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:
Water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.
Explanation:
Hello.
In this case, since no options are given we can infer from the statement that due to water's higher boiling point than acetone we can conclude that when they are in liquid state, water has stronger intermolecular forces which allow its particles to be held in a stronger way in comparison to the acetone's molecules, for that reason, more energy will be required in order to separate them and promote the boiling process, which is attained via increasing the temperature. Besides, less energy will be required for the separation of the acetone's molecules in order to boil it when liquid, therefore, a lower temperature is required.
In such a way, we can sum up that water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.
Regards.
Answer:
Explanation:
Hello!
In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:
Whereas solving for the final temperature T2, we get:
Thus, we plug in the given data (temperature in Kelvins) to obtain:
Best regards!
Answer:
100
Explanation:
M = mass/ number of mole
M = 3.5 g/0.035 mol = 100 g/mol - molar mass