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Chem 1AA3 Exam Review.docx

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Department
Chemistry
Course
CHEM 1AA3
Professor
Pippa Lock
Semester
Winter

Description
CHEM 1AA3 EXAM REVIEW - GOOD LUCK ON THE EXAM!! Note: A lot of the questions on these practice exams are recycled in practice tests for test 1 and 2.. somewhere you can find full solutions for a good number of them. IS THE EXAM IN THE SAME ROOM AS WHERE WE WRITE OUR MIDTERMS?? (it doesn’t say on Avenue) Yes, its in BSB 147 @ 6 PM today! :) Thanks GOOD LUCK TO EVERYONE! 2011 Exam: 1) B - A buffer must have limiting strong (acid/base) with excess weak (acid/base) [one must be an acid and one a base] or both weak that are conjugates of each other. 2) D - I got E? How do you incorporate sodium benzoate into the calculation? - Hmm i got A - I got A too! - How did you guys work out this question? Well, since they give you the Ka for benzoate, i assumed it was a buffer system (so buffer equation) with benzoate. So go through each of the reactants, they’re either going to increase or decrease the moles conjugate base or acid. pka +log[(moles benzoate - moles HCL +moles NaOH + moles NaBr)/(moles benzoate + moles HCl - moles NaOH - moles NaBr)]. Thats how i got A as an answer though, other people got different answers so i could be wrong. - Oh jeeze, that’s pretty complicated.. how did you know which moles increased and decreased? I got D: React the strong acid and base together first: see what moles of acid are left over: react that with the weak base and find your buffer pH (i got exactly 4.40 so i don’t know if he has multiple answers) - But when you react NaOH and HCl, you get NaCl, which is a salt, right? So how do you know what to react that with? - so: moles of NaOH= 0.02 mol and moles of HCl= 0.045 mol: when you do a before/after chart, you will have 0.025 mol remaining (the other moles will be neutralized they have no affect on your pH) - Yeah, I got that :) But then what? so then i just reacted the 0.025 mol of HCL with ur sodium benzoate (which will produce a buffer) - OHH, I see now! Thank you :D But how did you know the mol. formula for sodium benzoate? - not necessary: you have concentration and volume - what is used as the volume here to calculate the final concentrations of the HCl and sodium benzoate? the total v? - I think so.. if you use the total volume it doesn’t give you the answer...i’m confused on how to go from moles of the HCL and sodium benzoate to get the concentrations for the H-H equation :S - Don’t stress over it, there’s only 3 acid/base questions on the exam it’s frustrating! it seems so simple but i can’t get it haha - The approach you guys took to arrive to the answer D = 4.40 seems correct. But how did you guys solve it? The concentration of HCl that I carried over was (0.025 mol)/(0.145L) = 0.1724 M. However when I went about to solve it using HH, I didn’t get the right answer..I would greatly appreciate it if someone can explain it. Thank You! -how is there 0.025 moles of HCL left over? there’s more moles of NaOH than HCl isnt there? (0.200M)(0.100L) = 0.02 moles of NaOH and (0.045)(0.100L) = 4.5e-3 moles of HCl? Nevermind I’m an idiot, its 1.00L of HCl x 0.045M :S 3) E - Could someone explain their process for this question? From the titration, you know moles of NaOH = moles HCl at the equivalence point, right? So solve for the moles, divide that by the volume of initial HCl which is 100, that gives you the conc of HCl, multiply that by the volume used when the precipitation happened (3.34 mL) to find the moles, then divide that by the total volume in the protein beaker which is 3.34 + 225, that gives you the [H3O] which you can take the log of to get the pH. - Thank you! :) 4) C - Why? I think it’s E? Help? Yeah, pretty sure it’s C... iv is true I think it’s E, can’t be C (v) is true, CCl4 is non-polar, CHCl3 has a dipole moment making it polar i also think the answer is E, but when i google it, i found that hydrogen bonds are broken when water boils, which makes me confused. Notice how it says polarizable, not polar? What does this mean? - No one can figure that out :( Just my thought process, suggestions?: - (i) At a given temperature, HF has a lower vapour pressure than HCl True: HF has a lower vapour pressure than HCl because HCl wants to dissociate because it is a strong acid, leading it to have low IMF, a low boiling point, and a high vapour pressure - (ii) CH3OCH3 has a higher boiling point than CH3CH2OH False: CH3CH2OH has hydrogen bonding, and therefore has a higher boiling point - (iii) CH3CH2CH2CH2CH2OH has a higher viscosity than CH3CH2OCH3 True: The more IMF, the higher the viscosity - (iv) Hydrogen bonds are broken when water boils True: Googled it - (v) CCl4 is less polarizable than CHCl3 ***True/False?