This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
Not elementary school work
First, we need to get the number of moles:
from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+
∴ the number of moles n = 4
we are going to use this formula:
㏑K = n *F *E/RT
when K is the equilibrium constant = 4.98 x 10^-5
and F is Faraday's constant = 96500
and the constant R = 8.314
and T is the temperature in Kelvin = 298 K
and n is number of moles of electrons = 4
so, by substitution:
㏑4.98 x 10^-5 = 4*96500*E / 8.314*298
∴E = -0.064 V
I can’t see the picture where’s it at
