Calculate the molar mass of Mg(OH) Calculate how many moles are present in 34g of Ca(OH)

The formula indicates 1 Mg atom, 2 O atoms, and 2 H atoms.
$$
\text{Molar mass of Mg(OH)}_2 = (1 \times \text{Mg}) + (2 \times \text{O}) + (2 \times \text{H})
$$
$$
\text{Molar mass} = (1 \times 24.31 \, \text{g/mol}) + (2 \times 16.00 \, \text{g/mol}) + (2 \times 1.01 \, \text{g/mol})
$$
$$
\text{Molar mass} = 24.31 \, \text{g/mol} + 32.00 \, \text{g/mol} + 2.02 \, \text{g/mol}
$$
$$
\text{Molar mass} = 58.33 \, \text{g/mol}
$$
The molar mass of $\text{Mg(OH)}_2$ is $\boxed{\text{58.33 g/mol}}$.

The formula indicates 1 Ca atom, 2 O atoms, and 2 H atoms.
$$
\text{Molar mass of Ca(OH)}_2 = (1 \times \text{Ca}) + (2 \times \text{O}) + (2 \times \text{H})
$$
$$
\text{Molar mass} = (1 \times 40.08 \, \text{g/mol}) + (2 \times 16.00 \, \text{g/mol}) + (2 \times 1.01 \, \text{g/mol})
$$
$$
\text{Molar mass} = 40.08 \, \text{g/mol} + 32.00 \, \text{g/mol} + 2.02 \, \text{g/mol}
$$
$$
\text{Molar mass} = 74.10 \, \text{g/mol}
$$

The formula for the number of moles ($n$) is:
$$
n = \frac{\text{given mass (m)}}{\text{molar mass (M)}}
$$
$$
n = \frac{34 \, \text{g}}{74.10 \, \text{g/mol}}
$$
$$
n \approx 0.4588 \, \text{mol}
$$
Approximately $\boxed{\text{0.459 moles}}$ are present in 34g of $\text{Ca(OH)}_2$.