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Mathematics 22 Online
OpenStudy (anonymous):

Hi, I need some word problem help... See Atttached

OpenStudy (anonymous):

k

OpenStudy (anonymous):

@ankit042 @AravindG @Grazes

OpenStudy (anonymous):

hi

OpenStudy (aravindg):

Hi :) pH=-log[H+] -->e^pH=1/[H+] Use it for first question.

OpenStudy (anonymous):

ok how bout the next 2?

OpenStudy (aravindg):

For neutral solution pH=7 or [H+]=10^-7

OpenStudy (anonymous):

thankyou soo much

OpenStudy (aravindg):

yw :)

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