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Katen [24]
2 years ago
15

5. A sample of carbon dioxide (CO2) has a mass of 52.0 g. a. How many carbon atoms are present?

Chemistry
1 answer:
vladimir2022 [97]2 years ago
7 0

Answer:

52.0 gof CO2 contains 7.1 *10^23 molecules

1 molecule of CO2 has a mass of 7.3*10^-23 grams

Explanation:

Step 1: Data given

Mass of CO2 = 52.0 grams

Molar mass of CO2 = 44.01 g/mol

Number of Avogadro = 6.022 * 10^23 / mol

Step 2: Calculate moles of CO2

Moles CO2 = Mass CO2 / molar mass CO2

Moles CO2 = 52.0 grams / 44.01 g/mol

Moles CO2 = 1.18 moles

Step 3: Calculate molecules in 1.18 moles CO2

Number of molecules = 1.18 moles * 6.022*10^23 = 7.1 *10^23 molecules

1 molecule of CO2

Number of moles = 1 / 6.022*10^23

Number of moles = 1.66 *10^-24

Mass CO2 = 1.66*10^-24 moles * 44.01 g/mol

Mass CO2 = 7.3*10^-23 grams

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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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<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:

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In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

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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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