ESC103H1 : Chapter 1-6 Review - PCHEM 3491 - Dr. Haber .pdf

<span;>The Only source of BOD in a river is untreated wastes that are being discharge from a food processing plant. The characteristics of untreated waste is not reported for the environmental protection agency of the town . To identify the waste water property from the processing plant the authority measured the DO of the river at 10km and 40km away from the plant and found 5mg/l and 2.5mg/l respectively. The average flow rate of the river is 0.25m/s and average depth is 4.5m with bed activity coefficient of 0.2.assume saturation DO 10mg/l

<span;>Parameter            river     waste water  

<span;>Flow,m3/s         0.7            0.23

<span;>BOD @ 18°C        2.5           N/A

<span;>DO, mg /l             8             N/A

<span;>Temperature       20            20

<span;>K @ 20°c             -                  0.23

<span;>a. What is the DO and BOD of the untreated waste form the processing plant? 

<span;>b. What is the critical DO and where downstream does it occur?

<span;>c. If the state requirement for minimum DO is 2mg/l is assimilative of the river restricted or not ?

18. Which choice below will cause all of the values stored in the array data to be printed out in reverse order? int[] data = {22, 17, 95, 16, 103, -14, 95}; // Choice A for (int i = data.length-1; i >= 0; i--) { print (data[i] + " "); // Choice B for (int i = 0; i < data.length; i++) { print (data[data.length-l-i] + " "); C. Both A and B will print all of the elements in reverse order. D. Neither A nor B will print all of the elements in reverse order

57. Line 5 (A) for (int j = -i*2; j<= i*2; j++){ (B) for (int j = 0; j< i+2+1; j++){ (C) for (int j = -i; j> i; j++) { (D) for (int j = 0; j<= i*2; j++){ (E) for (int j = i; j<= i+2; j+i) {