12 x 10^23 atoms, i hope this helps
Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
