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Elan Coil [88]
3 years ago
11

The following questions are about ethyl bromide,n-propyl bromide, isopropyl bromide,t-butyl bromide, & neopentyl bromide

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

a) Watch the attaccment

b) Ethyl bromide is more reactive than n-propyl bromid, and this more than neopentyl bromide. Ethyl bromide has less steric hindrance than the others, to SN2 reactions.

c) t-butyl bromide is more reactive than  isopropyl bromide, and this more than ethyl bromide. t-butyl bromide structure stabilize the carbocation, better than the others.

Explanation:

Speed of SN2 reactions depends on steric hindrance, the less hindrance, the most reaction speed, meaning more reactivity. Then, those linear structures are more reactive to SN2 reactions.

In the other hand, speed of SN1 reactions depends on the stability of the carbocation formed. Structure with ramifications can stabilize better the carbocation, these structures are more reactive to SN1 reactions.

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Answer is c

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A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.58
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Answer:

\% O=27.6\%

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}  =0.218molC

- Moles of hydrogen are contained in the 3.922 grams of water:

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- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO

Finally, we compute the percent by mass of oxygen:

\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%

Regards.

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