What volume of a 0.300 M solution of aluminum fluoride would be required to obtain 14.0 g of aluminum fluoride
Molar mass of aluminium fluoride "=83.9767g\/mol"
"\\frac{V}{1000}\u00d70.3=\\frac{14}{83.9767}"
"V=555.71cm^3"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment