Chemistry 2019 Jamb Examination

From Faraday’s law of electrolysis, the amount of substance discharged is proportional to the quantity of electricity used and inversely proportional to its valency. 

Given that 10~g of $X^{3+}$ costs ₦20.00 to discharge, the cost per mole equivalent is the same for both metals under identical conditions. Let the cost be proportional to the mass over valency:\[ \text{Cost} \propto \frac{M}{n},\] where $M$ is mass and $n$ is the valency. Then for two metals $X$ and $Y$:\[ \frac{C_Y}{C_X} = \frac{(M_Y / n_Y)}{(M_X / n_X)} \times \frac{m_Y}{m_X}\] \[ \Rightarrow C_Y = 20 \times \frac{24/2}{27/3} \times \frac{6}{10} \]\[= 20 \times \frac{12}{9} \times 0.6 = 9.00~\text{₦}\] Hence, it would cost ₦9.00 to discharge 6~g of $Y^{2+}$.

Cost of discharge is directly proportional to \( \dfrac{m \times n}{M} \).
Given: for X (divalent metal), cost = ₦12.00 for 6 g.\[ \dfrac{C_Y}{C_X} = \dfrac{(m_Y n_Y / M_Y)}{(m_X n_X / M_X)} \] Substitute:\[ m_X = 6, n_X = 2, M_X = 63\] \[m_Y = 9, n_Y = 3, M_Y = 27 \]\[ C_Y = 12 \times \dfrac{(9 \times 3 / 27)}{(6 \times 2 / 63)}\]\[ = 12 \times \dfrac{1}{(12/63)} \]\[= 12 \times 5.25 = 63.00 \]Therefore, ₦63.00 is required to discharge 9.0 g of the trivalent metal Y.

For ethers (alkoxyalkanes), the general formula is \( R–O–R' \).
With molecular formula C₄H₁₀O, possible structures are:
1. CH₃CH₂CH₂OCH₃ (methoxypropane)
2. CH₃CH(OCH₃)CH₃ (ethoxyethane)
3. CH₃OCH₂CH₂CH₃ (same as methoxypropane rearrangement).
Hence, there are 3 structural isomers of alkoxyalkanes possible from C₄H₁₀O.

In this reaction, sulphur undergoes oxidation and reduction simultaneously.
In H₂S, sulphur has an oxidation state of -2, while in SO₂ it is +4.
In the products, sulphur appears as elemental sulphur (oxidation state 0).
Hence, sulphur in H₂S is oxidised (-2 → 0) and sulphur in SO₂ is reduced (+4 → 0).
This shows that H₂S acts as the reducing agent (reductant) and SO₂ acts as the oxidising agent (oxidant).
Therefore, the reaction is a redox process in which SO₂ is the oxidant and H₂S is the reductant.

When AgNO₃ reacts with Cl⁻, a white precipitate of AgCl forms, which dissolves in NH₃ due to the formation of [Ag(NH₃)₂]⁺ complex.
Similarly, when solution Y reacts with X giving a white precipitate that dissolves on heating, it suggests PbCl₂ formation since PbCl₂ is sparingly soluble in cold water but soluble in hot water.
Hence, X contains Cl⁻ ions and Y contains Pb²⁺ ions.

Electrons are found outside the nucleus of an atom, not within it.
The nucleus contains only protons and neutrons. Given that the atom has a mass number of 23 and 17 neutrons, the number of protons is 6 (23 − 17).
However, since the question specifically asks about electrons *in the nucleus*, the number is zero — none are found there.

Relative atomic mass = (fractional abundance × mass number) for each isotope.  
R.A.M = (0.8 × 16) + (0.2 × 18) = 12.8 + 3.6 = 16.4  

Therefore, the relative atomic mass of element Z is 16.4.

Ethyl ethanoate (CH₃COOCH₂CH₃) is an ester derived from ethanoic acid and ethanol.
When the alkyl groups are interchanged, the acid part (COO−) attaches to the other alkyl group, forming CH₃CH₂COOCH₃, which is methyl propanoate.
Hence, the new compound formed is methyl propanoate.

The energy available to do useful work in a thermodynamic system at constant temperature and pressure is the Gibbs free energy (G).
It is defined by the relation \( G = H - TS \), where \( H \) is enthalpy, \( T \) is temperature, and \( S \) is entropy.
It determines the spontaneity of a reaction — a negative ΔG means the reaction can occur spontaneously.

The rate of reaction depends on factors such as concentration and surface area.
Increasing the concentration of HCl increases the number of reacting particles per unit volume, while increasing the surface area of CaCO₃ allows more contact between the reactants.
Both actions speed up the rate of CO₂ evolution, hence both A and B are correct.