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
Distillation of a mixture
Hey there!
Cu(CN)₂
Find the molar mass.
Cu: 1 x 63.546 = 63.546
C: 2 x 12.01 = 24.02
N: 2 x 14.07 = 28.14
-----------------------------------
115.706 grams
The mass of one mole of Cu(CN)₂ is 115.706 grams.
We have 4 moles.
115.706 x 4 = 463
4.00 moles of Cu(CN)₂ has a mass of 463 grams.
Hope this helps!
You would get four moles of magnesium nitrate :) you would have to
“ ?molesmg(oh)2 = 8molmg(no3)2 x molmg(oh)2 / 2molhno3 = 4 moles of magnesium nitrate :))) hopefully this helps! <3
D. dishwashing Soap/Liquid