Answer:
![[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Explanation:
Hello there!
In this case, for the ionization of silver iodide we have:
![AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]](https://tex.z-dn.net/?f=AgI%28s%29%5Crightleftharpoons%20Ag%5E%2B%28aq%29%2BI%5E-%28aq%29%5C%5C%5C%5CKsp%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:
![[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D%5BH%5E%2B%5D%3D10%5E%7B-3.55%7D%3D2.82x10%5E%7B-4%7DM)
Now, we can set up the equilibrium expression as shown below:
Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:
![x=[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=x%3D%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Best regards!
The answer is 236.5 J/K
According to Δ G formula:
ΔG = ΔH - TΔS
when ΔG is the change in free energy (KJ)
and ΔH is the change in enthalpy (KJ)= ΔHvap * moles
= 71.8 KJ/mol * 1.11 mol
= 79.7 KJ
and T is the absolute temperature (K)= 64 °C + 273°C = 337 K
Δ S is the change in entropy KJ/K
by substitution:
when at equilibrium ΔG = 0
∴ΔS = ΔH / T
=79.7 KJ/ 337 K
= 0.2365 KJ/K
= 236.5 J/K
Answer:
See explanation
Explanation:
The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. Hence, chemical energy in a battery can be converted to electrical energy.
Usually, the conversion of energy from one form to another is not 100% efficient according to the second law of thermodynamics. Some energy is wasted in the process, sometimes as heat.
Hence, in an ideal situation where no heat energy is produced; all the chemical energy is converted to electrical energy (100% energy conversion). There will be no energy loss if no heat is produced.
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Ammonium is NH₄⁺ and Carbonate is CO₃⁻² => Ammonium Carbonate is (NH₄)₂CO₃
