User:Puja Mody/Notebook/Chem 571: Gold Nanoparticles/2012/11/13: Difference between revisions
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.05 x .05= .0025 moles | .05 x .05= .0025 moles | ||
.0025 moles x 268.07g/1mol = .6702g sodium phosphate in 50 mL H<sub>2</sub>0. Actual amount = .6705. pH of 7.4 was achieve by adding roughly 3 drops of 12M HCl from a glass pasture pipette. | .0025 moles x 268.07g/1mol = .6702g sodium phosphate in 50 mL H<sub>2</sub>0. Actual amount = .6705. pH of 7.4 was achieve by adding roughly 3 drops of 12M HCl from a glass pasture pipette. | ||
* | # 5mL of .1mM adenosine was made by adding 100mg of adenosine in 1mL of .05M phosphate buffer. This solution of adenosine was .00373M. .134mL of the .00373M adenosine solution was then diluted to a final volume of 5mL with .05M phosphate buffer giving a final concentration of .1mM adenosine. | ||
* '(0.001 g)/(268.02 g/mol) = 3.73×10<sup>-6</sup> mol (3.73×10<sup>-6</sup> mol)/0.001 L = 0.00373 M | |||
(0.00373 M)(V<sub>1</sub>)=(5 mL)(0.1×10<sup>-3</sup> M) V<sub>1</sub> = 0.134 mL | |||
100 mg of adenosine was dissolved in 1 mL of 0.05 M phosphate buffer, giving a solution of 0.00373 M adenosine. | |||
** 0.134 mL of 0.00373 M adenosine was diluted to a final volume of 5 mL with 0.05 M phosphate buffer, giving a solution of 0.1 mM adenosine. | |||
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Revision as of 12:16, 15 November 2012
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ADA Kinetics and moreGoals
Procedure
.05 x .05= .0025 moles .0025 moles x 268.07g/1mol = .6702g sodium phosphate in 50 mL H20. Actual amount = .6705. pH of 7.4 was achieve by adding roughly 3 drops of 12M HCl from a glass pasture pipette.
(0.00373 M)(V1)=(5 mL)(0.1×10-3 M) V1 = 0.134 mL
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