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Helen [10]
3 years ago
8

In the reversible reaction: 2NO2 (g) ⇌ N2O4 (g) the formation of dinitrogen tetroxide releases heat, and the formation of nitrog

en dioxide absorbs heat. If the reaction is at equilibrium and the temperature decreases, what will the effect be?
A. The equilibrium will shift so that there is more nitrogen dioxide.

B. Dinitrogen tetroxide releases heat and changes from gas to liquid.

C. Nitrogen dioxide releases heat and changes from gas to liquid.

D. The equilibrium will shift so that there is more dinitrogen tetroxide.
Chemistry
2 answers:
emmasim [6.3K]3 years ago
8 0

Answer:

The equilibrium will shift so that there is more dinitrogen tetroxide.

Explanation:

Answer for Educere/ Founder's Education

Fed [463]3 years ago
3 0

Answer:

D

Explanation:

Any change of temperature will subsequent in a change of the equilibrium position. With reference to Le Chaterlier's principle, this equilibrium shift will occur in a manner as to <em>neutralize the change</em> that the system has undergone.

In this case, the temperature is seen to have decreased. To neutralize this effect, the equilibrium position will <em>shift </em>in a manner as to <u>increase</u> the temperature back to its previous state.

This equilibrium shift will occur by favoring the reaction the reaction which results in the release of heat -which is the formation of dinitrogen tetrahydroxide.

This will result in the presence of more dinitrogren tetrahydroxide relative to the amount of nitrogen dioxide in the system.

<h3>Hope this helps</h3>
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Answer :

Charles's Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

V\propto T

                                Boiling water bath        Cool bath 1       Cool bath 2

Temperature (⁰C)                  99                              17                       2

Temperature (K)(T)    273+99=372             273+17=290      273+2=275

Volume of water                  0.0                             27.0                34.0

in cool flask (mL)

Volume of water=              135.8                           135.8               135.8

Air in flask (mL)

Volume of air                    135.8                           108.8               101.8

in cool flask (V)

\frac{V}{T}                                \frac{135.8}{372}=0.365             \frac{108.8}{290}=0.375         \frac{101.8}{275}=0.370

The graph volume versus temperature for a gas is shown below.

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3 years ago
Sugar dissolving in tea is it chemical or physical change?​
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5. The density of a substance is 3.4 X 10^3 g/mL. Mixing it with another substance increases the density by 1.5 X 10^1 times. Wh
JulijaS [17]

Answer:

the total density of the mixture = 3.415 x 10^3 g/mL

Explanation:

given:

density of a substance is 3.4 x 10^3 g/mL

Mixing it with another substance increases the density by 1.5 x 10^1 times.

find:

What is the density of the mixture?​

-------------------------------------------------

let density (a) = 3.4 x 10^3 g/mL

density(b) = 1.5 x 10^1 g/mL

since there is no specific density provided for the mixture, we then add both to increase the density,

total density = density(a) + density(b)

total density = 3.4 x 10^3 g/mL + 1.5 x 10^1 g/mL

total density = 3.415 x 10^3 g/mL

therefore,

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5 0
3 years ago
Read 2 more answers
2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas
IrinaK [193]

Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

<em>P = 0.163atm</em>

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3 years ago
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