JAMB UTME - Chemistry - 2025

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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\(^{2-}\).

Bayanin Amsa

The rate of reaction measures how quickly a reactant is consumed or a product is formed over a given period. Here, 5 g of iron filings reacted completely with excess dilute \(\text{H}_2\text{SO}_4\) in 10 minutes.

The rate of reaction (in terms of mass of reactant consumed per unit time) is:

Substituting the given values:

The question asks for the rate in grams per hour. Since there are 60 minutes in one hour:

The rate of reaction is 30 g/hr.

A common mistake is to forget the unit conversion from minutes to hours, which would give an incorrect answer of 0.5 g/min. Always check that the units in your final answer match what the question requests.

Bayanin Amsa

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.

Bayanin Amsa

Litmus is an acid-base indicator. Red litmus paper turns blue only in the presence of a base (alkaline solution), which has a pH greater than 7.

Examining the given pH values:

• pH 2.5 - strongly acidic (no colour change; red litmus stays red)
• pH 3.5 - acidic (no colour change)
• pH 7.0 - neutral (no colour change; litmus remains the same colour)
• pH 8.0 - basic/alkaline (red litmus turns blue)

Only the solution with pH 8.0 is alkaline, so it is the only one that will turn red moist litmus paper blue. Acidic and neutral solutions cannot cause this change.

Bayanin Amsa

In the laboratory preparation of chlorine gas, concentrated hydrochloric acid (HCl) is reacted with manganese(IV) oxide (MnO₂). The chlorine gas produced is contaminated with hydrogen chloride (HCl) vapour and water vapour.

The gas is first passed through a wash bottle containing water. The purpose of this step is to absorb the HCl gas. Hydrogen chloride is extremely soluble in water and dissolves readily, while chlorine is only slightly soluble. This allows the HCl impurity to be removed selectively.

After this, the gas is typically passed through concentrated sulphuric acid to remove the remaining water vapour (drying), and then collected.

The water wash does not dilute the chlorine (chlorine is a gas, not a solution being diluted), does not neutralize it (that would require a base, not water), and does not absorb the Cl₂ (chlorine has low solubility in water, and the goal is to collect it, not remove it).

Bayanin Amsa

The structural formula shows H₃C-CH₂-C(=O)-O-CH₂-CH₃, which is an ester. To name an ester using IUPAC nomenclature, identify two parts:

The acid component (to the left of the ester linkage -C(=O)-O-): There are three carbon atoms (CH₃-CH₂-C=O), which corresponds to propanoic acid. In the ester name, this becomes propanoate.

The alkyl component (to the right of the ester oxygen): There are two carbon atoms (-O-CH₂-CH₃), which is an ethyl group.

Combining both parts, the ester is named ethyl propanoate. The alkyl group name comes first, followed by the name derived from the parent carboxylic acid with the -ic acid suffix replaced by -ate.

Bayanin Amsa

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).

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

Carbon(II) oxide is the IUPAC-style name for carbon monoxide (CO). It is a colourless, odourless, and highly toxic gas that is a major air pollutant, especially in urban areas.

The primary source of carbon monoxide as an air pollutant is the incomplete combustion of carbon-containing fuels. When fuels such as petrol, diesel, kerosene, coal, or wood burn with an insufficient supply of oxygen, carbon is only partially oxidised to CO instead of fully oxidised to CO₂:

\[ 2\text{C} + \text{O}_2 \rightarrow 2\text{CO} \]

Vehicle exhaust emissions are the single largest contributor of CO to the atmosphere, along with industrial furnaces and domestic cooking fires that operate under oxygen-poor conditions.

Respiration produces carbon dioxide (CO₂), not carbon monoxide. Photochemical smog is a secondary pollution phenomenon caused by sunlight acting on nitrogen oxides and volatile organic compounds; it is not a source of CO itself. Decomposition of sewage releases gases such as methane (CH₄) and hydrogen sulphide (H₂S), not carbon monoxide.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

Surface tension is the property of a liquid that causes its surface to behave like a stretched elastic membrane. It arises because molecules at the surface of a liquid experience a net inward pull from neighbouring molecules below and beside them, but not from above. This inward force causes the surface to contract to the smallest possible area.

