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Keith_Richards [23]
3 years ago
12

Write a balanced half-reaction describing the reduction of gaseous dichlorine to aqueous chloride anions.

Chemistry
1 answer:
Annette [7]3 years ago
4 0

Answer:- Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

Explanations:- In reduction the electrons are accepted and so they are written on the reactant side. When an atom accepts electrons then it forms anion. Chlorine has 7 valence electrons and it needs one more electron to complete it's octet. Since, dichlorine has two Cl atoms and each Cl atom needs one more electron to complete it's octet, two electrons are accepted by dichorine to make aqueous chloride ion. For balancing the equation, there would be two chloride ions as the reactant side has two chlorine atoms.

Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

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6 0
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What is the relationship between chemical equilibrium and the rates of forward and reverse reaction?
Nataly_w [17]

Answer:

c

Explanation:

the rate of a forward process must be exactly balanced by the rate of the reverse process.

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Which type of solid can become an electrical conductor via chemical substitution?
Ludmilka [50]
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8 0
3 years ago
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At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration
wlad13 [49]

Answer:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.03901 mol/liter
  • [Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

  • Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
  • Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

  • Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
  • Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

            H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

<u />

<u>4. Equilibrium expression</u>

       K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}

     0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}

<u />

<u>5. Solve:</u>

            x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0

Use the quadatic formula:

x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}

The positive result is x = 0.00909

Thus the concentrations are:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
  • [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0
3 years ago
Guys please answer this
iren [92.7K]

Answer:

Mass = 96 g

Explanation:

Given data:

Number of moles of C = 8 mol

Mass of C in gram = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of C = 12 g/mol

8 mol = mass / 12 g/mol

Mass = 8 mol × 12 g/mol

Mass = 96 g

6 0
3 years ago
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