a sealable bag holds 3.79 l of gas. how many moles of h2 gas are needed to fill the bag at 1.0 atm and 293…

Explanation:

Step1: Identify given values

$P = 1.0\ atm$, $V=3.79\ L$, $T = 293\ K$, $R=0.0821\ \frac{L\cdot atm}{mol\cdot K}$

Step2: Rearrange ideal - gas law for $n$

The ideal - gas law is $PV = nRT$. Rearranging for $n$ gives $n=\frac{PV}{RT}$.

Step3: Substitute values into formula

$n=\frac{1.0\ atm\times3.79\ L}{0.0821\ \frac{L\cdot atm}{mol\cdot K}\times293\ K}$ $n=\frac{3.79}{0.0821\times293}$ $n=\frac{3.79}{24.0553}\approx0.157\ mol$

Answer:

$0.157$