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3 years ago

5

Assuming conditions of standard temperature and pressure, if you have a sample of a noble gas that weighs 12.69 g, in a containe

r with a volume of 14.09 L, which noble gas is it? (R = 0.0821 L∙atm/mol∙K)

1 answer:

VARVARA [1.3K]3 years ago

5 0

Answer:

Noble gas is Ne

Explanation:

Let's make the expression for the Ideal Gases Law at STP

Temperature = 273K

Pressure = 1 atm

1 atm . 1.409L = n . 0.0821 L∙atm/mol∙K . 273K

1.409 L.atm / (0.0821 L∙atm/mol∙K . 273K) = n

0.0628 mol = n

Let's calculate the molar mass to find out the noble gas

1.269 g / 0.0628 m = 20.2 g/m

BE CAREFUL BECAUSE THIS ARE WRONG IN THE STATEMENT.

12.69 g

14.09 L

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