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Plot on to graph moles of S2O8 vs time to achieve M/s. The initial concentration of S2O8: 0.05M (added is 2.5mL of 0.200M, but it was diluted so 0.200 x 2.5/10mL = 0.05M).  Below is the time (s) and moles consumed. 188s - 0.00002 moles 362s - 0.00004 moles 548s  - 0.00006 moles 740s - 0.00008 moles 1001s - 0.0001 moles I was told to plot concentration vs time then subtract the moles, but I'm confused how to do that.

Plot on to graph moles of S2O8 vs time to achieve M/s. The initial concentration of S2O8: 0.05M (added is 2.5mL of 0.200M, but it was diluted so 0.200 x 2.5/10mL = 0.05M).  Below is the time (s) and moles consumed. 188s - 0.00002 moles 362s - 0.00004 moles 548s  - 0.00006 moles 740s - 0.00008 moles 1001s - 0.0001 moles I was told to plot concentration vs time then subtract the moles, but I'm confused how to do that.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Plot on to graph moles of S2O8 vs time to achieve M/s.

The initial concentration of S2O8: 0.05M (added is 2.5mL of 0.200M, but it was diluted so 0.200 x 2.5/10mL = 0.05M). 

Below is the time (s) and moles consumed.

188s - 0.00002 moles

362s - 0.00004 moles

548s  - 0.00006 moles

740s - 0.00008 moles

1001s - 0.0001 moles

I was told to plot concentration vs time then subtract the moles, but I'm confused how to do that. 

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