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romanna [79]
3 years ago
12

Ammonia is one of the chemical constituents of industrial waste that must be removed in a treatment plant before the waste can s

afely be discharged into a river or estuary. Ammonia is normally present in waste water as aqueous ammonium hydroxide (NH4OH-). A two-part process is frequently carried out to accomplish the removal. Lime (CaO) is first added to the wastewater, leading to the reaction
CaO + H2O --> Ca2+ + 2(OH-)
The hydroxide ions produced in this reaction drive the following reaction to the right, resulting in the conversion of ammonium ions to dissolved ammonia:

NH4+ + OH- <--> NH3(g) + H2O(l)
Air is then contacted with the wastewater, stripping out the ammonia.

a. One million gallons per day of alkaline wastewater containing 0.03 mol NH3/mole ammonia-free H2O is fed to a stripping tower that operates at 68oF. Air at 68oF and 21.3 psi contacts the wastewater countercurrently as it passes through the tower. The feed ratio is 300 ft3 air/gal wastewater, and 93% of the ammonia is stripped from the wastewater. Calculate the volumetric flow rate of the gas leaving the tower and the partial pressure of ammonia in this gas.
b. Briefly explain in terms a first-year chemistry student could understand how this process works. Include the equilibrium constant for the second reaction in your explanation.

Engineering
1 answer:
Yanka [14]3 years ago
5 0

Answer:

Following is attached the solution or the question given.

I hope it will help you a lot!

Explanation:

You might be interested in
Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho
marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0
3 years ago
Following are several z-transforms. For each one, determine inverse z-transform using both the method based on the partial-fract
tigry1 [53]

Answer:

For now the answer to this question is only for partial fraction. Find attached.

6 0
3 years ago
You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC
aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

<h3>What is PMCS?</h3>

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

  • M1114
  • M1151
  • M1123

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

Read more on PMCS here: brainly.com/question/15720250

#SPJ1

4 0
1 year ago
Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s
djyliett [7]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

3 0
3 years ago
A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

    T = T_m + (T_0-T_m)e^{kt}

where,

T_0 = initial temp

T_m = temp of room

T = temp after t hours

k = how fast the temp is changing

t = time (hours)

T_0 = 31     because the body was initlally 31ºC when the police found it

T_m = 22   because that was the room temp

T = 30  because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

k=???   we don't know k so we must solve for this

rearrange the equation to solve for k

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k

plug in the numbers to solve for k

k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}

k = \frac{ln(\frac{30 - 22}{31-22})}{1}

k=ln(\frac{8}{9})

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

T_0 = 37     because the body was 37ºC right after being killed

T_m = 22   because that was the room temp

T = 31  because the body temp when the police found it

k=ln(\frac{8}{9})   we solved this earlier

t = ???   we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t

plug in the values

t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}

t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0
3 years ago
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