Answer:
Charge= -2.
Gains two electron into the 4p^4 to become 4p^6.
Explanation:
The element in the periodic table/chart that matches with the valence electron configuration is Selenium with full electron configuration of [Ar] 3d^10 4s^2 4P^4 which is a non-metal that is found in group 4 of the periodic table/chart.
Selenium can receive 2 more electrons on the 4p^4 to give a -2(minus 2) ion that is Se^2-.
Selenium can also loose 2 electron from 4s^2 to give a +2 ion that is Se^2+.
Selenium can also loose 2 electrons from 4s^2 and 2 electrons from 4p^4 to form Se^4+.
Selenium can also loose 2 electrons from 4s^2 and 4 electrons from 4p^4 to form Se^6+.
Thus, in order to form a monatomic ion with a charge(we will be making use of the most stable one). Thus, it will gain two more electron since this is easier to become 4s^2 4p^6.
It’s B the cu looses its 2 and passes it to the NH3 that needs a bracket to separate them. The NH3 doesn’t loose its 3 because it’s already a compound!
Hope this helps!
<span>Answer:
3 N2H4 (l) = 4 NH3 (g)+ N2 (g)
2 N2H4 (l) + N2O4 (g) =3 N2 (g)+ 4 H2O (g)</span>
Answer : The equilibrium concentration of 
at 
is, 
.
Solution : Given,
Equilibrium constant, 
Initial concentration of 
= 0.260 m
Let, the 'x' mol/L of are formed and at same time 'x' mol/L of are also formed.
The equilibrium reaction is,
Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get
By rearranging the terms, we get the value of 'x'.
Therefore, the equilibrium concentration of at is, .
