J.R. S. answered 10/15/20
Tutor
5.0
(145)
Ph.D. University Professor with 10+ years Tutoring Experience
pOH = -log[OH-]
[OH-] = 1x10-10.17
[OH-] = 6.76x10-11 M
Itzel R.
asked 10/14/20J.R. S. answered 10/15/20
Ph.D. University Professor with 10+ years Tutoring Experience
pOH = -log[OH-]
[OH-] = 1x10-10.17
[OH-] = 6.76x10-11 M
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