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Anna [14]
1 year 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]1 year 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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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

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pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

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OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

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