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bearhunter [10]
2 years ago
14

The dilution factor D for an unseeded mixture of wastewater is 0.05. The DO of the mixture is initially 8.0 mg/L, and after 5 da

ys it has dropped to 2.0 mg/L. The reaction rate constant K has been found to be 0.20 day? (i) What is the 5-day BOD of the waste? (ii) What would be the ultimate carbonaceous BOD?
Chemistry
1 answer:
enyata [817]2 years ago
6 0

Answer:

(i) 5-day BOD of the waste is 120 mg/l.

(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.

Explanation:

The dilution factor D is 0.05.

The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.

The BOD of the waste for an unseeded mixture is

BOD_5=(DO_5-DO_0)/D=(8-2)/0.05=6/0.05=120mg/l

The ultimate carbonaceous BOD (Lo) can be calculated as

L = L_o*10 ^{- k_1t} \\L_o=L/10^{- k_1t}=2/10^{- 0.2*5}=2/10^{- 1}=2*10=20mg/l

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Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

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Heat transfer is the phenomenon that occurs when the two objects are in the vicinity of each other and by increasing the area of their contact. Thus, option B is correct.

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