Answer:
The main difference between evaporation and boiling are : 1. Evaporation takes place at all temperatures, while boiling occurs at a particular temperature. 2. Evaporation takes place from the surface, whereas the entire liquid boils. 3. Evaporation can occur using the internal energy of the system, while boiling requires an external source of heat.
First write the molecular equation with states:
(NH4)2S (aq) + 2AgNO3(aq) → Ag2S (s) + 2NH4NO3
Now write a full ionic equation by separating into ions all substances that dissociate: anything (s) (g) or (l) does not dissociate
2NH4 + (aq) + S 2-(aq) + 2Ag+ (aq) + 2NO3- (aq) → Ag2S(s) + 2NH4 + (aq) + 2NO3- (aq)
To write the NET IONIC equation, inspect the full ionic equation above and delete anything that appears on both sides of the → sign:
Net ionic equation:
S 2-(aq) + 2Ag + (aq) → Ag2S(s)
Answer:
a. 3.72 [atm]
Explanation:
For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law: 
![(1.556[atm])(268.5[mL])=P_2(112.4[mL])](https://tex.z-dn.net/?f=%281.556%5Batm%5D%29%28268.5%5BmL%5D%29%3DP_2%28112.4%5BmL%5D%29)
![\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281.556%5Batm%5D%29%28268.5%5BmL%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%29%7D%7B112.4%5BmL%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%7D%3D%5Cdfrac%7BP_2%28112.4%5BmL%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%29%7D%7B112.4%5BmL%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%7D)
![3.716957[atm]=P_2](https://tex.z-dn.net/?f=3.716957%5Batm%5D%3DP_2)
It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a. 3.72
Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
Arranged in groups and periods.
Group 1 in the same place.
