Unit of mole fraction
Provide solution of this question. Unit of mole fraction is: (a) Moles/litre (b) Moles/litre2 (c) Moles-litre (d) Dimensionless
Which of the following solutions will have a lower vapour pressure and why? a) A 5% aqueous solution of cane sugar. b) A 5% aqueous solution of urea.
Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N
Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL
Can someone help me in going through this problem. The statement “When 0.003 moles of a gas are dissolved in 900 gm of water under a pressure of 1 atm, 0.006 moles will be dissolved under the pressure of 2 atm", signfies: (a)
Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5
Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20
Determine the correct regarding Henry’s law: (1) The gas is in contact with the liquid must behave as an ideal gas (2) There must not be any chemical interaction among the gas and liquid (3) The pressure applied must be high (4) All of these.
1. A solution of 0.100 M acetic acid is prepared. a) What is its pH value? b) If 20% of the initial acetic acid is converted to the acetate form by titration with NaOH, what is the resultant pH?
Addition of conc. HCl to saturated Bacl2 solution precipitates Bacl2 ; because of the following reason : (a) It follows from Le Chatelier's principle (b) Of common-ion effect (c) Ionic product (Ba++)(cl) remains constant in a saturated sol
Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi
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