*** CCl4 is nonpolar, and CHCl3 is polar- therefore...? This is true as well. So is iv. This question must be a mistake. ^ I agree, since polar molecules have dipole-dipole interaction while non-polar have just dispersion. - making the statement for (v) true. 5) C- Why? C makes sense theoretically, but when you calculate the pKa using the Henderson- Has. or by doing the -log Ka you get a pH of approx. 3.55. - you have excess KOH, pH can’t be less than 7 - But how did you get to the answer? (Educated guess, if there is a buffer solution the pH range is usually 4-10, but someone should share the real way) why can’t you just use the 3.55 pH? i don’t understand why this is incorrect - Me neither! - assume a 1L sample/ work in moles instead of [ ]? get a pH in the range of answer option C pH= 8.35 - But even if you work in moles, the ratio of [A-]/[HA] will be the same! there was a similar question to this one in test 1, #16 in version 2. i got it wrong though and the answer key doesn’t really show a full solution. could someone please post how to do these?! :) thanks using an indicator: therefore want to find pH at equivalence BEFORE: (assume a 1L sample total) acetic acid: 0.150M*1L= 0.150mol KOH: 0.150M*1L= 0.150mol < The concentration of KOH in the question is 0.20M. Why did you put 0.150M? (finding pH at equivalence weak acid and weak base have to be equal?- this process isn’t really right LOL, but it made the most sense - Lmao okay! AFTER: acetate: 0.150mol Find pH of 0.150mol/1L of acetate (given ka value therefore find kb) will calculate the pOH when using these values pH= 14-pOH pH= 8.35 Not too sure if this process is right, but logically at equivalence point it’s going to be weakly basic- so the other basic option is too basic? - It makes sense, but there’s a lot of odd assumptions and guessing haha :( i feel like we’re overlooking a very simple way to answer this question...i’m going to go with the educated guess because it’s taking way too much time to figure out considering there’s only 4 acid/base questions lol - True! What’s your thought process for the educated guess? well, the fact that there is leftover KOH basically says the pH cannot be lower than 7 as said before, and assuming the buffer range is 4-10, C makes the most sense. but i still don’t know why A isn’t right (just saying) - i usually trust math over my gut - Hmm.. okay :) Thanks for all your help! 6) A Equivalence point, meaning NH3 and HCl must have the same number of moles! Before and after table: NH3 + HCl → NH4+ + Cl- 0.01 0.01 0 0 0 0.01 Figure out how many litres of HCl: (0.0500 mol/L)(? L) = 0.01 mol = 0.20 L HCl Figure out the new concentration of NH4+ with the new volume: [NH4+] = 0.01 mol/ (0.2 + 0.1 L) = 0.033 mol/ L Now make an ICE table: NH4+ + H2O NH3 + H3O+ 0.033 (0.033 - x) +x +x Solve for x (which is the [H3O+]) by equating products/reactants to the Ka (Kb is given in the question), and solve for the pH. 7) A? I got a Pretty sure it’s A. E is true..., A is true as well. This isnt the proper graph for temp vs. pressure so E should be wrong... It is the proper graph... Increasing temperature at constant pressure results in more gas therefore higher vapour pressue. I am pretty sure it’s A as well. A is true. A Is the false statement. If you compress it, so you increase P you DO cross the boundary to become a liquid - Kind of tricky because the line is slanted, but all the options except A are clearly true. A - False - Start at 100K and 0 pressure, going straight up you will travel through gas to solid to liquid; thus you can compress a solid into a liquid 8) C (B - Transition states exists at maximas; intermediates exist at minimas. Also, Heterogeneous catalysis is when catalyst is different phase than reactants, so solid Pd with liquid cyclohexene) - The answer is B- reaction intermediates exist at local minimums, not local maximums. To the writing in the blue, how did you know that cyclohexene was in liquid form? Solid hydrocarbons are very uncommon (at lower carbon counts) 9) D - Use vo = (kcat [Eo][S])/ (Km + [S]), where kcat is constant for both situations and therefore can be ignored. 