<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span>
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration
b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.
Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.
Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:
Properties of alkali metals are: Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.
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You have to use Dalton's law of partial pressure for this question. Dalton's law of partial pressure basically states that the total pressure of the system is all of the partial pressures of the components added together. Therefore to answer the question you just need to add all the patial pressures together meaning that the total pressure would be 700+500+500=1700.
The answer would be 1700 torr.
I hope this helps. Let me know if anything is unclear or if you have any further questions.
Answer is C. 2, 3 and 5 shows how as the number of carbon increase, butane, pentane and hexane (4,5 and 6 carbons) the octane number does decrease
Given what we know, we can confirm that an example is a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided.
<h3>What are examples of the situations given?</h3>
- A number that is a multiple of 10 is 40, since 10 times 4 equals 40.
- In order to get a product of 10, we can multiply two and five.
- To result in a quotient of 10, we can divide one hundred by ten.
<h3>What are non-examples of the situations given?</h3>
- one non-example of a multiple of 10 would be to multiply three and seven.
- A non-example of a product of 10 is to multiply the number fifty by twenty-five.
- To result in a non-example for a quotient of 10, we can divide the number fifteen by three.
Therefore, given the definition of an example as a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided, we can confirm that the ones listed above are correct.
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