For a given volume of liquid, the shape with the smallest surface area is a sphere. Therefore, when water forms small droplets (such as raindrops or drops on a waxy surface), surface tension pulls the water into a spherical shape.

The other properties do not explain the spherical shape:

• Polarity explains why water molecules attract each other (through hydrogen bonding), which contributes to surface tension, but polarity itself does not directly cause the spherical shape.
• Viscosity is the resistance of a liquid to flow. It affects how quickly water moves, not the shape of its droplets.
• Density is mass per unit volume and determines whether a substance sinks or floats, not the shape of droplets.

Exam tip: Surface tension explains several everyday observations: water forming spherical drops, insects walking on water surfaces, and a needle floating when placed gently on water.

Bayanin Amsa

To name an organic compound using IUPAC nomenclature, follow these steps:

Step 1: Identify the structure. The compound is CH₃C≡CCH(CH₃)₂. Writing it out carbon by carbon:

• C-1: CH₃
• C-2: C (triple bond to C-3)
• C-3: C (triple bond to C-2)
• C-4: CH with two CH₃ groups attached

Step 2: Find the longest carbon chain containing the triple bond. The four carbons above give a chain of 4. However, one of the methyl groups on C-4 can extend the chain to 5 carbons: C-1, C-2, C-3, C-4, C-5 (incorporating one methyl into the main chain). The remaining methyl group on C-4 becomes a branch.

Step 3: Number the chain to give the triple bond the lowest possible locants. Numbering from the CH₃ end: the triple bond is at positions 2-3. This gives pent-2-yne.

Step 4: Name the substituent. The methyl branch is on C-4.

The complete IUPAC name is 4-methylpent-2-yne.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

Reduction is the gain of electrons (or a decrease in oxidation state). To identify which species is reduced, track the oxidation states of each element.

In the reaction \(2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}\):

• Sodium (Na): Oxidation state changes from 0 (in elemental Na) to +1 (in NaCl). This is an increase in oxidation state, meaning sodium is oxidised (it loses one electron per atom).
• Chlorine (Cl₂): Oxidation state changes from 0 (in elemental Cl₂) to -1 (in NaCl as Cl^-). This is a decrease in oxidation state, meaning chlorine is reduced (each Cl atom gains one electron).

The species that undergoes reduction is Cl₂, because it is the substance that accepts electrons and has its oxidation state lowered from 0 to -1.

Note that Cl^- is the product of the reduction, not the species that undergoes it. The question asks for the species that undergoes reduction, which is the reactant Cl₂.

Bayanin Amsa

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 \(\Delta S > 0\).

Bayanin Amsa

When 2-methylpropene reacts with HCl, the reaction follows Markovnikov's rule: the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms, and the halide (Cl) adds to the more substituted carbon.

2-Methylpropene has the structure CH₂=C(CH₃)₂. The double bond is between C-1 (which bears two hydrogens) and C-2 (which bears no hydrogens but has two methyl groups). According to Markovnikov's rule:

• H adds to C-1 (the less substituted carbon)
• Cl adds to C-2 (the more substituted carbon)

This produces 2-chloro-2-methylpropane, (CH₃)₃CCl. The reaction proceeds via a tertiary carbocation intermediate at C-2, which is the most stable carbocation possible in this molecule. This stability drives the regioselectivity.

The other options are incorrect: 1-chloro-2-methylpropane would result from anti-Markovnikov addition; 3-chloro-2-methylpropane does not correspond to a valid position on a three-carbon chain with a methyl branch; and 2-chloro-3-methylbutane would require a five-carbon skeleton that is not present in the starting material.

Bayanin Amsa

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.

Bayanin Amsa

The diagram shows the equation 2NaCl(l) → 2Na(l) + Cl₂(g) with electricity as the energy source. This is the electrolysis of molten sodium chloride to produce metallic sodium and chlorine gas.

This industrial process is known as the Downs process, named after J.C. Downs who patented the Downs cell in 1924. In the Downs cell, molten NaCl (often mixed with CaCl₂ to lower the melting point from 801°C to about 600°C) is electrolysed. At the cathode, Na⁺ ions are reduced to liquid sodium metal, while at the anode, Cl⁻ ions are oxidised to produce chlorine gas.

