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3 years ago

Oque é um hibrido de ressonancia?

Chemistry

2 answers:

Yuri [45]3 years ago

8 0

Answer:

Resonance hybrid: A weighted average of all significant resonance contributors depicting the true electronic structure of a molecule.

Explanation:

Translation: Híbrido de ressonância: uma média ponderada de todos os contribuintes de ressonância significativos, representando a verdadeira estrutura eletrônica de uma molécula.

Rainbow [258]3 years ago

4 0

Answer:

el híbrido de resonancia es un promedio ponderado de todos los contribuyentes de resonancia significativos que representan la verdadera estructura electrónica de una molécula.

Explanation:

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Select the compounds that cannot undergo friedel crafts alkylation. check all that apply. check all that apply. aniline phenol k

mr Goodwill [35]

Freidal craft reaction is the attack of a carbon or carbon chain on aromatic ring with the help of anhydrous AlCl3 to produce alkylated benzene ring.

Only ketone not be able to undergo friedal craft reaction as, it is not aromatic compound whereas all the given reactants are aromatic and gives friedal craft reaction.

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3 years ago

What is the molar mass of Mg(NO2)2?

qwelly [4]

Answer:

116g/mol

Explanation:

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3 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

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