User:Mary Mendoza/Notebook/CHEM572 Exp. Biological Chemistry II/2013/01/23

(Difference between revisions)

Revision as of 16:23, 23 January 2013

Project name

It was decided to prepare a 30 mM stock solution of adenosine. The calculation below shows the amount needed to obtain 30 mM of adenosine.

.030 $\frac{mol}{L}$ mol of adenosine × $\frac{267.24 g}{1 mol}$ = $\frac{8.0172 g}{L}$ × .010 L = 0.0802 g of adenosine

Revision as of 16:22, 23 January 2013 (view source) Mary Mendoza (Talk \| contribs) (→Calculation of reagents for ADA kinetic assay) ← Previous diff		Revision as of 16:23, 23 January 2013 (view source) Mary Mendoza (Talk \| contribs) (→Calculation of reagents for ADA kinetic assay) Next diff →
Line 9:		Line 9:
	* It was decided to prepare a 30 mM stock solution of adenosine. The calculation below shows the amount needed to obtain 30 mM of adenosine.		* It was decided to prepare a 30 mM stock solution of adenosine. The calculation below shows the amount needed to obtain 30 mM of adenosine.

-	.030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g	+	.030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g of adenosine