CHEMISTRY COMMUNITY

CMaduno_1L

"Balancing Redox Reactions in Acidic and Basic Conditions" by Professor Dave Explains had a nice and simplified explanation that really helped. I'll paste the link. Hope this helps and good luck!!
https://www.youtube.com/watch?v=N6ivvu6xlog&t=390s

CMaduno_1L

they are typically used when using the 2 of the 3 different forms of the entropy equation:

1. Constant volume (isochoric): dS = n*Cv*ln(T2/T1)

2. Constant pressure (isobaric): dS = n*Cp*ln(T2/T1)

CMaduno_1L

it is also important to know the difference between gradual expansion and sudden expansion, in which you would apply the appropriate formula. The questions I have seen typically state whether the process is occurring gradually or suddenly. Hope this helps!

CMaduno_1L

specifically, if temperature increases, then so does the rate constant. If temperature decreases, the rate constant decreases as well. In short, there is a direct relationship between temperature and the rate constant.

CMaduno_1L

OH- is only added when balancing a redox reaction in a basic solution. You add the same number of OH- as you have H+ to both sides of the equation.

CMaduno_1L

if you are referring to the reaction rate constant (lowercase k), then I believe that varying the temperature and solvent would affect its value.

CMaduno_1L

The only other name I can think of is reaction rate coefficient, but oftentimes it's just presented as k in a question.

CMaduno_1L

remember when given a table with the concentration of reactants and rates of different reactions, the order can be found by setting up a ratio of the concentration from (ex.) (Conc. Experiment 1 / Conc. Experiment 2)^n = (Rate Experiment 1 / Rate Experiment 2). Remember that the variables that repre...

CMaduno_1L

I believe so. If I'm understanding your question correctly, the balancing of a redox reaction in basic and acidic conditions follows the same steps. The only difference is that for a basic reaction you add OH- as you did H+ to both sides. The side with both H+ and OH- combines to H2O.

CMaduno_1L

also make sure not to confuse intermediates for catalysts, which are first consumed then produced in a later elementary step, as opposed to intermediates, which are first produced then consumed during a later elementary step.

CMaduno_1L

I've learned you balance in order of... Acidic: any element besides O and H, balance O by adding H2O, balance H by adding H+, add electrons to appropriate side Basic: any element besides O and H, balance O by adding H2O, balance H by adding H+, add the same number of OH- (as you did H+ in the previo...

CMaduno_1L

No they are not always acid and base reactions. For a reaction to be considered redox, the number of electrons an element has must change, which can occur in the absence of acids and bases.

CMaduno_1L

Keep in mind, however, that in order to answer the question correctly, you must write out the half-reactions for both part a and part b. This will allow you to determine how many moles of electrons were transferred, n. Otherwise, all the other needed pieces of information are given.

CMaduno_1L

I think the half reactions should be...

Reduction: 2Ce(4+) + 2e- --> 2Ce(3+)
Oxidation: 3I(1-) --> I3(1-) + 2e-

From this we know that the moles of electrons transferred, n, is 2. Hope this helps!

CMaduno_1L

In regards to the first part of your question, I believe that a negative value for standard electrode potentials indicates that the element or compound in question forms ions easily.

CMaduno_1L

a salt bridge is a device used to neutralize the charges of each half-cell in a galvanic cell. Because it prevents the buildup of charge in both the anode and cathode, the transfer of electrons can continue.

CMaduno_1L

I believe an electrode is the solid metal that is placed into the solutions. They function to provoke electrical conductivity (can someone confirm this please).

CMaduno_1L

Without a salt bridge, because the anode becomes more positive overtime and the cathode becomes more negative overtime, the cell would no longer cause a transfer of electrons. Electrons would have no reason to leave the anode to the cathode, which is negatively charged (like charges repel each other...

