Question

How many moles of cobalt III bromide are present in 4.34 grams of this compound?

Answer 1

Well, what is the formula mass of `CoBr_3`?

Explanation:

I make it `298.65*g*mol^-1`...could you calculate this yourself?

`n_"cobaltic bromide"=(4.43*g)/(298.65*g*mol^-1)~=0.015*mol`