There are 1.20 × 10²³ atoms of Bi are there in a 41.8 g sample.
First, we have to convert 41.8 g to moles using the molar mass of Bi (208.98 g/mol) as the conversion factor.
[tex]41.8 g \times \frac{1mol}{208.98g} = 0.200 mol[/tex]
Next, we will convert 0.200 moles to atoms using Avogadro's number (6.02 × 10²³ atoms/mole) as the conversion factor.
[tex]0.200 mol \times \frac{6.02 \times 10^{23} atoms }{mol} = 1.20 \times 10^{23} atoms[/tex]
There are 1.20 × 10²³ atoms of Bi are there in a 41.8 g sample.
You can learn more about Avogadro's number here: https://brainly.com/question/13302703
