CHEMISTRY COMMUNITY

Hazelle Gunawan 3G

It might be in a table of values somewhere else in the textbook? I don't remember doing this problem so I'm not sure, but I remember a lot of questions requiring values that weren't given directly in the problem.

Hazelle Gunawan 3G

I've seen it used in some textbook problems. The conversion is on the formulas sheet though so you don't have to worry about memorizing it!

Hazelle Gunawan 3G

In terms of actual problems you might have to solve on the exam I don't think there will be any huge difference; both pretty much function as a way for ions to move throughout the solution.

Hazelle Gunawan 3G

Multiply 525 W by 6.88min * 60 sec/min to get total heat in J. Then convert each of your substances to moles and calculate how many moles of each substance were vaporized. Take this number and divide it by the number you got for heat in J and that's your molar enthalpy of vaporization. Then divide t...

Hazelle Gunawan 3G

Definitely Chemical Equilibrium, that topic just clicked way faster for me than probably any other topic we've covered.

Hazelle Gunawan 3G

In this case K actually refers to how fast the reaction is proceeding. A smaller k means a slower reaction while a larger k means a faster reaction.

Hazelle Gunawan 3G

b) gr stands for graphite, if I recall correctly!
e) Sn2+ would make sense!

If it's being oxidized, you would write Ag--> Ag^+ + 2e-.

Hazelle Gunawan 3G

kt(1/2)= 0.693 would be your best bet, I think!

Hazelle Gunawan 3G

A reaction can only occur as fast as its slowest step is occurring, so any step that occurs faster is secondary to the slowest step!

Hazelle Gunawan 3G

If you have a group of people to study with, what really helps me is talking through some of the theoretical concepts with friends! I'm also planning on working through the outlines again and maybe rewatching lectures on 2x speed just to jog my memory?

Hazelle Gunawan 3G

A state property is the same no matter what path you take to get to it. Work and heat aren't considered state properties because they can change depending on what steps you take to get to the end product.

Hazelle Gunawan 3G

I think the water is mostly responsible for dissolving the metals in the galvanic cell and wouldn't become part of the reaction unless it's an acid/base, so I assume if you could find another liquid to do the same then that would work?

Hazelle Gunawan 3G

If I'm not mistaken I think it's because Zn is more electronegative than Cu. I'm not sure, though.

Hazelle Gunawan 3G

Delta G naught is for standard conditions, whereas delta G is for the actual conditions you're running the reaction under. At equilibrium they should both be 0. Hope this helps!

Hazelle Gunawan 3G

Usually positive entropy leads to negative Gibbs free energy and a spontaneous reaction, but this isn't true all the time and depends on the enthalpy of the reaction. If both enthalpy and entropy are positive, then the reaction will be favorable at higher temperatures. The opposite is true for lower...

Hazelle Gunawan 3G

I think you're correct that more ordered the system is the more disordered the surroundings become and vice versa! But I think since you're dividing - delta H(sys) by temperature, the change in entropy of the system and surroundings won't be equal and there will be a net increase overall.

Hazelle Gunawan 3G

I think the answer choices will include units! kJ is pretty standard for delta G and H, but when you're calculating entropy it usually gives you J, so don't forget to convert it to kJ!

Hazelle Gunawan 3G

Exactly as above! Specific heat capacity is intensive because it stays the same, but heat capacity is extensive because it depends on the mass present.

Hazelle Gunawan 3G

I didn't see it either, just the possible point totals for each question, so I don't think you're missing anything!

Hazelle Gunawan 3G

Isabella Chou 1A wrote:I think you are right, maybe try switching the answers so that spontaneous is in the top boxes, and Q<K / Q>K is in the bottom boxes. I had this same issue initially, but after switching them from top to bottom, it worked for me.

Oh this worked for me, thank you so much!!

Hazelle Gunawan 3G

I think it's because H2O still releases energy when it's formed, so this needs to be taken into account for when you're calculating overall Gibbs free energy of the reaction.

