User:Melissa Novy/Notebook/CHEM-572/2013/02/19
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(→ISE Study of PLA2002D + 5 wt% 100Ag-LMT) |
(→ISE Study of PLA2002D + 5 wt% 100Ag-LMT) |
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'''(3.53 × 10<sup>-5</sup> mol Ag<sup>+</sup>) ÷ (0.05062 g 100Ag-LMT) = x mol Ag<sup>+</sup> ÷ (0.025 g 100Ag-LMT)''' | '''(3.53 × 10<sup>-5</sup> mol Ag<sup>+</sup>) ÷ (0.05062 g 100Ag-LMT) = x mol Ag<sup>+</sup> ÷ (0.025 g 100Ag-LMT)''' | ||
'''4.47 × 10<sup>-8</sup>mol Ag<sup>+</sup> in 0.025 g 100Ag-LMT of 0.5 g PLA2002D + 5 wt% 100Ag-LMT film''' | '''4.47 × 10<sup>-8</sup>mol Ag<sup>+</sup> in 0.025 g 100Ag-LMT of 0.5 g PLA2002D + 5 wt% 100Ag-LMT film''' | ||
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Revision as of 14:21, 20 February 2013
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Objectives
PLA2002D Films
0.00622 g AgNO3 × (1 mol AgNO3/169.87 g AgNO3) × (1 mol Ag+/1 mol AgNO3) = 3.66 × 10-5 mol Ag+ [Ag+] in 5 mL 100Ag-LMT filtrate was 0.000253614 M 0.000253614 M Ag+ × 0.005 L = 1.2681 × 10-6 mol Ag+ Therefore, 3.66 × 10-5 mol Ag+ − 1.2681 × 10-6 mol Ag+ = 3.53 × 10-5 mol Ag+ was exchanged into 100Ag-LMT.
0.05 × 0.5 g PLA2002D + 5 wt% 100Ag-LMT = 0.025 g 100Ag-LMT (3.53 × 10-5 mol Ag+) ÷ (0.05062 g 100Ag-LMT) = x mol Ag+ ÷ (0.025 g 100Ag-LMT) 4.47 × 10-8mol Ag+ in 0.025 g 100Ag-LMT of 0.5 g PLA2002D + 5 wt% 100Ag-LMT film
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