Fractional Distillation is the answer
<u>Answer:</u> The 
for the reaction is -1406.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of 
is:
The intermediate balanced chemical reaction are:
(1) 
( × 6)
(2) 
( × 3)
(3) 
( × 2)
(4) 
The expression for enthalpy of formation of is,
![\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B6%5Ctimes%20%5CDelta%20H_1%5D%2B%5B3%5Ctimes%20%5CDelta%20H_2%5D%2B%5B2%5Ctimes%20%5CDelta%20H_3%5D%2B%5B1%5Ctimes%20%5CDelta%20H_4%5D)
Putting values in above equation, we get:
![\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B%28-74.8%5Ctimes%206%29%2B%28-185%5Ctimes%203%29%2B%28323%5Ctimes%202%29%2B%28-1049%5Ctimes%201%29%5D%3D-1406.8kJ)
Hence, the for the reaction is -1406.8 kJ.
A scientific law is a statement that is universally accepted and answers the "how?" question.
Hope this helps!
Answer: Water becomes gas at D-E.
Explanation:
In the flat line D-E water is absorbing energy to become gas, and when the water is completely gas, the temperature will increase again therefore the line will rise up.
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
