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)
Moles He = 7.83 x 10^24 / 6.02 x 10^23 =13.0
<span>mass He = 13.0 mol x 4.00 g/mol = 52.0 g</span>
Answer:
The physical states that are represented by each graph region are the liquid and the solid, the highest temperature is the liquid and as it freezes it becomes a solid. The particles change because when it's a liquid, it isn't that compact it's just spreading smootly but as it freezes the atoms start to stick together and become compact.
Explanation:
Hope that made sense!
Answer:
Explanation:
We need to use the formula for heat of vaporization.
Identify the variables.
- The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
- Q is the energy, in this problem, 50,000 Joules.
- m is the mass, which is unknown.
Substitute the values into the formula.
We want to find the mass. We must isolate the variable, m.
m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.
Divide. Note that the Joules (J) will cancel each other out.
Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.
The mass is about 22 grams, so choice B is correct.
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
