A persons lungs can hold 7.0 L of air at 315 K (normal body temperature) and 101 kpa (normal atmospheric pressure). Given that air is 25% oxygen, find the number of oxygen molecules in the person’s lungs.
N = 3.09 x 10 exp 20 particlesN = 4.09 x 10 exp 15 particlesN = 4.067 x 10 exp 25 particlesN = 1.99 x 10 exp 25 particles
pls help #7?
2020-11-03 1:21 pm
回答 (1)
2020-11-03 1:57 pm
✔ 最佳答案
Consider the air in the person's lung:Pressure, P = 101 kPa
Volume, V = 7.0 L
No. of moles, n = ? mol
Gas constant, R = 8.314 L kPa / (mol K)
Temperature, T = 315 K
Gas law: PV = nRT
Then, n = PV/(RT)
Moles of air, n = 101 × 7.0 / (8.314 × 315) mol = 0.27 mol
Avogadro constant = 6.022 × 10²³ /mol
No. of O₂ molecules = (0.27 mol) × (25%) × (6.022 × 10²³ /mol) = 4.065 × 10²²
The answer: None of the four options
收錄日期: 2021-04-25 14:30:15
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