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Katarina [22]
1 year ago
7

The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entr

opy in the system in j/k when 24.1 grams of c2cl3f3 vaporizes to a gas at the normal boiling point?
Chemistry
1 answer:
Goshia [24]1 year ago
7 0
According to this formula when:

ΔG = ΔH - TΔS = 0

∴ ΔS = ΔH/T

∴ ΔS = n*ΔHVap / Tvap

- when n is the number of moles = mass/molar mass 

when the mass = 24.1 g 

and the molar mass = 187.3764 g/mol

by substitution: 

∴ n = 24.1 / 187.3764g/mol

      = 0.129 moles

and ΔHvap is the molar enthalpy of vaporization is 27.49 kJ/mol

and Tvap is the temperature in Kelvin = 47.6 + 273 = 320.6 K

So by substitution, we will get the ΔS the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

      = 11 J/K
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If the composition of the reaction mixture at 400 k is [brcl] = 0.00415 m, [br2] = 0.00366 m, and [cl2] = 0.000672 m, what is th
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by considering this reaction:
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but this equation in the gaseous or aqueous states only.
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by substitution:
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Answer:

RbOH  → Rb⁺ +  OH⁻

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Explanation:

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Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

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Makovka662 [10]

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The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95. 

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