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pashok25 [27]
3 years ago
11

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b

oiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Chemistry
1 answer:
____ [38]3 years ago
6 0

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

Thus, with the proper units, we obtain:

T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C

Hence, answer is approximately 100 °C.

Best regards.

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How many moles of O2- ions are there in 0.450 moles of aluminum oxide, Al2O3?
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Answer:

                     1.35 moles of O²⁻

                     21.6 grams of O²⁻

Explanation:

We know that the charge on Aluminium ion is +3 (i.e. Al³⁺) while, the charge on Oxide ion is -2 (i.e. O²⁻). Therefore, the overall neutral Al₂O₃ compound has 2 Al³⁺ ions and 3 O²⁻ ions. Since, we can say that,

                 1 mole of Al₂O3 contains  =  3 moles of O²⁻ ions

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                     0.450 moles of Al₂O₃ will have  =  X g of O²⁻

Solving for X,

                      X =  0.450 mol × 3 mol ÷ 1 mol

                     X =  1.35 moles of O²⁻

As the mass of an atom is mainly due to the presence of protons and neutrons hence, the addition of two electrons (-ve 2 shows two gained electron) to Oxygen will make a negligible change to the atomic masss of Oxygen because electron is said to be almost 1800 times lighter than proton. Hence, the ionic mass of O²⁻ will be 16 g/mol and the mass of given moles is calculated as,

                     Mass  =  Moles × Ionic Mass

                     Mass  =  1.35 mol × 16 g/mol

                    Mass  =  21.6 g

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