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Potassium iodate, KIO3, was prepared by dissolving 2.023 g of KIO3 in a 500 mL volumetric flask. A 50 mL aliquot of the solution was treated in excess 1 g Kl and 15 mL of 0.5033 molar H₂SO4. a. Calculate the mmoles of 13 created on the reaction? Reaction: 103 + 1 → 13 + H₂O b. The triiodide formed reacted with 38.66 mL of standard Na2S2O3 solution. Calculate the concentration of the standard solution? Reaction: 13 + S2O3-2 → |- + S406-²

Potassium iodate, KIO3, was prepared by dissolving 2.023 g of KIO3 in a 500 mL volumetric flask. A 50 mL aliquot of the solution was treated in excess 1 g Kl and 15 mL of 0.5033 molar H₂SO4. a. Calculate the mmoles of 13 created on the reaction? Reaction: 103 + 1 → 13 + H₂O b. The triiodide formed reacted with 38.66 mL of standard Na2S2O3 solution. Calculate the concentration of the standard solution? Reaction: 13 + S2O3-2 → |- + S406-²

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Potassium iodate, KIO3, was prepared by dissolving 2.023 g of KIO3 in a 500 mL volumetric flask. A 50 mL aliquot of the solution was treated in excess 1 g Kl and 15 mL of 0.5033 molar H₂SO4. a. Calculate the mmoles of 13 created on the reaction? Reaction: 103 + 1 → 13 + H₂O b. The triiodide formed reacted with 38.66 mL of standard Na2S2O3 solution. Calculate the concentration of the standard solution? Reaction: 13 + S2O3-2 → |- + S406-²
Transcribed Image Text:Potassium iodate, KIO3, was prepared by dissolving 2.023 g of KIO3 in a 500 mL volumetric flask. A 50 mL aliquot of the solution was treated in excess 1 g Kl and 15 mL of 0.5033 molar H₂SO4. a. Calculate the mmoles of 13 created on the reaction? Reaction: 103 + 1 → 13 + H₂O b. The triiodide formed reacted with 38.66 mL of standard Na2S2O3 solution. Calculate the concentration of the standard solution? Reaction: 13 + S2O3-2 → |- + S406-²
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