10) E - E is false because vo = 1/a × d[A]/dt is missing a negative sign! 11) D - Use ln (k2/k1) = Ea/R x (1/T1 - 1/T2) 12) A/E? I think A- thoughts? I believe it’s E, look up structure for pyridine. agree with E too. i’m pretty sure the answer is A, the question says the lone pair electrons occupy this orbital. - I’m confused about this question, could someone explain it? How do the lone electrons on N in pyridine occupy sp2? Because there is a rotating pi bond going throughout the entire circle- this means that one of nitrogen’s pairs must be used in a pi bond to allow this to happen. Its other pair is therefore in an sp2 orbital to keep the structure of an aromatic compound. Dr. Hatala explained that to me yesterday... - Okay, I kind of get that. But what do you mean by the “circle” that the pi bonds go through? If alone then nitrogen is sp3, however since pyridine is an aromatic there is a huge driving force to form a sp2 orbital to achieve aromaticity and greater stability. There are 2 double bonds in pyridine technically but, the pyridine hybridizes to sp2 to add to the delocalized pi electrons of the 2 double bonded carbons in pyridine thereby making it aromatic. 13. A? I think it would be E, because Sn1 are tertiary cations, which are i, and iv ? which is E..anyone agree? If I remember right Sn1 deals with tertiary carbocations, and Sn2 with primary, so yes to E. This is right. The answer is E, i and iv are the only ones which are tertiary E as well Same here, i agree with E since i) and iv) are the only teriary ones. 14. E/C? Can someone please explain?- I think it is E, C does not have the correct molecular formula. You know that it needs to be a first degree alcohol and E is the only one that satisfies this. IT’S D, I confirmed w/ the TA. I can’t be an alcohol because it DOES NOT react w/ PCC...therefore alcohols are out... react w/ KmNO4 to produce a carboxylic acid which then reacts w/ the base to form a salt, switching out the OH group :) - Dr. Hatala went over it during class, the answer is D. Simply match up the compound to the mol. formula in the question! 15. E/D d but couldn’t you argue that technically phenol and alcohol are the same thing? And where is the amine you’re seeing.. I only see amides. The amine is on the fat left, it does not have a double bonded O directly beside it so it is an amine and not an amide. - There’s no alcohol, it doesn’t count if it’s attached to the phenol because the phenol is its own functional group! So what is the answer for this? I would say D So if it is count to be phenol, you won’t count it as alcohol again? yea, because the alcohol attached to a benzene ring is what is phenol and gives it its properties so yea treat it as Phenol I think the answer is alcohol because all of the OH groups are within other functional groups so i’m not sure if we can count them. also, amines can be found in 2 locations so it’s definitely not the amine. The amide is the structure on the far left attached to the C which is attached to the double bonded oxygen. And the phenol and carb. acid can be easily found, leaving alcohol as the only option. Its D, its specifically mentions phenol as functional group in the M/C so alcohol would be the more correct answer. Just as an aside, acetyl groups are not functional groups they are structural motifs. 16. A? Not too sure.. i think the answer is C.. try putting the double bond on the substituents How is the question even phrased though? That’s what’s messing me up...the wording...it is a little confusing but think about it this way.. the formula says C5H8 and there starting off with cyclopropene which only has 3 carbons in the structure so the other two carbons have to substituents. they can be methyl or ethyl- - Yes exactly, that’s how I got only 2 structures :( - I count 5 possibilities (C) Its an alkyne, general formula for an alkyne is CnH2n-2, Plug in n=5 and you would get C5H8. So there has to be a triple bond to satisfy the number of hydrogens.I think its C , not sure though, could be D as well, anyone wants to confirm it ? hopefully you understand the wordings triangle with a dbl bond and two methyl coming off the same carbon - But carbon can only make 4 bonds :O ? http://www.google.ca/imgres?q=11+di+methyl+cyclopropene&um=1&hl=en&sa=N &biw=1366&bih=677&tbm=isch&tbnid=wqAgs2TYgx0E1M:&imgrefurl=http://www .chemsynthesis.com/base/chemical-structure-29778.html&docid=r-ZuvgO3LcruJ M&imgurl=http://www.chemsynthesis.com/molimg/1/big/29/29778.gif&w=400&h= 300&ei=PWAZUKi6L4Gx6wHt9IHACg&zoom=1&iact=hc&vpx=375&vpy=158&dur =966&hovh=194&hovw=259&tx=92&ty=85&sig=109176886677571802855&page =1&tbnh=152&tbnw=241&start=0&ndsp=18&ved=1t:429,r:1,s:0,i:76 you see the cyclopropene on the left side.. and the two methyl groups on carbon one.. its like that - Oh my.. okay! Thank you :) Good im glad that helped! triangle with dbl bond and methyl coming out of carbon 1 and carbon 2 triangle with dbl bond and ethyl coming out of carbon 1 triangle NO DBL BOND but an ethyl with a dbl bond on it on carbon 1 triangle NO DBL BOND but ethyl with dbl on on the second carbon. i know thats a really weird way of explaining structures but maybe you may get it :S - Thank you! so is 5 isomers the answer? - Seems like it! 17. B - Why? There’s 3x the bonds, so why is it not 3x the strength? Because the first bond is a sigma bond while the second and third are pi-bonds. pi bonds have different strengths than sigma bonds, so it’s not as simple as just 3x - Ohhh I see, didn’t think of that! Thanks!! :) 18. C -I don’t think we can assume b is dilute h30 IS incredibly strong. I think it’s C. Benzene may not be very reactive, but it can be brominated (in the presence of a catalyst, i dunno if we’re supposed to assume that or not), then those bromines will be substituted for hydroxyls. it’s D: benzene will not break its aromaticity to react with a bromine - I think it’s D, because benzene will not willingly break its aromaticity and we assume the H3O+ is diluted, because we weren’t taught to treat it as a strong acid when dealing with functional group reactions. I think it’s D because the reaction ii and iii are both true. I only used process of elimination though, i’m not sure what’s going on in the i reaction 19. C I think it’s b, please explain otherwise?- in SN2 reactions just have one activated complex. SN1 reactions have a slow step followed by a fast step. You can see this in the reactions process vs. potential energy diagrams in the slides, therefore C is false. - C is false because it is SN1 reactions that have the fast and slow steps. The slow step is forming the carbocation, and the fast step is the nucleophile attack. He said the answer was C in class :) E is correct as well because you can a polar molecule? Technically neutral right? But E is true, you could have charged or neutral nucleophiles in SN2 reactions. The answer is C. People in Purple and Orange above have the right idea :p 20. D - Can someone explain this please. - Really don’t understand this.. if you google the structure, option C exists. How can C be an intermediate though? C is ethene, it’s a reactant. It’d D because, the polyethylene must always have the formula -[CH2-CH2]-n .. and D does not follow that - But neither does B, there’s no carbons in B at all! When you polymerize something you’re using R-O-O-R so I think all of them can exist except ethene, which as stated above is a reactant... I think its C I think the answer might be D, there is no way during a polymerization rxn that three carbons can exist like that as a radical formed from ethene :S the structure shown in C on the other hand is simply ethene which is used in propagation to continue the growth of the chain. Those are just my thoughts though!I That does make sense...I think by that logic, D seems most feasible, the radical chain is not a multiple of 2 (eth) According to our notes page 10 of the orgo slides D would be correct as it specifically shows that it needs to be a multiple of 2. Sorry do you mean D as in the correct answer, or that the D structure can exist? haha just confused by your sentence! D as the answer The Answer is D, tutorial TA told us. Its because this is a polymer of ethylene, which is two carbons, so you must always have a number of carbons that is a factor of two since you’re always adding two at a time. D has only 3 so its false. 21. B - yes 22. E - From lab 23. C Which C atoms do you look at for this question? Look at the two double bonds in the chain, not the double bonds in the rings. I got the second C from the left that one was (E)... How did you get Z for the first double bond? When you look at the benzene ring at the top and bottom they are pretty similar. If you move to a carbon in the ring you can see a fluorine on the top one and a carbon on the bottom one. This gives the top benzene priority. On the other side there is a carbon and hydrogen and the carbon gets the priority. So the 1-fluorobenzene and the carbon atom are on the same side therefore giving it a
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