CHEMISTRY COMMUNITY

Lesly Lopez 3A

Hello, so when work is done by the system, the final value would be negative. This is because the system is losing energy into the surroundings by doing work. When work is done on the system, the resulting value would end up being positive because the system gains energy from its surroundings. For s...

Lesly Lopez 3A

Hello!
From what I have learned, Gibbs Free Energy is a way to calculate the amount of work done at constant pressure and constant temperature. Within this concept, there is also enthalpy and entropy and it would be recommended to know how to define is a reaction is spontaneous or not.

Lesly Lopez 3A

From my own study habits, I think that going over sapling is good in order to understand the concept as a whole. However, if you really want to practice the material I would suggest going over the problems from the syllabus and the outlines provide by the prof himself.

Lesly Lopez 3A

Hey you guys, simply here to affirm the statements above. I believe it can be a zero reaction only if all the reactants in the reaction are zero-order although it is not often seen. Still, it is not impossible just not really seen.

Lesly Lopez 3A

From previous textbook exercises, I think it's usually in seconds. However, always be on the lookout if the problem asks for specific units. Also, make sure you convert in order to do the math right if needed.

Lesly Lopez 3A

IsaacLaw1E wrote: $k=e^{-\frac{E_{A}}{RT}}$
So k for the catalyzed = e^(-70*1000/8.314/291) = 2.72E-13 and k for the uncatalyzed = e^(-137*1000/8.314/291) = 2.56E-25
So kcat/kuncat = 2.72E-13/2.56E-25 = 1.06E12 J/mol

Thank you so much for this. It was super helpful.

Lesly Lopez 3A

Mmm I think the strongest reducing metal will have to be the one that is most likely to be oxidized. The reason being it might cause the reduction of another molecule later. I agree with my peers above.

Lesly Lopez 3A

Hi, I believe and hope that the slow step will be identified in the question. That way you will be able to know what rate and how to write the equation and such.

Lesly Lopez 3A

Before anything, you have to find out what type of order it is. From there, you can determine which equations to use as necessary. I have attached a picture maybe it helps.

Lesly Lopez 3A

Trust me I had a hard time with this concept. Before anything, add H2O as necessary to balance out the oxygen. You then have to add H+ to the opposite side because of the hydrogen in the water molecules from the water molecules you just added. Then, add OH- to both sides to balance the added H+. You...

Lesly Lopez 3A

AS mentioned before, the equilibrium constant, K, is determined by dividing the rate constant of the forward reaction, k1, by the rate constant for the reverse reaction, k−1.

K=k1/k−1

That is the only step you have to do divide the numbers provided.

Lesly Lopez 3A

According to the sapling question, when using a table to find the order, you practically divide two different experiments/trials where one reactant is constant and the other is changing. The change in the outcome of the reaction will show how each reactant affects or not affect each other.

Lesly Lopez 3A

Actually, I have a follow-up question to that. We are supposed to use the slow step. We also need to substitute HCIO and we use the reverse equation. But why do we not use Cl^- from the slow reaction? Shouldn't that be included?

Lesly Lopez 3A

From what I understand, it practically means that the rate of the reaction is not dependent. It does not depend on whether the rest of the reaction increases or decreases. As in d[A]/dt=-k[A]^0=-k, the rate is solely dependent on k. Hope that kinda clears things up.

Lesly Lopez 3A

Simply just to add on, I think we implement factors like Gibbs, enthalpy, entropy since they are known to be state functions. However, the state constant k is not a state function. Hence, there is a specific reason for which you would know which to multiply or add.

Lesly Lopez 3A

They will most likely be given to us but if not, they can be quickly derived from the rate law equations. They have to at least give us one of them to solve any kind of problem. Either way, you should be fine as long as you know how to use them.

Lesly Lopez 3A

Hi, just like my peers have mentioned, in the lecture professor said that it was not needed to memorize them. Still, I have tried to get familiar with the half-life equations because those will help you understand the concept a bit more.

Lesly Lopez 3A

Key thought: focus on the slow reaction
The slow step is practically the fast step but a bit more straight to the point if you will. Molecularity is telling you how much; the number of molecules that each step consists of.

Lesly Lopez 3A

From the practice question we have done I think that they ould give us that information so I do not think you should worry about that. If anything, know how to find conjugates or something along those lines. Acidic solution --> H+ ions in the overall reaction. Basic solution --> OH- ions in the over...

Lesly Lopez 3A

Hi, from my understanding, I think that's right. I believe molecularity describes the number of molecules that take part in an elementary step. For this reason, it should be equal to the sum of the stoichiometric coefficients of the reactants. Hope that kind of helps.

