CHEMISTRY COMMUNITY

Nathan Nakaguchi 1G

Dear Dr. Lavelle, Prior to attending UCLA I had very little chemistry experience and was afraid of what I had signed up for. I am very grateful I have had the opportunity to learn from you for the past 2 quarters and be well introduced to one of my now favorite branches of science. I really apprecia...

Nathan Nakaguchi 1G

E=E^0-\frac{RT}{nf}ln([P]/[R]) E and T were given and R is a constant (8.314 J/Kmol). E^0= 0 since this is a concentration cell. The question also provided that the cathode reaction had some concentration which for concentration cells is the [R] in Q. Considering that the redox reaction was...

Nathan Nakaguchi 1G

A linear plot is in the form y=mx+b and for second order 1/[A] = kt + 1/[A]0 so the plot is 1/[A] versus time.

Nathan Nakaguchi 1G

A catalyst is something that lowers the activation energy of a reaction and leaves K(equilibrium) the same, a catalyst should appear on both sides of the total net reaction, while an intermediate is an actual product between steps of the reaction that will not appear in the final net reaction.

Nathan Nakaguchi 1G

For balancing basic reactions, add H2O's to balance oxygens, after that add H+'s to balance the hydrogens and finally, since this is basic, for every H+ you have added add that same amount of OH- onto both sides and note that OH-+H+=H2O, so there will be some cancelling.

Nathan Nakaguchi 1G

G=Wmax always, but I think that Wmax=-nfE is only for standard conditions since that equation is from G^0=-nFE^0 which the knots imply standard conditions.

Nathan Nakaguchi 1G

Volts = Joules/Coulomb so in deltaG=-nFE the Coulombs cancel out from the multiplication between E (J/C) and F (C/mol), and n has no units so the final units will be J/mol.

Nathan Nakaguchi 1G

Q is the reaction quotient, [Products]/[Reactants] for concentration cells the lower concentration is the products. Or the cathode concentration is the reactants and the anode concentration is the products.

Nathan Nakaguchi 1G

You add an inert electrode like Pt(s) when the reaction does not have a conducting solid already involved.

Nathan Nakaguchi 1G

k and k' have different numerical values, it is not necessarily a derivative or anything like that. Its the same as just arbitrarily writing k(forward) and k(reverse) they are both rate constants but they aren't the same number.

Nathan Nakaguchi 1G

The equation to use is E^0=(RT/nF)lnK and not E since pH's are always recorded at equilibrium, but that's about all I know. I'm also really unsure why K=[H+]/[OH-] either... the only reaction I can think of relating the two is H + OH -> 2H2O but that doesn't work here.

Nathan Nakaguchi 1G

Lavelle said in an email that the test will include the 2nd page of outline 4 and all of outline 5 (which can be found on his website)

Nathan Nakaguchi 1G

You add xOH-'s to both sides after adding xH+'s. Notice that OH-+H+= H2O so that will likely lead to some cancelling. The general steps for balancing a redox reaction in a basic solution is 1.balance everything EXCLUDING H&O 2. balance O's by adding H2O 3. balance H's by adding H+ 4. add as many...

Nathan Nakaguchi 1G

n is the moles of electrons within the redox reaction, so for this question n=2 since your reduction reaction and oxidation reaction both have 2e-. Concerning the separate Nernst equations either is fine to use, however the -.05916V one always assumes temperature is 298k (this one might also be more...

Nathan Nakaguchi 1G

I ran into this too, I believe your right, it is supposed to be Cl- instead of Cl2 on the products side.

Nathan Nakaguchi 1G

Compare the reduction potentials (E^0) of Ni2+ + 2e- to the H2O reduction reaction provided at the top of the 6O exercises. The cathode will be determined by which of the two has the greatest reduction potential, the one with the greater reduction potential will be the cathode and the one with the l...

Nathan Nakaguchi 1G

Yes, add OH- to both sides and when you do also notice that OH- + H+ = H2O.

