Answer:
Es la cantidad de protones menos la cantidad de electrones.
Explanation:
Es la cantidad de protones menos la cantidad de electrones.
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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
No, it does not. I believe it's false.
The answer is the telephase.<span />
Explanation:
1st one will be cathodic reaction and reduction will occur since electrons are being gained and hydrogen goes from +1 to zero oxidation state.
2nd one will be anodix reaction and oxidation will take place because Iodide ion's oxidation number increases from -1 to 0
