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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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A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to d
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Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

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n =?

PV = nRT

n = PV /RT

n = (1.5x12)/(0.082x280)

n = 0.784mol

Next, we convert this mole ( i.e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

Therefore, 29.792g of F2 will react with Xg of NaCl i.e

Xg of NaCl = (29.792 x 117)/38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2

3 0
2 years ago
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Answer:

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