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Karl Casas · Answer 1 · 2020-10-28T17:29:31+0000

Answer:

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Step-by-step explanation:

Molarity of the solution = 2.20 M

$Molarity=(number\:of\:moles)/(Volume\:of\:Solution\:in\:L)\\\\Number\:of\:moles=Molarity*(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35*10^(-3)L\\\\Number\:of\:moles=2.20*35*10^(-3)=77*10^(-3)\:moles\:of\:HNO_(3)$

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

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