What is the molarity of a barium hydroxide solution if 93.9 mL is completely titrated by 15.3 mL of 0.247 M sulfuric acid soluti
1 answer:
You might be interested in
Answer:
the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.
Explanation:
Let state the given parameters;
Let A= solvent (hexane)
B= solid(inert soiid)
C= solvent(oil)
= mass of solvent + mass of oil (i.e A+C)
<u>Feed Phase:</u>
Total feed (i.e slurry of flakes soybeans)= 100kg
B= mass of solid =75 kg
F= mass of solvent + mass of oil (i.e A+C)
= 25kg
Mass ratio of oil to solution =
mass of oil (C) =25 × 0.1 wt = 2.5kg
mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg
therefore =
= 0.1
mass ratio of solid to solution =
=3
<u>Solvent Phase:</u>
C= Mass of oil= 0(kg)
A= Mass of hexane = 100kg
mass of solutions = A+C = 0+100kg
solvent= 100kg
<u>Underflow:</u>
underflow = L₁ = (unknown) ???
L₁ = E₁ + B
the value of N for the outlet and underflow is 1.5 kg
i.e N₁ =
solution in underflow E₁ = Mass (A+C)
<u>Overflow:</u>
Overflow = V₁ = (unknown) ???
solution in overflow V₁ = Mass (A+C)
This is because, B = 0 in overflow
Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).
solid in feed = solid in underflow = 75
75= E₁ × N₁
75 = E₁ × 1.5
E₁ = 50kg
Underflow L₁ = E₁ × B
= 50 + 75
=125kg
The Overall Balance: Feed + Solvent = underflow + overflow
100 + 100 = 125 + V₁
V₁ = 75kg