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Unformatted text preview: Total Volume = 0.00600 L = Volume of CuSO4 x 5H2O + H4EDTA V of CuSO4 x 5H2O = 0.00600 L - 0.001151L = 0.004848 L # mol of reagent 1 = M x Volume of reagent 1 #mol of reagent 1 = 5.0x10^ -3 M x 0.004848 L #mol of reagent 1 = 2.424x10^ -5 moles n = 2.424x10^ -5 moles / 1.1515x10^ -5 moles n = 2:1...
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This note was uploaded on 04/18/2008 for the course CHEM 102 taught by Professor Haworth during the Spring '08 term at Marquette.
- Spring '08