Question

the kinetic energy of 16 grams of oxygen at 27℃​

Answer 1

Answer:

K.E OD O2 gas = 1870.65 J

Step-by-step explanation:

K.E.= 3/2 nRT

where n=no. of mole

          R=universal gas constant=8.314 J/K/mol

           T= temperature(KELVIN)=(273+27)K=300K

           no. of mole=16/32=1/2=0.5

K.E. = 3/2*1/2mol*8.314J/K/mol*300K

      = 3*8.314*75 J

      = 24.942*75 J

      = 1870.65 J

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