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Astha pandey Asked a Question
January 24, 2021 3:17 pmpts 30 pts
(29.6) We have, AG-2.303 RT log K eq 1x10 =-2.303 x 8.314 x 310 x log 0.01 = - 5935.6 x log(1 x 10-5) = 5935.6x5 29,678 J/mol =29.6 kJ/mol
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  • Astha pandey
    plz explain this also. How 7/5 came??
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    Priyanshu kumar
    In scientific calculator you can calculate this value or this value is given in your question
  • Astha pandey
    Thank You 😇
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  • Priyanshu kumar thankyou
    see this
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    Priyanshu kumar
    please mark the answer