Dilute countercurrent immiscible extraction
A feed of 100.0 kg/min of a 1.2 wt % mixture of acetic acid in water is to be extracted with 1-butanol at 1 atm pressure and 26.7°C, we desire an outlet concentration of 0.1 wt % acetic acid in the exiting water. We have available solvent stream 1 that is 44.0 kg/min of pure 1-butanol and solvent stream 2 that is 30.0 kg/min of 1-butanol that contains 0.4 wt % acetic acid. Devise a scheme to do this separation, find the outlet flow rate and concentration of the exiting 1-butanol phase, and find the number of equilibrium contacts needed.
In Example we assumed that we were going to use all of the solvent available. There are other alternatives. Determine if the following alternatives are capable of producing outlet water of the desired acetic acid concentration.
a. Use only the pure solvent at the bottom of the extractor.
b. Mix all of the pure and all of the impure solvent together and use them at the bottom of the column.
c. Mix all of the pure and part of the impure solvent together and use them at the bottom of the column.

Answer:

i dob't know

Explanation:

Answer:

In  case whereby you  work for rev and check your work email on your mobile phone and You have decided to resign your job and take a position with a different company. Regarding rdm on your mobile phone, you should B. Take your phone to IT Services, and they will wipe it clean.

What is expected to do when leaving the company?

It is expected from you to return anything that belong to the company before leaving the company, but in the case whereby there is still a rdm on your mobile phone, then you can work to the IT services so they can wipe it away.

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Missing options:

A. Turn your phone to IT Services

B. Take your phone to IT Services, and they will wipe it clean

C. Download RDM to your personal computer

D. Call IT Services for support, and they will assist

E. Nothing. IT Services will automatically remove all data

Answer:true

Explanation:

It is true that around 50 percent of all sales of farm commodities are from dairy products.

Dairy products or milk products, also known as lacticinia, are milk-based foods.

Cows, water buffalo, nanny goats, and ewes are the most common dairy animals. Dairy products include common grocery store foods such as yogurt, cheese, and butter in the Western world.

Milk, yogurt, cheese, lactose-free milk, and fortified soy milk and yogurt are all part of the Dairy Group. The Dairy Group excludes milk-based foods with low calcium and a high fat content. Cream cheese, sour cream, cream, and butter are all examples of this.

Dairy products account for roughly half of all farm commodity sales.

Thus, the given statement is true.

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Answer:

What's that

Explanation:

Please brainlest me

Answer:

Presents the relevant parameters to be kept in mind when using the following types of motors: 1. AC induction motor 2. BLDC motor 3. Synchronous motorn

Explanation:

sorry

Answer: Hepatitis A

Explanation:

nagdudulot ng pamamaga ng atay

The most common use for encoders is to measure angular or linear distance, but encoders can also be used to perform speed or velocity measurements. In other words, as the encoder rotates faster, the pulse frequency increases at the same rate

Answer:

Liquids have less force of attraction between molecules than solids and more than gases; gases have a very weak force of attraction between the molecules which is the least amongst the three states of matter.

Explanation:

The person that is correct as regards statement made about stoichiometric ratio is; Tech B

In air-fuel mixture, when there is a mix air and fuel in the ideal amounts to make an engine to run efficiently,the ideal ratio of air to fuel is known Stoichiometric ratio.

This stoichiometric ratio is the comparison between the weight of the air relative to the weight of the fuel that it has been mixed with.

Now, weight is in lb and as such, tech B is correct because his ratio is 14.7lb to 1 lb.

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Answer:

Tech B

Explanation:

I took the test and got it right

Answer:

Mark me please as brainliest

Explanation:

A length of pipe will weigh the most when O A. part dimensions are such that the pipe is in the MMC. O B. the OD and ID are at the minimum allowable limit. OC. the OD and ID are at the maximum allowable limit. OD.part dimensions are such that the pipe is in the LMC

Answer:

Part dimensions are such that the pipe is in the MMC.

Explanation:

Answer:

yes i do.

Explanation:

it's pretty easy in my opinion but i dont know about you why do you ask?

Answer:

468

Explanation:

Based on the engineering analysis, the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF) between the output and the adjust terminal."

The LM1117 generally has a 1.2V at 800mA of load current.

The LM1117 is known to be a progression of low dropout linear voltage regulators. It is adjustable and be assigned between the output voltage of 1.25 to 13.8 V using just two external resistors.

Hence, in this case, it is concluded that the correct answer is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF), between the output and the adjust terminal.

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Answer: 8.33333333 or 6.1989778

Explanation:

Answer:

not sure

Explanation:

not sure

Answer:

30 1/2

Explanation:

Answer:

Elasticity

Explanation:

Cloud computing comprises of three (3) service models and these are;

1. Platform as a Service (PaaS).

2. Software as a Service (SaaS).

3. Infrastructure as a Service (IaaS).

9514 1404 393

Answer:

  (x, y) = (5.76, 1 5/7)

Explanation:

The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.

For the moment about the y-axis, we can define a differential of area as ...

  dA = (y2 -y1)dx

where y2 = √(x/k2) and y1 = k1·x^3

The distance of that area from the y-axis is simply x.

So, the x-coordinate of the centroid is ...

  [tex]\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\[/tex]

For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...

  [tex]a_x=115.2\\a=20\\c_x=5.76[/tex]

The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...

  [tex]\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143[/tex]

The centroid of the shaded area is ...

  (x, y) = (5.76, 1 5/7)

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 [tex]\frac{N}{kg}[/tex] · 1.50m

   = 35.316 [tex]\frac{N}{m}[/tex]

   = 35.316 W