CHEMISTRY COMMUNITY

Karan Thaker 2L

Can add Pt (s), or C (graphite)... Can also use Hg (l) if your ions / solids involve mercury.

Karan Thaker 2L

To do E cath - E anode = E cell, always keep the reaction in reduction form... if you reverse anything, you have to add.

Karan Thaker 2L

Important to note catalysts lower Ea, through altering and lowering the transition state of the reactants. They can affect orientation to idealize reactants' position for a reaction, or also weaken bonds of reactants to make them easier to break.

Karan Thaker 2L

I always convert to Kelvin to be safe... There are many instances, such as when you calculate for entropy when you divide two temperatures, and using Celcius here will mess up your ratio and give you an incorrect answer. You also know what to use based on the units and what you need to cancel out......

Karan Thaker 2L

I believe it is just a law. there can be no change in U if temperature does not change.

Karan Thaker 2L

Use the appropritate time conversions: 1 hr is 60 mins, 60 secs in 1 min, 24 hr in day, etc.Set up diagonal conversions to cancel units out and end in whatever time unit you need.

Karan Thaker 2L

Elementary steps are any steps that transition from products to reactants, and don't require additional steps to transfer the reactants to products...

Karan Thaker 2L

Yes, in many cases it's aqueous substances as well.

Karan Thaker 2L

Also, I thought of it as being an initial rate, which means products may be negliglble or very low, and allow us to focus our calculations on reacatants only.

Karan Thaker 2L

That's exactly how I think of it as well. Simply examine the equation and see what would happen. With the Arrhenius equation, it can be a little confusing to predict it right off the bat so plug in numbers to see what happens as well.

Karan Thaker 2L

think about it as bases. When we cancel a log, we do it by taking the base of "log" which is 10, and raising it to the power of the value. Here, it is the same principle, except the base is now "e' instead of 10.

Karan Thaker 2L

Not always, you could have a negative entropy. This would give you negative - positive, and if the entropy is large enough it could become larger than the negative value and make delta G positive. Always refer back to the equation G = H - TS

Karan Thaker 2L

It's the maximum nonexpansion work a system can do at reverseible conditions.

Karan Thaker 2L

Only necessary things I'd keep in mind from the previous quarter is drawing lewis structures and basics like limiting reacatant etc.

Karan Thaker 2L

You know how you have a -log(H+) to find pH, then to find H+ you raise 10 to the value of the pH? It's the similar thing... ln is just a log that has base e, so you take e and raise it to that power.

Karan Thaker 2L

The way I think of it, is that the anode repels the negatively charged electrons since it is negatively charged (loss of electrons). The cathode attracts and receives the electrons (gain of electrons).

Karan Thaker 2L

I write it out as Ka1, Ka2, Ka3... so on and so forth. Use the respective conjugate bases and acid each time you do the step.

Karan Thaker 2L

Exactly! Boyle's law shows that they are inversely related. V2/V1 is the same "ratio" as P1/P2.

Karan Thaker 2L

You have to approach it depending on what you want to cancel out within the context of the problem. Choose the R value with units that will cancel out whatever form of units you have already been given.

Karan Thaker 2L

I think theoretically we learn that it equals, but realistically, the maximum work that can be done may not equal delta G since it could be in the form of non usable work.

Karan Thaker 2L

The spring would release as the rubber band melts away I believe and the spring would release all its potential energy there. Not sure tho.

Karan Thaker 2L

Remember that standard conditions are 1 atm and 298 K, meaning it's constant.

Karan Thaker 2L

I think you're right... the work energy is thought of a value, rather than being gained or loss in this context.

Karan Thaker 2L

Think of the graph we examined in lecture... Lack of volume change would contribute to no area under the curve, making the word done equal to zero.

Karan Thaker 2L

I agree... There are other factors that affect delta U which can alter it such as q and w even if temperature is constant.

Karan Thaker 2L

Maybe we would only make current calculations on the system's state, but as far as dynamics go, there are none so ya we couldn't really do much.

