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Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.$\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(s) \longrightarrow \mathrm{Co}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$
$4 \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightarrow 2 \mathrm{Co}_{2} \mathrm{O}_{3}(s)+8 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$
Chemistry 102
Chemistry 101
Chapter 19
Transition Metals and Coordination Chemistry
Transition Metals
Drexel University
University of Maryland - University College
University of Kentucky
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consider the following reaction. First, let us separate the reaction in tow. Oxidation, half reaction oxidation have a reaction that iss she. Oh, and, uh, three to farms. So you two or three and started to see a three and reduction half reaction. Mm. Okay. And all three forms for and l do it toe balance. The oxygen atoms multiply oxygen with half. So the complete balance reaction is so CEO and not trade toe solid forms his c 02 03 solid plus for an or to gas lads. Uh huh. Or yes. Now, in order to remove fractional values off oxygen multiplied equation with to So the balance rations based to four CEO and how? Trade toe solid farms to see oh, two or three solid plus eight and two gas plus or two. Guess so. This is the balanced reaction
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