User:Klare Lazor/Notebook/Chem-496-001/2013/02/11: Difference between revisions
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** 0.369mols of H2SO4/L * 2mols HCl/1mol of H2SO4 = 0.7M HCl, which is equal to 6ml of 12 M HCl in 94 ml of H2O | ** 0.369mols of H2SO4/L * 2mols HCl/1mol of H2SO4 = 0.7M HCl, which is equal to 6ml of 12 M HCl in 94 ml of H2O | ||
* To neutralize a NaHCO3 solution will be made, 0.7 M NaHCO3 | * To neutralize a NaHCO3 solution will be made, 0.7 M NaHCO3 | ||
''' Switching to lower molecular weight: PVOH (MW: 22,000) + phenol + glutaraldehyde + 30 ml of water''' | |||
*22,000 MW PVOH → 44g/mol | |||
*1 gram PVOH/ 22,000g/mole = 4.57E-5 m oles PVOH | |||
* 22,000 g/mol PVOH / 44 g/mol = 500 OH groups per a molecule of PVOH | |||
* 4.54E-5moles PVOH * 5000 oh groups/molecule of PVOH * 6.002 E23 molecules/mol = 1.369 E 22 oh groups in 1 gram of PVOH 22000 mw | |||
* we want one OH in Phenol to react with 10%, 1%,.1% and 0.01% of oh groups in PVOH | |||
* used the 10% stock left over from last time and reduce by a factor of ten for each percentage | |||
** 1%= 1ml of stock | |||
**.1%=0.1ml of stock | |||
**.01%= 0.01 ml of stock | |||
** 10%= whatever ml left over will end up close to 10% will calculate after measuring how much stock is left over | |||
==Data== | ==Data== |
Revision as of 12:13, 11 February 2013
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ObjectiveDescriptionCrosslinking PVOH (MW:146,000-186,000) Films With 0.7M HCl and neutralizing .6 Molar sodium bicarbonate
Switching to lower molecular weight: PVOH (MW: 22,000) + phenol + glutaraldehyde + 30 ml of water
Data
NotesThis area is for any observations or conclusions that you would like to note.
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