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-
The surface between a liquid and a vapour distinguishes these fluids. The surface tension of liquids can be looked upon as that the property which draws a liquid together and forms a liquid vapour interface, therefore, distinguishing liquids from gases.<
How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g (ii) 6 g (iii) 40 g (iv) 60 g
Illustrate the Lewis Dot Structure for the CH4O.
living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?
why allyl halide and haloarenes are more reactive than alkyl halide towards nucleophilic substitution
Liquid solutions are obtained when the solvent is liquid. The solute can be a gas, liquid or a solid. In this section we will discuss the liquid solutions containing solid or liquid solutes. In such solutions the solute may or may not be volatile. We shall limit our d
The main objective of this particular aspect of Physical Chemistry is to examine the relation between free energies and the mechanical energy of electromotive force of electrochemical cells. The ionic components of aqueous solutions can be treated on the basis of the
give atleast two application of following colligative properties
I already did Materials and Methods section. I uploaded it with the instructions. Also, make sure to see Concept Questions and Thinking Ahead in the instructions that I uploaded. deadline is tomorow at 8 am
Give me answer of this question. The concentration of an aqueous solution of 0.01M CH3OH solution is very nearly equal to which of the following : (a) 0.01%CH3OH (b) 0.1%CH3OH (c) xCH3OH= 0.01 (d) 0.99MH2O (
18,76,764
1956580 Asked
3,689
Active Tutors
1428575
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!