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 electrode A is a cathode. This is because it is negatively charged electrode and it attracts cations, or we can say it is positively charged.
The reaction occurs at the cathode
Zn2+ +2e⇒Zn
Anode reaction is
2CL- ⇒CL2+2e-
The overall reaction will be Zncl2(l) which completely disassociates then it is zn∧2+(l)+2cl∧-(l).
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
