Chemistry 110B Quiz 2
Chemistry 110B Quiz 2 CHEM 110
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This 8 page Class Notes was uploaded by Mikaela Notetaker on Sunday April 3, 2016. The Class Notes belongs to CHEM 110 at West Virginia University taught by Melissa G. Ely in Spring 2016. Since its upload, it has received 16 views. For similar materials see Introduction to Chemistry in Chemistry at West Virginia University.
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Date Created: 04/03/16
Quiz 7 Molarity of ions in solution: Any soluble ionic compound really exists as ions in solution so you may want the concentration. Equation for Molarity: Moles/Leter = Molarity +3 - 1. 0.20 M Al(NO ) 3 3ou want M of Al and NO 3 a. Al : 0.20 mol Al(NO ) 3 3 1 mol Al +3 = 0.20 M Al +3 +3 1 mol Al(NO )3 3 (only 1 mole for Al because Al(NO )3 3s 1 Al) +3 - - b. Al : 0.20 mol Al(NO ) 3 3 3 mol NO 3 = 0.60 M NO 3 1 mol Al(NO )3 3 (3 mole for NO 3-because Al(NO )3 3s the 3 outside the parenthesis, the other 3 does not count since it is already a part of the compound) Conversion of Molarity to Moles: Volume (L)*Molarity = moles of solution 1. 75.0 mL (turn to L by moving the decimal three times to the left) of + 0.15 M K P3 so4ution, what is the moles of K + + a. 0.0750 L K PO3 4 0.15 M K P3 4 3 mol K = 0.034 mol K 1 L 1 mol K 3O 4 Dilution Problems: Adding pure solvent to a solution and, because you’re increasing the volume of the solution, the molarity will decrease. M = Mol (Number of moles does not normally change) / L (changes as volume Increases and molarity decreases) Basic Dilution equation: M V = M 1 1ith M 2 2olarity, V = volume + - Method: Since H is left over then it must have been the excess reagent. Thus OH - + was the limiter. Need moles of OH at the start to find moles H used up since it will + + equal the limiting reagent. Moles of H used = Moles H start – leftover.
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