Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
Explanation:
Hello,
In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:
![\% ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
![[H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B%5C%25%20ionization%2A%5BHA%5D%7D%7B100%5C%25%7D%20%3D%5Cfrac%7B4%5C%25%2A0.30M%7D%7B100%5C%25%7D%3D0.012M)
Therefore the Ka is computed by using the equilibrium expression:
![Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%5Cfrac%7B0.012M%2A0.012M%7D%7B0.30M-0.012M%7D%5C%5C%20%5C%5CKa%3D5x10%5E%7B-4%7D)
And the pH:
![pH=-log([H^+])=-log(0.012)\\\\pH=1.92](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29%3D-log%280.012%29%5C%5C%5C%5CpH%3D1.92)
Regards.
Answer: In essence, Boyle defined alkalies as substances that consume, or neutralize, acids. Acids lose their characteristic sour taste and ability to dissolve metals when they are mixed with alkalies. Alkalies even reverse the change in color that occurs when litmus comes in contact with an acid.
Explanation:
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Answer:
The empirical formula is CH3O2
Explanation:
From The question, number of moles of CO2= 4.89/44=0.111mol
Number of moles of H2O=3/18= 0.1667moles
Since 1mole of CO2 contains 1mole of carbon and 2 moles of O2,
Implies that there 0.111 moles of C and 0.222moles of O2
Similarly also 0.1667moles of water contains 2 moles of H2 hence = 0.333 moles of H2
Dividing by smallest number= 0.111mol
For C = 0.111/0.111= 1
For H = 0.333/ 0.111= 3
For O = 0.222/0.111= 2
Hence the empirical formula= CH3O2
