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Norma-Jean [14]
3 years ago
15

Lewis structures for the perchlorate ion (ClO4−) can be drawn with all single bonds or with one, two, or three double bonds. Dra

w each of these possible resonance forms, including any nonbonding electrons. Include the values of any nonzero formal charges. Use formal charges to determine the most important resonance structure and calculate its average bond order. slader

Chemistry
1 answer:
Vanyuwa [196]3 years ago
5 0

Answer:

The most important resonance structure is 4 (attached picture). Its bon order is \frac{7}{4} or 1\frac{3}{4}.

Explanation:

A picture with 4 forms of the perchlorate structure is attached. The first structure has simple bonds. The second structure contains a double bond, the third structure has two double bonds and the fourth structure has three double bonds.

Formal charge = group number of the periodic table - number of bonds (number of bonding electrons / 2) - number of non-shared electrons (lone pairs)

The formal charges in the first structure is +3 in chlorine and -1 in oxygen.

The formal charges in the second structure is +2 in chlorine, -1 in oxygen and 0 in the double bond oxygen.

The formal charges in the third structure is +1 in chlorine, -1 in the single bond oxygens and 0 in the double bond oxygens.

The formal charges in the fourth structure is 0 in chlorine, -1 in the single bond oxygen and 0 in the double bond oxygens.

The most important resonance structure is given by:

  • Most atoms have 0 formal charge.
  • Lowest magnitude of formal charges.
  • If there is a negative formal charge, it's on the most electronegative atom.

Hence, the fourth structure is the mosr important.

The bond order of the structure is:

Total number of bonds: 7

Total number of bond groups: 4

Bond order= \frac{7}{4} =1\frac{3}{4}

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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
Brilliant_brown [7]

Answer :  The rate of reaction is,

Rate=4.77\times 10^{-19}M/s

The appearance of NO_3 is, 4.77\times 10^{-19}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)

The rate law expression will be:

Rate=k[NO_2][O_3]

Given:

Rate constant = k=1.69\times 10^{-4}M^{-1}s^{-1}

[NO_2] = 1.77\times 10^{-8}M

[O_3] = 1.59\times 10^{-7}M

Rate=k[NO_2][O_3]

Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})

Rate=4.77\times 10^{-19}M/s

The expression for rate of appearance of NO_3 :

\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}

As, \text{Rate of reaction}=4.77\times 10^{-19}M/s

So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

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Alborosie

Answer:

32.8%

Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

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There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:

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Finally we calculate the percentage composition of Pb:

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brainly.com/question/8344791

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