Five-days BOD of a 10% diluted sample having Do = 6.7 mg/l, D = 2 mg/l and consumption of oxygen in blank = 0.5 mg/l, will be A 22 mg/l B 42 mg/l c 62 mg/l D 82 mg/l
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- Five-days BOD of a 10% diluted sample having Do = 6.7 mg/l, D = 2 mg/l and consumption of oxygen in blank = 0.5 mg/l, will be A 22 mg/l B 42 mg/l c 62 mg/l D 82 mg/lThe following dissolved-oxygen values were found after 5days of incubation in 310-mL BOD bottles: 7.7, 7.9 and 7.9 in three blank samples; 6.5, 4.0, and 0.5mg/L in bottles containing 2,5, and 10 mL of sample, respectively. The 0-day dissolved-oxygen of the sample was 0.0 mg/L. What is the 5-day BOD of the sample?Read the question carefully and give me all right solutions. If you don't know the solution please leave it but don't give me wrong solution. If the initial dissolved oxygen of a 3 mL sample is 7.5 mg/L and after day 5 is 5.5 mg/L, with a k rate determined to be 2 mg/L. what is the BOD5 ? Ultimate BOD ? BOD 2 ? BOD 10? . Create a BOD graph (time vs BOD) .
- Calculate the BOD, Initial Sample D.O. = 8.1 mg/L 5-Day Sample D.O. = 2.1 mg/L Vol. Of Sample in 300 mL Bottle 60 mL Select one: a. 120 mg/L O b. 30 mg/L O c. 60 mg/L O d. 90 mg/LH Gy Q#4:The table shows data of a wastewater sample collected from a river with a temperature of 30 °C analyzed for the BODs test. The rate constant is 0.2 day-¹ at the lab temperature (20 C). Find BODs at the river. Initial dissolved oxygen, mg/L 9 What will be the BOD7 at the river? Final dissolved oxygen, mg/L 3.5 Sample volume, ml 35Suppose the estimated BOD of an influent sample is expected to be 150 mg/L and the dissolved oxygen concentration of the oxygen saturated dilution water used in the BOD test is 8.5 mg/L. If you are using a 300 mL BOD bottle, estimate the maximum and minimum amount of sample you would add to the BOD bottle to ensure satisfactory test results. NOTE: Acceptable BOD test should comply with the two essential criteria: DO final ≥ 2 mg/L and DO initial – DO final ≥ 2 mg/L. Assume that the dilution water is the only source of oxygen.
- A column 3.00 cm in diameter and 32.6 cm long gives adequate resolution of a 72.4-mg mixture of unknowns, initially dissolved in 0.500 mL. (a) If you wish to scale down to 10.0 mg of the same mixture with minimum use of chromatographic stationary phase and solvent, what length and diameter column would you use? (b) In what volume would you dissolve the sample? (c) If the flow rate in the large column is 1.85 mL/min, what should be the flow rate in the small column?23.22 The following dissolved-oxygen values were found after 5 days of incubation in 310-mL BOD bottles: 7.7, 7.9, and 7.9 in three blank samples; 6.5, 4.0, and 0.5 mg/L in bottles containing 2, 5, and 10 mL of sample, respectively. The 0-day dissolved oxygen of the sample was 0.0 mg/L. What is the 5-day BOD of the sample?An Analysis of water from a surface stream yields the following results Ca+2=60mg/L HCO3-=115mg/L Mg+2=10mg/L SO4-2=96mg/L Na+=7mg/L NO3-=10mg/L K+=20mg/L CI-=11 mg/L If an error of 10 percent is acceptable, should the analysis be considered complete?
- For a BOD analysis, 30 ml of waste with a DO of zero mg/L, is mixed with 270 ml of dilution water with a DO of 9 mg/L. The sample is then put in an incubator. The final DO is measured after 7 d. The final DO is measured at 3.0 mg/L. However, it is discovered that the incubator was set at 30°C. Assume a k1 of 0.2 d–1 (base e) at 20°C and θ = 1.05. Determine the 5 d, 20°C BOD of the sampleDetermine the ultimate BOD of a waste water sample which was subjected to the BOD determination as follows: 6 mL of waste water containing no dissolved oxygen (D.O) was mixed with 294 mL, of dilution water containing 8.6 mg/L of D.O. After incubation at 20° C for 5 days, the D.O. of the mixture was 5.4 mg/L. The BOD rate constant, k to the base e is 0.25/d.Q5/ In a test to determine the bulk modulus of a liquid it was found that as the absolute pres- sure was changed from 15 to 3000 psi the volume decreased from 10.240 to 10.138 in.³ Determine the bulk modulus for this liquid.