DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

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Miguel Velasco 2J
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Ni3+ electron configuration

Postby Miguel Velasco 2J » Mon Nov 06, 2017 8:26 pm

For Ni3+, why is the electron configuration [Ar]3d7 as opposed to [Ar]4s^2 3d5

Chem_Mod
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Re: Lyndon's and Michael's Practice Midterm

Postby Chem_Mod » Mon Nov 06, 2017 8:59 pm

Here are some of the answers. Full solutions are not available to be posted. The practice midterm is no longer available for download.

Answers removed due to changes in course syllabus

Chem_Mod
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Re: Ni3+ electron configuration

Postby Chem_Mod » Mon Nov 06, 2017 10:36 pm

Miguel Velasco 3K wrote:For Ni3+, why is the electron configuration [Ar]3d7 as opposed to [Ar]4s^2 3d5


The 3d subshell is higher energy than the 4s subshell until the 3d subshell has an electron. Once the 3d subshell has an electron, it is now lower energy than the 4s subshell. For nickel, the neutral configuration is [Ar] 3d8 4s2. Since the 4s subshell is higher in energy, electrons are removed from there first. Therefore, Ni3+ has the configuration [Ar] 3d7.

Maria1E
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Re: DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

Postby Maria1E » Tue Nov 07, 2017 9:38 am

For number 1, how is the answer C24H42O21? Once I divided by the molar masses, I got 3.5967 moles of C, 6.2996 moles of H, and 3.15 moles of O. I divided all three by 3.15 to get the ratio 1.14 C, 2 H, and 1 O. Then I multiplied each number by 7 to get whole numbers and got C7H14O7 for the empirical formula. I found the molar mass of that to be 210.189 g/mol. Then I divided the molecular molar mass by the empirical molar mass and got 3, so I multiplied the empirical formula by 3 to get the answer C21H42O21. Why are there 24 C's in the molecular formula?

Alex Kashou
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Re: DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

Postby Alex Kashou » Tue Nov 07, 2017 10:14 am

Maria1E wrote:For number 1, how is the answer C24H42O21? Once I divided by the molar masses, I got 3.5967 moles of C, 6.2996 moles of H, and 3.15 moles of O. I divided all three by 3.15 to get the ratio 1.14 C, 2 H, and 1 O. Then I multiplied each number by 7 to get whole numbers and got C7H14O7 for the empirical formula. I found the molar mass of that to be 210.189 g/mol. Then I divided the molecular molar mass by the empirical molar mass and got 3, so I multiplied the empirical formula by 3 to get the answer C21H42O21. Why are there 24 C's in the molecular formula?


1.14x7=8. You just made a math error. So the empirical is C8H14O7. Which will give you the right answer when you multiply by 3.

Naama 1A
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Re: DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

Postby Naama 1A » Tue Nov 07, 2017 9:04 pm

Can someone help me out with problem 4b? I tried using the wavelength of the ejected electron to find velocity through the De Broglie equation. Then I found the total KE, added it to the work function (after converting from moles) and got the total photon energy. Then I used E=hv to solve for frequency, but I got a different value than the solution posted.

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Re: DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

Postby Chem_Mod » Tue Nov 07, 2017 11:06 pm

Naama 1A wrote:Can someone help me out with problem 4b? I tried using the wavelength of the ejected electron to find velocity through the De Broglie equation. Then I found the total KE, added it to the work function (after converting from moles) and got the total photon energy. Then I used E=hv to solve for frequency, but I got a different value than the solution posted.


You seem to have the right idea. Your options to figure out what went wrong are
1. Attach a picture of your work
2. Compare with a friend who has the solutions from the session.
3. Ask Lyndon directly on Wednesday (details above)

sofiakavanaugh
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Re: DOWNLOAD PRACTICE MIDTERM HERE: Lyndon and Michael's session

Postby sofiakavanaugh » Wed Nov 08, 2017 10:11 pm

Can someone please explain what the structure for 8a would look like and why? (draw the lowest energy lewis structure for (NH2)2CO2 )


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