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LUX equation is balanced since there are equal number of each type of both sides of the equation. Practice Exercises 8.2 Balance the reactions below using the change in oxidation number method. +2 -1₂ 0 1. PbO₂ + 2HBr →>>> PbBr₂ + O₂ + H₂O 2 2. PbO + NH, → Pb + N, +H,O 3. KBr + KClO₂ + H₂SO4 → Br₂ + KCl + K₂SO4 + H₂O 4. Br + CIO + H → Br₂ + Cl + H₂O 4+ 2+ 5. Sn²+ + VO³ + H → Sn¹+ + V²+ + H₂O Chapter 8 Electroche

LUX equation is balanced since there are equal number of each type of both sides of the equation. Practice Exercises 8.2 Balance the reactions below using the change in oxidation number method. +2 -1₂ 0 1. PbO₂ + 2HBr →>>> PbBr₂ + O₂ + H₂O 2 2. PbO + NH, → Pb + N, +H,O 3. KBr + KClO₂ + H₂SO4 → Br₂ + KCl + K₂SO4 + H₂O 4. Br + CIO + H → Br₂ + Cl + H₂O 4+ 2+ 5. Sn²+ + VO³ + H → Sn¹+ + V²+ + H₂O Chapter 8 Electroche

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter7: Reactions In Aqueous Solutions
Section: Chapter Questions
Problem 17ALQ: Write the balanced formula and net ionic equation for the reaction that occurs when the contents of...
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ons lost 8 Sotoms 1 equation is balanced since there are equal number of each type of om in 11 both sides of the equation. Practice Exercises 8.2 Balance the reactions below using the change in oxidation number method. +4 +2 -1₂ 0 1. PbO₂ +2HBr →>> PbBr₂ + O₂ + H₂O 2 2. PbO + NH, → Pb + N, +H,O 3. KBr + KClO₂ + H₂SO4 → Br₂ + KCl + K₂SO4 + H₂O 4. Br + CIO + H → Br₂ + Cl + H₂O 4+ 2+ 3- → Sn Sn+ + V²+ + H₂O 5. Sn + VO³ + H Chapter 8 Electrochemistry 191
Transcribed Image Text:ons lost 8 Sotoms 1 equation is balanced since there are equal number of each type of om in 11 both sides of the equation. Practice Exercises 8.2 Balance the reactions below using the change in oxidation number method. +4 +2 -1₂ 0 1. PbO₂ +2HBr →>> PbBr₂ + O₂ + H₂O 2 2. PbO + NH, → Pb + N, +H,O 3. KBr + KClO₂ + H₂SO4 → Br₂ + KCl + K₂SO4 + H₂O 4. Br + CIO + H → Br₂ + Cl + H₂O 4+ 2+ 3- → Sn Sn+ + V²+ + H₂O 5. Sn + VO³ + H Chapter 8 Electrochemistry 191
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