Karan Thaker 2L

I agree... specifics that he wants us to take into further consideration would be included on a test... I would just remember that it's isolated if bomb calorimeter is used.

Karan Thaker 2L

I'm not sure if it matters as long as you show the progression of the reaction correct?

Karan Thaker 2L

They're both logarithmic expression of completely different things. pKa is a logarithmic expression of Ka values while pH is a logarithmic expression of hydrogen ion concentration.

Karan Thaker 2L

If given pH, just take the inverse log of the pH and this will give you H30+ concentration. Follow steps on ice tables and determine other concentartions and you have your Ka value. The main difference is that pKa is a logartithmic expression of Ka values and pH is a logarithmic expression of hydrog...

Karan Thaker 2L

Doesn't K value (by itself) normally imply that you should concentrations thought?

Karan Thaker 2L

You need to know this for the test too. Adding heat changes the Q value for a reaction. If a reaction is endothermic that means heat is on the left side (reactants side) of the equation, while exothermic reactions have heat being released on the products side. You treat heat as a reactant or product...

Karan Thaker 2L

Alkaline I think has to do with elements from the second row (since they are called alkaline earth metals). They are basic, but not all bases have these alkaline earth metals.

Karan Thaker 2L

You could also increase the partial pressure of that gas by decreasing the moles of gas of the other gasses right?

Karan Thaker 2L

For strong acids I just memorize the list. For strong bases a good guideline is that first and second row elements elements (metals) hydroxides make strong bases. There are exceptions though so I would memorize most of the list above.

Karan Thaker 2L

Also usually carry a negative charge because they were just deprotonated.

Karan Thaker 2L

Ya they're always given (as of now)!

Karan Thaker 2L

I would probably just follow the way Dr. Lavelle taught us in lecture.

Karan Thaker 2L

Gases as well as aqueous substances!

Karan Thaker 2L

Ya you can't have the greater formal charge distributed along the less electronegative atom.

Karan Thaker 2L

That's the same methodology I use when I get confused ^ ... I often forget which atoms are capable of forming more than an octet and can have expanded shells and simply think of the possible values based on quantum number. It always ends up making sense and working out.

Karan Thaker 2L

small positively charged ions with high charge have the greatest polarizing power. No, ions increase in polarizing power as they get smaller, since the positively charged nucleus can interact with outer electrons of an anion more closely.

Karan Thaker 2L

By neutralization, assume that the strong acid / base cancel each other out. While this may not happen in reality, we assume this for this course to simplify our calculations process.

Karan Thaker 2L

The first four elements can form less than an octet because some of them don't even have p orbitals to fill 8 electrons with such as hydrogen. They also don't form octets because they have lower formal charges with less than an octet (more stable) or simply don't have enough electrons to form an oct...

Karan Thaker 2L

I think for the purposes of this class, we were told to just treat radicals as having the same effect as lone pairs (which they do), just keeping in mind that they have the effect at a lower magnitude than a lone pair.

Karan Thaker 2L

I don't think that was discussed in lectures OR the book. I doubt they would test us on something that detailed that we don't know. I'd just know the trend that you already listed because they'll probably only test us on different cations having EITHER varying size or charge.

Karan Thaker 2L

It's just a ligand that has multiple electron binding sites... I feel like Dr. Lavelle only uses it for ligands that have a lot of binding sites (over two or three).

Karan Thaker 2L

Because the electrons in lone pairs themselves come from orbitals too, you need to account for them while you generate your hybrid orbitals.

Karan Thaker 2L

Also must keep in mind that even though it has an incomplete octet, it is still stable though. We can examine the stability of a compound by calculating formal charges. In BF3, the formal charge on each atom is zero, indicating that the Boron is stabler in that state than if it were to form addition...

Karan Thaker 2L

No. Hybrid orbitals also still form sigma and pi bonds. Same rules apply: first bond is a sigma bond, additional bonds are pi.

Karan Thaker 2L

In fact, VSEPR says that bonds of different orders don't get the classification as having greater electron repulsion, so thinking of it as making the molecule take up additional space as the order of the bond increases may not be correct.

