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Nata [24]
3 years ago
6

1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the

Chemistry
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

Explanation:

Rubidium has a more negative reduction potential (-2.98 V) compared to strontium (-2.89 V).

Hence, in a redox reaction involving rubidium and strontium, rubidium will be oxidized while strontium is reduced.

The balanced redox reaction equation is obtained from;

Oxidation half equation;

2Rb(s) ---->2Rb^+(aq) + 2e

Reduction half equation;

Sr^2+(aq) + 2e ----> Sr(s)

Overall reaction equation;

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

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The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p
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Answer:

The frequency is  f =  2,01 * 10^{15} \  Hz

Explanation:

From the question we are told that

   The energy required to ionize boron is E_b  =  801 KJ/mol

Generally the ionization energy of boron pre atom is mathematically represented as

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Here  N_A is the Avogadro's constant with value N_A  =  6.022*10^{23}

So

      E_a  =  \frac{801}{6.022*10^{23}}

=>     E_a  =  1.330 *10^{-18} \  J/atom

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

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=>     hf  =  E_a

Here h is the Planks constant with value h = 6.626 *10^{-34}

So

       f =  \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}

=>      f =  2,01 * 10^{15} \  Hz

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