Problem on making solutions
The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g (b) 1g (c) 2g (d) 10g Choose the right answer from above.
The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g (b) 1g (c) 2g (d) 10g
Choose the right answer from above.
Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N (b) 19.8 N (c) 39.6 N (d) 98
Why oxidising character of oxoacids of halogens decreases as oxidation number increases?
: 1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to
The two solutions which are having equivalent osmotic pressure are called isotonic solutions. The isotonic solutions at the same temperature also have same molar concentration. If we have solutions having different osmotic pressures then the solution having different
Select the right answer of the question. How much water is required to dilute 10 ml of 10 N hydrochloric acid to make it exactly decinormal (0.1 N): (a) 990 ml (b) 1000 ml (c) 1010 ml (d) 100 ml
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
2.0gram of dolomite is heated to a constant weight of 1.0g. Calculate the total volume of CO2 produced at STP by this reation
Choose the right answer from following.The relative lowering of the vapour pressure is equal to the ratio between the number of: (a) Solute moleules and solvent molecules (b) Solute molecules and the total molecules in the solution (c) Solvent molecules and the tota
Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M
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
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