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Dmitry [639]
3 years ago
5

Arrange the following molecules according to the strength of their dispersion forces. Organize in order from largest to smallest

.Dispersion Forceslargest dispersionCH3CH(CH3)C(CH3)2CH2CH31CH3CH2(CH2)4CH2CH32CH3CH2CH2CH2CH33CH3C(CH3)2CH34CH3CH2CH35CH46
Chemistry
1 answer:
ki77a [65]3 years ago
3 0

Explanation:

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Therefore, more is the surface area occupied by the carbon chain more will be the dispersion forces present in it. Hence, less is the surface area occupied by a molecule less will be the dispersion forces present in it.  

Hence, the given molecules are organized from largest to smallest dispersion forces as follows.

CH_{3}CH(CH_{3})C(CH_{3})_{2}CH_{2}CH_{3} > CH_{3}CH_{2}(CH_{2})_{4}CH_{2}CH_{3} > CH_{3}CH_{2}CH_{2}CH_{2}CH_{3} > CH_{3}C(CH_{3})2CH_{3} > CH_{3}CH_{2}CH_{3} > CH4

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55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?
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Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1},

where

  • V_2 is the final volume,
  • V_1 is the initial volume,
  • T_2 is the final temperature in degrees Kelvins.
  • T_1 is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K.

T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K.

Apply Charles's Law to find the new volume of this gas:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L.

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