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 .
You have to do a report :
So
the truck efficiency is 38%.
Answer:
The veins that carry oxygenated bloof back into the heart are the pulmonary arteries.
Explanation:
Oxygen-rich blood flows from the lungs back into the left atrium (LA), or the left upper chamber of the heart, through four pulmonary veins. Oxygen-rich blood then flows through the mitral valve (MV) into the left ventricle (LV), or the left lower chamber.
Answer:
The concentration of the solution is 1.364 molar.
Explanation:
Volume of perchloric acid = 29.1 mL
Mass of the solution = m
Density of the solution = 1.67 g/mL
Percentage of perchloric acid in 48.597 solution :70.5 %
Mass of perchloric acid in 48.597 solution :
= 
Moles of perchloric acid = 
In 29.1 mL of solution water is added and volume was changed to 250 mL.
So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)
The concentration of the solution is 1.364 molar.
