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nikklg [1K]
3 years ago
5

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

f oxygen in this mixture?
a. 140.0 mm hg
b. 320.0 mm hg
c. 373.0 mm hg
d. 480.0 mm hg
Chemistry
1 answer:
klasskru [66]3 years ago
7 0
Hello!

<span>We have the following statement data:
</span>
Data:
P_{Total} = 800 mmHg
P\% N_{2} = 60\%
P\% O_{2} = 40\%
P_{partial} = ? (mmHg)

<span>As the percentage is the mole fraction multiplied by 100:

</span>P =  X_{ O_{2} }*100

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture? 
</span>
X_{ O_{2} }  =  \frac{P}{100}
X_{ O_{2}} =  \frac{40}{100}
\boxed{X_{ O_{2}} = 0.4}


<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to O_{2} :

\frac{P O_{2} }{P_{total}} = X_O_{2}
\frac{P O_{2} }{800} = 0.4
P_O_{2} = 0.4*800
\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark
<span>
Answer:
</span><span>b. 320.0 mm hg </span>
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