Being three-dimensional structures is a feature of giant covalent substances (like diamond), which actually have very high melting points.
• Possessing shared electron pairs is what makes them covalent, but the shared pairs (covalent bonds) are not broken during melting or boiling, so they do not account for the low melting/boiling points.

To find the oxidation state of nitrogen in each compound, use the rule that oxygen has an oxidation state of \(-2\) and the sum of oxidation states in a neutral compound is zero.

For a compound of the form N\(_2\)O\(_x\):

\[2(\text{oxidation state of N}) + x(-2) = 0\]

\[\text{oxidation state of N} = \frac{2x}{2} = x\]

Wait - more precisely: \(\text{oxidation state of N} = \frac{2x}{2} = +x\). Let me calculate each:

• N\(_2\)O\(_5\): \(2n + 5(-2) = 0 \Rightarrow n = +5\)
• N\(_2\)O\(_3\): \(2n + 3(-2) = 0 \Rightarrow n = +3\) ✔
• N\(_2\)O\(_4\): \(2n + 4(-2) = 0 \Rightarrow n = +4\)
• N\(_2\)O: \(2n + 1(-2) = 0 \Rightarrow n = +1\)

The compound in which nitrogen has an oxidation state of +3 is N\(_2\)O\(_3\) (dinitrogen trioxide).

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.

Nitrogen gas (N₂) makes up about 78% of the atmosphere and has several important industrial uses. One of the most significant is the production of nitric acid (HNO₃).

The industrial production of HNO₃ occurs in two stages:

1. Haber process: Nitrogen reacts with hydrogen at high temperature and pressure in the presence of an iron catalyst to produce ammonia:
   \[\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\]
2. Ostwald process: Ammonia is then oxidised over a platinum-rhodium catalyst to produce nitrogen monoxide, which is further oxidised and dissolved in water to form nitric acid:
   \[4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}\]
   \[2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\]
   \[3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}\]

The other options are incorrect: margarine production uses hydrogen (not nitrogen) for hydrogenation of vegetable oils; nitrogen is not used in manufacturing oil; and while liquid nitrogen can serve as a coolant, the question refers to nitrogen as a component of air used in chemical production.

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.

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.

Sulphuric acid (H₂SO₄) is a diprotic acid, meaning each molecule donates two hydrogen ions when it dissociates completely in water:

\[\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{H}^+(aq) + \text{SO}_4^{2-}(aq)\]

Step 1: Find [H⁺]

Since each mole of H₂SO₄ produces 2 moles of H⁺:

\[[\text{H}^+] = 2 \times 0.06 = 0.12 \text{ mol dm}^{-3} = 1.2 \times 10^{-1} \text{ mol dm}^{-3}\]

Step 2: Find [OH^-]

Using the ionic product of water at 25 °C:

\[K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\]

\[[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-1}}\]

\[[\text{OH}^-] = \frac{1.0}{1.2} \times 10^{-14+1} = 0.833 \times 10^{-13} = 8.3 \times 10^{-14} \text{ mol dm}^{-3}\]

Therefore [H⁺] = \(1.2 \times 10^{-1}\) mol dm^-3 and [OH^-] = \(8.3 \times 10^{-14}\) mol dm^-3.

A common mistake is forgetting that sulphuric acid is diprotic and using [H⁺] = 0.06 instead of 0.12, which would give \(1.2 \times 10^{-2}\) instead of \(1.2 \times 10^{-1}\) and an incorrect OH^- concentration of \(8.3 \times 10^{-13}\).

Producer gas is a mixture of carbon monoxide (CO) and nitrogen (N₂), while water gas is a mixture of carbon monoxide (CO) and hydrogen (H₂). Both gases contain CO, which is toxic and must be removed for certain industrial applications.

Ammoniacal copper(I) chloride solution (CuCl dissolved in ammonia) is the reagent used to selectively absorb CO from these gas mixtures. The CO molecules form a coordination complex with the copper(I) ions in solution:

\[ \text{CuCl} + \text{CO} + 2\text{NH}_3 \rightarrow [\text{Cu(CO)(NH}_3\text{)}_2]\text{Cl} \]

This reaction is reversible: gentle heating releases the absorbed CO and regenerates the ammoniacal CuCl solution for reuse.

The other gases in these mixtures are not absorbed by this reagent. Hydrogen (H₂) does not form stable complexes with Cu⁺ under these conditions. Nitrogen (N₂) is chemically inert at room temperature. Carbon dioxide (CO₂) would be absorbed by alkaline solutions such as NaOH or KOH, not by ammoniacal CuCl.

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.

The basicity of an acid is the number of replaceable hydrogen ions (\(\text{H}^+\)) that one molecule of the acid can donate in a reaction with a base.

The compound \(\text{C}_2\text{H}_2\text{O}_4\) is oxalic acid (also called ethanedioic acid). Its structural formula is:

\(\text{HOOC-COOH}\)

Oxalic acid contains two carboxyl groups (\(-\text{COOH}\)). Each carboxyl group carries one hydrogen atom that can be released as \(\text{H}^+\) during a neutralisation reaction. The remaining hydrogen atoms in the molecule are bonded to carbon and are not ionisable.

Since there are two replaceable hydrogen atoms, the basicity of oxalic acid is 2. This means it is a dibasic acid (also called a diprotic acid).

The neutralisation reaction with sodium hydroxide confirms this:

Two moles of NaOH are required to completely neutralise one mole of oxalic acid, confirming a basicity of 2.

When determining basicity, count only the hydrogen atoms bonded to oxygen in carboxyl or hydroxyl groups, not those bonded directly to carbon.

