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m_a_m_a [10]
3 years ago
5

Key Questions and Terms Notes

Chemistry
1 answer:
Elanso [62]3 years ago
5 0

Answer:

Following are the solution to the given question:

Explanation:

  • Its interactions among light waves as well as different media were also called light products.
  • Even before sound rays hit a medium, the light rays bounce off.
  • Even at a similar angle with which the material initially struck its medium, sunlight reflecting.
  • When the light hits a mirror at such an angle of forty degrees, everything just rebonds at an angle of 40 degrees.
  • When light waves pass via a transparent medium, those who bend as well as change their direction.
  • Light is called transparent if it can pass via an object.
  • Whenever light waves strike a platform, this is neither re-elected nor refracted, nor expected to move.
  • The light of interacts with these materials are:

Reflection Mirror:

Water glass: refractive

Dark fabricate: Absorption 

  • Sound waves and waves of water are examples of matter waves.
  • Light waves and X-rays are electromagnetic waves.
  • There is no need for sound radiation to pass through a medium/requirement to communicate with other particles to travel like waves of matter.
  • No, even though sound waves didn't reach through the space vacuum.
  • Matter waves should interact with solid, liquid, or gas particulates to travel.

The light waves pass thru the air more quickly than sound waves.

Sound waves almost always go through a metal frame.

Light waves travel faster than electrical signals even though they don't have to crash in electromagnetic waves to particles.

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8.324 Hope this helps! =)

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When Kristen was five, she became sick with chickenpox and then recovered. What claim explains why Kristen is unlikely to get ch
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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que
boyakko [2]

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

  • Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
  • The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
  • Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
  • An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

#SPJ4  

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