Correct Answer: A) NH₃

Explanation

(a) The three properties are:-

(i) The reaction occurring in the system is reversible.

(ii) the rate of the forward reaction is equal to that of the reverse reaction

(iii) The value of the free energy change of the system is zero

(b)(i) Since the forward reaction is exothermic, increase in temperature will not favour the forward reaction. The yield of ammonia will be lower.

(c)(i) The ore is powdered and dissolved under pressure in a hot concentrated solution of sodium hydroxide. Sodium aluminate solution is formed Al\(_2\)O\(_3\) + 2NaOH \(\to\) 2NaAIO\(_2\) + H\(_2\)O. The impurities do not dissolve and filtered off. The sodium aluminium hydroxide to precipitate the hydroxide which is filtered off, washed, dried and heated strongly to form pure alumina.

2A1(OH)\(_3\) \(\to\) Al\(_2\)O\(_3\) + 3H\(_2\)O

Al\(_2\)O\(_{3(s)}\) + 2NaOH\(_{(aq)}\) + 3H\(_2\)O --> 2NaAl(OH)\(_4\)

Then; NaAI(OH)\(_{4(aq)}\) --> Al(OH)\(_{3(s)}\) + NaOH\(_{(aq)}\)

(ii) At the ANODE: 2O\(^{2-}\) - 4e\(^-\) \(\to\) O\(_2\) OR 2O\(^{2-}\) ---> O\(_2\) + 4e\(^-\)

At the CATHODE: Al\(^{3+}\) + 3e\(^-\) -> Al

(iii) The function is to act as a solvent or as a flux.

Correct Answer: D) 4.05g

Explanation

During electrolysis, the amount of substance deposited at the cathode is directly proportional to the amount of electric charge passed through the electrolyte. To find the mass of aluminum deposited, we can use the concept of Faraday's laws of electrolysis.

First, let's determine the Faraday's constant (F) which is the electric charge per mole of electrons:

\[F = 96500\,\text{C/mol}\]

Next, we need to determine the number of moles of calcium (n) that were deposited:

\[n_\text{Ca} = \frac{\text{mass of Ca}}{\text{molar mass of Ca}} = \frac{9\,\text{g}}{40\,\text{g/mol}} = 0.225\,\text{mol}\]

Now, let's calculate the number of Faraday's (F) used to deposit this number of moles of calcium:

\[F_\text{Ca} = n_\text{Ca} \times F = 0.225\,\text{mol} \times 96500\,\text{C/mol} = 21712.5\,\text{C}\]

Since the same current was passed through both cells, the same amount of charge was used for the deposition of aluminum. We can determine the number of moles of aluminum (n) deposited by dividing the charge used by the Faraday's constant:

\[n_\text{Al} = \frac{F_\text{Al}}{F} = \frac{21712.5\,\text{C}}{96500\,\text{C/mol}} = 0.225\,\text{mol}\]

Note that the molar ratio of calcium to aluminum deposited is 2:3. So, we need to multiply the number of moles of aluminum by 3/2 to account for this ratio:

\[n_\text{Al}^\text{corrected} = n_\text{Al} \times \frac{3}{2} = 0.225\,\text{mol} \times \frac{3}{2} = 0.3375\,\text{mol}\]

Finally, we can find the mass of aluminum deposited by multiplying the corrected number of moles by the molar mass of aluminum:

\[\text{mass of Al} = n_\text{Al}^\text{corrected} \times \text{molar mass of Al} = 0.3375\,\text{mol} \times 27\,\text{g/mol} \approx 4.05\,\text{g}\]

So, the mass of aluminum deposited in the second cell is approximately 4.05g, which corresponds to Option D.

Correct Answer: D. neoprene

Correct Answer: B. 3

Correct Answer: B. Dative

Correct Answer: A. K₂CO₃

Correct Answer: C) have duplet or octet electronic configuration in the outermost shells in their atoms

Correct Answer: D. 30g

Correct Answer: B)

SO\(_2\) and NH\(_3\)

Explanation

The question is asking about the identification of two gases labelled A and B based on their chemical properties. Gas A is acidic to litmus paper and turns acidified potassium dichromate solution green. This means that gas A is an acid and can react with a solution of potassium dichromate in the presence of acid to turn it green. On the other hand, gas B turns red litmus paper blue and forms dense white fume with hydrogen chloride. This means that gas B is a base and can react with hydrogen chloride to form white fumes.

From the given options, the correct identity of gas A and gas B respectively are option B: SO\(_2\) and NH\(_3\). Sulfur dioxide (SO\(_2\)) is acidic and can turn acidified potassium dichromate solution green. On the other hand, ammonia (NH\(_3\)) is basic and can turn red litmus paper blue and form dense white fumes with hydrogen chloride. Therefore, option B is the correct answer to this question. It is important to note that understanding the properties of acids and bases is crucial in identifying their chemical properties. Acids are substances that can donate hydrogen ions (H\(^+\)), while bases are substances that can accept hydrogen ions. Litmus paper is a common indicator used to identify whether a solution is acidic or basic. A solution that is acidic turns blue litmus paper red, while a solution that is basic turns red litmus paper blue. To learn more about acid-base properties and their identification, please read the relevant sections of the recommended textbooks.