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Calculate kinetic energy of `4g N_(2)` at `-13^(@)C`. |
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Answer» We know, `KE=(3)/(2)xxnRT` Number of moles of `4 g N_(2)=(4)/(28)g mol^(-1)` `T=-13^(@)C=260.15K` `R=8.314 J K^(-1) mol^(-1)` `:. K.E.=(3)/(2)xx(4)/(28)xx8.314xx260.15` `=463.47 J mol^(-1)` |
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