Boric anhydride is a non-flammable substance. When it reacts with a coloured metallic salt, it produces a distinctive coloured metal metaborate... View Article
It is a reversible & isothermal process. n = 0.1 molP1 = 1P2 = 10T = 100 Work done = -nRT ln (P1/P2)= -(0.1) x 8.314 x 100 x ln(1/10)= +... View Article
One sigma and one pi bond are found in alkene bonds. One sigma and one pi bond are also found in benzene. In the case of benzene, however, the pi... View Article
Molecular Mass of CH4=16 Moles of CH4 = x/16 Molecular Mass of O2=32 Moles of O2 = x/32 Total moles = (x/16) + (x/32) = 3x/32 Mole fractions of... View Article
SO2 and Cl2 pairs has bleaching action. Sulfur dioxide and chlorine are both effective bleaching agents. Chlorine's bleaching action is based on... View Article
1st half- Life → 50% decay 2nd half- Life → 75% decay The half-life period of the radioactive element, if 75% of it disintegrates in 40 min, is... View Article
We can say that at equilibrium, the rate of dissolution is the forward reaction and the rate of crystallisation is the backward reaction. For a... View Article
According to the law of definite proportion, a given compound always has the same proportion of constituents by weight. Case I 2.8 g of CaO... View Article
For Cu(OH)2, the expression of solubility product is: Ksp = [Cu2+][OH-]2 = (s)(2s)2 = 4s3 Substitute Ksp = 1.6 × 10–19 in the above equation.... View Article
Flocculation, also known as coagulation is the process of precipitation of a colloidal solution by the addition of the excess of an electrolyte.... View Article
The relation between lead nitrate and potassium iodide is an example of precipitation reaction. Pb(No3)2 + 2KI ⟶ PbI2 + 2KNO3 (yellow ppt) The... View Article
t=0 activity is N0 then after time t the activity is given as 100=1600×(1/2)^n so n=4 n=4 and t=8s we get t1/2 = 2s So activity after 2 seconds... View Article
Mole fraction of NO2=2x/(1−x+2x)=2x/(x+1)=1/2 solving, we get 4x=x+1 3x=1 x=1/3 =(4/9)/(2/3)=4X3/9X2=2/3 Therefore, 2/3 is the answer.... View Article
Any metal will reduce the oxide of other metals above it on the Ellingham diagram because the free energy difference will become more negative,... View Article