CHEMISTRY COMMUNITY

Julie_Reyes1B

These are just from his unit outlines (I included industrial and environmental as well): Unit 1 (Chemical Equilibrium): ATP hydrolysis, osmotic pressure Unit 2 (Acids and Bases):Carbon dioxide reacts with water to form carbonic acid, which makes carbonated drinks acidic, acidic rain, rivers, and lak...

Julie_Reyes1B

It is important to note that when you look at the elementary reactions, you can see that the catalyst is consumed at the beginning and then regenerated as a product later in the experiment

Julie_Reyes1B

You can also look at units of the rate constant if it is given and be able to tell what the order is. Yes! for zero order reactions the units are M/s. First order is s^-1, and second order is 1/M.s. The way that I remember this is that rate is in the units M/s, then when you solve for k you have to...

Julie_Reyes1B

This is more of a test of whether we understand how coefficients are used in the unique reaction rate formula, which is the one that it like 1/a (dA/dt)= 1/b (dB/dt)...etc.

Julie_Reyes1B

Anyone know where to find any past exam questions for practicing? Really need to work on some actual problems! I've found that the practice exam Lyndon creates is extremely accurate to what Dr. Lavelle asks. Besides that, I'm not really sure since Dr. Lavelle does not make his exams available in th...

Julie_Reyes1B

I think one of the best parts of the review session is that Lyndon is able to provide conceptual techniques/hints for the types of problems Dr. Lavelle tends to ask on exams. So if I were you I would definitely try to go! If you are really unable to make it, you can PM me and I can send you a detail...

Julie_Reyes1B

I believe that there is no net change in the amount of catalyst in the final reaction. So, it will show up in the first elementary step (since it is put into the reaction) and then show up again in the products of the last elementary step. Overall, it will cancel out. Socratic.com gave this example:...

Julie_Reyes1B

To add on, this explanation is usually called the experimental method. Dr. Lavelle went over this in the lecture on 2/28, using an example with NH₄⁺ and NO²_-. If you need extra help, I can explain that specific example step by step :)

Julie_Reyes1B

I don't think we have the skills necessary to do this. We may be given some proposed reaction mechanisms and have to choose one and explain why it could work. That would involve showing that the reaction mechanism satisfies two criteria: A. Sum of elementary steps= overall reaction B. Mechanism must...

Julie_Reyes1B

A pH meter is an electrode that can solve for Q in order to calculate the H⁺ concentration of a solution. Ecell° is already solved for by calibrating the pH meter, I believe (by measuring Ecell of a solution with a known pH).

Julie_Reyes1B

I think that it is probably fair game for test 2, because one of the main topics on Outline 5 is "Use and apply the above concepts and equations in industrial (e.g., batteries, rust, pH electrodes) and biological (e.g., electron transfer) examples". I would do all of the homework just in c...

Julie_Reyes1B

I think it doesn't just have to do with the number of moles, but with which reaction will be the strongest and thus produce the largest temperature change. Since b. has the highest molarity in the smallest volume, it will react the most (I think this is because the particle density is really high, s...

Julie_Reyes1B

3I - to I 3 - would be an example of oxidation. The total charge on the left side is -3 (since there is a coefficient of 3) and the total charge on the right is -1. In order to balance both sides, 2 electrons must be added to the right side. Since the reaction shows that electrons are lost, you know...

Julie_Reyes1B

You can find the electron transfer by writing the half reactions and seeing how many electrons are transferred. The reduction one is a little bit easier. 2Ce +4 +2e- ---> 2Ce +3[sup]. The charge of left side is +8, and the charge of right side is +6, so 2 electrons are needed on the left side to bal...

Julie_Reyes1B

So basically what I did was use the Nernst equation (the one with the logarithms) with log(H⁺) for log Q. Since log(H⁺)=-pH, you can replace that in the equation. Then, I solved for Ecell°. To find the pH for the unknown concentration, replace Ecell° with the value you found and solve for pH.

