CHEMISTRY COMMUNITY

Zane Mills 1E

Isn't it ∆G=-nFE so that the -VE shows spontaneity (-VE of ∆G) and thus +VE E for the cell's capacity to do work?

Zane Mills 1E

Activated complex model = transition state model. He said it would be sufficient to state "As temp increases, so does the avg. # of molecular collisions. In order for a rxn to occur, the molecules must not only collide but must collide with the correct orientation. This is given by A (pre expon...

Zane Mills 1E

An easier way to think of it is just match up your concentrations and their exponents with the rate (which you know is M/s or mol/L*s)

Zane Mills 1E

Because k must be positive (positive slope in the graph), answering your second question (yes).

Zane Mills 1E

Enthalpy estimations can vary based on how you find them (i.e. using Hess's law, bond enthalpies, etc.) so they were probably just calculated different ways. It will explicitly state how to calculate it / which one to use on the test so don't worry about the difference.

Zane Mills 1E

From there, because q=0, ∆U=w so if you go back to step 1 and see that w=-158J, so is ∆U from step 1. Step two tells you that it releases heat so q=-73J and using ∆U=q+w you can find that w=231J. Then you just use w=-p∆v to find the final volume, keeping in mind that ∆V = V2 - V1

Zane Mills 1E

You always want a positive E˙ so usually you can just look at the half rxns and see which one you should flip to get a positive Ecell (b/c they're intensive properties). If this doesn't work, simply go about balancing the rxn as usual and see which are reduced/oxidized and this will show you which o...

Zane Mills 1E

An aldehyde always ends in a hydrogen while a ketone has a carbon on both sides. Think of aldehyde as "hiding the hydrogen"

Zane Mills 1E

Strong oxidizing agents cause other molecules to lose e-'s and thus are highly EN and become more stable with the addition of the e-'s (O2, O3, F2, etc.) Strong reducing agents donate their e-'s to become more stable and thus give them away easily (Li, Na, Mg, etc.) They're all ranked on standard re...

Zane Mills 1E

Is the answer here correct? Seems to me that they added Ti in as 20*C instead of the 25* it says it started at.

Zane Mills 1E

How do you make a 32 Molar Soln? Put your grandma's dentures in water.

Zane Mills 1E

What is a cation afraid of ? A dogion.

Zane Mills 1E

Yep b/c they're just pseudo and real eq constants so just showing the direction favored units don't matter

Zane Mills 1E

I think it actually is given on the eqn sheet if you're referring to the ideal gas U.. its on the one in the course reader

Zane Mills 1E

It doesn't necessarily, you would have to consider it with the nernst equation to see if it changes Q

Zane Mills 1E

Probably not necessary because you'll have the linear version on the test and can easily derive it from that if needed.. only time I can think of that you could definitely need it is radioactive decay problems

Zane Mills 1E

When ∆G = 0 this means the rxn is at equilibrium.. phase changes are equilibrium between two phases (i.e. liquid and gas) thus it is a straight line on a ∆G vs time plot and ∆G would be zero.

Zane Mills 1E

Ethane (C2H6) has a higher heat capacity than Ethene (C2H4) because it has only single bonds connecting the carbons, not double bonds, so using the equipartition theorem we can see that there are more DOF which gives it a higher heat capacity.

Zane Mills 1E

Anyone have any useful Mnemonics for electrochem other than the old 'OIL Rig', maybe some for the ∆ø's and their signs depending on electron flow or other useful things to memorize?

Zane Mills 1E

Not really sure what you mean by optimum but activation energy is essentially an energy barrier between reactants and products. It's not so much collisions as it is stability, as catalysts help to stabilize intermediates and bring down the 'barrier' between products and intermediates. I guess optimu...

Zane Mills 1E

The Michaelis constant is specific to the enzyme based on the equation where the substrate concentration allows the enzyme to perform at half Vmax. It's essentially how good your enzyme is at forming the ES complex in the binding portion of general michaelis menten eqn.

