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A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)
Correct answer is between '1000,1005'. Can you explain this answer?
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A 2 % sewage sample (in distilled water) was incubated for 3 days at 2...
°C. After incubation, the sample was diluted 1:10 and 1 ml of the dilution was then spread on a Petri dish with agar. The number of colonies that grew on the plate was 50.
To calculate the number of bacteria in the original sewage sample, we can use the following formula:
Number of bacteria = Number of colonies / Dilution factor
In this case, the dilution factor is 1:10, which means that the original sample was diluted 10 times before spreading it on the Petri dish. Therefore, the dilution factor is 10.
Number of bacteria = 50 colonies / 10 dilution factor = 5 bacteria/ ml
Since only 1 ml of the diluted sample was spread on the Petri dish, the number of bacteria in the original sewage sample can be calculated by multiplying the number of bacteria/mL by the total volume of the original sample.
Let's assume the original sample volume is V mL.
Number of bacteria in the original sample = 5 bacteria/mL * V mL
To calculate V, we need the initial volume of the sewage sample and the dilution factor.
If the initial volume of the sewage sample was 100 mL, then the total volume after dilution would be 100 mL * 10 dilution factor = 1000 mL.
Number of bacteria in the original sample = 5 bacteria/mL * 1000 mL = 5000 bacteria
Therefore, there are 5000 bacteria in the original sewage sample.
To calculate the number of bacteria in the original sewage sample, we can use the following formula:
Number of bacteria = Number of colonies / Dilution factor
In this case, the dilution factor is 1:10, which means that the original sample was diluted 10 times before spreading it on the Petri dish. Therefore, the dilution factor is 10.
Number of bacteria = 50 colonies / 10 dilution factor = 5 bacteria/ ml
Since only 1 ml of the diluted sample was spread on the Petri dish, the number of bacteria in the original sewage sample can be calculated by multiplying the number of bacteria/mL by the total volume of the original sample.
Let's assume the original sample volume is V mL.
Number of bacteria in the original sample = 5 bacteria/mL * V mL
To calculate V, we need the initial volume of the sewage sample and the dilution factor.
If the initial volume of the sewage sample was 100 mL, then the total volume after dilution would be 100 mL * 10 dilution factor = 1000 mL.
Number of bacteria in the original sample = 5 bacteria/mL * 1000 mL = 5000 bacteria
Therefore, there are 5000 bacteria in the original sewage sample.
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Community Answer
A 2 % sewage sample (in distilled water) was incubated for 3 days at 2...
= 500 mg/l
BOD3 (27°C) = L0(1 - e-kt)
500 = L0 (1 - e-0.23 x 3)
L0 = 1003 mg/l
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A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer?
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A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer?.
A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day-1 (at base e). The ultimate BOD (in mg/L) of the sewage will be___________ . (round off to the nearest integer)Correct answer is between '1000,1005'. Can you explain this answer?.
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