Answer:
Relative volume of ether to water that should be used for the extraction = 1.205
Explanation:
The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).
K = (C₁/C₂)
Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).
If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.
Let the volume of ether required be x mL.
Let the volume of water required be y mL.
Relative volume of ether to water that should be used for the extraction = (x/y)
Mass concentration of the organic substance X in ether = (0.94/x)
Mass concentration of organic substance X in water = (0.06/y)
The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.
13 = (0.94/x) ÷ (0.06/y)
13 = (0.94/x) × (y/0.06)
13 = (15.667y/x)
(x/y) = (15.667/13) = 1.205
Hope this Helps!!!
Optimization helps you make better choices when you have all the data, and simulation helps you understand the possible outcomes when you don’t. Frontline Solvers enable you to combine these analytic methods, so you can make better choices for decisions you do control, taking into account the range of potential outcomes for factors you don’t control.
Answer:
Decrease
Explanation:
The lower you go, the colder because the sun can't reach lower levels of the ocean.
Answer:
Depending on the thermometer, it may have the ability to go as high or low as melting, freezing or boiling point for water. Just make sure you know the boiling, melting and freezing points in Celsius, Fahrenheit and/or Kelvin and read your thermometer accordingly.
Explanation:
As far as I remember, the needed formula for squaric acid is C4H2O4.
According to this one mole should be 114.06 g., which means we have <span>0.015mol of this acid.
Then we can easly calculate : </span><span>4(0.015) = 0.06 mol for both for Carbon and Oxygen and </span><span>0.03 mol of Hydrogen.
</span><span>To get more clear answer, we multiply by avogadros :
</span><span>6.022 x 10^23. Hope everything is clear! regards</span>
