CHEMISTRY COMMUNITY

Posted: **Fri Oct 26, 2018 9:58 am**

Eventually, my posts always get lost.
The title says "Garlic Bread" so that you can search "Garlic Bread" and easily find this post!

This is Lyndon Bui, a UA, and I have created many helpful practice problems for your midterm. The questions are modeled after previous test and exam questions, but it may not have everything you need to know. This covers material for the midterm.

Please complete these problems before my review session. My review session this quarter (F18) will be Saturday, Nov3 in CS24: 3-6PM I will go over each problem in detail during my review session. The answers, but not full solutions or work, will be posted here at this post after the review session. Any errors in the test will be posted below as well.

Printed copies will NOT be provided at the review session.

Link to Download Problems: Problems no longer available. Quarter has ended. See Course website and syllabus for most up-to-date material.

***This is not indicative of the structure, length, or format of the actual midterm. Treat these as extra practice problems. ***

Errors: None Found Yet (I'm too good)

Link for Answers: No Longer Available

Happy Studying and Good Luck!
-Lyndon Bui, UA

Posted: **Fri Oct 26, 2018 10:00 am**

Thank you so much for the help!

Posted: **Sun Oct 28, 2018 2:15 pm**

Thank you so much! Are you going to post more midterm practice worksheets? Where can we get more practice problems from?

LYNDON: I will not be posting more practice worksheets. I do hand out paper copies of practice at my normal session workshops. However, you can only get previous ones from students who attended.

Posted: **Sun Oct 28, 2018 3:53 pm**

Thank you!

Posted: **Sun Oct 28, 2018 3:55 pm**

Thank you so much for the extra practice!!!!

Posted: **Mon Oct 29, 2018 4:33 pm**

Hi, is the review session in Young Hall where our lectures are or is it CS24?

Posted: **Mon Oct 29, 2018 6:58 pm**

Chan Mee Lee wrote:Hi, is the review session in Young Hall where our lectures are or is it CS24?

Hey, the session is in CS24, which is the same place where the 2 pm lecture is held. If yours is in CS50 it will be right next door!

Posted: **Mon Oct 29, 2018 9:57 pm**

Is this the only review session you will be having?

Lyndon: I have my regular UA sessions 6-8pm Tuesday night in Covel 210 where I review material from each week. This Saturday is my only midterm review session, but it is 3 hours and should be plenty!

Posted: **Tue Oct 30, 2018 2:44 pm**

Does anyone know how to do question 13 a and b? I though I could do 13a but thirteen b said that what she did (which is what I did) was wrong.

Posted: **Thu Nov 01, 2018 12:23 pm**

Thanks for the midterm practice problems!

Posted: **Thu Nov 01, 2018 10:03 pm**

Can anyone help me with 9b? It asks for the electron configuration of Chromium. I got an answer of [Ar]4s2 3d4 but I looked it up online and it says that its' actually [Ar]4s1 3d5. Why is one of the 4s electrons actually in the 3d orbital? I looked it up and it mentioned something about stability? So does this always happen with elements in the 3d block or is it an exception for chromium? Thanks

Posted: **Thu Nov 01, 2018 11:07 pm**

Chromium is an exception, because it will be more stable to have half of the d block filled and only one electron in the s block, [Ar]3d5 4s1 instead of [Ar]3d4 4s2. Copper is also an exception, and its electron configuration will be [Ar]3d10 4s1. From its placement on the periodic table, you would expect it to be [Ar]3d9 4s2 (like for Chromium you would expect [Ar]3d4 4s2) but it is more stable having an entirely full d block and one electron in its valence s shell.

Posted: **Fri Nov 02, 2018 12:54 am**

Thanks for the practice midterm problems!

Posted: **Fri Nov 02, 2018 2:23 pm**

Thank you very much!! This is so helpful

Posted: **Fri Nov 02, 2018 2:30 pm**

can anyone please help me with number 2? I got to dividing the molar mass of 667 by 30.024 and got 22? but i think its wrong because its such a large number. Thus my molecular formula so far is C22H44O22 ?? pls help

Posted: **Fri Nov 02, 2018 2:42 pm**

I got a molar mass of 30g/mol as well. I believe that the calculations are right.

