Question

1. how many grams of bromine are there in 1.00 x 10^9 bromine atoms?

Explanation:

Step1: Recall Avogadro's number

Avogadro's number ($N_A$) is $6.022\times 10^{23}$ atoms/mol.

Step2: Calculate the number of moles of bromine

The number of moles ($n$) of bromine is calculated using the formula $n=\frac{N}{N_A}$, where $N = 1.00\times 10^{9}$ atoms. So $n=\frac{1.00\times 10^{9}}{6.022\times 10^{23}}$ mol.
$$n=\frac{1.00\times 10^{9}}{6.022\times 10^{23}}\text{ mol}\approx1.66\times 10^{-15}\text{ mol}$$

Step3: Find the molar mass of bromine

The molar mass ($M$) of bromine ($Br$) is approximately $79.904$ g/mol.

Step4: Calculate the mass of bromine

The mass ($m$) of bromine is calculated using the formula $m = n\times M$. So $m=(1.66\times 10^{-15}\text{ mol})\times(79.904\text{ g/mol})$.
$$m=(1.66\times 10^{-15})\times79.904\text{ g}\approx1.32\times 10^{-13}\text{ g}$$

Answer:

$1.32\times 10^{-13}$ g