The Bosch process produces hydrogen gas from water gas. The Chlor-alkali process electrolyses aqueous (not molten) NaCl to give NaOH, Cl₂, and H₂. The Browning process is not a standard industrial chemistry term in this context.

Bayanin Amsa

In qualitative analysis, ions are classified as either acid radicals (anions) or basic radicals (cations).

• Acid radicals are the negative ions (anions) derived from acids. Examples include CO₃^2- (from H₂CO₃), SO₄^2- (from H₂SO₄), NO₃^- (from HNO₃), PO₄^3- (from H₃PO₄), and Cl^- (from HCl).
• Basic radicals are the positive ions (cations) derived from bases. Examples include Na⁺, K⁺, Mg²⁺, Ca²⁺, and Fe²⁺.

The question asks which group contains only acid radicals. Examining each option:

• CO₃^2-, PO₄^3-, Mg²⁺: contains Mg²⁺, which is a basic radical (cation).
• CO₃^2-, SO₄^2-, Na⁺: contains Na⁺, which is a basic radical (cation).
• CO₃^2-, SO₄^2-, NO₃^-: all three are anions derived from acids. This group contains only acid radicals.
• CO₃^2-, SO₄^2-, K⁺: contains K⁺, which is a basic radical (cation).

The correct answer is the group containing CO₃^2-, SO₄^2-, and NO₃^-, as all three are acid radicals.

Bayanin Amsa

The equation is:

\[2\text{X} + 2\text{HCl} \rightarrow 2\text{XCl} + \text{H}_2\]

The product formed is XCl, which tells us that element X combines with chlorine in a 1:1 ratio. This means X has a valency of +1 and forms a monovalent chloride.

Examining the options:

• K (potassium): Group 1 element with a valency of +1. It forms KCl. The equation becomes \(2\text{K} + 2\text{HCl} \rightarrow 2\text{KCl} + \text{H}_2\), which balances correctly.
• Mg (magnesium): Group 2 element with a valency of +2. It forms MgCl₂, not MgCl. Does not match.
• Ca (calcium): Group 2 element with a valency of +2. It forms CaCl₂, not CaCl. Does not match.
• Ba (barium): Group 2 element with a valency of +2. It forms BaCl₂, not BaCl. Does not match.

Only potassium (K) has a valency of +1 and forms a chloride with the formula XCl, making it the correct identity of X.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. It occurs in gases, liquids, and solids, though it is slowest in solids because the particles are closely packed and vibrate in fixed positions.

When a gold bar and a silver bar are pressed firmly together and left for years, gold atoms gradually migrate into the silver bar (and vice versa). This happens because the metal atoms, though fixed in a lattice, vibrate continuously. Over very long periods, some atoms acquire enough energy to move into neighbouring lattice positions, slowly spreading through the other metal. This is solid-state diffusion.

Brownian movement describes the random, erratic motion of microscopic particles suspended in a fluid (liquid or gas) caused by collisions with the surrounding fluid molecules. It does not apply to atoms within a solid lattice. Displacement is a chemical reaction in which a more reactive element replaces a less reactive one in a compound. Osmosis is the movement of water molecules through a semipermeable membrane from a dilute to a concentrated solution. Neither of these describes the mixing of atoms between two solid metals in contact.

Bayanin Amsa

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).

Bayanin Amsa

Freons are a group of chlorofluorocarbons (CFCs) - synthetic compounds containing chlorine, fluorine, and carbon. They were widely used as propellants in aerosol cans, as refrigerants in air conditioners and refrigerators, and as solvents in industrial cleaning.

When released into the atmosphere from these sources, freons rise to the stratosphere where ultraviolet radiation breaks them down, releasing chlorine atoms. These chlorine atoms catalytically destroy ozone molecules, contributing to the depletion of the ozone layer.

Fossil fuel combustion releases carbon dioxide, sulphur dioxide, and nitrogen oxides, but not freons. Photosynthesis is a biological process that produces oxygen and consumes carbon dioxide. Organic decay releases methane and carbon dioxide. None of these processes involve freons.

The Montreal Protocol (1987) restricted the production and use of CFCs, leading to a gradual recovery of the ozone layer.

