a student reacted 40 drops of an unknown amine with 20 drops of iodomethane in 2 ml of a suitable solvent mixture. after 15 minu
1 answer:
The compound formed is the quaternary ammonium salt N,N,N-trimethyl-1-phenylmethanaminium iodide.
Quaternary ammonium salts are formed when methyl iodide reacts with an amine. The quaternary ammonium salt is an ionic substance which often has a high melting point. Remember that the melting point of a substance can be used to identify the substance.
In this case, a student reacted 40 drops of an unknown amine with 20 drops of iodomethane in 2 ml of a suitable solvent mixture. after 15 minutes of reaction, a white precipitate formed. the precipitate was isolated and its melting point was recorded to be 178 °c. This precipitate must be N,N,N-trimethyl-1-phenylmethanaminium iodide (Benzyltrimethylammonium iodide) because it is a quaternary ammonium salt whose melting point is 178 °c.
Learn more: brainly.com/question/5325004
You might be interested in
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
where,
= equilibrium constant at 244°C =
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature =
= final temperature =
Putting values in above equation, we get:
Hence, the rate constant at 324°C is