What is the most significant intermolecular force acting between molecules of ch3cl

Dipole-dipole interactions.

There are three classes of intermolecular interactions: London dispersion force, Dipole-dipole interaction, and hydrogen bonds.

Hydrogen bonds are observed only in molecules with hydrogen atoms bonded directly to atoms of fluorine, oxygen, or nitrogen that are of high electronegativity. Chloromethane contains no such bonds. Therefore, there would be no hydrogen bonds between molecules of .

Dipole-dipole interactions are weaker than hydrogen bonds in most molecules. Dipole-dipole interactions exist only between polar molecules. Chloromethane contains polar carbon-chlorine bonds. Chloromethane molecules are asymmetric, such that polarity due to the presence carbon-chlorine bonds does not cancel out within the molecule. Chloromethane molecules are therefore polar, and would experience dipole-dipole interactions.

London distribution forces attribute to the movement of electrons within a molecule. Such interactions are transient and thus weaker than both hydrogen bonds and dipole-dipole interactions in small molecules such as .

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year