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# What volume is occupied by 0.35 mol of helium gas at SATP?

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#### Explanation

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#### Explanation:

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1
Maxwell Share
Jul 18, 2017

$\text{7.84 L}$

#### Explanation:

I will assume that you mean $\text{STP (Standard Temperature and Pressure)}$.

Since we know that $\text{1 mole}$ of an ideal gas occupies $\text{22.4 L}$ at $\text{STP}$, then can establish the following relationships:

color(white)(------)color(orange)[("22.4 L")/("1 mole") or ("1 mole")/("22.4 L")

Knowing that, we can solve for the volume occupied by the helium gas,

(0.35 cancel"mole")/(1) * ("22.4 L")/(1 cancel"mole") = color(blue)"7.84 L"

as $\textcolor{b l u e}{\text{7.84 L}}$

$- - - - - - - - - - - - - - - - - - - -$

The answer makes sense because if $\text{1 mole}$ occupies $\text{22.4 L}$, then $\text{0.35 moles}$ (roughly a third of the moles) will occupy a third of "22.4 L" ("if we round down 22.4 L to 21 L")

$\textcolor{w h i t e}{- - - - - - - -} \frac{21}{3} \approx \text{7 L}$

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