Answer:
Here's what I get.
Explanation:
According to Markovnikov's rule, the H will add to a terminal carbon, generating three resonance stabilized carbocations.
The Br⁻ ion will add to any of the three carbocations.
There are three possible products:
- 5-bromo-2,5-dimethylhexa-1,3-triene (1)
- 3-bromo-2,5-dimethylhexa-1,4-triene (<em>2</em>)
- 1-bromo-2,5-dimethylhexa-2,4-triene (3)
Answer:
14.99 mols of CO2
Explanation:
STP= 273 k , 1 atm , .0821 gas constant
N=PV/RT
N= 1x336/.0821x273
14.99 mols of CO2
659.6 grams of CO2
Answer:
I am not understand bro a question
<u>Answer:</u> The potential of electrode is -0.79 V
<u>Explanation:</u>
When zinc is dipped in zinc sulfate solution, the electrode formed is 
Reduction reaction follows: 
To calculate the potential of electrode, we use the equation given by Nernst equation:
![E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}](https://tex.z-dn.net/?f=E_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D%3DE%5Eo_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5D%7D%7B%5BZn%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = -0.76 V
n = number of electrons exchanged = 2
![[Zn]=1M](https://tex.z-dn.net/?f=%5BZn%5D%3D1M)
(concentration of pure solids are taken as 1)
![[Zn^{2+}]=0.1M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.1M)
Putting values in above equation, we get:
Hence, the potential of electrode is -0.79 V
Answer:
The solutions should be added in this order NaCl > Na2SO4 > Na2S
Explanation:
Silver is insoluble as a chloride, so the silver ions get precipitated on addition of chloride ion as silver chloride. This means Ag+ would be removed the first.
So we will add NaCl in the first step.
The following reaction will occur.
Ag+ + Cl- → AgCl(s)
Both, Pb2 and Ni are soluble as chlorides. (lead chloride is soluble as a hot solution but will ppt when colder).
When we add Na2SO4, Pb2+ will get precipitated (because it's insoluble) as PbSO4 and Ni will remain soluble as NiSO4 is soluble in water.
The reaction that will occur is:
Pb^2+ + SO4^2- → PbSO4(s)
Nickel is insoluble as a sulfide. So when we will add Na2S, nickel will be precipitated as sulfide and be able to separate and be collected.
The solutions should be added in this order NaCl > Na2SO4 > Na2S
