Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
= 0.8
The rate-out
= 
= 
We can say that:
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
Integration of the above linear equation =
so we have:
![\frac{d}{dt}[e^{\frac{1}{3}t}A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Be%5E%7B%5Cfrac%7B1%7D%7B3%7Dt%7DA%5D)
∴ 
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= 
%
= 0.0456667 %
= 0.046% to three decimal places
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The answer is c but it might be b it’s be
What is the name of CuC2H3O2?
The answer is Copper (I) Acetate. (:
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
![rate = k*[O_{3}][NO]](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D%5BNO%5D%20)
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
![rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D_%7B0%7D%5BNO%5D_%7B0%7D%20%3D%203.91%20%5Ccdot%2010%5E%7B6%7D%20M%5E%7B-1%7Ds%5E%7B-1%7D%2A2.35%5Ccdot%2010%5E%7B-6%7D%20M%2A7.74%20%5Ccdot%2010%5E%7B-5%7D%20M%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%20)
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