Julie_Reyes1B

Because pH>pKa, the acid will mostly deprotonate. Since it deprotonates it will become the conjugate base, ChH3COO-. The conjugate base will have a net charge of -1.

Julie_Reyes1B

This is the way I did it, but I'm not quite sure if it is correct. So I used the equation ΔG°= ΔH- TΔS°. So for G and H to be equal, TΔS°=0. B, C, and D would be wrong because there is a pretty clear change in entropy for each one.

Julie_Reyes1B

For that problem, it does look like Zinc is a reactant and the zinc ion is the product. It seems like you got that part right! Maybe you made a mistake with the Q term; for that problem, make sure it is [Zn 2+ ]/[Ni 2+ ]. In terms of deciding which is the anode and cathode, the locations of the subs...

Julie_Reyes1B

Hi! Just wondering the difference in the two equations listed in the book for Gibbs free energy: ΔG=-nFE cell ΔG r =-n r FE cell They look very similar but the book states the second is in "molar form", so that the units are KJ/mol. Not really sure what this means or when to use one instea...

Julie_Reyes1B

Tomorrow after class :)

Julie_Reyes1B

The points at the end may be adjusted based on the midterm and final averages. For example, everyone may get an extra 10-20 points (out of 500) to boost the grades a little bit.

Julie_Reyes1B

In the book, K without a subscript refers to Kp, whereas in Dr. Lavelle's class, both terms will be identified using the subscripts.

Julie_Reyes1B

I'm not really sure of the reason why, but I don't think it is possible to have too many/too little electrons for a reaction. If you balance it correctly, all of the electrons should go somewhere, and they should cancel when you combine the half reactions as well. I always know that I did some part ...

Julie_Reyes1B

I probably speak for most people when I say that the midterm was difficult. There were a lot of really tough problem-solving heavy questions that I struggled with even though I felt prepared. During the midterm, it was really hard for me to keep focused and calm when I saw questions I felt like I co...

Julie_Reyes1B

Certain substances can act as either oxidizing or reducing agents based on what the reaction is, but to my knowledge never at the same time. In all the reactions we will be given, it should be pretty clear which is being reduced and which is being oxidized. I know it is not possible for something to...

Julie_Reyes1B

I think there may be an error in the solutions manual, but it has not yet been added to Dr. Lavelle's "Errors" page if that is the case. I looked at the problem and researched your question and I am pretty positive that the reaction is always spontaneous when delta G is negative. I couldn'...

Julie_Reyes1B

Oxidation reaction: 6OH - +3H 2 O+Br - --->BrO 3 - +6H 2 O+6e - Reduction reaction: 3H 2 O+3O 3 +6e - --->3O 2 +6OH - This one is a little bit tricky, but I think it's because reduction is defined by gaining electrons, which you can see in the reduction half reaction even if the actual O molecules d...

Julie_Reyes1B

One thing that will give you an idea as to whether the grades will be adjusted at the end of the quarter or not is the average of the midterm, which he will state in class on Wednesday. Last quarter the averages for the midterm and the final were in the mid 70's so he did not really adjust the grade...

Julie_Reyes1B

Adding to Lily, I found that this video in particular helped me get a good grasp of all of the rules- https://www.youtube.com/watch?v=iSAwDJTLIKY. Toolbox K.1 in Fundmentals K in the textbook also has a concise list of all of the rules!

Julie_Reyes1B

Why does delta H always equal Q, but Q does not always equal delta H? Lyndon was mentioning something about this in the review session and I didn't really understand it. Thank you!

Julie_Reyes1B

A sudden expansion indicates an irreversible reaction, while a gradual expansion indicates a reversible reaction

Julie_Reyes1B

Also, for thermodynamics, the midterm will not cover the second half of the book notes and problems. That is the section involving K constant and equilibrium as they relate to thermodynamics.

Julie_Reyes1B

I like to do problems from the end of the chapter as well as doing the especially difficult/confusing ones on the Sapling modules. Beyond that, I think it really helps to base your studying around the bullet points on each chapter outline. Sometimes they include things like "know these specific...

