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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Objective

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

This area is for any observations or conclusions that you would like to note.


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