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:
the answer is c. their atomic masses are different clearly because an atom of gold has 79 protons and the atom can be divided multiple times. An atom of silver has an atomic number of 47. 47 electrons. Clearly different. Hope it helps :)
Explanation:
Answer:
The answer is
<h3>2.53 × 10²⁴ molecules</h3>
Explanation:
The number of molecules present can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 4.21 × 6.02 × 10²³
We have the final answer as
<h3>2.53 × 10²⁴ molecules</h3>
Hope this helps you
Answer:
<u>= 2.2 g pf S. produced</u>
Explanation:
Balanced Reaction equation:
→ 
1 mole of H2S - 34.1g
? moles - 3.2g
= 3.2/34.1 =<u> 0.09 moles of H2S</u>
Also,
1 mole of S02 - 64.07 g
? moles - 4.42g
= 4.42/64.07 <u>= 0.069 moles of SO2</u>
<u />
<em>Meaning SO2 is the limiting reagent</em>
Finally, 3 moles of S - 32g of sulphur
0.069 mole = ? g of Sulphur
= 0.069 x 32
<u>= 2.2 g pf S.</u>
Answer:
it has made the study of chemistry systematic and easy. it acts as an aid to memory
