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tatuchka [14]
2 years ago
9

How do you know that some substances in coffee beans are soluble?

Chemistry
1 answer:
Fantom [35]2 years ago
8 0

Answer:

When you make coffee, some of the compounds from the ground coffee end up in the water, giving it the distinctive color, aroma, and flavor that we associate with coffee. The solubility of some coffee constituents varies depending on the temperature of the water.

Explanation:

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Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of
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21 KJ/mol

Explanation:

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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!

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3 years ago
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