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slega [8]
4 years ago
11

How many moles of sodium iodide are needed to remove 5.95×10−6 mol of o3?

Chemistry
1 answer:
igor_vitrenko [27]4 years ago
8 0
You have to find the reaction involving sodium iodide and ozone or O₃. If you search in the internet, the reaction would be

O₃<span>(g) + 2 NaI(aq) + H</span>₂O(l) → O₂<span>(g) + I</span>₂<span>(g) + 2 NaOH(aq) 
</span>
Using the stoichiometric coefficients in the reaction, the solution would be

5.95×10⁻⁶ mol O₃ * (2 mol NaI/1 mol O₃) = 1.19×10⁻⁵ mol NaI
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pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

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More about buffer: brainly.com/question/4177791

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1 year ago
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True

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