Answer:
There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Explanation:
Molarity of the solution = 2.20 M
[tex]Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}[/tex]
Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
