Category: Chemical Engineering homework help

 
 You are an engineer at a mining company which sells a mineral ore in three quality grades: Low, Medium, and High. The company owns three mines which produce this ore: Mine A, Mine B, and Mine C. Each mine can produce different quantities (in tons) of each grade of ore per day as shown below: 
 Where xy are the last two digits of your URN. For example, for the URN 6835725, xy = 25. In this case Mine A produces 60+25 = 85 tons of Low-quality ore per day. Please write your value of xy at the top of your solution. The company has a contract to deliver 800 tons of Low-quality ore, 900 tons of Medium-quality ore, and 1000 tons of High-quality ore. The set of equations which relates the number of days (a, b, c) each mine (A, B, C) must operate to the produce required quantities of ore is: (60 + ????????)???? + 15???? + 10???? = 800 70???? + 10???? + (30 + ????????)???? = 900 20???? + 50???? + 35???? = 1000 
 a) (i) Write the system of equations in matrix form 
 (ii) Using matrix methods, solve for the number of days each mine must operate to meet this contract with no excess ore (fractions of a day are ok). 
 b) If one mine is not available, discuss the mathematical implications for computing the number of days the remaining mines must operate (you do not need to solve). Also, discuss the consequences for the quantity of ore produced to meet the contract 

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 You are an engineer at a mining company which sells a mineral ore in three quality grades: Low, Medium, and High. The company owns three mines which produce this ore: Mine A, Mine B, and Mine C. Each mine can produce different quantities (in tons) of each grade of ore per day as shown below: 
 Where xy are the last two digits of your URN. For example, for the URN 6835725, xy = 25. In this case Mine A produces 60+25 = 85 tons of Low-quality ore per day. Please write your value of xy at the top of your solution. The company has a contract to deliver 800 tons of Low-quality ore, 900 tons of Medium-quality ore, and 1000 tons of High-quality ore. The set of equations which relates the number of days (a, b, c) each mine (A, B, C) must operate to the produce required quantities of ore is: (60 + ????????)???? + 15???? + 10???? = 800 70???? + 10???? + (30 + ????????)???? = 900 20???? + 50???? + 35???? = 1000 
 a) (i) Write the system of equations in matrix form 
 (ii) Using matrix methods, solve for the number of days each mine must operate to meet this contract with no excess ore (fractions of a day are ok). 
 b) If one mine is not available, discuss the mathematical implications for computing the number of days the remaining mines must operate (you do not need to solve). Also, discuss the consequences for the quantity of ore produced to meet the contract 

 A drug called Kamagra Oral Jelly 100 mg is used to treat male sexual issues such as premature ejaculation and erectile dysfunction. It is a medication that is included in the PDE5 inhibitor class and aids in boosting blood flow to the penile area. When it comes to treating erectile dysfunction, Kamagra Oral Jelly 100 mg is a more convenient option than tablets and pills. It is available in many flavors, such as strawberry, orange, pineapple, and butterscotch.  
 
It is critical to understand that sexually transmitted infections (STDs) are not treated by this medicine. To ensure that sexual activity is done responsibly and to prevent any accidents, always use a condom.  
See below for all the information regarding this prescription drug.  
Concerning the medication 100 milligrams of Kamagra Oral Jelly  
A medication called Kamagra Oral Jelly 100 mg has been clinically studied for the goal of helping men who struggle to achieve and maintain an erection during sexual activity. It can also help sustain a strong erection for longer than rounds and treat premature ejaculation.  
The main ingredient of Kamagra Oral Jelly 100 mg is sildenafil. The active ingredient sildenafil salt is also found in other well-known erectile dysfunction drugs as Cenforce and Viagra. By inhibiting the phosphodiesterase type 5 (PDE5) enzymes, it helps you achieve an erection by causing the smooth muscles of your penis to relax.  
Men have an increase in blood flow to their penile regions as a result, causing them to develop a powerful erection.  
 
It’s important to keep in mind nevertheless that Kamagra Oral Jelly 100 mg is not a cure-all for erectile dysfunction (ED). The patient should partake in sexual activities that arouses them even more after taking the drug.  
After taking Kamagra Oral Jelly, we suggest indulging in foreplay as a sensuous form of self-tickling.
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Hydrogen-free carbon in the form of coke is burned (a) with complete combustion using theoretical air,
(b) with complete combustion using 50% excess air, or (c) using 50% excess air but with 10% of the
carbon burning to CO only. In each case calculate the gas analysis that will be found by testing the flue
gases on a dry basis.

Hydrogen-free carbon in the form of coke is burned (a) with complete combustion using theoretical air,
(b) with complete combustion using 50% excess air, or (c) using 50% excess air but with 10% of the
carbon burning to CO only. In each case calculate the gas analysis that will be found by testing the flue
gases on a dry basis.