Julie_Reyes1B

I actually figured out how to get the 14B course. I had to go to the purchase page and use store credit given by Sapling, which made the course free. Thank you for your help!

Julie_Reyes1B

I think what Dr. Lavelle was trying to show is that when a phase change occurs, it is possible no temperature change occurs. For example, in labs they add heat really slowly to melt ice to liquid water. The actual act of breaking hydrogen bonds to melt the ice requires energy in the form of heat. Th...

Julie_Reyes1B

In the textbook, they use K to indicate Kp. So the two columns are Kp and Kc. Dr. Lavelle always prefers to use subscripts, which makes it a lot more clear for me. But many resources just assume you know K means Kp

Julie_Reyes1B

So looking at NO specifically, the initial would be 1.0, and the change would be -2p. The two comes from the 2 coefficient in the reaction. So the equilibrium concentration is 1.0-2p. The whole quantity is squared in the K equation because of the 2 coefficient as well.

Julie_Reyes1B

When the entire reaction is multiplied by a value, such as 2 in this case, the K is raised to that power. So in this case, 41 is squared, which is 1681. Since 41 has two sig figs, the answer is rounded to 1700, also written as 1.7x10³.

Julie_Reyes1B

Yes, it can be asked in terms of Le Chatelier's principle. I.e. what happens when you raise the temperature on an endothermic/exothermic reaction. They might give you a reaction and then ask whether it is endothermic or exothermic, then ask questions about Le Chatelier's. For this, they would have t...

Julie_Reyes1B

It's also important to double check that the x value is in fact less than 5% of the initial concentration or the approximation is invalid. I always do that at the end of every problem in which I approximate, especially when the K value is only slightly smaller than 10^-3

Julie_Reyes1B

Hi everyone, I bought the multi-term bundle recommended by Dr. Lavelle during 14A. I used Sapling last quarter and wanted to use it again this quarter. However, the Sapling website says I need to enroll in Chem 14B which is an additional 65 dollars. Does anyone know if I have to pay this or if there...

Julie_Reyes1B

Adding on, if there is a mix of gases and aqueous solutions, you would use Kc instead of Kp so you would need concentrations rather than partial pressures.

Julie_Reyes1B

This would be endothermic because the bonds are being broken which requires energy. That means that the reaction must take in energy in order to proceed, thus it is endothermic.

Julie_Reyes1B

You can approximate by considering x insignificant any time it seems like x is extremely small compared to the initial concentration. Dr Lavelle gave an example where the K value was around 10^-4. Since the inital concentration was 0.10, it would be fair to assume the x in the equation would be real...

Julie_Reyes1B

If Q <K, the reaction will proceed towards products. If Q>K, the reaction will proceed towards reactants. This makes sense if you think about the math. Q will always change until the reaction reaches equilibrium, that is when Q=K. If Q>K, for example, then Q must become a smaller number so that Q=K....

Julie_Reyes1B

I believe if the problem requires an ICE chart, you should use a chart to show each step in order to get full credit. You should ask your TA to be sure, but I would say yes to be safe.

Julie_Reyes1B

It basically states that if a system in equilibrium is disturbed in some way, the system will shift to counteract the change and return to equilibrium. Dr. Lavelle’s main example for Friday’s lecture involved adding/removing product/reactants. For example, if you were to add more reactant, Le Chatli...

Julie_Reyes1B

It is interesting to note that in the textbook, K automatically means Kp, and if they want you to use Kc they will specifically say Kc. Dr. Lavelle will usually specify which is helpful.

Julie_Reyes1B

I feel like really interacting with Chemistry Community makes a difference. It’s easy to answer 5 questions to fill the quota each week but I’ve found it is such a helpful resource to browse and use whenever possible. For me, answering questions helps reinforce how well I understand it. Beyond that,...

Julie_Reyes1B

It’s also important to keep in mind that when you have a mix of both aqueous solutions and gases, you may have to convert partial pressure to concentration. You can do this by rearranging PV=nRT. You would then use Kc to calculate the equilibrium constant.

