Correct Answer: C) heptane to propene and butane

Correct Answer: A) sodium dioxonitrate(III) and ammonium chloride

Correct Answer: B. x belongs to group 1, Y belongs to period 4

Correct Answer: C. Charcoal

Correct Answer: D) 4g

Explanation

NaOH= 40.

Vol. 500/1000=0.5.

0.5

Correct Answer: A. Aluminium

Correct Answer: D. Saponification

Explanation

(a)(i) Aluminium is preferred to copper for making overhead electric cables because aluminium: is lighter/has a lower relative density

-is more resistant to corrosion.

(ii) Extraction of aluminium from purified bauxite. A solution of the pure alumina in molten cryolite (Na\(_3\)AIF\(_6\)) is placed in the electrolytic cell made of iron lined with graphite which serves as the cathode. The anode consists of graphite rods dipping into the electrolyte (solution of alumina in molten cryolite). The cell is maintained at high temperature 900 ± 100°C. At the cathode: Al\(^{3+}\) + 3e\(^-\)m ---> Al ; At the anode: 2O\(^{2-}\) ---> O\(_2\) + 4e\(^-\). The molten aluminium collects at the bottom of the cell and is tapped off at intervals.

(b) Illustration of the cell used for the purification of copper.

(i) Electrode I increases in mass during the electrolysis, because Cu\(^{2+}\) migrate to it. The Cu\(^{2+}\) is reduced to copper/Cu\(^{2+}_{(aq)}\)+ 2e --> Cu\(_{(s)}\) and the copper is deposited, leading to an increase in its mass.

(ii) Electrode II is the site of oxidation. Oxidation is electron loss and the reaction at this electrode is Cu -> Cu\(^{2+}\) + 2e\(^{2+}\)

(iii) Ill is CuSO\(_{4(aq)}\)/any soluble copper (II) salt. Its colour does not change in intensity because Cu\(^{2+}\) which is responsible for the colour is removed by the cathode (i) from the solution and is replaced by the anode.

(c) To calculate the current that will deposit 8.00g of calcium from fused CaCl\(_2\) in 1 hr. 15mins.

Ca\(^{2+}\) + 2e\(^{-}\) -> Ca.

40g of calcium will be deposited by (2 x 96500)C

8.00g of calcium will be deposited by \(\frac{8 \times 38600}{40}\) = 38600 C ; Let x represent 386000C

Q = It xc = 1 x 75 x 60 ;

Current (I) = \(\frac{x}{75 \times 60}\) ; Current = 8.58

Correct Answer: D) of its high reactivity

Correct Answer: C) 75cm³

Explanation

To answer this question, we need to first understand the balanced chemical equation of the reaction between butane and oxygen. Butane (C₄H₁₀) reacts with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O).

The balanced chemical equation for this reaction is:

2C₄H₁₀ + 13O₂ ? 8CO₂ + 10H₂O

From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen. To determine the volume of the mixture, we need to find the volume of the unreacted oxygen after the reaction.

We have 10 cm³ of butane, and since the ratio of butane to oxygen is 2:13, we can calculate that 10 cm³ of butane will react with 65 cm³ of oxygen.

Since we initially had 100 cm³ of oxygen, the remaining unreacted oxygen is:

100 cm³ - 65 cm³ = 35 cm³

Now, let's find the volume of the products of the reaction. From the balanced equation, we can see that the reaction produces 8 moles of CO₂ and 10 moles of H₂O for every 2 moles of butane. Since we have 10 cm³ of butane, the volume of CO₂ produced is:

\(\frac{8}{2}\) * 10 cm³ = 40 cm³

The volume of H₂O produced is:

\(\frac{10}{2}\) * 10 cm³ = 50 cm³

However, the water produced will condense at room temperature and pressure, so we only need to consider the volume of CO₂ and unreacted oxygen in the mixture.

Therefore, the total volume of the mixture when brought back to the original conditions of measurements is:

35 cm³ (unreacted oxygen) + 40 cm³ (CO₂) = 75 cm³

The correct option isOption C: 75 cm³.