Answer:
The answer is 0.20231 mol BeS
Explanation:
we look for the molecular weight of beryllium and sulfur in the periodic table:
molecular weight of Be = 9.01 g/mol
molecular weight of S = 32.065 g/mol
therefore, the molecular weight of beryllium sulfide is equal to:
molecular weight of BeS = 9.01 + 32.065 = 41.075 g/mol
The number of moles will be equal to:
[tex]n BeS = 8.31 g BeS x\frac{1 mol BeS}{41.075 g BeS}=0.20231 mol BeS[/tex]
