Distribution law
Help me to go through this problem. The distribution law is applied for the distribution of basic acid between : (a) Water and ethyl alcohol (b) Water and amyl alcohol (c) Water and sulphuric acid (d) Water and liquor ammonia
Choose the right answer from following. When 6gm urea dissolve in180gm H2O . The mole fraction of urea is : (a)10/ 10.1 (b)10.1/10 (c)10.1/ 0.1 (d) 0.1/ 10.1
The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea (b) BaCl2 > AlCl3 > Urea (c) Urea > BaCl2<
Briefly describe the properties of the carbohydrates?
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
Choose the right answer from following. The molarity of pure water is: (a) 55.6 (b) 5.56 (c)100 (d)18
Average covalent radii can be assigned on the basis of molecular structures. The accumulation of structural data by spectroscopic studies and both electron and x-ray diffraction studies allows one to investigate the possibili
When the catalyst exists in a different phase than that of reactants, it is said to be heterogeneous catalyst, and the catalysis is called heterogeneous catalysis. For example, SO2 can be oxidized to SO3
The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A
The colligative properties of a solution depend on: (a) Nature of solute particles present in it (b) Nature of solvent used (c) Number of solute particles present in it (d) Number of moles of solvent only
If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M (c) 0.3M (d) 0.4M Choose the right answer fron above.
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