CHEMISTRY COMMUNITY

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Hi! the equations are still valid for exothermic you will just get a negative number, although because exothermic reactions happen spontaneously I think it is fairly rare to see them reversed.

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I think in general it has a small effect (so conceptually yes) but in reality it is pretty negligible and can be disregarded.

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Hi! Enthalpy doesn't change because it is a state function (so it only matters where it starts and ends not the path it takes) the best way for me to think of catalysts is everything stays the same it just takes a path with a lower activation energy, so it does not affect the enthalpy.

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Hi! personally I have found myself using the one that solves for k more frequently than lnk, but its pretty easy to go from one to the other.

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Thank you for this! Very helpful post, and I think your steps look correct

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Hi! I think the biggest difference is that if something is thermodynamically stable it is spontaneous (so it tells if it will happen or not) and then kinetically stable refers more to the speed of the reaction.

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Hi! I agree with what was said above that reaction rates will always be negative in this class, I would think that if a reaction rate was positive than there would need to be a catalyst or enzyme invlolved.

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Hi! Yes we want to use that when we are solving for half life because you will most likely with a ln that you want to cancel out. I believe half life was the only time we used it in lectures thus far though.

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The best way for me to approach kinetics problems is to just start with what you know and try to figure out how those things relate to find what you want to find and then trying to match that to one of the equations.

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Hi I think that all reactions have a certain amount of time they take to occur based on outside factors so one is always "naturally" faster than the other.

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I think this is covered in sapling, but the general format is anode ion/ ion compound//ion compound/ ion cathode

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I agree with Eliana, I kept switching the side of the cathode and anode on sapling and it took me a while to figure out that the order mattered.

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Hi! As I understand it the purpose of the salt bridge is to allow for the electrons to flow freely from the cathode.

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I don't think that there can be a perfect isolated system, however I'm pretty sure that we are able to get fairly close and therefore in these cases can make the assumption that it will follow the same equations and patterns as a perfectly isolated system.

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Hi! I think a species can be both if it is the only thing on the left side of the equation and is then broken up in the reaction

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I think you know by either the subscript (or superscript in some cases) of 0, but you can also tell if its delta H of formation is 0.

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Hi! I think if if was a monatomic or linear Gas we would be told as well as given what value we should use a R.

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Yes! I think this is because if there was a change in temperature the reaction would favor one side over the other and therefor it would change into an irreversible reaction.

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Hi! I think for this question you just need to add the change that occurs when you go from a solid to a liquid and then the change that occurs as a result of the increase in temperature (I think Hess's Law makes that possible)

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Hi! like above I don't think you would ever have to integrate because he never went over it in class, I think the whole point of him showing that was just to show where the equation came from. Also yes in real life a system is never perfectly in equilbrium!

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Yes I agree with above that delta u=q+w and then we derived different equations from that in lecture 11 I believe

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I think the equation should be -deltaH/T. I kept getting this question wrong because I was forgetting to convert the units, and then I believe you just add your previous two answers for the third one.

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I think that normally the units are in KJ but if you are taking into account moles then you use KJ/mol

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I agree, I can't think of a situation where the signs would be opposite for the phase changes. I thought that was a bit of an odd sapling question too.

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Hi! I agree with above that it is often a case by case basis, but a pretty clear sign is if it asks for it in terms of moles

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I think I started by converting the mass of ice to moles and then multiplying to get q1 and then using the equation q1+q2=-q3 to solve the rest

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Hi! just to add on to what has already been said, my TA in discussion section said that you can pretty much assume that the reversible reaction will do more work every time than the same reaction that is irreversible.

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I agree with above that it dose not matter what the compound is if you use the change in temperature and information from part a then you should be able to solve the question.

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To add on to above, I think the most confusing thing is that the phases are the lines that have a slope and the flat lines are when the phase is changing so make sure to note that!

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Hi! I think he meant that because endothermic reactions need heat to happen that when heat is added the products are favored and the reaction shifts right (I think of it like adding heat is like adding a reactant), so I think all he means is the addition of heat favors more products being formed

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Hi! I agree with above that because endothermic reactions require heat it is like adding a reactant (even though that is not a perfect analogy) so the reaction will favor the products and K will increase. I'm not entirely sure how it would effect exothermic

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hi, like above I think they are used alot because they appear commonly in nature and can bond to water. When you see a COOH group it is a good sign you are dealing with a weak acid

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You only have to divide by L if you are given moles instead of molarity (you need to use molarity for the ICE table)

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Hi! I agree with above that you should have the most common strong acids memorized, but that will probably happen in general with the more practice problems you do and recognize the patterns that they share

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I think you should use whatever units are given in the problem, even though it is usually kj/mol I think they would specify if they wanted kj/mol or j/mol

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I agree with above that you use the table and add that to the enthalpy of the phase change which I assume would be given, if not does anyone know how to calculate that?

