gg425h3a - 4.1E-08 4.1E-08 2.06E-04 98 2.06E-04 98-20...

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GG425 Fall 2004 - HW#3 problem 1 solution given M 1.1E-08 100% Total PO4 @ 100% O2 saturation O2 saturation Thames River M 1.6E-07 Total NO3 @ 100% O2 saturation Redfield gain in PO4 = loss in O2/138 M 2.10E-04 O2 conc For part B, you use the relationship: (change in O2)/138 = - (change in P) note: Change in P is that which must be added to the standing crop in the water as O2 is utilized O2 initial = saturation, O2 final = the observed value P initial = P left at O2 saturation (a very low amount) 1.1E-08M + (change in O)/138 P final = Pintial + Paddedby respiration -----> PART B PART A 1950s 1893 Tot PO4 Tot PO4 1950s O2 conc 1950s O2% 1893 O2 conc 1893 O2% Dist below London Bridge
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Unformatted text preview: 4.1E-08 4.1E-08 2.06E-04 98 2.06E-04 98-20 1.4E-06 6.3E-07 2.52E-05 12 1.24E-04 59-10 1.5E-06 1.1E-06 1.05E-05 5 5.67E-05 27 1.5E-06 1.2E-06 2.10E-06 1 5.25E-05 25 10 1.5E-06 1.1E-06 2.10E-06 1 6.30E-05 30 20 1.4E-06 5.0E-07 2.10E-05 10 1.43E-04 68 30 5.6E-07 2.4E-07 1.34E-04 64 1.79E-04 85 40 Notice the inverted shape of the P curve relative to O 1.0E-06 5.1E-05 1.0E-04 1.5E-04 2.0E-04 2.5E-04 O2 (mole/L) 0.00E+00 2.00E-07 4.00E-07 6.00E-07 8.00E-07 1.00E-06 1.20E-06 1.40E-06 1.60E-06 PO4 (mol/L)-20 -10 0 10 20 30 40 Distance Below London Bridge O2 1893 O2 1950s PO4 1893 PO4 1950s Thames River...
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This note was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Yoshimoto during the Fall '05 term at Hawaii.

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