Answer:
(a) 
(b) 
Explanation:
Hello,
(a) In this case, for the given chemical reaction, the law of mass action becomes:
![Kc=\frac{[C6H5CHO][H2]}{[C6H5CH2OH]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC6H5CHO%5D%5BH2%5D%7D%7B%5BC6H5CH2OH%5D%7D)
In such a way, as 1.20 g of benzyl alcohol are placed into a 2.00-L vessel, the initial concentration is:
![[C6H5CH2OH]_0=\frac{1.20g*\frac{1mol}{108.14g} }{2.00L} =5.55x10^{-3}M](https://tex.z-dn.net/?f=%5BC6H5CH2OH%5D_0%3D%5Cfrac%7B1.20g%2A%5Cfrac%7B1mol%7D%7B108.14g%7D%20%7D%7B2.00L%7D%20%3D5.55x10%5E%7B-3%7DM)
Hence, by writing the law of mass action in terms of the change
due to equilibrium:

Solving for
by using a quadratic equation one obtains:

Thus, the equilibrium concentration of benzyl alcohol is computed:
![[C6H5CH2OH]_{eq}=5.55x10^{-3}M-5.50x10^{-3}M=5x10^{-5}M](https://tex.z-dn.net/?f=%5BC6H5CH2OH%5D_%7Beq%7D%3D5.55x10%5E%7B-3%7DM-5.50x10%5E%7B-3%7DM%3D5x10%5E%7B-5%7DM)
With that concentration the partial pressure results:
![p_{C6H5CH2OH}=[C6H5CH2OH]_{eq}RT =5x10^{-5}\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*523K \\p_{C6H5CH2OH}=2.14x10^{-3}atm](https://tex.z-dn.net/?f=p_%7BC6H5CH2OH%7D%3D%5BC6H5CH2OH%5D_%7Beq%7DRT%20%3D5x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%20%2A0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A523K%20%5C%5Cp_%7BC6H5CH2OH%7D%3D2.14x10%5E%7B-3%7Datm)
(b) Now, the fraction of benzyl alcohol that is dissociated relates its equilibrium concentration with its initial concentration:

Best regards.
The reasonable ground-state electron configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 4d8
Answer: ?
Explanation: I translated this and It makes no sense.
Tôi đã dịch điều này và nó không có ý nghĩa.
Answer:
Europa
Explanation:
it is though that to have a subsurface liquid of water