Answer:
10. mol NaOH
Explanation:
2.5 M = 2.5 mol/L
2.5 mol/L * 4 L = 10. mol
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:
823.7g
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
Using the information given in this question as follows:
Q = 6,400 J
m = ?
c of soil = 0.840 J/g°C
∆T = 9.25°C
Using Q = mc∆T
m = Q ÷ c∆T
m = 6,400 ÷ (0.840 × 9.25)
m = 6400 ÷ 7.77
m = 823.7g
Answer:
the answer is D i believe. i am not shure but im like 70% sure
Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)