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Chemistry 12 Online
OpenStudy (anonymous):

Calculate the root mean square velocity of nitrogen molecules at 25 C. a. 515 m/s b. 149 c . 729 got wrong d. 297 m/s

OpenStudy (aaronq):

\[\sqrt{\frac{3 RT }{ M }}\] M is molar mass

OpenStudy (anonymous):

Okay so T = 298 K, M = 14 g and r = .08

OpenStudy (anonymous):

i get something like 2.25

OpenStudy (abb0t):

nitrogen is N2

OpenStudy (anonymous):

Got it 1) 23 degree c = 23+273 =296 Kelvin 2) Nitrogen (M 28) mass in this equation ( /1000) = 0.028 Kg 3) RMS velocity of N at 23 degrees c = √ (3 x R x T/M ) √ (3 x 8.31 x 296/0.028) 515

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