<span>The root mean square velocity of gas molecules as a function of temperature is (M is molecular weight):
v²rms = 3RT/M,
and kinetic energy is given by:
KE = ½Mv ²
We can make the substitution:
KE = ½M3RT/M, or
KE = (3/2)RT.
Notice the molecular weight variable is gone; that means the answer will be the same for F2, Cl2, and Br2:
KE = (3/2) (8.314 J/K*mole) (290 K) = 3617 J/mole.</span>
Well that depends on the central atom. Remember each bond indicates another electron that is being shared with the central atom.
So CO4, carbon would be your central atom, with the 4 oxygen around it. There would be no double bonds because carbon only requires the 4 electrons provided by the oxygens, because of its location on the periodic table.
Nitrogen on the other hand is different. It's found in group 5 (5V) and has 5 valence electrons on its outer shell. So NO4 would have a central Nitrogen atom, with 3 oxygens providing a single bond, and one oxygen providing a double bond, because Nitrogen needs 5 electrons!
We learned it like this, the atom brought x amount of electrons to the party, x representing the valence electrons, and must leave with x electrons. It's silly but it works!
I hope I was able to help and clarify a few things up!
Because phosphorus is more flammable than sulfur!
Have a nice day! :)
The answer to the question Retraction. Its because the light goes around it
Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:
Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
