<span>Ionization means pulling off an electron. So the first IE means removing 1 electron from lithium:
Li --> Li+ + 1 e-
The product is a + ion because it now has one more proton than electron.
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
Li+ --> Li2+ + 1e- </span>
Zn(s) is the correct answer
Answer:
answer is C
Explanation:
encourage the release of carbon dioxide from the stems
Answer:
Hydrogen and Cobalt
Explanation:
Break up each individual element
- Hope that helps! Please let me know if you need further explanation.
Answer : The concentration of
and
at equilibrium is, 0.0031 M and 0.0741 M respectively.
Explanation : Given,
Moles of
= 0.166 mol
Volume of solution = 2.15 L
First we have to calculate the concentration of ![PCl_5](https://tex.z-dn.net/?f=PCl_5)
![\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DPCl_5%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DPCl_5%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
![\text{Concentration of }PCl_5=\frac{0.166mol}{2.15L}=0.0772M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DPCl_5%3D%5Cfrac%7B0.166mol%7D%7B2.15L%7D%3D0.0772M)
Now we have to calculate the concentration of
and
at equilibrium.
![PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)](https://tex.z-dn.net/?f=PCl_5%28g%29%5Crightleftharpoons%20PCl_3%28g%29%2BCl_2%28g%29)
Initial conc. 0.0772 0 0
At eqm. 0.0772-x x x
The expression for equilibrium constant is:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
![1.80=\frac{(x)\times (x)}{(0.0772-x)}](https://tex.z-dn.net/?f=1.80%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.0772-x%29%7D)
By solving the term, we get the value of 'x'.
x = 0.0741
Thus,
The concentration of
at equilibrium = (0.0772-x) = (0.0772-0.0741) = 0.0031 M
The concentration of
at equilibrium = x = 0.0741 M