Karan Thaker 2L

Also, you want to place more EN atoms around the less EN central atom, so that the more EN atoms pull the electrons towards them and sort of "even out" the dipole moments. This just makes the compound/molecule more stable.

Karan Thaker 2L

Sigma bonds also always form first; formation of a pi bond always comes after a sigma bond exists.

Karan Thaker 2L

It also allows you to differentiate between electron-bond interactions and bond-bond interactions. Variances in bond angle allow us to further conclude whether there are just bonds that exist, or whether there are electron lone pairs.

Karan Thaker 2L

Pi bonds always form after the formation of a sigma bond, so that is why we always say sigma bonds form first.

Karan Thaker 2L

You always take the average and add it to the bond lengths, because double or single bonds in resonance structures are equally shared among the compound. So for a molecule with a triple bond and two single bonds, the average bond length would be 1 and 1/3.

Karan Thaker 2L

All I remember him saying in today's lecture was that non polar means that the electrons are bonded and set up in a way that cancels out all dipole moments.

Karan Thaker 2L

Also, an important distinction to make is that lewis structures can determine shape but within the context of a 2-D model. VSEPR effectively works for 3-D context, which is the correct model.

Karan Thaker 2L

Dipole moments develop due to variances in electronegativity. The atom that is more electornegative has a negative value, and less electronegative is positive value. The direction can be shown by drawing an arrow from positive to negative atom, since the negative atom pulls electrons towards itself ...

Karan Thaker 2L

It's because electric attraction is stronger between oppositely charged ions, compared to the bond formed by just sharing electrons in covalent interactions.

Karan Thaker 2L

I agree. In our context it is frequently used in electron kinetic energy calculations and for relating debroglie principles to the photo electric effect.

Karan Thaker 2L

I thought they have the same number of valence electrons as the group number they fall in?

Karan Thaker 2L

Usually due to ions being created or charged molecular compounds that results in an oftentimes odd number of electrons can do that.

Karan Thaker 2L

also important to note that the Plancks constant and momentum units are a bit odd, but they always work themselves out by cancelling, especially when we solve for indeterminacy in position. may be worth practicing working through the units and looking at how they cancel out, because this becomes rea...

Karan Thaker 2L

exactly, it's in the format of minimum, or greater

Karan Thaker 2L

I thought it would be the opposite? Because dividing planks constant by a larger number would lead to a smaller wavelength value? Because don't they have an indirect relationship?

Karan Thaker 2L

I thought it only has energy and momentum, and negligible mass? Can someone clarify?

Karan Thaker 2L

So are valence electrons and non valence electrons that are still bound to the nucleus both considered bound electrons?

Karan Thaker 2L

that's a confusing way to look at it Anna. I'm not sure on the specifics of how electrons exactly "hold" their energy, but I'd just think of the electron itself containing the high state of energy as it buzzes around, and I think for our purposes at this level that should be sufficient. As...

Karan Thaker 2L

this is understandable as a detached electron was likely excited and is also free to interact with other atoms (thus giving it high PE). A bound electron also has high PE, but is still bound to the atom and is being held there by stable bonds (so less energy).

Karan Thaker 2L

the meaning and significance of it, is that it relates wave frequency and levels of energy. (J/S)

Karan Thaker 2L

Agreed. the line that shows up is spectrum emission I think, because you can't translate absorption into a visible graph but can do this with emission. Absorption is related to what wavelengths are being taken in by a given atom. For colors and energy give off think emission.

Karan Thaker 2L

The -1 after the mol just means the mol goes in the denominator, but has been brought up and assigned a negative exponent.

Karan Thaker 2L

Ya he said he'd give us partial credit, and my TA said she would take some points off for errors in significant figures. My TA seemed somewhat lenient about it but still follow it correctly if you don't want to lose any points.

Karan Thaker 2L

I feel like for what we've done so far just always go with grams unless the question asks for kilograms. Chemistry involves measuring and calculating things in such small scale that using kilograms instead of grams seems unlikely. Of course, if the question specifies than go for kilogram but I think...

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