Question

Whole resting, an average 70 kg human male
consumes 16.628L of oxygen per hour at
27°C and 100KPa. Number of moles of
oxygen are consumed by the 70 kg man
while resting for 1 hour is 6.7x10-P where P =​

Answer 1

Answer:

Pv=nRT

n=PV over RT

n=0.987 atm (14l) over 0.082 l atom over mol k and (298.15k)

n=0.565 moles

Explanation:

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Answer 2

Answer:

pv=nRt

n=pv over RT

n=0.987 atm (14) over 0.082 I atom over mol k and (298.15km)

n=0.565moles

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