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:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places
C. You should ALWAYS ask the teacher if you don't get something; your friends could be wrong, don't guess it, and NEVER cheat. Hope this helps!!
The answer to this item is TRUE. This can be explained through the Graham's law. This law states that the rate at which gases diffuse is inversely proportional to the square root of their densities which is also related to their molecular masses.
337.92 moles of Ozone will be produced
1 Oxygen atom is 8 g
1 mole of ozone, O3 = 8 * 3 = 24 g
7.92 * 1024 = 8110.08 g
1 mole = 24 g
? moles = 8110.08 g
? = 337.92 moles
Read more on moles here:
brainly.com/question/15356425
Hope it helps