Molarity of acid solution
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 Choose the right answer fron above.
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
Choose the right answer fron above.
Write a short note on the IUPAC name of the benzene?
Which one of the following pairs of solutions can we expect to be isotonic at the same temperature:(i) 0.1M Urea and 0.1M Nacl (ii) 0.1M Urea and 0.2M Mgcl2 (iii) 0.1M Nacl and 0.1M Na2SO4 (iv) 0.1M Ca(NO3<
Which solution will have highest boiling point:(a) 1% solution of glucose in water (b) 1% solution of sodium chloride in water (c) 1% solution of zinc sulphate in water (d) 1% solution of urea in waterAnswer: (b) Na
The solubility of a gas in water depends on: (a) Nature of the gas (b) Temperature (c) Pressure of the gas (d) All of the above. Can someone help me in finding out the right answer.
25 ml of a solution of barium hydroxide on titration with 0.1 molar solution of the hydrochloric acid provide a litre value of 35 ml. The molarity of barium hydroxide solution will be: (i) 0.07 (ii) 0.14 (iii) 0.28 (iv) 0.35
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
Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)
What do you mean by the term Organic Chemistry? Briefly define the term?
Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
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