Asample of a radioactive nuclide has a mass of 10 g. how many half lives must transpire until less than 21% of the original sample is present?

3 half-lives.

Explanation:

Half-life time is constant for first-order reaction.

21% of the sample (10 g) represents (10.0 x 0.21 = 2.1 g).

10 g (first half-life) → 5 g (second half-life) → 2.5 g (third half-life) → 1.25 g.

∴ 3 half-lives are needed for the sample to reach less than 21% (<2.1 g) of its initial value.

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.

answer; capital earns interest; entrepreneurship receives a profit or incurs a loss;