Question

4-how many atoms of oxygen are present in 300 g of calcium carbonate[CaCO3]. Ca=40 u. C=12 u. I=16 u *
2 points
54.207*10^23
6.207*10^23
12.207*10^23
22.2*10^23​

Answer 1

Explanation:

1 mole contains------> 3 oxygen atoms

3 moles contains------> 9 oxygen atoms

so the answer is =9×6.023×10²³

=54.207×10²³