How do I do this net ionic equation in acidic solution?
Cr2+(aq) + (Cr2O7)^2- (aq) --->
Please explain it to me step-by-step.
Thanks
2007-04-18 14:29:47 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Cr2+ ----> Cr3+ + e-
(Cr2O7)2- + 14H+ + 6e- -----> 2Cr3+ + 7H2O
Multiply the first equation by 6 and add it to the second one, missing out the electrons.
You should find that you now have 8Cr3+ on the right hand side, for example.
2007-04-18 20:10:19 · answer #1 · answered by Gervald F 7 · 0⤊ 0⤋
LiCl(aq) + AgNO3(aq) --> LiNO3 + AgCl(s) --All nitrates are soluble in water, yet silver chloride isn't. Li+ + Cl- + Ag+ + NO3- --> Li+ + NO3- + AgCl(s) The Li+ and NO3- cancel on the two aspects, leaving you with a internet ionic equation of Ag+ + Cl- --> AgCl(s) NaNO3(aq) + KI(aq) = Na+(aq) + NO3-(aq) + ok+(aq) + I-(aq) --> NaI(s) + KNO3(aq) --returned, all nitrates are soluble in water. in this occasion, NaI(s) + ok+ + NO3- is your result, the ok+ and NO3- cancel with the reactants, and you have a internet ionic of: Na+(aq) + I-(aq) --> NaI(s) Make experience?
2016-12-20 18:33:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