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lara [203]
2 years ago
15

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

in liquid water according to the equation 2HgO(s)+H2O(l)+2Cl2(g)⇌2HOCl(aq)+HgO⋅HgCl2(s) What is the equilibrium-constant expression for this reaction? View Available Hint(s) The acid (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation What is the equilibrium-constant expression for this reaction? K=[HOCl]2[HgO⋅HgCl2][Cl2]2[H2O][HgO]2 K=[HOCl]2[Cl2]2[H2O] K=[Cl2]2[HOCl]2 K=[HOCl]2[Cl2]2
Chemistry
1 answer:
skelet666 [1.2K]2 years ago
4 0

Answer:

k = \frac{[HOCl]^2}{[Cl]^2}

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium

The global reaction is:

2 HgO (s) + H₂O (l) +2 Cl₂ (g) ⇌ 2 HOCl (aq) + HgO⋅HgCl₂ (s)

Thus, equilibrium-constant expression is:

<em>k = \frac{[HOCl]^2}{[Cl]^2}</em>

You don't include HgO nor HgO⋅HgCl₂ because are pure solids nor water because is pure liquid.

I hope it helps!

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3. What's the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?
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Answer:

Empirical formula is  Li₂CO₃.

Explanation:

Percentage of oxygen= 65.0%

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Percentage of carbon= 16.3%

Empirical formula = ?

Solution:

Number of gram atoms of C = 16.3/12 = 1.4

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Number of gram atoms of O = 65.0/ 16 = 4.1

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2.7/1.4      :       1.4/1.4         :   4.1/1.4

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2 years ago
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Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in BeCl_2  is, sp

(2) The hybridization of central atom nitrogen in NO_2  is, sp^2

(3) The hybridization of central atom carbon in CCl_4  is, sp^3

(4) The hybridization of central atom xenon in XeF_4  is, sp^3d^2

Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

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(1) The given molecule is, BeCl_2

\text{Number of electrons}=\frac{1}{2}\times [2+2]=2

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(2) The given molecule is, NO_2

\text{Number of electrons}=\frac{1}{2}\times [4]=2

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, sp^2.

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(4) The given molecule is, XeF_4

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