Answer:
The concentration of COF₂ at equilibrium is 0.296 M.
Explanation:
To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00 - 2x x x
Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".
![Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852](https://tex.z-dn.net/?f=Kc%3D8.30%3D%5Cfrac%7B%5BCO_%7B2%7D%5D%20%5Ctimes%20%5BCF_%7B4%7D%5D%20%7D%7B%5BCOF_%7B2%7D%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%282.00-2x%29%5E%7B2%7D%20%7D%20%5C%5C8.30%3D%28%5Cfrac%7Bx%7D%7B2.00-2x%7D%20%29%5E%7B2%7D%20%5C%5C%5Csqrt%7B8.30%7D%20%3D%5Cfrac%7Bx%7D%7B2.00-2x%7D%5C%5C5.76-5.76x%3Dx%5C%5Cx%3D0.852)
The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M
Depends on what the problems are, you are going to have to be a bit more specific, Young Padawan
Answer:
Ionization and dissociation
There is no element in 2s3
First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>