User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/11/13
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| - | .0082 g of adenosine × <math>\frac{1 mol}{267.24 g} = 3.06840 × 10<sup>-5</sup> ÷ .010 L = .00307 M = 3.07 mM | + | .0082 g of adenosine × <math>\frac{1 mol}{267.24 g}</math> = 3.06840 × 10<sup>-5</sup> ÷ .010 L = .00307 M = 3.07 mM |
Revision as of 04:51, 9 December 2012
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ADA Kinetic Assay Preparations
V1 = 0.08934 mL = 89.3 μL in 5 mL of buffer
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= .0025 mol of Na2HPO4 ×
= 0.6702 g
= .000005596 mol ÷ .001 L = .005596 M = 5.596 mM
= 3.06840 × 10-5 ÷ .010 L = .00307 M = 3.07 mM