Nathan Nakaguchi 1G

Platinum is not very reactive, so it is just a means of getting electrons from one source to the next for the actual electron transfer to occur.

Nathan Nakaguchi 1G

R is a constant, so refer to the constants and equations sheet on Lavelle's website and choose whichever has the correct units you need. T is provided in the question.

Nathan Nakaguchi 1G

E is the electromotive force of the system at any condition where $E^{0}$ is the electromotive force of a system at standard conditions 1M, 1atm, 273K.

Nathan Nakaguchi 1G

You have to use the H's and S's instead of G's because you have to calculate G for temperatures different than 25C. So you have to use $\Delta G=\Delta H-T\Delta S$ where T= the different temperature.

Nathan Nakaguchi 1G

Off the top of my head I think something like 4I.9 and 4.43 were homework questions very similar to two on the midterm.

Nathan Nakaguchi 1G

When a system is isobaric that means that the pressure is constant, which implies that $w = -P_{ext}\Delta V$ . I think this is all you can assume from this statement. Dr. Lavelle didn't really talk specifically about the terms isobaric and isochoric, so I wouldn't worry too much about it.

Nathan Nakaguchi 1G

\Delta G=\Delta H - T\Delta S If delta G is negative then the reaction is spontaneous. So, using the formula above you can see that in most cases when \Delta H is negative and T\Delta S is positive or T\Delta S< \Delta H then delta G will be a negative value. There are generally more scenarios of d...

Nathan Nakaguchi 1G

STP is 0 Celsius, 273.15K and 1.00 atm. The equations should work the same except for when your dealing with things like \Delta G_{formation}^{0} where typically those constants are calculated from 298 K. Its probably best to make sure the constants you're dealing with explicitly say it was calculat...

Nathan Nakaguchi 1G

$\Delta S_{total}=\Delta S _ {system} + \Delta S_{surrondings}$

Usually if its just $\Delta S$ then it is the change in entropy of the system, the total change in entropy refers to both the change in entropy by the system and the surroundings.

Nathan Nakaguchi 1G

\Delta U = 0 when it is isothermal, meaning temperature does not change so there is no net addition or loss of energy to the system. \Delta U = q + w, but since w can be written as w = - P_{ext}\Delta V in the cases where \Delta V = 0 (the volume is constant) then \Delta U = q + 0, thus \Delta U = ...

Nathan Nakaguchi 1G

I'm not very sure, but my guess is that since its standard Gibbs free energy the only thing you need to do to calculate it is to do a Hess's approach only involving standard G's of formation.

Nathan Nakaguchi 1G

Recall that S= Cln(T2/T1) so that's where heat capacities comes into play. Additionally since S= q/T then S= deltaHvap/T Those will be the formulas you will use for this question. Also, I think you have misread the question, you're reading 4F.18, 4F.17 uses water (however they both are the same proc...

Nathan Nakaguchi 1G

For an isothermal process there is no addition or loss of energy, so the change in internal energy is 0. Since delta U = q +w and the definition of isothermal is no addition or loss of energy from the surroundings 0 = q + w. Doing some algebra can yield q = -w (remember that usually work= -number, a...

Nathan Nakaguchi 1G

If you were to compare the change in entropy of 1 mole of a monatomic ideal gas to 1 mole of atoms making up diatomic molecules by increasing both their temperatures, why would the 1 mole of monatomic ideal gas have a greater change in entropy? Problem 4H.9 in the book says something like this as an...

Nathan Nakaguchi 1G

Say A->B is exothermic, meaning that B is a lower energy state than A. B is therefore more stable to have, since it is lower energy meaning that the reaction will spontaneously go towards B (until it reaches equilibria) The less stable A requires more energy to be maintained. So if you add more heat...

Nathan Nakaguchi 1G

Bomb Calorimetry is the measure the amount of heat given off by a combustion reaction with constant volume.

