CHEMISTRY 104

Cambridge College

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Exam (elaborations)  CHEMISTRY 104
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    Exam (elaborations) CHEMISTRY 104

  • Question 1 9 / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molar solubility (mol/L) of Mg(OH)2, Ksp of Mg(OH)2 = 1.8 x 10-11 . Your Answer: ksp= [ M g + 2 ] [ O H − ] 2 = (s ) (2s) 2 s 3 = 1.8 x 10 − 11 / 4 s= 3 ( 1.8 x 10 − 11 4 ) s= 1.65 x 10 − 4 mol/L Mg(OH)2 (s) Mg+2 (aq) + 2 OH-1 (aq) (s) (s) (2s) 1.8 x 10-11 = [Mg+2] x [OH-1] 2 1.8 x 10-11 = [s] x [2s]2 This study source was downloaded b...
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