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Originally Posted by redXI
A 46.2-mL, o.568 M calcium Nitrate [Ca(NO3)2] solution i smixed with 80.5 mL of 1.396 M calcium nitrate solution. What is the concentration? :shock:
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use c = n/V to calculate the amount of mmol Ca(NO3)2 of the 46.2 mL 0.568 M solution and the same way for the 80.5 mL 1.396 M Ca(NO3)2 solution.
Add the mols and volumes and use again c = n/V to calculate the final concentration Ca(NO3)2. The answer will be 1.09 M
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Originally Posted by redXI
If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?
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First: what is the balanced reaction equation ?
Then you need to calculate the amount of mol CaCl2 and AgNO3 using c = n/V.
Then determin which is the limiting reagent and use stoichiometry to calculate the amount of mol AgCl formed.
--> convert amount of mol AgCl to g. The answer will be 0.21 g.