In standardization of KMNO, againstNa,C,04 (134 g/mol), if 35 mL of the 0.02 MKMNO, is used find the mass of Na,C,042MN0, + 5 H,C,04 + 6 H* ==> 2 Mn2 + 10co, + 8 H20a. 0.0375 gb. 0.0523c. 0.101 gd. 0.235 gIn the reactionCe +Fe s>+]Knowing that Fe +e => Fe2 E 0.767 VCe +e sa> Ce E= 1.70 VAnd Hydrogen reference electrode was used.In titrating 25 ml of 0.2 M Fe against 0.1 M ofCe, it the volume of titrant is half theequivalence volume, then the2O a Ereaction = E of Fe*O b. E reaction = 0OC.E reaction E" of Ce4+O d. E reaction 1//2 (E of Fe + E of Ce

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In standardization of KMNO, against NazC,04 (134 g/mol), if 35 mL of the 0.02 M KMNO, is used find the mass of Na2C204 2Mno, + 5 H2C204+6 H* ==> 2 Mn2+ + 10 co, + 8 H,0 1 a. 0.0375 g b. 0.0523 C. 0.101 g d. 0.235 g

In standardization of KMNO, against NazC,04 (134 g/mol), if 35 mL of the 0.02 M KMNO, is used find the mass of Na2C204 2Mno, + 5 H2C204+6 H* ==> 2 Mn2+ + 10 co, + 8 H,0 1 a. 0.0375 g b. 0.0523 C. 0.101 g d. 0.235 g

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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Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

In standardization of KMNO, against
Na,C,04 (134 g/mol), if 35 mL of the 0.02 M
KMNO, is used find the mass of Na,C,04
2MN0, + 5 H,C,04 + 6 H* ==> 2 Mn2 + 10
co, + 8 H20
a. 0.0375 g
b. 0.0523
c. 0.101 g
d. 0.235 g
In the reaction
Ce +Fe s>
+]
Knowing that Fe +e => Fe2 E 0.767 V
Ce +e sa> Ce E= 1.70 V
And Hydrogen reference electrode was used.
In titrating 25 ml of 0.2 M Fe against 0.1 M of
Ce, it the volume of titrant is half the
equivalence volume, then the
2
O a Ereaction = E of Fe*
O b. E reaction = 0
OC.E reaction E" of Ce4+
O d. E reaction 1//2 (E of Fe + E of Ce

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