Answer:
3.6 × 10²⁴ atoms of O
Explanation:
Let's consider the molecular formula of silver nitrate: AgNO₃.
We can establish the following relations:
The atoms of oxygen in 2.0 moles of silver nitrate are:
[tex]2.0molAgNO_{3}.\frac{6.02\times10^{23}moleculeAgNO_{3} }{1molAgNO_{3}} .\frac{3atomO}{1moleculeAgNO_{3}} =3.6\times10^{24}atomO[/tex]