how much alum product would be lost to the crystallization solution if you had 42.5 ml of solution after filtration and the solubility of alum is approximately 2.63 g alum in 100 ml of 0 degrees celcius acidic water

Total amount of alum lost = 0.5122 grams

Explanation:

Let the total volume of the solution be 100 mL

In 100 mL of solution, there is 2.63 gram of alum.

Out of this 100 mL solution, 42.5 mL is remaining.

Amount of alum in 42.5 mL solution is

grams

Now the amount of alum lost is grams

the answer is 0.8m.

by: nicholas

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.