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Anna [14]
7 months ago
7

Select the correct answer. which substance in this redox reaction is the oxidizing agent? cu 2agno3 → 2ag cu(no3)2 a. n b. agno3

c. cu d. no3− e. cu(no3)2
Chemistry
1 answer:
lana66690 [7]7 months ago
8 0

AgNO₃ will act as the oxidising agent.

<h3><u>For the given chemical equation:</u></h3>

Cu + 2AgNO₃ →  2Ag + Cu(NO₃)₂

Half reactions for the given chemical reaction:

<u>Reducing agent:</u>

Cu → Cu²⁺ + 2e⁻

Copper is a reducing agent because it is losing 2 electrons, which causes an oxidation process.

<u>Oxidising Agent</u>:

Ag⁺ + e⁻ → Ag

The silver ion undergoes a reduction process and is regarded as an oxidizing agent since it is acquiring one electron per atom.

Hence, AgNO₃ is considered as an oxidizing agent and therefore the correct answer is Option B.

<h3><u>Oxidising and Reducing agents</u></h3>
  • An oxidizing agent is a substance that reduces itself after oxidizing another material. It passes through a reduction process in which it obtains electrons and the substance's oxidation state is decreased.
  • A reducing agent is a chemical that oxidizes after reducing another material. It passes through an oxidation process in which it loses electrons and the substance's oxidation state increases.

To know more about the process of Oxidation and Reduction, refer to:

brainly.com/question/4222605

#SPJ4

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

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\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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Explanation:

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The complete question is

If the compound below is oxidized, the resulting product is ___.

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