What are the no of molecules of Iron iii in 285 g of FePO4
Answers
Answered by
0
Answer:
1.137 ×10^24 Fe3+
Explanation:
150.85 gm FePO4= 6.02×10^23 Fe3+
285 gm FePO4 = (6.02×10^23×285) /150.85 Fe3+
= 1.137 ×10^24 Fe3+
Similar questions