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umka21 [38]
2 years ago
6

When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 169.5 kJ are

absorbed and PΔV for the vaporization process is equal to 14.5 kJ.
What is the change of internal energy?


What is the change of enthalpy?


Is the change in entropy positive or negative for this process?


If the pressure would be only ½ atm, what are the changes of internal energy and enthalpy during vaporization (approximately)?


How much energy is absorbed?


Write the definition of the Gibbs Free Energy.


At the lower pressure, will the boiling point be higher or lower.

Chemistry
1 answer:
NNADVOKAT [17]2 years ago
3 0

Answer:

∆E=155kJ

Explanation:

Check attachment

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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

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C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

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Is true. Nitrogen gas behaves more like an ideal gas as the temperature increases. Under normal conditions such as normal pressure and temperature conditions , most real gases behave qualitatively as an ideal gas. Many gases such as air , nitrogen , oxygen ,hydrogen , noble gases , and some heavy gases such as carbon dioxide can be treated as ideal gases within a reasonable tolerance. Generally, the removal of ideal gas conditions tends to be lower at higher temperatures and lower density (that is at lower pressure ), since the work made by the intermolecular forces is less important compared to the kinetic energy<span> of the particles, and the size of the molecules is less important compared to the empty space between them. </span><span>The ideal gas model tends to fail at lower temperatures or at high pressures, when intermolecular forces and intermolecular size are important.</span>

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