Correct Answer: A. hydrogen and oxygen

Correct Answer: D. esterification

Correct Answer: C. 5.6dm³

Correct Answer: A. AI₂CI₆

Correct Answer: C) -300KJ mol^?1

Explanation

To determine the heat of combustion of ethanol, we first need to find the amount of heat released by burning 0.92g of ethanol, and then convert that to kJ/mol.

First, we can find the amount of heat released using the formula:

\(\Delta Q = mc\Delta T\)

Where \(\Delta Q\) is the heat released, \(m\) is the mass of water, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the change in temperature.

Using the given values: \(m = 100g\), \(c = 4.2 Jg^{-1}K^{-1}\), and \(\Delta T = 312.3K - 298K = 14.3K\), we can calculate the heat released:

\(\Delta Q = (100g)(4.2 Jg^{-1}K^{-1})(14.3K) = 6006 J\)

Now we need to determine the number of moles of ethanol in 0.92g. The molecular formula of ethanol is \(C_2H_5OH\), so its molar mass is:

\(2(C) + 6(H) + 1(O) + 1(H) = 2(12) + 6(1) + 16 + 1 = 46 g/mol\)

To find the number of moles, we divide the mass by the molar mass:

\(0.92g / 46 g/mol = 0.02 mol\)

Finally, we can find the heat of combustion per mole of ethanol by dividing the heat released by the number of moles:

\(\frac{6006 J}{0.02 mol} = 300300 J/mol \approx 300 KJ/mol\)

Since the heat of combustion is exothermic, the value should be negative. Therefore, the correct answer is:

Option C: -300KJ mol\(^{-1}\)

Correct Answer: D. React with metals to produce hydrogen gas

Explanation

Acids are substances that can donate protons (H+) in aqueous solutions. When acids react with certain metals, they can release hydrogen gas (H2) as one of the products. This is a common behavior of many acids and can be used to distinguish them from other substances.

Correct Answer: B. X/2 g

Correct Answer: C. Their oxides are acidic

Explanation

(a)(i) Heat of neutralization is defined as the amount of heat evolved when one mole of hydrogen ions (H\(^+\)/H\(^3\)O\(+\)) from an acid reacts with one mole of hydroxide ions (OH\(^-\)) from an alkali to form one mole of water.

(ii) Neutralization can be used to prepare only soluble salts, hence CuCl\(_2\) which is soluble can be prepared by this method while PbC\(_2\) which is insoluble cannot.

(b)(i) Manufacture of HNO\(_3\) in outline by the catalytic oxidation of NH\(_3\). Ammonia reacts with air or oxygen in the presence of platinum-rhodium catalyst and a high temperature (of about 550 - 850°C) to form Nitrogen (11) oxide and steam. 4NH\(_3\) + 5O\(_3\) ---> 4NO + 6H\(_2\)O. Nitrogen (II) oxide reacts with more or excess air to give nitrogen (IV) oxide which is dissolved' in hot water in the presence of more air to give trioxonitrate (V) acid.

2NO + O\(_2\) —> 2NO\(_2\); 4NO\(_2\) + 2H\(_2\)O + O\(_2\) —> 4HNO\(_3\)

(ii) Oxygen, NaNO\(_3\) are obtained when NaNO\(_3\) is heated strongly.

(c) Reactions of Mg with nitrogen.

(i) Equations of the reaction 3Mg + N\(_2\) --> Mg\(_3\)N\(_2\).

(ii) Calculation of the amount of Mg\(_3\)N\(_2\) that can be obtained from 3.0g of Mg. From the equation, (3 x 24)g of Mg gives 1 mole of Mg\(_3\)N\(_2\).

3g of Mg will give 3 x \(\frac{1}{3 \times 24}\) mole of Mg\(_3\)N\(_2\) = 0.0417 mole.

Correct Answer: C) Dissolution of sodium metal in water