CHEMISTRY COMMUNITY

Mitch Mologne 1A

I think it is just delta G, but that being said, wmax=deltaG when pressure and temperature are constant.

Mitch Mologne 1A

If you are referring to reversible expansion, in our case, almost always will reversible reactions be isothermal. Usually tests and problems state them together.

Mitch Mologne 1A

On the anode side, write the element that is getting reduced first, then its reduced state. On the cathode side, write the element getting oxidized, and then its oxidized state.

Mitch Mologne 1A

I believe that would be correct!

Mitch Mologne 1A

Conceptually, if a metal has a lower specific heat, it should take less time to get to the same temperature as a metal with a higher one, giving it a steeper slope.

Mitch Mologne 1A

a. The entropy of the inside is increasing due to the fact that heat is getting pumped into the system. The entropy of the outside is decreasing as it is losing heat. The inside change is greater due to higher change in temperature (the outside does not change in temperature by much).

Mitch Mologne 1A

I would think they would give you details about using a pseudo first order.

Mitch Mologne 1A

Seeing as it hasn't been covered too in depth in lecture, I would not think we need to know the half lives.

Mitch Mologne 1A

If you apply a catalyst to a system that has both a forward and a reverse reaction, does the catalyst have an effect on the activation energy for both sides?

Mitch Mologne 1A

Many times, a problem will give you mechanisms, and in turn tell you explicitly which step is the slow mechanism.

Mitch Mologne 1A

To elaborate, essentially if you take two experiments and only change one initial concentration of reactant, you can observe how the rate of the rxn is affected, thus determining the order in respect to the reactant you altered.

Mitch Mologne 1A

You could also relate this to our delta g equation last chapter as the higher the temperature is with a positive delta s, the more favorable a reaction is.

Mitch Mologne 1A

I am pretty sure I heard lavelle reference that we would learn that at a later date.

Mitch Mologne 1A

Au becomes a reactant because you flip the anode half reaction.

Mitch Mologne 1A

A good way to remember is that an open system is 'open to the surrounding' meaning the system is able to lose some of its particles, or gain some from the surroundings.

Mitch Mologne 1A

I think usually it can be implied because one of two is usually changing, thus you would use the other.

Mitch Mologne 1A

It is a closed system because no matter can leave/enter the system, however, energy can, thus, it is a closed system.

Mitch Mologne 1A

When you expand against a vacuum, can there be energy still transferred in the form of heat?

Mitch Mologne 1A

When you are finding the boiling point of water at room temperature, why do you have to cool the water back down to room temperature?

Mitch Mologne 1A

No, the bond enthaply method is less accurate because it is an average energy released between the bond between two atoms across a multitude of molecules.

Mitch Mologne 1A

deltaH is equal to q at constant pressure.

Mitch Mologne 1A

You can still calculate it by taking values and equations you know already. Because deltaS is a state function, you can take multiple steps to get there rather than just one. Just because water is not boiling at room temp, you can still find it the entropy of vaporization.

Mitch Mologne 1A

If the pressure is constant, you known the process is irreversible.

Mitch Mologne 1A

Not necessarily. There can be work done.

Mitch Mologne 1A

I would assume that we will be given that information as he has given us other similar information on prior tests.

Mitch Mologne 1A

I believe he said it is more commonly used in the form that he stated, and that it is more applicable.

Mitch Mologne 1A

So if Q=K, then a system is at equilibrium. If Q>K, then the system will favor the reverse because that means there is too much product (higher the value the more it favors the products. If Q<K, then the system will favor the forward reaction.

Mitch Mologne 1A

Delta U represents the internal energy of a system, whether that is in the form of heat or work while Delta S represent the variability of different states that a molecule can be in (relating to degeneracy). Furthermore, Delta U uses strictly Jules as units while Delta S uses J/degree K.

Mitch Mologne 1A

Delta G represent free energy which is essentially saying the energy that is able to perform work. Thus the max amount of work a system can do corresponds to the free energy of the system.

Mitch Mologne 1A

One thing I would say is to wait to round with sig figs until your final answer!

Mitch Mologne 1A

Because neither energy nor matter can be added to system, nor can it leave the container. Just because a system is isolated does not mean that it cannot have heat or energy already in the system, but rather that it cannot be lost/gained.

Mitch Mologne 1A

In 14a, Dr. Lavelle made an announcement before the midterm stating which exact topics would be covered on the midterm.

Mitch Mologne 1A

I believe it is because KCl is a very ordered crystal, giving it a lower entropy than KClO4 and KClO3 has less variability in where the Cl can be than KClO4, thus also resulting in lower entropy.

Mitch Mologne 1A

I was under the impression that we really only care about systems at equilibrium because biological systems are at equilibrium. I believe that he just made the reference that if Gn=Gp then the equilibrium is not at equilibrium.

Mitch Mologne 1A

An isolated system just means that energy can not be released/gained in the system. That does not mean that there can't be energy already in the system, and in this case, that energy is in the form of heat.

Mitch Mologne 1A

The more particles the molecule has, the higher the molar entropy. Because HBr is a bigger molecule (thus more particles) it will have the higher molar entropy.

Mitch Mologne 1A

I believe it is because the starting equation is deltaS=nRTln(V2/V1) with volumes instead of pressure. If we apply Boyle's law that Pressure and Volume are inversely proportional (V1P1=V2P2) then we would be able to inverse the volumes to be pressures, thus deltaS=nRTln(P1/P2). To my knowledge, that...

