User:Elizabeth Ghias/Notebook/Experimental Biological Chemistry/2011/08/31: Difference between revisions

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Need 0.0015 mM of BSA
Need 0.0015 mM of BSA


0.0015 mmol/L x 1L/1000 mL x 10 mL = 0.015mmol/1000 = 1.5 x 10<sup>-5</sup> mol x 66776 g/mol = 0.001 g BSA
0.0015 mmol/L x 1L/1000 mL x 10 mL = 0.015 mmol/1000 = 1.5 x 10<sup>-5</sup> mol x 66776 g/mol = 0.001 g BSA




Need 0.25 mM of HAuCl<sub>4</sub>
Need 0.25 mM of HAuCl<sub>4</sub>


0.25 mmol/L x 1L/1000 mL x 10 mL = 0.0025 mol  
0.25 mmol/L x 1L/1000 mL x 10 mL = 0.0025 mmol/1000 = 2.5 x 10<sup>-6</sup> mol x 393.83g/mol = 0.000985 g HAuCl<sub>4</sub>
 




Revision as of 10:50, 31 August 2011

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Objective

To synthesize gold nanoparticles (AuNPs) from bovine serum albumen (BSA) and HAuCl4

Description

  1. A 10 mL aqueous mixture of 0.001 g BSA (0.0015 mM) and 0.001 g HAuCl4 (0.25 mM) was created using a 50 mM acetate buffer with a pH of 4.3.
  2. The reaction tube was then heated at 80°C for ______
  3. Every 30 minutes the UV-vis Spectrometer was used to assess the progress of the reaction. Measurements were also taken of the buffer and the reaction mixture at the beginning of the reaction. A wavelength range between 200 and 900 nm was used.

Data

  • Add data and results here...

Notes

Calculations

Need 0.0015 mM of BSA

0.0015 mmol/L x 1L/1000 mL x 10 mL = 0.015 mmol/1000 = 1.5 x 10-5 mol x 66776 g/mol = 0.001 g BSA


Need 0.25 mM of HAuCl4

0.25 mmol/L x 1L/1000 mL x 10 mL = 0.0025 mmol/1000 = 2.5 x 10-6 mol x 393.83g/mol = 0.000985 g HAuCl4



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