Difference between revisions of "User:Melissa Novy/Notebook/CHEM-572/2013/03/27"

From OpenWetWare
< User:Melissa Novy‎ | Notebook‎ | CHEM-572‎ | 2013‎ | 03
Jump to: navigation, search
(Autocreate 2013/03/27 Entry for User:Melissa_Novy/Notebook/CHEM-572)
 
Line 15: Line 15:
 
<!-- ##### DO NOT edit above this line unless you know what you are doing. ##### -->
 
<!-- ##### DO NOT edit above this line unless you know what you are doing. ##### -->
  
==Entry title==
+
==Objectives==
* Insert content here...
+
* Calculate the amount of silver-loaded clay necessary to achieve theoretical leaching of 2.5 μmol Ag<sup>+</sup> in 50 mL of solution.
 +
 
 +
  '''MW Ag<sup>+</sup>: 107.87 g/mol'''
 +
  '''MW AgNO<sub>3</sub>: 169.87 g/mol'''
 +
  '''Given that 3.662 μmol Ag<sup>+</sup> was exchanged into 1.49711 g LMT, (3.662 μmol Ag<sup>+</sup>)/(1.49711 g LMT) = (2.5 μmol Ag<sup>+</sup>)/(x g LMT)'''
 +
  '''x = 1.02206 g LMT'''
 +
 
 +
* Therefore, 2.5 μmol Ag<sup>+</sup> is present in 1.02206 g of 100AgLMT.  The corresponding control film must be 3 g PLA2002D with 1.02206 g LMT.
 +
 
 +
PLA2002D + 33wt% LMT
 +
* Actual mass PLA2002D: 3.03753 g
 +
* Actual mass LMT: 1.02205 g
  
  

Revision as of 11:02, 2 April 2013

<!-- sibboleth --><div id="lncal1" style="border:0px;"><div style="display:none;" id="id">lncal1</div><div style="display:none;" id="dtext"></div><div style="display:none;" id="page">User:Melissa Novy/Notebook/CHEM-572/2013/03/27</div><div style="display:none;" id="fmt">yyyy/MM/dd</div><div style="display:none;" id="css">OWWNB</div><div style="display:none;" id="month"></div><div style="display:none;" id="year"></div><div style="display:none;" id="readonly">Y</div></div>

BDLlogo 300.png <sitesearch>title=Search this Project</sitesearch>

Customize your entry pages <html><img src="/images/a/aa/Help.png" border="0" /></html>

Objectives

  • Calculate the amount of silver-loaded clay necessary to achieve theoretical leaching of 2.5 μmol Ag+ in 50 mL of solution.
  MW Ag+: 107.87 g/mol
  MW AgNO3: 169.87 g/mol
  Given that 3.662 μmol Ag+ was exchanged into 1.49711 g LMT, (3.662 μmol Ag+)/(1.49711 g LMT) = (2.5 μmol Ag+)/(x g LMT)
  x = 1.02206 g LMT
  • Therefore, 2.5 μmol Ag+ is present in 1.02206 g of 100AgLMT. The corresponding control film must be 3 g PLA2002D with 1.02206 g LMT.

PLA2002D + 33wt% LMT

  • Actual mass PLA2002D: 3.03753 g
  • Actual mass LMT: 1.02205 g