Posted: **Fri Nov 02, 2018 5:00 pm**

For number 6 part A, we need to use the DeBroglie wavelength equation right? I used it and got 1.5547 x 10^/34 m but im not sure if I used the right equation? and I also think I messed up the units so pls correct me if needed

Posted: **Fri Nov 02, 2018 5:06 pm**

Hello Everyone,

This is Lyndon! Regarding questions posted about the practice problems, don't worry! It will all be addressed tomorrow CS24 3-6pm (Saturday Nov 3).
After the review session, feel free to come talk to me to clarify anything you're not comfortable with!

See you tomorrow!

Posted: **Sat Nov 03, 2018 8:16 am**

Hello, I was just wondering if you will still be posting solutions online after the review session? On the practice midterm I know it says there won't be detailed solutions but the answers will still be posted? Thanks!

Posted: **Sat Nov 03, 2018 12:04 pm**

Thank you so much!

Posted: **Sat Nov 03, 2018 1:09 pm**

thank you!

Posted: **Sat Nov 03, 2018 4:19 pm**

Karla_Ocampo 4E wrote:can anyone please help me with number 2? I got to dividing the molar mass of 667 by 30.024 and got 22? but i think its wrong because its such a large number. Thus my molecular formula so far is C22H44O22 ?? pls help

Same... :(

Posted: **Sat Nov 03, 2018 4:27 pm**

DavidEcheverri3J wrote:
Karla_Ocampo 4E wrote:can anyone please help me with number 2? I got to dividing the molar mass of 667 by 30.024 and got 22? but i think its wrong because its such a large number. Thus my molecular formula so far is C22H44O22 ?? pls help

Same... :(

Actually, I just solved it. ill walk you through it:

first you want ti divide the percentages by the molar mass. (C - 3.6 ; H - 6.3 ; O - 3.15). Next, you want to divide by the smallest number (3.15). (C - 1.14 ; H - 2; O - 1). Now, you want to multiply all the numbers until the one with a decimal becomes a whole number. So multiply by 7. You will be left with C8H14O7. After this, you find its molar mass (approx. 222 g/mol), and divide 667 by it. Then you multiply the empirical by 667/222 to get C24H42O21. Ah-hah!

Good luck!

Posted: **Sat Nov 03, 2018 6:25 pm**

For #6, how are you supposed to convert the molar mass of GarBreadium to kg in order to use the de broglie equation?

Posted: **Sat Nov 03, 2018 6:45 pm**

Daisylookinland4B wrote:For #6, how are you supposed to convert the molar mass of GarBreadium to kg in order to use the de broglie equation?

You change it to kg you divide the molar mass of GarBreadium by Avogadro's constant, 6.02*10^23 which gives you 5.244*10^-24g then you divide again by 1000 to get the value in kg.

Posted: **Sat Nov 03, 2018 8:18 pm**

Thank you so much! This was really helpful!

Posted: **Sat Nov 03, 2018 9:12 pm**

Hello! Does anyone know where we can find the solutions sheet. Thank you!

Posted: **Sat Nov 03, 2018 9:39 pm**

hello! does anyone know how soon the solutions sheet will be up?

Posted: **Sat Nov 03, 2018 9:41 pm**

Will someone explain 8b? I know that we need to use all of the information given in the question but I am not sure how wavelength corresponds to work functions. Thank you!

Posted: **Sat Nov 03, 2018 10:11 pm**

Thank you Lyndon for pushing through today's review session. Feel better and good luck on your own midterms :)

Posted: **Sat Nov 03, 2018 10:29 pm**

Thank you so much for all your help today Lyndon!! You are incredibly generous with your time! Good luck on your midterms and get well soon!!

Posted: **Sat Nov 03, 2018 10:29 pm**

Elle_Mendelson_4I wrote:Will someone explain 8b? I know that we need to use all of the information given in the question but I am not sure how wavelength corresponds to work functions. Thank you!

So the main equation you need is Eph = (work function of metal) - Ek (energy of electron ejected) This is not correct. See post below.

Everything has a wavelength and this can give you the energy for the work function equation.

The equation that you need to find Ek is ---> λ = h / m v

Input the wavelength into that equation 1.10 nm (must be in meters) and find the Velocity of the ejected electron.