Nathan Nakaguchi 1G

When the K is less than 10^-3 you can assume that something like (.15-x) in the denominator will become just .15 since the x is so small it does practically nothing to that amount. But in the numerator you will still have an x present like x^2.

Nathan Nakaguchi 1G

I think that the only difference between using the two is their corresponding heat capacities (in q=mCdeltaT). So for qp you'd use Cp and for qv you'd use Cv in your calculations.

Nathan Nakaguchi 1G

For the most part delta H will be in kJ/mol, as it represents energy absorbed or released per mole. I think that the only time it would not include "/mol" is probably when units in an analysis cause moles to cancel out. To be safe, I would stick with kJ/mol as it is what I've seen for majo...

Nathan Nakaguchi 1G

State properties only depend on the final and initial states. They only care about the destinations not the journey. So for deltas like enthalpy (which is state dependent), to find it you do: Final state value - Initial state value. State dependent properties are only involved with total changes fro...

Nathan Nakaguchi 1G

For Delta H= +, then the reaction is endothermic meaning it needs input For Delta H= -, then the reaction is exothermic and is spontaneous/needs no input. The way I like to think about these is that for endothermic reactions, they are going from generally a more stable form to a less stable form thu...

Nathan Nakaguchi 1G

Only liquids and solids are supposed to be ignored in K expressions. While it is H2O we're talking about (which tends to be liquid and therefore ignored), if it is in the gas phase it should be used in the expression.

Nathan Nakaguchi 1G

This one is a little tricky, you have to recall that Al^3+ is a central metal ion in chelation. So for AlCl3(aq)+H2O(l) the Cl3- ion and Al^3+ will depart and Cl3- doesn't affect the pH so it can be ignored. With the Al^3+ in water 6 H2O molecules will bond to make an octahedral Al(H2O)^3+. So the r...

Nathan Nakaguchi 1G

Partial pressure just means that it is a part/contributing to the total pressure of the system. Total pressure is all of the components of a reaction that would cause pressure combined (added), while partial pressure is just the pressure created by one aspect or compound.

Nathan Nakaguchi 1G

Le Chatelier's Principle says that a reaction will counteract a change in order to stay at equilibrium. If you increase pressure, P=(nRT)/V then you are essentially altering n/v (either making v smaller or n larger) also known as the concentration, and if the concentration of everything in the react...

Nathan Nakaguchi 1G

Q can be calculated at any time (including when it has already reached equilibrium) during the reaction and is compared to K to see the direction of the reaction, if Q equals K then that just means the reaction is at equilibrium.

Nathan Nakaguchi 1G

From my ebook it says 1.20 milli-mol SO2, I think you may have misread it or there was a typo.

Nathan Nakaguchi 1G

It is partial pressure because each compound contributes to the whole pressure, so the pressure caused by one compound would be called partial.

Nathan Nakaguchi 1G

I think its cause .10M is the initial value of HClO2, I think the way that Ka and Kb work is similar to regular K in that the system must be at equilibrium. So while there was .10M of HClO2 at the start, .06 of it became the product(s) thus the final HClO2 you would put into the calculations would b...

Nathan Nakaguchi 1G

In question 5i.29 of the textbook, the solution has you omit x in the denominator (or treat it as 0) since the provided K is 3.2x10^-34. For what range of K's are you allowed to say is small enough to treat x as a 0? And what is the justification for treating the denominator's x like this?

Nathan Nakaguchi 1G

Problem 5i13 part c for reference, asks to determine which dissociation reaction is more thermodynamically stable, dichloride to chlorine vs difluorine to fluorine. I know that a reaction reaches its equilibrium constant so that its ratio of products to reactant is as stable as possible, but how can...

Nathan Nakaguchi 1G

How can you tell if more products or more reactants will be formed in a reaction based on the Q compared to K? Is there a rule? Like if Q<K it makes products?

