Explanation:
The given data is as follows.
Solvent 1 = benzene, Solvent 2 = water
= 2.7, 
= 100 mL
= 10 mL, weight of compound = 1 g
Extract = 3
Therefore, calculate the fraction remaining as follows.
![f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5B1%20%2B%20K_%7Bp%7D%28%5Cfrac%7BV_%7BS_%7B2%7D%7D%7D%7BV_%7BS_%7B1%7D%7D%7D%29%5D%5E%7B-n%7D)
= ![[1 + 2.7(\frac{100}{10})]^{-3}](https://tex.z-dn.net/?f=%5B1%20%2B%202.7%28%5Cfrac%7B100%7D%7B10%7D%29%5D%5E%7B-3%7D)
= 
= 
Hence, weight of compound to be extracted = weight of compound - fraction remaining
= 1 -
= 0.00001
or, = 
Thus, we can conclude that weight of compound that could be extracted is .
Answer:
Carbon forms the large numbers of compound due to the following reasons
<em>Answer</em><em>:</em>
<em>Glycolysis</em>
<em>E</em><em>xplanation</em><em> </em><em>:</em>
Glycolysis is the first step in the breakdown of glucose to extract energy for cell metabolism.Many living organisms carry out glycolysis as part of their metabolism. Glycolysis takes place in the cytoplasm of most prokaryotic and all eukaryotic cells.
Answer: Water because if we say aqueous it means it resembles water or related to water.
Answer:
All cells have structural and functional similarities. Structures shared by all cells include a cell membrane, an aqueous cytosol, ribosomes, and genetic material (DNA). All cells are composed of the same four types of organic molecules: carbohydrates, lipids, nucleic acids, and proteins.
Explanation:
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