Calculate the number of atoms in 37.1 grams of LiBr.

- Use [tex]$6.022 \times 10^{23} \, \text{mol}^{-1}$[/tex] for Avogadro's number.
- Report your answer in scientific notation, using the multiplication symbol, [tex]$\times$[/tex], not the letter [tex]$x$[/tex].
- Your answer should have three significant figures.

To calculate the number of atoms in 37.1 grams of LiBr, follow these steps:

1. Find the Molar Mass of LiBr:
- Lithium (Li) has an approximate atomic mass of 6.94 grams per mole.
- Bromine (Br) has an approximate atomic mass of 79.90 grams per mole.
- Therefore, the molar mass of LiBr is:
[tex]\[
6.94 \, \text{g/mol} + 79.90 \, \text{g/mol} = 86.84 \, \text{g/mol}
\][/tex]

2. Calculate the Number of Moles of LiBr:
- Use the formula:
[tex]\[
\text{Moles of LiBr} = \frac{\text{mass of LiBr}}{\text{molar mass of LiBr}}
\][/tex]
- Substitute the given values:
[tex]\[
\text{Moles of LiBr} = \frac{37.1 \, \text{g}}{86.84 \, \text{g/mol}}
\][/tex]
- The calculated moles of LiBr is approximately 0.427 (rounded to three significant figures).

3. Calculate the Number of LiBr Molecules:
- Use Avogadro's number, [tex]$6.022 \times 10^{23} \, \text{mol}^{-1}$[/tex], to find the number of molecules in the calculated moles:
[tex]\[
\text{Number of LiBr molecules} = 0.427 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1}
\][/tex]
- The number of molecules is approximately [tex]$2.57 \times 10^{23}$[/tex] (rounded to three significant figures).

4. Calculate the Total Number of Atoms in LiBr:
- Each molecule of LiBr consists of 2 atoms (one lithium atom and one bromine atom).
- Therefore, the total number of atoms is:
[tex]\[
\text{Number of atoms} = 2 \times \text{Number of molecules}
\][/tex]
- So, the number of atoms is:
[tex]\[
2 \times 2.57 \times 10^{23} = 5.15 \times 10^{23} \text{ atoms}
\][/tex]

So, the total number of atoms in 37.1 grams of LiBr is [tex]$5.15 \times 10^{23}$[/tex].