Vant Hoff factor
The Van't Hoff factor of the compound K3Fe(CN)6 is: (a) 1 (b) 2 (c) 3 (d) 4 Answer: (d) K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3-
The Van't Hoff factor of the compound K3Fe(CN)6 is:
(a) 1 (b) 2 (c) 3 (d) 4 Answer: (d) K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3-
Vander Waal's radii can be assigned to the atoms of molecules on the basis of the closeness of approach of these atoms in crystalline substances. Diffraction studies of crystals give information about hoe molecules can approach each other and can pack
Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M
An extension of the kinetic molecular theory of gases recognizes that molecules have an appreciable size and deals with molecule-molecule collisions. We begin studies of elementary reactions by investigating the collisions b
What do you mean by the term tripod? Also state its uses?
Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc
The equation S = k in W relates entropy to W, a measure of the number of different molecular level arrangements of the system.In the preceding developments it was unnecessary to attempt to reach any "explana
Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The
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?
Illustrate the reason, why acetic has less conductivity than Hcl?
The diazonium salts are represented by the general formula ArN2 +X where X- ion may be anion such as (Cl) ¨, B ¨r, HSO
18,76,764
1945145 Asked
3,689
Active Tutors
1437270
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!