CHEMISTRY COMMUNITY

AArmellini_1I

Observed rate is the rate that an experiment would find, the experimental rate is the rate we would expect (such as the rate of the slow step). These rates always equal one another but it may not look like they do. When this happens you have to look at the equilibrium constant of the previous faster...

AArmellini_1I

He hasn't specified but from the sounds of it, it may be flexible.

AArmellini_1I

Thank you so much for all your help! I know it may be a lot to ask but I really would appreciate it if you could livestream or record a review session. They help me so much for the exam

AArmellini_1I

Do you mean the lower the activation energy the greater k and the higher the activation energy the smaller k is?

AArmellini_1I

Also, does anyone know if review sessions will still be held?

AArmellini_1I

It's because the solid on the anode side is iodine which is not a metal. You need a solid metal to conduct electricity and thus allow for the movement of electrons

AArmellini_1I

This goes back to Chem14A. You can find an elements oxidation level by looking at the periodic table. For examples, group 1 elements like to lose one electron so they are +1, and halogens like to gain one electron so they are typically -1. But when you are dealing with a compound with multiple eleme...

AArmellini_1I

It's because the whole point of your solid electrode is that it can conduct electricity (movement of electrons) which only metals can effectively do. So will you do have a solid, iodine cannot be the electrode because it is not a metal

AArmellini_1I

I would always check what is spontaneous because that will help you to confirm the cathode and anode

AArmellini_1I

If it is part of the reaction yes, because a compound can be both oxidizing and reducing in a concentration cell

AArmellini_1I

You include H+ ions if they don't cancel out in the final balanced redox reaction. I don't see why it would not be the same for water as well

AArmellini_1I

Use H+ for acidic solutions and OH- for basic :)

AArmellini_1I

Also consider electronegativity when dealing with multiple group 5, 6 ,7 elements such as oxygen and sulfur. While both prefer gaining two electrons, oxygen is more electronegative so if the two were to react S would likely be +6 rather than -2

AArmellini_1I

You can reference the periodic table and see what elements have however many electrons and whether they would prefer to gain or lose a specific number of electrons to fulfill the octet rule.

AArmellini_1I

I believe it because you know the anode is an oxidation. And since you only have O2 and H+ for the anode the only reasonable reaction you can make were oxidation occurs would be having 2H2O --> 4H+ + 2OH- + 4e-. And same for reduction in the cathode. Since you only have OH- and O2 the only reasonabl...

AArmellini_1I

You want to use log when dealing with pH and [H+] [OH-] concentrations

AArmellini_1I

What was Avogadro's favorite sport?
Golf, because he always got a mole-in-one!

AArmellini_1I

I had the same question, but I believe we may have completed it, or at the very least through 6N (I think we covered 6O too though)

AArmellini_1I

I think you have to assume that both half reactions are interacting with water because water is the only solvent we use in redox reactions in regards to this class

AArmellini_1I

I was a bit curious about that question too. The best reason I can come up with is that the solid included in the reaction is a product, not a reactant. Therefore it can't serve as the electrode initially because it hasn't been produced yet, so you have to have an inert electrode to initiate the rea...

AArmellini_1I

Ag(s) doesn't have a positive charge because it is not interacting with another element and therefore cannot be gaining or losing an electron

AArmellini_1I

Thanks, if we were told that a gas expansion is isothermal, do you think that we should assume it's an ideal gas if not specified (so that we can use the fact that delta U=0)? Yeah I believe so! But I don't think whether or not it is an ideal gas influences delta U = 0. I'm pretty sure that stateme...

AArmellini_1I

For constant pressure q = delta H not delta S. So you would use delta S = q/T or delta H / T (you may also be able to use delta S = -n R ln (V2/V1)?
For constant volume you also still use q/T but couldn't use S = -n R ln (V2/V1)
If the system is isothermal use delta S = n C ln (T2/T1)

AArmellini_1I

Yep! If you have changing temperature then you use delta S = n C ln (T2/T1)

AArmellini_1I

The only real difference is how the work was done. In an irreversible reaction the expansion occurs instantaneously, where as in a reversible reaction the expansion occurs over infinitely small expansions.

AArmellini_1I

Then it's based on your conditions regarding the system. Is there constant pressure? How about constant temperature? etc.

