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aivan3 [116]
3 years ago
15

Which substance loses electrons in a chemical reaction? the one that is oxidized, which is the oxidizing agent the one that is r

educed, which is the reducing agent the one that is oxidized, which is the reducing agent the one that is reduced, which is the oxidizing agent
Chemistry
2 answers:
victus00 [196]3 years ago
7 0

Answer:

The one that is oxidized or the reducing agent

Explanation:

Oxidation results in loss of electrons or an increase in oxidation state.

Reduction results in gain of electrons or  decrease in oxidation state.

The element that undergoes oxidation is said to oxidised, similarly the element that undergoes reduction is said to be reduced.

In a redox reaction, both oxidation and reduction takes place.

If a substrate undergoes oxidation or is oxidised, it is also responsible for reduction of the other species, as the total number of electrons should always be conserved.

The substance that undergoes oxidation, releases some electrons, these electrons are taken by the other substrate and it undergoes reduction.

Hence the substance that undergoes oxidation is called the reducing agent, as it is responsible for reduction of other substrate, when oxidizing itself.

Similar thought works for a substance that undergoes reduction or is reduced, works as an oxidizing agent.

LenKa [72]3 years ago
7 0

Answer:

the one that is oxidized, which is the reducing agent

Explanation:

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PtichkaEL [24]
<h3>Answer:</h3>

100 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   2 mol CH₄

[Solve]   x g O₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol CH₄ → 2 mol O₂

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 2 \ mol \ CH_4(\frac{2 \ mol \ O_2}{1 \ mol \ CH_4})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Multiply/Divide:                           \displaystyle 128 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

128 g O₂ ≈ 100 g O₂

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3 years ago
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