Chemistry help.....

Question

What concentration of NO3_ results 871 mL of 0.769 M NaNO3 is mixed with 919 mL of 0.505 M Can(NO3)2

Answer 1

1) 871 ml x 1 L/1000 ml x 0.769 mol NO₃^-_/L = 0.6698 moles NO₃^-

2) 919 ml x 1 L/1000 ml x 0.505 mol Ca(NO₃)₂/L x 2 mol NO₃^-/mol Ca(NO₃)₂ = 0.9282 mol NO₃^-

3) Total moles NO₃^- = 0.6698 mol + 0.9282 mol = 1.598 mol NO₃^-

4) Final volume = 871 ml + 919 ml = 1790 ml = 1.790 L

5) Final [NO₃^-] = 1.598 mol NO₃^-/1.790 L = 0.893 M (to 3 significant figures)