I really don't know what the answer is
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 .
A. A force that acts on a body moving in a circular path
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
Answer:
20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED
Explanation:
M1 V1 = M2 V2
M1 = Molarity of sodium hydroxide = 3 M
V1 = volume of sodium hydroxide = 40 mL
M2 = Molarity of hydrochloric acid = 6 M
V2 = Volume of hydrochloric acid = unknown
Rearranging the equation, we have:
V2 = M1 V1 / M2
V2 = 3 * 40 mL / 6
V2 = 120 / 6
V2 = 20 mL
To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.