Mitch Mologne 1A

More particles usually results in a higher entropy, which would explain why HBr (being the bigger molecule) has the higher entropy.

Mitch Mologne 1A

Because there are only F atoms around the B, there is no variability in how the molecule is structured. However, with COF2, the Oxygen atom can be in one of three places, which causes a higher molar entropy.

Mitch Mologne 1A

I am fairly certain that the signs are just flipped.

Mitch Mologne 1A

I think a helpful tip to avoid this confusion is to always convert to Kelvin when working with entropy!

Mitch Mologne 1A

A gas prefers to be at a lower pressure, and thus a higher volume. If it is able to do work to get to a lower pressure, it will stay there, making the process irreversible.

Mitch Mologne 1A

For the most part, in thermochem we use a change in temperature, which means either kelvin or celsius will work considering they use the same scale. That being said, the ideal gas law uses kelvin.

Mitch Mologne 1A

part b

Mitch Mologne 1A

I am struggling a bit on problem 8.31. Could someone explain how to approach this question?

Mitch Mologne 1A

Put more simply, the specific heat of water will be the same in a 1g sample and a 1000g sample. Essentially, the property does not differ due to the amount of sample present.

Mitch Mologne 1A

Put more simply, the specific heat of water will be the same in a 1g sample and a 1000g sample. Essentially, the property does not differ due to the amount of sample present.

Mitch Mologne 1A

Those are correct!

Mitch Mologne 1A

Also to add, extensive properties are just things that we can alter about the substance not just the amount.

Mitch Mologne 1A

The length of the sloped lines will just represent the time it takes(horizontal) or the amount of temperature change after a phase change(vertical)

Mitch Mologne 1A

I would suggest doing homework in pen to practice for tests and finals, that do require pen, however, I do not think points would be taken off if you used pencil.

Mitch Mologne 1A

Try to study the shapes in groups of the amount of bonding regions, that way when you substitute a lone pair for a bond, you will know the still know the shape.

Mitch Mologne 1A

An easy way to tell is if the molecule has both an acidic functional group and an acidic functional group. For example, most amino acids will have both. Another easy way to tell is if when a conjugate base to an acid still has a hydrogen in the molecule. HCO3^- can both accept a proton and give on a...

Mitch Mologne 1A

I believe that is because H2PO4^2- is a more stable ion than H3PO4 is. Thus, the atom would be more likely to donate, so it would gain more stability.

Mitch Mologne 1A

Yes you can consider lone pairs as a region of electron density! For example, H2O has two lone pairs but still has a hybridization of sp3.

Mitch Mologne 1A

I believe that is just a mistake in the book because the hybridization of the molecule is sp, implying that the molecule is linear.

Mitch Mologne 1A

I believe we are going to need to know the most common ligates from memory.

Mitch Mologne 1A

They are essentially the same thing. If you are given just K, you can usually infer whether to use concentration or partial pressure by what you are given in the problem.

Mitch Mologne 1A

Just to emphasize, polydentate refers to molecules that connect to the central atoms at multiple points.

Mitch Mologne 1A

I would venture to say you will never have to verify the derivation of the molecular structure, but rather just what it is.

Mitch Mologne 1A

However, these elements will have varying atomic radii.

Mitch Mologne 1A

I believe there is a link on Lavelle's website about the midterm and possible partial credit.

Mitch Mologne 1A

Yes and this is due to the molecule having a non-hybrid p orbital.

Mitch Mologne 1A

I would be prepared to identify both, as both have been covered.

Mitch Mologne 1A

The f orbitals are extremely complex, with some shapes looking like a double sided pacifier.

Mitch Mologne 1A

They are able to expand their valence shells by adding electrons to the d orbital!

Mitch Mologne 1A

I think if there are possible resonance structures, the potential question will instruct you to include how many or at least inform you there are multiple structures.

Mitch Mologne 1A

Personally, I think it would be smart to model your Lewis structure to match the VSEPR shape just to incorporate both in your answer, but I don't think it would be incorrect to just put the lone pair on top, with four surrounding bonds for SF4.

Mitch Mologne 1A

In essence and put simply, all the four different orbitals have different shapes relating to their axes. S has very simplistic orbitals, while f has very complex ones.

Mitch Mologne 1A

When using the uncertainty equation, you in fact use the uncertainty given rather than the certainty. So in this case, you would use the 0.1

Mitch Mologne 1A

From what I understand, a blackbody is an object that absorbs all frequencies of visible light, and does not reflect any of them. An example would be glass. It is useful because we are able to study light with them.

Mitch Mologne 1A

The question asked for the energy for a single atom, not for a mole of sodium, while the work function was given in KJ/mol. In order to find the energy for a singular atom, divide the work function by Avogadro's number (6.022* 10^23)

Mitch Mologne 1A

Zeroes are only considered significant figures when they are "sandwiched" between two non-zero numbers or a non-zero number and decimal point ( ex. 505, 9.09, 50.); or if it is at the end of decimal showing precision (34.0).

Mitch Mologne 1A

When doing a problem that does not have any number values in the question but uses molar masses that extend to 4 sig figs, should you use the standard 3 sig figs in your answer or 4 sig figs coming from the molar masses?

Search found 74 matches