Brass is an alloy of

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

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)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

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 $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

=3

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

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

5 0

3 years ago

Of the choices below, which is true for the relationship shown? it is ka for the acid h3p2o72â. it is kb for the acid h3p2o72â.

Semenov [28]

A For PLATO. just did it

3 0

3 years ago

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What is the summary at the end of an experiment that explains the results

jasenka [17]

Answer:

The conclusion

Explanation:

6 0

3 years ago

A ______________________ is the distance from one compression to the next compression in a longitudinal wave.

svetoff [14.1K]

Answer: D

Explanation:

8 0

3 years ago

Consider the reaction: 2 H2O (g)-->2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag

denis-greek [22]

DE = dH - PdV

2 H2O(g) → 2 H2(g) + O2(g) 

You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. 

1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. 

Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas 

2 moles reacts to form 3 moles 

The gas equation is 

PV = nRT 
P = pressure 
V = volume (unknown) 
n = moles (1) 
R = gas constant = 8.314 J K^-1 mol^-1 
- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) 
T = temp in Kelvin (kelvin = deg C + 273.15 
So T = 403.15 K 

Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say 

dE = dH -dnRT (because PV = nRT) 

Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules 

dH = 483.6 kJ/mol = 483600 Joules/mol 

dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) 
dE = 483600 J/mol - 3351.77 J 
dE = 480248.23 J/mol 
dE = 480.2 kJ/mol

5 0

3 years ago

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Brass is an alloy of​

Brass is an alloy of