Element Y whose atomic number 11 react with chlorine gas to form a compound. a) Name the group and period to which Y belongs. b) Write an equation for the reaction.

Sulphur (IV) oxide is $\text{SO}_2$.
Molar mass of $\text{SO}_2$ ($M_{\text{SO}_2}$) = $32 (\text{S}) + 2 \times 16 (\text{O}) = 32 + 32 = 64 \text{ g/mol}$.
Rate of diffusion of $\text{SO}_2$ ($R_{\text{SO}_2}$) = $25 \text{ cm}^3/\text{minute}$.
Rate of diffusion of Gas G ($R_G$) = $26.26 \text{ cm}^3/\text{minute}$.
Graham's Law of Diffusion states:

$$\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}$$

Let Gas 1 be $\text{SO}_2$ and Gas 2 be Gas G.

$$\frac{R_{\text{SO}_2}}{R_G} = \sqrt{\frac{M_G}{M_{\text{SO}_2}}}$$ $$\frac{25 \text{ cm}^3/\text{min}}{26.26 \text{ cm}^3/\text{min}} = \sqrt{\frac{M_G}{64 \text{ g/mol}}}$$

Square both sides of the equation:

$$\left(\frac{25}{26.26}\right)^2 = \frac{M_G}{64}$$ $$(0.95194)^2 \approx \frac{M_G}{64}$$ $$0.90619 \approx \frac{M_G}{64}$$ $$M_G = 0.90619 \times 64$$ $$M_G \approx 57.996$$

The molecular mass of Gas G is approximately $\boxed{\text{58 g/mol}}$.

6.

a) Element Y has an atomic number of 11. This means it has 11 electrons.
The electron configuration is $1s^2 2s^2 2p^6 3s^1$.
The highest principal energy level is 3, so it belongs to Period 3.
It has 1 valence electron in the s-orbital, so it belongs to Group 1.

b) Element Y (Sodium, Na) reacts with chlorine gas ($\text{Cl}_2$) to form sodium chloride (NaCl).

$$2\text{Na}(\text{s}) + \text{Cl}_2(\text{g}) \longrightarrow 2\text{NaCl}(\text{s})$$

3 zimekamilika, zimebaki 2 leo. Unapiga hatua.