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
<span>2H2 + O2 → 2H2O</span>
<span>
</span>
<span>okay???</span>
<span>
</span>
Answer:
0.085 moles of N₂O₅ are needed
Explanation:
Given data:
Mass of NO₂ produces = 7.90 g
Moles of N₂O₅ needed = ?
Solution:
2N₂O₅ → 4NO₂ + O₂
Number of moles of NO₂ produced :
Number of moles = mass/ molar mass
Number of moles = 7.90 g/ 46 g/mol
Number of moles = 0.17 mol
now we will compare the moles of NO₂ with N₂O₅.
NO₂ : N₂O₅
4 : 2
0.17 : 2/4×0.17 = 0.085 mol
Thus, 0.085 moles of N₂O₅ are needed.
Answer:
Your babe is in your school
Explanation:
go ask babes out