AArmellini_1I

Because isothermal means energy/heat and matter cannot leave or enter the system, therefore the internal energy of the system (delta U) is constant

AArmellini_1I

I don't have my homework on me right now but on thing I messed up on was using the wrong R, check your units and make sure everything cancels and you get the correct units. Different R's have different units

AArmellini_1I

Because for infinitesimally small expansion, work must be done and equilibrium must be regained. This repeats over and over and this creates a decreasing curve where as irreversible expansion occurs immediately so work occurs instantly only once and equilibrium is only regained once (this creates a ...

AArmellini_1I

You have to cool it because in order to vaporize you must be at 100C. So first you have to increase the temperature to 100, then vaporize, then reduce the temp back down to 85degrees in order to satisfy the question

AArmellini_1I

Not that I am aware of but usually they email the class when they do

AArmellini_1I

Thank you so much for posting a part 2! I was so excited to see it. I was wondering how you found opportunities (work, internships, volunteering, really any experience). As you probably are well aware pre-med programs at UCLA are tough to get into especially as a first year. Do you have any tips or ...

AArmellini_1I

I believe it just means that its a closed system (like a piston) so while volume may change the pressure will never change because the system can expand and contract

AArmellini_1I

Thats because you only need to break:
1 O-H (463kJ/mol) and 1 C=C (612kJ/mol)
And you form:
1 C-H -1(412) 1 C-O -1(360kJ/mol) and 1 C-C -1(348kJ/mol)
Add that all up and you get -45kJ/mol

AArmellini_1I

Standard reaction enthalpy is the enthalpy of the P and R at their most stable point. Standard enthalpy of formation is the summation of all the R and P standard reaction enthalpy [I think but I may have messed this up]

AArmellini_1I

4E.5
3(837)kJ/mol - 6(518)kJ/mol = -597 kJ/mol
4E.7
a)
1(944)kJ/mol + 3(158)kJ/mol - 6(270)kJ/mol = -202kJ/mol
b)
1(463)kJ/mol + 1(612)kJ/mol - 1(412)kJ/mol - 1(360)kJ/mol - 1(348)kJ/mol = -45kJ/mol
c)
1(412)kJ/mol + 1(242)kJ/mol - 1(338)kJ/mol - 1(431)kJ/mol = -115 kJ/mol

Hope this helps!

AArmellini_1I

Actually I think its 4E and maybe 4D, I'm pretty sure Lavelle said we are going backwards

AArmellini_1I

Because the total enthalpy you end with is not dependent on the process you took to get to that point (like if I boil a pot of water and tell you it initially was 80 degrees and now its 150 degrees. You would tell me the temperature increased by 70 degrees, you wouldn't need to know how long I boile...

AArmellini_1I

Because the more energy provided the more the substance starts to become the next phase, but this energy must go to maintaining that phase so increasing heat won't increase temperature until the substance has completely achieved the next phase

AArmellini_1I

Ok so this might be hard to explain without the picture but basically while converting a liquid to a gas, by increasing the heat you don't actually change the temperature of the substance itself because all that extra heat is being used to hold the newly developing phase. Temperature of a substance ...

AArmellini_1I

Because delta H = enthalpy of final(or product) - enthalpy of initial(or reactant). Exothermic reactions release energy, therefore the enthalpy of the product is less than the enthalpy of the reactants. A small number - a bigger number results in a negative value for delta H. And vice versa for endo...

AArmellini_1I

Basically if K is less than 10^3, meaning that the acid/base barely disassociates, then we can assume that the dissociation affects the initial condition of your acid/base so little that it's insignificant. This means that when calculating your x, or the change in concentration, instead of have your...

AArmellini_1I

I really like Lydon's (spelt his name wrong) workshop on Monday's in Covel. He gives you a worksheet and eventually the answers and it usually covers most (if not all) the material on the test

AArmellini_1I

If your dealing with acids and bases you can always plug your x back into your equation or convert your answer back into pKa, pH, etc to see if your answer makes sense

AArmellini_1I

K changes in terms of temperature but not by increasing/decreasing the concentration of your products or reactants. Q just tells you what direction you reaction is moving (forward/reverse) as it approaches equilibrium or K.

AArmellini_1I

So because it is a very weak acid/base it only dissociates by a small concentration. Therefore, since your KA or B is [products] / [reactants] and only a small concentration will dissociate into the product, Ka or B is a very small number divided by a much bigger number which = a very small number :...

AArmellini_1I

basically the only thing the remains unchanged in the equilibrium constant, and since d is the equilibrium constant and e is the reciprocal, they won't change. And while c does have products/reactants, it lacks the correct exponents so a change in concentration of a reactant or product would result ...

AArmellini_1I

If it's talking about gases used Kp for partial pressure. If it's talking about aqueous solutions use Kc for concentration.

