There are two naturally occurring isotopes of chlorine. 35 Cl has a mass of 34.9689 u. 37 Cl has a mass of 36.9659 u. Determine the abundance of each isotope.

Abundance for chlorine:

76.9% is 35 Cl and 23.1% is 37 Cl.

Explanation:

the average mass of one mole of chlorine is 35.453g.

Let x be the fraction of the chlorine that is number 35.

1 – x is the fraction that is number 37.

Since they have equal mass number, we equate the two expressions :

34.9689 * x + 36.9659 * (1 – x) = 34.9689 * x + 36.9659 – 36.9659 * x

34.9689 * x + 36.9659 – 36.9659 * x = 35.453

1.997 * x = 1.5359

x = 1.5359 ÷ 1.997

The decimal fraction is approximately 0.769. This means 76.9% is 35 Cl and 23.1% is 37 Cl.

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 is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.