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jok3333 [9.3K]
3 years ago
9

Complete the equations showing how 2−phenylethanol (c6h5ch2ch2oh) could be prepared from 2−phenylethanoic acid (c6h5ch2co2h) as

the starting material.

Chemistry
1 answer:
avanturin [10]3 years ago
7 0
We are asked to provide an equation for the transformation of 2-phenylethanoic acid to 2-phenylethanol. This type of a reaction is converting a carboxylic acid to an alcohol, which is classified as a reduction reaction since we are decreasing the number of bonds to oxygen in the molecule. In order to reduce a carbonyl to an alcohol, we need a source of hydride, H⁻. Reducing the carboxylic acid once will convert it to the aldehyde. However, we need to reduce the functional group all the way down to an alcohol, which is another reduction step after aldehyde formation. Therefore, the hydride source of choice is lithium aluminum hydride, LiAlH₄.

A reaction scheme is provided to show the reaction of the reduction of carboxylic acid to alcohol. The first step is addition of lithium aluminum hydride which does the reduction, and the second step is a work-up of acid which protonates the alcohol to get the final product.

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Vera_Pavlovna [14]

Answer: Molar mass of Al = 26.98 g/mol

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mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

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hope this help boo❤️❤️❤️

Explanation:

3 0
3 years ago
a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas
jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

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Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

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pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

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pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

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3 years ago
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Answer:

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Explanation:

This  should  be right hopefully it is!

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