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Vlad1618 [11]
4 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]4 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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3 years ago
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Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693.\overline{3168} grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃)  is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}

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The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}

\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693.\overline{3168} grams

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3 years ago
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