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Andrew [12]
3 years ago
15

A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume

is 6 L, and the final pressure is 8atm, what was the initial temperature of the argon? Please give answer WITHOUT the unit of KELVIN and round to the tenth position (two places to the right of the decimal). Example: 25.14 not 25.4K
Chemistry
1 answer:
jenyasd209 [6]3 years ago
4 0
Volume of Argon V1 = 5.0 L  
Pressure of Argon P1 = 2 atm 
Final temperature T2 = 30 C = 30 + 273 = 303 K 
Volume at final temperature V2= 6 L 
Pressure at final temperature P2 = 8 atm  
We know that (P1 x V1) / T1 = (P2 x V2) / T2  
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48 
Initial Temperature T1 = 3030 / 48 = 63.12 
Initial Temperature = -209. 8 C
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Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

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rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

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dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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XH_{3}
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How many molecules of CF₂Cl₂ are in 45.7 grams of CF₂Cl₂? (Show work)
posledela

Answer:

The answer to your question is: 2.20 x 10 ²³ molecules

Explanation:

Data

mass = 45.7 g

molecules of CF₂Cl₂ = ?

Process

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2.- Use the Avogradro's number to solve the problem

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