7.0 E 2 of a 50:50 mixture on n-propanol (CH3CH2CH2OH) and n-butanol (CH3CH2CH2CH2OH) are discharged into...

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7.0 E 2 of a 50:50 mixture on n-propanol (CH3CH2CH2OH) andn-butanol (CH3CH2CH2CH2OH) are discharged into a body of watercontaining 1.0 E 10 L of H2O. What is the BOD of this water?

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Dear Friend please read the following and you will findthe way how to calculate BOD With the provided datarange of BOD is in between 414The biochemical oxygen demand BOD is an empirical test in whichstandardised laboratory procedures are used to estimate therelative oxygen requirements of wastewaters effluents and pollutedwaters Micro organisms use the atmospheric oxygen dissolved inthe water for biochemical oxidation of organic matter which istheir source of carbon The BOD is used as an approximate    See Answer
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