Hazelle Gunawan 3G

Hi! The question I'm struggling with is: The diagram shows the free energy change of the reaction A(g)+B(g)↽−−⇀C(g) The reaction progress starts on the left with pure reactants, A and B, each at 1 atm , and moves to pure product, C, also at 1 atm on the right. Place the statements in the appropriate...

Hazelle Gunawan 3G

Since q = u + pv, and you have delta u already, you can multiply pressure by change in volume and add it to your u value to find q. Hope that helps!

Hazelle Gunawan 3G

When it does work on the surroundings, the system itself loses heat/energy, so the value is negative. Hope this helps!

Hazelle Gunawan 3G

I think that's correct, yeah!

Hazelle Gunawan 3G

For this one it should be [heat required to melt ice] + [heat required to bring ice temperature up to equilibrium] = [heat water releases when temperature drops]. What was especially tricky for me was keeping track of the change in temperature, so while it doesn't matter for the ice since it's at 0 ...

Hazelle Gunawan 3G

First convert kcal to kJ, then subtract how much heat it releases from the work done to it!

Hazelle Gunawan 3G

You can think of it as steam being hotter than liquid water, too! Since the water is changing from a hotter form to a less hot form, it's giving off heat and is therefore exothermic.

Hazelle Gunawan 3G

I felt that doing the textbook problems on the outlines prepared me pretty well for the exam, and I actually ended up having lots of time to double check my answers. I don't want to say anything yet for fear of jinxing myself but I felt like it was a fair exam.

Hazelle Gunawan 3G

Do you have an example of the problem? From what I'm thinking it might be [enthalpy of formation of reactants] - [enthalpy of formation of products] but it would help if you could show the problem.

Hazelle Gunawan 3G

For me ideally it would about 3 hours for lecture + 1 hour for discussion + 2-3 hours to do textbook problems + 1 hour for sapling homework. If I go to any of the workshops or step-up sessions that's another 2-4 hours depending on how many I attend... in total it adds up to anywhere from 7 to 12 hou...

Hazelle Gunawan 3G

Since H2SO4 is a strong diprotic acid, the first dissociation will be a direct 1:1 ratio so you'd end up with [H+] = 0.15M. After that, I don't have the problem pulled up but if they give you the Ka for the second dissociation, you do the ICE box like you'd do for any weak acid. Hope that helps!

Hazelle Gunawan 3G

pKa is the acid dissociation constant and it represents how much of a weak acid dissociates in solution. pH represents the concentration of H+ ions in solution. If you're given pH and the initial concentration of the acid you can convert pH to [H+] and use that to find Ka, then convert it to pKa.

Hazelle Gunawan 3G

Exactly as above! As far as I know the Ka is small enough compared to the initial concentration that you can disregard x in the denominator.

Hazelle Gunawan 3G

It's because NH4+ is the conjugate acid of a weak base! The ClO4 should basically become null if I'm remembering correctly, so the NH4+ becomes the predominant ion in solution that disassociates to form H30+ and NH3.

Hazelle Gunawan 3G

If you remember this from 14A, it has to do with the IMFs! Since I2 is such a large molecule it forms really strong London Dispersion forces with other I2 molecules and thus requires more energy to be broken apart. Hope this helps!

Hazelle Gunawan 3G

From what I can remember it says it's a monoprotic acid, so you know that the weak acid will dissociate into H3O+ and the conjugate base. Since we don't actually know what the acid is you can use A as a placeholder.

Hazelle Gunawan 3G

I think it has to do with the angles of the amine groups... I want to say it's because the angle of 3-Hydroxyaniline is wide enough to allow it to disassociate while the 180 deg angle of 4-Hydroxyaniline makes it hard to h-bond to other molecules, but I'm really not sure so I'd be interested in seei...

Hazelle Gunawan 3G

The values we've been getting for initial concentration have been reliably around 0.1 so far, so I don't really see that changing on the exams... I would just be on the lookout and use your best judgment.

Hazelle Gunawan 3G

Use the smaller one! The bigger one is way larger than your existing concentrations, so if you tried to subtract it from what you already have you'd end up with a negative number.

