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
.
C ( Porcupines use there’s sharp quills to defend themselves from larger predators!)
Lord dragon you are wrong.
The answer is A: Incandescent light bulbs are almost 100 percent efficient. Because b-d do not make any sense.
Hey There!:
Molar Mass KI => 166.003 g/mol
* number of moles:
n = mass of solute / molar mass
n = 49.8 / 166.003
n = 0.3 moles KI
Therefore:
M = n / V
M = 0.3 / 1.00
M = 0.3 mol/L
hope this helps!
PH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg[OH⁻]
pH=14+lg(8.7*10⁻¹²)=14-11.06=2,94