(4x10^-9)^5

Given that Kw = 2.4x10^-14 at 37 celsius, compute the PH of a neutral aqueous solution at 37 celsius...

if i have: 31x^5=1.4x10^-85 what would x equal?

26.3 x 10^3 + 4x10^3 = 26.3+4= =30.3x10^3

6.4x10^3+1.4x10^4+7.5x10^3

4x10^2

(5.2x10^2)x(4x10^4)=

1. There are about 1.4x10^9 cubic kilometers of water on Earth. About 2.5% of water is fresh water.H...

2 x 10^2 4x10^1

6.4x10^3

3+24/:4x10

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