Molarity
Give me answer of this question. 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
Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules
Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M
Can someone please help me in getting through this problem. Which of the given concentration factor is affected by the change in temperature: (1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction
Select the right answer of the question. "The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called: (a) Henry's law (b) Raoult's law (c) Ostwald's law (d) Arrhenius's law
The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery. Q : Molarity in Nacl The molarity of 0.006 The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(
The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(
Differentiate between the modern periodic table and Mendeleevs table?
The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter
What do you mean by the term enzymes? Briefly illustrate it.
A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu
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