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butalik [34]
3 years ago
5

Consider the assumption that the four molecules of interest have approximately the same molar mass and very similar polarizabili

ties such that the magnitude of the dispersion forces should be about the same. this assumption allows us to use the data for pentane to account for the dispersion forces in the other three samples. based on your data and observations, is the assumption valid? why or why not? how has this assumption affected your data evaluation of intermolecular forces in the summary section? discuss polarizability, polarity, and relative imfs as part of your answer
Chemistry
1 answer:
Igoryamba3 years ago
5 0
The supposition is substantial on the grounds that, for these whose enthalpies is more prominent than n-pentane, they have a positive ΔH(polar), which implies there are more polar than the n-pentane. for these whose enthalpies is littler than n-pentane, they have a negative ΔH(polar), which implies there are less polar than the n-pentane. This supposition has an impact when I consider the predominant IMFS and different IMFS in the outline segment.
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Consider the elements bromine and chlorine; which element has a larger ionic radius?
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A 8.95 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.373 atm
Amanda [17]

Answer:

The total pressure is 1.616 atm

Explanation:

First of all we say:

In a closed system, sum of partial pressures is the value for the total pressure.

In our system, we have gas A and B

Total pressure without the third gas is = 0.373 atm + 0.650 atm = 1.023 atm

As we add a third gas, with no change in volume or T°, let's find out by the Ideal Gases Law, its pressure:

P . V = n . R . T

P = ( n . R . T) / V

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6 0
3 years ago
How many joules of heat are needed to change 50.0 grams of ice at 0 C to boil at 120.0 C? Show work
Tcecarenko [31]
This can be done in the following way;
1 determining the heat required to convert 0° C ice to 0°C water
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2. Determining the heat required to raise the temperature of water from 0° C to 100°C.
Specific heat of water is 4.18 J/g°C
Change in temperature is 100°C
Therefore; Heat = 50 g × 4.18 J/g°C × 100 = 20900 J
3. Determining the heat required to convert 100 ° C water to 100°C vapor
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Heat = Mass of water × heat of vaporization
Heat = 50 g × 2257 = 112850 J
4. Determining the heat required to go from 100° C to 120° vapor
specific heat of vapor = 2.09 J/g°C
Heat = mass × Specific heat of vapor × change in temperature
        = 50 g × 2.09 ×(120-100) = 2090 J

Therefore the total heat required is 
 = 16700 J + 20900 J + 112850 J + 2090 J =  152540 J or 152.54 kJ

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