User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/16: Difference between revisions
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* The concentration of luminol has been diluted by the addition of 15 mL of water. | * The concentration of luminol has been diluted by the addition of 15 mL of water. | ||
Molarity of diluted luminol = <math>\frac{10 mM * 6 mL}{21 mL}</math> = 2.85714 mM of luminol | Molarity of diluted luminol = <math>\frac{10 mM * 6 mL}{21 mL}</math> = 2.85714 mM of luminol | ||
* Since there was limited amounts of the solid form of luminol, it was decided to take 6 mL of the 2.86 mM luminol stock solution and then add the appropriate amount of luminol to the 6 mL volume. The molarity of the 6 mL solution was very minute; the molarity was approximated as 2 mM. By making this assumption, .00106 g of luminol would make a 1 mM solution in 6 mL of water; this amount was multiplied by 8. The product was 0.00848 g of luminol was needed to be added into the 2 mM solution of luminol to increase the molarity to 10 mM. | |||
Revision as of 23:48, 25 October 2012
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PCR mutation
0.46 mg = 0.46E6 ng
Continuation of Chemiluminescence
[math]\displaystyle{ \frac{1.91 g}{15 mL} }[/math] × [math]\displaystyle{ \frac{1 mol}{105.9784 g} }[/math] = [math]\displaystyle{ \frac{0.00120 mol}{mL} }[/math] × [math]\displaystyle{ \frac{1 mL}{1E(-3) L} }[/math] = [math]\displaystyle{ \frac{1.20 mol}{L} }[/math] = 1.20 M of sodium carbonate
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