CMaduno_1L

In short, what is lost by one half reaction (oxidation) is gained by the other reaction (reduction), therefore representing a transfer of electrons. This is the number that should be inputted for n. in the equation: G = -nFE

CMaduno_1L

For example: if given two reduction reactions, the higher the voltage, the more likely the reaction is to occur. Therefore, we know that the lower number is the reaction that we must flip to become an oxidation reaction (because it is less likely to occur as a reduction reaction relative to the othe...

CMaduno_1L

If you have fractional coefficients, multiplying to yield whole number coefficients would affect every component of the reaction and therefore the ratio should remain the same and so should your answers. I think you should be okay.

CMaduno_1L

I'm not completely sure, but I think for this question, since the given pressures would be used to find Qp, the units of bars would cancel out since Q has no units (and because it's products/reactants, so units would cancel) , so as long as the two pressures have the same units, I don't think it mat...

CMaduno_1L

Additionally, I believe that a system being isothermal suggests that the change in internal energy for the system is 0, meaning that
dU = q + w ==> q = -w

CMaduno_1L

To add to the statement above, many questions in the textbook describe systems as isothermally reversible and irreversible, so it is not necessarily correct to consider an isothermal system to be only reversible.

CMaduno_1L

During Lyndon and Matt's Pizza Rolls review session, Matt made a comment that might help with remembering the meaning of isobaric. He emphasized the BAR in isoBARic and compared it to BARometer, which is an instrument used to measure atmospheric pressure. Hope this helps!

CMaduno_1L

To answer the question above, I believe it's safe to expect to see both variations. We should know the different conditions that correspond to each type of calorimeter because both types show up in the homework problems. Hope this helps!

CMaduno_1L

In regards to calorimeters, it is also important to remember that -q = qcal. In other words, heat that is lost from the system is gained by the calorimeter. Hope this helps.

CMaduno_1L

Oftentimes, the question will also state whether the gas in question "expands reversibly" or "expands irreversibly", so be on the look out for key terms when reading the question!

CMaduno_1L

dU = w whenever there is no each exchanged with the surroundings (a.k.a q = 0). This condition is known as adiabatic.

CMaduno_1L

Entropy is associated with maximization of the number of microstates or number of arrangements. Spontaneous processes are also characterized by maximizing microstates, and therefore maximizing entropy.

CMaduno_1L

I'm pretty sure it's just to understand what is going on conceptually. I think the function of the integral was to clearly show the derivation of the reversible isothermal expansion equation: w = -nRT ln(Vf - Vi) .... (I would focus more on knowing how and when to use this equation. Essentially, the...

CMaduno_1L

Since no heat is exchanged with the surroundings in the presence of an adiabatic wall, q=0.

If we apply this fact to the $\Delta U = q + w$ equation, we find that $\Delta U = w$ when adiabatic

CMaduno_1L

Good question! I'm not entirely sure, but I understand that a closed system has a constant internal energy, since there is no gain or loss of energy/heat. Since internal energy is the heat plus work, I am assuming the definition accounts for work as well. Can someone please double check this? Thanks...

CMaduno_1L

A counter example would be in an isolated system, where neither heat nor matter can be exchanged, dU = q + w.

CMaduno_1L

Although both are structured differently, it is important to remember that they serve the same function-- to measure specific heat capacity (often referred to as specific heat).

CMaduno_1L

I'm not too sure about entropy, but some of the enthalpy unit terms we have are: enthalpy itself, which refers to the study of heat released or absorbed in chemical reins and physical changes...it can also refer to the amount of heat released or absorbed at a constant rate if reaction gives a net re...

CMaduno_1L

According to my notes, the reaction was: N2 +O2 --> 2 NO dHrxn = 180 kJ + 2 NO + O2 --> 2 NO2 dHrxn = -112 kJ as you can see, the product of the first reaction is found in the reactants of the second reactions. That means that it is an intermediate component and can be cancelled out. Additionally, t...

