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Burka [1]
3 years ago
8

(15 points) A BOD test is run using 50 mL of treated wastewater mixed with 250 mL of pure water. The initial DO of the mix is 9.

0 mg/L. After 5 days, the DO is 4.5 mg/L. After a long period of time, the DO is 2.5 mg/L, and it no longer seems to be dropping. Assume there is no nitrogenous BOD and that all BOD is carbonaceous. a. What is the 5-day BOD of the wastewater
Chemistry
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

27 mg/L

Explanation:

From the given information:

To determine the 5-day BOD using the formula:

BOD_5 = \dfrac{DO_i -DO_f}{P} \ \ ---(1)

where;

BOD = biochemical oxygen demand

DO_i = initial dissolved oxygen

DO_f = final dissolved oxygen

P = dilution factor

The dilution factor P = \dfrac{V_w}{V_m}

where;

V_w = \text{volume of waste water} \\ \\  V_m = \text{volume of mixture}

P = \dfrac{50}{300} \\ \\  P = 0.167

replacing the value of P into equation  (1);

BOD_5 = \dfrac{(9.0 - 4.5)mg/L}{0.167}

BOD_5 = \dfrac{(4.5)mg/L}{0.167}

\mathtt{BOD_5 = 26.946 \ mg/L}

\mathtt{BOD_5 \simeq 27 \ mg/L}

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In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

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Calculate the volume of one mole of a gas at 1.00 atm pressure and 0 °C.
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Answer:

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