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lakkis [162]
3 years ago
6

You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate

with neat hexanes and found that only one spot moved a small distance up the plate. What might you expect to see if you ran a TLC plate in a 50/50 mixture of hexanes and ethyl acetate
Chemistry
1 answer:
goldenfox [79]3 years ago
3 0

Answer:

we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

Explanation:

when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC  plate in a 50/50 mixture of hexanes and ethyl acetate

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261 nm to millimeters
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Metabolic activity in the human body releases approximately 1.0 x 104kJ of heat per day. Assuming the body is 50 kg of water, ho
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Answer:

The body temperature would rise by 47.85 °C

The  amount of water the body evaporates is  4.15 kg.

This makes sense because firstly the value obtained is positive  then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature

Explanation:

Considering the relationship (between the heat released and the mass of the object)  as shown below

              q = msΔT    

where q is the heat released per day = 1.0 * 10^{4} kJ

              m is the mass of the body  = 50 kg

              ΔT  is the temperature rise = ?

               s is the specific heat  of water =  4,18kJ/kg^{o} C

              substituting values we have

                              1.0 * 10^{4} kJ  = (50kg) (4.18kJ/kg^{o}C ) ΔT

                              ΔT = \frac{1.0 * 10^{4}kg}{(50kg)(4.18kJ/kg^{o} C)}    

                                   =  47.85°C  

To maintain the normal  body temperature (98.6F = 37°C) the amount of heat released by metabolism activity must be utilized for evaporation of some amount of water

Hence

        Amount of water that must be evaporated =\frac{heat released per day}{heat of vaporization of water}

                                                                      = \frac{1.0* 10^{4}kJ}{2.41kJ/g} \\= 4149.38g\\=4.15kg  

          Note  (1 kg = 1000 g)

This makes sense because firstly the value obtained is positive  then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature

7 0
3 years ago
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