Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:

You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
![Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_%7B3%7D%20%5D%20%2A%5BCl_%7B2%7D%20%5D%20%7D%7B%5BPCl_%7B5%7D%20%5D%7D)
The concentrations are:
- [PCl₃]=

- [Cl₂]=

- [PCl₅]=

Replacing:

Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
Answer:D.
Explanation:i used photo math. It is really helpful when I'm doing math homework etc...
Answer:
anion
Explanation:
anion undergoes reduction.
cation undergoes oxidation.
The answer is 28.02.
The chemical formula for cadmium nitrate is Cd(NO₃)₂.
There are in total 2 nitrogen atoms in Cd(NO₃)₂. If <span>the atomic weight of nitrogen is 14.01</span>, the mass <span>of two nitrogen atoms in one mole of cadmium nitrate is 28.02:
2 </span>· 14.01 = 28.02