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:
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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Answer:
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- <em>So, the ΔH in kJ of heat released per mole of NH₃ formed is (100.4 kJ/2) = - 50.2 kJ.</em>
Answer:
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