CHEMISTRY COMMUNITY

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I find it helpful to wake up with enough time to maybe just do a quick look through my notes for a little refresher but to keep cramming right before the test to a min just because that can lead to just more anxiety over little things

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I like doing the practice problems and also looking back at homework problems because they have the detailed explanations of what they're doing so it makes it nice and clear

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I think activated complex is another way of saying transition state or when the reaction is at the top of the activated energy if you were to look at an energy diagram

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you get rid of an intermediate by using the pre equilibrium approach so you create a K equation (K=[prod]/[react]) for the step before the slow step that has the intermediate as a product. Then you solve that equation to get it in terms of the intermediate and can substitute it into the rate law equ...

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I think it would still be an increase in entropy just because you are going from having a solid to all gas which would be quite an increase in disorder

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I think when you do T2-T1 you need a negative out in front of the whole thing so you would get ln(k1/k2)=-EA/R (1/T2-1/T1)

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We know that the units of a first order rate constant are s^-1 and that is shown in the problem when it says the rate constant is 9.63 x 10-6 s^-1 that we''l be dealing with first order

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It is super clear once you draw an energy diagram and it'll show what way you need to add or subtract Ea and the enthalpy

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A is known as the frequency factor and it takes into account the fraction of collisions that happen that actually have enough energy and the right orientation to overcome Ea barrier. I'm not super sure the way to find it but I don't think we need to know that for this class

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the arrhenius equation is k=Ae^(-Ea/RT) and I think one big point of it is to help show how changes in temp are going to affect your rate constant.

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oooh that makes sense thank you!

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Thank you all so much!

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It just depends on if you are using substances that are solid and conductive so if there is no solid and conductive thing there already then the pt has to be added

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I think you used some first order equations for a second order reaction

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Hi I'm curious which question is this referring to?

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I think you would have to also have the Ka whether it be given or you have to look it up

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A catalyst will be present at the start and will come back out and the opposite goes for an intermediate that will be both created and used within the reaction

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i think you have to test each step and find each steps rate law and see which one matches the actual rate law

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So in what situations should we know to include [OH] in the equation even if it's an intermediate? or do we always include it

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what is the point of unique rate laws like what do they help us accomplish?

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no but if u have the unique rate then the coefficients will have some meaning

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A negative delta H signifies exothermic and release of energy while positive is endothermic and requires energy

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yup k' is one way people show reverse reactions but I think that can change depending on who is writing it

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I would say that enthalpy is more related to gibs free energy because you have G=H-TS

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isn't k equal to the slope unless that changes when u get past 2nd order

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I think you can use steady state when the step before is slower than the step after and pre equilibrium when the second step is slower

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Maybe you are thinking of the fact that you cant ln an already negative number which checks out because k can never be negative

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i dont think its true that ln cant be a negtive number because ln of a decimal is negative

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I would say to look at each situation on it's own and look at what is actually doing the work or getting work done on it and then being able to flip between the two as the problem calls for. Ask yourself what is considered the surroundings and whats being considered the system

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Yes you just add each order up so for example for rate=k[A]^2[B] would be a third order because 2+1=3

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Usually Pt will be on the furthest out of whatever side whether it be the anode or cathode

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It has that setup because you want a linear line to be able to interpret what is happening if you make changes. also if you compare it to a y=mx+b setup you have lnk=-Ea/RT+lnA so if the slope is determined by -Ea/R that leaves 1/T as our "x" from y=mx+b and lnk as the"y" and -Ea...

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Hi thats because if we have the limiting step that means its the slowest step, so since the step before it is faster, its creating a buildup almost of reactants that the slower step can't keep up with

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since it's concentration cell you know standard cell potential is zero so you can use E=-.05916*logQ/n

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Hi it seems that the way to balance redox reactions varies depending on the problem and the conditions and Im just confused on where and when to start with these problems and was wondering if someone could tell me just the basics rules kind of for starting depending on the conditions of the rxn

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Thank you great relaxer!

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Happy val day in march <3

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Kinetic has control at lower temps because even if the other product is thermodynamically favored, at low temps there isn't enough energy to overcome the activation energy needed for the thermodynamically stable one, so kinetics will then be predominant.

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Thank you!

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Ya i'm almost positive that all Ka vals and such will be given when needed

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adsorption is when molecules adhere to the surface of something instead of intermixing with it and an example is when water droplets adsorb onto the beads of the silica gel packets so the moisture is removed from the area because the water molecules are drawn to and stuck on the silica beads

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I'm not entirely sure but I do know that one way they connect and that the equation can only be as fast as its slowest step so you can look at the elementary reactions and find the slowest step and know that that will be determining your rate law

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I don't think rate constants are ever supposed to be expressed as negative

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one way that I know for sure is just comparing different experimental results if they're given. If u remember from class to look at the data they give you and then look at what reactants stay constant and compare what happens to the rate as each reactant is changed individually. I found this vid to ...

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I cant say for sure but since we did spend such a large chunk on it I'd say expect a good amount bc I believe my Ta said that the number of questions on a topic usually correlates to time spent on it in class

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For problem where u have a K and you manipulate the equation there are a few things to remember. If you reverse the reaction that means you inverse K, if you add coefficients you put K to the power of whatever the equation was multiplied by.

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There's a list of the 8 strong bases you can look at so if the cation is the same as one of the cation from the strong base list then you know. If it's not one of the 8 then it's a weak base. https://sites.google.com/site/chempendi ... cids-bases

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A photon checks into a hotel and is asked if he needs any help with his luggage. He says, "No, I'm traveling light."

