Answer:
where is the figure . u can simply add them
Complete Question
A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?
Answer:
The mass of caffeine extracted is 
Explanation:
From the question above we are told that
The K value is 
The mass of the caffeine is 
The volume of water is 
The volume of caffeine is 
The number of times the extraction was done is n = 3
Generally the mass of caffeine that will be extracted is
![P = m * [\frac{V}{K * v_c + V} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%20m%20%20%2A%20%20%5B%5Cfrac%7BV%7D%7BK%20%2A%20%20v_c%20%2B%20V%7D%20%5D%5E3)
substituting values
![P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%2040%20%20%20%2A%20%20%5B%5Cfrac%7B2%7D%7B4.6%20%2A%20%202%20%2B%202%7D%20%5D%5E3)

The law of conservation of mass<span> states that </span>mass<span> in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.
</span>
Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M