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Fudgin [204]
2 years ago
12

Calculate the Gibbs energy, entropy, and enthalpy of mixing when 1.00mol C6H14 (hexane) is mixed with 1.00mol C7H16 (heptane) at

298K. Treat the solution as ideal.
Chemistry
1 answer:
Norma-Jean [14]2 years ago
5 0

With Hexane of 1.00 ml and 1.00 ml of hectane at 298, the calculated

  • Gibbs energy = 3.43
  • Entropy = -3.43
  • Enthalpy = 0

<h3>How to solve for the Gibbs energy </h3>

C6H14 = C6H14/C6H14+C7H16

1mol/1mol+1mol = 1/2 = 0.5 mol

1+1 x 8.3145J x 298 x 0.5ln0.5 + 0.5ln0.5

= 4955.442 0.6932

= 3435J

convert to KJ

G mix = 3.43

The Gibbs energy is therefore  3.43KJ/mol

<h3>The entropy of the solution</h3>

-(1+1 x 8.3145J x 298 x 0.5ln0.5 + 0.5ln0.5)

S mix = -3.43 KJ/mol

3.43/298 = 11.5

The enthalpy of the solution

3.43-3.43 KJ/mol

H mix = 0

Read more on gibbs solution here: brainly.com/question/17310317

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What is oxidized and what is reduced <br> C2H4 + 2O2 → 2CO + 2H2O<br> C2H4 + 3O2 → 2CO2 + 2H2O
Anika [276]

C2H4 is oxidized and O2 is reduced in both reactions.

<h3>What is oxidation/reduction?</h3>

Oxidation is defined in several ways. Some of the definitions are:

  1. The addition of oxygen or removal of hydrogen
  2. Increase in the oxidation number of atoms
  3. Addition of electronegative or the removal of electropositive elements

Reduction, on the other hand, is defined as:

  1. Removal of oxygen or addition of hydrogen
  2. Decrease in the oxidation number of atoms
  3. Addition of electropositive elements or the removal of electronegative elements.

In the two reactions, oxygen is being added to C2H4. Thus, C2H4 is being oxidized.

The oxidizing agent is O2. In oxidation reactions, the oxidizing agents usually get reduced. Thus, O2 is reduced in both reactions.

More on oxidation and reduction can be found here: brainly.com/question/3867774

#SPJ1

7 0
2 years ago
Which pair will for an ionic compound
vovikov84 [41]

Answer: A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.

Explanation:

A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.

Explanation:

The chart below shows monatomic ions formed when an atom loses or gains one or more electrons, and the ionic compounds they form. You can check your periodic table to see that the cations are monatomic ions formed from metals, and the anions are monatomic ions formed from nonmetals.

3 0
2 years ago
Which SI unit would be most appropriate for expressing the mass of this animal
Ede4ka [16]

Answer:

Kilograms, I think.

Explanation:

8 0
3 years ago
How many carbon atoms are in 26-hydrogen alkyne
navik [9.2K]
An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)
5 0
3 years ago
The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.
Kitty [74]

Answer:

a) The theoretical yield is 408.45g of BaSO_{4}

b) Percent yield = \frac{realyield}{408.45g}*100

Explanation:

1. First determine the numer of moles of BaCl_{2} and Na_{2}SO_{4}.

Molarity is expressed as:

M=\frac{molessolute}{Lsolution}

- For the BaCl_{2}

M=\frac{1.75molesBaCl_{2}}{1Lsolution}

Therefore there are 1.75 moles of BaCl_{2}

- For the Na_{2}SO_{4}

M=\frac{2.0moles[tex]Na_{2}SO_{4}}{1Lsolution}[/tex]

Therefore there are 2.0 moles of Na_{2}SO_{4}

2. Write the balanced chemical equation for the synthesis of the barium white pigment, BaSO_{4}:

BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the BaCl_{2}:

\frac{1.75}{1}=1.75

- For the Na_{2}SO_{4}:

\frac{2.0}{1}=2.0

As the BaCl_{2} is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = \frac{realyield}{theoreticalyield}*100

Percent yield = \frac{realyield}{408.45g}*100

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0
3 years ago
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