 
Start a game on Impossible.
When 3 or more enemy units show up on the field, use your special ability. Make sure it kills at least 3 enemies.
Build the $200 turret so you can take advantage of how quickly it attacks and how much damage it does to more advanced units.
From now on, train Clubmen over and over again until almost the end of the game. You MUST do it as soon as you can afford it.
Use your ability as often as you can to help clear.
When new turret slots open up, buy them as soon as you can if you have the money, and then fill them with $200 turrets again.
Don’t change into the next age. This may sound crazy, but the way the game is coded, your army of Clubmen and your turrets can easily kill an enemy unit in later ages. Plus, when enemy units reach the end of your territory, you can do a lot of damage with your Clubmen’s army (causing all your clubmen to attack resulting in huge damage). With this setup, it should only take a minute or two to kill a tier 3 unit in the last age. The price of Clubmen is only $15, so every kill you make makes you money.
Keep doing these steps until you’ve made $200k and 400k experience. To get to the newest age right away, trade in two old turrets for two $100k turrets. You’ve pretty much won the game, so you can finally let go of your hand.
From now on, this is just like any other Age of War game: you keep farming to buy $100k turrets to replace the rest of your turrets. All of your units will die before they can even damage your base. Farm for $600k-$750k.
The rest of your money should go into Super Soldiers. Watch as your army destroys everything in its way, including its troops, vehicles, and base.
Congratulations, you have completed  age of war on Impossible!

1. Sulfate (molecular weight = 96g/mole) from an acid mine treatment pond leaves via an
outlet stream at a concentration of 275 mg/L. The outlet stream flows into a small river
with a flow of 30 cfs upstream, 37 cfs downstream, and a background sulfate
concentration of 17 mg/L. Please answer the following.
a. What is the concentration of outlet stream sulfate in moles/L and equivalents/L?
b. What is the concentration of sulfate in the river immediately downstream of the
confluence with the outlet stream in mg/L?
2. One oxidation treatment process uses light and hydrogen peroxide to generate the
highly reactive hydroxyl radical (OH•), which can be used to degrade certain PFAS
“forever compounds”, but not others. A very recent paper by Zhang (2023
Environmental Science and Technology: so recent no volume has been assigned) report a
second order rate constant between OH• and the PFAS, perfluoro(2-
ethoxyethane)sulfonic acid (PFSA with a k = 1.2 x 10 7 L/mole-second), while this radical is
less reactive with 4.8-dioxa-3H-perfluorononanoic acid (DPFA with a k = 5 x 10 5 L/mole-
second). Suppose your UV-hydrogen peroxide system generates a constant output
(steady-state) of this radical at a concentration of 3 x 10 -14 moles/L, please answer the
following.
a. What is the pseudo-first order rate constant for the degradation of both
compounds at this OH• concentration?
b. What are the half-lives for these two compounds?
c. Suppose you have to treat these two compounds such that they both must not
exceed the 50 nM water quality standard. If both compounds have a starting
concentration of 1µM how long (in hours) will it take each to reach this
concentration? For the slower reacting compound is the time needed to achieve
the water quality standard a reasonable time from a design perspective?

1. Sulfate (molecular weight = 96g/mole) from an acid mine treatment pond leaves via an
outlet stream at a concentration of 275 mg/L. The outlet stream flows into a small river
with a flow of 30 cfs upstream, 37 cfs downstream, and a background sulfate
concentration of 17 mg/L. Please answer the following.
a. What is the concentration of outlet stream sulfate in moles/L and equivalents/L?
b. What is the concentration of sulfate in the river immediately downstream of the
confluence with the outlet stream in mg/L?
2. One oxidation treatment process uses light and hydrogen peroxide to generate the
highly reactive hydroxyl radical (OH•), which can be used to degrade certain PFAS
“forever compounds”, but not others. A very recent paper by Zhang (2023
Environmental Science and Technology: so recent no volume has been assigned) report a
second order rate constant between OH• and the PFAS, perfluoro(2-
ethoxyethane)sulfonic acid (PFSA with a k = 1.2 x 10 7 L/mole-second), while this radical is
less reactive with 4.8-dioxa-3H-perfluorononanoic acid (DPFA with a k = 5 x 10 5 L/mole-
second). Suppose your UV-hydrogen peroxide system generates a constant output
(steady-state) of this radical at a concentration of 3 x 10 -14 moles/L, please answer the
following.
a. What is the pseudo-first order rate constant for the degradation of both
compounds at this OH• concentration?
b. What are the half-lives for these two compounds?
c. Suppose you have to treat these two compounds such that they both must not
exceed the 50 nM water quality standard. If both compounds have a starting
concentration of 1µM how long (in hours) will it take each to reach this
concentration? For the slower reacting compound is the time needed to achieve
the water quality standard a reasonable time from a design perspective?