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ahrayia [7]
2 years ago
15

Draw the Lewis structure (including resonance structures) for diazomethane (CH2N2)(CH2N2). For each resonance structure, assign

formal charges to all atoms that have formal charge.

Chemistry
1 answer:
Tanya [424]2 years ago
8 0

Answer : The Lewis-dot structure and resonating structure of CH_2N_2 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 CH_2N_2.

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, CH_2N_2

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in CH_2N_2 = 4 + 2(1) + 2(5) = 16

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 H}=1-0-\frac{2}{2}=0

\text{Formal charge on H}=1-0-\frac{2}{2}=0

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

For structure 2 :

\text{Formal charge on H}=1-0-\frac{2}{2}=0

\text{Formal charge on H}=1-0-\frac{2}{2}=0

\text{Formal charge on C}=4-0-\frac{8}{2}=0

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Dominik [7]
I hope this helps you.

5 0
2 years ago
Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli
Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
3 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
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Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

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2 years ago
A 1.34 mole sample of LiCl dissolves in water, The volume of the final
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Answer:

1.56 \,\,mol/L

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Molarity = volume of solution in litres/number of moles of solute dissolved in solution

Volume of solution in litres = 0.86 L

Also, 1.34 mole sample of LiCl dissolves in water

So,

Molarity of the Solution = \frac{1.34}{0.86}=1.56 \,\,mol/L

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2 years ago
A cough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarit
Reil [10]
<span>You are given a cough syrup that contains 5.0% ethyl alcohol, c2h5oh, by mass and its density of the solution is 0.9928 g/ml. The molarity of the alcohol in the cough syrup is 21.55.</span>
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2 years ago
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