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katrin [286]
3 years ago
14

The ozone "hole" over Antarctica formed as the result of a series of chemical reactions. The loss occurs in part through three c

onsecutive steps in which ozone is broken down: Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen. Step 2: Chlorine monoxide molecules combine to form ClOOCl gas Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen. a) Write a balanced equation for each step. b) Write an overall balanced equation for the sequence
Chemistry
1 answer:
mixer [17]3 years ago
7 0

Answer:

A)

Cl + O3 --> ClO + O2

2ClO --> ClOOCl

ClOOCl --> 2 Cl + O2

B)

Cl2 + 4O3 -> 2 ClO + 5O2

Explanation:

Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen

Cl + O3 --> ClO + O2

Step 2: Chlorine monoxide molecules combine to form ClOOCl gas

2ClO --> ClOOCl

Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen

ClOOCl --> 2 Cl + O2

B) Overall Balanced equation

Cl2 + 4O3 -> 2 ClO + 5O2

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The answer is: mass is 40.17 kilograms.

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
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Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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