Chemistry 2004 Jamb Examination

The correct answer is C. Carbon.

Carbon is commonly used as the anode in the electrolysis of brine due to its resistance to chlorine, which is produced during the process.

The correct answer is A. a decrease in pressure.

In the given endothermic reaction:

\( N_2O_4(g) \; ⇌ \; 2NO_2(g) \)

There is an increase in the number of moles of gas on the right side (2 moles) compared to the left side (1 mole). According to Le Chatelier's principle, a decrease in pressure will favor the side with more moles of gas, which is the products side (NO₂).

The correct answer is C. +7

To determine the oxidation state of chlorine in \( \text{HClO}_4 \):

- Hydrogen (H) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

Let the oxidation state of chlorine (Cl) be \( x \).

The sum of the oxidation states in the compound \( \text{HClO}_4 \) should equal 0:

\[ +1 + x + 4(-2) = 0\]

\[ +1 + x - 8 = 0\]

\[ x - 7 = 0\]

\[ x = +7\]

So, the oxidation state of chlorine in \( \text{HClO}_4 \) is +7.

The correct answer is C. HI.

Entropy is a measure of disorder or randomness. Among the hydrogen halides, entropy generally increases with the size and mass of the molecules. Since iodine is the largest and heaviest of the halogens listed, HI (hydrogen iodide) has the highest entropy value.

To determine the mass of silver deposited, you can use Faraday's laws of electrolysis. The formula to calculate the mass \( m \) of a substance deposited is:

\[ m = \dfrac{Q \cdot M}{n \cdot F} \]

Where:
- \( Q \) is the total charge (in coulombs),
- \( M \) is the molar mass of the substance (silver, in this case),
- \( n \) is the number of electrons transferred per ion (for silver, \( n = 1 \) because silver ions gain one electron to become silver metal),
- \( F \) is Faraday's constant.

First, calculate the total charge \( Q \) using the formula:

\[ Q = I \times t \]

where:
- \( I \) is the current (10 A),
- \( t \) is the time (4830 s).

So,

\[ Q = 10 \, \text{A} \times 4830 \, \text{s} = 48300 \, \text{C} \]

Now, use the formula for mass:

\[ m = \dfrac{Q \cdot M}{n \cdot F} \]

Substitute the values:

- \( M \) (molar mass of silver) = 108 g/mol,
- \( n = 1 \),
- \( F = 96500 \, \text{C/mol} \).

\[ m = \dfrac{48300 \, \text{C} \times 108 \, \text{g/mol}}{1 \times 96500 \, \text{C/mol}} \]

\[ m = \dfrac{5216400 \, \text{g}}{96500} \approx 54 \, \text{g} \]

So, the mass of silver deposited is 54.0 g.

The correct answer is A. H₂S.

H₂S (hydrogen sulfide) can act as both a reducing agent and an oxidizing agent. In reactions, hydrogen sulfide can be oxidized to sulfur (or sulfur dioxide) and can also reduce other substances, making it versatile in its redox behavior.

For the other options:

• CO₂ (carbon dioxide) is typically a product of oxidation and does not usually act as both a reducing and oxidizing agent.
• H₂ (hydrogen gas) is primarily a reducing agent.
• SO₂ (sulfur dioxide) can act as a reducing agent but is less commonly a strong oxidizing agent compared to H₂S.

The correct answer is B. \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \).

This reaction involves an equal number of moles of gases on both sides of the equation (2 moles of reactants and 2 moles of products). Therefore, changing the volume of the system has little to no effect on the equilibrium position because the number of gas molecules is the same on both sides.

Energy change (ΔH) = 2ΔHCO₂ - (2ΔHCO + ΔHO)
= 7x - 393 - (7 * 110.4 + 0)
= -393 + 110.4 = -282.6 kJ

In the reaction:

\[ \text{ZnO} + \text{CO} \rightarrow \text{Zn} + \text{CO}_2 \]

Zinc in \(\text{ZnO}\) is in the form of \(\text{Zn}^{2+}\). During the reaction, zinc is converted from \(\text{Zn}^{2+}\) to \(\text{Zn}\), which means it gains electrons. This process is known as reduction.

Therefore, zinc has been:

C. reduced.

To solve this, we'll first need to write the balanced chemical equation for the reaction between calcium trioxocarbonate (IV) (CaCO₃) and hydrochloric acid (HCl):

\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\]

From the equation, 1 mole of calcium trioxocarbonate (IV) reacts to produce 1 mole of carbon dioxide (CO₂).

Next, we'll find the molar mass of CaCO₃:\[= 40 + 12 + (16 \times 3) = 100 \text{ g/mol}\]

Now, we calculate the number of moles of CaCO₃ in 2 g:\[ \text{Moles of CaCO}_3 = \dfrac{2 \text{ g}}{100 \text{ g/mol}} = 0.02 \text{ moles}\]

Since 1 mole of CaCO₃ produces 1 mole of CO₂, 0.02 moles of CaCO₃ will produce 0.02 moles of CO₂.

Finally, using the molar volume of a gas at s.t.p (22.4 dm³ or 22400 cm³ per mole):\[ \text{Volume of CO}_2 = 0.02 \times 22400 \text{ cm}^3\]\[ = 448 \text{ cm}^3\]

So, the correct answer is D. 448 cm³.