goalieboy
  • goalieboy
will post in attachments
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
goalieboy
  • goalieboy
1 Attachment
Michele_Laino
  • Michele_Laino
again, we have to compute the moles of sodium. So we have to do this computation: \[n = \frac{{115}}{{23}} = ...?\] since \(23\) is the atomic weight of sodium, pleas check the periodic table
Michele_Laino
  • Michele_Laino
please*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

goalieboy
  • goalieboy
that would be 5
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
now, looking at the right side of the chemical rection, we note that number of moles of sodium chloride is equal to number of moles of sodium, so the quantity of produced sodium chloride is: \[m = 5 \cdot 58.5 = ...?\] \(58.5\) being the molecular weight of sodium chloride
goalieboy
  • goalieboy
292.5
Michele_Laino
  • Michele_Laino
so, what is the right option?
goalieboy
  • goalieboy
C
goalieboy
  • goalieboy
I am going to open up a new question
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
ok!

Looking for something else?

Not the answer you are looking for? Search for more explanations.