D=m/v ⇒ m=d*v
d=density
m=mass
v=volume
d(ether)=0.71 gr/cm³=0.71 gr/ ml
v=130 ml
m=d*v
m=0.71 gr/ml*(130 ml)=92.3 g
Solution: m=92.3 g
There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
Learn more: brainly.com/question/9743981
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