Chemistry 2025 Jamb Examination

Amines are classified based on the number of carbon atoms directly attached to the nitrogen atom.
In the compound \(CH_3CH(NH_2)CH_2CH_2CH_3\) (2-pentanamine), the nitrogen atom is bonded to only one carbon atom (the second carbon of the pentane chain). Therefore, it is classified as a primary amine. A secondary amine would have the nitrogen bonded to two carbons, and a tertiary amine would have it bonded to three.

Nitrogen is a key industrial starting material. One of its primary large-scale uses is in the Haber process to produce ammonia (\(NH_3\)), which is then used in the Ostwald process to manufacture nitric acid (\(HNO_3\)).
While liquid nitrogen is used as a cooling agent, the context of 'production' in chemical exams usually refers to its role as a chemical feedstock for synthesis.

Reduction is defined as the gain of electrons or a decrease in oxidation state.
In this reaction:
1. Sodium (\(Na\)) starts with an oxidation state of 0 and ends as \(Na^+\) in \(NaCl\) (oxidation state +1). It loses electrons and is oxidised.
2. Chlorine (\(Cl_2\)) starts with an oxidation state of 0 and ends as \(Cl^-\) in \(NaCl\) (oxidation state -1). It gains electrons and is reduced.
Therefore, \(Cl_2\) is the species undergoing reduction.

Sublimation is an endothermic phase transition where a substance moves directly from the solid state to the gaseous state without passing through the liquid phase.
Breaking the intermolecular forces holding the solid lattice together requires the input of heat from the surroundings, which is why it involves energy absorption.

Ammonium trioxonitrate(V), also known as ammonium nitrate (\(NH_4NO_3\)), decomposes when heated gently. The chemical equation for this reaction is:\[NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(g)\]The products are dinitrogen oxide (laughing gas) and water vapour. High temperatures or shock can lead to more explosive decomposition, but the standard academic product is \(N_2O\).

We can identify the element \(X\) by looking at the stoichiometry of the balanced equation. The reaction shows that 2 moles of \(X\) react with 2 moles of \(HCl\) to produce 1 mole of \(H_2\). This implies that 1 mole of \(X\) replaces 1 mole of Hydrogen.
This means \(X\) must have a valency of +1. Looking at the options:
- K (Potassium) is in Group 1 and has a valency of +1.
- Mg, Ca, and Ba are in Group 2 and have a valency of +2.
If \(X\) were a Group 2 metal, the equation would be \(X + 2HCl \rightarrow XCl_2 + H_2\). Therefore, \(X\) must be Potassium (K).

Both alkanols and alkanoic acids can form hydrogen bonds. However, alkanoic acids are capable of forming stable 'dimers' where two molecules are held together by two hydrogen bonds.
This doubling of hydrogen bonding between molecules creates significantly stronger intermolecular forces compared to the single hydrogen bonds found in alkanols, leading to much higher boiling points for alkanoic acids of similar molecular mass.

Surface tension is the property of a liquid surface that causes it to behave like a stretched elastic membrane.
Molecules at the surface are pulled inward by cohesive forces from other water molecules. To minimise the surface area for a given volume, the liquid assumes a spherical shape, as a sphere has the smallest surface-to-volume ratio.

Brine is a concentrated solution of sodium chloride (\(NaCl\)) in water. It contains \(Na^+\), \(Cl^-\), \(H^+\), and \(OH^-\) ions.
At the anode (positive electrode), negative ions are attracted. These are \(Cl^-\) and \(OH^-\). In concentrated brine, the concentration of \(Cl^-\) is much higher than \(OH^-\), so chloride ions are preferentially discharged to form chlorine gas (\(Cl_2\)).

We use Faraday's first law of electrolysis. Copper is deposited from \(CuSO_4\) as \(Cu^{2+}\), requiring 2 moles of electrons (2F) per mole of copper.
First, calculate the number of moles of Copper:\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{64} \approx 0.07031 \text{ mol}\]The total charge \(Q\) required is:\[Q = n \times z \times F = 0.07031 \times 2 \times 96500 \approx 13570 \text{ C}\]Since \(Q = It\), we solve for time \(t\):\[t = \frac{Q}{I} = \frac{13570}{2.5} = 5428 \text{ sec}\]