So the equation EK = 1/2 m v^2 is used to find the energy of the ejected electron, Input the mass of electron and the velocity to find Ek

So now use Eph = (work function of metal) - Ek (energy of electron ejected)

Eph is the energy of the laser pointer.

Use E = h ν to to find frequency.

Frequency will be in hertz

Don't get mix up v and ν. One means frequency and the other means velocity. Idk you just have to memorize which is which.

Posted: **Sat Nov 03, 2018 10:43 pm**

Javier_Ochoa_DIS_3J wrote:
So the main equation you need is Eph = (work function of metal) - Ek (energy of electron ejected)

The above post is not quite correct.

The correct version is: Energy of photon - work function (threshold energy) = Kinetic energy (of ejected electron)

Posted: **Sat Nov 03, 2018 10:45 pm**

Thank you!

Posted: **Sat Nov 03, 2018 10:55 pm**

When will you post the answers?

Posted: **Sat Nov 03, 2018 10:57 pm**

Answers are now posted. Scroll up.

Posted: **Sat Nov 03, 2018 11:35 pm**

Hello!

For question 3, I keep getting 37.2 g NO for my answer, not 31.7g NO. Is this an error in the solutions, or did I do something wrong?

Equation: 4NH3 + 5O2 --> 4NO + 6H2O

(21.1 g NH3/17.04 g/mol NH3) x (4 mol NO/4 mol NH3) (30.01 g/mol NO) = 37.16 g NO, rounded to 37.2 g NO.

Please let me know if something is wrong with my stoichiometry. Thanks for putting this review worksheet together! Garlic bread was good review AND a good meme :)

Posted: **Sat Nov 03, 2018 11:39 pm**

Katie_Duong_3B wrote:Hello!

For question 3, I keep getting 37.2 g NO for my answer, not 31.7g NO. Is this an error in the solutions, or did I do something wrong?

Equation: 4NH3 + 5O2 --> 4NO + 6H2O

(21.1 g NH3/17.04 g/mol NH3) x (4 mol NO/4 mol NH3) (30.01 g/mol NO) = 37.16 g NO, rounded to 37.2 g NO.

Please let me know if something is wrong with my stoichiometry. Thanks for putting this review worksheet together! Garlic bread was good review AND a good meme :)

I just realized that I used the wrong limiting reagent. O2 is the limiting reagent, not NH3. I have 31.7 g NO now.

Posted: **Sun Nov 04, 2018 1:30 am**

Katie_Duong_3B wrote:
I just realized that I used the wrong limiting reagent. O2 is the limiting reagent, not NH3. I have 31.7 g NO now.

Good! Remember that the amount of product formed is determined by the limiting reactant. Therefore, the reactant that produces less product is limiting and will be used to determine theoretical yield.

Posted: **Sun Nov 04, 2018 1:27 am**

Could someone explain 13c to me?

Posted: **Sun Nov 04, 2018 9:03 am**

sonalivij wrote:Could someone explain 13c to me?

-13c is a Heisenberg Uncertainty Principle problem. The equation for this problem is ΔPΔX=>h/4pi. For this equation we need the mass, the velocity(these two values multipled together are ΔP), and we need ultimately to find ΔX.
-We are given the mass which is 2.8 grams so we have to divide the number by 1000 in order to obtain the SI unit of Kg, and you will get .0028 Kg. This number will be used for our M in ΔP.
-We are also given the speed and in order to find ΔV we have to multiply .34 and 2 together because that is the total uncertainty of the speed, negatively and positively. So our ΔV will be .68 m/s.
-With all of these values that we found we only have one unknown value which is ΔX, so we simply just plug the values into the equation. (.0028 Kg)*(.68 m/s)*(ΔX)>=h/4pi
Using some simple algebra you will get ΔX>=2.8*10^-32 meters. Remember to do significant figures too!

Posted: **Sun Nov 04, 2018 10:06 am**

The review session on Saturday that went over this practice midterm really helped prepare me for the exam, thank you.

Posted: **Sun Nov 04, 2018 10:49 am**

It was so crowded but I appreciate the practice problems and review session so much!!

Posted: **Sun Nov 04, 2018 10:58 am**

Hello, for number 6a I keep getting 9.36X10 ^-46 instead of 9.36*10^-11 will someone explain what I am doing wrong? I have looked at another students work and we did the exact same thing but I got a different answer. Please help!!
Thank you!

