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nikitadnepr [17]
3 years ago
11

During the workup portion of the reaction of alkenes with HBr as described in the experiment provided, a student transferred the

reaction mixture to a separatory funnel, rinsed the reaction flask with diethyl ether, and added the ether rinses to the separatory funnel. The student then added sodium bicarbonate to the separatory funnel. Extremely vigorous bubbling occurred. What did the student do wrong
Chemistry
1 answer:
inn [45]3 years ago
6 0

Answer:

Explanation:

Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.

Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.

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Reactioneaza total 22 kg calcar de puritate 80 % cu o solutie de acid clorhidric de concentratie 30 %. Stiind ca rezulta clorura
WINSTONCH [101]

Answer:

De la Alex;)(da alex ăla de e din cls cu tn:)

7 0
3 years ago
the hemiacetal below is treated with 18o-labeled methanol (ch3o*h) and acid. where will the label appear in the products?
ioda

The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

<h3>Position of 18o-labeled methanol in the products</h3>

The 18O label will appear at position b in the product as indicated in the image.

This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).

While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.

Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

Learn more about methanol here: brainly.com/question/17048792
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8 0
2 years ago
In a combustion reaction, one of the reactants is _____.
zepelin [54]
The answer is oxygen
3 0
3 years ago
Read 2 more answers
PLZZZZZZZZZZZZZZZZZZ15. As much as 90 percent of the oxygen in our atmosphere is the result of
vladimir2022 [97]

Answer:

The correct answer is - option D. photosynthesis.

Explanation:

It is shown by the study that most of the atmospheric oxygen comes from the photosynthesis by plants as oxygen is the byproduct of the photosynthesis. Photosynthesis is the process that uses light energy, carbon dioxide, and water to produce food or glucose/sugar and release oxygen as the byproduct.

Many scientists believe that oceanic phytoplankton that releases oxygen by the photosynthesis process makes 80 to 85% of the total oxygen of the atmospheric oxygen.

7 0
3 years ago
Read 2 more answers
For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of
viktelen [127]

Answer:

\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\

Now, we compute the atomic percentages as shown below:

\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%

Best regards!

4 0
2 years ago
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