A water pollutant is any substance or condition that degrades the quality of a water body and harms aquatic life or makes the water unsuitable for its intended use.

Oxygen gas is not a water pollutant. In fact, dissolved oxygen is essential for aquatic life. Fish and other aquatic organisms depend on dissolved oxygen for respiration. A water body with adequate dissolved oxygen levels is considered healthy.

The other options are all recognised water pollutants:

• Warm water effluent (thermal pollution): Discharge of heated water from power plants or factories raises the temperature of natural water bodies, which reduces the solubility of dissolved oxygen and disrupts aquatic ecosystems.
• Inorganic fertilizers: Nitrates and phosphates from agricultural runoff cause eutrophication, leading to excessive algal growth, oxygen depletion, and death of aquatic organisms.
• Biodegradable waste: Organic matter such as sewage increases the biological oxygen demand (BOD) of the water, depleting dissolved oxygen as microorganisms decompose the waste.

Since oxygen gas is a natural and beneficial component of water, it is not classified as a pollutant.

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.

In the nitrogen cycle, when organisms die, their proteins and other nitrogenous compounds are broken down by decomposing bacteria in a process called ammonification (or decay). The first product of this decomposition is ammonia (NH₃).

The process occurs in stages:

1. Proteins in dead material are broken down into amino acids by proteolytic enzymes from decomposing bacteria and fungi.
2. The amino acids are further deaminated (the amino group -NH₂ is removed), releasing ammonia (NH₃).

After ammonia is produced, nitrifying bacteria can convert it further: first to nitrites (dioxonitrate(III), NO₂^-) by Nitrosomonas, then to nitrates (trioxonitrate(V), NO₃^-) by Nitrobacter. However, the question asks specifically about the first conversion product of nitrogenous compounds in dead materials, which is ammonia.

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.

Crude oil (petroleum) is a complex mixture of hydrocarbons with different boiling points. To separate it into useful fractions (such as petrol/gasoline, kerosene, diesel, lubricating oil, and bitumen), fractional distillation is used.

In fractional distillation, crude oil is heated in a furnace until most of it vaporises. The vapour enters a tall fractionating column that is hot at the bottom and cool at the top. As the vapour rises through the column:

• Hydrocarbons with higher boiling points condense lower in the column (e.g., bitumen, heavy fuel oil).
• Hydrocarbons with lower boiling points rise higher before condensing (e.g., petrol, refinery gas).

The column contains trays at different heights where each fraction is collected.

The other separation methods are not suitable:

• Steam distillation is used for heat-sensitive compounds (like essential oils) that would decompose at their normal boiling points.
• Fractional crystallisation separates solids from solution based on differences in solubility at different temperatures.
• Precipitation is used to separate insoluble solids from a solution by adding a reagent.

The energy profile diagram shows the energy changes during a chemical reaction. The reactants (A+B) start at an energy level of approximately 30 units, while the products (C+D) end at approximately 50 units. The activation energy peak reaches about 80 units.

Since the products have a higher energy level than the reactants, the reaction has absorbed energy from the surroundings. This net gain in energy by the reacting system is the defining characteristic of an endothermic reaction. The energy difference between products and reactants (\ (\Delta H\)) is positive, confirming that heat was taken in rather than released.

An exothermic reaction would show products at a lower energy level than reactants, indicating a release of energy. Here, the upward shift from reactants to products clearly indicates energy absorption.

The boiling point of a substance depends on the strength of the intermolecular forces that must be overcome to convert it from liquid to gas. Stronger intermolecular forces mean higher boiling points.

Alkanoic acids (carboxylic acids, R-COOH) have higher boiling points than alkanols (alcohols, R-OH) of comparable molecular mass because alkanoic acids form stronger hydrogen bonds.

The -COOH group in a carboxylic acid can form two hydrogen bonds simultaneously with another carboxylic acid molecule. This is because the -COOH group has both a highly polar O-H bond and a C=O group that can act as a hydrogen bond acceptor. Carboxylic acids commonly exist as dimers, with two molecules linked by a pair of strong hydrogen bonds:

\[\text{R-COOH} \cdots \text{HOOC-R}\]

In contrast, the -OH group in an alkanol can only form one hydrogen bond per molecule. The hydrogen bonding in alcohols is therefore less extensive than in carboxylic acids.

The other options are incorrect:

• Saying alkanoic acids are more volatile would imply they have lower boiling points, which contradicts the question.
• Alkanols are not more polar than alkanoic acids; carboxylic acids are more polar.
• For molecules of comparable chain length, the size difference is not significant enough to explain the boiling point gap.

Esters (alkanoates) are the class of organic compounds responsible for the pleasant, fruity scents found in many fruits and perfumes. Esters are formed by the condensation reaction between an alkanol (alcohol) and an alkanoic acid (carboxylic acid) in the presence of a concentrated acid catalyst:

\[\text{Alkanoic acid} + \text{Alkanol} \xrightarrow{\text{H}^+} \text{Alkanoate (ester)} + \text{H}_2\text{O}\]

For example, ethyl ethanoate (CH\(_3\)COOC\(_2\)H\(_5\)) has a fruity smell resembling nail polish or pear drops. Different combinations of acids and alcohols produce esters with distinct aromas - banana, pineapple, apple, and many others.

The other options are different functional group classes with different characteristic properties:

• Alkanamides (amides, R-CO-NH\(_2\)) are generally odourless solids.
• Alkanones (ketones, R-CO-R') have sharp, often solvent-like odours - not fruity.
• Alkanals (aldehydes, R-CHO) tend to have pungent, irritating smells at low molecular weight.