Question

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Answer 1

Answer:

$V_2=435.4L$

Explanation:

Hello there!

In this case, for this gas problem, we can use the Avogadro's law as a way to understand the volume-mole behavior as a directly proportional relationship:

$\frac{V_2}{n_2}=\frac{V_1}{n_1}$

Thus, given the initial and final moles and the initial volume, we obtain the following final volume:

$V_2=\frac{V_1n_2}{n_1} =\frac{100.L*14.15mol}{3.25mol}\\\\V_2=435.4L$

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