****TEST # 2 ***** ***During Regularly Scheduled Class***
Answers for Exam 2: ExamView – Test2, 2009
Announcements:
1)The material covered for this exam was more difficult than that of the first exam – it is normal for the scores to be a little lower. Average = 64, as calculated below. I will post a histogram tomorrow.
2) The exam scores were calculated as follows: (# correct)x(4) + 4 = score.
3) one question was worded poorly, so I gave everybody 4 points for that question (see above)
4) the exam answer key had a mistake on each exam form: The correct answer to ion radius is Br- (Form A, #14; varies from form to form). This mistake was corrected prior to calculating the scores above.
5) In order to help normalize scores, I will do the following: Your percentage score on the final exam will replace that of your lowest midterm, provided that this will improve your score. (eg: MT = 62/100, final = 90/120. Your MT score would get dropped, and replaced by a “75”.)
Johnathan Aprile…the Answer to your question is B)Millikan!
how are wavelength and frequency related?
You disregard the 3 moles of oxygen because delta H for oxygen is 0
O, Oxygen….what is its oxidation number?
@Nick
That would be “Fe” which is iron
Jason/Andrew:
The answer for number 12 is actually c) -1212.2.
You have to subtract the total molar enthalpies of fomation of the reactants from the total molar enthalpies of formation of the products…
This is hard to explain in words, but basically it would be like:
((2x(130.5))+(6x(-285.9)))-((2x(-46.2))+(2x(-74.9)))= -1212.2 kJ.
I’m not sure why we disregard the 3 moles of oxygen in the reactants…can someone explain why that is?
Is work equal to 0 when there is constant pressure or constant volume?
Chris –
Remember, L = n – 1 (L can be no larger than n – 1). So for the shell n = 1, L must be 0. Therefore, only one subshell is possible, which is the s subshell, and an orbital in that subshell is the s orbital. When n = 2, L can be 0 or 1. Since two values of L are possible, there are two subshells for n = 2, but L = 1 represents the p orbital. This pattern continues, and as L goes up one in value with n, so does the subshell. So L = 2 = d subshell, L = 3 = f subshell.
Does this help? It makes more sense when you include m sub L, the magnetic quantum number. M ranges from +L to -L so when n = 2, L = 1, and the values can be -1, 0, and +1 which represent 3 orbitals. P is the only subshell with 3 orbitals (s has 1, d has 5). I hope I didn’t just confuse you more!
Fe, Iron.
what is the element with the electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d6
bonus point
Hi, can someone explain to me how to do number 17 on the practice exam? I still do not know the steps in solving for the answer.
Thank You!
is question #24 on the practice test right? and if so how do you figure it out?
Chris,
to figure what orbital when l=#: s=0 p=1 d=2 f=3 so l=2 would be d
When L =# how do you figure out what orbital it is like s,p,d or f
Andrew, i think of bomb calorimeters almost the same as coffee cup, just instead of something hot being dropped in it, it is something being burned inside of it. By measuring how much the water changed and the calorimeter changed in temperature along with their respective heat capacities, you can find out how much energy the substance that was burned contained. Hope this helps
*Sm+3
Derek Melzar:
Sm’s electron configuration is: [Xe] 6s2 4f3
For Andy Ng’s Question on Hess’s Law:
The person that answered it didn’t explain that Hess’s Law dictates that each electron box needs to be filled with one electron (same spin) before they get completely filled up. Example:
You can’t do this: [up] [up] [up/down] [up/down]
It should be: [up/down] [up/down] [up] [up]
I know this is sort of last minute but if anyone could offer an explanation to 5.6 MAS – bomb calorimetry, that would be nice… I can more or less do the problems, but I’m not really sure of the concept behind it… Thanks!
Jason,
Think no more of it than just a simple addition problem…
NH3 2(-46.2) -92.4
CH4 2(-74.9) -149.8
HCN 2(130.5) 261
H2O 6(-285.9) -1715.4
=-1696.6 (d)
Dear Robert,
There was a practice test given out at the SI session, but other than that I’m pretty sure the only practice exam is the one given to us on this site, however it’s from a different teacher and semester, so the topics vary a bit.
