No, a precipitate will not form when barium nitrate, Ba(NO3)2, reacts with potassium hydroxide, KOH, in aqueous solution because both products are soluble in water.
Answer:
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Answer:
pOH = 0.3
Explanation:
As KOH is a strong base, the molar concentration of OH⁻ is equal to the molar concentration of the solution. That means that in this case:
With that information in mind we can<u> calculate the pOH </u>by using the following formula:
I think you divide something
Na: +1 (sodium)
S: -2 (sulfur)
=> Na2S