Question

1. the deposition of ammonium chloride crystals on the walls of a test - tube after heating is due to

a. evaporation.
b. sublimation.
c. recrystallization.
d. fractional precipitation.

1. the product of decarboxylation of ch₃ch₂ch₂coona is

a. propanoic acid.
b. propene.
c. butane.
d. propane.

Explanation:

1. For question 13: Sublimation is the process where a substance changes directly from solid to gas and then back to solid upon cooling. Ammonium chloride undergoes sublimation. Evaporation is liquid - to - gas transition, recrystallization involves dissolving and then reforming crystals from solution, and fractional precipitation separates different substances based on their solubility differences. So the deposition of ammonium chloride crystals on test - tube walls after heating is due to sublimation.
2. For question 14: Decarboxylation of a carboxylate salt ($R - COONa$) involves the removal of the $-COONa$ group as $CO_2$ and $Na^+$ and the formation of an alkane with one less carbon atom. For $CH_3CH_2CH_2COONa$, upon decarboxylation, propane ($CH_3CH_2CH_3$) is formed. Propanic acid is a carboxylic acid, propene is an alkene, and butane has one more carbon atom than the product of this decarboxylation reaction.

Answer:

1. B. sublimation
2. D. propane