### Problem Statement:Calculate the concentration of \(\text{ClO}_3^-\) when 979 mL of 0.289 M \(\text{AgClO}_3\) is mixed with 651 mL of 0.273 M \(\text{Mn(ClO}_3\))_2?#### Solution:To determine the final concentration of \(\text{ClO}_3^-\), use the following steps:1. **Calculate the moles of \(\text{ClO}_3^-\) from \(\text{AgClO}_3\):** - Volume = 979 mL = 0.979 L - Molarity = 0.289 M - Moles of \(\text{ClO}_3^-\) = Volume × Molarity = 0.979 × 0.289 = 0.283031 moles2. **Calculate the moles of \(\text{ClO}_3^-\) from \(\text{Mn(ClO}_3)_2\):** - Volume = 651 mL = 0.651 L - Molarity = 0.273 M - Moles of \(\text{ClO}_3^-\) = Volume × Molarity × 2 (since there are 2 \(\text{ClO}_3^-\) ions per formula unit of \(\text{Mn(ClO}_3)_2\)) = 0.651 × 0.273 × 2 = 0.355386 moles3. **Total moles of \(\text{ClO}_3^-\):** - Total moles = 0.283031 + 0.355386 = 0.638417 moles4. **Total volume of the mixture:** - Total volume = 979 mL + 651 mL = 1630 mL = 1.630 L5. **Calculate the concentration of \(\text{ClO}_3^-\) in the mixture:** - \([\text{ClO}_3^-]\) = Total moles / Total volume = 0.638417 / 1.630 = 0.3916 MThe concentration of \(\text{ClO}_3^-\) in the final

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Answered: What concentration of CIO, results when…

What concentration of CIO, results when 979 mL of 0.289 M AgCIO, is mixed with 651 mL of 0.273 M Mn(CIO,),? [CIO,] = %3D M

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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Question

### Problem Statement:
Calculate the concentration of \(\text{ClO}_3^-\) when 979 mL of 0.289 M \(\text{AgClO}_3\) is mixed with 651 mL of 0.273 M \(\text{Mn(ClO}_3\))_2?

#### Solution:

To determine the final concentration of \(\text{ClO}_3^-\), use the following steps:

1. **Calculate the moles of \(\text{ClO}_3^-\) from \(\text{AgClO}_3\):**
- Volume = 979 mL = 0.979 L
- Molarity = 0.289 M
- Moles of \(\text{ClO}_3^-\) = Volume × Molarity = 0.979 × 0.289 = 0.283031 moles

2. **Calculate the moles of \(\text{ClO}_3^-\) from \(\text{Mn(ClO}_3)_2\):**
- Volume = 651 mL = 0.651 L
- Molarity = 0.273 M
- Moles of \(\text{ClO}_3^-\) = Volume × Molarity × 2 (since there are 2 \(\text{ClO}_3^-\) ions per formula unit of \(\text{Mn(ClO}_3)_2\)) = 0.651 × 0.273 × 2 = 0.355386 moles

3. **Total moles of \(\text{ClO}_3^-\):**
- Total moles = 0.283031 + 0.355386 = 0.638417 moles

4. **Total volume of the mixture:**
- Total volume = 979 mL + 651 mL = 1630 mL = 1.630 L

5. **Calculate the concentration of \(\text{ClO}_3^-\) in the mixture:**
- \([\text{ClO}_3^-]\) = Total moles / Total volume = 0.638417 / 1.630 = 0.3916 M

The concentration of \(\text{ClO}_3^-\) in the final

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