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.
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
in bioremediation gibberella fusarium is used to break down____?
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
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 support your answer
Presence of small concentrations of appropriate electrolyte is necessary to stabilize the colloidal solutions. However, if the electrolytes are present in higher concentration, then the ions of the electrolyte neutralize the charge on the colloidal particles may unite
Liquid solutions are obtained when the solvent is liquid. The solute can be a gas, liquid or a solid. In this section we will discuss the liquid solutions containing solid or liquid solutes. In such solutions the solute may or may not be volatile. We shall limit our d
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 various elements
Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<
IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN
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