Julie_Reyes1B

To add on to John, my high school chemistry teacher gave us a good way to remember which way the reaction will go. If you write K and Q in alphabetical order, then the less-than and greater-than symbols will point in the direction of the reaction. For example K>Q looks like it is pointing to the rig...

Julie_Reyes1B

For the scope of this class, K A is just a measure of the relative strength of an acid. The higher the K A , the stronger the acid. The pK A is the -log K A . The lower the pK A , the stronger the acid. There is also K B and pK B (the same definitions but for a base instead. Usually, you will be gi...

Julie_Reyes1B

For the scope of this class, K A is just a measure of the relative strength of an acid. The higher the K A , the stronger the acid. The pK A is the -log K A . The lower the pK A , the stronger the acid. There is also K B and pK B (the same definitions but for a base instead. Usually, you will be gi...

Julie_Reyes1B

A compound is "oxidized" when its oxidation state increases due to a reaction with an oxidizing agent. The compound will lose electrons in order to become oxidized. This process is called a redox reaction. I don't think we will learn these terms formally until Chem 14B, however :)

Julie_Reyes1B

1. Draw the Lewis structure of the molecule. 2. Pick a "central atom" to focus on. 3. Determine the electron arrangement around that central atom. 4. # of electron regions= # of hybrid orbitals (linear is 2 sp orbitals, trigonal planar is 3 sp 2 orbitals, etc.) When drawing a diagram, reme...

Julie_Reyes1B

For the information you are given, it seems like you need an ICE chart to solve for the pH. If you were given the [H + ] concentration, it would be pretty simple to solve for the pH. But since you aren't, there is no way to solve the problem using the methods from Chem 14A. You will not be given any...

Julie_Reyes1B

To find the charge on the iron, you have to work backwards. Since the entire compound is neutral, the cation and anion charges must cancel each other out. Since each chloride has a -1 charge and there are two, the cation (in this case, the entire coordination complex), must have a charge of +2. So t...

Julie_Reyes1B

The porphyrin refers to the chelating ligand in the heme complex only, while the heme complex is the porphyrin ligand attached to the Fe (II). Porphyrin is a chelating ligand that is also tetradentate (meaning it bonds to the Fe at four different sites). The heme complex is square planar shaped. Bey...

Julie_Reyes1B

Ethylenediamine on its own is C 2 N 2 H 8 because each N is bonded to a C and two hydrogen atoms. However, in the EDTA complex, the two hydrogen atoms on each nitrogen are replaced by acetates. So therefore the hydrogen count is 4 lower than what you would expect it to be. It is a little confusing b...

Julie_Reyes1B

For the scope of this class, K A is just a measure of the relative strength of an acid. The higher the K A , the stronger the acid. The pK A is the -log K A . The lower the pK A , the stronger the acid. There is also K B and pK B (the same definitions but for a base instead. Usually, you will be giv...

Julie_Reyes1B

I don't think the 2 is completely necessary, probably because there is an assumption that the bonding electrons are from the outermost shell (in this case, 2). For example, when Dr. Lavelle did examples of drawing hybrid orbitals (C and NH 3 ), he omitted the principal quantum number. I would still ...

Julie_Reyes1B

Denticity measures how many bonding sites a specific ligand has, whereas chelating just means if the ligand forms a ring-shape around the metal ion. All chelates are polydentates, but not all polydentates are chelates.

Julie_Reyes1B

I don't think we would have to calculate K_A or K_B without being given pK_A or pK_B because that requires ICE calculations (which we will learn in 14B). However, if you are given K_A and asked to find K_B or vice versa, you can do that with the equation K_W=10^-14=K_AK_B

Julie_Reyes1B

I don't believe we need to do any calculations with the formula he gave us. Just know that because of the r 6 in the denominator, the attraction decreases exponentially as distance decreases. Thus, the formula is very distant dependent. The higher the charge and smaller the cation, the more polarizi...

Julie_Reyes1B

Because the molecule was tetrahedral, in the 3d space the polarities would not cancel each other out. The only time a tetrahedral molecule is nonpolar is when the outer atoms are all the same. The textbook has good 3d pictures to show why they would not cancel out.

