Problem on molality
Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05
why pentahalids are more covalent than tetrahalids
Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. The phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.
At 20°C the solubility of solid naphthalene in hexane is 0.09 mol/mol of solution. Use this information and the data below to estimate the following for this system: a) The mol fraction of naphthalene in the vapour phase in equ
The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(
Can you please illustrate that HCl is polar or non-polar? Briefly illustrate it.
When the forces of attraction existing between adsorbate and adsorbent are van der Waal's forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or van der Waal's adsorption. Since the forces existing between adsorbent and adsorbate are very w
Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H
Sulfur trioxide (SO3) is manufactured by the gas-phase oxidation of SO2 over a platinum catalyst: SO2 + ½ O2 à SO3 The catalyst is a non-porous ext
Describe the net charge of a non-ionized atom?
Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g
18,76,764
1949426 Asked
3,689
Active Tutors
1420979
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!