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A tuman lung at maximum capacity has a volume of 3,0 liters. if the partial pressure of oxygen in the air is.21 1 kiag is 29g k. how many moles of oxygen are in the lung?

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Answer:

To calculate the number of moles of oxygen in the lung, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

We are given that the volume of the lung is 3.0 liters, or 0.003 m^3. We can assume that the pressure is equal to atmospheric pressure, which is approximately 101.3 kPa. The temperature can be converted from Celsius to Kelvin by adding 273.15, so T = 20.0 + 273.15 = 293.15 K.

We can rearrange the ideal gas law to solve for n:

n = PV/RT

Substituting the values we have, we get:

n = (101.3 kPa)(0.003 m^3)/(8.314 J/(mol K))(293.15 K)

n = 0.000372 mol

Therefore, there are approximately 0.000372 moles of oxygen in the lung.

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