Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

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Chem_Mod
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Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Chem_Mod » Sat Feb 09, 2019 4:06 pm

The title is called "Hot dog" so that you can search "Hotdog" and easily find this post!

This is your UA, Lyndon Bui, with information on the much anticipated HOTDOG Review Session! As always I am creating a packet of practice problems that stem from past exam questions and course outlines. You are covering some major topics for the midterm this quarter so there is no possible way to cover everything. Definitely use course outlines to make sure you are ready! I am doing my best to cover ALL TOPICS for the midterm.

Please complete these problems before my review session. My review session this quarter (W19) will be Monday, Feb 11, 7-10pm, in Franz 1178.
I will go over each problem in detail during my review session. Due to private concerns/engagements, please do not expect answers to be posted. Any errors in the test will be posted below.

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. ***


[b][u]This contains few questions about acid/base equilibria. I will go over concepts and problem solving for various acid/base type problems in detail at the review session but you should find more acid/base problems for practice.


[i]Errors:
#11 should say M for the units of the solution concentration

Happy Studying and Good Luck!
-Lyndon Bui, UA


Printed copies will NOT be provided at the review session.

AnnaYan_1l
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby AnnaYan_1l » Sun Feb 10, 2019 6:02 pm

For 4C, are we supposed to find/know the enthalpy of formation for the reactant O2, or should that be given?

Thanks!

Venya Vaddi 1L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Venya Vaddi 1L » Sun Feb 10, 2019 6:34 pm

The enthalpy of formation of O2 is zero since it is in its standard state.

gabbym
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby gabbym » Sun Feb 10, 2019 6:45 pm

Thank you for doing this for us! Greatly appreciated :)

AnnaYan_1l
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby AnnaYan_1l » Sun Feb 10, 2019 8:27 pm

Venya Vaddi 1L wrote:The enthalpy of formation of O2 is zero since it is in its standard state.


Thank you!!!

annabel 2A
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby annabel 2A » Sun Feb 10, 2019 10:15 pm

How do you calculate internal energy for #6 without knowing the temperature?

1K Kevin
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby 1K Kevin » Sun Feb 10, 2019 10:23 pm

Thank you thank you for all this :D

Angela Cong 3C
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Angela Cong 3C » Sun Feb 10, 2019 10:30 pm

Yah for number six, im confused as well. Internal energy is dependent on temperature and the work function for a reversible reaction has temperature for one of the unknown variables. Is there a way around this?

Emmaraf 1K
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Emmaraf 1K » Sun Feb 10, 2019 11:00 pm

Temperature is supposed to be given for #6 I believe.
EDIT: Okay the review system I went to did end up giving us the temperature but you can calculate it n your own using PV=nRT where pressure and volume is the pressure and volume of the system after the isochoric step so V=10l and P=10atm
Last edited by Emmaraf 1K on Sun Feb 10, 2019 11:58 pm, edited 1 time in total.

Ethan Baurle 1A
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Ethan Baurle 1A » Sun Feb 10, 2019 11:22 pm

You can solve for temp using PV=nRT. I keep getting a wack answer for this question, however. I'm assuming the overall delta U and delta S will be zero because they are state functions, but I am not getting the right q and w quantities. Can someone who feels confident put down their values so I can compare?

Chem_Mod
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Chem_Mod » Sun Feb 10, 2019 11:30 pm

There is great discussion here! There is no mistake for #6, the temperature does not need to be given.

MichelleTran 1I
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby MichelleTran 1I » Mon Feb 11, 2019 12:01 am

Thanks for posting this again, Lyndon! See you tomorrow at the review session!

Chem_Mod
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Chem_Mod » Mon Feb 11, 2019 11:33 am

See you all 7pm Franz 1178!

Phil Timoteo 1K
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Phil Timoteo 1K » Mon Feb 11, 2019 5:23 pm

Thank you for posting, see you soon.

aisteles1G
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby aisteles1G » Mon Feb 11, 2019 5:54 pm

Hello, Unfortunately I won't be able to attend the review, will answers be posted? If not could someone please upload a pic of the answers after the review?? I would greatly appreciate it!! Thank you!

Desiree1G
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Desiree1G » Mon Feb 11, 2019 6:23 pm

Thank you, I had such a hard time finding it haha!

Jack Hewitt 2H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Jack Hewitt 2H » Mon Feb 11, 2019 9:28 pm

I will not be able to make the review session tonight. Can someone post the answers given?

Megan_Ervin_1F
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Megan_Ervin_1F » Mon Feb 11, 2019 11:02 pm

when will you post the answers to hotdog?

Mya Majewski 1L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Mya Majewski 1L » Tue Feb 12, 2019 1:31 am

For #6, how did we get w=9.119 x 10^3 J?

katherinemurk 2B
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby katherinemurk 2B » Tue Feb 12, 2019 10:30 am

Mya Majewski 1L wrote:For #6, how did we get w=9.119 x 10^3 J?

