The answer is Condensing or condensation.
This question may only be ansewered by frequent mattrrs
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
Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Answer:
[SO2Cl2] = = 0.015 M
[SO2] = = 0.0027 M
[Cl2] = = 0.0027 M
Q = = = 4.8 × 10−4
No. Q < Kc, so reaction will shift to the right.
Explanation: