Answer:
copper(I) bromide: CuBr
copper(I) oxide: Cu₂O
copper(II) bromide: CuBr₂
copper(II) oxide: CuO
iron(III) bromide: FeBr₃
iron(III) oxide: Fe₂O₃
lead(IV) bromide: PbBr₄
lead(IV) oxide: PbO₂
I hope this helped you! Brainliest would be greatly appreciated.
1 mol of Br = 79.9 g
15.7 g / 79.9 g = 0.196 moles of atoms
1.27moles of K3PO4 can be formed.
