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Nadya [2.5K]
3 years ago
6

When 11.12 g of neon is combined in a 100 L container at 80oC with 9.59 g of argon, what is the mole fraction of argon?

Chemistry
1 answer:
PolarNik [594]3 years ago
3 0
The correct answer for the question that is being presented above is this one: "<span>0.3."

Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
</span>
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Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

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Density (D) = mass (m) / Volume (V)

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11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

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V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

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Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

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