Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Hello!
The center of an atom is called the nucleus; it is comprised of protons and neutrons, collectively known as nucleons. The electrons float around the nucleus in a cloud.
Answer:
A) Protons and neutrons are in the center of the atom, and electrons move around in the cloud.
Yes I think & I Belive it moves across the surface
I would think that b would be the right answer
Answer:
4KNO3 ==> 2K2O + 2N2 + 5O2
Explanation:
It's a decomposition, but not a simple one.
KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3
4KNO3 ==> 2K2O + 2N2 + 5O2
Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2
That should do it.