For the first part it says you preform an isobaric compression to 10L. Isobaric is constant pressure so work is equal to -PdV. So pressure is 1.00 atm and the change of V is 10-100 so -90. This comes to -90 L.atm you then convert this to Joules

ariana_apopei1K
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby ariana_apopei1K » Tue Feb 12, 2019 11:24 am

Megan_Ervin_1F wrote:when will you post the answers to hotdog?

He mentioned that he won't be posting answers this time

705192887
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby 705192887 » Tue Feb 12, 2019 1:50 pm

If he said he isn't gonna post the answers, could someone post them so I can check my answers?

dgerges 4H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby dgerges 4H » Tue Feb 12, 2019 2:32 pm

what was the full answer for w on #6? I got that part of it is 2.33KJ from the w=-nrtln(v2/v1)

dgerges 4H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby dgerges 4H » Tue Feb 12, 2019 2:34 pm

I also got that the other part of w=9.12 from w=-pdv. so i got a total of 11.45kj or 1.145x10^5j for w. is this correct?
Last edited by dgerges 4H on Tue Feb 12, 2019 2:35 pm, edited 1 time in total.

dgerges 4H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby dgerges 4H » Tue Feb 12, 2019 2:35 pm

with a corresponding q=-11.45kj?

MichaelMoreno2G
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby MichaelMoreno2G » Tue Feb 12, 2019 3:11 pm

Can someone please explain why 3E is false?

Jennifer Su 2L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Jennifer Su 2L » Tue Feb 12, 2019 4:37 pm

MMoreno3K wrote:Can someone please explain why 3E is false?


Because heat is required during melting or boiling (phase change transition), temperature of a sample can remain constant even though heat is being supplied.
For example, if you have boiling water, it will remain at 100 C even though heat is being added, until it has enough heat/energy to go through vaporization.

005199302
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby 005199302 » Tue Feb 12, 2019 6:29 pm

for #6, how do we know that change in entropy is zero?

Reva Kakaria 1J
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Reva Kakaria 1J » Tue Feb 12, 2019 7:31 pm

005199302 wrote:for #6, how do we know that change in entropy is zero?


Since the system starts and ends at the same state, the entropy doesn't change.

He whose name cannot be spoken
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby He whose name cannot be spoken » Tue Feb 12, 2019 8:07 pm

Help with question 2c)? I understand that since we're cooling it kc will also drop because it's dependent on temperature, but what is the detailed explanation?

Philipp_V_Dis1K
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Philipp_V_Dis1K » Tue Feb 12, 2019 9:12 pm

can someone explain how to do #5

lukezhang2C
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby lukezhang2C » Tue Feb 12, 2019 10:19 pm

Are there answers to these problems?

Kevin Tang 4L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Kevin Tang 4L » Tue Feb 12, 2019 10:48 pm

Philipp_V_Dis1K wrote:can someone explain how to do #5


For this problem, we have to find the change in entropy, delta S. To do this we want to change the Helium and Krypton to mols.
We get 2.25mol He and 1.49mol Kr. Then we can plug them into the equation DeltaS=nRln(V2/V1).

For the Helium, it is placed in a compartment that is 1/3 the size of Kr, so lets assume Volume that He is placed in is 1L and Volume that Kr is placed in is 3L. So when they are combined the final Volume is 4L.

Now we can finish the DeltaS of He and DeltaS of Kr:

He: DeltaS1= (2.25mol)(8.3145 J*mol-1*K-1)(ln 4L/1L)= Use ur calc im lazy :P
Kr: DeltaS2= (1.49mol)(8.3145 J*mol-1*K-1)(ln 4L/3L)= Use ur calc im lazy :P

However, not only did the volume change, the temperature changed too, so we need to use the equation DeltaS=Cp,m(ln T2/T1) where Cp,m= 3/2(8.3145 J*mol-1*K-1). This can be found on the equation sheet. We need to add the mols of He and Kr together as they are not independent reactions.

DeltaS3= (2.25+1.49mol)(3/2)(8.3145 J*mol-1*K-1)(ln 348K/323K)= Use ur calc im lazy :P

Now we add DeltaS1+DeltaS2+DeltaS3= Total DeltaS of the whole system which should get u 32.97J*K-1 or 33.0J*K-1 (sig figs).

Hope this helps. Correct me if I'm wrong!!

Kevin Tang 4L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Kevin Tang 4L » Tue Feb 12, 2019 11:14 pm

lukezhang2C wrote:Are there answers to these problems?


He said he wouldn't be posting answers for this practice test.

pamcoronel1H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby pamcoronel1H » Tue Feb 12, 2019 11:24 pm

Could someone explain for 4a how he got 5.0x10^3?? When I worked it out on my own I got -498 and all my elements cancelled out.

