BOD5 = (DOinitial - DOfinal) * Dilution factor
Dilution factor = final volume after dilution / initial volume of sample
dilution factor = 1/P = 100
DOinitial = 8.15
DOfinal = 7.9
Therefore
BOD5 = (8.15 - 7.9) *100 = 25 mg/l
Given the following BOD data: Time (days) Diluted sample DO (mg/l). 7.95 3.75 3.45 2.75 2.15...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
0. What is the BODs of the wastewater sample if the DO values for the blank and diluted samples afher 5 days are R1 mg/L and 3.5 mg/L, respectively, and the wastewater is diluted from 2 miL to 200 milL? (10 points) If the 20 day BOD (assume that this is the ultimate BOD) is 700.0 mg/L determine the BOD rate constant k (in base e). (10 points) 0. What is the BODs of the wastewater sample if the DO...
5. The following data is for a BOD determination run on a raw sewage sample incubated for 5 days at 20°C in the dark; calculate the BOD5,20. Raw sewage contains a sufficient population of bacteria to conduct a BOD test. Therefore, this is an unseeded sample. Data: BOD bottle volume mL raw sewage added to BOD bottles DO, in bottle containing waste and dilution water DO in bottle containing waste and dilution water 300 mL 10.00 mL 7.70 mg/L 2.35...