Answer:
12.1 kJ
Explanation:
For this problem, we are going to use the Arrhenius Equation.
![ln(\frac{k_{2} }{k_{1}})=\frac{E_{a} }{R} (\frac{1}{T_{1}} -\frac{1}{T_{2}} )\\ln(\frac{2}{1} )=\frac{E_{a} }{8.314} (\frac{1}{300}-\frac{1}{350} )](https://tex.z-dn.net/?f=ln%28%5Cfrac%7Bk_%7B2%7D%20%7D%7Bk_%7B1%7D%7D%29%3D%5Cfrac%7BE_%7Ba%7D%20%7D%7BR%7D%20%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%20-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20%29%5C%5Cln%28%5Cfrac%7B2%7D%7B1%7D%20%29%3D%5Cfrac%7BE_%7Ba%7D%20%7D%7B8.314%7D%20%28%5Cfrac%7B1%7D%7B300%7D-%5Cfrac%7B1%7D%7B350%7D%20%29)
We are going to solve for activation energy,
.
kJ
Answer:
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Explanation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).
HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.
Any added H⁺ will immediately combine with the chromate ions according to the reaction
H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻
thereby removing chromate ions from solution.
According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺ removes.
The overall equation for the reaction is
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + <em>CrO₄²⁻(aq)
</em>
<u>H⁺(aq) + </u><em><u>CrO₄²⁻(aq)</u></em><u> ⟶ HCrO₄⁻(aq)
</u>
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Answer:
2, Argon PLEASE VOTE 5.0 MARK ME BRAINLIEST AND THANK ME
Explanation:
2,Argon because its only logic answer PLEASE VOTE 5.0 MARK ME BRAINLIEST AND THANK ME