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alexdok [17]
3 years ago
6

Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Cu(s)

+2AgNO3(aq)⟶2Ag(s)+Cu(NO3)2(aq)
Chemistry
1 answer:
spin [16.1K]3 years ago
4 0

<u>Answer:</u> Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

X^{n+}+ne^-\rightarrow X

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)

The half reactions for the above reaction are:

<u>Oxidation half reaction:</u>  Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-

<u>Reduction half reaction:</u>  2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

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How many molecules are there in 21.4 grams of Ca(OH)2?
nata0808 [166]

The answer would be 0.288827452320528

4 0
3 years ago
During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that
bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0
3 years ago
Can someone help me with my science it’s super hard :(
Lunna [17]

Answer: here is a description of what they look like since I cannot see the diagram.

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5 0
2 years ago
Help me on this please !
Nat2105 [25]

Answer: your answer is A

Explanation:

5 0
3 years ago
The rate constant for the reaction is 0.660 M − 1 ⋅ s − 1 0.660 M−1⋅s−1 at 200 ∘ C. 200 ∘C. A ⟶ products A⟶products If the initi
Irina-Kira [14]

<u>Answer:</u> The concentration of A after 865 seconds is 0.00125 M

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.660M^{-1}s^{-1}

t = time taken = 865 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.00440 M

Putting values in above equation, we get:

0.660=\frac{1}{865}\left (\frac{1}{[A]}-\frac{1}{(0.00440)}\right)

[A]=0.00125M

Hence, the concentration of A after 865 seconds is 0.00125 M

5 0
3 years ago
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