Final Exam Review - University of California, Irvine

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Transcript Final Exam Review - University of California, Irvine

Applicable concepts/equations E

total

= E

photon

* # photons

𝐸 = β„Žπ‘ l 𝐸 π‘β„Žπ‘œπ‘‘π‘œπ‘› = 𝐸 (6.626π‘₯10 π‘β„Žπ‘œπ‘‘π‘œπ‘› βˆ’34 422 )(2.9979π‘₯108) π‘₯10 = 1.99π‘₯10 #π‘β„Žπ‘œπ‘‘π‘œπ‘›π‘  = 𝐸 𝐸 π‘‘π‘œπ‘‘π‘Žπ‘™ π‘β„Žπ‘œπ‘‘π‘œπ‘›

=

40.0

𝐽 𝑠𝑒𝑐 βˆ—2𝑠𝑒𝑐 1.99π‘₯10 βˆ’25 𝐽

= 4.03x10

26

photons

βˆ’9 π‘š βˆ’25 𝐽

Can a wave do this?

yes

Can a Particle do this?

yes yes no yes no yes no no yes

****Things we discussed in this course.

https://www.youtube.com/watch?v=DfPeprQ7oGc

Double Split Experiment

Does this show that light has wavelike or particle like properties? Why?

What would the results look like if they had particle like properties?

Wave-like: It creates an interference pattern. Two bright lines behind the splits no interference pattern.

Photoelectric Effect

Does this show that light has wavelike or particle like properties? Why? What would the results look like if they had only wave-like properties? Particle: 1 photon=1 electron, which must be of high enough energy.

High total energy still won’t eject an electron if each isn’t of sufficient energy. What is the effect of increasing the intensity of a laser of a frequency less than the threshold frequency?

None: it must be OVER the threshold frequency to eject electrons. What is the effect of increasing the intensity of a laser of a frequency greater the threshold frequency?

The rate of the ejected electrons is increased. http://phet.colorado.edu/en/simulation/photoelectric

E

k = β„Žπ‘£ βˆ’ 𝑀 1.3x10

βˆ’19 = (6.626x10

βˆ’34 )𝑣 βˆ’ (7.53x10

βˆ’19 ) 𝑣 = 1.33π‘₯10 βˆ’15 Hz Ke Y-intercept = work function Slope=h

E

k = β„Žπ‘£ βˆ’ 𝑀

y = m x + b

frequency Threshold Frequency

Ξ¨ Ξ¨ 2

Particle in a box (1d) Schrodinger equation solutions Particle in a box (2/3D) Hydrogen Atom Didn’t cover

Multi electron atoms= many types of approximation, no exact solutions

Subject of Current research: We saw result, of appoximations

For each of the previous neutral electron configurations, give an excited state. Keep same number of electrons, move at least one up in energy: Many many many correct answers. Sb: [Kr]5s 2 4d 10 5p 3 [Kr]4d 10 5p 5 ect…..

οƒ  [Kr]5s 1 4d 10 5p 4 or Cu: [Ar]4s 1 3d 10 [Ar]3d 10 4p 1 οƒ  ect…..

[Ar]4s 2 3d 9 or

-1,0,1 -1,0,1

9+4=13 orbitals 2 electrons= 13*2=26 electrons

Effective Nuclear Charge Electronegativity Ionization Energy Electron Affinity Atomic Radius

Effective Nuclear Charge Electronegativity Ionization Energy Electron Affinity Atomic Radius

Write the electron configurations for C, N, and O. Place in order of increasing ionization energy, increasing electron affinity and increasing electronegativity. For each characteristic, do these follow the trend? Why or why not? Wrong Answer 1: All: C, N, O Yes Because the trend goes up and to the right. Wrong Answer 2: Variety of wrong/right answers No Because the trend goes up and to the right. But hydrogen is half filled.

Correct: I.E. C, O, N : No, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize. E.A. N, C, O: No, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize. E.N. C, N, O: Yes. Because it is speaking of atoms in a stable bond, electronegativity is only based on effective nuclear charge which increases as you go to the right.

Hint: what is the p block exceptions to electron affinity?

Half filled subshells are more stable and therefore have a lower electron affinity. Hint: What would be the equivalent of that with the dblock?

Half filled subshells are more stable and therefore have a lower electron affinity.

ns 2 np 5 and also ns 1 np 5

Which group do these elements belong to?

First is always smallest, The β€œjump” indicates stable configuration

I1 I2

Element 1 0.605 1.110

Element 2 0.203

3.215

I3 I4

1.45 5.10

3.89 4.42

1 st , 2 nd , 3 rd come off, then big jump: so group 3 Big jump after 1 st electron, so group 1

Why do we see the sun as different colors?

Which lights depicted in this diagram are emitted, which are scattered.

What causes the events depicted by the blue arrows in the atmosphere.

C CN 2p N E [ He ]s 2s 2 s 2s *2 p 2p 4 diamagnetic Bond order= Β½ (6-2)=2 Bonding orbitals are a lower energy than antibonding and the original atomic orbitals. Most of the electron density in bonding orbitals are between the nuclei rather than outside of it. This adds to the bond order, rather than subtracts.

Yes!

+0 Except sp 3 Tetrahedral 109.5

sp 3 Trig. Planar 120

s s s

sp

3

sp

3

sp

3

sp

3

p sp

2

sp

2

p p p sp

2

p p

HO O OH

CH3OH can be synthesized by the reaction shown. What volume of H 2 required to synthesize 25.8 g CH 3 OH? 𝐢𝑂 𝑔 + 2𝐻 2 𝑔 β†’ 𝐢𝐻 3 𝑂𝐻 gas (in L), at 748 mmHg and 86 o C, is Grams product moles product moles reactants Volume Reactant conversion conversion Ideal gas law 25.8𝑔 𝐢𝐻3𝑂𝐻 1π‘šπ‘œπ‘™ 𝐢𝐻 3 𝑂𝐻 32.04 𝑔 𝐢𝐻 3 𝑂𝐻 2 π‘šπ‘œπ‘™ 𝐻 2 1π‘šπ‘œπ‘™ 𝐢𝐻3𝑂𝐻 = 1.610 mol H 2 𝑉 = 𝑛𝑅𝑇 𝑃 = (1.610 π‘šπ‘œπ‘™βˆ—0.0821

π‘™βˆ—π‘Žπ‘‘π‘š π‘šπ‘œπ‘™βˆ—π‘˜ βˆ— 86+273 𝐾 748π‘šπ‘šπ»π‘”/760 π‘šπ‘šπ»π‘” π‘Žπ‘‘π‘š =48.2 L H2 Part 2: If the reaction is known to only be 70.0% efficient (aka has a 70% yield) what volume is required (at the same conditions)? π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ βˆ— 100% 𝑔𝑖𝑣𝑒𝑠: 48.2

= 0.700 π‘€β„Žπ‘–π‘β„Ž 𝑔𝑖𝑣𝑒𝑠 π‘₯ = 68.8 𝐿 π‘₯