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Unformatted text preview: ( ) =(.0282L)(0.0966M)(0.5)(106g) =0.1444g Mass of Na2Co3 in unknown= ( ) =(.0277L)(0.0966M)(0.5)(106g) =0.1418g 2. %Na2CO3= ( )*100% =( )*100% For M1=40.70% For M2= 40.63% Conclusion : Average % of Na2CO3 is 40.7% with a standard deviation 0.1%...
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This note was uploaded on 09/01/2009 for the course CHEM 316 taught by Professor Soriaga during the Spring '08 term at Texas A&M.
- Spring '08