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
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
<u>= 34.72 g</u>