User:Allison K. Alix/Notebook/CHEM-581/2012/10/26: Difference between revisions
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<div class="center" style="width:auto; margin-left:auto; margin-right:auto;">Preparation of 50μM Fe<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub></div> | <div class="center" style="width:auto; margin-left:auto; margin-right:auto;">Preparation of 50μM Fe<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub></div> | ||
<math>\frac {278.01 g} {mol} \times \frac {50 x 10^-3 mol} {L} \times \frac {1 L} {1000 mL} \times 10mL = 0.138 g | <math>\frac {278.01 g} {mol} \times \frac {50 x 10^-3 mol} {L} \times \frac {1 L} {1000 mL} \times 10mL = 0.138 g</math> | ||
0.138 g Fe<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub> in 10mL H<sub>2</sub>O to get 50mM. Dilute to 50μM by using 0.01 mL (10μL) in 9.99mL (9990μL) H<sub>2</sub>O | |||
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Revision as of 13:30, 6 November 2012
Experimental Chemistry | <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page <html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>Previous entry<html> </html>Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html> |
Objective
Description
DataPreparation of 50μM H2SO4
[math]\displaystyle{ \frac {50 x 10^-3 mol} {L} \times \frac {98.07 g} {mol} \times 10mL \times \frac {1L} {1000 mL} \times \frac {1mL} {1.84 g} = 0.0266mL }[/math] 0.0266mL = 26.65μL Preparation of 50μM Fe2(SO4)3
[math]\displaystyle{ \frac {278.01 g} {mol} \times \frac {50 x 10^-3 mol} {L} \times \frac {1 L} {1000 mL} \times 10mL = 0.138 g }[/math] 0.138 g Fe2(SO4)3 in 10mL H2O to get 50mM. Dilute to 50μM by using 0.01 mL (10μL) in 9.99mL (9990μL) H2O NotesThis area is for any observations or conclusions that you would like to note.
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