Dipole attractions-London dispersion forces-hydrogen bonding
Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
Expert
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.
Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately
In some two component, solid liquid systems, a eutectic mixture forms.Consider, now a two component system at some fixed pressure, where the temperature range treated is such as to include formation of one or more solid phases. A simple behavior is shown b
Can someone please help me in getting through this problem. 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute HCL for complete reaction. The strength of the HCL solution is given by: (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N
Provide solution of this question.10N/and 1/10N solution is called: (a) Decinormal and decanormal solution (b) Normal and decinormal solution (c) Normal and decanormal solution (d) Decanormal and decinormal solution
Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M
IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN
In the manufacture of sulphuric acid by the contact process, S02 is oxidized to SO3 over a vanadium catalyst: The reactor is adiabatic and operates at atmospheric pressure. The gases enter the reactor at 410&d
C-X bond length in halobenzene less then C-X bond lengthin CH3-x
Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre
During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is
18,76,764
1933476 Asked
3,689
Active Tutors
1428982
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!