Lesly Lopez 3A

For the first one: the phases of each species present should correspond to their phases as originally specified in the equation given in the question: MnO4- is (aq) S2- is (aq) H2O is (l) MnO2 is (s) OH- is (aq) S is (s) For the second one: First begin by writing out the half-reactions. This is in ...

Lesly Lopez 3A

I believe you would balance the equation then look at the different charges for the red or ox. You would have to take a look and figure out the the change in charge. Then add e- to one side to make sure the charges are even and not have a 'heavier' side on the charges.

Lesly Lopez 3A

Hi you guys! I am having trouble with this. Identify the oxidizing and reducing agents in the skeletal (unbalanced) reaction. Then, balance the reaction, including the phase (solid, liquid, etc.) of each species. The reaction takes place in basic aqueous solution. MnO−4(aq)+S2−(aq)⟶S(s)+MnO2(s) I th...

Lesly Lopez 3A

I would have to agree with my peers above. The reduction half‑reaction should contain Cl2 as the reactant and Cl− as the product. A coefficient of 2 is needed in front of the product to balance the number of chlorine atoms. Finally, two electrons are needed on the more positive side to balance the c...

Lesly Lopez 3A

I do not think there is not a way to solve this without knowing the reduction potentials. We would have to be given the chart or the values to solve this. In this case there was a chart and if we were to be tested on something about this concept I think there would be a chart.

Lesly Lopez 3A

In the shorthand notation, the anode is written on the far left and the cathode on the far right. The different phases in the cell are separated by a vertical line, and the phases are listed in the order in which they are in contact. The salt bridge is represented by a double vertical line. The aque...

Lesly Lopez 3A

Hey okay so I put the exact same thing and It worked. The only difference is that wrote everything out. That means that I even wrote out the bullet point by doing shift+8. For some reason that was the only way it accepted my answer. Do not click on the little table that comes up when you are writing...

Lesly Lopez 3A

Yep, it sure does :)

Lesly Lopez 3A

From experience I think it's safe to always use Kelvin unless told otherwise. It is a simple step but can be easily forgotten.

Lesly Lopez 3A

It is possible for which an exothermic reaction could be non-spontaneous/ This could happen when $\Delta$ S is negative. Another reason would be that the temperature of the reaction is high.

Lesly Lopez 3A

The negative sign is important because it is telling you that the reaction has released or lost energy. The relationship between E and $\Delta$
G is that when E is positive, $\Delta$ G should be negative meaning the reaction is spontaneous.

Lesly Lopez 3A

I think that looking back to the lecture, we don't multiply the E value by the same value we multiplied the equation by. The E value should stay the same so nothing should be done to that value.

Lesly Lopez 3A

I believe the state property should be the same no matter how you get to the final part. Taking this into account, work and heat are not considered state properties. This is because their values can change depending on what certain steps that are made take to get to the end value.

Lesly Lopez 3A

The sign can tell you the most important thing. In most cases negative meaning non-spontaneous while positive means the reaction is spontaneous.

Lesly Lopez 3A

I am not really sure about this topic but, standard reduction potentials are actually used to measured how much a compound wants to get reduced. Positive: compound is likely to gain electrons and get reduced. Negative: compound is likely to loose electrons and get oxidized.

Lesly Lopez 3A

When I did this problem, I think the first way to approach this would be to use the formula w=-P∆V. In this case ∆V would be found by finding the volume of the cylinder using the given diameter. Then since we have the pressure as well, we could plug the values in and get w.

Lesly Lopez 3A

From what I understand:
Degeneracy (W) = (# of positions)^(# of molecules). Then you can just plug this into the S=KblnW equation.

Lesly Lopez 3A

The unites for delta G would be kilojoules. Follow up question: this is what units for an answer right? Like if it was in J that it is for sure that you need to convert to kj right?

Lesly Lopez 3A

You have to pay really close attention on your units. It is something that I have been working on as well. hen calculating delta G or delta S, you will use 8.314 J.K^-1.mol^-1. However, when using the ideal gas law use 8.206x10^-2 L.atm.K^-1.mol^-1.

Lesly Lopez 3A

The entropy value measures the disorder of the system. Usually, if S is positive then that means that entropy is increasing. When S is negative it is more likely that the entropy id decreasing.

Lesly Lopez 3A

I believe that all are considered spontaneous fro the reason that they release heat, they might not occur quickly.

Lesly Lopez 3A

I think a closed system means that matter cannot be transferred in or out, but energy still can. A beaker is sealed, it is a closed system.

Lesly Lopez 3A

The system that absorbs energy will carry a positive q value and in terms of temperature, it will absorb heat because it will increase in temperature. The system that releases energy will carry a negative q value and the higher initial temperature will release heat.

Lesly Lopez 3A

Out of all the four equations that you have, you have to create the desired reaction that is at the bottom of the problem. The point of this is that if you have to flip the equation then you’re enthalpy will have to change signs whether that is from positive to negative or negative to positive. Anot...

Lesly Lopez 3A

Hey, so to my understanding they cannot be negative because it is how much energy has to be gained in order to raise a substance's temperature by a certain amount. This I entire process is endothermic therefore it must be positive. Hope that clears it up.

Lesly Lopez 3A

Going from liquid to gas means the system is heated --> increase in enthalpy I am not sure what lecture it is but there is a graph that the prof shows in one of his lecture that helped me a lot since I am a visual person.

Lesly Lopez 3A

Smaller the pKa <--> stronger the acid
Ka of x 10-1 becomes a pKa of 1, which is stronger than an acid with a Ka of x 10-6 that becomes a pKa of 6

Lesly Lopez 3A

In 14A, Dr. Lavelle would tell us the number of questions on the midterm before it was administered which helped me with timing. I am also a little slow in calculating and I still hd a bit of time to look over my answers.

Lesly Lopez 3A

Hey, I think you have the wrong equation to begin with. You should redo you ICE table and then manipulate your equation so that it is a quadratic equation: 0.032x^2+x-0.154. In the end your percentage has to be small so you would ignore the negative x value. You take that value and divide it by 0.15...

Lesly Lopez 3A

I have a quick question, how did you know you had to use 3/2 in the equation for delta U?

Lesly Lopez 3A

I was also confused about that. If he were to ask us a conceptual question, how would we know to use the generalization or when it is like the problems in sapling? What are the explicit rules for the relationship between pH and acid/bases.

Lesly Lopez 3A

As many of our peers have said when a compound creates H3O+, the compound is an acid therefore, we use Ka. If a compound forms OH- , it is a base and we use Kb. Technically, we know this as Kw = Ka x Kb.

Lesly Lopez 3A

The point of this problem is that you have to flip, add, multiply certain equation for you to end up with the equation shown in the problem. For examples, the first equation should be multiplied by 3 in order to get the right coefficients as the result equation. You have to choose equations that wou...

Lesly Lopez 3A

Lesly, I did the same thing you did, and I got a different answer. When I kept your numbers into my calculator I got 0.0234. (0.12^2 x 0.06)/(.192^2)=0.0234. Oh I see what I was doing wrong. I was placing the initial value of SO3 into the Kc formula. Thank you, I was focused on the table I didn't e...

Lesly Lopez 3A

Hi, so I am having a similar problem. At a certain temperature, 0.780 mol SO3 is placed in a 2.50 L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.150 mol O2 is present. Calculate Kc. For this I divided 0.780 by 2.5 for molarity, 0.312 M. Then I set up my ICE table. With SO3 having the initia...

Lesly Lopez 3A

Thank you guys! That helped so much :)

Lesly Lopez 3A

Hi you guys, In the lecture today, in the example, the question gives us the equilibrium concentrations of N 2 , H 2 , and NH 3 . How did they find the concentrations? If any case, the problem does not give us the concentrations at equilibrium, how are we suppose to find them? And will those always ...

Lesly Lopez 3A

Hi,

Hannah explained it well. Overall, favored in this case is 'more being made.' Whether that is more products being made or more reactants are being made.

Lesly Lopez 3A

Hi, solids or liquids do not have concentrations. Reason being, a concentration is the relative amount of a given substance contained within a solution. With that being said, solids are not diluted in liquid and pure liquid is to be by itself. So technically, they do not have a concentration. And as...

Lesly Lopez 3A

Wait we have outlines for 14B? I thought those were only for 14A...

Lesly Lopez 3A

Hi,

They are different because ONO is nitrito but NO2 is nitro.

Lesly Lopez 3A

To be honest I tried to do 3 minutes per question but at times I would just try to answer the question and I would see I took about 10 minutes to answer it. It was difficult to manage my time and try to understand the questions.

Lesly Lopez 3A

If they are posted on Saturday, I am assuming there is no debate about our final grade. What we see is the final final grade huh?

Lesly Lopez 3A

Straight Answer: Yes, all of the ones that are part of group one and two are able to form oxides/hydroxides.