Julie_Reyes1B

You can use the equation pOH+pH=14 if you are given one and asked to find the other!

Julie_Reyes1B

pOH can be found by taking the negative logarithm of the OH concentration, or using the equation pOH+pH=14. Anything with a pOH under 7 will be a base, and anything with a POH over 7 will be an acid. pH is more commonly used, however.

Julie_Reyes1B

Dr Lavelle will release the schedules for his final review sessions tomorrow in class. We will finish the syllabus by Wednesday and begin review at that time. That is probably when he will describe the format of the final.

Julie_Reyes1B

Most acids will have an H attached, or a COOH attached if it is a biological acid in the carboxyl group. Likewise, bases will generally have either an O or OH attached. If you look at a reaction, you can often figure out if a compound is an acid or base based on rather it gives a proton (acid) or ac...

Julie_Reyes1B

I think it is because the question referred to calcium as an ion rather than an atom. Calcium ion thus refers to Ca²⁺

Julie_Reyes1B

Yes, you are completely correct. Electron density regions, which affect molecular shape, only refer to the electrons directly surrounding the atom.

Julie_Reyes1B

They are both the same shape- trigonal pyramidal-. Because the lone pair forces the bonding electrons closer together, the angles between the bonding pairs are smaller than 109°. This is true for both. The only difference is the O-S-O bond angles are 106° and the H-N-H bond angles are 107°. However,...

Julie_Reyes1B

Hi, I would love to form a study group! Let me know by emailing me: juliereyes0508@gmail.com

Julie_Reyes1B

Hi! I'm not sure if bond order will be on the midterm but for this class' purpose it is a pretty simple concept. For the molecules in this class, bond order is just the number of bonds between two atoms. If there is a single bond between two carbons, the bond order is 1. If there is a double bond be...

Julie_Reyes1B

Noble gases are generally not included in the periodic table trends. Because noble gases already have a full valence shell (completed octet), it would be unfavorable for them to gain any more electrons. Because of this, noble gases tend to not react with other compounds

Julie_Reyes1B

Draw Lewis structure and use that to determine if the molecule is polar. For many of them, there will be negative charges on opposite sides that will cancel each other out.

Julie_Reyes1B

Yes, that is all correct. The photon energy would then equal the work function. You have to convert the work function into the units J/e- and then solve for whatever variable you need.

Julie_Reyes1B

CCl₄ has stronger dispersion forces than CH₄. The stronger dispersion forces are because the chlorine atoms are electron-rich compared to the hydrogen atoms.

Julie_Reyes1B

For midterm question 1, you had to convert CO₂ to C using the mol to mol ratio 1 mol CO₂=1 mol C. Do that for all 3 to find grams of C, N, and H. Then subtract all of the grams from grams of compound to get grams of oxygen. Then you can proceed with the empirical formula analysis as usual.

Julie_Reyes1B

Yes, the least electronegative atoms are typically in the center because they are more likely to share electrons than the outer, more electronegative electrons. Compounds are also usually symmetrical so that is a good hint as well.

Julie_Reyes1B

Hi, Effective nuclear charge is basically the charge an electron in an atom experiences after taking into an account the shielding effect. Non-valence electrons, known in this case as shielding electrons, repel the valence electrons and somewhat lessen the strength of the positive nucleus. Therefore...

Julie_Reyes1B

Hi! There are two main factors that can help you guess which element is in the center of a structure. a. Symmetry- Most compounds tend to be as symmetrical as possible. For a compound such as CH 4 , C would be the central atom rather than any of the four H atoms. b. Electronegativity- The element wi...

Julie_Reyes1B

Hi, Electron affinity simply tells you the amount of energy that is released when an electron is added to a neutral atom. A positive value means the atom releases energy, while a negative value means energy is required for the atom to accept the electron. It basically tells you how favorable it is f...