Nathan Nakaguchi 1G

Heterogenous equations contain compounds with different (hetero) phases while homogenous equations contain the same (homo) phases. For example a chemical equation with all of the compounds in the gas phase is homogenous.

Nathan Nakaguchi 1G

I believe it is true, the reaction will reach equilibrium ending with satisfying the ratio of products to reactants (and/or reactants to products ratio) that makes up the equilibrium constants k and k^-1.

Nathan Nakaguchi 1G

It is probably best to memorize the periodic trends that electronegativity has. As you go up the table EN increases, and as you go right from the table EN increases. I don't believe we will have to know the exact hierarchy/numbers for electronegativity, we just need to be able to work with the idea ...

Nathan Nakaguchi 1G

Oxoacids are a category of acids that contain oxygen(s) which hold/bond to the soon-to-be-donated hydrogen/proton. HClO3 falls into the oxoacids category.

Nathan Nakaguchi 1G

S character of a bond refers to how much S is relatively in the hybridization, so for example sp3 has less S character than sp2 which will have less S character than sp (ie sp will have the most S character). Notice the corresponding molecular shapes that one can make with the central atom being sp3...

Nathan Nakaguchi 1G

I understand that electron withdrawers would make a molecule more likely to lose a proton and electron donors less likely to lose a proton. Like in 6C 21 how acetic acid is a weaker acid than formic acid since the methyl group -CH3 on the acetic acid is a electron donor to the molecule making it sli...

Nathan Nakaguchi 1G

How does one determine which base is stronger than the other. For example, in 6C 17 we are given BrO- and C17H19O3N (morphine). My answer would be that BrO- is more basic because it has a full negative charge to pull in protons compared to morphine's possible lone pairs (I haven't attempted to creat...

Nathan Nakaguchi 1G

kA is the ionization constant which lets you know how ionized/dissociated a solution is. The more ionized the stronger it is, the less ionized the weaker.

Nathan Nakaguchi 1G

Sigma and Pi bonds are overlapping of orbitals/ sharing electrons. Ionic bonds do not share electrons, so they do not include sigma or pi bonds.

Nathan Nakaguchi 1G

I remember it as a base will take electrons while an acid will receive electrons (I use bronsted definition).

Nathan Nakaguchi 1G

I believe it is primarily the naming rules for coordinate compounds, which involves memorization of ligands. Dr. Lavelle sent a email with a pdf for the naming rules and etc around 2 weeks ago. I would go off of that.

Nathan Nakaguchi 1G

I'm currently doing the Acids and Bases HW and a handful of questions require knowledge of ate/ite nomenclature which I don't think we have covered thus far. For the final will it be necessary that we fully know "-ate" nomenclature? Thanks.

Nathan Nakaguchi 1G

A= central atom, X= bonded atoms, E= lone pairs.

T shape: AX3E2

trigonal pyramid: AX3E

Nathan Nakaguchi 1G

I'm not sure specifically. I believe the fact that pH impacts function in biological systems is probably the biggest takeaway. I'll just list what other parts i remember that where mentioned: Copper, Zinc, and Nickel contribute to enzyme function (which speed up reactions), Iron is important in elec...

Nathan Nakaguchi 1G

I believe cisplatin bonds to two nucleotides preventing the DNA to unravel itself and therefore can't be replicated.

Nathan Nakaguchi 1G

I think that the amount of electron densities a molecule has corresponds to the kind of hybridization shells may be present so VSEPR may be helpful in that if you know its shape you know the amount of electron densities. .

Nathan Nakaguchi 1G

A ligand is a ion or molecule that creates a coordinate bond to a metal atom.

Nathan Nakaguchi 1G

Usually A will not be accompanied by a number as it represents the central atom of the molecule. Whatever number follows X represents how many atoms are bonded to the central atom A. The number that comes with E represents how many lone pairs there are on the molecule's central atom.

