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EastWind [94]
3 years ago
8

What does the photoelectric effect prove?

Chemistry
1 answer:
Elza [17]3 years ago
6 0

Answer:

<em>By showing that changing the frequency of light causes the emission of faster electrons. </em>

Explanation:

<em>The photoelectric effect happens when light strikes a metal surface causing the emission of electrons from it (photoelectrons). </em>

<em>If you increase the intensity of the light you get, as acresult, more electrons emitted but their kinetic energy does not increase. </em>

<em>If you increase the frequency of the incident light the number of photoelectrons emitted does not increase while the velocity, and so their kinetic energy, increases...the emitted electrons are more...energetic! </em>

<em> </em>

<em>This can be explained considering the incident light as a shower of particle-like packets of energy (photons); if you increase the intensity you simply increase the number of packets (all with the same energy) hitting the metal; these can be used by a lot of electrons to escape. </em>

<em>On the other hand if you increase the frequency the number of packets remains the same (emitting fewer electrons perhaps) but the energy carried by each of them increases. </em>

<em>Each packet carries an energy directly proportional to the frequency.</em>

<em />

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Answer:

Protons and neutrons which packed tightly into central core of the atom.

Explanation:

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An enzyme is a catalyst used by living things to promote and regulate chemical reactions. What is the most likely enzyme for the
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3 years ago
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Orlov [11]
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Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o
anzhelika [568]
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.139M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.139-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or 

≈5.03% dissociation.
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3 years ago
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