to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9 How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3? 500 mmol 250 mmol 25 mmol 12.5 mmol What is the concentration of the excess reagent? 0.1125 M 0.3219 M 0.1195 M 0.2375 M What is the molar solubility of Ba(IO3)2 in this solution? 04x10^-5 M 07x10^-5 M 15x10^-5 M 30x10^-5 M
to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9 How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3? 500 mmol 250 mmol 25 mmol 12.5 mmol What is the concentration of the excess reagent? 0.1125 M 0.3219 M 0.1195 M 0.2375 M What is the molar solubility of Ba(IO3)2 in this solution? 04x10^-5 M 07x10^-5 M 15x10^-5 M 30x10^-5 M
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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In order to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3.
Ksp= 1.57x10^-9
How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3?
- 500 mmol
- 250 mmol
- 25 mmol
- 12.5 mmol
What is the concentration of the excess reagent?
- 0.1125 M
- 0.3219 M
- 0.1195 M
- 0.2375 M
What is the molar solubility of Ba(IO3)2 in this solution?
- 04x10^-5 M
- 07x10^-5 M
- 15x10^-5 M
- 30x10^-5 M
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