Lesly Lopez 3A

Hi,

You could say that but it is not always. There are a lot of factors aside from the number of bonds that determine bond length, so it is hard to say or apply that for every case.

Lesly Lopez 3A

Hi,

pKa can be solved with -log (Ka).
I actually had a question. pKa, is it dependent on something or is it something separate?

Lesly Lopez 3A

Hi,
All late but for the conjugate acid, it's a base that turns into a conjugate acid. try to reason it, when doing the reverse of the chemical reaction it would go back to being a base. It donates an H+ to go back to being a base. This is why for the conjugate acid you would add an H+.

Lesly Lopez 3A

Hi,

From what I know, it is just covalent bonds. This is because bonded ions will not be able to share electrons as much as they have the ability to exchange them.

Lesly Lopez 3A

Hi, From my understanding, a conjugate base has one less H in the formula and a decrease in charge by one unit. If the formula is HSO 4 - , then the base would be SO 4 -2 . As for the conjugate acid, there is one more H added to the formula and an increase in charge by one unit. Following the rule t...

Lesly Lopez 3A

Hi, this was very helpful and wanted to add in my compound for others to see a different example. Compound: [Co)NH 3 ) 5 Cl]Cl 2 Practically the same steps as Idella, Since NH3 is my ligand and there are five, I said Penta ammine. Next is the one chlorine, so I did not put a prefix just "chlor...

Lesly Lopez 3A

Hi,

The reason why it is more acidic is because it makes a stronger Oxoacids. Reason being it contains more polarized O-H bonds. The more electronegative the central atom is, the more polarized the O-H bond will be. In this case HBrO4 has more bonds. Hope that was not too confusing.

Lesly Lopez 3A

Hi,

I agree with Margaret. My TA also said that we will not be able to use any other resources expect for the usual: periodic table and equation sheet. For this reason, I think that it is definitely for the best if you try to memorize them. Just in case.

Lesly Lopez 3A

Hi, this was very helpful and wanted to add in my compound for others to see a different example. Compound: [Co)NH 3 ) 5 Cl]Cl 2 Practically the same steps as Idella, Since NH3 is my ligand and there are five, I said Penta ammine. Next is the one chlorine, so I did not put a prefix just "chloro...

Lesly Lopez 3A

Hey! I agree with everyone above. He did not specify that we needed to know to such extent. We would just put tetrahedral or square planar for the compound shape or whatever you are looking at. With this shape the coordination number is usually 4 but I would still draw it out just in case there migh...

Lesly Lopez 3A

Hi,

I was also thinking about this but I think Alvin is right. There are definitely 3 pi bonds but instead of 6 sigma bonds you have to take into considerations the extra bonding the carbons have with the hydrogen atoms. Therefore the total number of sigma bonds would be 12.

Lesly Lopez 3A

Hi! Correct, you are right. From my understanding, NH3 is neutral therefore the charge of two chlorine outside would be -2. The compound charge must then be +2 for everything to make sense. That is why, in this case, Cl has a charge of -1 you set this equation: X -1 = +2. It may sound confusing but ...

Lesly Lopez 3A

Hi,

Yeah Dr. L said that we should not worry much about it since we did not go much into depth but just in case you really want to know I believe ferrate will always be used. From my understanding some metals, iron being one of them, use Latin names, as someone mentioned above.

Lesly Lopez 3A

Hello, This is an interesting question. To be completely honest, I am not sure. However, I think you should be okay if you are able to figure out how the ligands bond to the TM for each complex. the concept itself is not hard but the answer to your questions is interesting. Hope someone can answer it.

Lesly Lopez 3A

Hi!

They are both correct but in this class I think that we would be using -o much more than the other option.

Lesly Lopez 3A

Hi! This is a hard concept for me too because I feel like I can never do it right lol. I draw the one that is the best and then from there I simply try to make either double or triple bonds. Still, you have to keep in mind that some of them will not be the most stable or even plausible. Make sure yo...

Lesly Lopez 3A

Hi!
Having more bonds does make the London Dispersion Forces stronger of the molecule stronger. Even though LDF's are the weakest, when stacked upon each other they have great strength. Hope tis helps.

Lesly Lopez 3A

It can since it tells you how it is formed. The theory says that repulsion among the pairs of electrons on a central atom (whether bonding or non-bonding electron pairs) will control the geometry of the molecule. Hope that helps!

Lesly Lopez 3A

The way I understood this was that since F is the largest element(look at the periodic table and follow the atomic radius trend), it would have a greater atomic radius. It will require a greater bond length from the end of the atom to the other atoms. However, since the length is greater the bond wo...

