23.Compute the root-mean-square speed of O2 molecules in a sample of oxygen gas at a temperature of 212°C.

The root-mean-square speed-vrms = 615 m/s

Given

O₂ molecules

T = 212 + 273 = 485 K

Required

The root-mean-square speed

Solution

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  = 0.032 kg/mol

Input the value :

v rms = √(3x8.314 J/mol K x 485 K)/0.032

vrms = 615 m/s

pure substance: iron,copper

impure substance: water,charcoal

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.