Naturally occurring indium has two isotopes, indium-113(112.9040580 amu) and indium-115 (114.9038780 amu. the atomic mass of indium is 114.82 amu. what is the percent of indium-115? a)4.32% b)48.4% c)95.7% d)0.973%

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu)  ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂)  / 100

%₁ = x I established this is an equation

%₂ = 100 - x

Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100    and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x  /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

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.