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suter [353]
4 years ago
9

The most stable conformation of 1,2-dibromoethane is:

Chemistry
1 answer:
zhenek [66]4 years ago
4 0
I have attached an image that shows the most stable conformation of 1,2-dibromoethane. The top image shows the line drawing of the structure and the bottom image shows the Newman project of the structure. The Newman project shows up the perspective of looking down the C1-C2 bond of the compound. From this perspective, we can see the relationship between the two bromo-substituents.

Since the bromo-substituents are the largest groups substituted on each carbon atom, these groups will want to be as far away from each other in space as possible. The result is this anti- relationship shown, where the two bromo atoms are as far away from each other as possible within the molecule to reduce the amount of steric interactions. Therefore, the anti-conformation is the most stable conformation of 1,2-dibromoethane.

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A living cell with a tonicity (solute concentration) equivalent to 0.9% NaCl is placed in a solution containing 2% NaCl. Assume
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Explanation:

This illustration refers to an hypertonic solution. Hypertonic solution is a solution in which the surrounding solution has a higher solute concentration (2% of NaCl) than the cell's cytosol (0.9% of NaCl). In hypertonic solution, the solution outside the cell (with higher concentration) pulls the water from the cell's cytosol (via osmosis) causing the cell to shrink.

4 0
3 years ago
According to the website of the National Aeronautics and Space Administration (NASA), the average temperature of the universe is
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Explanation :

The conversion used for the temperature from Kelvin to degree Celsius is:

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As we are given the temperature in Kelvin is, 2.7

Now we have to determine the temperature in Kelvin.

^oC=K-273.15

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3 0
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The activation energy for the gas phase isomerization of dimethyl citraconate is 105 kJ.
Mamont248 [21]

Answer:

k_2=9.06x10^{-4}/s

Explanation:

Hello there!

In this case, since the activation energy, rate law and temperature, when variable, are related to each other as shown below:

ln(\frac{k_2}{k_1} )=\frac{-Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Thus, when solving for the rate constant at 682 K, we will obtain:

ln(\frac{k_2}{2.77x10^{-4}/s} )=\frac{-105000J/mol}{8.3145\frac{J}{mol*K}}(\frac{1}{682K} -\frac{1}{641K} ) \\\\ln(\frac{k_2}{2.77x10^{-4}/s} )=1.184\\\\\frac{k_2}{2.77x10^{-4}/s}=exp(1.184)\\\\k_2=2.77x10^{-4}/s*3.269\\\\k_2=9.06x10^{-4}/s

Best regards!

8 0
3 years ago
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