CHEMISTRY COMMUNITY

Ian Marquez 2K

Like said above, microstates and talk about degeneracy should indicate that the Boltzmann equation would be at play.

Ian Marquez 2K

Reversible expansion will always do more work. It is also clear to make this distinction by looking at some of the graphs and the area underneath the curve when comparing reversible and irreversible expansion.

Ian Marquez 2K

Also, the UA's post worksheets here and its really helpful to review those on the topics we covered throughout the quarter. The review sessions are also really helpful because they usually cover past exam problems.

Ian Marquez 2K

This exponential curve is modeled as a linear function because of the ln[A] on the left side of the equation. If you were to solve this equation for just [A], you would see that the function would be a decreasing exponential curve. Since the left is not solving for [A] and is in terms of ln[A], this...

Ian Marquez 2K

Usually for exams, deriving equations, such as rate laws, is fair test material. Also, it is helpful and a bit easier to know how to use them if you know where they come from and how to derive them.

Ian Marquez 2K

Like the other responses, thermodynamics gives no information based on time or the rate of the reaction. This information involves activation energy, which is a study included in kinetics. Kinetics focused on the rate of the reaction itself whereas thermodynamics can say if a reaction were to be spo...

Ian Marquez 2K

Also, it is useful to refer to the equation that says Eºcell = Eºcathode - Eºanode. Since Eºcell should be positive, the more positive Eº should be denoted as the cathode and the other as the anode. Since it is - Eºanode, it is essentially flipped in this equation with all the elements switching sid...

Ian Marquez 2K

Like said above, in most cases, water is not included. Since it is not used to necessarily balance charges and is not being reduced or oxidized, it does not need to be in the cell diagram. Since the other parts of the diagram are indicated as aqueous, water can be assumed to be present.

Ian Marquez 2K

Like stated above, if the redox reaction that is given is already balanced, you can just look at either half-reaction to see how many electrons cancel out since everything is already properly balanced.

Ian Marquez 2K

Also, on Professor Lavelle's website there is a specific announcement that explains what will be covered on the test.

Ian Marquez 2K

Like said above, if you look at the individual factors that determine what delta G is in the equations, inferences can be made such as higher temperatures leading to a greater chance of a reaction being spontaneous or a negative enthalpy change.

Ian Marquez 2K

Delta G, or just Gibbs free energy, is a state function. This means that you can apply the properties state functions have such as the properties you use for enthalpy and entropy.

Ian Marquez 2K

It probably would be helpful to be aware of the strong acids and bases and realize that these are 100% dissociated in water.

Ian Marquez 2K

This reasoning would be correct for state functions like delta S (entropy) and delta U (internal energy). However, w (work) and heat (q) are path functions and are not just dependent on starting and end values.

Ian Marquez 2K

Like the replies above, delta S of the universe, or when negative delta S of the system is equal to delta S of the surroundings, is 0 when the reaction is at equilibrium and is reversible. Also, delta U would be 0.

Ian Marquez 2K

Like said before, when there is a constant volume, energy cannot be released through work. Therefore, the heat capacity at constant pressure is higher because this added energy through heat can be given off through work and expansion.

Ian Marquez 2K

Also, it is helpful to know which equations apply for which problems so that your time is used efficiently to understand and set up the problems. Make sure you know the general equations needed to solve different problems (specific heat capacity, internal energy, constant volume/pressure, etc.).

Ian Marquez 2K

Also, in order to find the energy density per liter, you have to multiply the density which is given by 1.65g/cm^3 by 10^3cm^3/L so your answer and units are correct.

Ian Marquez 2K

For thermodynamics, a system should always be able to fit one of these three descriptions. However, for isolated systems, these are typically a lot harder to replicate in nature or in labs because its hard to make sure no energy or matter can get in or out. Examples like a thermos that is insulated ...

Ian Marquez 2K

Like everyone else said, I also had to jump around the 6th edition to find problems that related to the lecture order. However, if you read or skim through the chapter then it isn't too hard to do the earlier homework problems on work, etc.

Ian Marquez 2K

Also, it is helpful to recall the graph Professor Lavelle showed during lecture that shows that the added heat is contributing to the phase change and not raising the temperature. Like said above, the energy added is going into the phase change reaction. This added heat can also be observed when dis...

Ian Marquez 2K

Also, it is important to keep in mind that x is not actually 0 because if you plug in 0 for all x's it will be impossible to solve for x. However, it is useful in approximations to make the calculations more simple as the other replies have stated.

Ian Marquez 2K

For Chem 14B during the ICE tables and chemical equilibria, I believe we have been using the Bronsted acids and bases definition.

