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Trava [24]
3 years ago
6

Redox 1/2 reaction Cu(s) + 2 AgC2H3O2(aq) = Cu(C2H3O2)2(aq) + 2 Ag(s)

Chemistry
1 answer:
Anna11 [10]3 years ago
4 0

Explanation:

 Reaction:

              Cu  +     2AgC₂H₃O₂     →    Cu(C₂H₃O₂)₂    +     2Ag

The problem is to split the reaction into oxidation and reduction halves:

 The oxidation half is the sub-reaction that undergoes oxidation

  The reduction half is the one that undergoes reduction:

The ionic equation:

 Cu   +  2Ag⁺   +   2C₂H₃O₂⁻   →   Cu²⁺  +  2C₂H₃O₂⁻   +  2Ag

Oxidation half:

     Cu →  Cu²⁺ + 2e⁻

 

Reduction half:

    2Ag⁺ + 2e⁻ → 2Ag

C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.

learn more:

Oxidation state brainly.com/question/10017129

#learnwithBrainly

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the abbreviation form of full name is called symbol.

the smallest unit of cimpound is called molecule.

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Which object forms when a supergiant explodes
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3 years ago
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How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

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3 0
1 year ago
If metal ions in a solution were reduced, what would you expect to see?
marysya [2.9K]

Answer:

Solid metal

Explanation:

The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:

Na⁺ + e- → Na(s)

8 0
2 years ago
Please help me no links <br> if good answer you get brainliest
Olin [163]
The crystals will not be oxidised since it will be underground
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2 years ago
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