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Vinil7 [7]
2 years ago
8

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

tes of reaction, a white precipitate formed. the precipitate was isolated and its melting point was recorded to be 178 °c. what could this precipitate be?
Chemistry
1 answer:
lutik1710 [3]2 years ago
8 0

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

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Answer:

C_{vm} of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}

Where:

C_{pm} = Specific heat of gas at constant pressure = 75.3 J/mol·K

C_{vm} = Specific heat of gas at constant volume = Required

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Therefore,

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C_{vm}  = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} } - 75.3 = 256.834 J/mol·K.

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