Draw the conjugate acid of ethanol
1 answer:
Answer:
H₃O⁺.
Explanation:
Hello,
In this case, ionization of ethanol yields hydrogen cations and ethoxide anions as follows:
Which is at equilibrium as ethanol is a weak acid. Thus, since the conjugate acid is formed by the outgoing H⁺ and the water, we say that the conjugate acid is then the hydroxonium ion, H₃O⁺ whereas the conjugate base is the remaining anion CH₃COO⁻.
Best regards.
You might be interested in
Answer:
<u><em></em></u>
- <u><em>Because the x-intercet of the graph represents volume zero, which indicates the minimum possible temperature or absolute zero.</em></u>
Explanation:
Charle's Law for ideal gases states that, at constant pressure, the <em>temperature</em> and the <em>volume</em> of a sample of gas are protortional.
That means that the graph of the relationship between Temperature, in Kelivn, and Volume is a line, which passes through the origin.
When you work with Temperature in Celsius, and the temperature is placed on the x-axis, the line is shifted to the left 273.15ºC.
Meaning that the Volume at 273.15ºC is zero.
You cannot reach such low temperatures in an experiment, and also, volume zero is not real.
Nevertheless, you can draw the line of best fit and extend it until the x-axis (corresponding to a theoretical volume equal to zero), and read the corresponding temperature.
Subject to the experimental errors, and the fact that the real gases are not ideal, the temperature that you read on the x-axis is the minimum possible temperature (<em>absolute zero</em>) as the minimum possible volume is zero.
Answer:
The answer is "Choice A and Choice B"
Explanation:
The Zero-Order reactions are usually found if a substrate, like a surface or even a catalyst, is penetrated also by reactants. Its success rate doesn't depend mostly on the amounts of the various reaction in this reaction.
Let the Rate = k
As doesn't depend on reaction rate, a higher reaction rate does not intensify the reaction.
By the rate the created based and the reaction rate is about the same.