Question

1. a 4.0 - liter container has two gases inside, neon and argon. it is known that at 18 °c, the total pressure of the combined gases is 0.850 atm. if it is known that there are 0.100 moles of neon in the container, how many moles of argon is in the container?

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a) 0.042 moles argon
b) 0.142 moles argon
c) 4.98 moles argon
d) 6.93 moles argon

1. how many moles of oxygen must be placed in a 3.00 - liter container in order to exert a pressure of 2.00 atmospheres at 25 °c?

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a) 0.245 moles
b) 4.08 moles
c) 146.8 moles
d) 21778 moles

Explanation:

Step1: Convert temperature to Kelvin

$T = 18^{\circ}C+273 = 291K$ for the first - problem. For the second problem, $T = 25^{\circ}C + 273=298K$. The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant ($R=0.0821\ L\cdot atm/(mol\cdot K)$).

Step2: For the first problem, find the total number of moles

Using $PV = nRT$, we have $n=\frac{PV}{RT}$. Given $P = 0.850\ atm$, $V = 4.0\ L$, $T = 291K$, and $R = 0.0821\ L\cdot atm/(mol\cdot K)$. Then $n=\frac{0.850\times4.0}{0.0821\times291}\approx0.142\ mol$.

Step3: Find the moles of argon

We know there are $n_{Ne}=0.100\ mol$ of neon. Let $n_{Ar}$ be the moles of argon. Since $n = n_{Ne}+n_{Ar}$, then $n_{Ar}=n - n_{Ne}=0.142 - 0.100=0.042\ mol$.

Step4: For the second problem, find the moles of oxygen

Using $PV = nRT$, and given $P = 2.00\ atm$, $V = 3.00\ L$, $T = 298K$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$. Then $n=\frac{PV}{RT}=\frac{2.00\times3.00}{0.0821\times298}\approx0.245\ mol$.

Answer:

1. A. 0.042 moles argon
2. A. 0.245 moles