## FINAL PRACTICE - Lyndon's Churro Review Session [ENDORSED]

Any general non-science questions (software questions/problems, etc.,) and class announcements.

SydBenedict2H
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Double and triple bonds count as single regions of electron density correct? Im trying to explain them to my friend and she's convinced they count as 2 and 3 respectively.

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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

bonnie_schmitz_3K wrote:I think I understand the basis of how to do number 34, but I keep getting a pH of 2.10. Did anyone else make the error same/know what my error is?

I'm making that error, but I don't know what the issue is.

Posts: 72
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Can someone please explain how to do number 34? I understand that they neutralize each other, but I'm not exactly sure on what steps to use to find the answer

Max Kwon 1J
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

for number 34 you have to find the limiting reagent of the acid and base essentially. CaO dissociates to Ca2+ and O2-, and think of O2- + H2O -> 2OH- . So 1 mol CaO is equal to 2 mol OH-, maybe that's your error?

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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Max Kwon 1J wrote:for number 34 you have to find the limiting reagent of the acid and base essentially. CaO dissociates to Ca2+ and O2-, and think of O2- + H2O -> 2OH- . So 1 mol CaO is equal to 2 mol OH-, maybe that's your error?

Okay.. so you don't subtract the CaO concentration from the HCl concentration? you just use the limiting concentration (CaO), and then multiply that by 2?

Max Kwon 1J
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

oh you still do subtract, you should have 1.04961466x10^-2 mol H+ and 5.24x10^-3 mol OH-, u subtract so you are left with 5.2497 x10^-3mol H+. Then you divide by 1L to get molarity (same number) then -log[H+], you should get a pH of 2.28

Manas_Varma_4B
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Thank you so much! That was very helpful.

Natalie Liu 4I
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

For 39, why can't PH3 have hydrogen bonds?

Chem_Mod
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

BenJohnson1C wrote:So when calculating the pH of an acid and base reaction, would you need to put both into one equation or write equations separately for each?

Answer: When finding pH, you have to figure out concentrations of hydronium or hydroxide ions. The equation is for your own understanding of mole to mole ratios.

904837647 wrote:Will the final be similar to this practice final?

As noted above and on the test, these are practice problems designed to help you as much as possible, but is not meant to be indicative of the length, structure, or format of your test.

Sydney Takeda 1G wrote:For #31, would it also be correct to put NO2 in the complex or does it have to be ONO?

Personally, I find it better to write ONO to signify the O as the actual atom binding to the transition metal.

Ahmet_Dikyurt_3L wrote:Can someone explain 33b?

You can use the concentration of strong base to determine concentration of hydroxide ions. THen you can find pOH and ultimately pH.

bonnie_schmitz_3K wrote:For question 18, I'm having trouble finding the 180˚ bond. Which bond would it be?

The C is triple bonded to the N, creating a linear structure with the N triple bonded to the C which is singled bonded to another C.

Daniela Alvarado 3B wrote:for #24, why can't we find the oxidation state of Fe by finding the nitrogen' formal charges while attached?

The oxidation state of a ligand is inherent of the ligand before binding. Just as halogens have oxidation state -1, when they bind they become neutral in terms of formal charge.

marg44 wrote:Can someone explain 41c? I'm confused as to why the nitrogen with the lone pair can accept the proton but not the oxygen.

The oxygen is less likely to accept a proton because it already has a double bond, and since it is more electronegative, it will not like the resulting positive formal charge from accepting a proton.

bonnie_schmitz_3K wrote:I think I understand the basis of how to do number 34, but I keep getting a pH of 2.10. Did anyone else make the error same/know what my error is?

You need to make sure to understand how metal oxides generate OH-. Metal oxides dissociate into O2- which reacts with water to form 2 moles of OH- so you have to make sure to keep this ratio of 1 mole base to 2 mole OH- in mind.

Ligand names come first in alphabetical order ignoring prefixes. THen the transition metal name comes with its oxidation state. Then whatever is outside the brackets follows. See other posts and textbook which go in detail on naming coordination compounds.

Ashley Odibo Dis3E wrote:What is a porphyrin ligand and will we be asked about ligands like this on the test?

