User:Melissa Novy/Notebook/CHEM-572/2013/02/20
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'''V<sub>1</sub> = 0.005 mL of 0.1 M AgNO<sub>3</sub>''' | '''V<sub>1</sub> = 0.005 mL of 0.1 M AgNO<sub>3</sub>''' | ||
| - | * AgNO<sub>3</sub> solutions made on [[User:Melissa_Novy/Notebook/CHEM-572/2013/02/05|2013/02/05]] and [[User:Melissa_Novy/Notebook/CHEM-572/2013/02/13|2013/02/13]] | + | * AgNO<sub>3</sub> solutions made on [[User:Melissa_Novy/Notebook/CHEM-572/2013/02/05|2013/02/05]] and [[User:Melissa_Novy/Notebook/CHEM-572/2013/02/13|2013/02/13]] will be added to the LB media solutions in the following amounts: |
{| {{table}} | {| {{table}} | ||
| align="center" style="background:#f0f0f0;"|'''Final [Ag<sup>+</sup>]''' | | align="center" style="background:#f0f0f0;"|'''Final [Ag<sup>+</sup>]''' | ||
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| - | * Note that for each volume of AgNO<sub>3</sub> added to the LB media, the same volume of LB media | + | * Note that for each volume of AgNO<sub>3</sub> added to the LB media, the same volume of LB media will first be removed from the flask to maintain the final concentration of Ag<sup>+</sup>. |
==ISE Study of PLA2002D + 5 wt% 100Ag-LMT== | ==ISE Study of PLA2002D + 5 wt% 100Ag-LMT== | ||
Revision as of 15:29, 20 February 2013
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Objectives
Calculations
1 μM Ag+ in 50 mL of LB media (0.0105 M Ag+) × V1 = (1×10-6 M) × (50 mL) V1 = 0.00476 mL of 0.0105 M AgNO3 10 μM Ag+ in 50 mL of LB media (0.1 M Ag+) × V1 = (10×10-6 M) × (50 mL) V1 = 0.005 mL of 0.1 M AgNO3
ISE Study of PLA2002D + 5 wt% 100Ag-LMT
(4.47 × 10-8mol Ag+0 ÷ 0.051 L = 8.76 × 10-7 M or 0.876 μM
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