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Soloha48 [4]
3 years ago
11

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture

containing mostly a dextrorotatory ketone A and a trace of an optically inactive isomer B. Treatment of A with zinc amalgam and HCl affords an optically active, dextrorotatory hydrocarbon C. Identify compounds A, B, and C, including stereochemical configurations. Be sure to explicitly draw H and both wedge/dash bonds at any configuration center.
Chemistry
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

use google and use the first link

Explanation:

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Since vinegar is 5% acetic acid and 95% water, find the mole ratio of acetic acid to water in 100 g of vinegar? Hint: Instead of
Oksanka [162]

The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.

<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>

The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.

The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.

In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.

Moles = mass/molar mass

molar mass of acetic acid = 62 g/mol

molar mass of water = 18 g/mol

moles of vinegar = 5/62 = 0.08

moles of water = 95/18 = 5.28

total moles = 5.36

Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36

Mole ratio of vinegar to water = 0.015 : 0.985

In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.

Learn more about mole ratio at: brainly.com/question/19099163

#SPJ1

7 0
2 years ago
What example of mixture could be seperated using two or more techniques? Explain your answer
STatiana [176]
You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
8 0
3 years ago
Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate t
LenaWriter [7]
The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He. 
To fill a volume of 6.3 L the mass of He required 
    = 1.79 x 10⁻⁴ g/mL * 6300 mL
   = 11 277 * 10⁻⁴ g
  Therefore mass of He required = 1.1277 g of He
8 0
3 years ago
Read 2 more answers
Mechanism with a small activation energy or one with large activation energy​
tangare [24]

Answer:

Rate depends on the rate constant. The rate constant depends on temperature and activation energy. If you have lower activation energy the rate will be higher. This is why catalysts are added since catalysts provide an alternate pathway that requires lower activation energy and catalysts are added to increase the rate of reaction.

Explanation:

This is only the answer if you were asking:

"Which corresponds to the faster rate: a mechanism with a small activation energy or one with a large activation energy?"

Thats what I understood about your question.

8 0
4 years ago
How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f
BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

A=\text{ 50 }\times2^{\frac{-t}{14.3}}

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}

So, after 84 days the P-32 remaining will be 0.85 mg

4 0
1 year ago
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