Nathan Nakaguchi 1G

I think of regions of electron density just as another way to say a general area of a group of electrons, so on a molecule wherever either lone pairs or bonds are is a general area of a group electrons therefore regions of electron density.

Nathan Nakaguchi 1G

Personally I refer to it as angular for the same reasons as the post before me. I think either is acceptable, but double check with your TA to be sure.

Nathan Nakaguchi 1G

Unless on Monday Dr. Lavelle teaches hybridization, I don't think it will show up on test two. He said for test two it will be everything between the midterm up until Monday (which I believe he will teach sigma and pi bonds).

Nathan Nakaguchi 1G

polarizing power is how powerful a specific ion can polarize another ion (pull/distort a electron cloud towards it)

polarizability is how much a ion can be polarized (how distorted it can become)

Nathan Nakaguchi 1G

In general, what range of time should we use as a rule of thumb to determine what section of homework we should turn in for that week? I'm guessing 2 but I'm not sure if thats too long.

Nathan Nakaguchi 1G

As you go up and right electronegativity increases. Using this rule Florine is more electronegative than Bromine so the differences are EN(F) - EN(C) versus EN(Br) - EN(C) since you know Florine's EN will be larger the difference between F and C is greater than C and Br. The more significant the ele...

Nathan Nakaguchi 1G

Yes you are correct in both aspects. The way I think of it is a covalent bond is a sharing of electrons and ionic bond is an anion stealing an electron from a cation. Some ionic bonds show covalent character because as an ionic bond is polarized some electrons (or the electron cloud) "stretch&q...

Nathan Nakaguchi 1G

When energy is absorbed the system is gaining energy so that is a positive delta E going from a lower n to a higher n.

Nathan Nakaguchi 1G

For the midterm will we need to know how to do electron configurations for the f shell?

Also, do we have to memorize what general ranges the different types of electromagnetic radiation are in?

Nathan Nakaguchi 1G

I'm not 100% on this but I believe Dissociation energy is always positive because breaking a bond will always require an input of energy, if it were to release energy dissociation energy would be negative but that can't happen because breaking a bond is going from a lower potential energy to a highe...

Nathan Nakaguchi 1G

Currently working on Homework 2A #15, essentially it asks to find a most likely charge for a given element if it were to become an ion (like for S or Ga). I'm not sure how to approach this, do I focus on ionization energy and electron affinity? How does one determine how many electrons gained or los...

Nathan Nakaguchi 1G

If you go onto Dr. Lavelle's website under the class websites tab, scroll down slightly, on the right you should see a section with the title "Exam information." Click "midterm review sessions and rooms" and it should bring you to a pdf with all the info you need about the midter...

Nathan Nakaguchi 1G

Planck's Constant's units are J * s (joules times seconds) not necessarily just joules, thats why the Planck's units are (kg m^2 s^-2) * (s) = kg m^2 s^-1.

Nathan Nakaguchi 1G

Formal Charge is assigned to an atom by following the formula: FC= V - (L+S/2) where V is the amount of valence electrons the atom has, L are the lone pairs it has (on a Lewis structure it would be the pairs of dots like ":"), and s being the bonds the atom has.

Nathan Nakaguchi 1G

Roman numerals next to atoms/metals just indicate the specific charge it has since metals can have a variety of different charges for the same thing. So this one the (III) indicates a +3 charge

Nathan Nakaguchi 1G

The atomic radius is half the distance between the centers of neighboring atoms, I don't think we will have to ever calculate a measurement for this (it would probably be given). I believe to answer this question you go based off of the periodic trends of atomic radii, as you go down and left of the...

Nathan Nakaguchi 1G

For me, I don't memorize the order I just have memorized the exceptions like when you get to Cr the D shell can have 10 electrons and Cu's 4s shell can only hold 1 electron. The main thing I try to do when finding electron configuration is to build up from the shells (counting the electrons) and to ...

