Explanation:
The given data is as follows.
Partition coefficient, K = 6.2
Volume of phase 1, 
= 74.0 mL
Volume of phase 2, 
= 17.0 mL
So, after one extraction fraction of solute remaining is given as follows.
q = 
After 3 times extraction, fraction of S remaining is as follows.
q = ![[\frac{V_{1}}{V_{1} + KV_{2}}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BV_%7B1%7D%7D%7BV_%7B1%7D%20%2B%20KV_%7B2%7D%7D%5D%5E%7B3%7D)
Putting the given values into the above formula as follows.
q =
= ![[\frac{74.0 ml}{74.0 ml + 6.2 \times 17.0 ml}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B74.0%20ml%7D%7B74.0%20ml%20%2B%206.2%20%5Ctimes%2017.0%20ml%7D%5D%5E%7B3%7D)
= ![[\frac{74.0 ml}{179.4 ml}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B74.0%20ml%7D%7B179.4%20ml%7D%5D%5E%7B3%7D)
= 0.0699
Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.
Answer:
15 mL of the solute
Explanation:
From the question given above, the following data were obtained:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution is simply defined as:
Solution = solute + solvent
With the above formula, we can easily obtain the solute in the solution as follow:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution = solute + solvent.
50 = solute + 35
Collect like terms
50 – 35 = solute
15 = solute
Solute = 15 mL
Therefore, 15 mL of the solute is required.
It equals 0.005 but plz mark this as the brainliest question plz
Answer is: empirical formula is H₄N₂O₂.
m(unknown compound) = 8,5 g.
m(N₂) = 2,37 g.m(H₂O) = 3,04 g.
n(N₂) = m(N₂) ÷ M(N₂).
n(N₂) = 2,37 g ÷ 28 g/mol.
n(N₂) = 0,0846 mol.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 3,04 g ÷ 18 g/mol.
n(H₂O) = 0,168 mol.
n(H₂O) : n(N₂) = 0,168 mol : 0,0846 mol.
n(H₂O) : n(N₂) = 2 : 1.
Compound has four hydrogen, two oxygen and two oxygen.
It will be easier to answer if you show the choices.
