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3 years ago

What is the mass of a silver coin that contains 1.69x10^23 silver atoms? Report your answer in units of grams

Chemistry

1 answer:

Margaret [11]3 years ago

3 0

Answer: 30.27 g

Explanation:

Since the silver is in units of atoms, we have to use Avogadro's number and molar mass.

Avogadro's number: 6.022×10²³ atoms/mol

Molar Mass: 107.87 g/mol

Now that we have everything we need, we can convert atoms to grams.

$1.69*10^{23} atoms*\frac{1mol}{6.022*10^2^3 atoms} *\frac{107.87 g}{1mol} =30.27 g$

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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

natima [27]

Answer:

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Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

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-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

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Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

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Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse 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