general solution of the differential equation

Definitions of solute and solvent The dissolving process as a result of random molecular motion Temperature, pressure, and surface area - their affect the dissolving process Concentration of a solute in terms of grams per liter, molarity, parts per million and percent co..

is a substance that can dissolve in a solvent, to form a solution. Examples: Salt, sugar, et..

(a) 38% solution, means 38 g of HCl in 100 g of solution. Then, Mass of the solution= 100 g Molar mass of HCl = 36.5 g mol - 1 (b) The molarity of conc. HCl sample = 12.38 mol/L Molarity of HCl solution to be prepared = 0.10 mol/L Volume of HCl ..

Introduction - A Solution is a homogeneous mixture of two or more chemically non-reacting substances called components. The composition of the components can be varied with in certain limit..

Heat capacity of solution = Mass of solution x Specific heat capacity Total mass of solution = 100 + 100 = 200 ml Heat capacity of solution = 200 x 4.2 = 840 JK - 1 Heat change in the reaction = Heat capacity x Rise in temperature = (840 JK - 1 ) (298.006..

Gold being a noble metal occurs in its native form. It is further purified by the process of leaching or precipitiation called hydrometallurgy. The native gold in its complex form is first precipitated with potassium cyanide solution. It is then recovered from the solution by a..

So if we calculate the volume of O 2 , which is liberated by 30.36 g of H 2 O 2 at NTP, this will correspond to volume strength of H 2 O 2 . As per the equation: 2 x 34 = 68 g of H 2 O 2 will liberate 22.4 L of O 2 at NTP 68 g of H 2 O 2 will produce O 2 at NTP = 22.4 L The give..

When mixed, the total volume gets doubled and hence the effective concentrations of the ions would be half of the initial concentration, i.e., in solution [Mg 2 + ] =(0.01/2)=0.005 mol/L [C 2 O 4 2 - ] = (0.02/2) = 0.01 mol/L These ions would react to form sparingly soluble salt MgC 2..

According to Henderson's equation: ..

According to Rydberg equation: For the first line in Balmer series, n 1 = 2, n 2 = 3 For the second line in Balmer series, n 1 = 2, n 2 = 4 Dividing equations (i) by (ii..