Kevin Tang 4L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Kevin Tang 4L » Tue Feb 12, 2019 11:28 pm

pamcoronel1H wrote:Could someone explain for 4a how he got 5.0x10^3?? When I worked it out on my own I got -498 and all my elements cancelled out.


Your answer is correct. He gave the answer DeltaHrxn= -5.0*10^5 Joules which is the same thing as -498kJ. Just make sure you use your sig figs as there are only 2 sig figs in the DeltaH of Vodka+Sweat->2 Honey. :)

paytonm1H
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby paytonm1H » Tue Feb 12, 2019 11:42 pm

can someone explain the reasoning behind the three steps to solve for enthalpy of #12B

Chem_Mod
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Chem_Mod » Tue Feb 12, 2019 11:45 pm

Please feel free to come in at 1pm to CS50 on Wednesday (tomorrow) and Lyndon will be one of the UA's overseeing the drop in session

005199302
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby 005199302 » Tue Feb 12, 2019 11:46 pm

For #10, should both heat capacities (of ice and of water be the same), or should we use the heat capacity of ice on the left side of the equation. I remember at the review he used 4.184 for both, but shouldn't we take into account C of ice as well?

Kevin Tang 4L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Kevin Tang 4L » Wed Feb 13, 2019 12:11 am

paytonm1H wrote:can someone explain the reasoning behind the three steps to solve for enthalpy of #12B


So we are trynna calculate the DeltaHrxn(total) of the human body at 37C and we know the DeltaHrxn=-2756kJ at 200C

First, we need to heat up the reactants C6H12O6 and 6O2 to 200 which will be your DeltaH1
Then we use the given DeltaHrxn at 200C which will be your DeltaH2. Now we have the products instead of the reactants.
Then we need to cool down the products, 6 CO2 and 6 H20 to 37 which will be your DeltaH3

DeltaHrxn at 37C=DeltaH1+DeltaH2+DeltaH3.

Kevin Tang 4L
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Kevin Tang 4L » Wed Feb 13, 2019 12:15 am

005199302 wrote:For #10, should both heat capacities (of ice and of water be the same), or should we use the heat capacity of ice on the left side of the equation. I remember at the review he used 4.184 for both, but shouldn't we take into account C of ice as well?


Because the Ice is at 0C, it will melt into water first, then start heating up using the constant 4.184J*mol-1*K-1.
So in this problem, we are actually not at any point using heat capacity of ice because we don't need to heat it up to the temperature it vaporizes, because it is already at 0C. All we need to use is the DeltaHfusion= 6.01kJ/mol

Now... if the Ice was at -50C, we would need to use the specific heat capacity of ice in order to heat it up to 0 degrees C and then use the DeltaH of fusion.

Hope this helps

Gillian Murphy 2C
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Gillian Murphy 2C » Wed Feb 13, 2019 12:27 am

He whose name cannot be spoken wrote:Help with question 2c)? I understand that since we're cooling it kc will also drop because it's dependent on temperature, but what is the detailed explanation?


Since the reaction is breaking bonds, we know that it will be endothermic. When we learned how temperature affects a system, we learned that heating an endothermic reaction will favor the forward reaction. Therefore, if we are cooling an endothermic system, it will favor the reverse reaction.

CaminaB_1D
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby CaminaB_1D » Wed Feb 13, 2019 4:01 am

Can anyone share the answers to questions 1-3 on Lyndon's review? I was an hour late to the review session bc I had a midterm

bonnie_schmitz_1F
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby bonnie_schmitz_1F » Wed Feb 13, 2019 1:27 pm

Can anyone explain 3b?

Becky Belisle 1A
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Becky Belisle 1A » Wed Feb 13, 2019 1:35 pm

For 3b: the first step is to calculate the amount of heat need to raise the temperature of all the ice cream to 0 degrees C. Then, you subtract this value from the total heat that was given. Next, you set the value for heat from the subtraction equal to the half the mass of the ice cream multiplied by the enthalpy of fusion for the ice cream. Then just solve for the enthalpy of fusion.

Hovik Mike Mkryan 2I
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Hovik Mike Mkryan 2I » Wed Feb 13, 2019 2:32 pm

Does anyone have the solutions?

904936893
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby 904936893 » Wed Feb 13, 2019 3:34 pm

dgerges 4H wrote:I also got that the other part of w=9.12 from w=-pdv. so i got a total of 11.45kj or 1.145x10^5j for w. is this correct?

I also got that for work

MichelleTran 1I
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby MichelleTran 1I » Thu Feb 14, 2019 10:39 pm

Thanks again Lyndon for the practice exam! Helped a lot! C:

Maayan Epstein 14B
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby Maayan Epstein 14B » Sat Mar 16, 2019 10:25 pm

Can anyone explain 1D?
Thanks!

avshi10b
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Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Postby avshi10b » Sun Mar 17, 2019 7:45 pm

Thank you


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