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Question #7acfa

↳Redirected from "What is the oxidation state of copper in #"CuSO"_4#?"

1 Answer

Soumalya Pramanik

Jan 29, 2018

#4.643856#

Explanation:

#log 25/log 2#

= #(log_(10) 25)/(log_(10) 2)#

= #log_2 25# [Base changing rule of logarithms.]

= #4.643856# (approx)
Hope this helps.

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