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-
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please solve both questions, I will like answer with positive remarks

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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