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3 years ago

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A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

al pressure is 525 mmHg?
PLEASE HELP, will mark brainliest!!!

1 answer:

kondor19780726 [428]3 years ago

7 0

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

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Then:

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3 years ago

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3 years ago

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$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

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$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago