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11 days ago

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A gas mixture called heliox, 6.11% o2 and 93.89% he by mass, is used in scuba tanks for descents more than 65 m below the surfac

e. calculate the mole fractions of he and o2 in this mixture. he is 0.9980

2 answers:

castortr0y [3K]11 days ago

4 0

Result: The mole fraction for He in this mixture is 0.992.

The mole fraction for $O_2$ in the mixture equals 0.008.

Explanation:

The mixture, known as heliox, consists of 6.11% $O_2$ and 93.89% helium by mass.

The mass percentage of oxygen is 6.11%

Thus, in a total of 100 grams, 6.11 grams represent oxygen: $\frac{100\times 6.11}{100}=6.11 grams$

The mass percentage of helium amounts to 93.89%

This means that in a 100-gram mixture, 93.89 grams corresponds to helium: $\frac{100\times 93.89}{100}=93.89 grams$

The mole fraction for component 'x' in a mix of 'x' and 'y' equals $=\chi _x=\frac{n_x}{n_x+n_y}$

$n_x$ = moles of gas 'x'= $\frac{\text{mass of gas 'x'}}{\text{Molar mass of gas 'x'}}$

$n_y$ = moles of gas 'y'= $\frac{\text{mass of gas 'y'}}{\text{Molar mass of gas 'y'}}$

With 6.11 grams of oxygen in the mixture, we can find the moles of oxygen gas:

$n_{O_2}=\frac{6.11 g}{16g/mol}=0.3818 moles$

Likewise, with 93.89 grams of helium, we find the moles for that gas:

$n_{He}=\frac{93.89 g}{2g/mol}=46.945 moles$

$=\chi _{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{He}}=\frac{0.3818 mol}{0.3818 mol+46.945 mol}=0.008$

$=\chi _{He}=\frac{n_{He}}{n_{O_2}+n_{He}}=\frac{46.945 mol}{0.3818 mol+46.945 mol}=0.9919\approx 0.992$

The mole fraction of helium in the mixture is 0.992.

Conversely, the mole fraction of $O_2$ in the mixture is 0.008.

Alekssandra [3K]11 days ago

3 0

098.9 that’s what I think it is

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