Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. The average specific heat of gaseous ethanol is about 1.80 J/g-K. a. How much heat is required to convert 35.0 g of ethanol at 27 °C to the vapor phase at 120 °C? b. How much heat is required to convert the same amount of ethanol at –120 °C to the vapor phase at 120 °C?

Answers

Answer 1

Answer:

First question

       [tex]Q = 36826 \ J[/tex]

Second  question  

       [tex]Q = 52299.7 \ J[/tex]

Explanation:

From the question we are told that

     The melting point of Ethanol is  [tex]T_m = -114 ^oC[/tex]

      The boiling point of Ethanol is  [tex]T_b = 78^ oC[/tex]

       The enthalpy of fusion of Ethanol is [tex]F = 5.02 \ kJ / mol = 5.02 *10^{3}\ kJ / mol[/tex]

        The enthalpy of vaporization  of Ethanol is [tex]L = 38.56 \ kJ / mol = 38.56 *10^{3} \ J / mol[/tex]

         The specific heat of solid Ethanol is  [tex]c_e = 0.97 \ J/ g \cdot K[/tex]

          The specific heat of liquid  Ethanol is [tex]c_l = 2.3 \ J / g \cdot K[/tex]

           The mass of the Ethanol given is  [tex]m = 35.0 \ g[/tex]

Considering the first question

           The initial  temperature is [tex]T_i = 27^oC[/tex]

             The final  temperature is  [tex]T_f = 120^oC[/tex]

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       [tex]Q_1 = m * c_l * (T_b - T_i)[/tex]

=>      [tex]Q_1 = 35.0 * 2.3 * ( 78 - 27)[/tex]

=>      [tex]Q_1 =4106 \ J[/tex]

Genially the number of moles of Ethanol given is mathematically represented as

         [tex]n = \frac{m}{Z}[/tex]

Here Z  is the molar mass of Ethanol  with value  [tex]Z = 46 g/mol[/tex]

So

         [tex]n = \frac{35}{46 }[/tex]

=>      [tex]n = 0.7609 \ mol[/tex]

Generally the heat of vaporization of the Ethanol is mathematically represented as

         [tex]Q_2 = n * L[/tex]

=>        [tex]Q_2 =0.7809 * 38.56 * 10^{3}[/tex]

=>        [tex]Q_2 =29339 \ J[/tex]

Generally the heat required too raise the Ethanol from  its boiling point to  [tex]T_f[/tex]  is  mathematically represented as

       [tex]Q_3 = m * c_l * (T_f - T_b)[/tex]

=>     [tex]Q_3 = 35 * 2.3 * (120 - 78 )[/tex]

=>     [tex]Q_3 = 3381 \ J[/tex]

Generally the total heat required is  

     [tex]Q = Q_1 + Q_2 + Q_3[/tex]

=>   [tex]Q = 4106 + 29339 + 3381[/tex]

=>   [tex]Q = 36826 \ J[/tex]

Considering the second question

           The initial  temperature is [tex]T_i = -120^oC[/tex]

             The final  temperature is  [tex]T_f = 120^oC[/tex]

Generally the heat required too raise the Ethanol to its melting  point is mathematically represented as

       [tex]Q_1 = m * c_e * (T_m - T_i)[/tex]

=>      [tex]Q_1 = 35.0 * 0.97 * ( -114 - (- 120) )[/tex]

=>      [tex]Q_1 = 203.7 \ J[/tex]

Generally the heat of fusion  of the Ethanol is mathematically represented as

                 [tex]Q_2 = n * F[/tex]

=>        [tex]Q_2 =0.7809 * 5.02 *10^{3}[/tex]

=>        [tex]Q_2 =3920 \ J[/tex]

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       [tex]Q_3 = m * c_l * (T_b - T_m)[/tex]

=>      [tex]Q_3 = 35.0 * 2.3 * ( 78 - (- 114) )[/tex]

=>      [tex]Q_3 =15456 \ J[/tex]

Generally the heat of vaporization of the Ethanol is mathematically represented as

         [tex]Q_4 = n * L[/tex]

=>        [tex]Q_4 =0.7809 * 38.56 * 10^{3}[/tex]

=>        [tex]Q_4 =29339 \ J[/tex]

Generally the heat required too raise the Ethanol from  its boiling point to  [tex]T_f[/tex]  is  mathematically represented as

       [tex]Q_5 = m * c_l * (T_f - T_b)[/tex]

=>     [tex]Q_5 = 35 * 2.3 * (120 - 78 )[/tex]

=>     [tex]Q_5 = 3381 \ J[/tex]

Generally the total heat required is  

     [tex]Q = Q_1 + Q_2 + Q_3+Q_4 + Q_5[/tex]

=>   [tex]Q = 203.7 + 3920 + 15456 +29339+3381[/tex]

=>   [tex]Q = 52299.7 \ J[/tex]


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1.957 × 10²⁴ molecules

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Group of answer choices

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Answers

Answer:

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Explanation:

How many grams are in 7.5 moles of C₆H₁₂?

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Answers

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Answers

Answer:

No mass loss

Explanation:

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Answers

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Answers

Answer:

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Explanation:

M-A=N

Here is an example.

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So, we have;

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1 Neutron

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Answers

Answer:

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Answer:

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Answers

Answer:

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Answers

Answer:

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Explanation:

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Answers

The molecular mass​ : 81.72 g/mol

Further explanation

In general, the gas equation can be written  

[tex]\large {\boxed {\bold {PV = nRT}}}[/tex]

where  

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V = volume, liter  

n = number of moles  

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P = 760 mmHg=1 atm

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Answer:

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Explanation:

Its right there????

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Answers

Answer:

I believe its A. Mass

Explanation:

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Answer:

B. Magnetism

Explanation:

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Answer:

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DRAW A PEDIGREE

Read the following information and ON NOTEBOOK PAPER, construct a pedigree using the symbols we went over Tuesday: Scott is married to Christa. They have 3 children, Blake (a son), Peyton (a daughter), and Ashton (a daughter). Blake is married to Allie and they have 2 children, Henry (a son) and Harper (a daughter).

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Answers

Answer:

Here you go

Explanation:

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Answers

Answer:

See explanation

Explanation:

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Answers

Answer:

117g

Explanation:

Given parameters:

Number of moles = 2moles

Unknown:

Mass of NaCl  = ?

Solution:

To solve the problem, we need to use the expression below;

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So;

Insert the parameters and solve;

     Mass of NaCl  = 2 x 58.5  = 117g

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