I can say it oxidizes gas pollutants such as hydrocarbons and carbon monoxide. It also reduces nitrogen oxides into water, hydrogen, and carbon dioxide.
A tank with a volume of 40 cuft is filled with a carbon dioxide and air mixture. The pressure within the tank is 30 psia at 70oF. It is known that 2 lb of carbon dioxide was placed in the tank. Assume that air is 80% nitrogen and 20% oxygen and use the ideal gas laws. Calculate ,
The correct responses are;
(i) Weight percent of nitrogen: 58.6%
Weight percent of oxygen: 14.65%
Weight percent of carbon dioxide: 29.59%
(ii) Volume percent of nitrogen: 64.38%
Weight percent of oxygen: 14.1%
Weight percent of carbon dioxide: 21.53%
(iii) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(iv) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(v) The average molecular weight is approximately 32.02 g/mole
(vi) At 20 psia, 60 °F, the density is 0.11275 lb/ft.³
At 14.7 psia, 60 °F, the density is 0.0844 lb/ft.³
At 14.7 psia, 32 °F, the density is 0.0892 lb/ft.³
(vii) The specific gravity of the mixture 0.169
Reasons:
The volume of the tank = 40 ft.³ = 1.132675 m³
Content of the tank = Carbon dioxide and air
The pressure inside the tank = 30 psia = 206843 Pa
The temperature of the tank = 70 °F ≈ 294.2611 K
Mass of carbon dioxide placed in the tank = 2 lb.
Percent of nitrogen in the tank = 80%
Percent of oxygen in the tank = 20%
(i) 2 lb ≈ 907.1847 g
Molar mass of carbon dioxide = 44.01 g/mol
Number of moles of carbon dioxide = [tex]\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles[/tex]
Assuming the gas is an ideal gas, we have;
[tex]\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588[/tex]
The number of moles of nitrogen and oxygen = 95.7588 - 20.613 = 75.1458
Let x represent the mass of air in the mixture, we have;
[tex]\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458[/tex]
Solving gives;
x ≈ 2158.92 grams
Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g
[tex]\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}[/tex]
ii.
[tex]\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles[/tex]
[tex]\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles[/tex]
Number of moles of carbon dioxide = 20.613 moles
Sum = 61.65 moles + 13.5 moles + 20.613 moles ≈ 95.763 moles
In an ideal gas, the volume is equal to the mole fraction
[tex]\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}[/tex]
(iv) The partial pressure of a gas in a mixture, [tex]P_A = \mathbf{X_A \cdot P_{total}}[/tex]
Partial pressure of nitrogen, [tex]P_{N_2}[/tex] = 0.6438 × 30 psia ≈ 19.314 psia
Partial pressure of oxygen, [tex]P_{O_2}[/tex] = 0.141 × 30 psia ≈ 4.23 psia
Partial pressure of carbon dioxide, [tex]P_{CO_2}[/tex] = 0.2153 × 30 psia ≈ 6.459 psia
(v) The average molecular weight is given as follows;
[tex]\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole[/tex]
(vi) At 20 psia 70 °F, we have;
Converting to SI units, we have;
[tex]\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84[/tex]
The number of moles, n ≈ 63.84 moles
The mass = 63.84 moles × 32.02 g/mol ≈ 2044.16 grams ≈ 4.51 lb
[tex]\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247[/tex]
The mass = 47.8247 moles × 32.02 g/mol ≈ 1531.35 grams = 3.376 lb.
[tex]\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 and 32 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482[/tex]
The mass = 50.5482 moles × 32.02 g/mol ≈ 1618.55 grams = 3.568 lb.
[tex]\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}[/tex]
(vii) [tex]\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}[/tex]
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Question 10 of 25: Select the best answer for the question. 10. In a series RC circuit, if the resistance is 8,000 ohms and X is 6,000 ohms, what is the impedance of the circuit O A. 8,600 ohms O B. 10,000 ohms O C. 14,000 ohms O D. 9,200 ohms Mark for review (Will be highlighted on the review page) << Previous Question Next Question >> M LL 5 - 1080 acer
Answer:
B. 10,000 ohms
Explanation:
|8000 +j6000| = √(8000² +6000²) = 1000√(64 +36) = 1000√100 = 10,000
The impedance is 10,000 ohms.
Find the mean deviation of the set of numbers
(a) 12, 6, 7, 3, 15, 10, 18,5
Answer:
4.25
Explanation:
The mean deviation is how far, on average, all values are from the middle. To do that, we subtract each element from the mean of the elements, take their absolute values, and then divide by the number of elements in the set. Here's the formula: [tex]\sum\frac{|x-\mu|}{n}[/tex]
Here the mean is 9.5, the count n = 8, so the mean deviation is 4.25
A person has driven a car 180 m in 40 seconds. What is the car’s speed?
Answer:
4.5 m/s = 16.2 km/h
Explanation:
The speed is the ratio of distance to time:
speed = (180 m)/(40 s) = 4.5 m/s
Chemical engineering is one of the simpler fields in engineering.
True
False
Answer:
False
Explanation:
What is the first step in the decision-making process?
Answer: identify your decision. relize you need to make a decision.
Explanation:
how many times greater is the value of the 2 of the 270413 than the valuce of the 2 in 419427?
I need help finding a good selling for a 71 Ford mustang hardtop coupe price This isn't really a school question but it's the only place I know we're I can ask questions so I'll give it a go
Answer:
you know you could use like quora or yahoo answers
what is the value of slip in the block rotor test? and why?
Technician A says that to start a fuel-injected car, the accelerator should be depressed
once. Technician B says that to start a car with a carburetor, do not touch the
accelerator pedal until the engine is running. Who is correct?
Technician A
Technician B
Both A and B
Neither Anor B
Answer:
Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A
Explanation:
Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A
The part of the circuit that converts electrical energy into other forms
Answer:
Alternator
Explanation:
A device which converts mechanical energy into electrical energy.
I hope it helps.
Describe the first case where the power of synthesis was used to solve design problems.
The first case where the power of synthesis was applied was in Chile when they had to put 100 families in houses around 40m².
It should be noted that the power of architecture is the fact that it can synthesize a complex entry to a problem.
Chilean architect Alejandro used this in building more than several houses for the poorest communities in Chile. His rigorous and innovative design approach used a social framework that laid a precedent within the profession.
During this period in Chile, a normal middle-class home was about 80m² but he built his in 40m² and it was a good home that managed the resources that were available.
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Answer:
oh this happened with me to ,this will be alright just wait a bit