Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction
of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq)<span>.
</span>s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.
Answer:
The answer to your question is P = 1.64 atm
Explanation:
Data
Volume = 2.5 x 10⁷ L
Temperature = 22°C
Pressure = ?
Moles = 1.7 x 10⁶
R = 0.082 atm L/ mol°K
Process
1.- Convert temperature to °K
Temperature = 22 + 273
= 295°K
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for P
P = nRT / V
- Substitution
P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷
- Simplification
P = 41123000 / 2.5 x 10⁷
- Result
P = 1.64 atm
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
Potassium metal + Chlorine Gas -------->
Potassium Chloride<span>
The chemical equation using symbols and formula is
<span>K (s) + </span></span><span><span><span><span>Cl</span>2</span> </span><span>(g) ---------> 2KCl (s)</span></span>