Answer:
Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer
Answer:
At equilibrium, the concentration of 
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+ 
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor present. Immediately, and are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: []=0 ; [ ]= 0 ; []=0.60M
C: []=+x ; [ ]= +x ; []=-2x
E: []=0+x ; [ ]= 0+x ; []=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 
At equilibrium, the concentration of is going to be 0.30M
Answer
D
Explanation
<em>the</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>D</em>
Answer:
1s2 2s2 2p6 3s2 3p4
Explanation:
For ground state atoms you need to fill the orbitals in order.
Orbital s has max 2 electrons
Orbital p has max 6 electrons
Level 1 only has s orbital
Level 2 has s and p orbitals
Level 3 has s, p and d orbitals
You fill them in order until you reach 16 electrons:
1s2 2s2 2p6 3s2 3p4
