Whether water is frozen, in a gaseous state, or is a liquid, it is still H2O. So the chemical composition does stay the same.
PbI(ii) ionization in the solution of PBI(ii) into water is:
<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻
If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.
Therefore,
<span>Ksp=<span>[Pb</span></span>²⁺][I-]²
Plugging the values:
1.4×10⁻⁸ = x ⋅ (2x)²
1.4×10⁻⁸ = 4x³
x³ = {1.4×10⁻⁸}÷4
x³ = 0.35 x 10⁻⁸
or
x³ = 3.5 x 10⁻⁹
x = 1.51 x 10⁻³
Hence,
Concentration of iodide ions in the solution:
2x = 3.02 x 10⁻³
The simplest particles are atoms.
Answer:
Oxidized : CO
Reduced: Fe
Oxidizing agent: Fe2O3
Reducing agent: CO
Explanation:
Loss of electrons is oxidation while gaining of electrons is reduction.
CO is a reducing agent, Fe2O3 is an oxidizing agent.
Each carbon atom is oxidized in CO. CO is the reducing agent
Each Fe atom in FeO3 is reduced. FeO3 is the oxidizing agent
Answer:
0.24
Explanation:
We are given that
Rate constant for A=
Rate constant for B,k'=0.0750/s
We have to find the value of equilibrium constant for the reaction
Equilibrium constant, for k=
Using the formula
Hence, the value of the equilibrium constant for the reaction
at this temperature=0.24
