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48.0 g LICI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AGNO3 in 424.0 mL of solution. M

48.0 g LICI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AGNO3 in 424.0 mL of solution. M

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
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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int 48.0 g LiCI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AgNO3 in 424.0 mL of solution. M WE
Transcribed Image Text:int 48.0 g LiCI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AgNO3 in 424.0 mL of solution. M WE
edu A chemist wishes to make a solution using 4.21 g of CoCl3 in which the total concentration of CI ions is equal to 0.333 mol/L. What must the final volume of the solution be? Let's start by understanding the relationship between [CoCl3] and [CI-]. In order for [CI-] to equal 0.333 M, what must [COCI3] equal? moles solute Molarity is defined as So we need to calculate the moles of CoCl3. How many moles of CoCl3 are equivalent to L solution 4.21 g of CoCl3? moles CoCl3 Finally we can calculate the final volume of the solution. What final volume will produce a solution in which [CI ] = 0.333 M, using 4.21 g of CoCl3? L solution
Transcribed Image Text:edu A chemist wishes to make a solution using 4.21 g of CoCl3 in which the total concentration of CI ions is equal to 0.333 mol/L. What must the final volume of the solution be? Let's start by understanding the relationship between [CoCl3] and [CI-]. In order for [CI-] to equal 0.333 M, what must [COCI3] equal? moles solute Molarity is defined as So we need to calculate the moles of CoCl3. How many moles of CoCl3 are equivalent to L solution 4.21 g of CoCl3? moles CoCl3 Finally we can calculate the final volume of the solution. What final volume will produce a solution in which [CI ] = 0.333 M, using 4.21 g of CoCl3? L solution
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