CMaduno_1L

Examples of other state properties include: energy, pressure, volume, temperature, density, and heat capacity

In contrast, work and heat both depend on the path taken to from its initial value to its final value, preventing them from being considered state functions

CMaduno_1L

This is very similar to how when reactions took multiple steps to reach equilibrium we would multiply the equilibrium (K) values from each step. When you put products over reactants of each step and multiplied, any intermediate components were cancelled out. Kind of similar to Hess's law, so we've s...

CMaduno_1L

All of the above in addition to the fact that water vapor is at a higher temperature when it is in the form of steam can be visualized by looking at the cooling/heating chart that professor Lavelle went over during lecture. Hope this helps with understanding!

CMaduno_1L

It's also important to understand that partial pressure is different from total pressure, in which the total number of moles on each side of the reaction would be considered.

CMaduno_1L

How do we know when to use K_c and when to use Partial Pressure Notation? Kc notation is used when the reactants and products of a homogeneous reaction are in the aqueous phase, while partial pressure notation is used when the reactants and products of a homogeneous reaction are in the gas phase. K...

CMaduno_1L

Why are solids and liquids not included when calculating K values? I believe solids and liquids are not included when calculating K values because their concentrations don't change significantly over the course of a reaction. Can anyone confirm this? I'm not exactly sure if this is the right explan...

CMaduno_1L

In regards to what the difference is between Kc and Kp:

Kc represents the equilibrium concentrations of reactants and products, while Kp suggests the equilibrium partial pressures of the reactants and products

CMaduno_1L

A method I use to remember the meanings of when Q>K and Q<K is to think of Q wanting to approach K. This can be visualized as a number line with K placed in the center. If Q is less than K, then the reaction must proceed in the forward direction in order to reach K. Conversely, if Q is greater than ...

CMaduno_1L

In regards to the activity of each ion, I believe they are referring to the ion's respective concentration. I hope this helps!

CMaduno_1L

Essentially, this principle applies to changes in all physical parameters, as stated in lecture

CMaduno_1L

Le Chatelier's Principle tells us that chemical reactions adjust to minimize the effect of changes, but because solids and liquids are negligible in equilibrium equations, adding or removing them will not have an effect on the equilibrium.

CMaduno_1L

Try searching for a free (or even cheaper) download for the textbook PDF. It would likely be easier to use and you'd save money!

CMaduno_1L

He usually will let us know in lecture. I assume everything in outline 1 and all other material covered during lecture up until the test, unless otherwise stated. Your TAs might also be able to provide a more definite answer. Hope this helps!

CMaduno_1L

Knowing these comes in handy when given their values and you have to rearrange them in order to find a solutions concentration, for example. Notably, concentration (or n/V) = P/RT.

CMaduno_1L

It is also important to remember that boiling does not break bonds. So when explaining the effect that boiling point has, my TA suggested avoiding using the phrase "break bonds" and use "overcome intermolecular forces" instead.

CMaduno_1L

Yes, it is the weakest relative to the others and its strength (compared to other molecules with only LDFs present) increases with size.

CMaduno_1L

Yes, it is also important to understand that hydrogen atoms bonded to carbon atoms are not counted when determining possible hydrogen bonding sites.

CMaduno_1L

^I understand that this is true for conventional numbers, however, for pH specifically, my TA mentioned that each number after the decimal point is a significant figure. Please double check with another source, though.

CMaduno_1L

Yeah! This video (https://youtu.be/7N0lRJLwpPI) helped me with this topic. Hope this is what you're looking for!

CMaduno_1L

I think in terms of this unit, we just need to know that boiling point and melting point are when a molecule changes its phase. If the intermolecular forces are stronger in one molecule as opposed to another, it will have a higher boiling point and higher melting point relative to the other molecule...

CMaduno_1L

To add to this, among molecules that contain only London dispersion forces, the way to tell whether one is stronger than another is by the size of the entire molecule. The larger the molecule is, the greater the strength of LDFs.

