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1. Solubility in water that in part one of the lab you determined that [S203= 0.145 M. Titration of 1.00 mL Lets say calcium


1 mol 103 iodate and thio sulfate: 6 mol S203
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Answer #1

ns,os & ImL = 1 2 3 x mol = 0.145M TUO 0.39mL 0.00039 L = 0.39 ணெ = 3-9x104 x = 5-655x10 mol (b) 105 = 1 non = x= 5.6558105 N

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