Q) The following data.......

Asked by  | 11th Oct, 2009, 09:30: PM

Expert Answer:

           2.303

K =----------------------- log   p0/   2po  -  pt

 

 

        2.303

K =----------------------- log   0.5/   20.5 -  0.6

           100

    =2.2318 X 10-3s-1

now p0   +x    = 6.5

x=  0.15

pso2cl= p0  - x

                =.35

required rate = 2.2318 x 10-3     x        0.35

                          = 7.18  x  10-4  atms-1

 

         t

Answered by  | 12th Oct, 2009, 09:55: AM

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