A 46.2-ml, .568 M calcium nitrate solution is mixed with 80.5 ml of 1.396 M calcium nitrate[Ca(NO3)2] solution. calculate the concentration of the final solution.
can someone explain me the steps to this problem. i would greatly appreciate it.
2006-08-28 05:57:43 · 2 answers · asked by Paul C 1 in Science & Mathematics ➔ Chemistry
when u mix solutions the type that you can use is
C1*V1 +C2*V2=C3*V3
C1= the concentration of the first solution=.568M
V1= the volume of the first solution=46.2ml
C2=the concentration of the second solution=1.396M
V2=the volume of the second solution=80.5ml
V3=the volume of the final solution=V1+V2
C3=?
So you have all the information u need to replace in the equation and find C3
If u need more help add a detail and i will help u!
2006-08-28 06:12:26 · answer #1 · answered by girl24gr 3 · 0⤊ 0⤋
Concentration is moles/volume, so you need to calculate the total number of moles of Calcium Nitrate present, then divide by the total solution volume. In each of the initial solutions, the number of moles of Calcium Nitrate is:
mol = Conc * Vol
To finish, just add the number of moles from the two inital solutions, then divide by the total volume. Since there is no reaction taking place, there is no stoichiometry needed.
2006-08-28 13:09:09 · answer #2 · answered by Duluth06ChE 3 · 0⤊ 0⤋