H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
Everything has chemical properties it depends on the reactivity and the reactivity of the other element and what form it is in
<span>IF WE TAKE N=1 IT IS CALLED GROUND STATE. THEN THE OTHER FOLLOWING HIGHER STATES ARE CALLED EXCITED STATES. IF THE ELECTRON IN AN ATOM JUMPS FROM A STATE TO A LOWER STATE, IT LOSES ENERGY. FROM THE GIVEN STATEMENT, THE WAY TO FIND THE ENERGY RELEASED IS GIVEN BY THE FORMULA, E(n)=(-13.6 eV)/n^2. FIRST TO FIND E(5)=(-13.6 eV)/(5)^2, WE GET E(5)=-0.544 eV. E(3)=(-13.6 eV)/(3)^2, WE GET E(3)=-1.5111 eV. THEN WE HAVE TO FIND THE ENERGY TRANSITION LEVEL. ON SUBTRACTING WE GET 0.967eV. THIS ENERGY HAS TO BE CONVERTED IN JOULES. SO WE MAKE E=0.967*(1.60*10^(-19)) J/eV, WHICH CORRESPONDS TO 0.15472*10^(-18) J. WE NEED TO FIND TO THE WAVELENGTH. THE CORRESPONDING FORMULA E = hf = hc/λ, λ = hc/E. BY SUBSTITUTING THE KNOWN VALUES, WE GET THE ANSWER TO BE 1285.548 NM.</span>