AArmellini_1I

It really doesn't have a ton of application in real-life scenarios so you really just need to understand it as it applies to the homework

AArmellini_1I

So far Lavelle has only really touched on PV=nRT

AArmellini_1I

I would check your assigned TA again, there may have been a change the morning of.

AArmellini_1I

I believe so because both variable uses [products]/[reactants] so I would assume the trends are the same

AArmellini_1I

If you get the right answer no but if you don't get the right answer you may not get as many partial credit points that you would if you did it Lavelle's way

AArmellini_1I

I believe it is because it reacts with water?

AArmellini_1I

I think I'd still with 6A and 6B to be safe and reserve 6C and 6D (maybe a little of 6B) for HW 10

AArmellini_1I

I doubt so but if you know the book lists from - to + it wouldn't hurt to mimic it! Also it may confuse you when looking at what planes (x,y,z) your orbital is in but I don't think he really expects us to know that anyways

AArmellini_1I

For Binary acids: Across a period: more polar the HA bond the stronger the acid (weak bonds) Down a group: the weaker the HA bond the stronger the acid (so greater atomic diameter means a weak bond which means a stronger acid) For OxoAcids: The more O bonded to the central atom the stronger the acid...

AArmellini_1I

It's all about how strongly a base will bond with a H+ so I recommend looking at what characteristics (such as EN and size) will affect it

AArmellini_1I

And while they are different you won't (at least according to my TA) won't be docked an entire point for it

AArmellini_1I

Lavelle said in an email that we finish the syllabus Wednesday and start lecture that same day. So I'd assume Friday is also review :)

AArmellini_1I

I first use the outlines Lavelle provides us. Then I do practice problems and for things I forgot I create a quizlet. Then it's just more practice problems in the areas I struggle in

AArmellini_1I

What I do first is draw the lewis structures. The number of electron dense regions corresponds with the hybridization orbital. For hybridization orbitals it go s, p1, p2, p3, d1, d2, d3, d4, d5 and so on (I believe)

AArmellini_1I

I have the same problem especially in this class. What I'm trying to do is not hyping up the test. Just view as it as an opportunity to prove your knowledge where you will earn as many points as possible as opposed to thinking about the points you may lose.

AArmellini_1I

I believe by polydenate it means the ligand can form three of more chelated rings

AArmellini_1I

So I know Lavelle has discussed naming in terms of ligands but do we need to know the names for acids and bases? I only ask because in fundamentals J questions 7 and 9 the questions gives us compounds such as phosphoric acid and phosphate and we are asked to write the balanced chemical equations. Do...

AArmellini_1I

Ramneet Sandhu 3D wrote:A head-on overlap is formed by sigma bonds. Pi bonds are able to form side-to-side overlaps.

Which is why sigma bonds are stronger than pie bonds

AArmellini_1I

Flash Cards (especially with things that require memorization such as molecular shape) and I like making outlines of sections so I know what I need to know and can test myself to find out what I need to focus on (I pull a lot from Lavelle's outlines and my own notes)

AArmellini_1I

It helps to picture the molecule and look at how many other angles a lone pair will affect in each region

AArmellini_1I

Because in a sigma bond the orbitals can overlap to a greater extent where as with pie bonds they are overlapping indirectly which means they can't overlap as much as sigma bond. The more they over lap, the stronger the pull

AArmellini_1I

Yes, ion-dipole interactions are stronger than hydrogen bonds because the ion has a greater charge

AArmellini_1I

This isn't quite what you are asking for but I'm memorizing them by imaging the electron density shape then removing atoms and replace them with lone pairs. Then you can picture what the molecule looks like and make it easier to identify its shape. For shapes not consisting of lone pairs, I'm memori...

AArmellini_1I

I don't believe Lavelle has mentioned any so in terms of his class, I don't think you have to worry

AArmellini_1I

I think he mentioned something about how since xenon has a low electronegativity, its electrons are more easily distorted by atoms with stronger electronegativity, like fluorine. Exactly! But he was careful to add that this cannot occur with some noble gases, specifically He and Ne, and only at ext...

AArmellini_1I

If you would please give us the actual question. Then I think more people would be able to help! :)

AArmellini_1I

It basically considering how the number of atoms bonded to the central atom will affect it's shape and how the number of electron dense regions (so bonded atom AND lone pairs) affect the bond angles of the initial shape. These two factors ultimately determine the shape of the molecule

AArmellini_1I

Polar bonds are stronger than non-polar bonds due to their dipole movements. As for coordinate covalent compounds I am not sure

AArmellini_1I

The final is cumulative. Even if it wasn't though, a lot of the concepts we are going to learn will likely build off of topics and concepts we learned pre-midterm.

