User:Anh Do/Notebook/CHEM 481 Polymer/2016/10/19: Difference between revisions

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(Autocreate 2016/10/19 Entry for User:Anh_Do/Notebook/CHEM_481_Polymer)
 
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==Entry title==
==Testing Beads' filtration ability - Heavy Metals==
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1. Preparation of solution for 20 ppm concentration




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==Preparation of Stock Solution==
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1. ZnCl<sub>2</sub> - 120 ppm
* 0.00266 g of ZnCl<sub>2</sub> in 10 mL
** 10mL*1L/1000mL*120mg Zn/1L*138.315 g/mol ZnCl<sub>2</sub>/65.38 g/mol Zn = 2.66 mg ZnCl<sub>2</sub>
2. CuCl2 - 100 ppm
* 0.00266 g of CuCl<sub>2</sub> in 10 mL
**10mL*1L/1000mL*100mg Cu/1L*134.45 g/mol CuCl<sub>2</sub>/63.54 g/mol Cu = 2.116 mg CuCl<sub>2</sub>


__NOTOC__
3. Pb(Ac)<sub>2<sub/> - 100 ppm
* 0.00266 g of PbAc<sub>2</sub> in 10 mL
**10mL*1L/1000mL*100mg Pb/1L*325.29 g/mol PbAc/207.2 g/mol Cu = 1.57 mg PbAc<sub>2</sub>

Revision as of 06:07, 19 October 2016

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Testing Beads' filtration ability - Heavy Metals

1. Preparation of solution for 20 ppm concentration


Preparation of Stock Solution

1. ZnCl2 - 120 ppm

  • 0.00266 g of ZnCl2 in 10 mL
    • 10mL*1L/1000mL*120mg Zn/1L*138.315 g/mol ZnCl2/65.38 g/mol Zn = 2.66 mg ZnCl2

2. CuCl2 - 100 ppm

  • 0.00266 g of CuCl2 in 10 mL
    • 10mL*1L/1000mL*100mg Cu/1L*134.45 g/mol CuCl2/63.54 g/mol Cu = 2.116 mg CuCl2

3. Pb(Ac)2 - 100 ppm

  • 0.00266 g of PbAc2 in 10 mL
    • 10mL*1L/1000mL*100mg Pb/1L*325.29 g/mol PbAc/207.2 g/mol Cu = 1.57 mg PbAc2
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