Answer:
The moles of O₂ which are present in the sample are 0.433. (option D)
Explanation:
We should apply the Ideal Gases Law, to solve this:
P . V = n. R. T
1.38 atm . 7.51L = n . 0.082 . 292K
(1.38 atm . 7.51L) / (0.082 . 292K) = n
0.433 moles = n