2.สารละลายชนิดหนึ่งมี [H3O+] = 1x10

2.สารละลายชนิดหนึ่งมี [H3O+] = 1x10-2 mol/dm3 [OH–] จะมีค่าเท่าใด
วิธีทำ Kw = [H3O+] [OH–]
1x10-4 = 1x10-2 [OH–]
[OH–] = 1x10-14 / 1x10-2
= 1x10-12 mol/dm3

ผลลัพธ์ (อังกฤษ) 1:

2. one of the solution is [H3O +] = 1x10-2 mol/dm3, the [OH-] will have any value.How do the Kw = [H3O +] [OH –] 1x10-4 = 1x10-2 [OH–] [OH–] = 1x10-14 / 1x10-2 = 1x10-12 mol/dm3

ผลลัพธ์ (อังกฤษ) 2:

2. One solution is [H3O +] = 1x10-2 mol / dm3 [OH-] is worth, how much
do Kw = [H3O +] [oh]
= 1x10-2 1x10-4 [oh]
[oh]. = 1x10-14 / 1x10-2
= 1x10-12 mol / dm3.

ผลลัพธ์ (อังกฤษ) 3:

2. Solution of a [] = H3O 1x10-2 mol / dm3 [] will be OH - how much value
how to do Kw = [H3O] [-]
OH 1x10-4 = 1x10-2 [-]
OH OH [-]. = 1x10-14 / 1x10-2
= 1x10-12 mol / dm3


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