How many oxygen atoms are in 1.40 g of quartz?

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

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

katovenus [111]

Answer: The final temperature of the coffee is 43.9°C

Explanation:

To calculate the final temperature, we use the equation:

$q=mC(T_2-T_1)$

where,

q = heat released = $1.0\times 10^3J=1000J$

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

$T_2$ = final temperature = ?

$T_1$ = initial temperature = 20°C

Putting values in above equation, we get:

$1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC$

Hence, the final temperature of the coffee is 43.9°C

6 0

3 years ago

In a liquid, the energy required to increase the surface of the area by a unit amount is called :a.dipole-dipole force. b.viscos

Sedaia [141]

Answer:

D. Surface tension.

Explanation:

Surface tension is defined as the energy required to increase the surface area of a liquid by a unit amount.

The surface tension of a liquid results from an imbalance of intermolecular attractive forces, the cohesive forces between molecules:

A molecule in a liquid experiences cohesive forces with other molecules in all directions while molecules at the surface of a liquid experiences only net inward cohesive forces.

5 0

3 years ago

Which of the following lists contain only elements?

N76 [4]

In which elements are belonging to? I will grant that option B ; ti's the best answer.
In which thou might contain the elements zinc , gold , aluminium , and last by not least oxygen.

I truly hope ti's answer helps thou.

7 0

3 years ago

If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K and then I raise the pressu

mixas84 [53]

Answer:

The answer to your question is V2 = 29.6 l

Explanation:

Data

Pressure 1 = P1 = 12 atm

Volume 1 = V1 = 23 l

Temperature 1 = T1 = 200 °K

Pressure 2 = 14 atm

Volume 2 = V2 = =

Temperature 2 = T2 = 300°K

Process

1.- To solve this problem use the Combine gas law.

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (12)(23)(300) / (200)(14)

3.- Simplification

V2 = 82800 / 2800

4.- Result

V2 = 29.6 l

5 0

3 years ago