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](https://tex.z-dn.net/?f=BOD_5%3D%28DO_5-DO_0%29%2FD%3D%288-2%29%2F0.05%3D6%2F0.05%3D120mg%2Fl)
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](https://tex.z-dn.net/?f=L%20%3D%20L_o%2A10%20%5E%7B-%20k_1t%7D%20%5C%5CL_o%3DL%2F10%5E%7B-%20k_1t%7D%3D2%2F10%5E%7B-%200.2%2A5%7D%3D2%2F10%5E%7B-%201%7D%3D2%2A10%3D20mg%2Fl)
0.761 mol of
would you have if you have 76.36 grams.
<h3>What is a mole?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
Mass=76.36 grams
![Moles = \frac{mass}{molar \;mass}](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20%5C%3Bmass%7D)
![Moles = \frac{76.36 grams}{100.21 g/mol}](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7B76.36%20grams%7D%7B100.21%20g%2Fmol%7D)
Moles = 0.761
Hence, 0.761 mol of
would you have if you have 76.36 grams.
Learn more about moles here:
brainly.com/question/8455949
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Answer:
O3 => CLO + O2
Explanation:
ozone gas yields one Chlorine Monoxide gas, and two Oxygen gas. i forgot how to explain it well, but I know how to do it and will help in the future if needed :)