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Alex17521 [72]
3 years ago
10

2. Water is a great example of a molecule with polar covalent bonds. How does this bond affect the

Chemistry
2 answers:
Rzqust [24]3 years ago
8 0

Answer :

Example of polar covalent molecules H-O-H(water), ammonia

Explanation:

The presence of intermolecular Hydrogen bonding makes the boiling point of water unexpectedly high, and the polar covalent nature makes it dissolve polar solute/compound

dimulka [17.4K]3 years ago
6 0

Answer:

Explanation:

The constituent of a water molecule is one oxygen atom and two hydrogen atom. Oxygen is highly electronegative ( ability to pull electron closer to itself in a covalent bond) compare to hydrogen and as such it creates two poles with oxygen been partially negative and hydrogen been partially positive.  This attributes allow water to dissolve polar substances such sodium chloride, sodium hydroxide. When this substance are added to water, they are pull apart with positive sodium ion attracted to the partially positive oxygen and the chloride ions attracted to the partially positive hydrogen.

The polarity of water ice is able to float in water because of the polar nature as the molecules is extended as far as possible and are held together by the hydrogen bond. Water expands when its freezes and it  density decreases

The polar nature also of its molecules also contributes to its high boiling, freezing point and strong surface tension.

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3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

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 ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Select the correct image.
masya89 [10]

Answer:

A

Explanation:

In all the other options either Carbon has more than 4 atoms attached or less than 4.

Carbon can not form more than four bonds. It can only share 4 electrons .

In other option C, No H-atom is linked in the right most Carbon.

whereas according to the definition of Hydrocarbon it should have been a Hydrogen atom.

3 0
4 years ago
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