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Chemistry 6 Online
OpenStudy (anonymous):

PLEASE HELP ! . Calculate the mass of KI in grams required to prepare 5.00 x 102 mL of a 2.80 M solution.

OpenStudy (doc.brown):

Does M mean mol/L?

OpenStudy (anonymous):

yes

OpenStudy (doc.brown):

\[K = 39.0983\frac{g}{mol}\]\[I = 126.90447\frac{g}{mol}\]\[5.00\times10^{2}mL\times\frac{10^{-3}L}{mL}\times\frac{2.80mol}{L}\times\frac{KIg}{mol}\]

OpenStudy (anonymous):

yes thank you ! that set up is really helpful

OpenStudy (doc.brown):

Post the answer when you get it.

OpenStudy (anonymous):

232.40 grams

OpenStudy (doc.brown):

Right! \[2.32\times10^{2}g\] three sig.figs from the 5.00 and 2.80 in the question well done

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