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: 6.64 moles of carbon.
Explanation:
Given data:
Number of moles of C = ?
Number of moles of CCl₂F₂ = 6.64 mol
Solution:
In one mole of CCl₂F₂ there is one mole of carbon two moles of chlorine and two moles of fluorine are present.
In 6.6 moles of CCl₂F₂ :
Moles of carbon = 6.64 × 1 = 6.64 moles of carbon.
Moles of chlorine = 6.64× 2 = 13.28 moles of chlorine
Moles of fluorine = 6.64× 2 = 13.28 moles of fluorine
Read more on Brainly.com - brainly.com/question/15602143#readmore
2021 hoped I helped you god bless you
Answer: 
Explanation:Bond energy of H-H is 436.4 kJ/mole
Bond energy of C-H is 414 kJ/mol
Bond energy of C=C is 620 kJ/mol
Bond energy of C≡C is 835 kJ/mol

= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

