I'm pretty sure the answer is the first one. The fact that it doesn't have a full valence shell means that electrons from other atoms can react and bond with the sodium atom.
Hope this helps x
Don’t worry, you’re right :)
It’s b for sure that’s the answer
What Can you be more clear --Molecules change it from being less dense
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
