1.What are the final concentrations of all the ions when the following are mixed? 50 mL of 0.12 M Fe(NO3)3\text{Fe(NO}_3)_3Fe(NO3)3, 100 mL of 0.10 M FeCl3\text{FeCl}_3FeCl3 and 100 mL of 0.26 M Mg(NO3)2\text{Mg(NO}_3)_2Mg(NO3)2.a.[Fe3+]=0.064[\text{Fe}^{3+}] = 0.064[Fe3+]=0.064 M, [NO3−]=0.28[\text{NO}_3^-] = 0.28[NO3−]=0.28 M, [Cl−]=0.12[\text{Cl}^-] = 0.12[Cl−]=0.12 M, [Mg2+]=0.04[\text{Mg}^{2+}] = 0.04[Mg2+]=0.04 Mb.[Fe3+]=0.064[\text{Fe}^{3+}] = 0.064[Fe3+]=0.064 M, [NO3−]=0.28[\text{NO}_3^-] = 0.28[NO3−]=0.28 M, [Cl−]=0.12[\text{Cl}^-] = 0.12[Cl−]=0.12 M, [Mg2+]=0.104[\text{Mg}^{2+}] = 0.104[Mg2+]=0.104 Mc.[Fe3+]=0.04[\text{Fe}^{3+}] = 0.04[Fe3+]=0.04 M, [NO3−]=0.28[\text{NO}_3^-] = 0.28[NO3−]=0.28 M, [Cl−]=0.12[\text{Cl}^-] = 0.12[Cl−]=0.12 M, [Mg2+]=0.104[\text{Mg}^{2+}] = 0.104[Mg2+]=0.104 Md.[Fe3+]=0.064[\text{Fe}^{3+}] = 0.064[Fe3+]=0.064 M, [NO3−]=0.8[\text{NO}_3^-] = 0.8[NO3−]=0.8 M, [Cl−]=0.12[\text{Cl}^-] = 0.12[Cl−]=0.12 M, [Mg2+]=0.104[\text{Mg}^{2+}] = 0.104[Mg2+]=0.104 MLogin to continueOnly logged in users canattempt or see the solution.