Hope that helps
I need the extra point
Hey,
I was just doing a quick review and I’m still having difficulty doing number 12 on the practice exam?
Thanks
Chris,
I think by “j” you mean specific heat capacity (Csp). If that is the case, then yes– water at different phases has different specific heat capacities. At liquid it is 4.184 j/g*K, at gas it is 1.97 j/g*K, and at solid (or ice) is it 2.06 J/g*K.
I believe the same is true (that different states of a substance have different heat capacities) for other substances as well, or at least some of them.
are there any other practice tests available other than the ones given out at the SI session and online?
Debra, p is a subshell and there are orbitals in the subshell which =(2l+1) so the subshell p would have 3 orbitals.
does the j of water change per phase. Like water in liquid form is 4.184, is ice also 4.184 or is it differnt?
GOOD LUCK ON THE EXAM EVERYONE
Debra, I think there are 3 subshells/shapes..
nevermind lol
Hi Debra,
There are actually 3 subshells in the p orbital !
Hey can anyone tell me how many subshells fit on the p orbital?
Can anyone help me to find the electron configuration of Sm^3+^ please!!
So much appreciated!
@ Justin…
If you use the equation delta E= (Ef-Ei), the answer would be 2.14 x10^-18 kJ/mol. I looked high and low for how to convert kJ/mol in Hz but I couldn’t find anything. Also this table — http://mccammon.ucsd.edu/~dzhang/energy-unit-conv-table.html — shows an empty space for that particular conversion, which leads me to believe that it can’t be done….
But please let me know if I’m wrong!
Justin, delta E = 2.13E-18 and since E=hv, v=3.22E15!!! So, the answer is:
b) 3.22E15 !!!!
Well Sarah, the quantum number are n, l, ml, ms. So n should be greater than 1, I is found by (N-1), ml is -/+(L), and ms can only be -/+ 1/2
for example,
3,0,1 – invalid
3,2,0 – valid
It’s elementary my dear Sarah
Justin, you need to use two eqautions to find this, they are all right next to each other on the test suppliment sheet. The first is to find the change in energy, so delta E = -Rhc(1/7^2-1/1^2) Then use E in the 2nd equation down to get v, or frequency. E=hv. Note that h, R, and c are all constants also given on that sheet.
Hope this helps!
Ben
Hey another question!
How do you find out if a quantum number is invalid?
n is energy level
L is the shape of the orbital
mL is the spatial orientation
ms is the spin
I know what the quantum numbers are, n, L, mL, ms, but what are they?
Hey does anyone know why only copper and chromium are more stable with a half full shell?
here is a question:
According to Bohr theory , when an electron moves from n = 1 to n = 7 a photon of what frequency in Hz is absorbed?
a) 2.14 x 10^8
b) 3.22 x 10^15
c) 410
d) 4.89 x 10^10
Can anyone help me through the process of solving study question 71 in chapter 5?
Thanks!
To Alyssa’s question…
I think the interchapters are the ones with the two experiments…the Millikan Oil Drop and the Cathode Ray Tube, and also he said that some of Ch. 21 will also be on there.
Alyssa, for #8… the heat capacity is the specific heat capacity divided by the mass, or 179.4J/oC/345g=(b) 0.520 J/goC. hope this helps
There is a link in the blog so you can look at an example of this exam’s information page so as long as you know when to apply it you should be all set.
Metals with unpaired electrons are paramagnetic. Not sure about the other half of the question
“Diamagnetic metals have a very weak and negative susceptibility to magnetic fields. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Diamagnetic materials are solids with all paired electron resulting in no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron orbits under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.
Paramagnetic metals have a small and positive susceptibility to magnetic fields. These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron orbits caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.
Ferromagnetic materials have a large and positive susceptibility to an external magnetic field. They exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains”(http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Physics/MagneticMatls.htm).
how do i do #8?