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Charge Me Up, Scotty!
(10/10 points)
Determine the potential (V) by which a proton must be accelerated so as to assume a particle wavelength of 0.0293 nm.
0.9544 - correct
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PES S+P 500
(10/10 points)
Shown below are the photo electron spectra for phosphorous and sulfur
The PES spectra for phosphorous and sulfur are shown above. Answer the following questions concerning these spectra.
Why is the phosphorous peak at 1.06 MJ/mole greater in energy than the sulfur peak at 1.00 MJ/mole?
paired electron easier to remove - correct
Why is the phosphorous peak at 1.95 less energy than the sulfur peak at 2.05?
sulfur has larger Z - correct
What should be the ratio of intensities of the phosphorous peak at 1.06 to the phosphorous peak at 13.5?
1/2 - correct
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/* 1-5 */
You'll Learn About Electrons, I Guarantee It
(10/10 points)
Which of the following orbital diagrams is/are incorrect for all electrons in the lowest-energy levels of an atom?
1,2,4 - correct
Which of the following is the correct electronic configuration for the bromide ion, Br-?
[Ar]4s23d104p5
[Ar]4s24p5
[Ar]4s23d104p6 - correct
[Ar]4s23d104p65s1
Which of the following orders of filling orbitals is/are incorrect? They should be in the exact order that they would be filled in.
3s, 4s, 5s - correct
5s, 5p, 5d - correct
5s, 4d, 5p
6s, 4f, 5d
6s, 5f, 6p - correct
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CaF2, Great for Teeth!
(10/10 points)
Calculate the lattice energy of CaF2 given the information below (in kJ/mol):
Ca(g)⟶Ca(s) ΔH=−178kJ/mol
Ca(g)⟶Ca+(g)+e− ΔH=589.8kJ/mol
F(g)+e−⟶F−(g) ΔH=−328.2kJ/mol
Ca2++e−⟶Ca+(g) ΔH=−1145.4kJ/mol
F2(g)⟶2F(g) ΔH=139.0kJ/mol
Ca(s)+F2(g)⟶CaF2(s) ΔH=−1219.6kJ/mol
- 2615.4 - correct
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L6Z3RS DOOD
(10/10 points)
In an effort to purify the output from a catalytic chemical reaction, C-Cl bonds are to be broken in a gaseous compound by means of LASER radiation. Determine the longest wavelength of radiation still capable of breaking the C-Cl bond. Express your answer in meters.
[Given: BEC-C = 334 kJ/mol; BECl-Cl = 243 kJ/mol]
3.729*10^-7 - correct
Cyanohydrins can be made from carbonyl compounds by generating CN– ions from HCN in the
presence of a weak base.
In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases.
Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong
base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3
Ans: C
I have no idea how to obtain the answer!
Please explainnnn!
Thank you in advance :)
1 answer