Answer: 45.983 g CBr₄
Explanation:
To convert from moles to grams, you know that we will need molar mass and Avogadro's number.
Avogadro's number: 6.022×10²³ molecules/mol
Molar mass: 331.627 grams/mol
Now that we have what we need, you can use these to solve for grams. [tex]8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g[/tex]
Our final answer is 45.983 g CBr₄.
