Tumia English please

The molar mass of $NH_3 = 14 + (3 \times 1) = 17$ g/mol.
Graham's Law states:

$$\frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}}$$

Where $Rate_1$ and $Rate_2$ are the rates of diffusion, and $M_1$ and $M_2$ are the molar masses.

We are given that $Rate_{NH_3} = 1.41 \times Rate_{XH_3}$. So, $\frac{Rate_{NH_3}}{Rate_{XH_3}} = 1.41$.
The molar mass of $XH_3$ is $M_{XH_3} = X + (3 \times 1) = X + 3$.

$$1.41 = \sqrt{\frac{X + 3}{17}}$$

Square both sides of the equation:

$$(1.41)^2 = \frac{X + 3}{17}$$ $$1.9881 = \frac{X + 3}{17}$$

Multiply both sides by 17:

$$1.9881 \times 17 = X + 3$$ $$33.7977 = X + 3$$

Subtract 3 from both sides:

$$X = 33.7977 - 3$$ $$X = 30.7977$$

The atomic mass of X is approximately 31.

The atomic mass of element X is $\boxed{\text{31}}$.

6a) An iron ore was found to contain 7g of iron and 3g of oxygen. (Fe=56, O=16). Find its empirical formula.

For iron (Fe):

$$\text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Atomic mass of Fe}} = \frac{7\, \text{g}}{56\, \text{g/mol}} = 0.125\, \text{mol}$$

For oxygen (O):

$$\text{Moles of O} = \frac{\text{Mass of O}}{\text{Atomic mass of O}} = \frac{3\, \text{g}}{16\, \text{g/mol}} = 0.1875\, \text{mol}$$

The smallest number of moles is 0.125 mol.
For Fe:

$$\frac{0.125}{0.125} = 1$$

For O:

$$\frac{0.1875}{0.125} = 1.5$$

To eliminate the 0.5, multiply both ratios by 2.
For Fe: $1 \times 2 = 2$
For O: $1.5 \times 2 = 3$
The ratio of Fe:O is 2:3.

The empirical formula is $\boxed{\text{Fe}_2\text{O}_3}$.

6b) Write a balanced equation for the reaction between magnesium and steam.

The reaction between magnesium (Mg) and steam (H$_2$O(g)) produces magnesium oxide (MgO) and hydrogen gas (H$_2$).

$$\text{Mg(s)} + \text{H}_2\text{O(g)} \longrightarrow \text{MgO(s)} + \text{H}_2\text{(g)}$$

This equation is already balanced, with one Mg, one O, and two H atoms on both sides.

The balanced equation is $\boxed{\text{Mg(s)} + \text{H}_2\text{O(g)} \longrightarrow \text{MgO(s)} + \text{H}_2\text{(g)}}$.

7a) Carbon (iv) Oxide can undergo the changes below.
CO$_{2(s)}$ $\xrightarrow{A}$ CO$_{2(g)}$
CO$_{2(g)}$ $\xrightarrow{B}$ CO$_{2(s)}$
What are processes A and B?

A: The process of a solid changing directly into a gas without passing through a liquid phase is called sublimation.
B: The process of a gas changing directly into a solid without passing through a liquid phase is called deposition (or reverse sublimation).

A: $\boxed{\text{Sublimation}}$
B: $\boxed{\text{Deposition}}$

7b) Suggest one use of carbon (iv) oxide that utilizes processes A and B.

One use of carbon (iv) oxide that utilizes both processes A and B is as a refrigerant or cooling agent, particularly in the form of dry ice. Dry ice is formed by depositing gaseous CO$_2$ (process B) and then used to cool items by subliming into a gas (process A).

One use is $\boxed{\text{As a refrigerant (e.g., dry ice)}}$.