Font (1) If the BOD5 of a waste is 220.0 mg/L and the ultimate BOD is 320.0 mg/L, what is the rate constant?

A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial DO of 8.55 mg/L; after five days, it has a final DO of 2.40 mg/L. Another bottle containing just the seeded dilution water has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L. Find the five-day BOD of the waste.

1. A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial Do of 8.55 mg/L; after 5 days, it has a final DO of 2.40 mg/L. Another bottle containing just the seeded dilution water has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L. Find the 5-days BOD of the waste.

2-14. Using the moisture contents (wet basis) of the waste components given in Table P2-14, calculate the overall moisture content of the waste having the composition also shown. This calculation can be simplified by assuming 100 lb of waste. Table P2-14 Moisture content, Percent by Component percent (wet basis) weight Food 70 10 Paper 6 33 Cardboard 5 8 Plastics 2 5 Textile 10 4 Rubber 2 3 Yard waste 60 18 Metals 3 10 Miscellaneous 6 9 2.Calculate the moisture content of the waste in Problem 2-14 on both a wet and a dry basis. Why the difference? Which number makes more sense to you, and why? 3.Determine the composition of the waste in Problem 2-14 if yard waste is sepa-rately collected. How would this affect the moisture content?