Zane Mills 1E

The SN2 mechanism is a neuclophilic substitution where the slow step involves two components -- nucleophile attacks and makes a bond and LG leaves with pair of e-'s.

Zane Mills 1E

If it was covered in lecture then yes most likely

Zane Mills 1E

First order is -k slope with time vs ln[A]. Time vs just concentration [A] represents zero order (also -k slope).

Zane Mills 1E

No it's longer, he usually does pts=minutes testing so the final will be like 170 minutes for 170 pts while the midterm was 110 min for 110 pts.

Zane Mills 1E

On this question why is the oxidation # for O2 zero but oxygen in other atoms is -2? can you explain in terms of e- pairs?

Zane Mills 1E

In the soln manual the answer for question 15.3 C is actually for B, so I just wanted to make sure I am doing it correctly, it is still 3.3x10^-3 mol/L.s right? Using the unique eqn and the coefficients from the provided rxn

Zane Mills 1E

∆S=q/T for reversible processes and ∆S > q/T for irreversible. Because some heat is lost you can't use the same equation because technically the heat lost + the heat of the irreversible step = the reversible heat there would be in a reversible system if that makes sense.

Zane Mills 1E

viewtopic.php?t=14502

Zane Mills 1E

I assume you mean upcoming test 1. Because calculations on work and internal energy aren't involved, I'm assuming he will be heavy on enthalpy calculations using the three methods w/ state changes and whatnot

Zane Mills 1E

Hydrogenation is just to react with hydrogen (H2) which usually takes away double bonds in the process. For this you should just be able to use the given enthalpy changes in the balanced equations (reverse the seccond, add the first, and double the third eqn) then when you add up the entalpies in th...

Zane Mills 1E

As i'm sure you know a kj is just 1000 joules, the main reason you would use one vs the other would depend on the constants in the equations you're using and the units they're measured in so that they would cancel out correctly and give you an accurate answer.

Zane Mills 1E

I had this exact question in office hours and my TA sent me this link to help me understand it, hopefully it'll be helpful to you too http://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf it explains exactly how the dof are calculated and what they represent.

Zane Mills 1E

This is just the change in enthalpy of one mol of the substance in standard state. Standard state just means at 1atm of pressure and 298K from its 'pure' elements.

Zane Mills 1E

Not necessarily, the enthalpy of vaporization (and that of melting and sublimation) are only necessary if there is a state change which would be made obvious in the written reaction. I unfortunately don't know how to draw on this forum but it would essentially be benzene (haha I spelled it wrong ear...

Zane Mills 1E

Chapter 8 problem 39? I have this as a heat problem that is not multiple choice so I'm not sure what you're referring to.

Zane Mills 1E

For (a) this is the case because the water in it's gaseous form was vaporized to become so (from liquid) therefor if you are to revert it back to it's liquid form you can just take this difference I believe in order to find the correct enthalpy. As for (b), C(gr) is the solid graphite form of carbon...

Zane Mills 1E

Two scientists walk into a bar. "I'll have H2O," says the first. "I'll have H2O, too," says the second. Bartender gives them both water because he is able to distinguish the boundary tones that dictate the grammatical function of homonyms in coda position, as well as pragmatic co...

Zane Mills 1E

For this problem on the last test, how do you set up your ICE table since there are four values? Could you walk me through this question please and thanks!

Zane Mills 1E

Does anyone know if the correct name for [PtCl4(NH3)2] on test #4 is bisammine-tetrachloro-platinum(IV)?

Zane Mills 1E

This essentially just means the electron density used in bonding is spread out due to residency features that allow for bonds to change and make the overall molecule more stable. Delocalized pi bonding is best explained using phenyl examples like benzene. Because there are three internal pi bonds co...

Zane Mills 1E

yes, just as when you're calculating eq constants, change in pressure/conc doesn't impact solids or gasses, thus you only take into account the gasses that are compressed by increase in pressure/decrease in volume. (moving it to the left after you calculate Q)

Zane Mills 1E

Well as it says the bold atom is the 'central' atom or the one it's asking about regarding orbitals used in bonding. If you draw out each lewis structure you will see the number of atoms connected to the bold-faced atom and thus be able to determine if it is sp, sp2, sp3 etc. These problems are espe...

