Dalton's law states that the total pressure, $P_{\text{t}\text{o}\text{t}\text{a}\text{l}},$ of

a mixture of gases in a container equals the sum of

the pressures of each individual gas:

$P_{\text{t}\text{o}\text{t}\text{a}\text{l}}=P_1+P_2+P_3+$dots

The partial pressure of the first component, $P_1,$ is

equal to the mole fraction of this component, $x_1,$

times the total pressure of the mixture:

$P_1=x_1{P}_{\text{t}\text{o}\text{t}\text{a}\text{l}}$

The mole fraction, $x,$ represents the concentration

of the component in the gas mixture, so

$x_1=\frac{\text{m}\text{o}\text{l}\text{e}\text{s}of\text{c}\text{o}\text{m}\text{p}\text{o}\text{n}\text{e}\text{n}\text{t}1}{\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{o}\text{l}\text{e}\text{s}in\text{m}\text{i}\text{x}\text{t}\text{u}\text{r}\text{e}}$

Three gases $(8\mathrm{.}00g$ of methane, $C{H}_4,18\mathrm{.}0g$ of ethane, $C_2{H}_6,$ and an unknown amount of propane, $C_3{H}_8)$ were

added to the same $10\mathrm{.}0-L$ container. At $23\mathrm{.}0C,$ the total pressure in the container is $4\mathrm{.}40$atm. Calculate the partial

pressure of each gas in the container.

Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of

methane first, then ethane, then propane.

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Part B

A gaseous mixture of $O_2$ and $N_2$ contains $33\mathrm{.}8\%$ nitrogen by mass. What is the partial pressure of oxygen in the

mixture if the total pressure is $645mmHg?$

Express you answer numerically in millimeters of mercury.

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$P_{\text{o}\text{x}\text{y}\text{g}\text{e}\text{n}}=,mmHg$