Lesly Lopez 3A

Hi, The way I do it is to look at the differences in electronegativity between the atoms on the periodic table. When the atoms with a large difference in electronegativity compared to each other then this will create a dipole moment. This is what will cause the molecules to be polar. The large diffe...

Lesly Lopez 3A

later in this problem it asks us for the oxidation number of chlorine, how do we find this? Hi, the sum of the oxidation numbers of the atoms in ion must equal the charge on the ion, -1. When oxygen is bonded to a nonmetal other than F, it has an oxidation number of -2. This is since there are four...

Lesly Lopez 3A

Hi, so one possible resonance structure is by changing one of the C-O to a double bond and make the double bond C-O single bond. This means that the -1 formal charge will be on the other O atom than in the picture on the question. That would be one resonance structure. The other resonance would have...

Lesly Lopez 3A

Hi, The formal charges on the atoms should add up to zero because the HCNO molecule is neutral. The most stable structure has the fewest number of, and smallest magnitude of, formal charges, and with the negative formal charges on the more electronegative elements. Both resonance structures B and C ...

Lesly Lopez 3A

Hi! These London dispersion forces are often found in the halogens (e.g., F2 and I2), the noble gases (e.g., Ne and Ar), and in other non-polar molecules, such as carbon dioxide and methane. As long as there is no polarization between the atoms in the compound then you should be fine. London forces ...

Lesly Lopez 3A

Hi, You draw the Lewis structure and look for a pair of electrons that can form a bond to another atom. A Lewis base is a molecule or ion that can donate a pair of electrons to form a bond. An atom, ion, or molecule with a lone-pair of electrons can be a Lewis base. I like to draw out my compounds t...

Lesly Lopez 3A

Omg this is an amazing thread! So far I loved New Girl, Supernatural, The Office, and Friends. I am currently watching Criminal Minds and so that is what I am up to right now. Have a great week you guys.

Lesly Lopez 3A

Hi! There are multiple ways to represent a formula using lewis structures. As long as the atom with the highest electronegativity is the one with the formal charge that is not zero it is okay to have an atom with a formal charge. The whole point is to make sure that most of the atoms do have a forma...

Lesly Lopez 3A

Hi! I understood that, electron distortion is what you call when electrons of the anion are then attracted to the cation. This causes the shape of the electron density to have a slight change. For ionic bonds, the most straightforward concept we tend to think of is a bond where the more electronegat...

Lesly Lopez 3A

Hi! From what I understood, electron distortion is what you call when electrons of the anion are then attracted to the cation. This causes the shape of the electron density to have a slight change. For ionic bonds, the most straightforward concept we tend to think of is a bond where the more electro...

Lesly Lopez 3A

I cannot figure out how to determine the number of spectral for question #6? Is there a different formula to use other than En=-hR/n^2 or DeltaE=Ef-Ei? Hi, for that it is simply 5. Reason being you got form 6-1, 6-2, 6-3, 6-4, and 6-5. Or the other way around. In that case there are just 5 spectral...

Lesly Lopez 3A

Hi! always try to look back at the reference sheet. I was having trouble too until Someone told me it was on the reference sheet. All the masses are there in units of kg.

Lesly Lopez 3A

nayha a 1L wrote:Can someone explain to me how we find the value for m? What would we plug in there?

Hi there, I am not sure where this number came about but someone in another question on this website said that the mass od an electron was: 9.109 x 10^-31 kg. Hope that helps!

Lesly Lopez 3A

I calculated the wl in meters, and you got those right. But you just got the conversion wrong. When converting m to nm, you don't divide by 10^9 (what you did), you multiply the number in meters by 10^9. Intuitively, you should have a bigger number in nm than in m. Hope this resolves it Thank you f...

Lesly Lopez 3A

I think when you tried to convert the wavelength from meter to nm, you didn’t use the correct conversion. 1 nm = 10^{-9} m. So 9.4\cdot 10^{-8} m converted to nm would be 94. Same thing for the second one. 7.46 \cdot 10^{-6} m converted to nm would be 7460 nm. The wavelength range would then be fro...

Lesly Lopez 3A

Hi guys! I have been having o much trouble with this question. Please help The electron in a hydrogen atom is excited to the n=6 shell and emits electromagnetic radiation when returning to lower energy levels. Determine the number of spectral lines that could appear when this electron returns to the...

Lesly Lopez 3A

It does depend on the TA. There are some that are only willing to give you the back of the periodic table and the formula sheet. However, there are others, like my TA, that said to show him the piece of paper before to make sure there is no previous writing and he will allow us to use that paper. Ei...

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