CMaduno_1L

My TA told me that generally if there is at least one atom that is not the same (amongst the outside 4 atoms of a tetrahedral) then the molecule as a whole is polar.

CMaduno_1L

In response to Matthew ILG 1L, I believe the term "hydrogen-bonding sites" refers to the places where hydrogen bonds can potentially form as well as where they already exist, which is why you would include the H-bonds already present in order to find the total number of bonding sites.

CMaduno_1L

I believe the bond formed between an s-orbital and an s-orbital is a sigma bond.

CMaduno_1L

I believe this only applies to transition metals, as the central atom of a coordination compound is a metal and the surrounding atoms are nonmetals/ligands. I think it would be different for other compounds. Can someone please confirm this though?

CMaduno_1L

I believe this is because you are counting the number of atoms directly bonded to the central atom, not necessarily the number of bonds they contain. Hope this makes sense!

CMaduno_1L

^I agree. It may help to know that ligands directly attached to the central atom make up the coordination sphere, so it makes sense that the number of bonds attached to the central atom are what comprise the molecule's coordination number (this is how I try to remember it).

CMaduno_1L

I also believe that we have to use these principles for naming ligands in 14B, so it would be helpful to have them memorized, as later chapters likely build upon this knowledge

CMaduno_1L

I believe it is CN-, meaning it can form the coordinate covalent bond. However, I'm not exactly sure why the total charge is 4-.

CMaduno_1L

I believe that H, He, Be, and Li are also exceptions in that they are content with having less than 8 valence electrons.

CMaduno_1L

I believe so, this is according to the example he gave during lecture regarding cis-DICHLOROETHENE (polar) v.s. trans-DICHLOROEHTENE (non-polar). The arrangement of the Cl atoms influences the polarity it seems.

CMaduno_1L

Sorry I put square pyramid twice, I meant square planar for that last one

CMaduno_1L

To add on to these responses, I believe this includes: linear, trigonal planar, bent/angular (<120 degrees), tetrahedral, trigonal pyramid, bent/angular (<<109 degrees), trigonal bipyramid, seesaw, T-shape, octahedral, square pyramid, square pyramid (So about 12 shapes, double check just in case)

CMaduno_1L

In terms of molecules containing only Van Der Waals interactions: the larger the size of the molecule, the more interactions there are, meaning that the interactions in that given molecule are stronger relative to one with a smaller size/less interactions.

CMaduno_1L

It is also important to keep in mind that in addition to all of this, the number of protons remains the same in the given cation or anion. Isoelectronicity means that the number of valence electrons is the same between two different elements, not the number of protons.

CMaduno_1L

I don't believe that knowing the equation is necessary. I think it's just important to know that angular momentum is denoted by the letter "l" and that it can be calculated by subtracting the principle quantum number, n, by 1.

In other words, l = n-1

CMaduno_1L

I find watching videos very helpful. It's a change of pace from lecture since we usually have to copy down stuff. Videos give me a chance to just listen. If the modules that are provided aren't enough, then I suggest watching Crash Course or Bozeman Science videos on YouTube.

CMaduno_1L

Yes, but this should not be confused with the ionization energy. Ionization energy is the energy required to remove an electron from a different state of matter, which is not the same as removing it form a metal surface.

CMaduno_1L

The delta x is the uncertainty in position and the delta p is uncertainty in momentum. The product of these two are greater than or equal to Planck's constant divided by 4*pi.

If we know the position, then the momentum is uncertain, and vice versa.

CMaduno_1L

I believe the rule is to multiply the uncertainty by two

CMaduno_1L

In response to Brennayoung's question, I believe that we will only be applying this to hydrogen for now. This principle is used when measuring position and momentum of a particle.

CMaduno_1L

Crash Course has a quantum mechanics series out on YouTube, which is pretty helpful as well!

