Question

9.11 g of CuBr2 contains how many bromide (Br-) ions?

Answer 1

Cu --> 63.5 g/mol

Br --> 80 g/mol
63.5 + 160 = 223.5 g/mol
so
# of moles = 9.11/223.5 * 6*10^23
= .245 * 10^23 molecules
twice that many Br- ions so
.49 * 10^23 or 4.9*10^22 ions