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AveGali [126]
3 years ago
12

explain the following observation. When 3-methylbutan-2-ol is treated with HBr, a single alkyl bromide is isolated, resulting fr

om 1,2, shift. When 2-methylpropan-1-ol is treated with HBr no rearrangement occurs to form alkyl bromide

Chemistry
1 answer:
babunello [35]3 years ago
5 0

Answer:

1. For the first reaction, there is Carbonation and rearrangement

2. For the second reaction, No Carbonation, No rearrangement.

Explanation:

See attached screenshot below

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A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase
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The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final pressure (P₂) = 13.5 atm

Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

  • The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298}  = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

V_{2} = \frac{58752}{4023}\\\\

<h3>V₂ = 15 L </h3>

Therefore, the final volume of the gas is 15 L

Learn more: brainly.com/question/25547148

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