Hello, 3Coli Here!
Your Answer is Here:
I would recheck, redo, and revise my answers, if necessary. Also, I can look at the lesson again, if I forgot something. So that's what I would do.
Hopefully, this helps!
Ask your question below!
<span>electrons hope this helps</span>
Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
Answer:
QP
Explanation:
P has 9 electrons.
Electronic Configuration : 2, 7
Valence electrons : 7
P needs 1 electron to get stable electronic configuration.
Q has 3 electrons.
Electronic Configuration : 2, 1
Valence electrons : 1
P needs to loose 1 electron to get stable electronic configuration.
Q donates 1 electron,
Q -----> Q+ + 1 e-
P gains 1 electron,
P + 1 e- -----> P-
Q+ + P- -----> QP
This is an ionic compound.
The factors that affect geometry of a molecule are
> The number of bonding electron pairs around the central atom.
> The number of pairs of non-bonding ("lone pair") electrons around the central atom.
