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hichkok12 [17]
3 years ago
14

Which member of each pair is the stronger base? part a ethylamine or aniline ethylamine or aniline ethylamine aniline request an

swer part b ethylamine or ethoxide ion ethylamine or ethoxide ion ethylamine ethoxide ion?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.

Part 1: 
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.

Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.

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Why are pipes bursting in the in extremely cold weather?
avanturin [10]

Water is the only common substance that when you freeze it, it's volume INCREASES.

When the pipe originally held the "all full" volume and the the water expanded, it put a tremendous amount of pressure on the pipe. Enough pressure and the pipe would burst.

8 0
3 years ago
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0
3 years ago
What is the coefficient for CO2?
Stells [14]
1?....................
8 0
3 years ago
Dissolution of wax in kerosene
yanalaym [24]

Answer:

the first answer is correct don't forget you can use quizzlet app to

8 0
3 years ago
A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

\frac{V}{T} =k

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

\frac{V1}{T1} =\frac{V2}{T2}

<h3>Temperature of the gas in this case</h3>

In this case, you know:

  • P1= 1800 psi
  • V1= 10 L
  • T1= 20 C= 293 K (being 0 C= 273 K)
  • P2= 1800 psi
  • V2= 6 L
  • T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

<u><em>T2=163.8 K= -109.2 C</em></u>

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

7 0
3 years ago
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