--%>

Dipole attractions-London dispersion forces-hydrogen bonding

Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

E

Expert

Verified

They all are forces of attraction which is employed to keep the molecules altogether. As the molecules and atoms bonded altogether, there are no electrons, or not sufficient, left over to bond with more atoms. The outcome would be trillions of minute molecules floating about. Rather, each of such kinds of attractions draws the molecules altogether into solids, liquids and gases.

   Related Questions in Chemistry

  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

  • Q : Reactivity of allyl and benzyl halides

    why allyl halide and haloarenes are more reactive than alkyl halide towards nucleophilic substitution

  • Q : Chem Explain how dissolving the Group

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Define Bond Energies - Bond Charges

    Energy changes in some chemical reactions can be used to deduce the energies of chemical bonds. Our understanding of the molecular basis of thermodynamic properties is extended when we ask why the enthalpy change for a reaction is what it is. We deduce,

  • Q : Define alum Illustrate alum?

    Illustrate alum?

  • Q : Chem Silicon has three naturally

    Silicon has three naturally occurring isotopes. 28Si, mass = 27.976927; 29Si, mass = 28.976495; 30Si, mass = 29.973770 and 3.10% abundance. What is the abundance of 28Si?

  • Q : Reducible Representations The number of

    The number of times each irreducible representation occurs in a reducible representation can be calculated.Consider the C2v point group as described or Appendix C. you can see that (1) sum of

  • Q : Law of vapour pressure Select the right

    Select the right answer of the question. "The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called: (a) Henry's law (b) Raoult's law (c) Ostwald's law (d) Arrhenius's law

  • Q : Coagulation what is the meaning of

    what is the meaning of fourth power of valency of an active ion?

  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace