Calculate the volume in mL of 1.043 M acetic acid (CH3COOH) that is needed to prepare 50.0 mL of 0.10 M acetic acid. Show your work and give your answer to the correct number of significant figures.

4.8 mL of 1.043 M acetic acid that is needed to prepare 50.0 mL of 0.10 M acetic acid

Explanation:

According to the dilution law,

where,

= molarity of stock solution = 1.043 M

= volume of  stock solution = ?

= molarity of diluted solution = 0.10 M

= volume of diluted solution = 50.0 ml

Putting in the values we get:

Therefore, volume  in mL of 1.043 M acetic acid (CH3COOH) that is needed to prepare 50.0 mL of 0.10 M acetic acid is 4.8

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

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.