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
A Lewis base is characterized as any species that can donate a lone pair of electrons the Lewis acid (which has a vacant orbital for the lone pair to bond with)
Answer:
The order of reactivity towards electrophilic susbtitution is shown below:
a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene
b. p-cresol>p-xylene>toluene>benzene
c.Phenol>propylbenzene>benzene>benzoic acid
d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene
Explanation:
Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.
For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.
The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.
Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.
Answer : The molar mass of the solute would be low.
Explanation :
Formula used for depression in freezing point is:
where,
= change in freezing point
= freezing point of solution
= freezing point of water
i = Van't Hoff factor
= freezing point constant
m = molality
= mass of solute
= mass of solvent
= molar mass of solute
From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.
Hence, the molar mass of the solute would be low.
Answer:
true this is correct its your econimically
