Correct Answer: A. Calcium trioxocarbonate ( iv ) chips a 20°C

Correct Answer: A. Activation energy

Correct Answer: A) N27.02

Explanation

First, let's consider the relationship between the amount of deposited metal and the cost of electricity. In electrolysis, the amount of deposited metal is directly proportional to the number of moles of electrons involved, which in turn is directly proportional to the cost of electricity.

To find the cost of depositing 24g of copper, we need to determine the number of moles of electrons involved in depositing 1g of aluminum and 24g of copper. We can then use the given cost of electricity to find the cost for depositing the copper.

For aluminum (A), the relative atomic mass is 27. Therefore, 1 mole of aluminum weighs 27g. In the electrolytic process, 3 moles of electrons are required to deposit 1 mole of aluminum, as its valency is +3. Hence, for 1g of aluminum, the number of moles of electrons is:

\[\frac{1}{27} \times 3 = \frac{1}{9}\]

For copper (Cu), the relative atomic mass is 64. Therefore, 1 mole of copper weighs 64g. In the electrolytic process, 2 moles of electrons are required to deposit 1 mole of copper, as its valency is +2. Hence, for 24g of copper, the number of moles of electrons is:

Now, we need to find the cost ratio between the number of moles of electrons required to deposit 1g of aluminum and 24g of copper. The ratio is:

Given that the cost of electricity required to deposit 1g of aluminum is ?4.00, the cost to deposit 24g of copper is:

So, the cost to deposit 24g of copper is ?27.00, which is the correct answer,Option A.

Correct Answer: C) production electons

Correct Answer: C. hygroscopic

Correct Answer: C) CaCl2

Correct Answer: C. C4H10

Correct Answer: C. Silicon

Correct Answer: D) chemical symbols and number of atoms

Explanation

It is important to note that a molecular formula is different from an empirical formula, which shows the simplest whole number ratio of the atoms in a compound. For example, the empirical formula of glucose is CH2O, which represents the ratio of one carbon atom, two hydrogen atoms, and one oxygen atom in the compound.

Understanding molecular formulas is crucial in Chemistry as it helps in identifying and naming compounds, determining their properties, and predicting their behavior in different chemical reactions. To learn more about molecular formulas and their applications, please read the relevant sections of the recommended textbooks.

Explanation

(a) (i) X - oxygen

Y - sulphur

(ii) I. - S(s) + O\(_{2(g)}\) \(\to\) SO\(_{2(g)}\) (1)

II. - 2SO\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2SO\(_{3(g)}\) (2)

III. - SO\(_{3(g)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) (1)

IV. - H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\) 2H\(_2\)SO\(_{4(aq)}\) (2)

(iii) Vanadium (V) oxide (1)

(iv) It is because the reaction is exothermic

(v) - In bleaching wood pulp, wool silk, hay

- as germicide and fumigant/ disinfectant

- as refrigerant / refrigerating gas/coolant

- as preservative for grains and fruits

- manufacture of H\(_2\)SO\(_4\)

- as a reducing agent in metallurgy

- for dechlorination

(b) From the equation,

2 mol of H\(_2\) reacts with 1 mol of O\(_2\)

hence 40 cm\(^3\) of H\(_2\) would react with 20 cm\(^3\) of O\(_2\)

Volume of unused oxygen = (30 – 20) cm\(^3\)

= 10 cm\(^3\)

(c) (i) CaCO\(_3\) \(\to\) CaO + CO\(_2\)

(ii) CaCO\(_3\) = 40 + 12 + (16 x 3) = 100 g

CaO = 40 + 16 = 56 g

100 g of CaCO\(_3\) \(\to\)&