User:Klare Lazor/Notebook/Chem-496-001/2013/02/11

(Difference between revisions)
 Revision as of 15:04, 11 February 2013 (view source) (→Description)← Previous diff Revision as of 15:13, 11 February 2013 (view source) (→Description)Next diff → Line 20: Line 20: ** 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 15:13, 11 February 2013

Biomaterials Design Lab Main project page
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Description

Crosslinking PVOH (MW:146,000-186,000) Films With 0.7M HCl and neutralizing .6 Molar sodium bicarbonate

• To high of a molecular weight will remake with with a lower molecular weight
• MW: 22,000 will be used instead for this....higher molecular weight is to thick for this type of reaction with crosslinking with glutaraldehyde
• 20ml of 100% H2So4 in 1 L of H20 was use for the fixant will determine amount of H+ ions in solution to determine concentartion of HCL
• 20ml* 1.84g/ml ÷ 98.079g/mol = 0.375moles of H2SO4 ÷ 1 Liter= 0.369 Molar H2SO4
• assume complete dissociation
• 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

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

• Add data and results here...

Notes

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