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)
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The clay soil holds the least amount if water.
Answer:
ΔG°rxn = +50.8 kJ/mol
Explanation:
It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:
<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>
In the reaction:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)
ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)
ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}
ΔH°rxn = 136.5kJ/mol
And S°:
S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)
ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)
ΔH°rxn = 0.2875kJ/molK
And replacing in (1) at 298K:
ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK
<em>ΔG°rxn = +50.8 kJ/mol</em>
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Answer:
Al2(SO4)3 and Mg(OH)2
Explanation:
1. Al has a charge of 3-, and SO4 of 2-
when you cross multiply the charges you get
Al2 and (SO4)3
*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*
Hence, Al2(SO4)3
2. Mg has a charge of 2- and OH of 1-
again cross multiply
Mg (you dont need to add the 1) and (OH)2
again, the bracket around OH means the charge appiles to Oxygen AND hydrogen
hence, Mg(OH)2
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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