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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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For a reaction that follows first-order kinetics, the half-life is related to the rate constant by the formula:

\[t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}\]

Given \(k_1 = 3.85 \times 10^{-3}\, \text{s}^{-1}\):

\[t_{1/2} = \frac{0.693}{3.85 \times 10^{-3}}\]

\[t_{1/2} = \frac{0.693}{0.00385}\]

\[t_{1/2} = 180\, \text{s}\]

The half-life of radioactive phosphorus is 180 s.

An important feature of first-order kinetics is that the half-life is independent of the initial concentration - it depends only on the rate constant. This is why radioactive decay (which always follows first-order kinetics) has a constant half-life regardless of how much of the substance remains.

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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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Magnesium tetraoxosulphate(VI) is the systematic name for magnesium sulphate (MgSO\(_4\)). In its hydrated form, MgSO\(_4\)\(\cdot\)7H\(_2\)O, it is commonly known as Epsom salt.

Epsom salt is widely used in medicine as a laxative. When taken orally, magnesium sulphate draws water into the intestines by osmosis (it is poorly absorbed), which softens the stool and stimulates bowel movement. This makes it an effective saline laxative.

The other options do not match:

• A disinfectant kills microorganisms - examples include sodium hypochlorite or phenol, not MgSO\(_4\).
• A plasticizer is added to polymers to increase flexibility - these are typically organic esters such as dioctyl phthalate.
• While magnesium and sulphur are plant nutrients, MgSO\(_4\) is not the standard compound referred to as a fertilizer at examination level. Its primary and best-known use is as a laxative.

Bayanin Amsa

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.

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According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass:

\[\text{Rate} \propto \frac{1}{\sqrt{M}}\]

This means lighter gases diffuse faster and heavier gases diffuse slower. To arrange gases in order of increasing rate of diffusion, we arrange them from heaviest (slowest) to lightest (fastest).

Calculate the molar masses of the gases that appear in the options:

Arranging from heaviest to lightest (i.e., increasing rate of diffusion):

SO₂ (64) → O₂ (32) → NH₃ (17) → H₂ (2)

This matches the sequence SO₂, O₂, NH₃, H₂. The heaviest gas (SO₂) diffuses most slowly, and the lightest gas (H₂) diffuses most rapidly.

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The equation shows propane (\(C_3H_8\)) reacting with chlorine gas (\(Cl_2\)) in the presence of ultraviolet light to produce chloropropane (\(C_3H_7Cl\)) and hydrogen chloride (\(HCl\)).

In this reaction, a hydrogen atom on the propane molecule is replaced by a chlorine atom. This is the hallmark of a substitution reaction, specifically a free-radical substitution. The UV light provides the energy needed to break the \(Cl-Cl\) bond homolytically, generating chlorine free radicals that then attack the alkane.

It is not neutralization (no acid-base reaction), not polymerization (no repeating monomer units are joined), and not oxidation in the classical sense used here. The defining feature is the direct replacement of one atom (H) by another (Cl) in the organic molecule.

Bayanin Amsa

A reaction is spontaneous when the Gibbs free energy change is negative, that is, \(\Delta G < 0\). The Gibbs equation relates enthalpy, entropy, and temperature:

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

For the decomposition of phosphorus pentachloride:

\[\text{PCl}_5(g) \rightarrow \text{PCl}_3(s) + \text{Cl}_2(g)\]

Consider the entropy change. On the reactant side there is 1 mole of gas, and on the product side there is 1 mole of solid and 1 mole of gas. Since a solid has much lower entropy than a gas, the total entropy of the products is lower than that of the reactant. Therefore \(\Delta S\) is negative.

With \(\Delta S < 0\), the term \(-T\Delta S\) becomes positive, which adds to \(\Delta G\). For \(\Delta G\) to still be negative (spontaneous), \(\Delta H\) must be sufficiently negative to overcome the positive \(-T\Delta S\) contribution:

\[\Delta G = \Delta H - T\Delta S < 0\]

\[\Delta H < T\Delta S \quad (\text{where } \Delta S < 0, \text{ so } T\Delta S < 0)\]

This means \(\Delta H\) must be negative. An exothermic reaction (\(\Delta H\) is negative) releases enough energy to drive the process forward despite the unfavourable entropy change.

Exam tip: When \(\Delta S\) is negative, only a sufficiently negative \(\Delta H\) can make \(\Delta G\) negative, and such reactions tend to be spontaneous only at low temperatures.