Nathan Nakaguchi 1G

Uncertainty accounts for both the positive and negative, so if your v was 103 +or- 3 then the uncertainty would be 3(2)= 6.

Nathan Nakaguchi 1G

For conversions involving angstroms use the ratio: 10^-10m/Angstrom. For example 100pm*(10^-12m/pm)*(1Angstrom/10^-10m)= 100*10^-2 Angstroms.

Nathan Nakaguchi 1G

I'm not entirely sure on this either, but here is my two cents: 3d orbitals are lower energy than 4s orbitals because the n=3 on the 3d and n=4 on the 4s and we know n refers to energy and the electron configuration is written out in increasing order (so 4s2,3d10 is not correct because the order of ...

Nathan Nakaguchi 1G

Linear momentum is p, where its formula is composed of p=mv or Momentum = Mass (usually kg) * Velocity (usually m/s).

Nathan Nakaguchi 1G

+1/2 refers to spin up and -1/2 refers to spin down. I believe it is a reference to what would happen if you performed an experiment similar to the one Dr. Lavelle showed in lecture Oct 16th (the silver atoms one). Some electrons end up going upwards after going through a magnetic field while some o...

Nathan Nakaguchi 1G

Personally, I would use whatever number is more accurate. In test scenarios I would go for every digit given on the constants sheet. It shouldn't be too much of a big deal though to opt for 6.63 so long as it is clear that you're using that number and your answer is pretty much the same.

Nathan Nakaguchi 1G

For E=hv, h has the unit J*s and v has hz or s^-1 so the answer should come out to be in just joules J (s*s^-1=1), however for this question the v calculated is per photon so it is ok to make it J per photon.

Nathan Nakaguchi 1G

Yes, it is always true that shorter wavelengths lead to higher energy because based off of c=wavelength *frequency if wavelength decreases the frequency has to increase to satisfy the equation and if frequency goes up energy goes up based off of the equation E= h * frequency.

Nathan Nakaguchi 1G

After I've commented (or I look at one of my comments) when I click on the box on the right with my name I see that it says I have posted 10 times, I click on it and it shows me my posts, but I can only find 9 of the 10. I remember commenting on a photoelectric effect post this week but I can't find...

Nathan Nakaguchi 1G

To find the wavelength of Krypton-86 I would use the ratio that they gave you so: (1 meter/1,650,763.73 krypton wavelengths) * (1 Krypton wavelengths) to convert to krypton wavelength in meters (or nanometers). From there I would use the formula c= Wavelength * Frequency, solve for frequency, and pl...

Nathan Nakaguchi 1G

Decreasing frequency of light leads to less energy per photon, (E=hv). So I think that the extent of the change in the electrical field at a given point decreases because the energy of the light (electromagnetic radiation) has decreased. Or another way is that the electric field vector's height have...

Nathan Nakaguchi 1G

I believe it goes: (Molar Mass Given)/(Empirical molar mass)= Number you need to multiply the empirical formula to get the molecular formula.

Nathan Nakaguchi 1G

I'm not entirely sure, I think that you're only required to write the correct amount of sig figs, but there can be cases were scientific notation would be more practical. I would just make sure that the reader can understand the sig figs in your answer.

Nathan Nakaguchi 1G

So there is not really a step by step process for a reaction like this? Do you just kinda have to use your intuition to figure out problem like this one?

Nathan Nakaguchi 1G

Usually I have no issues balancing reactions, but the second part of problem H13 broke my usual solving steps. I usually balance from least frequent to most frequent but when I try to do that on NO + O2 = NO2 it just confuses me. I go from least frequent to most, so I first balance the N's which is ...

Nathan Nakaguchi 1G

Moles can be used for anything because it is like using the term dozen. If there are 12 eggs present you can say there are a dozen eggs, if there were 6 eggs you apply the the dozen ratio: (12 things = 1 Dozen things) and say there are .5 dozen eggs. In the chemistry world chemists like to use moles...

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