a6 1Smoke inhaled by a cigarette smoker contains roughly 003

a=6

1-Smoke inhaled by a cigarette smoker contains roughly 0.03 mg per cigarette of benzo(a)pyrene. For a (75-a) kg individual who smokes (20+a) cigarettes per day for (40+a) years, estimate the lifetime risk of cancer caused by that benzo(a)pyrene.

2-Estimate the cancer risk for a 70-kg individual consuming (300+10a) g of fish for dinner once every week for (60+a) years from a stream with (20+a) ppb of DDT.

3-A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial DO of (8.50 + 0.01a) 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 (8.70 + 0.01a) and a final DO of 8.53 mg/L. Find the five-day BOD of the waste.

Solution

1.consumption of 13096.2mg of benzo a pyrene in 46 yrs

3. BOD of the sample = (D₁-D₂) (Volume of diluted sample/Vol of undiluted sewage sample)

                                 =(D₁-D₂) x(dilution factor)

when pure distilled water is not used for dilution of the sewage sample, and the aged domestic water or polluted water containing some organic matter is used, then evidently some of the DO shall be consumed by the diluting water. the BOD₅ of sewage sample can in such a case be calculate by using the modified equation, given as

D₁=8.56mg/l , D₂=2.40 mg/l

B₁=8.76mg/l, B₂=8.53 mg/l

f=240/270 (ratio of dilution water volume in waste water test to the dilution water volume in BOD test on dilution water)

dilution factor= 270/30

BOD₅=[(D₁-D₂)- (B₁-B₂)f] x dilution factor

      =[(8.56-2.40) - (8.76-8.53) 240/270]270/30

      = 53.604 mg/l