Ian Marquez 2K

Like said above, sig figs are a little bit different when dealing with pH and pOH because of what the log function does. A good rule is to just count sig figs in pH and pOH as the figures after the decimal point. This means that a pH of 1.2 has only one sigfig and 1.23 would have two. This rule only...

Ian Marquez 2K

Like said above, you can use both when calculating for your ICE table. The thing to keep in mind though is what the question asks for in the final answer. If in partial pressures, you may be asked to convert to molarity, or concentration (n/V).

Ian Marquez 2K

The test will most likely cover chemical equilibrium as well as acids and bases up to the lecture this Friday. It is also useful to refer to Outline 1 and Outline 2 on the course website to see which topics we are expected to know for the upcoming tests.

Ian Marquez 2K

Shifting can be a confusing term because it can give the impression that the reaction is moving. In fact, "shifting," "lie to," and "favored" all mean the same thing. This term either denotes that a reaction favors the reactants or products depending on a condition. The...

Ian Marquez 2K

For the two options, if both are positive, plug it into the ICE table and make sure that your equilibrium concentration is still positive. If x is too large then this will be negative which cannot happen. In any case, it is good to check both solutions if unsure, especially if both are positive valu...

Ian Marquez 2K

As long as the homework problems are relevant to the content covered in lecture in the past week, you should be fine. Like said above, the main focus should probably be acids and bases as well as the ICE tables in chemical equilibria.

Ian Marquez 2K

Although these equations are given on the constants and equations sheet, it is helpful in terms of time and ease to memorize equations used often such as PV=nrT.

Ian Marquez 2K

K is the equilibrium constant so the numbers you use are the equilibrium concentrations of the products and reactants. This tells you that this ratio will be approached when a reaction reaches equilibrium from its initial concentrations.

Ian Marquez 2K

Like the replies above, it is very helpful to view these changes visually. Similarly, it's helpful to think of adding reactants or products to a reaction at equilibrium as tipping a scale or seesaw. In order to react to this increase, either the forward or reverse reaction is favored to even everyth...

Ian Marquez 2K

At the end of lecture, Professor Lavelle briefly stated that you should use atmosphere as the unit for partial pressure.

Ian Marquez 2K

I left a bright, large blue hydroflask with stickers in the La Kretz exam room. It wasn’t there when I went back so please let me know if you’ve seen it or took it. Thank you!!

Ian Marquez 2K

When looking at an equilibrium, it is helpful to note whether the reaction other than water is a base or acid. This tells you that water has to act the opposite in this equilibrium reaction because the water either loses a proton to donate to the base (acid) or gains a proton from the acidic donor (...

Ian Marquez 2K

There are diploe moments present with C and H so nothing is ignored. However, in molecules such as methane (CH4), the symmetrical tetrahedral shape has each of these dipole vectors canceling out, making the whole molecule nonpolar. These dipole moments are still in play though when considering the m...

Ian Marquez 2K

Also, it is important to note that Lewis acids and bases are more general than Bronsted acids and bases. In the textbook, it says all Bronsted acids and bases are Lewis acids and bases but not the other way around.

Ian Marquez 2K

When listing down what intermolecular forces molecules have, just always note that London dispersion forces occur no matter what. In some molecules, this is the only force that is acting.

Ian Marquez 2K

Also, like we learned in the past, make sure to account for formal charge and make sure the Lewis structure is the most stable possible. After, proceed with VSEPR and this should tell you all you need to answer the question.

Ian Marquez 2K

Professor Lavelle refers to the shape as "bent" and the textbook refers to it as "angular." Either should be accepted but I would lean toward bent for the naming since that's what Professor Lavelle has said during lecture.

Ian Marquez 2K

With that, any bond formed between two atoms, the sigma bond is formed first and there can only be one, with the double and triple bonds being composed of one sigma bond and one and two pi bonds.

Ian Marquez 2K

To reiterate, here there are 4 areas of electron density around the oxygen. The two lone pairs are not shown but have to be present to complete the octet. It is a tetrahedral configuration with the VSEPR equation of AX2E2, meaning that it is of bent shape with about a 109.5 degree angle.

Ian Marquez 2K

Moreover, this is just another example of covalent character within ionic bonds. Ions which cause large distortion, such as AlI, are described as being highly polarizable. This causes the ionic bond to have more covalent bond character.

Ian Marquez 2K

When it comes to molecules like this, draw the Lewis structure as instructed before. The molecular geometry, as stated above, would be different around the carbon atoms and with respect to the C-Be-C bond. According to the VSEPR model, just note the areas of electron density about these atoms.

Ian Marquez 2K

Lewis structures are needed for the VSEPR model to determine the areas of electron density around the central atom (bond pairs and lone pairs). From this information obtained by drawing a Lewis structure, the molecular shape can be determined.

