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Svetradugi [14.3K]
3 years ago
13

A ____________--(also called a reductant or reducer) is an element or compound that loses (or "donates") an electron to an elect

ron recipient (oxidizing agent) in a redox chemical reaction.
A reducing agent is thus oxidized when it loses electrons in the redox reaction. Reducing agents "reduce" (or, are "oxidized" by) oxidizing agents. Oxidizers "oxidize" (that is, are reduced by) reducers.
Chemistry
2 answers:
gladu [14]3 years ago
4 0

Answer:

A <u><em>reducing agent </em></u>is an element or compound that loses (or "donates") an electron to an electron recipient (oxidizing agent) in a redox chemical reaction.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

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

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

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

Electrons are transferred from one atom to another in this type of reaction.

Reducing agent is defined as those substance which reduces other chemical compound and oxidizes itself by donating electrons.

Oxidizing agent is defined as those substance which oxidizes other chemical compound and reduces itself by accepting electrons.

PSYCHO15rus [73]3 years ago
4 0

Answer:

A reducing agent

Explanation:

A reducing agent--(also called a reductant or reducer) is an element or compound that loses (or "donates") an electron to an electron recipient (oxidizing agent) in a redox chemical reaction.

A reducing agent thus oxides( loses an electron) itself to electron acceptor. This mean that oxidation and reduction occur in pair. One element is oxidized and the other is reduced.

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What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Etha
KatRina [158]

Answer:

percent yield = 56.6 %

Explanation:

Data given:

mass of  Ethylene gas (C₂H₄) = 140 g

actual yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 280 g

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = ?

Reaction Given:

                        C₂H₄ + Cl₂ -------> C₂H₄Cl₂

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                       C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

                      1 mol                         1 mol

As 1 mole of C give 1 mole of CH₄

Convert moles to mass

molar mass of C₂H₄ = 2(12) + 4(1)

molar mass of C₂H₄ = 24 + 4

  • molar mass of C₂H₄ = 28 g/mol

molar mass of C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

  • molar mass of C₂H₄Cl₂ = 99 g/mol

Now

                       C₂H₄     +       Cl₂    -----—>     C₂H₄Cl₂

                1 mol (28 g/mol)                        1 mol (99 g/mol)

                          28 g                                          99 g

28 grams of Ethylene gas (C₂H₄) produce 99 grams of C₂H₄Cl₂

So

if 28 g of C₂H₄ produce 99 g of C₂H₄Cl₂ so how many grams of C₂H₄Cl₂ will be produced by 140 g of C₂H₄.

Apply Unity Formula

                        28 g of C₂H₄ ≅ 99 g of C₂H₄Cl₂

                        140 g of C₂H₄≅ X of C₂H₄Cl₂

Do cross multiply

                      mass of C₂H₄Cl₂ = 99 g x 140 g / 28 g

                      mass of C₂H₄Cl₂ = 495 g

So the Theoretical yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 495 g

Now Find the percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                percent yield = 280g / 495 g x 100 %

                 percent yield = 56.6 %

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = 56.6 %

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The mechanism of a reaction consists of a pre-equilibrium step with forward and reverse activation energies of 27 kJ mol1 and 35
Marina CMI [18]

Explanation:

The given data is as follows.

       E_{a}(a) = 27 kJ/mol  

       E_{a}(b) = 35 kJ/mol

and,  E_{a}(c) = 15 kJ/mol

Activation energy of the overall reaction will be the sum of given activation energies as follows.

          E_{a} = E_{a}(a) + E_{a}(b) + E_{a}(c)

                           = (27 + 35 + 15) kJ/mol

                           = 77 kJ/mol

Thus, we can conclude that the overall value of activation energy is 77 kJ/mol.

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Balance the Chemical Equations<br><br> MgF2 + Li2CO3---&gt; MgCO3 + LiF <br><br> Please Help
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3<span> NaOH + H3PO4 = 3 H2O + Na3PO4</span>



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3 years ago
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