The given equation is balanced in acidic condition. The steps to balance in basic condition would be: CIO3¹ +6 Br¹ + 6 H+¹ →CI-¹ + 3 Br₂ + 3 H₂O -1 A. Add OH´ to both sides of the equation. B. Add 6 OH to the left side of the equation. Then, subtract the waters from both sides of the equation. C. Add 6 OH-1 to the right side of the equation and add the H₂O with the OH´to make more water. -1 to both sides of the equation and add to make 6 H₂O. Then, add to D. Add 6 OH make more water. E. Add 6 OH-¹ to both sides of the equation and add to make 6 H₂O. Then, subtract the waters. B

Chemistry for Engineering Students
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Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.25PAE
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D
The given equation is balanced in acidic condition. The steps to balance in basic
condition would be:
CIO3-¹ + 6 Br¹ +6 H¹ →CI-¹ + 3 Br₂ + 3 H₂O
-1
A. Add OH to both sides of the equation.
B. Add 6 OH to the left side of the equation. Then, subtract the waters from both
sides of the equation.
C. Add 6 OH¹ to the right side of the equation and add the H₂O with the OH to
make more water.
D. Add 6 OH to both sides of the equation and add to make 6 H₂O. Then, add to
make more water.
to both sides of the equation and add to make 6 H₂O. Then, subtract
E. Add 6 OH
the waters.
B
Transcribed Image Text:D The given equation is balanced in acidic condition. The steps to balance in basic condition would be: CIO3-¹ + 6 Br¹ +6 H¹ →CI-¹ + 3 Br₂ + 3 H₂O -1 A. Add OH to both sides of the equation. B. Add 6 OH to the left side of the equation. Then, subtract the waters from both sides of the equation. C. Add 6 OH¹ to the right side of the equation and add the H₂O with the OH to make more water. D. Add 6 OH to both sides of the equation and add to make 6 H₂O. Then, add to make more water. to both sides of the equation and add to make 6 H₂O. Then, subtract E. Add 6 OH the waters. B
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