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Vlad1618 [11]
3 years ago
6

Consider the following standard reduction potentials: Zn2+(aq) + 2 e− LaTeX: \longrightarrow⟶ Zn(s) Eo = −0.76 V Mg2+(aq) + 2 e−

LaTeX: \longrightarrow⟶ Mg(s) Eo = −2.37 V Ag+(aq) + e− LaTeX: \longrightarrow⟶ Ag(s) Eo = +0.80 V Which is the strongest oxidizing agent? Group of answer choices
Chemistry
1 answer:
zavuch27 [327]3 years ago
8 0

<u>Answer:</u> The strongest oxidizing agent is silver.

<u>Explanation:</u>

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions and reduction reaction is the reaction in which an atom gains electrons.

The substance having highest positive reduction E^o potential will always get reduced and will undergo reduction reaction easily.

We are given:

Zn^{2+}(aq.)+2e^-\rightarrow Zn(s);E^o =-0.76V

Mg^{2+}(aq.)+2e^-\rightarrow Mg(s);E^o =-2.37V

Ag^{+}(aq.)+e^-\rightarrow Ag(s);E^o =+0.80V

As, silver has the highest positive reduction potential. So, it will be the strongest oxidizing agent.

Hence, the strongest oxidizing agent is silver.

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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?
olga_2 [115]

Answer:

pH =3.8

Explanation:

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.

4 0
3 years ago
Question 4
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Answer:

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Explanation:

In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  the major species present are H(aq) + NO3 (aq) + HF(aq).

On the reaction of  HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  , it will give a polar solution and will form a homogenous mixture.

Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".

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Answer:

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