Problem based on molarity
Choose the right answer from following. The molarity of a solution of Na2CO3 having 10.6g/500ml of solution is : (a) 0.2M (b)2M (c)20M (d) 0.02M
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
The nuclear states produced by a magnetic field are studied in nuclear magnetic resonance spectroscopy.The frequency of the radiation that corresponds to the nuclear magnetic energy level spacings and the weakness of the radiation absorption that must be e
The important methods for the preparation of alcohol on large-scale are given below:  
The pressure-temperature relation for solid-vapor or liquid vapor equilibrium is expressed by the Clausis-Clapeyron equation.We now obtain an expression for the pressure-temperature dependence of the state of equilibrium between two phases. To be specific,
discuss practical uses of coordination compounds
Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g
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?
Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The
The distribution law is exerted for the distribution of basic acid among: (i) Water and ethyl alcohol (ii) Water and amyl alcohol (iii) Water and sulphuric acid (iv) Water and liquor ammonia What is the right answer.
18,76,764
1945961 Asked
3,689
Active Tutors
1446946
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!