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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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Explanation:

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Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

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