Ian Marquez 2K

In the VSEPR model, only refer to the central atom when talking about electron density. The presence of lone pairs on the central atom will change the molecular shape of the molecule. Note that molecular shape and electron geometry are different as electron geometry just notes the areas of electron ...

Ian Marquez 2K

As stated above, e=hv would be sufficient for this problem. The only thing you have to watch out for is converting kJ (10^3) to J. Other than that, this is the energy of a single photon so you would just multiply this number with Planck's constant and apply proper significant figures.

Ian Marquez 2K

Also, it has been experimentally observed that half-filled subshells are more stable than predicted. The electron configuration exception of copper allows it to have half-filled subshells making it its most stable ground state and proper electron configuration.

Ian Marquez 2K

Just to reiterate, after all electrons are accounted for and single bonds are made, formal charges can then be calculated. This will indicate whether a double bond or even a triple bond should be made to lower formal charges. As you do more Lewis structures, patterns with certain elements, such as o...

Ian Marquez 2K

Summed up, use De Broglie when dealing with anything involving a rest mass and use the light equations when dealing with photons that have no resting mass.

Ian Marquez 2K

Another important thing to note is the formal charges present within these structures. If you are trying to determine if something has resonance, check to see if the formal charges make sense when changing a bond. For example, a formal charge of +1 for chlorine in a possible resonance hybrid is high...

Ian Marquez 2K

Since d-orbitals have 5 orbitals, they can hold 10 electrons as each orbital can have a spin up and spin down pair. This means that, with the d-orbital in play, there can be up to 18 valence electrons as s-orbitals can have 2 electrons and p-orbitals can have 6 electrons.

Ian Marquez 2K

The speed of light is the fastest speed that anything can go. In fact, light and photons are the only things that can travel at the speed of light. It is not physically possible for things with rest mass, like electrons, to move at or faster than this speed. Therefore, this theory is proved incorrec...

Ian Marquez 2K

It would be good to know the order of the EM spectrum and that visible light is present from wavelengths of 400-700nm. The order from shortest to longest wavelength (conversely highest to lowest frequency): gamma rays, x-rays, ultraviolet, visible, infrared, microwave, radio.

Ian Marquez 2K

Also, if you look at the shape of an s-orbital when graphed out, you see on the axis that there are no spots on any of the axis planes where electrons cannot be present. Nodal planes denote where electrons cannot be observed, but since the s-orbital is a sphere that encompasses all the axes, there a...

Ian Marquez 2K

Also, once you apply the equation E=hc/λ to find how much energy is produced by each individual photon, multiply the energy found from this equation with the total number of photons. This will cancel photons so your final answer will just be in just joules instead of joules/photon.

Ian Marquez 2K

Also, putting Joules in terms of SI units can be especially handy when seeing if units make sense and cancel correctly. For example, in calculating wavelength for light, the kg, seconds, and one m cancel leaving your answer in m (meters).

Ian Marquez 2K

In Lecture, Dr. Lavelle said anything less than 10^-15 or so would be undetectable, like said above. Pico as a prefix is 10^-12 so anything much smaller than that can be qualified as undetectable.

Ian Marquez 2K

Like said before, enter the fraction instead of rounding something such as 1/3 to .333 ensure a more accurate result. When it comes to rounding, try to hold off on any rounding as much as possible until you get to the final piece of data the question is asking for. Then apply the rules of sig figs.

Ian Marquez 2K

When using Einstein's equation, given MHz, do you have to convert this to Hz for the formula to work? I just want to make sure, but I would assume so since Hz is the base unit and M is just a prefix.

Ian Marquez 2K

Also, it is important and useful to know that 'E' will be in joules since that is the standard measurement of energy and is taken from Planck's constant.

Ian Marquez 2K

During the lecture today, Professor Lavelle stated that for computing and converting to moles or grams, just use the significant figures for the molar masses provided to you by the given periodic table. At the end of the problem, proceed to include the necessary significant figures. Basically, this ...

Ian Marquez 2K

What do you call a fish made of two sodium atoms?

2Na (Tuna)!

Ian Marquez 2K

Does anybody know if you must maintain a correct number of significant figures while working out, for example, a molarity problem, or if you would simply apply the correct number of significant figures once you have found your solution? When working through a problem, I believe that rounding to mai...

Ian Marquez 2K

When you calculate a percent yield problem, make sure to convert the limiting reactant into moles. From there, using the stoichiometric coefficients, you can calculate the moles of the theoretical yield of the product (N2H4). Since the ratio of the coefficients is 2:1, there will be half as many mol...

Ian Marquez 2K

Yes, I believe that you receive a point everytime you participate in the forums as either a response to a thread or posing a question. Also, I believe these are checked and marked down by your TA.

Search found 66 matches