Correct Answer: C. 0

Correct Answer: B. alkaline

Correct Answer: A. Benzene

Correct Answer: D) 3 - propylheptane

Correct Answer: B. sodium, magnesium, aluminium, phosphorus, chlorine

Correct Answer: C. C₃H₆

Explanation

(a)(i)

(ii) Application of electrolysis

- For purification of elements e.g. copper, gold, silver

- For protection of metals from corrosion / electroplating

- For manufacture of chemicals e.g. NaOH, NaCIO\(_{3}\)

- For extraction of metals e.g Al, Ca, K, Na, Mg, etc.

(iii) Electrolysis of dilute NaCl(\(^{aq}\)) using carbon electrodes.

At the Cathode H\(^{+}\) + e\(^-\) —> H

H + H —> H\(^2\) Or 2H\(^{+}\) + 2e\(^-\) —> H\(_2\)

At the anode OH\(^-\) —> OH + e\(^-\)

or 4OH\(^- \) ---> 2H\(^2\)O + O\(_2\) Or 4OH\(^-\) —> 2H\(_2\)O + O\(_2\) + 4e\(^{-}\).

(b)(i) An electrochemical cell is a device for producing an electric current by chemical reaction.

(ii) Examples of primary cell: Daniell cell, Leclanche cell, cadmin / weston mercury cell.

(iii) Mg + Fe\(^{2+}\) --> Mg\(^{2+}\) + Fe ;

Half cell reactions: Mg --> Mg\(^{2+}\) + 2e\(^{-}\)

Fe\(^{2+}\) + 2e\(^-\) --> Fe or Mg - 2e\(^-\) --> Mg\(^{2+}\)

(c)(i) To calculate the quantity of electricity produced by a current of 0.72 A in 3 hours 20 mins.

Q = It

t = (3 x 60 + 20) min x 60 sec = 12000 Sec.

Q = (0.72 x 12000) C = 8640 C.

(ii) If 1 dm\(^3\) of gas was evolved at the cathode during electrolysis of acidified water, volume of gas evolved at the anode will be 0.5 dm\(^3\) or 500cm\(^3\).

(d)(i) Calculation of excess CuO if 20g was warmed with 0.05 mole H\(_2\)SO\(_4\). CuO + H\(_2\)SO\(_4\) --> CuSO\(_4\) + H\(_2\)O. From the equation 0.05 mole of H\(_2\)SO\(_4\) reacts with 0.05 mole of CuO.

Molar mass of CuO = 64 + 16 = 80g.

Hence 0.05 mole CuO = (80 x 0.05)g = 4.0g.

Mass of CuO in excess = 20 - 4 = 16g.

(ii) Type of reaction involved in the reaction in (d)(i) is neutralisation / acid - base reaction.

Correct Answer: B) electro-negativity increases regularly from left to right across a period

Correct Answer: D. Zinc

Explanation

Brass is an alloy made by combining copper with zinc. It is a versatile material with a golden appearance and is used in various applications, including musical instruments, plumbing fittings, and decorative items.

Correct Answer: A) 0.094M

Explanation

In this question, we are looking at a simple acid-base titration. The acid in this case is hydrochloric acid (HCl) and the base is sodium hydroxide (NaOH). The concept of molarity (M) is fundamental here, which is a measure of the concentration of a solute in a solution, or of any chemical species, in terms of amount of substance in a given volume.

The balanced chemical equation for the reaction between NaOH and HCl is:

\(NaOH + HCl \rightarrow NaCl + H_2O\)

From the balanced equation, we can see that the ratio of NaOH to HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.

Since molarity (M) is defined as the moles of solute per liter of solution, we can use the formula:

\[M = \frac{moles}{volume}\]

In a titration, the point where the reaction is complete is called the endpoint. At this point, the moles of the acid will be equal to the moles of the base. So, we can say that:

\[moles_{NaOH} = moles_{HCl}\]

We know the molarity and volume of the NaOH solution, so we can calculate the moles of NaOH using the formula:

\[moles = Molarity \times Volume\]

Substituting the given values:

\[moles_{NaOH} = 0.15M \times 24.83cm^3\]

Note that the volume should be in liters, so we convert from cm^3 to L by dividing by 1000:

\[moles_{NaOH} = 0.15M \times \frac{24.83cm^3}{1000} = 0.003725 moles\]

Since \(moles_{NaOH} = moles_{HCl}\), we also have 0.003725 moles of HCl. We can now find the molarity of the HCl solution using the formula \(M = \frac{moles}{volume}\). The volume of HCl is given as 39.45 cm^3, which we convert to liters:

\[M_{HCl} = \frac{0.003725 moles}{\frac{39.45cm^3}{1000}} = 0.094M\]

So, the molarity of the HCl solution is 0.094M, which corresponds to Option A.