Answer:
After three hours, concentration of C₂F₄ is 0.00208M
Explanation:
The rate constant of the reaction:
2 C2F4 → C4F8 is 0.0410M⁻¹s⁻¹
As the units are M⁻¹s⁻¹, this reaction is of second order. The integrated law of a second-order reaction is:
![\frac{1}{[A]} =\frac{1}{[A]_0} +Kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2BKt)
<em>Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time of the reaction (In seconds).
</em>
3.00 hours are in seconds:
3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds
Initial concentration of C2F4 is:
0.105mol / 4.00L = 0.02625M
Replacing in the integrated law:
![\frac{1}{[A]_0}= \frac{1}{0.02625} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]_0}=480.9M^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%3D%20%5Cfrac%7B1%7D%7B0.02625%7D%20%20%2B0.0410M%5E%7B-1%7Ds%5E%7B-1%7D%2A10800s%5C%5C%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%3D480.9M%5E%7B-1%7D)
[A] = 0.00208M
<h3>After three hours, concentration of C₂F₄ is 0.00208M</h3>
Answer:
Answers may vary
Explanation:
Mol is the amount of molecules in something. Mol are similar to donuts, if you have a dozen of something you have 12 of them. If you have a mol of Hydrogen gas you 6.02•10^23 molecules of Hydrogen gas. If you are taking a chemistry class you will probably receive questions involving converting grams to mol and vice versa.
Converting Grams to Mol:
grams of x•(1/molecular mass of x)
note: you can find molecular mass of x by looking at the masses of the elements you need to add up in the periodic table
Converting Mol to Grams:
mol of x•(molecular mass of x)
When atoms of an element combine they form a A-Compound
-Seth
Greetings!
I believe the answer is true.
While precipitation does require a lot of energy, evaporation takes longer, and requires more energy from heat, whereas precipitation doesn't really require any.
Hope this helps you!
>>DL<<