Hazelle Gunawan 3G

Using the initial equilibrium concentrations they give you, you can find Kc. From there, you can set that value equal to your new expression for the equilibrium constant, which will be [initial NO concentration - 2x]^2 / [initial N2 or O2 (they're the same) concentration - x]^2 . Once you get x you ...

Hazelle Gunawan 3G

Squaring is just for the equilibrium constant! If you're just trying to find the concentration, you would plug whatever you got for x back into your expression for that reactant/product. For me it was 3.32-2x, so I just plugged my x value back into that to find the concentration of NO2.

Hazelle Gunawan 3G

If you're only given the concentration of one reactant/product and you can assume everything else is at 0, then whatever you have will decrease and whatever you don't have will increase. If you're given multiple concentrations you can look at the Kc and see if it's really large or small --> if it he...

Hazelle Gunawan 3G

It depends! If you're reversing the reaction, then you would take the reciprocal of Kc. If you're multiplying the number of reactants by x, then you would set Kc to the x power. And if you're combining reactions, you would multiply your Kc's by each other.

Hazelle Gunawan 3G

ICE tables are used when you're given the initial concentration of a compound and you want to find the equilibrium concentration!

Hazelle Gunawan 3G

From my notes I have the problem saying 500 mL and the ratio saying 0.500 L, with the final result being 3 mol/L, so you might have misread?

Hazelle Gunawan 3G

An inert gas is another name for noble gas, if I remember correctly! They aren't part of the equilibrium reactions so their presence doesn't really change anything.

Hazelle Gunawan 3G

If you reply that counts as a post too! So you can create topics or reply to people's topics, and every separate post/reply is counted.

Hazelle Gunawan 3G

My understanding is that they don't dissolve into the solution so their concentrations don't change, so they don't have any effect on the equilibrium reaction.

Hazelle Gunawan 3G

The impression I got was that it should be during your lecture time now.

Hazelle Gunawan 3G

In this case it's because the halides are not bonded to the H atoms but the O atoms. Because Cl is more electronegative than Br, it pulls the O atom's electrons towards itself and 'delocalizes' the electrons, thus making the O-H bond weaker and more likely to dissociate. Hope that helps!

Hazelle Gunawan 3G

I remember one of the answers being charged and the other being neutral? Not sure though, I don't have it pulled up. For a general explanation, the way I thought of it was that pKa refers to the acidity of the species you're putting into the solution, and pH is the acidity of the solution itself. If...

Hazelle Gunawan 3G

Looking at the latter half of this complex, my thought process was that since CN has a 1- charge and this is a coordination compound, it'll be providing both electrons. 6/2=3 which would make Cr have an oxidation number of 3+. I'm not sure though so I'd be interested in seeing other answers!

Hazelle Gunawan 3G

My assumption before reading this post was that weak acids generally tended to correspond with weak conjugate bases, so it's interesting that you said the opposite! Would be interested in seeing what anyone else has to say on this.

Hazelle Gunawan 3G

I don't really think it matters seeing as the test is multiple choice, but I generally tend to preserve at least 4 decimal places between calculations. Hope that helps!

Hazelle Gunawan 3G

Oh this is a really good question! I always thought that it was mostly hydroxides that were strong bases and not anything else, but this is interesting to think about. My only thought is that it might be more difficult for metal oxides to dissociate than metal hydroxides? I'm really not sure though ...

Hazelle Gunawan 3G

I believe what Dr. Lavelle was explaining in that section is that the ideal geometry for ligands that can form more than one bond to the same transition metal will consist of lone pair-spacer-spacer-lone pair. Hi, thanks so much for the super thorough answer! I'm still a little confused on what spa...

Hazelle Gunawan 3G

In this case the s-character increases as the p-character decreases, so increasing s-character would cause regions of electron density to decrease. Because an sp2 orbital has less regions of e- density than an sp3 orbital, the bond angles increase. Hope that helps!

Hazelle Gunawan 3G

I also normally have issues with memory retention over breaks, so I definitely feel you there! My suggestion would be to randomly pick problems out of the outlines and (re)do them just to give yourself a refresher? Or take a textbook problem and change some things about it and try to answer it after...

