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beks73 [17]
3 years ago
13

Write a balanced half-reaction describing the reduction of gaseous dioxygen to aqueous oxide anions.

Chemistry
1 answer:
maksim [4K]3 years ago
8 0

<u>Answer:</u> The balanced half reaction is written below.

<u>Explanation:</u>

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

When oxygen gas is reduced to oxide ions, the number of electron transferred are 2

The chemical equation for the reduction of oxygen gas to oxide ions follows:

O_2+2e^-\rightarrow 2O^{2-}

Hence, the balanced half reaction is written above.

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How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?​
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Answer:

3.72 mol Hg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Density = Mass over Volume

Explanation:

<u>Step 1: Define</u>

D = 13.6 g/mL

54.8 mL Hg

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

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Find</u>

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

<u>Step 4: Convert</u>

<u />745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} ) = 3.71544 mol Hg

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.71544 mol Hg ≈ 3.72 mol Hg

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When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When
3241004551 [841]

Answer:

See explanation

Explanation:

The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.

On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.

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