Posted: **Sun Nov 04, 2018 11:30 am**

Posted: **Sun Nov 04, 2018 11:31 am**

Hi can anyone explain 8a? I don't know how to correctly word my answers...

Posted: **Sun Nov 04, 2018 1:11 pm**

Daisylookinland4B wrote:For #6, how are you supposed to convert the molar mass of GarBreadium to kg in order to use the de broglie equation?

First you must find the weight of a single atom of GarBreadium -> (3.257g/1mol) x (1mol/6.02x20^23atoms) = 5.24x10^-24 g/atoms. Then you multiply by 1000 (1000g in 1 kg). Lastly you plug that in as your m in the de Broglie equation (lambda = h/mv). Ah-hah!

Hope this helped!

Good luck!

Posted: **Sun Nov 04, 2018 1:21 pm**

For problem 9b, the electron configuration of Chromium is [Ar]3d5 4s1. I thought that the 4s orbital was filled before the 3d orbital. I know Copper and some other elements also follow this trend. Why do these elements only fill one of the two electrons in the 4s or 5s orbitals?

Posted: **Sun Nov 04, 2018 1:55 pm**

Thank you so much for the review session! It really helped clarify so many topics for me.

Posted: **Sun Nov 04, 2018 1:56 pm**

He really is too good. THE MVP!!!

Posted: **Sun Nov 04, 2018 1:59 pm**

@Kevin Tang, This is because Chromium strives to fulfill the half-filled orbital energy, making it more stable. So instead of 3p4 4s2, it is 3p5 4s1. This is (I think) the only exception when it comes to electron configuration. (as well as all elements in the same group.)

Posted: **Sun Nov 04, 2018 1:59 pm**

@Kevin Tang, This is because Chromium strives to fulfill the half-filled orbital energy, making it more stable. So instead of 3p4 4s2, it is 3p5 4s1. This is (I think) the only exception when it comes to electron configuration. (as well as all elements in the same group.)

Posted: **Sun Nov 04, 2018 2:17 pm**

For question 4, the answer is given with only 2 sig figs (the answer is 9.9 mL of 9.9x10^-3 L, even though the question uses 3 sig figs. The question is: What volume of 0.0380 M KMnO4 is needed to prepare 250 mL of 1.50 x 10-3 M KMnO4?
(Hint: In these types of problems, the volume containing solute is diluted with water to the final
volume, in this case 250mL). Would 3 sig figs as my answer be acceptable?

Posted: **Sun Nov 04, 2018 2:25 pm**

Why do we multiply by 7 in number 1?

Posted: **Sun Nov 04, 2018 2:37 pm**

juleschang16 wrote:Would 3 sig figs as my answer be acceptable?

250 mL is only 2 sig figs. The number of sig figs you use should depend on the value that has the least sig figs. Therefore, 2 sig figs is correct .

Posted: **Sun Nov 04, 2018 2:40 pm**

Thank you so much for your time and help, Lyndon!

Posted: **Sun Nov 04, 2018 2:42 pm**

For number 5, I keep getting .127 M after setting the problem up like (.211 M)(.150 L) = M2(.250 L). But in my notes, I believe Lyndon got 1.69 x 10^-2. I was wondering if anyone knows what I am doing wrong. Thanks!

Posted: **Sun Nov 04, 2018 3:43 pm**

Bijan Mehdizadeh 1E wrote:For number 5, I keep getting .127 M after setting the problem up like (.211 M)(.150 L) = M2(.250 L). But in my notes, I believe Lyndon got 1.69 x 10^-2. I was wondering if anyone knows what I am doing wrong. Thanks!

Hi!
Since the problem stated that only 20ml was removed to the 2nd flask, you put 0.02 as your L1 instead of 0.150 L. Now you should get the right answer.

Posted: **Sun Nov 04, 2018 4:01 pm**

How is that for the Lewis structure of N2O (question 12c), N can have more than 3 bonds?

Posted: **Sun Nov 04, 2018 5:37 pm**

For 8b, why can't you use Ek = hc/λ to find the kinetic energy then proceed to solve for frequency using this value + the work function all over h?