Hazelle Gunawan 3G

Yep, that's correct! Be careful when you're dealing with transition metals, though, because (for example) generally the 4s electrons are added/removed before the 3d electrons because they have lower energy. This tripped me up on the last midterm so I just wanted to mention it!

Hazelle Gunawan 3G

Agreed with the above answer! I'd also like to add from an answer I saw elsewhere that if you have a weak acid and a weak base, they would also likely form a neutral salt.

Hazelle Gunawan 3G

If you think about the compounds in terms of their chemical formulas, e.g. potassium bromide would be KBr, you can separate them into their component ions and then identify acids/bases that contain those ions. Ex: K+ and Br-, KOH is a strong base that contains K+ and HBr is a strong acid that contai...

Hazelle Gunawan 3G

I was also confused on this, and would be interested in seeing any answers anyone else had!

Hazelle Gunawan 3G

I'm not sure how often they'll appear on the final, but cis is basically when you have the same element on the same side of the compound and trans is when they're on opposite sides (e.g. diagonal from each other).

Hazelle Gunawan 3G

I'm not sure, but generally Cl is considered to be more electronegative than O.

Hazelle Gunawan 3G

My process is something like:
- identify the strength of the acid that the anion comes from
- identify the strength of the base that the cation comes from
- compare the two and see which one wins out

Regarding a weak acid and a weak base, I think it would just be neutral, but I'm not sure.

Hazelle Gunawan 3G

I'm not exactly sure what it means by conjugate system, but B isn't electronegative enough to pull electrons towards itself to cause that delocalization, so it would be more difficult to remove the H+ ion as opposed to HClO, for example. Hope that helps!

Hazelle Gunawan 3G

This is also what I thought, so unless we both happen to be wrong I think you're correct! Good luck on the final!

Hazelle Gunawan 3G

I'm not really sure what you're referring to for part a, but for part f it would be an acidic salt because the NO3 part is from HNO3, which is a strong acid. Meanwhile Cu is a transition metal, not a group 1/2 metal, so it wouldn't be from a strong base (I think).

Hazelle Gunawan 3G

I think if you're counting the d subshell, then technically it has 8 valence electrons? I was actually discussing this in a review session earlier, and the people in the chat said that for transition metals you would count the d subshell as part of the valence electrons. But I think Fe usually exist...

Hazelle Gunawan 3G

You're correct in regards to pKb, which is directly based off the concentrations of the initial base and its dissociated ions. A lower pKb means more of the base has dissociated, which would make it stronger. I don't think you would need the pKb to compare the two bases in this case? BrO- is just a ...

Hazelle Gunawan 3G

I also thought the same for part b, so I think you're correct there! For part d, I don't see a charge on formic acid, so I'm not sure what you're referring to?

Hazelle Gunawan 3G

Yep, exactly as above! 6f doesn't exist as far as we know right now, but it's still theoretically possible.

Hazelle Gunawan 3G

Was hoping someone could give me a quick rundown on how to check for updates for the lockdown browser again? I've already forgotten since last time, haha.

Hazelle Gunawan 3G

Thanks all!!

Hazelle Gunawan 3G

I think it was in Wednesday's lecture that Dr. Lavelle mentioned a compound having ideal geometry due to the presence of a spacer? I didn't understand either of these concepts at all let alone how they work in relation to each other so I'd appreciate any clarification on this!

Hazelle Gunawan 3G

I was also wondering about this! I wasn't sure what exactly constituted an 'electron-rich species' or how it would replace water.

Hazelle Gunawan 3G

Is it a coordinate covalent bond that connects a ligand to a metal ion in a coordination complex? If so, is the ligand the one contributing both e- in the bond?

Hazelle Gunawan 3G

The coordination number is simply the number of bonds that the metal ion at the center of the coordination complex is making. So, to classify compounds by coordination number, you must count the number of ligands and consider each ligands "-dentate" (monodentate, bidentate, tridentate) wh...

Hazelle Gunawan 3G

You're correct that longer bonds are weaker! However, I think the boiling point is mostly in reference to intermolecular forces and not the bonds within individual molecules themselves. Since boiling isn't about breaking bonds, but instead exciting molecules to the point where they change phases, la...

