2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.
Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.
A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.
Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.
Full question :
Q. Which reactant is unlikely to produce the indicated product upon strong heating?
- A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
- B) 2-Ethylpropanedioic acid Butanoic acid
- C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
- D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
- E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone
Hence, option (D) is correct.
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First, we determine the energy released by the reaction using the heat capacity and change in temperature as such:
Q = cΔT
Q = 32.16 * 0.42
Q = 13.51 kJ
Next, we determine the moles of ammonia formed as the heat of formation is expressed in "per mole".
Moles = mass / molecular weight
Moles = 5/17
Moles = 0.294
Heat of formation = 13.51 / 0.294
The heat of formation of ammonia is 45.95 kJ/mol
Answer:
a) 300K
b) 373K
c) 273K
Explanation:
to go from °C to K all you have to do is add 273.
Answer:
2m/s²
Explanation:
When an object starts or at its state of rest it has an Initial speed U = 0
Final speed = 6m/s
total time taken for the acceleration = 3s
Acceleration =?
Acceleration is the change in velocity (speed) with time
OR
Time rate of change of velocity
Acceleration = <u>Change in Speed(velocity)</u>
Time taken
Hence,
Acceleration = <u> </u><u> </u><u>V - </u><u>U</u><u> </u><u> </u>
t
a = <u>6</u><u> </u><u>-</u><u> </u><u>0</u>
3
a = <u>6</u><u> </u><u> </u>
3
a = 2m/s²
Answer:
here
Explanation:
0.000141 to kilowatt-hours. hope this helped