Posted: **Sun Nov 04, 2018 5:47 pm**

Sydney Takeda 1G wrote:For 8b, why can't you use Ek = hc/λ to find the kinetic energy then proceed to solve for frequency using this value + the work function all over h?

The variable c incorporates the speed of light into the equation, but ejected electrons cannot travel at the speed of light. Therefore this cannot work. You can only use the speed of light when working with photons.

Posted: **Sun Nov 04, 2018 6:07 pm**

Sara Sadrolsadat 1G wrote:How is that for the Lewis structure of N2O (question 12c), N can have more than 3 bonds?

Nitrogen must have an octet. We will often see Nitrogen with 3 bonds and one lone pair so that it has an octet and it has 0 formal charge, but sometimes it can have 4 bonds and no lone pairs, giving it a +1 formal charge.

Posted: **Sun Nov 04, 2018 6:09 pm**

Sydney Takeda 1G wrote:For 8b, why can't you use Ek = hc/λ to find the kinetic energy then proceed to solve for frequency using this value + the work function all over h?

The formula you have presented is NOT for kinetic energy. That formula is only for energy of photons (light). Since the problem gave only the wavelength of the electron, and since an electron is not a photon, your method will not work.

Posted: **Sun Nov 04, 2018 6:32 pm**

Can anyone explain the reasoning for why 6d is true.

Posted: **Sun Nov 04, 2018 6:43 pm**

Samantha Chang wrote:
sonalivij wrote:Could someone explain 13c to me?

-13c is a Heisenberg Uncertainty Principle problem. The equation for this problem is ΔPΔX=>h/4pi. For this equation we need the mass, the velocity(these two values multipled together are ΔP), and we need ultimately to find ΔX.
-We are given the mass which is 2.8 grams so we have to divide the number by 1000 in order to obtain the SI unit of Kg, and you will get .0028 Kg. This number will be used for our M in ΔP.
-We are also given the speed and in order to find ΔV we have to multiply .34 and 2 together because that is the total uncertainty of the speed, negatively and positively. So our ΔV will be .68 m/s.
-With all of these values that we found we only have one unknown value which is ΔX, so we simply just plug the values into the equation. (.0028 Kg)*(.68 m/s)*(ΔX)>=h/4pi
Using some simple algebra you will get ΔX>=2.8*10^-32 meters. Remember to do significant figures too!

In the problem it says that the speed is 373.34 +/- 0.34 m/s. Where are you getting the 2 from?

Posted: **Sun Nov 04, 2018 6:47 pm**

For 8B I understand that we need to use E(photon) - work function = kinetic energy. We are given the wavelength and work function. However, I don't know where to use the wavelength since E(photon)=h*frequency and we have the given work function and Kinetic energy= 1/2 Mass* Velocity(unknown). Can someone please explain how to solve this? Thank you!

Posted: **Sun Nov 04, 2018 6:52 pm**

Thank you so much Lyndon!!!

Posted: **Sun Nov 04, 2018 6:55 pm**

For number 11, part b why would delta E be a negative value? If the problem is asking for the change in energy from n = 6 to n = 4, and the formula is
E = -hR(1/n1^2 - 1/n2^2), the energy released from n = 6 to n = 4 equals 7.5689 x 10^-20 J. So, why is the answer negative?

Posted: **Sun Nov 04, 2018 7:31 pm**

Hilda Sauceda 3C wrote:Can anyone explain the reasoning for why 6d is true.

As discussed at the review session, a particle with a debroglie wavelength larger than 10^-18 exhibits measurable wavelike properties

Posted: **Sun Nov 04, 2018 7:33 pm**

juleschang16 wrote:In the problem it says that the speed is 373.34 +/- 0.34 m/s. Where are you getting the 2 from?

Uncertainty is the range of values that we are unsure about. This means taking the highest possible speed, then subtracting the lowest possible speed. This will give you the uncertainty. You will find that doing so is the same as multiplying 0.34 by 2.

Posted: **Sun Nov 04, 2018 7:34 pm**

Anand Narayan 1G wrote:For number 11, part b why would delta E be a negative value? If the problem is asking for the change in energy from n = 6 to n = 4, and the formula is
E = -hR(1/n1^2 - 1/n2^2), the energy released from n = 6 to n = 4 equals 7.5689 x 10^-20 J. So, why is the answer negative?

