1. The solubility of lead II chloride is 1.06 g/100 ml. Explain how you would make 1 liter of a saturated solution. Mass 2. Explain how you would make 500 ml of a 0.5M solution of potassium chromate. lations: lculate the percent by 3. Write the balanced equations and solubility product expressions for the solubility equilibria of the following compounds: a)Mn3(PO4)2 b) AuCl3 c) ZnC204

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Pre-Lab for Solubility Produce Constant
1. The solubility of lead II chloride is 1.06 g/100 ml. Explain how you would make 1 liter
of a saturated solution.
1. Mas
Mass
2. Explain how you would make 500 ml of a 0.5M solution of potassium chromate.
3. Write the balanced equations and solubility product expressions for the solubility
equilibria of the following compounds: a)Mn3(PO4)2 b) AuCl3 c) ZnC2O4
the % composition
Transcribed Image Text:Pre-Lab for Solubility Produce Constant 1. The solubility of lead II chloride is 1.06 g/100 ml. Explain how you would make 1 liter of a saturated solution. 1. Mas Mass 2. Explain how you would make 500 ml of a 0.5M solution of potassium chromate. 3. Write the balanced equations and solubility product expressions for the solubility equilibria of the following compounds: a)Mn3(PO4)2 b) AuCl3 c) ZnC2O4 the % composition
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