The student's handwritten calculation for the moles of MCO₃ is correct. Calculate the relative molecular mass of MCO₃ given the mass is 0.100 g.

The concentration of hydrochloric acid is $0.1 \text{ M}$ and the volume is $23.8 \text{ cm}^3$.
First, convert the volume from $\text{cm}^3$ to $\text{dm}^3$:

$$\text{Volume} = \frac{23.8 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3} = 0.0238 \text{ dm}^3$$

Now, calculate the moles using the formula: $\text{Moles} = \text{Concentration} \times \text{Volume}$

$$\text{Moles of HCl} = 0.1 \text{ mol/dm}^3 \times 0.0238 \text{ dm}^3 = 0.00238 \text{ mol}$$ $$\text{Moles of HCl} = 2.38 \times 10^{-3} \text{ mol}$$

The student's handwritten calculation is correct.
The number of moles of hydrochloric acid is $\boxed{\text{2.38} \times 10^{-3} \text{ mol}}$.

The balanced chemical equation is:

$$\text{MCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$$

From the equation, the mole ratio of MCO₃ to HCl is $1:2$.
So, moles of MCO₃ = $\frac{1}{2} \times \text{moles of HCl}$

$$\text{Moles of MCO}_3 = \frac{1}{2} \times 2.38 \times 10^{-3} \text{ mol} = 1.19 \times 10^{-3} \text{ mol}$$

The number of moles of MCO₃ is $\boxed{\text{1.19} \times 10^{-3} \text{ mol}}$.

The mass of MCO₃ is $0.24 \text{ g}$ and the moles of MCO₃ are $1.19 \times 10^{-3} \text{ mol}$.
The relative molecular mass ($M_r$) is calculated as: $M_r = \frac{\text{Mass}}{\text{Moles}}$

$$M_r \text{ of MCO}_3 = \frac{0.24 \text{ g}}{1.19 \times 10^{-3} \text{ mol}} \approx 201.68 \text{ g/mol}$$

Rounding to a reasonable number of significant figures (e.g., 3 sig figs based on 0.24 g):

$$M_r \text{ of MCO}_3 \approx 202$$

The relative molecular mass of MCO₃ is $\boxed{\text{202}}$.

The relative molecular mass of MCO₃ is 202.
The atomic mass of Carbon (C) is 12.0.
The atomic mass of Oxygen (O) is 16.0.
In MCO₃, there is one M atom, one C atom, and three O atoms.

$$M_r(\text{MCO}_3) = A_r(\text{M}) + A_r(\text{C}) + 3 \times A_r(\text{O})$$ $$202 = A_r(\text{M}) + 12.0 + 3 \times 16.0$$ $$202 = A_r(\text{M}) + 12.0 + 48.0$$ $$202 = A_r(\text{M}) + 60.0$$ $$A_r(\text{M}) = 202 - 60.0 = 142.0$$

The relative atomic mass of the metal M is $\boxed{\text{142.0}}$.

The theory used to explain the shapes of simple covalent molecules and ions is the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion, thus determining the molecule's geometry.

The theory is $\boxed{\text{Valence Shell Electron Pair Repulsion (VSEPR) theory}}$.