Calculate the volume of 0.6M sulphuric (VI) acid solution needed to neutralize 30cm³ of 0.2M potassium hydroxide.

Sulphuric (VI) acid is $\text{H}_2\text{SO}_4$. Potassium hydroxide is KOH.

$$\text{H}_2\text{SO}_4(\text{aq}) + 2\text{KOH}(\text{aq}) \longrightarrow \text{K}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$$

From the equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 2 moles of KOH. So, the mole ratio $n_A:n_B = 1:2$.

The formula for neutralization is $C_A V_A n_B = C_B V_B n_A$.
Given:
$C_A = 0.6 \text{ M}$ ($\text{H}_2\text{SO}_4$)
$V_A = ?$
$C_B = 0.2 \text{ M}$ (KOH)
$V_B = 30 \text{ cm}^3$
$n_A = 1$
$n_B = 2$

Substitute the values into the formula:

$$(0.6 \text{ M}) \times V_A \times 2 = (0.2 \text{ M}) \times (30 \text{ cm}^3) \times 1$$ $$1.2 V_A = 6 \text{ cm}^3$$ $$V_A = \frac{6 \text{ cm}^3}{1.2}$$ $$V_A = 5 \text{ cm}^3$$

The volume of 0.6M sulphuric (VI) acid solution needed is $\boxed{\text{5 cm}^3}$.

9.

a) $\text{NH}_3$ (Ammonia)
Nitrogen has 5 valence electrons (represented by dots). Each Hydrogen has 1 valence electron (represented by crosses). Nitrogen forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons.

$$\begin{array}{c} \text{H} \\ \times \\ \text{H} \cdot \stackrel{\bullet}{\text{N}} \cdot \text{H} \\ \times \end{array}$$

b) $\text{NH}_4^+$ (Ammonium ion)
The ammonium ion is formed when a hydrogen ion ($\text{H}^+$) accepts the lone pair of electrons from the nitrogen in ammonia, forming a dative (coordinate) covalent bond. All four N-H bonds are equivalent once formed. The overall charge is +1.

$$\begin{bmatrix} \begin{array}{c} \text{H} \\ \times \\ \text{H} \cdot \stackrel{\bullet}{\text{N}} \cdot \text{H} \\ \times \quad \times \\ \quad \text{H} \end{array} \end{bmatrix}^+$$