CMaduno_1L

I believe .9 and above is a safe range for rounding up. Anything lower I would multiply to get the numbers behind the decimal point to 0.9 or higher. Hope this helps.

CMaduno_1L

Yes, I agree. SI units are a great way of checking whether or not your are on the right track. If the units, when cancelled out, leave you with the desired units, you can assume that your answer is right (given the math is done correctly).

CMaduno_1L

I believe so, as that's what Lyndon did during Dino Nuggets

CMaduno_1L

I wouldn't rule it out completely, especially if we are learning hybridization soon, as it is known to have an influence on molecular geometry. But for now I can assume it's a no.

CMaduno_1L

That being said, some structures are more stable than others, even though the substance is the same. This is where the calculation of formal charges comes into play. Substances that have more atoms with a formal charge of 0 will be more stable.

CMaduno_1L

MichelleRamirez_2f, yes. The formal charge helps us indicate which structure is the most stable (more formal charges of 0), which is favored.

CMaduno_1L

From what I have seen, the topics covered in lecture correlate to what appears on the test. If anything, the textbook may go into more detail than need, which isn't necessarily a bad thing. However, in terms of maximizing studying, I would rely more on the notes from lecture. Hope this helps!

CMaduno_1L

I'm not sure what the likelihood of this is, but if you still have your notes from chemistry in high school (perhaps in a binder), revisiting those may help clear up any misunderstandings

CMaduno_1L

In addition to these, it may be relevant for you to know about the paschen (n=3), brackett (n=4), and pfund (n=5) series.

CMaduno_1L

Also, if none of the above works, you can always resort to YouTube! Hearing it repeatedly could help retain information. Plus, if you're a visual learner, YouTube is a nice and convenient method that doesn't require access to a textbook or notes. Hope this helps!

CMaduno_1L

The AAP program also provides peer learning facilitators who function the same. If you are eligible to join, I would recommend doing so, as they tend to consist of less people (typically around 10 max).

CMaduno_1L

To add to this most recent comment, an example would be in combustion reactions, where oxygen gas is always by itself on the reactant side of the chemical equation.

CMaduno_1L

Personally, simply reviewing the tables and charts in the textbook helped me sharpen up on high school chemistry. YouTube videos are also a great resource!

CMaduno_1L

Yes, your last question was correct. For example, if you were adding 3.5 and 2.11, you'd get 5.61; however, in order for the answer to be rounded to the correct number of significant figures, you'd have to make your final answer 5.6. This is because the final answer must take the same amount of numb...

CMaduno_1L

I typically read each sentence and underline as I go. Although it sounds pretty elementary, I've found that it helps me better pick out the relevant information, while also ensuring that I don't miss anything that can be used to help solve the problem. Breaking it down sentence by sentence helps tre...

CMaduno_1L

If given the values of the other terms, you could simply solve the equation for the missing value. For example: if solving for actual yield, the question would likely provide the values of theoretical yield and percent yield, so you can solve for actual yield by multiplying the theoretical yield by ...

CMaduno_1L

No, all compounds will be given in terms of their formulas.

CMaduno_1L

Yeah, this week's test only consists of everything up until Quanta and Photons; however, for the homework we are able to turn in problems relating to the Quanta and Photons lessons in addition to those relating to the high school chemistry review lessons.

CMaduno_1L

Knowing the prefixes will be useful when using conversion factors to get to a final answer, so I do think memorizing prefixes will be necessary. In regards to fento, atto, and angstrom, I'm not too sure. I would focus more on the angstrom, since that's what he touched on the most during lecture. Hop...

CMaduno_1L

I don't think the delta has anything to do with the removal of products, it just signifies that the reaction requires a high temperature in order to occur. For example, the conversion of limestone into quicklime takes place at about 800 degrees Celsius (found in the textbook).

CMaduno_1L

I'm going to assume that the questions will resemble the practice we do during discussion, so free response. That way they can [censored] our ability to work through the problems, use sig figs etc..

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