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Calculate kinetic energy of 5 moles of N...

Calculate kinetic energy of 5 moles of Nitrogen at `27^@C`.

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Kinetic energy = `(3)/(2)nRT`
Where `n=5` moles, `R=8.314 "mol"^(-1)k^(-1)`
`T=27^(@)C+273=300k`
kinetic energy
`E_(k)=(3)/(2)xx5"mol"xx8.314"J mol"^(-1)K^(-1)xx300k=18706.50` J
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