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
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.
Answer:
<u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.
Correct answer - C
Explanation:
A solid sodium metal reacts with gaseous form of chlorine gas to form a solid sodium chloride.
This formation of sodium chloride arises due to transfer of one valence electron from sodium to chlorine atom and the both atoms get stable octet electronic configuration.
As a result, an ionic compound solid sodium chloride is formed.
The chemical reaction of formation sodium chloride is as follows.
Therefore, <u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.
Answer:
D.
The concentration of reactants and the concentration of products are constant.
Explanation:
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Number of Atoms in Gold for given mass can be calculated using following formula,
# of Moles = Number of Atoms / 6.022 × 10²³
Or,
Number of Atoms = Moles × 6.022 × 10²³ ------- (1)
Calculating Moles,
As,
Moles = Mass / M.mass
So,
Moles = 4.25 g / 196.96 g/mol
Moles = 0.0215
Putting value of mole in eq.1,
Number of Atoms = 0.0215 × 6.022 × 10²³
Number of Atoms = 1.299 × 10²²
Result:
4.25 g of Gold Nugget contains 1.299 × 10²² Atoms.
