Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:
- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:
Entonces se obtiene:
a) 72 °F= =22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F= = 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F
Answer:
Explanation:
Given parameters:
Initial temperature T₁ = 25.2°C = 25.2 + 273 = 298.2K
Initial pressure = P₁ = 0.6atm
Final temperature = 72.4°C = 72.4 + 273 = 345.4K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:
where P and T are temperatures, 1 and 2 are initial and final temperatures.
Input the parameters and solve;
P₂ = 0.7atm
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
Thus, the concentration of ethylamine in solution is:
Now, we can also infer that some salt is formed, and has the following concentration:
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
Finally, the pH turns out to be:
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
Answer:
It is the process in which a known amount of solution of known concentration is added to the concentration of the another solution to determine the concentration of unknown solution.
Explanation: