The density of ice is 917 kg/m3. The density of salt water is 1029 kg/m3. A mass of ice is 2.55 x 1013 kg. Calculate the volume
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Answer:
21 KJ/mol
Explanation:
For this question, we have to start with the <u>linear structure</u> of 2-methylbutane. With the linear structure, we can start to propose all the <u>Newman projections</u> keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).
Additionally, we have several <u>energy values for each interaction</u> present in the Newman structures:
-) Methyl-methyl <em>gauche: 3.8 KJ/mol</em>
-) Methyl-H <em>eclipse: 6.0 KJ/mol</em>
-) Methyl-methyl <em>eclipse: 11.0 KJ/mol</em>
-) H-H <em>eclipse:</em> 4.0 KJ/mol
Now, we can calculate the energy for each molecule.
<u>Molecule A</u>
In this molecule, we have 2 Methyl-methyl <em>gauche </em>interactions only, so:
(3.8x2) = 7.6 KJ/mol
<u>Molecule B</u>
In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:
(11)+(6)+(4) = 21 KJ/mol
<u>Molecule C</u>
In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:
3.8 KJ/mol
<u>Molecule D</u>
In this molecule, we have three Methyl-H <em>eclipse </em>interaction, so:
(6*3) = 18 KJ/mol
<u>Molecule E</u>
In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:
3.8 KJ/mol
<u>Molecule F</u>
In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:
(11)+(6)+(4) = 21 KJ/mol
The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).
I hope it helps!
