When ethyl ether is heated with excess HI for several hours, the only organic product obtained is ethyl iodide. T/F
1 answer:
Answer:
True
Explanation:
Ethers react with HI to form the corresponding alcohols and alkyl iodides.
Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes reaction with HI to form ethyl iodide.
<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>
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Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So, the pH of this buffer solution is 10,04
Have a nice day!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the Henderson-Hasselbach's equation in the following way:
So, the pH of this buffer solution is 10,04
Have a nice day!