Calculate [h3o+] and [s2−] in a 0.10 m solution of the diprotic acid hydrosulfuric acid. (for hydrosulfuric acid ka1 = 9.0 × 10−8 and ka2 = 1.0 × 10−17.) enter your answers in scientific notation.

The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸ 
Construct ICE table and obtain their equilibrium concentrations:
                  H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M):             0.1                            0              0
C (M):            -x                            +x            +x
E (M):        0.1 -x                            x              x
So:
9.0 x 10⁻⁸ = 
x = 9.4 x 10⁻⁵ 
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
                    HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M):          9.4 x 10⁻⁵                   0              9.4 x 10⁻⁵
C (M):            -x                            +x            +x
E (M):     9.4 x 10⁻⁵ -x                    x              9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ = 
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na