Delta E is the change in energy. When you go from 6 to 4, you are going from a higher energy state to a lower energy state. When going from high to low, the change is negative. The electron therefore lost energy. The energy lost takes the form of a photon, and photons always have positive energy.

Posted: **Sun Nov 04, 2018 7:36 pm**

Anand Narayan 1G wrote:For number 11, part b why would delta E be a negative value? If the problem is asking for the change in energy from n = 6 to n = 4, and the formula is
E = -hR(1/n1^2 - 1/n2^2), the energy released from n = 6 to n = 4 equals 7.5689 x 10^-20 J. So, why is the answer negative?

Remember that the change in energy is that of the final energy (n=4) minus the initial energy (n=6). This should give you a negative change in energy. Hope this helps!

Posted: **Sun Nov 04, 2018 7:48 pm**

What determines the energy of a Lewis structure? Formal charges of all the atoms added together?

Posted: **Sun Nov 04, 2018 8:14 pm**

Yvonne Du wrote:
Bijan Mehdizadeh 1E wrote:For number 5, I keep getting .127 M after setting the problem up like (.211 M)(.150 L) = M2(.250 L). But in my notes, I believe Lyndon got 1.69 x 10^-2. I was wondering if anyone knows what I am doing wrong. Thanks!

Hi!
Since the problem stated that only 20ml was removed to the 2nd flask, you put 0.02 as your L1 instead of 0.150 L. Now you should get the right answer.

I'm confused as to why .02 L would be the initial volume, if when we calculate the initial molarity (mol/L of KMnO4) in the compound we use .150 L, rather than .020L?

Posted: **Sun Nov 04, 2018 8:15 pm**

does anyone know how to solve 8b? i am having trouble with it.

Posted: **Sun Nov 04, 2018 8:27 pm**

Thank you so much, Lyndon. These are going to help me a lot as I'm last minute cramming.

Posted: **Sun Nov 04, 2018 8:33 pm**

juleschang16 wrote:
Samantha Chang wrote:
sonalivij wrote:Could someone explain 13c to me?

-13c is a Heisenberg Uncertainty Principle problem. The equation for this problem is ΔPΔX=>h/4pi. For this equation we need the mass, the velocity(these two values multipled together are ΔP), and we need ultimately to find ΔX.
-We are given the mass which is 2.8 grams so we have to divide the number by 1000 in order to obtain the SI unit of Kg, and you will get .0028 Kg. This number will be used for our M in ΔP.
-We are also given the speed and in order to find ΔV we have to multiply .34 and 2 together because that is the total uncertainty of the speed, negatively and positively. So our ΔV will be .68 m/s.
-With all of these values that we found we only have one unknown value which is ΔX, so we simply just plug the values into the equation. (.0028 Kg)*(.68 m/s)*(ΔX)>=h/4pi
Using some simple algebra you will get ΔX>=2.8*10^-32 meters. Remember to do significant figures too!

In the problem it says that the speed is 373.34 +/- 0.34 m/s. Where are you getting the 2 from?

You have to take the +/- 0.34 m/s and multiply by two to get the change in velocity. For example, if the velocity is (100 m/s +/- 1 m/s) then delta velocity is 2 m/s. That is, we took 1 * 2. You can also see it as 99-101 m/s which is also a delta velocity of 2 m/s. Hope this helps!

Posted: **Sun Nov 04, 2018 8:39 pm**

For question 12c, why is it better to have a triple bond with the two nitrogen atoms instead of double bonds on nitrogen and oxygen? The charge is both the same.

Posted: **Sun Nov 04, 2018 8:54 pm**

For question 13A, can someone please explain why the answer is 2.681x10^-26? I got the 2.681, but for some reason I got x10^-23.

Posted: **Sun Nov 04, 2018 8:59 pm**

Olivia L 3D wrote:For question 12c, why is it better to have a triple bond with the two nitrogen atoms instead of double bonds on nitrogen and oxygen? The charge is both the same.

It is better to have a triple bond as opposed to a double bond because Oxygen is the more electronegative atom.

Posted: **Sun Nov 04, 2018 9:04 pm**

Thank you so much this was very helpful!

