Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
The answer is
<h3>2.53 × 10²⁴ molecules</h3>
Explanation:
The number of molecules present can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 4.21 × 6.02 × 10²³
We have the final answer as
<h3>2.53 × 10²⁴ molecules</h3>
Hope this helps you
Answer:
First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of 
.
Explanation:
In presence of NaCl, 
forms an insoluble precipitate of AgCl.
Reaction: 
In presence of 
, AgCl gets dissolved into solution due to formation of soluble ![[Ag(NH_{3})_{2}]^{+}](https://tex.z-dn.net/?f=%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D)
complex.
Reaction: ![AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)](https://tex.z-dn.net/?f=AgCl%28s%29%2B2NH_%7B3%7D%28aq.%29%5Crightarrow%20%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D%28aq.%29%2B%20Cl%5E%7B-%7D%28aq.%29)
In presence of , complex gets destroyed and free 
again reacts with free 
to produce insoluble AgCl
Reaction: ![[Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)](https://tex.z-dn.net/?f=%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D%28aq.%29%2B2H%5E%7B%2B%7D%28aq.%29%2BCl%5E%7B-%7D%28aq.%29%5Crightarrow%20AgCl%28s%29%2B2NH_%7B4%7D%5E%7B%2B%7D%28aq.%29)
Ranking of the atom from highest to lowest is as follows:
Highest
Arrow = from outer edge to center
2nd Highest
Arrow = second closest ring to the outer edge to center
3rd Highest
Arrow = middle circle to center
Lowest
Arrow = outer edge to middle circle
Answer:
just subrracted them and find out whats m
Explanation:
