CHEMISTRY COMMUNITY

vsyacoubian2A

Cell structures are written anode on the left and cathode on the right. Since H+ is part of the cathode half reaction, it is written on the right.

vsyacoubian2A

Draw out the lewis structure of the molecule and see what places the attached atoms can fall into. Its like trading places: the maximum number of positions available for the connected elements to go to. Draw out the molecule and see for yourself.

vsyacoubian2A

The two equations are the same. The only reason the log equation exists is because it becomes easier to transcribe your answer to ph, since ph is on a log scale. You can use both formulas for ln and log. Your choice.

vsyacoubian2A

I believe they are only for aqueous solutions for both galvanic and electrolytic cells

vsyacoubian2A

For the part of your question asking why you don't use 2500K to find you specific heat capacity, since you would only use temperature to find the thermal energy needed to break the shell in part A. For part B, you could use the mcAT method to find the specific heat capacity by doing (thermal energy ...

vsyacoubian2A

If you have an acid/base reaction, one of your half reactions will probably be a simple, one element to one element +electron reaction, while another will involve compounds. For the more complicated compound, just write the half reaction with H+ and H20 added to it looking at the main reaction.

vsyacoubian2A

I have a step by step attempt at the problem posted as an attachment.

Hope that helps!

vsyacoubian2A

I believe that both Free expansion, or expansion against a vacuum, do not do work. The system must be applying a force over a distance on something to be doing work.

vsyacoubian2A

1) Find what the delta H would be when you have 49.7 g of PbO? -Basically, find the number of moles and multiply that by the delta H to get a new delta H 2) Use the formula q=mCdeltaT to figure out the mass needed. In the workbook, you will only need to go up to 100 degrees, and since there was no i...

vsyacoubian2A

You will not have to know any derivations for the quiz. Dr. Lavelle goes over the derivation to give students better understanding on how the final formula is obtained and how individual elements of the derivations affect the final formula.

vsyacoubian2A

The system can't be doing any work if you want to calculate free energy just using q. If a system is kept at constant volume, it can't do any work. That way, delta U =q.

vsyacoubian2A

Sure. The more complex a structure is, the more entropy it will have. With entropy we can have an idea of what kind of orientation a substance can be.

vsyacoubian2A

There will always be some small amount of residual entropy, but as the system approaches 0K, the residual entropy will approach 0, but will never actually reach 0. Reaching 0K is currently physically impossible.

vsyacoubian2A

Standard Pressure is 1 atm, so anything under standard pressure will be itself

vsyacoubian2A

Note that delta H and delta S for standard delta G should be standard as well. Both methods should be viable. A question will most likely give you information to do one method over the other.

vsyacoubian2A

There are different R values depending on the units. Double check what kind of R you are using.

vsyacoubian2A

The Delta S should be for whatever you are calculating delta G in. So if delta G is for the system, delta S would need to be for the system. The same applies for the surroundings. Entropy tends to be a chain reaction. If one reaction is spontaneous, the creation of entropy in the surroundings can cr...

vsyacoubian2A

Delta S=q/T only if q is reversible. An adiabatic system does not necessarily have to be reversible, so you wouldn't use that delta S formula in that situation. You might want to use other formulas like delta S=nRlnV2/V1 for example

vsyacoubian2A

The answer should be delta U= 0. Conceptually, if I put 5 kj of energy in a bank, and I didn't touch it for 100 years, I will still have 5 kj of energy in the bank. Since there is no change in energy, then delta U=0

vsyacoubian2A

Please note that L*atm is NOT joules. 1 L*atm is 101.33 Joules. You want to convert Volume to liters and keep pressure at atm to get your work in units of L*atm. Then you can convert L*atm into Joules

vsyacoubian2A

Entropy is less at smaller temperatures, because the more heat you give to a gas, for example, the more excited and random the atoms become. Think about it like heating popcorn. At low temperatures, you are pretty sure where the popcorn kernels will be. But at higher temperatures, when the popcorn p...

vsyacoubian2A

I would highly recommend going to any one of the office hours available. The UAs and the TAs do a great job in reviewing all the essential formulas and material you need to know.

vsyacoubian2A

You might conceptually need to understand degeneracy in understanding entropy, but I believe that it is not a big picture topic and will probably not be asked.

vsyacoubian2A

When something diffuses from high concentration to low concentration, it is not doing work. For example, if it were to push on a piston during diffusion and move it, it would be doing work. Diffusion is an irreversible process. You can go from high to low concentrations but not low to high without t...

vsyacoubian2A

You can check on Dr. Lavelle's website and if, you have the course reader, it should be there. First quiz will be week 4 in your discussion section. Midterm is Wednesday Feb 15 from 6 to 8pm. This info is on page 8 of the course reader.

vsyacoubian2A

You need to use two formulas 1. q=mC(delta)T (m=mass, C=heat capacity, T=Temp) 2. q=mL (m=mass, L=latent heat) a) You should go in increments and every time you jump temperatures you should calculate q=mC(delta)T b) If that increment also includes a phase change (solid-liquid, or liquid-gas) you nee...

vsyacoubian2A

Write down all the "givens" you have on one side and try and determine what formula to use with that information. Other than that I think its just a matter of memorizing formulas and practice to increase familiarity of what formula goes where.

vsyacoubian2A

Compression means work is being done on the system. Since the system is not exerting energy to do work and is getting free work, the value for work is positive. In the formula U=q+w, w is positive. If q is constant, you can say that the system gains energy. Work is affected by compression.

vsyacoubian2A

A state function can be defined by a start and end point. Work is quantified based on the pathways you take. For example, if two people exert a force over a certain distance to the same destination, but take different paths with varying distances, the path matters. So you can't quantify work with on...

vsyacoubian2A

Yes, for these types of problems, you have to assume that the temperature change is the same for both.

vsyacoubian2A

Based on the question, it should be clear which methods for solving Enthalpy you should use. For example, uses hess's law will be based on the question and whether or not you use enthalpy of formation or bond enthalpy will also be dependent on the problem.

vsyacoubian2A

Check Appendix 2A in the back of the book. Also, you could go online to check enthalpies of formation.

vsyacoubian2A

Internal energy= q+w

Internal energy is 982 and heat absorbed is 492. This means that the work will be positive, suggesting that work is done on the system. Negative work suggests that the system does work and thus expends energy.

vsyacoubian2A

Enthalpy changes are additive, because they only depend on the final and initial conditions. Nothing else. In fact it can be extremely difficult to map out the middle and even requires calculus to try. But in essence, state functions have deltas which represents the final state minus the initial sta...

vsyacoubian2A

From what I have heard, every week ends on Sunday and starts Monday. So if you post on Saturday, it will belong to the same week you posted on.

vsyacoubian2A

Given an exothermic reaction, for example, the system releases heat to its surroundings after product formation. If the heat released by the reaction is -q, then the heat absorbed by the surroundings can be +q. Conversely, an endothermic reaction will take heat from its surroundings to supplement a ...

vsyacoubian2A

I would stick with the appendix value for the problem. On future quizzes and exams, there will be a table of enthalpy values for you to reference.

vsyacoubian2A

All elements have their natural, most stable state. Look at the periodic table to see the state of matter each element is at room temperature. Metals will generally be solids, and Bromine and Mercury are liquids at room temperature. In terms of having an enthalpy of formation value of 0, all single ...

vsyacoubian2A

I believe they are to make sure people who don't have the lecture for that time (or none at all) come into the halls and take up seats. The scene at the lecture hall on Monday was packed, with people outside of the enrolled lecture coming in. You should probably have a copy of your study list if you...

vsyacoubian2A

A positive delta E would correlate to an increase in energy, while a negative delta E would be a decrease in energy. Different permutations of q and w can contribute to this, but if both q and w are positive, then the system gains energy. If q and w are both negative, the system gains energy. This m...

vsyacoubian2A

I think that you can use either bar or atm. I would use atm since the problems we have encountered thus far use atm for solutions.

vsyacoubian2A

You need to calculate Q in order to know which way the reaction will go in order to reach equilibrium. That way, when you are setting up your ice box, you will know whether to put + or - x for reactants and products for the change (c).

vsyacoubian2A

Lewis structures can really help in these situations to see which atoms lose or gain electrons. Try creating a stable lewis structure for NO+. You will see that oxygen loses an electron and obtains a 1+ charge while being triple bonded to nitrogen. Both have a lone pair of electrons.

vsyacoubian2A

I don't think there is a particular order for writing ligands in a chemical formula. Formulas don't follow the alphabetical rule.

vsyacoubian2A

I don't believe order of the ligands matters when writing down the molecular formula from the name. However, when writing the name, ligands appear in alphabetical order.

vsyacoubian2A

Z stands for atomic number. So it means that elements that have an atomic number less than 8 or greater than 8 have a certain molecular orbital structure.

vsyacoubian2A

When naming coordination compounds, transition metals will be indicated with roman numerals since they can have more than one oxidation state. If you were to look at a periodic table with labeled oxidation states on the elements, there are some transition metals that stick to a particular oxidation ...

vsyacoubian2A

You should check out previous year midterms as a good indicator of what kind of questions the midterm will be asking. Past midterms have many lewis-structure questions that asks to construct the shape of a molecule and answer some questions about its characteristics. The shape could be lewis, but vs...

vsyacoubian2A

Double bonds can be made if it the formal charge allows for it (ie. makes the atom more stable by doing so) and if the atom has enough electrons to do so. (hydrogen can't double bond) (Period 2 elements can't go over octet) In terms of calculating formal charge, you would have to calculate it every ...

vsyacoubian2A

You should know general electronegativity trends in order to figure that out.

vsyacoubian2A

Anything less than 0.5 electronegativity difference is a non-polar covalent bond.
From 0.5 to 2.1, it is a polar-covalent bond.
From 2.1+ is ionic bond.

vsyacoubian2A

Ionization energy is the energy required to remove one electron from an element in the periodic table. So the energy required to remove an electron from ground state and the ionization energy is essentially the same thing.

vsyacoubian2A

The d orbital should be listed first in the given scenario. The p orbital occurs in the p block, whereas Zr, for example, hasn't reached the p block in the 5th energy level. Zr is in the d block, past the s block. In general, just follow the blocks of the periodic table from left to right, and it sh...

vsyacoubian2A

For Quiz 1 the formula and name of all compounds will be provided.

vsyacoubian2A

The Self-Prep Workbook Quiz is a great indicator of what kind of content will be on the quiz. Quiz 1 should be on Fundamentals and Chapter 1. The Heisenberg Uncertainty Equation should be the last concept that will be on the quiz. Talks about, for example, orbital structures will not be on the quiz....

vsyacoubian2A

Page 14 of your course reader should be able to help you out. But to explain, when you purchased your 2016 course reader, it included a tan 2016 workbook that includes practice quizzes and other helpful materials from previous years. Self-Prepatory quiz 1 that does not have the answers in the back n...

vsyacoubian2A

The Roman Numeral (I) indicates the positive charge of the metal cation. Metals are always positive, so copper will be positive. The (I) will indicate its charge, so copper will be Cu1+. Reason for the Roman Numerals is that some transition metals can have different charges (i.e. 2+ or 3+) and the n...

vsyacoubian2A

The workbook comes with the 2016 Course Reader for the class. The workbook will be turned in to the TA before or on the day of your quiz and will be graded in the workbook. That being said, I believe the doing the self-prep quiz directly from a workbook may be the only way to get credit. You may be ...

vsyacoubian2A

Page 14 of the new 2016 Course Reader has a description.

As far as turning them in, they can be turned in on the day of your quiz to your TA.

vsyacoubian2A

I think there might be some select even-numbered textbook problems on Chemistry Community.

I have seen a couple already posted.

vsyacoubian2A

I believe that Dr. Lavelle has said to do the sig figs at the end of the problem. I have also talked to a couple TAs and they have said this as well.

Hope this helps!

vsyacoubian2A

C=v(lambda) doesn't really describe the photoelectric effect since the photoelectric effect describes light behaving like a particle, not a wave. The equation is applicable to the wave-like nature of light and would explain why scientists performing the photoelectric experiment initially thought tha...

vsyacoubian2A

I do not think the problem is correct. I may be wrong, but it seems that one part of the problem is from one question and another part is from another. H13, which is the question I think you're referring to, is as follows: When nitrogen and oxygen gases react in the hot exhaust environment of an aut...

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