Zane Mills 1E

Yes, the formal charges actually refer to the individual atoms and their electrons though, so you would use this number to find the overall charge on the molecule given the individual formal charges.

Zane Mills 1E

A bond dipole moment is created when you have polar bonds (such as in a molecule like Boron Trifluoride). There has to be a separation of negative charges from positive charges that act like vectors. It also depends on the distance between atoms. In the examples on the test, you can draw the lewis s...

Zane Mills 1E

This is also why you don't include solids in ICE tables. Although they may be mentioned in a rxn, when determining Kc for a given rxn, the concentrations of non aqueous/gaseous compounds is completely irrelevant (as mentioned above). A good example of this is found at https://chem.libretexts.org/Cor...

Zane Mills 1E

It is in fact basic. Because hydroxide ions are formed from the reaction of BaO with water, you know that it is a basic oxide. If it were to form a bronsted acid instead of these hydroxide ions, you would then know that it is instead an acidic oxide. One easy way to remember this is that metal oxide...

Zane Mills 1E

If you know the formal charges by heart then you should know the lone pairs needed to form the complex thus you should be able to determine the prefix necessary. The lewis structures just help with visualization of potential dentate formation for chelating complexes.

Zane Mills 1E

Why must the regions of electron density occupied by lone pairs be on the same side?

Zane Mills 1E

One way it was easier for me to determine hybridization is by drawing out s___ p___ ____ ____ and filling these pseudo-orbitals with excited electrons. Ex: if you have CH4 you put one in the S and one in each Px, Py, and Pz as opposed to trying to think about how many bonding orbitals are being used...

Zane Mills 1E

In most cases no, unless it is specified in the question. Most times when asking for molecular geometry they simply require the geometry of the molecule as is

Zane Mills 1E

Why is it that molecules such as water are regarded as 'bent' when they have two regions of electron density occupied by lone pairs and two by sigma bonds? Why don't the lone pairs repel eachother enough to go on either side of the oxygen molecule and make it linear?

Zane Mills 1E

Just like you stated, Lewis acids (not to be confused w bronsted acids) accept electrons and lewis bases donate them. An easy way to determine this is to look at formal charges on molecules. Positive formal charges indicate electron deficiency and thus a lewis acid, while negative formal charges ind...

Zane Mills 1E

They don't have enough electrons to fill an octet so they're considered 'electron deficient'. The valence electrons available simply form bonds as in hydrogen. Boron follows the 'sextet rule' which is similar except 6 electrons instead of 8 because of its valence deficiency. Meanwhile, beryllium can...

Zane Mills 1E

Expanded octet is simply elements w atomic numbers greater than 10 that can take on the excess electrons in order to accommodate these electrons. Because of this, I don't think there are any cases where there are excess electrons that aren't either in bonding or lone pairs on their respective atoms.

Zane Mills 1E

This is due to the overlap of orbitals. Single bonds such as C-C is an overlap of the sp3 orbitals, and is just one sigma bond. As the bonds get stronger (double = sigma + pi, triple = sigma + 2pi) the bond length gets shorter, representing the overlap of further orbitals as well as the heightened s...

Zane Mills 1E

Adding to this, when drawing resonance structures if there are multiple structures with identical formal charges the more electro-negative elements should have the negative formal charge, while less EN elements could hold the positive charge (if necessary). In the above example, it would likely just...

Zane Mills 1E

When calculating the frequency of a wavelength from a given energy (emitted electron eqn) do you used the total energy given at the end of the equation or do you strictly use the kinetic energy because that is the actual energy once the photon leaves?

Zane Mills 1E

^ this is exactly correct. It is important to note when writing the different energy occupied orbitals that the HIGHER energy orbitals are written last, thus the most vulnerable to lose a valence electron.

