User:Puja Mody/Notebook/Chem 571: Gold Nanoparticles/2012/09/25: Difference between revisions
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1mol/L x .025= .025 mol | 1mol/L x .025= .025 mol | ||
.025 mol x 121.14 g/mol = 3.0285 g of tris | .025 mol x 121.14 g/mol = 3.0285 g of tris | ||
*'''[[User:Abigail E. Miller|Abigail E. Miller]] 11:01, 7 October 2012 (EDT)''':you need to note that this is a 1.0 M Tris stock. | |||
To get the necessary concentrations of tris into the gold/BSA solutions, the M<sub>1</sub>V<sub>1</sub>=M<sub>2</sub>V<sub>2</sub> formula was used to calculate the volume of tris needed. From there, the value of the volume of tris was subtracted from 1 in order to determine the amount of water that would be added to make the desired concentration. | To get the necessary concentrations of tris into the gold/BSA solutions, the M<sub>1</sub>V<sub>1</sub>=M<sub>2</sub>V<sub>2</sub> formula was used to calculate the volume of tris needed. From there, the value of the volume of tris was subtracted from 1 in order to determine the amount of water that would be added to make the desired concentration. |
Revision as of 08:01, 7 October 2012
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Separate ProteinsGoals
Procedure
DATA/Calculations
to make a 25mL stock solution of the tris buffer at a pH of 10 1mol/L x .025= .025 mol .025 mol x 121.14 g/mol = 3.0285 g of tris
link for excel File:TrisvaryingmolarUVVis.xlsx Conclusions
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