for 5 stars please show ALL WORK and CLEARLY (a suggestion; solvtion could be pr
ID: 1920049 • Letter: F
Question
for 5 stars please show ALL WORK and CLEARLY
(a suggestion; solvtion could be presented as a picture of solution on paper and upload if possible)
Consider the following reaction representing the oxidation (combustion) of benzene by oxygen: C6H6 + O2 = CO2 + H2O Balance the equation. Note here that for any "hydrocarbon" compound, each C will be oxidized to one CO2 and so need one molecule of O2, and every two H atoms will be oxidized to one H2O and so will need 1/4 molecule of O2. Say you have water that contains 1 ppm (1 mg/L) of dissolved benzene. What is the BOD of that water in units of mg/L of O2? Find the KH (mols/L-atm) for O2 at 10 degree C and compute the level of dissolved oxygen (DO) in mg/L at that temperature for saturation with atmospheric levels of O2. You will need the fact that the MW of O2 is close to 32,000 mg/mol. Compare the results from parts b and c and comment about the likelihood that fish in the subject water will be facing unacceptably low DO levels if re-aeration as a source of O2 does not keep up with the expected degradation of benzene. Note here that benzene has been around in the environment "forever", and so bacteria etc. have long ago learned how to oxidize this compound and thereby use it as a source of energy.Explanation / Answer
C6h6 + 15/2o2 --> 3h2o + 6co2
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