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Chemistry 18 Online
OpenStudy (anonymous):

A chemist requires 0.387 mol Na2CO3 for a reaction. How many grams does this correspond to?

OpenStudy (anonymous):

@zepp

OpenStudy (zepp):

y u call me when dad calls me to go eat >.>

OpenStudy (anonymous):

i got this, but just one thing...would the total mass be Na2 + C + O2 together?

OpenStudy (anonymous):

what? :O nooo

OpenStudy (zepp):

>Get the molar mass of the thingy >Multiply this mass by the # of mole in the problem >Fun brb

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

\[.387molNa_2CO_3*\frac{2(23)+12+2(16)gramsNa_2CO_3}{1molNa_2CO_3}=.387*90\]

OpenStudy (anonymous):

\[.387*90=34.83gramsNa_2CO_3\]

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