This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and 
and 
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.
A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.
The actual electronic configuration of antimony is as follow:
[Kr] 4d10 5s2 5p3
B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.
The actual electronic configuration of Cesium is as follow:
[Xe] 6s1
Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
learn more about Noble gas:
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Answer: Benzene is less reactive than methylbenzoate and more reactive than Nitrobenzene
Explanation:
This is because the methyl group on the benzene ring is an electron donating group leading to the activation of the ring and subsequently leading to more canonical resonance structure at the intermediate stage of the reaction enhancing the faster reactivity
However for the Nitrobenzene the nitro group is an electron withdrawing group leading to a slower activation and less resonance canonical structure at the reaction intermediate leading to a slower reaction than the reaction of benzene without the nitro group
This is known as polymerisation
The answer is C Becuz D is a compound and that is the only answer that says it
