If anyone could please help me i'd greatly appreciate it! I have tried over and over again and can not figure this out.
What is the percent ionization of H₂NNH₂ in a solution with a concentration of a 0.490 M? (Kb = 1.3 × 10⁻⁶)

%Ionization = 1.63%

Explanation:

Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.

H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶

So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.

               HONHNHOH => HONHNH⁺  +  OH⁻

C(i) =>              0.490M                  0M           0M

ΔC =>                  -x                         +x             +x

C(eq) =>         0.490 - x                    x               x

                      ≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100

Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]

1.3 x 10⁻⁶ = x²/0.490

=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴

=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%

the answer is 1 i believe

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.