contestada

1. The total translational kinetic energy of the molecules of a sample of gas at
437 K is 32300 J. How many moles does the sample comprise?

moles of the sample:_____mol

2. Find the average translational kinetic energy of a single molecule.

average translational kinetic energy:_____J

Respuesta :

msakhile

Answer:

1. KE = [tex]\frac{3}{2}[/tex]nRT

    32300 J = [tex]\frac{3}{2}[/tex] (n)×[tex]\frac{8.3145 J}{mol.K}[/tex]×437K

    32300 J  = [tex]\frac{5449.827J(n)}{mol}[/tex]

                n = [tex]\frac{32300Jmol}{5449.827J}[/tex]

                n = [tex]5.9 mol[/tex]

2. KE =  [tex]\frac{3}{2}[/tex]KT

     [tex]= \frac{3}{2}(\frac{1.38x10^{-23}J }{K.molecules})(437K)[/tex]

   [tex](\frac{9.0459x10x^{-21}J }{molecule} ) (1 molecule)[/tex]

    [tex]= 9.0459x10^{-21}J[/tex]

Explanation:

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