Molarity of acid solution
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.
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.
Polymers are the chief products of modern chemical industry which form the backbone of present society. Daily life without the discovery and varied applications of polymers would not have been easier and colourful. The materials made of polymers find multifarious uses and applications in all walk
The ionic radii of Rb+ and I- respectively are 1.46 Å and 2.16Å. The very most probable type of structure exhibited by it is: (a) CsCl type (b) ZnS type (c) Nacl type (d) CaF2 type Q : What do you mean by the term dipole What do you mean by the term dipole moment? Briefly describe it.
What do you mean by the term dipole moment? Briefly describe it.
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
The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop
The Debye Huckel theory shows how the potential energy of an ion in solution depends on the ionic strength of the solution.Except at infinite dilution, electrostatic interaction between ions alters the properties of the solution from those excepted from th
What is the normal amount of glucose in 100ml of blood (8–12 hrs after meal) is: (i) 8mg (ii) 80mg (iii) 200mg (iv) 800mg Choose the right answer from above.
The volume of water to be added to 100cm3 of 0.5 N N H2SO4 to get decinormal concentration is : (a) 400 cm3 (b) 500cm3 (c) 450cm3 (d)100cm3
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
Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of moles of solute/ Mass of solvent in kg (d) no. of moles of any
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