Zane Mills 1E

In lecture, professor spoke of 'electron shielding' effects. I was wondering how the effects he described correspond to NMR shielding effects and if there is any connection between the two?

Zane Mills 1E

In a (surely) rare scenario where two of the reactants are equal in their ability to limit the reaction, how would you determine which is the true 'limiting reactant', or would you just deem them both to be 'limiting'?

Zane Mills 1E

Correct ^ you always use the total amount of liquid in the calculation, as to have an accurate representation of the molarity of the soln and the concentration of solutes. If you were to use 1 that would not be representative of the actual M1V1, thus producing an inaccurate answer. The final volume ...

Zane Mills 1E

When diluting a concentrated stock soln in order to obtain a soln with a predetermined molarity, it is important to go about it in a way such that you use the least possible number of steps regarding transferring the stock soln and previously diluted soln's, as to keep the relative room for error as...

Zane Mills 1E

thank you

Zane Mills 1E

I mentioned that post in my post, I understand how to get that far but when solving for k you have
rate=k[B]^1[C]o^2[A]o^0=ko[B]^1

k=(ko[B]^1)/([B]^1[C]o^2[A]o^0)

so then I assume your molarities for b would cancel out and you'd have k=(ko)/([C]o^2[A]o^0) and not a numerical value?

Zane Mills 1E

"Suppose you somehow synthesized an unknown compound X from 3 reactants A,B, and C. The chemical equation is given as: A+B+C->X. The three plots below show the relationship between concentrations of each reactant vs time given that the other two reactants are in large excess. Find the rate cons...

Zane Mills 1E

This is confusing me as well, so the 'k' is just -k for a 1st order rxn you dont have to find a numerical value?

Zane Mills 1E

yewww so pre eq approach is only when the fast step is first and slow (rate determining) is second [which we have here]. so you have the observed rate of rate = k [F2] [CHF3] (and by looking at the RXN's, you can determine that your intermediate is 2F. Basically what you're trying to do is find the ...

Zane Mills 1E

The combustion of ethane (C2H6) is represented by the equation: 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(l) In this reaction: (a) the rate of consumption of ethane is seven times faster than the rate of consumption of oxygen. (b) the rate of formation of CO2 equals the rate of formation of water. (c) wa...

Zane Mills 1E

In professor Lavelle's lecture, he mentioned that the butterfly conformation constantly oscillated in order to maintain less strain and a staggered conformation, is this true for envelope and chair conformations as well?

Zane Mills 1E

Why exactly is 1,3,5-cyclohexene considered a nucleophile and is it the same as benzene?

Zane Mills 1E

Like on question #5, I understand that the slow step in an addition reaction is the limiting step, but will the slow step always be the 1st step as well?

Zane Mills 1E

A pseudo first rate law is when you treat a 2nd order RXN like it was a 1st order RXN (n=1) and Pseudo 2nd order rate law is when (n=2). check this out for a better description http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions/Pseudo-1st-order_reactio...

Zane Mills 1E

So is the ∆n found by balancing the equation and taking the mols product - mols reactant?

Zane Mills 1E

How do you calculate the work associated with a combustion reaction (of butane) when you are given the moles and a constant pressure?? I found the initial volume but since there is only one you can't use w=-nRTln(v2/v1)

Zane Mills 1E

Video by Zane Mills (304463207 Discussion 4F) and Connor Olsen (804650584 Discussion 1I).

Zane Mills 1E

Why do the bonds (Å) get shorter as you increase in s character (%) during hybridization?

Zane Mills 1E

It would take forever to type out and make it completely accurate, but check out: http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Valence_Bond_Theory/Resonance"onclick="window.open(this.href);return false; There's a complete description of lewis structures, how and why to form the...

Zane Mills 1E

suppose 68.5kg of CO is reacted with 8.60kg of H2....calculate the percent yield of methanol if 3.57 x 10^4g of CH3OH is ACTUALLY produced. I got everything leading up to this question I am just super confused as to what it is asking and how to get there.

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