Which are the physical properties of water

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Ответ:

shelbyetheridge1011s

04.02.2021

Chemistry

$m_{air}=0.947g$

$n_{O_2} =0.00686molO_2$

Explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

$T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol$

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

$m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g$

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

$n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2$

Best regards.

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