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If the poh of vinegar is 9.45, what is its [oh− ]? round to the nearest hundredth. × 10n m n =

Sagot :

Abidemiokinidn

The [OH] of vinegar to the  nearest hundredth is 2.82 * 10^-5

How to calculate the [OH] of a substance:

The sum of the pH and pOH of a substance is 14. Mathematically

[OH−] = Antilog (pOH)

Given the following parameters

[OH−] = Antilog (9.45)

Substitute

[OH−] = 2.82 * 10^-5

Hence the [OH] of vinegar to the  nearest hundredth is 2.82 * 10^-5

Learn more on pH and pOH here:https://brainly.com/question/13557815

H00dmadeasia66idn

Answer:

The person above me is incorrect

1st blank is 3.55

2nd blank is -10

Step-by-step explanation:

Look at the image (bottom left corner)

View image H00dmadeasia66
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