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3 years ago

Which observation would best support that the bond present was a covalent bond?

Chemistry

1 answer:

DIA [1.3K]3 years ago

4 0

Answer:

A covalent bond is present between two non-metals that share electrons in their molecular structure.

Explanation:

Examples of covalent bonds may be between carbon dioxide or water. And there may be three types of covalent bonds such as ionic, covalent and polar. In the first type, there is a transfer of an electron resulting in a gain for one of the atoms. In the second type, there is a sharing of electrons between the atoms. The third type is characterized by a difference in electronegativity between atoms.

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Read 2 more answers

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

Mademuasel [1]

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of propane at $25.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

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3 years ago

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4 years ago

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Explanation:

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