User:Elizabeth Ghias/Notebook/Experimental Chemistry/2012/10/05: Difference between revisions
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==Description== | ==Description== | ||
# 0.0258 g CuSO<sub>4</sub>·5H<sub>2</sub>O was dissolved in 0.5 L water in a volumetric flask to create a 13.23 ppm solution of CuSO<sub>4</sub>·5H<sub>2</sub>O. | |||
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*See below for calculations | *See below for calculations | ||
Revision as of 10:23, 8 October 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> |
ObjectiveTo remake the CuSO4·5H2O and Ha2SO4 standards Description
DataNo data was collected today.
20 ppm SO42- = 20 mg/L SO4 · 0.5 L = 10 mg SO42- · .1 g / 1000 mg = 0.01 g SO42- 0.01 g SO42- ÷ 96.06 g/mole = 1.041 × 10-4 moles SO42- · 1 mole CuSO4·5H2O / 1 mole SO42- = 1.041 × 10-4 moles CuSO4·5H2O 1.041 × 10-4 moles CuSO4·5H2O · 249.68 g/mole CuSO4·5H2O = 0.0260 g CuSO4·5H2O 1.041 × 10-4 moles CuSO4·5H2O · 1 mole Cu2+ / 1 mole CuSO4 = 1.041 × 10-4 moles Cu2+ 1.041 × 10-4 moles Cu2+ · 63.546 g/mole Cu2+ = 0.0066 g Cu2+ · 1000 mg / 1 g = 6.615 g Cu2+ ÷0.5 L: = 13.23 mg/L = 13.23 ppm Cu2+
NotesThis area is for any observations or conclusions that you would like to note.
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