AArmellini_1I

Does anyone know the class average on the midterm from last year?

AArmellini_1I

I always like to remind myself of the units for J before doing a calculation because for most, if not all, equations the values you use for mass should be in kilograms, distance should be in meters, and time should be in seconds

AArmellini_1I

I think he is referring to the common way in which people write lewis structures for organic compounds where they leave out H atoms because it is implied? I think he just wants use to be comfortable with them and be able to understand what they imply

AArmellini_1I

Basically you are going to first fulfill the octet rule for all the atoms in the compound. I personally like to add my extra lone pairs to the central atom in order to form the expanded octet first then check the formal charge for each of the atoms. That way you know if you need double or triple bonds

AArmellini_1I

If Lavelle does give us a question asking us to compare the electronegativity between two elements he will have to provide us a periodic table with the corresponding electronegativity values or the values in general

AArmellini_1I

Natalie Benitez 1C wrote:So in terms of delocalization can we say this occurs when atoms form bonds with each other because in a way they are spreading out their electrons through sharing them?

Delocalization just represents a compound's typical behavior of dispersing charges to exterior elements

AArmellini_1I

Just mean that when an electron drops from a certain energy level to another, that electron releases certain energy or light due to the conservation of energy. That light or energy can be calculated Exactly, which is why this change in energy is often represented by a negative value because it's th...

AArmellini_1I

Black bodies won't be tested

AArmellini_1I

I don't know why specifically in this case this occurs, but Lavelle mentioned that the greater the period the more exceptions to our basic rules.

AArmellini_1I

705198479 wrote:what is the formula for wavelength ?

Wavelength = Speed of Light / Frequency

AArmellini_1I

Also cations and anions have positive/negative charges that can affect the orientation of a molecule (think of water) and affect what molecules they may be attracted too.

AArmellini_1I

Yep! With that being said you can use the periodic table to determine what bond two elements will likely make

AArmellini_1I

You're correct! It's actually an error in the textbook. Lavelle has the explanation under the tab "Solution Manual Errors 7th Edition". With that being said, because Δv = 10.0 m.s-1 then answer is actually Δx = 6.7 x 10-37 m

AArmellini_1I

I believe we are covering this either Friday or next week in class!

AArmellini_1I

Electron spin is just the direction in which the electron rotates (there are two options: up [+1/2] and down [-1/2]). You can almost think of an electron as the earth rotating around the sun, while the Earth rotates around the sun on its orbit in also rotates on its axis. As for what determines the ...

AArmellini_1I

The stern and Gerlach experiment basically consisted of emitted a beam of the same element through a magnetic field and rather than all the atoms continuing on a straight path, some of the electrons curved upwards and the others downwards. Because the n,l, and m(l) variables were all the same it mea...

AArmellini_1I

So first you need to calculate the wavelength of both the proton and electrons with there respective mass. The formula you would use in this case would be: wavelength = h (planck's constant) / (mass x velocity). You kind find in the back of the book that the mass of a proton is (1.673 x 10^-27)kg an...

AArmellini_1I

Lavelle also wanted to emphasized how we are subtracting the lower energy level from the higher energy level (such as n=2 - n=1), and how the positive value is expressed as emitted light at unique frequencies, which in emphasized by using his formula.

AArmellini_1I

Yes both light and electrons have wave like properties. As for defraction, it refers to the wave nature of both electrons and light. Lavelle used the example in class of a filter with holes in it and if you shine a beam of electrons/light through the filter the detector on the other side will detect...

AArmellini_1I

It's like Lavelle described in lecture, we assume water flowing from a facet in continuous but when we zoom in and reduce the flow, eventually gaps begin to form because you're not going to have half a water molecule so to be quantized refers to minimal whole values (n = 1,2,3...)

AArmellini_1I

You most likely won't but if you know the state it won't hurt you to include it!

AArmellini_1I

I had a similar question. I am assuming that because the orbital changes so does both n and the energy so to calculate the energy difference you would use the formulas provided in today's lecture? (I believe we go over orbitals later in this unit so hopefully it will make more sense then)

AArmellini_1I

So after sitting through Wednesday's lecture regarding the Atomic Spectra, I just was wondering if all questions regarding the topic would be mainly focused on calculating the change of energy when an electron moves from say n=2 to n=1. Or is there another large part I am missing? Thanks guys!

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