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wel
3 years ago
14

Multiple answers are allowed.

Chemistry
2 answers:
AveGali [126]3 years ago
7 0

Answer:

There are two correct choices:

  • <em>Real gas particles have significant volume</em>
  • <em>Real gas particles have more complex interactions than ideal gas particles.</em>

Explanation:

<em>Ideal gases </em>are not real; ideal gases are a theoretical model used to confere a better understanding of gas properties.  This model permits to predict the behavior of the gases using the ideal gas law, which is valid under certain conditions (mainly low pressure and high temperature).

Kinetic molecular theory states these basic assumptions for gases:

  •    Gas particles are in constant, random motion.
  •    Gas particles  move in a straight line until they collide with another particle or the walls of the container.
  •    Gas particles are much smaller than the distance between particles, so most of the volume of a gas is therefore empty space, and the particles are considered to occupy no space.
  •    There is no force of attraction acting on gas particles nor between them
  •    Collisions are perfectly elastic, meaning that no energy is lost durign collisions.

Those conditions are not perfected matched by real gas particles, since real gas particles do occupy a volume and interact with each other, this is r<em>eal gas particles have significant volume and  have more complex interactions than ideal gas particles.</em>

Marizza181 [45]3 years ago
5 0
<h2>Answer </h2>

Option 1 - Real gases are always hotter than the ideal gases.

<u>Explanation </u>

Gases that are different from ideality are called real gases because they are always hotter than the ideal gases as they are flying past each other at an extremely high speed that creates kinetic energy. Whereas idea gases have absolutely elastic collisions, this is as it has a valuable theory because it embraces the perfect concept of original gas law.

<h2>Answer </h2>

Option 2 - Real gases particles have significant volume.

<u>Explanation </u>

Real gases have significant volume because they have a high temperature and have a higher volume as compared to ideal gases. There is a general gas theory that is composed of several randomly crossing point particles. Several gases, for example, nitrogen, oxygen, and hydrogen can be manipulated like ideal gases within a generous immunity of the idea gas.

<h2>Answer </h2>

Option 3 - Real gases particles are smaller than an ideal gas.

<u>Explanation </u>

Real gas particles are smaller than ideal gas particles since they have a volume and are made up of molecules or atoms that typically take up some space than the ideal atoms. The ideal gas pattern manages to break at moderate temperatures level or greater pressures point when intermolecular energies and molecular mass enhances to be significant.

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Archy [21]

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}.

The \text{K}_b value for ammonia is small. The value of x will be so small that at equilibrium, 1.0 - x \approx 1.0 and 2.5- x \approx 2.5.

\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}.

\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}.

\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}.

Again, \text{K}_w = 1.0\times 10^{-14} at room temperature.

\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}

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3 years ago
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