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I agree with above that heat is not a state property because we cannot only consider the initial and final temperatures, unless it is a specific closed system. I think the best way to think of this is the way professor described it in class where you're hiking and one person might take a longer path...

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Hi! like said above I think it's different than the photoelectric effect. I believe that any phase change would require energy but I don't think we have to memorize these values.

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Hi! I think anything listed in a state not a gas (liquid or solid) it will be given to us because I don't think he ever went over how to calculate those values on our own

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Hi! I would always check with the outlines Lavelle puts on his website for what is expected for us to know for exams. I agree with above that it never hurts to know the formula.

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Hi! I think we should be able to recognize weak acids and bases based on similarities and patterns (for example COOH often indicates a weak acid) which does take a little bit of memorization, but with the more practice you get they become easier to see.

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I don't think gasses cannot be pure, I think we just cannot assume that they are where as he told us that we could assume liquids are pure.

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Yeah, I believe he said in lecture for the purpose of this class we only need to think of partial pressure applied to gasses

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Yes as everyone has been saying above you cross multiply the denominator by the value given then move all your terms to one side to get a quadratic equation. He also went through one example where there was an x^3 term which is what you might have been thinking of.

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Hi! I think the first step is always to balance the equation, but because that is more of a 14a concept I have noticed he has been giving us the equations already balanced most of the time.

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The above response is correct, because they can be used interchangeably in some instances I think it will be made clear in other ways what it is asking you to do.

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Lavelle normally posts detailed outlines that include what material this course covers and posts it on his website, if there is no mention of it there it's pretty safe to assume that you will not need to know it for this course.

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On sapling you can find a pdf version you can download which is probably easiest, they also have all the odd answers in the back of the book.

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I agree with the above answers that just comparing the q and K values cannot tell us about the stability of the reaction only which there is more of.

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Hi! I agree with what was said above that because the volume does not change with the addition of inert gas the concentration stays the same

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I agree I think the easiest way to think about it is when it reacts with water the positive end of the water molecule with mix with the negative end of the molecule (and vice versa) and will pull the ionic compound apart.

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Hi! I agree with what was said above, I think it is easiest to tell if you draw out the molecule because then you can clearly see that it will accept electrons.

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I agree with what was said above I think expanded octets are the last resort and it is more stable for oxygen to have a charge than sulfur

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Hi! like said above it's easiest to determine if a molecule is not symmetric than it is probably polar, if it is symmetric than you can look to see if the dipole moments cancel

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I think it could also depend on which series you're using but I think n=2 would be a good rule of thumb

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I think we should have them memorized, I would recommend flashcards for this because they can start to get a little confusing once you stray from the basic ones.

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Hi! Lone pairs are more repulsive than bonded atoms because they are held closer to the nucleus meaning they are not "stretched" out as far and their repulsion is more concentrated.

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Hi! like said above the more oxidation states available the less tightly electrons are held and can move more easily making these good for electron transfer

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I think flashcards are really helpful, even just the step of sitting down and making the flashcards always helps me remember information better.

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When the coordination complex has an overall negative charge you add ate (and drop the ide I think)

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Hi! Like said above it is probably helpful to have a basic understanding of nomenclature, however I do not think this would be a topic of focus. I always go off of what Professor posts in his outlines for what is needed to know for exams and I do not recall seeing that on the list.

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When a molecule has resonance the electrons are delocalized because there most stable state is somewhere between the lewis structures we can draw and most likely double bonds are present, these structures would have delocalized pi bonds.

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Hi! axial bonds are formed on the vertical axis (think of the earths axis, or the y axis in math) while equatorial are on the plane of the atom ( Think of the earths equator). If you have lone pairs than they both should be axial because that will create a 180 degree bond which is most stable, howev...

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Hi! I agree with what has already been said, I think that there would still be an effect and repulsion but it would have a smaller electron cloud and therefor less of an effect on the shape.

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Hi! I have personally found that making flashcards with the number of bonds, lone pairs, bond angles and shape have really helped me learn this content, I always prefer to make my own because I feel like it "sticks" better than just looking at a chart. Hope this helped!

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The dipole moment is how we can measure the difference in charge between to atoms or molecules and are typically represented by an arrow showing the way electrons are being pulled. This is an important concepts because this is one way molecules and ions are able to interact, although it is a fairly ...

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This is really helpful, thank you! I typically try to just use what I know about atomic radius and how that would effect the other trends, but this is probably much faster.

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Hi! Like other's said electron density is just the probability that an electron would be found there so a higher electron density would mean a higher probability of an electron being in the area. The term itself is a little confusing because it does not mean what we typically think of density meaning.

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I think dissociation energy is the energy needed to break a bond so weaker bonds are easier to break therefor they require less energy to break. Hope this helped!

