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6. What volume of 1.50 M NaCl is needed for a reaction that requires 146.3 g of NaCl? Lor

6. What volume of 1.50 M NaCl is needed for a reaction that requires 146.3 g of NaCl? Lor

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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I need help number 6 questions, pelase

5. How does increasing the temperature of a solid affect its solubility? What about for a gas? As the temperature of solution increases, Solids become soluble. As the temperature of solution increases gases become less soluble, 6. What volume of 1.50 M NaCl is needed for a reaction that requires 146.3 g of NaCl? dod 1.67 L 7. What is the molarity of a solution composed of 8.2 g of potassium chromate, K₂CrO4 dissolved in enough water to make 500. mL of solution? .084M 8.2g KaCrOy 1 mol 17% Soo MLIL 1000m2=0.500 L 8. What is the % (w/w) of a solution containing 21 g KCl in 125 grams of solution? 1194.2g k₂cr04 219 Tasg X100% = 16.8% 17% Wa 1x 22,99 = 22.99 CI IX35.45- +33.45 56.44 ках 39,10= 78.2 Crix52.00 = 52.00 04 x 16.00+ 64. 194.2 10.084 mol/l 10.500LL 40.084M = 0.042mol 0.042 mol
Transcribed Image Text:5. How does increasing the temperature of a solid affect its solubility? What about for a gas? As the temperature of solution increases, Solids become soluble. As the temperature of solution increases gases become less soluble, 6. What volume of 1.50 M NaCl is needed for a reaction that requires 146.3 g of NaCl? dod 1.67 L 7. What is the molarity of a solution composed of 8.2 g of potassium chromate, K₂CrO4 dissolved in enough water to make 500. mL of solution? .084M 8.2g KaCrOy 1 mol 17% Soo MLIL 1000m2=0.500 L 8. What is the % (w/w) of a solution containing 21 g KCl in 125 grams of solution? 1194.2g k₂cr04 219 Tasg X100% = 16.8% 17% Wa 1x 22,99 = 22.99 CI IX35.45- +33.45 56.44 ках 39,10= 78.2 Crix52.00 = 52.00 04 x 16.00+ 64. 194.2 10.084 mol/l 10.500LL 40.084M = 0.042mol 0.042 mol
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