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7nadin3 [17]
3 years ago
6

For fatty acids with the same number of carbon atoms, how does the melting point change as the number of double bonds in the fat

ty acid changes?
a. The melting point of the fatty acid decreases as the number of double bonds increases.
b. There is no relationship between the melting point of a fatty acid and the number of double bonds.
c. The melting point of the fatty acid is unchanged as the number of double bonds changes.
d. The melting point of the fatty acid decreases as the number of double bonds decreases.
Chemistry
1 answer:
Nikitich [7]3 years ago
7 0

The answer is <em>a. "The melting point of the fatty acid decreases as the number of double bonds increases" </em>because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.

In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.

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What is the number of moles of ammonia gas formed when 0.5 mole of nitrogen gas is reacted with excess of hydrogen gas?​
Troyanec [42]

Answer: 1.5mol NH3

Explanation: 1N2+3H2= NH3

                      0.5 mol N2x 3NH3/1molN2= 0.5x3= 1.5mol NH3

7 0
1 year ago
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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