**Question:**A mixture was found to contain 4.36 g of Na₂CO₃, 6.21 g of S₄O₃, and 1.89 g of Li₂Se. What is the percentage of Lithium selenium in the mixture?**Explanation:**To find the percentage of Lithium selenium (Li₂Se) in the mixture, use the formula:\[\text{Percentage of Li}_2\text{Se} = \left(\frac{\text{Mass of Li}_2\text{Se}}{\text{Total mass of mixture}}\right) \times 100\]1. **Calculate the total mass of the mixture:** \[ \text{Total mass} = 4.36\, \text{g (Na}_2\text{CO}_3) + 6.21\, \text{g (S}_4\text{O}_3) + 1.89\, \text{g (Li}_2\text{Se}) = 12.46\, \text{g} \]2. **Calculate the percentage of Li₂Se:** \[ \text{Percentage of Li}_2\text{Se} = \left(\frac{1.89\, \text{g}}{12.46\, \text{g}}\right) \times 100 \approx 15.17\% \]**Conclusion:**The mixture contains approximately 15.17% Lithium selenium.

Given : Mass of lithium selenium i.e Li2Se = 1.89 g Mass of Na2CO3 = 4.36 g Mass of S4O3 = 6.21 g The total mass of mixture = mass of Na2CO3 + mass of S4O3 + mass of Li2Se => Total mass of mixture = 1.89 + 4.36 + 6.21 = 12.46 g Hence the mass percent of lithium selenium is given by, => Mass percent of Li2Se = Mass of Li2Se X 100Total mass of mixture = 1.89 X 10012.46 = 15.17 approx. Hence the answer is 15.17 % approx.

Answered: A mixture was found to contain 4.36 g…

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A mixture was found to contain 4.36 g of Na2CO3, 6.21g of S403 and 1.89 g of LizSe. What is the percentage of Lithium selenium?

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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