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Hi! Posted above are great ways to find the oxidation number however for this course I don't think it is necessary to know. It is not listed in Professor Lavell's guidelines which I feel are an accurate way to know what is necessary to know.

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I'm pretty sure it refers to all unbonded electrons, if you go by the number of lone pairs you will only count half the number you need to

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I don't think you need to worry too much about having to draw all the resonance structures for an exam, as long as you know how to identify them and then compare (for example which is the most stable etc. like the sapling problem) I think you will be in a good spot.

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Hi adding onto everything else said I think there are some that do not need 8 to be stable for example B and Be can be stable with less than 8

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I think you should always include the formal charge if it is not equal to 0. That is at least what sapling wants so I would assume it would be the same for an exam

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Hi! Like others have said I think we will be required to know/ memorize the bond angles and names. Most of them are pretty straightforward/ can be figured out pretty easily. I'm not sure how much more complex we will get in this class but I assume the ones gone over in class are fair game.

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Hi! Yes you need to take lone pairs into consideration because they repel other electrons (especially other lone pairs) more than bonds. For this reason they effect the bond angles and shape of the molecule. Hope this helped.

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I think this would be a fair question to ask because it can be figured out using what we know about periodic trends.

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I think being able to recognize them is more important then memorizing them, based on how this class has worked so far I feel like it has been more about application than memorization. Also I think if it was something to memorize it would have been listed on the outline.

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If it helps you can use periodic trends to help answer this one! Ionization energy increases across a row and we know that Nitrogen and Fluorine are to the right of Oxygen. You could also use the trend of atomic radius which we know decreases as you go across a row because the electrons are being he...

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Hi! I always use the atomic radius to remember the rest so then I only have to memorize one. Atomic radius decreases across a period and increases down the group. If you then think of how increasing or decreasing the radius it effects the other principals (I think they are almost always inversely re...

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Hi! As others have said, I don't see it on the outline posted on the class website so I would not worry too much about it for this unit. While I think sapling is helpful sometimes I think can be a little off of what we are actually required to know.

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I am also confused by this, on sapling whenever this is given I just ignore whatever comes after the +/- and have always gotten the correct answer that way.

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Hi! yep you are correct, the lines refer to the distinct changes of energy possible so you thought process of 7-6,7-5,7-4,7-3,7-2,7-1 is right!

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I've noticed the same thing, I've always just used the equation on our equation sheet and process done in class because that's how we should do it on the exams, but you get the right answer either way so I don't think it really matters which process you use.

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Hi! Like said above I think its 5 times per week, you can also see how many times you've posted if you click on your name in the top right corner and go to "your posts" hope this helped!

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Hi! I've also had increased jitters when it comes to taking exams online! Some things that I have found helpful include scheduling in breaks for myself, waking up early that day after getting a good nights sleep and getting lots of exercise and time outside so I can get a little time away from scree...

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Hi! I'm not entirely sure but I would say that if you have a good idea of the concepts that would be fine and that you would not need to necessarily know the specifics of the experiments.

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I think because we covered it in our lectures before when he indicated it's fair game for the midterm.

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Hi! I would assume that because we provided with a formula sheet and the midterm is multiple choice this would not be a question that would come up.

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I think it is always recommended that you use the whole number until the end then round using sig fig rules. This should not be an issue for the exam because it's multiple choice.

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They found out from the photoelectric experiment that the kinetic energy of the ejected electrons equals the energy of the photon minus the work function

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I have always found outlining to be really helpful for me! I also think that during this weird time it is even more important to schedule out time to study and time for breaks so you are not cramming at the last minute or drive yourself crazy studying all the time. Hope this helped :)

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Hi, I do not think there is a curve but if there is it will only help your grade never lower your grade to encourage students to work together.

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Yes! Like everyone else has said they adjusted the brightness. I believe if they would have increased the wavelength there would have been a change.

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For the exams do we need to know the names of the equations or is it okay if we just know the equatiom?

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Hi! Sig Figs are used to make sure we are accurately presenting how precise the data we used is. For example if we measured something with a beaker that only had dashes for every 10 cm we could not be as precise as if we were measuring with something that has dashes at every cm. sig figs takes this ...

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Yes like others have said theoretical yield is assuming that there is no mass lost due to error (like sticking to the tube, etc.) When you compare the theoretical yield to what was actually produced you are most likely finding the percent yield.

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Hi! While it's correct to have a fraction, I think it's preferred to change it to a whole number. If it's written as a fraction you just have to multiply through by the denominator which does not take alot of time. I would go better safe than sorry and change to a whole number

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Hi! While it's correct to have a fraction, I think it's preferred to change it to a whole number. If it's written as a fraction you just have to multiply through by the denominator which does not take alot of time. I would go better safe than sorry and change to a whole number

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I think I used 100g/mol because the math worked out a little nicer and the answer was right.

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