Answer to Question #92001 in General Chemistry for jimbo
How many grams of solute are present in
845 mL
of
0.730 M KBr?
1
2019-07-26T03:16:56-0400
V(s) = 845 ml = 0.845 l
c(KBr) = 0.730 M
m(KBr) - ?
n(KBr) = c*V =0.617 mol
m(KBr) = 73.4 g
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