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On a far away planet elements M, Y, Cc and Bm form oxides solutions. Create a balanced redox equation from these two half reactions in acid solution, in question #1 and #2. Oxygen behaves as it does on earth, thus, you can solve for the oxidation numbers of M, Y, Bm and Cc in the oxides and anions. There are more questions on the back #1 M Y2O3 #2 BmO3 Cc²+ -> 1 MO₂ Y → Bm - CCO4²-

On a far away planet elements M, Y, Cc and Bm form oxides solutions. Create a balanced redox equation from these two half reactions in acid solution, in question #1 and #2. Oxygen behaves as it does on earth, thus, you can solve for the oxidation numbers of M, Y, Bm and Cc in the oxides and anions. There are more questions on the back #1 M Y2O3 #2 BmO3 Cc²+ -> 1 MO₂ Y → Bm - CCO4²-

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.18QP
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Please solve both questions, I will like answer with positive remarks
On a far away planet elements M, Y, Cc and Bm form oxides and react in redox reactions in acid solutions. Create a balanced redox equation from these two half reactions in acid solution, in question #1 and #2. Oxygen behaves as it does on earth, thus, you can solve for the oxidation numbers of M, Y, Bm and Cc in the oxides and anions. There are more questions on the back. M #1 #2 Y₂O3 BmO3 Cc²+ -> MO₂ -> Y →>> Bm - CcO4²-
Transcribed Image Text:On a far away planet elements M, Y, Cc and Bm form oxides and react in redox reactions in acid solutions. Create a balanced redox equation from these two half reactions in acid solution, in question #1 and #2. Oxygen behaves as it does on earth, thus, you can solve for the oxidation numbers of M, Y, Bm and Cc in the oxides and anions. There are more questions on the back. M #1 #2 Y₂O3 BmO3 Cc²+ -> MO₂ -> Y →>> Bm - CcO4²-
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