Hazelle Gunawan 3G

This was for the homework due this past week, but I jotted it down while I was working on it and wanted to get some clarification before taking the midterm. For the various resonance structures of ClO4-, does Cl have the same oxidation number of 7+ regardless of what formal charge it has?

Hazelle Gunawan 3G

This was for the homework due this past week, but I jotted it down while I was working on it and wanted to get some clarification before taking the midterm. I was confused by what it meant when it asked which Lewis structures "contributed substantially to the overall structure" of the ion,...

Hazelle Gunawan 3G

There was a question on the homework due last week about drawing out the structures for nitrate, nitrate, phosphite, phosphate, and I was just wondering if we'll be expected to have memorized the formulas that correspond to the names of each ion on the midterm?

Hazelle Gunawan 3G

This was for the homework due this past week, but I jotted it down while I was working on it and wanted to get some clarification before taking the midterm. For the question about the resonance structures of HCNO, why would the structure with a +1 charge on N and a -1 charge on O be favored over the...

Hazelle Gunawan 3G

Darren1e wrote:Could someone explain the differences in molecular structure between increasing anion vs increasing cation size? Also, How does polarizing power relate to distortion?

i'd also be interested in hearing an answer for this!

Hazelle Gunawan 3G

agreed w/ the above! it's important to keep in mind molecular geometry because ex: h2o has a bent shape to it which allows it to be polar

Hazelle Gunawan 3G

can someone explain how elements with a d orbital have an expanded octet? i'm a little confused on how that translates to the overall placement of electrons

Hazelle Gunawan 3G

I watched a video on youtube, which basically stated that there are certain atoms where the oxidation number is pretty much always the same. In oxygen's case, it is -2. From there, the video said that the oxidation number of the whole molecule should be equivalent to adding the oxidation numbers of...

Hazelle Gunawan 3G

Thanks for organizing this!

Hazelle Gunawan 3G

Is this something that will be covered in 14B or did Dr. Lavelle just bring it up as a hypothetical?

Hazelle Gunawan 3G

Is the energy necessary to excite an electron related to the energy necessary to remove an electron? Would that also be a possible explanation for why 4s has a lower energy than 3d?

Hazelle Gunawan 3G

Since this is in the spectrum of visible light, we know that it'll end up at n=2 following the Balmer series. From there you can use the Rydberg equation and input your values for frequency and ending energy level to find the beginning energy level.

Hazelle Gunawan 3G

My questions aren't in the same order as yours but I remember having this problem! I eventually figured out it was because of the conversions between units and not any mistake on Sapling's part.

Hazelle Gunawan 3G

I agree with the previous reply! I'd also add that you should make sure your mass is in kilograms, not grams, because I know I always forget that.

Hazelle Gunawan 3G

Does anyone have any input on which textbook questions for this week they personally felt were the most helpful or most important to complete?

Hazelle Gunawan 3G

I've been taking notes on separate documents for each lecture, but some of the topics are split between lectures so I was wondering if people were splitting up documents by topics or just compiling everything into one document for the week or even the entire course. Additionally, does anyone have an...

Hazelle Gunawan 3G

Avogadro's number is the number of any object in a mole of that object, so you'd only need Avogadro's number when trying to find the number of moles of something.

Hazelle Gunawan 3G

My understanding is that since the electron still has kinetic energy after being ejected, it means that the photon actually had more energy than was necessary to remove the electron, which means you would have to add the KE of the electron to the baseline level of energy needed to eject the electron...

Hazelle Gunawan 3G

I was also wondering about this! I vaguely remember reading something a long while back that the wave-particle duality of light could be treated as a sort of Schrodinger's cat situation? As in the wave model and the particle model coexist until the moment you try to measure the light in question, in...

Hazelle Gunawan 3G

This is a really good question! I thought that the two concepts were directly linked and the photoelectric effect was just an extension of atomic spectra, and I assumed that the threshold energy for displacing an electron would have to overcome the attraction between the nucleus and the outermost or...

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