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Chemistry 14 Online
OpenStudy (goalieboy):

will post in attachments

OpenStudy (goalieboy):

OpenStudy (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

OpenStudy (michele_laino):

please*

OpenStudy (goalieboy):

that would be 5

OpenStudy (michele_laino):

correct!

OpenStudy (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

OpenStudy (goalieboy):

292.5

OpenStudy (michele_laino):

so, what is the right option?

OpenStudy (goalieboy):

C

OpenStudy (goalieboy):

I am going to open up a new question

OpenStudy (michele_laino):

that's right!

OpenStudy (michele_laino):

ok!

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