7.4. Determine the distribution coefficient required to extract 90 percent of a pollutant using no more than five extractions and a total solvent volume of 5 percent of the wastewater volume.
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- A lake with a surface area of 525 acres was monitored over a period of time. During a one-month period the inflow was 30 cubic-feet/second (cfs), the outflow was 27 cfs, and a 1.5-in. seepage loss was measured. During the same month, the total precipitation was 4.25 in. Evaporation loss was estimated as 6.0 in. Estimate the storage change for this lake during the month.An environmental study was performed to measure the transport of a DDT (a banned pesticide) throughout the globe. A lake in northern Canada was sampled to see if low levels of DDT were present there in spite of it not being administered within 1000 miles of that location. 1.000 L of the lake water was spiked with 100.0 uL of a 2.451 ug/L solution of tetrachlorobenzene, which acted as the internal standard. The entire 1.000 L spiked solution was concentrated onto an SPE column. After washing the SPE column with ~5 mL of pure water, the analyte and internal standard were eluted from the SPE column using ~5 mL of CH2Cl2 solvent. The eluted sample was evaporated under a stream of nitrogen to a final volume of 1.000 mL. 1 uL of this final solution was analyzed using GC-MS, and gave a signal of 24835 for DDT and 11597 for tetrachlorobenzene. To calculate the response factor, a standard solution was prepared consisting of 1.482 ug/L of DDT and 2.451 ug/L of tetrachlorobenzene.…Ethyl alcohol (C2H5OH) may be prepared by the fermentation of glucose (C6H12O6) as indicated by the equation: yeastC6H12O6 ----> C2H5OH + CO2 74.12 mL of ethyl alcohol (specific gravity = 0.790) was collected by this fermentation pro- cess. What mass of glucose was used?
- An orange juice processing plant now produces essential oil from orange peels. one It is known that 250 kg of peel comes out of 1 ton of oranges and 2.5 g of essential oil comes out of 1 kg of peel. In a laboratory study, 250 g of bark was treated with hexane solvent and 0.548 g of essential oil was obtained in the sample cup of the rotary evaporator. Accordingly, the rotary Calculate the separation efficiency obtained in the evaporator?Ethyl alcohol (C2H5OH) may be prepared by the fermentation of glucose (C6H12O6) as indicated by the equation: yeastC6H12O6 ----> C2H5OH + CO2 74.12 mL of ethyl alcohol (specific gravity = 0.790) was collected by this fermentation pro- cess. What mass of glucose was used? SET-UP: Answer:A stock solution of 75% v/v ethanol was diluted with distilled water to prepare 1L of 42% v/v ethanol. (Density of ethanol = 0.789 g/mL; MW of ethanol = 46 g/mol). Determine the initial volume (in mL) of 75% v/v ethanol used in the preparation. Final answer should NOT contain any unit. Type the VALUE only. Please observe 0 decimal places.
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