If the poh of vinegar is 9.45, what is its [oh− ]? round to the nearest hundredth. × 10n m n =
1 answer:
The [OH] of vinegar to the nearest hundredth is 2.82 * 10^-5
<h3>How to calculate the [OH] of a substance:</h3>
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:brainly.com/question/13557815
You might be interested in
Answer:
5
571
x 38
---------
4,568 <------571×
+3,213x <------571x
------------
36,698
Answer:
20 degree angle
Step-by-step explanation:
Answer: I think its b
Step-by-step explanation:
F(x)= 650(1+.12)^x
f(x) = 650(1.12)^x
