Answer:
The student's conclusion is not correct
Explanation:
Activation energy is the minimum amount of energy required for a reaction to occur. All reactions require there activation energy to be met before the reaction can proceed. When the temperature of a reaction is increased, the kinetic energy of the reactant molecules increases; colliding more with each other, which makes them "surmount" the activation energy of the reaction faster as compared to a lower temperature.
In combustion, there is burning of an hydrocarbon (in this case propane) in excess oxygen. The burning assists in increasing the kinetic energy of the reactant particles which in turn easily surmounts the activation energy of the reaction by colliding (effective collision) more with oxygen. So, the reaction has an activation energy but the activation energy has been met and passed and hence the reaction is proceeding faster.
Increasing the temperature of a reaction is one of the ways of increasing the rate of a chemical reaction.
<span>Hydroelectric energy uses the movement of water to spin a turbine and produce electricity. Coal is a fossil fuel that is burned as an energy source, which results in emissions such as carbon dioxide, nitrogen, and sulfur. Coal use is associated with waste products both from the mining process to acquire the coal and also from the actual use of coal itself. Hydroelectric energy does not involve waste, but does pose potential problems to waterways. Coal use is more common than the use of hydroelectric energy.</span>
Answer:
it is a transverse wave. Does that help?
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
