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Portage Learning CHEM 104 Module 3 Exam
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Portage Learning CHEM 104 Module 3 ExamQuestion 1 Show the calculation of the molar solubility (mol/L) of BaF2, Ksp of BaF2 = 1.0 x 10-6. Your Answer: Ksp = [Ba+2] [F-]2 = 1.0x10-6 1.0x10-6 = [s] [2s]2 1.0x10-6 = 4s3 s = 6.3x10-3 mol/L Question 2 Show the calculation of the Ksp of MnS if the solubility of MnS is 0.0001375 g/100 ml. MW of MnS = 87.01 MnS (s) Mn+2 (aq) + S-2 (aq) Your Answer: molar solubility of MnS = (0.0001375 g/100 ml) x (1000 mL) x (1 mol/87.01 g MnS) molar solubility of MnS = 1.580x10-5 Ksp = [Mn+2] [S-2] Ksp = (1.580x10-5) (1.580x10-5 ) = 2.496x10-10
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Portage Learning CHEM 104 Module 3
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Portage Learning CHEM 104 Module 3
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portage learning chem 104 module 3 exam
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