Answer:
See explanation and images attached
Explanation:
a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.
b) In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.
Answer:
A)
1. Reaction will shift rightwards towards the products.
2. It will turn green.
3. The solution will be cooler..
B) It will turn green.
Explanation:
Hello,
In this case, for the stated equilibrium:
In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:
A)
1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.
2. The formation of the green complex is favored, therefore, it will turn green.
3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.
B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.
Regards.
Answer:
C₆H₈O₇+ 3NaHCO₃ --› Na₃C₆H₅O₇ + 3CO2 + 3H₂O
Explanation:
The reaction occuring in lava lamp is acid base reaction.
When you drop tablet into water the citric acid reacts with sodium bicarbonate and forms water, a salt, and bubbles of carbon dioxide gas.
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
I will list them from alkaline with the lowest boiling point and alkaline with the highest.
1. C2H6
2. C9H20
3. C11H24
4. C16H34
5. C20H42
6. C32H66
7. C150H302
I have taken a quiz similar to this before and can assure you this is correct and is primarily because of the number of Carbons and Hydrogens within this. More Carbons and Hydrogens causes Boiling Points to increase because of stronger bonds.
