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Alex787 [66]
3 years ago
15

Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless. 2NO2(g

) ⇌N2O4(g) When the mixture was moved from room temperature to a higher temperature, the mixture turned darker brown. Which of the following conclusions about the mixture is true? The forward reaction is exothermic. The backward reaction has a negative net enthalpy. The colorless gas has a higher enthalpy than the brown gas. The equilibrium between the two forms of the gas is disturbed at high temperatures.
Chemistry
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

The forward reaction is exothermic.

Explanation:

  • Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
  • When the mixture turned darker brown, this means that the reaction is shifted towards the left direction (reactants side).
  • The temperature is increased and the reaction shifted to the reverse direction, this means that the forward direction is exothermic.
  • Exothermic reaction releases heat and when increasing the temperature, the reaction will be shifted to the reverse direction to suppress the effect of increasing the temperature.
  • <em>So the right choice is: The forward reaction is exothermic. </em>

<em></em>

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4 0
3 years ago
Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n
lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239

According to Dalton's Law of partial pressure-

P_i=\chi_i\times P_{total}

Where,

P_i=The pressure of the gas component in the mixture

\chi_i= Mole fraction of that gas component

P_t= The total pressure of the mixture

P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

n=\frac{PV}{RT}

n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole

Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

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4 0
1 year ago
Consider the oxidation of sodium metal to sodium oxide described by the balanced equation:
mixas84 [53]

Answer:

1116 g.

Explanation:

The balanced equation for the reaction is given below:

4Na + O₂ —> 2Na₂O

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Next, we shall determine the theoretical yield of Na₂O. This can be obtained as follow:

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Therefore, 9 moles of O₂ will react to produce = 9 × 2 = 18 moles of Na₂O.

Finally, we shall determine the mass in 18 moles of Na₂O. This can be obtained as follow:

Mole of Na₂O = 18 moles

Molar mass of Na₂O = (23×2) + 16

= 46 + 16

= 62 g/mol

Mass of Na₂O =?

Mass = mole × molar mass

Mass of Na₂O = 18 × 62

Mass of Na₂O = 1116 g

Thus, the theoretical yield of Na₂O is 1116 g.

3 0
2 years ago
What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10
max2010maxim [7]
<span>The pH is given by the Henderson - Hasselbalch equation:
              pH = pKa + log([A-]/[HA])
              pH = -log(</span><span>1.3 x 10^-5) + log(0.50/0.40)
              pH = 4.98
The answer to this question is 4.98.
</span>
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A _____ works by lighting the slide containing the specimen from below with a light bulb.
lesya692 [45]
Condenser Lens - This lens system is located immediately under the stage and focuses the light on the specimen.
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