User:Klare Lazor/Notebook/Chem-496-001/2013/04/12: Difference between revisions
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==Description== | ==Description== | ||
'''NOTE: Mass of whole hydrogel: 12.1419 grams, piece of hydrogel 0.8 grams, so instead of 0.193 grams to occupy all the | '''NOTE: Mass of whole hydrogel: 12.1419 grams, piece of hydrogel 0.8 grams, so instead of 0.193 grams to occupy all the butayne-1-ol sites(0.000017mols) an amount of 0.014 grams of dye is needed.''' | ||
*the stock solution of dye is 5mg/ml=5g/L therefore the amount of grams of dye in vial, as well as assuming 1:1 reaction between butane-1-ol and dye then successful reaction will have a percent reacted of | *the stock solution of dye is 5mg/ml=5g/L therefore the amount of grams of dye in vial, as well as assuming 1:1 reaction between butane-1-ol and dye then a successful reaction will have a percent reacted ~ equal to the percentages below. Percentages below are equal to the amount of dye in mols added to the film/the amount of mols that should react with the butane-1-ol times 100. | ||
**'''1µL of Dye'''= 0.000005g= 8.9E-9 mol/0.000017mols= 0.0523% | **'''1µL of Dye'''= 0.000005g= 8.9E-9 mol/0.000017mols= 0.0523% | ||
**'''10µL of Dye'''= 0.00005g=8.9E-8 mols/0.000017= 0.523% | **'''10µL of Dye'''= 0.00005g=8.9E-8 mols/0.000017= 0.523% |
Revision as of 17:55, 17 April 2013
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Objective
DescriptionNOTE: Mass of whole hydrogel: 12.1419 grams, piece of hydrogel 0.8 grams, so instead of 0.193 grams to occupy all the butayne-1-ol sites(0.000017mols) an amount of 0.014 grams of dye is needed.
Data
NotesThis area is for any observations or conclusions that you would like to note.
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