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:
![\frac{d}{dt}[e^{\frac{1}{3}t}A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Be%5E%7B%5Cfrac%7B1%7D%7B3%7Dt%7DA%5D)
∴ 
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= 
%
= 0.0456667 %
= 0.046% to three decimal places
Answer:
0.64 L
Explanation:
Recall that
n= CV where n=m/M
Hence:
m/M= CV
m= given mass of solute =152g
M= molar mass of solute
C= concentration of solute in molL-1 = 1.5M
V= volume of solute =????
Molar mass of potassium permanganate= 158.034 g/mol
Thus;
152 g/158.034 gmol-1= 1.5M × V
V= 0.96/1.5
V= 0.64 L
Im pretty Sure its B Tuesday And Wensday
Answer:
You must divide the grams of your actual yield by the grams of the theoretical yield and multiply by 100 in order to obtain percent yield
Explanation:
Answer:
H₂SO₄
Explanation:
Given data:
Number of moles of H₂SO₄ = 15 mol
Number of moles of Fe = 13 mol
Which reactant is limiting reactant = ?
Solution:
Chemical equation:
3H₂SO₄ + 2Fe → Fe₂(SO₄)₃ + 3H₂
now we will compare the moles reactant with product.
H₂SO₄ : Fe₂(SO₄)₃
3 : 1
15 : 1/3×15 = 5
H₂SO₄ : H₂
3 : 3
15 : 15
Fe : Fe₂(SO₄)₃
2 : 1
13 : 1/2×13 = 6.5
Fe : H₂
2 : 3
13 : 3/2×13 = 19.5
Number of moles of product formed by H₂SO₄ are less thus it will act as limiting reactant.
