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Irina18 [472]
3 years ago
13

Bond energies do not account for the energy associated with the formation of aqueous solutions. Explain what energy is not accou

nted for AND how this energy contributes to the difference in calculated ∆Hrxn values using ∆Hf versus bond energy data ( -524 kJ/mol bond energy and -162.59 kJ/mol ∆Hf)
Chemistry
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

Why do we all not know the answer to this on the practical

Explanation:

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The term _____ describes a substance that can act as both an acid and a base.
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The term amphoteric describes a substance that can act as both an acid and a base.

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An isotope has 15 protons, 16 neutrons, and 15 electrons. What is the isotopic notation of the isotope
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Answer:

^{31}_{15}P is the isotopic notation of the atom

Explanation:

The isotope notation is:

^a_bX

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^a_bP

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Which periodic table groups have elements that cannot be found unbounded in nature?
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Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ
Len [333]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

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