To
determine the percent ionization of the acid given, we make use of the acid
equilibrium constant (Ka) given. It is the ration of the equilibrium
concentrations of the dissociated ions and the acid. The dissociation reaction
of the HF acid would be as follows:<span>
HF = H+ + F-
The acid equilibrum constant would be expressed as follows:
Ka = [H+][F-] / [HF] = 3.5 x 10-4
To determine the equilibrium concentrations we use the ICE table,
HF
H+ F-
I 0.337 0
0
C -x +x
+x
---------------------------------------------
E 0.337-x x
x
3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x] </span>
Solving for x,
x = 0.01069 = [H+] = [F-]
percent ionization = 0.01069 / 0.337 x 100 = 3.17%
Answer:
This is not an answer its just to get points
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
No..since research can take different format