As discussed in class, it is the ligand around the Fe that forms a heme complex.

SydBenedict3H wrote:Double and triple bonds count as single regions of electron density correct? Im trying to explain them to my friend and she's convinced they count as 2 and 3 respectively.

You are correct!

Madelyn Cearlock wrote:I'm making that error, but I don't know what the issue is.

Natalie Liu 4I wrote:Can someone please explain 40B?

Because of resonance, the real structure is a hybrid or average of the resonance structures. Therefore the bond lengths are the same and are an average of a single bond length and a double bond length.

Ivan Tadeja 4B wrote:For 39, why can't PH3 have hydrogen bonds?

The requirement is for the H to be covalently bound to an electronegative atom like F, O, or N

Eruchi Okpara 2E
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Thank you so much!!!

Michael Nirula
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Wait so regarding #21/the Porphyrin question does that mean Fe is not covalently bonded to the Ns. Why I'm still confused is that it looks like there are 2 Ns with 3 covalent bonds and one pair and 2 Ns with a double bond, single bond and a lone pair so in that case the overall formal charge of the ligand would be +2.

Lina Petrossian 1D
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

bonnie_schmitz_3K wrote:I think I understand the basis of how to do number 34, but I keep getting a pH of 2.10. Did anyone else make the error same/know what my error is?

I'm getting that same answer :(

Haowen_Liang_3E
Posts: 20
Joined: Fri Apr 06, 2018 11:03 am

### Re: FINAL PRACTICE - Lyndon's Churro Review Session

for 33a, how do we set up the equation with 1.5mol/L and 0.1155mol/210ml to find the mL require for the 1.5M solution?

Haowen_Liang_3E
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Lina Petrossian 3D wrote:
bonnie_schmitz_3K wrote:I think I understand the basis of how to do number 34, but I keep getting a pH of 2.10. Did anyone else make the error same/know what my error is?

I'm getting that same answer :(

The 1st step is to write out the acid and base equation,
[HCl] + [H2O] =>[H3O^+] + [OH^-] and [CaO] + [H2O] => [Ca^+] + 2[OH^-]
2nd, calculate the concentration of HCl and CaO, 1.05x10^-2 mol HCL and 2.62x10^-3 mol CaO.
3rd, convert reactants to products, 1.05x10^-2 mol H3O^+ and 5.24x10^-3 mol OH^-.
4th, acid and base will neutralize and make water, so find the difference of H3O^+ and OH^-, left with 5.26x10^-3 mol H3O^+.
5th, plug into the equation, pH=log10(H3O^+)

Henry_Phan_4L
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Finally it makes sense thank you!

ariana_apopei1K
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Haowen_Liang_3E wrote:for 33a, how do we set up the equation with 1.5mol/L and 0.1155mol/210ml to find the mL require for the 1.5M solution?

1.5 x L1 = .55 x 210 then solve and you should get 77 mL

paytonm1H
Posts: 74
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### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Once again, thank you so much Lyndon! You are a life saver!!

Lina Petrossian 1D
Posts: 77
Joined: Fri Oct 05, 2018 12:16 am

### Re: FINAL PRACTICE - Lyndon's Churro Review Session

Haowen_Liang_3E wrote:
Lina Petrossian 3D wrote:
bonnie_schmitz_3K wrote:I think I understand the basis of how to do number 34, but I keep getting a pH of 2.10. Did anyone else make the error same/know what my error is?

I'm getting that same answer :(

The 1st step is to write out the acid and base equation,
[HCl] + [H2O] =>[H3O^+] + [OH^-] and [CaO] + [H2O] => [Ca^+] + 2[OH^-]
2nd, calculate the concentration of HCl and CaO, 1.05x10^-2 mol HCL and 2.62x10^-3 mol CaO.
3rd, convert reactants to products, 1.05x10^-2 mol H3O^+ and 5.24x10^-3 mol OH^-.
4th, acid and base will neutralize and make water, so find the difference of H3O^+ and OH^-, left with 5.26x10^-3 mol H3O^+.
5th, plug into the equation, pH=log10(H3O^+)

Thank you so much!!!