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2 years ago

50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.

Chemistry

1 answer:

photoshop1234 [79]2 years ago

7 0

Limiting reactant : O₂

Mass of N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

$\tt =\dfrac{50}{44}=1.136$

mol O₂ :

$\tt =\dfrac{50}{32}=1.5625$

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136 1.5625

1.0416 1.5625 1.0416

0.0944 0 1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

$\tt =mol\times MW=1.0416\times 92=95.83~g$

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2 years ago

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$H_{2}Te$ < $H_{2}Se$ < $H_{2}S$ < $H_{2}O$

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2 years ago

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Answer: Option (c) is the correct answer.

Explanation:

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As a result, a partial positive charge will develop on hydrogen atom and a partial negative charge will develop on oxygen atom.

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2 years ago

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2. $Cl^-+BCl_3\rightarrow BCl_4^-$ Lewis acid : $BCl_3$ , Lewis base : $Cl^-$

3. $K^++6H_2O\rightarrow K(H_2O)_6$ Lewis acid : $K^+$ , Lewis base : $H_2O$

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. $H^++OH^-\rightarrow H_2O$

As $H^+$ gained electrons to complete its octet. Thus it acts as lewis acid. $OH^-$ acts as lewis base as it donates lone pair of electrons to electron deficient specie $H^+$ .

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As $BCl_3$ is short of two electrons to complete its octet. Thus it acts as lewis acid. $Cl^-$ acts as lewis base as it donates lone pair of electrons to electron deficient specie $BCl_3$ .

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2 years ago

Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr

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Answer:

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Explanation:

Number of double bonds of the given compound is calculated using the below formula.

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$N_{db}$ =Number of double bonds

$N_{c}$ = Number of carbon atoms

$N_{H}$ = Number of hydrogen atoms

$N_{N}$ = Number of nitrogen atoms

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The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

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Hence, the probable structures III and IV are given as follows.

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2 years ago