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It depends on if the reactant is an anion/weak base or a cation/weak acid. So for a you have NaC6H5O or you can also say C6H5O-. This is an anion and weak base so it will take a proton from the water to and leave OH-. For c you have C6H5O+ which is a cation, therefore weak acid in this case so it wi...

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Since hydrochloric acid is a strong acid it will completely disassociate in water so there will be no bonds between the chlorine and hydrogen

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The Na doesn't affect Ph and is just kind of there so it's not super necessary to have when solving

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Lexis symbols are the diagrams that just show with the dots the valence electrons of a molecule. The structure shows the arrangment of the valence electrons in an actual molecule.

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There is a list of the 7 strong acids and 8 strong bases https://sites.google.com/site/chempendi ... cids-bases . Anything not on that list is considered a weak acid or base

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A conjugate acid is what's formed when a proton gets added to a base and a conjugate base is what is formed when you remove the proton from an acid

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A polyprotic acid can donate more than one proton or hydrogen. And a polyprotic base can accept more than one H.

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I think that because of it's shape it would be impossible for all four oxygens to bind to the same metal. Only two would be able to bind to the same metal molecule.

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When you decrease the intensity of the light you are just decreasing the amount of photons that are coming out. So the actual energy of an individual photon isn't being changed so the KE stays the same, but the number of electrons that will be ejected will decrease because it is being hit with less ...

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I believe there are only the 8 strong bases LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 which are all alkali metals with OH

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https://i.pinimg.com/550x/36/3d/50/363d502c732e687a8089f2a230e16caa.jpg

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What kind of fish is made out of 2 sodium atoms?
2 Na

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That sounded like complaining I mean I love 80 degrees but I especially love getting all four seasons:)

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I will hopefully be skiing lots and enjoying the snow before I come back to the 80 degrees

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Ya I would say just forcing myself to leave my dorm and the hill even when I really don't want to can instantly change my day

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go to as many review sessions as possible

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The fact that all I need is to go hard for this next week and half and then I'll be fully done is what's getting me through

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Most up to date is 9:30 -11am

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Anna Turk 1D wrote:Does anyone know if it will be mainly what was covered after midterm 2?

I don't think so I believe the number/variation will be based off of how much time we spent on the topic in class

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H3O+ is just what you get when the H+ actually joins with the H20. but it's only H+ thats actually coming from the acid but i think they are used pretty interchangeably

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is coordination compound naming on the final?

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i guess i should add it's the same in the sense of the lockdown browser, the multiple choice, and no notes but the final is 90 minutes and has more than 15 questions

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it is the same format as the midterm

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this is because when a weak acid doesn't react to completion so it partially ionizes in water it will turn into some H+ and some A- where the A- will be able to form back with and take the H+ which would leave the solution with more OH- causing it to be more basic

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not entirely sure if you are talking about this but if you are talking about identifying if a salt is basic acidic or neutral then you see if the cation came from a strong base and the anion came from a strong acid then it will be neutral. If the cation is conjugate acid from a weak base then the sa...

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The key is to know the strong bases and acids

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You have to know the strong acids and bases so you can recognize if the cation is coming from a strong base or anion from a strong acid.

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I believe there are only 4 regions of electron density because 7 valence e times 4 is 28. Plus the 4 valence from Si is 32 electrons which you get by doing each F with a single bond

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is it saying you have incorrect formatting in the feedback?

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I think it might have to do with the oxygen being added and making the bond more polarized so the electron density is even further from the o-h bond so it releases the h easier

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Its also fun to use household ingredients for cooking to do experiments such as elephants toothpaste where you use something like yeast with hydrogen peroxide so you get to see the rapid decomposition of the hydrogen peroxide. You also get the visual of the energy being released with the steam that ...

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the conjugate base is what you get after it donates its proton which in this case would make it SO4 2-. and the conjugate acid is what is there after something accepts the proton which is why you get H2SO4

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this is when it can donate more than one proton

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i believe the conjugate base or acid is what is leftover after the acid donates a proton or base accepts a proton

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idk about being a molecules that can be a conjugate acid and base but i know water can act as a acid or base

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I'm not entirely sure about that but my ta did say there would be more or less questions based off of how long we spent on the topic in class

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It can be helpful to remember F as the most electronegative and base the other elements off of that

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Another thing to remember is it's not always black and white and both ionic and covalent character can be exhibited

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I don't have an answer but thank you for sharing that place and it seems like it is based off the way the older and newer crystals expand and react to the weather patterns

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Ya based off what lavelle said in class it seems like it just depends on where you and varies from textbook to textbook because there is the also induced dipole-induced dipole

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Another helpful thing to remember is that depending on how many points of electron density are there will depend on if its the axial or equatorial atoms that will be removed first if necessary

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Also helpful to remember that pi bonds needs the double bond to form and do not allow rotation

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I think he'll definitely send out an email clarifying the options that will be available

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It seems like it would go back to how they were before covid, which i believe were not the online multiple choice tests but you never know

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Another thing not to forget that can be confusing is that london dispersion forces can be called many things like induced dipole-induced dipole, disperion, or van der waals

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midterm grades are up and im especially thankful for my parents right now

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If you look at past posts there is another similar question in which they answered that it's monodentate because both of the lone pairs are on only one atom.

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Yes I believe they decided all will be in person for next quarter.

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