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Lilit [14]
3 years ago
11

Alkynes are reduced to trans alkenes by a process called dissolving metal reduction. The reaction uses sodium or lithium metal a

s the reducing agent and liquid ammonia as the solvent. The method is specific in the formation of trans alkenes from alkynes. The method involves two successive transfers of single electrons from the alkali metal to the triple bond, with abstraction of protons from the ammonia solvent.Draw curved arrows to show the movement of electrons in this step of the mechanism.Arrow-pushing Instructions

Chemistry
1 answer:
NISA [10]3 years ago
6 0

Answer:

Explanation:

The movement of the electrons is illustrated in the picture attached to this answer. It is a four-step reaction mechanism.

First STEP: The first step involves the transfer of an electron from sodium to form a radical anion.

Second STEP: This radical anion then removes a proton/hydrogen from ammonia in a bid to neutralize itself (hence the hydrogen becomes bonded to the anion).

Third STEP: The sodium (from NaNH₂ formed) transfers an electron again to produce a vinyl carbanion.

Fourth STEP: The carbanion then removes a proton/hydrogen from ammonia (like in the second step) to form a neutral trans-alkene.

NOTE: The circled numbers denote each step while the mechanism on the left represents the use of any alkyl group (R and R') while the mechanism on the right assumes both alkyl groups are methyl. Hence, 2-butyne started the reaction and the final product was trans-2-butene.

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If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

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3 0
3 years ago
How many moles are in 5.67x10^24 atoms of RbCl?
Mademuasel [1]

Answer:

<h2>9.42 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.67 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 9.418604...

We have the final answer as

<h3>9.42 moles</h3>

Hope this helps you

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