Problem on moles of solution
The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75
A pressure composition diagram for a liquid vapor system can be used to show the composition of the liquid and equilibrium vapor.Vapor equilibrium data are useful in the study of distillations. It is of value to have diagrams showing not only the vapor pre
The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at
Illustrate the 3 facts on evaporation?
Select the right answer of the question .The molarity of a 0.2 N N2Co3 solution will be: (a) 0.05 M (b) 0.2 M (c) 0.1 M (d)0.4 M
Laws of Chemical Combination- In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chem
Ideal mixing properties can be recognized in the formation of an ideal gas mixture from ideal gases. Consider the formation of a mixture of gases i.e. a gaseous solution, from two mixtures of pure gases. A useful characterization of an ideal mixture, or soluti
This classification of polymers is based upon how the monomeric units are linked together. Based on their structure, the polymers are classified as: 1. Linear polymers: these are the polymers in which monomeric units are linked together to form long straight c
order of decreasing basicity of urea and its substituents
The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of benzene = 5.5oC) &n
Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<
18,76,764
1937398 Asked
3,689
Active Tutors
1435223
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!