Posted: **Sun Nov 04, 2018 9:06 pm**

Thank you for this review it has really been helping me understand things better!

Posted: **Sun Nov 04, 2018 9:09 pm**

I don't understand why in order of increasing ionization energy, it goes C O N F. I thought it would be C N O F. Anyone know why?

Posted: **Sun Nov 04, 2018 9:30 pm**

I don't understand why in order of increasing ionization energy, it goes C O N F. I thought it would be C N O F. Anyone know why?

Nitrogen is an exception (so is phosphorus) to the trend in that it has a half-filled subshell, 2p^3, which makes it harder for an electron to be removed.

Posted: **Sun Nov 04, 2018 9:32 pm**

For question 13A, can someone please explain why the answer is 2.681x10^-26? I got the 2.681, but for some reason I got x10^-23.

Did you remember to convert the mass into kg?

Posted: **Sun Nov 04, 2018 9:57 pm**

Thank you for these practice problems!

Posted: **Sun Nov 04, 2018 10:19 pm**

How do you go about solving the first problem?

Posted: **Sun Nov 04, 2018 10:39 pm**

I tried the first problem and got it very very wrong. How would I solve it correctly?

Posted: **Sun Nov 04, 2018 10:46 pm**

Lina Petrossian 3D wrote:How do you go about solving the first problem?

https://wps.prenhall.com/wps/media/obje ... h0305.html

See for exact solution

Posted: **Sun Nov 04, 2018 10:55 pm**

Thank you so much this is really helpful!!

Posted: **Sun Nov 04, 2018 11:13 pm**

can someone further explain number 6c? I'm not sure how the wavelengths correspond with lower frequency

Posted: **Sun Nov 04, 2018 11:30 pm**

I wasn't able to make it to the review session, can someone who did go post their notes? It'd be super helpful!

Posted: **Sun Nov 04, 2018 11:33 pm**

9c) I know the answer is sodium but why don't we put the positive 1 charge :)

Posted: **Sun Nov 04, 2018 11:41 pm**

Thank you for the extra practice!

Posted: **Sun Nov 04, 2018 11:55 pm**

Question 6b is asking if a Helium atom or a "new" element GarBreadium have longer wavelengths. The answer key says the GarBreadium will have a longer wavelength but isn't it He because it is less heavy, therefore the denominator in de Broglie's equation $\lambda =h/m\cdot v$ be smaller, corresponding to a bigger wavelength ? Is this a typo ?

Posted: **Mon Nov 05, 2018 12:19 am**

Could someone explain 8b to me. Please tell me where I went wrong here. I found:
wavelength: 1.10 * 10^-9m
P=7.29 * 10^-43
V=8.00 * 10^-13
Work Function= 2.50 * 10^-19 J/atom
Kinetic Energy= 7.00 * 10^-55
Final Frequency= 5.94 * 10^-10

Posted: **Mon Nov 05, 2018 12:23 am**

Jk final frequency I get 3.77 * 10^14. Why is this off

Posted: **Mon Nov 05, 2018 12:34 am**

304981930 wrote:Jk final frequency I get 3.77 * 10^14. Why is this off

This is because you didn't add the KE with the threshold value. You only did 2.5x^-19 =hv
v=3.8x10^14
If you add the KE, you should get the answer 6.8*10^14 Hz

KE= 1/2mv^2

Posted: **Mon Nov 05, 2018 12:36 am**

404817859 wrote:can someone further explain number 6c? I'm not sure how the wavelengths correspond with lower frequency

Refer to the equation c=(lambda)(v), you can rewrite this as c/(lambda)=v or c/v=(lambda). You can see they are inversely related; when wavelength increases, the frequency decrease and vice versa.

Posted: **Mon Nov 05, 2018 12:37 am**

Imelda Mena 3I wrote:Question 6b is asking if a Helium atom or a "new" element GarBreadium have longer wavelengths. The answer key says the GarBreadium will have a longer wavelength but isn't it He because it is less heavy, therefore the denominator in de Broglie's equation $\lambda =h/m\cdot v$ be smaller, corresponding to a bigger wavelength ? Is this a typo ?

Helium actually weights more because it is roughly 4g/mol whereas GarBreadium is 3.157g/mol. I made this mistake too, (you have to add the neutrons and protons) 2+2=4

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