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Ludmilka [50]
3 years ago
6

Please help me with this thank you !!

Chemistry
1 answer:
Sindrei [870]3 years ago
4 0
The answer is I think is c
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1. Draw a wedge/dash structure for trans-1,2-dimethylcyclohexane.
Angelina_Jolie [31]

Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

6 0
3 years ago
.One ballon is filled with Hydrogen (H2) gas, the other balloon is filled with methane (CH4). Both balloons have the same temper
Neko [114]

Answer:

C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2

Explanation:

Based on combined gas law, gases under the same pressure, temperature and volume have the same number of moles. With this information we can say the rigth statement is:

<h3>C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2</h3>
7 0
3 years ago
Perform the following mathematical operation, and report the answer to the correct number of significant figures. 0.34 x 0.568=?
slavikrds [6]

The correct answer to the problem is 0.193 which is three significant figures.

<h3>What are significant figures?</h3>

The term significant figures has to do with the figures that have a mathematical meaning. We know that the result has to correspond to the highest number of significant figures.

Hence, If we multiply 0.34 x 0.568 the result ought to be recorded as 0.193 which is three significant figures.

Learn more about significant figures:brainly.com/question/14804345

#SPJ1

6 0
1 year ago
An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit
Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of triprotic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL

Putting values in above equation, we get:

3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0
3 years ago
Which of the following best describes an atom?A)Protons and electrons grouped together in a random pattern.B) A core of electron
madreJ [45]

Answer: B

Explanation:

5 0
3 years ago
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