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melomori [17]
2 years ago
11

Alcohols can be oxidized by chromic acid derivatives. One such reagent is pyridinium chlorochromate, (C5H5NH )(ClCrO3–), commonl

y known as PCC. Draw the proposed (neutral) intermediate and the organic product in the oxidation of 1-butanol by PCC when carried out in an anhydrous solvent such as CH2Cl2."

Chemistry
1 answer:
arsen [322]2 years ago
7 0

Answer:

See figure attached.

Explanation:

Pyridinium chlorochromate (PCC) is a weaker oxidizing agent when compared with oth chromic acid derivatives as dichromate, for example. This results in the fact that PCC wil oxidize a primary alcohol transforming it to a aldehyde.

The entire mechanism of the reaction of 1-butanol with PCC in an anhydrous solvent is shown in the figure attached.

The neutral intermediate and the organic product (1-butanal) are both represented in the figure.

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I think is 29'flies
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2 years ago
Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden
LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

5 0
2 years ago
A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coeff
ra1l [238]

Answer:

activity coefficient \mathbf{\gamma =0.809}

activity coefficient \mathbf{\gamma = 0.791}

The change in pH in part A = 0.092

The change in pH in part B =  0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = 10^{-1.092}

[a] = 0.0809 M

Recall that [a] = \gamma × C

where;

\gamma  = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

\gamma = \dfrac{[a]}{C}

\gamma = \dfrac{0.0809 \ M}{0.100  \ M}

\mathbf{\gamma =0.809}

B.

The pH of a solution of HCl and KCl = 2.102

[a] = 10^{-2.102}

[a] = 0.00791 M

activity coefficient \gamma = \dfrac{0.00791 \ M}{0.01 \ M}

\mathbf{\gamma = 0.791}

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

3 0
2 years ago
Your task is to measure the amount of energy evolved during the combustion of some hydrocarbon. Which of the following would be
Dahasolnce [82]

Answer:

c. Bomb calorimetry

Explanation:

The hydrocarbons are combustibles, it means that they can react in a combustion reaction to release energy. To measure this amount of energy, it's necessary equipment that the reaction can be placed in a controlled way. The bomb calorimeter is this equipment, which is an adiabatic vessel, with water. The heat is calculated based on the increase in the water temperature.

The coffee-cup calorimetry is used to measure the heat of a dissolution reaction and the bomb manometry is used to measure the pressure.

7 0
2 years ago
How would each of the following procedural errors affect the results to be expected in this experiment? Give your reasoning in e
Digiron [165]

Answer:

a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.

b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.

c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.

d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling

temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.

7 0
2 years ago
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