Chemistry, 19.12.2019 21:31 lkipjjjjjjjjjj2272

Identify the possible resonance structures for cno−. select all that apply.

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Answer from: nghtcll

Answer : The resonating structure of CNO^- is shown below.

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of CNO^- .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, CNO^-

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in CNO^- = 4 + 5 + 6 + 1= 16

According to Lewis-dot structure, there are 8 number of bonding electrons and 8 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

For structure 1 :

$\text{Formal charge on O}=6-6-\frac{2}{2}=-1$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 2 :

$\text{Formal charge on O}=6-4-\frac{4}{2}=0$

$\text{Formal charge on C}=4-4-\frac{4}{2}=-2$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 3 :

$\text{Formal charge on O}=6-2-\frac{6}{2}=+1$

$\text{Formal charge on C}=4-6-\frac{2}{2}=-3$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

The resonating structure of CNO^- is shown below.

Identify the possible resonance structures for cno−. select all that apply.

Answer from: Quest

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation: if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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Identify the possible resonance structures for cno−. select all that apply....