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lina2011 [118]
3 years ago
14

Liquid water is about 800 times denser than dry air under standard conditions. Use kinetic molecular theory to explain this diff

erence in the densities of the two substances.
Chemistry
1 answer:
Savatey [412]3 years ago
4 0

<em>Kinetic molecular Theory points:</em>

1. Gases molecules are always in random motion.

2. Gases molecule collide with each other and with the walls of container.

3.  Gases molecules total volume is negligible as compare to container.

4. There is no attractive forces between the gases particles

5. K.E is directly proportional to absolute temperature.

<em>Why liquid are denser than gases according to K.M.T:</em>

1. As gases are in random motion, so they have weak forces/ no attractive forces (Acc. to KMT) due to this they are apart as compare to liquid which have stronger forces are closer to each other. So volume of gases will be greater. d= m/v

2.  K. E of gases are greater than liquid due to this gases move randomly, apart from each another. Due to this their volume will be greater. d= m/v

3. Gases molecules total volume is negligible as compare to container which show that gases molecules have large spaces between them which causes lower in density.

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When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/m
Anni [7]

Answer:

The heat of vaporisation of methanol is "3.48 KJ/Mol"

Explanation:

The amount of heat energy required to convert or transform  1 gram of liquid to vapour is called heat of vaporisation

When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.

Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is =   \frac{8.7}{2.5}KJ

 = 3.48 KJ

So, the heat of vaporization \delta H_{vap} = 3.48KJ/Mol

Therefore, the heat of vaporization of methanol is 3.48KJ/Mol

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2 years ago
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

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