Julie_Reyes1B

Hi, The answer in the text book is 750 W. First, use the equation c= c=\lambda v to solve for frequency. Then, use the equation E=hv to solve for the energy of one photon. Then, multiply your answer by the number of photons indicated in the problem. Since this is the amount of energy released each s...

Julie_Reyes1B

Hi!
From Dr. Lavelle's website, it looks like VSEPR models will be discussed in our next unit, Unit 4: Molecular Shape and Structure. I found this by looking at the outlines for each unit.
Hope this helps!

Julie_Reyes1B

Hi! For this specific example, I don't think you have to calculate the bond length numerically. I think each portion of the problem includes a single, double, and triple bond. Single bonds are the longest while triple bonds are the shortest. For each section, draw Lewis diagrams of each compound and...

Julie_Reyes1B

Hi! When we look at the four quantum numbers, it can help explain why only n=4 and higher can support an f-subshell. For whatever n is, l (angular momentum number) can be any integer up to n-1. For n=4, l can be at most 3. L=3 corresponds to an f-subshell. Any principal quantum number below 4 could ...

Julie_Reyes1B

Hi!
I believe the d5 and d10 exceptions apply to any element whose electron configuration would be lower energy with a half-filled or filled d-orbital rather than a filled s-orbital. I think for the purposes of this class, we are only responsible for chromium and copper.
Hope this helps!

Julie_Reyes1B

Hi!
These elements have access to d- orbitals, allowing them to possess more electrons than the traditional octet. Usually an element will do this in order to make a formal charge closest to 0 in a bond, which makes it more stable.
Hope this helps!

Julie_Reyes1B

Hi! Simply put, when an atom is in “ground state configuration”, it just means that all electrons in the atom are in the lowest possible energy levels. A ground state oxygen atom, for example, would have two electrons in the 1s subshell, two in the 2s subshell, and 4 in the 2p subshell. The opposite...

Julie_Reyes1B

Hi! My TA made it clear that until we learn how to name compounds, the tests will always give you both the name and the formula. If I had to guess, I would say that we are learning it next unit, which includes ionic and covalent bonds as one of the topics.

Julie_Reyes1B

Hi again! Without the l value, we would not know which subshell we are talking about within the shell. So n=5 tells us we are at the fifth shell, but we don’t know if we are in the s, p, d, or f subshell unless we have the l value.

Julie_Reyes1B

Hi! I read and take textbook notes in advance before Dr. Lavelle’s lectures. It helps me reinforce the topics while he is talking about them. During textbook notes and lecture I write down possible questions or confusing topics for my TA’s office hours. Throughout the week I also look at every assig...

Julie_Reyes1B

Hi! The problem is asking you to identify a specific subshell based on the principal quantum number and angular momentum quantum number. So for (a),for example, n=5 tells us we are in the fifth shell. l=2 then indicates the subshell is "d" (l=0 is s, l=1 is p, etc). So the answer would be ...

Julie_Reyes1B

I really don't think there is a specific reason other than the fact that h-bar(ℏ), which represents h/2π, is a common and recognizable symbol used in quantum mechanics.

Julie_Reyes1B

The Rydberg equation is used specifically for hydrogen atoms. When given the initial and final quantum numbers, you can determine the velocity of the radiation emitted/absorbed. Dr. Lavelle really discouraged the use of this equation because he says students too often plug in random numbers. I would...

Julie_Reyes1B

Hi! In short, photons (light particles) are considered massless. They have energy and momentum, but no measurable mass. This was determined experimentally. Here is a really interesting article that goes into depth about some of the findings: http://www.desy.de/user/projects/Physics/Relativity/SR/lig...

Julie_Reyes1B

Hi! The reason why the uncertainty principle is only significant with smaller objects is similar to why the de Broglie equation only yields noticeable results for smaller objects. Both equations involve a particle’s wavelike properties, which really only applies to subatomic particles. Dr. Lavelle’s...

Julie_Reyes1B

Hi! To my understanding the principle just tells you the degree of uncertainties of the location and momentum of a subatomic particle. Because location and momentum are complementary in this equation, if the range of uncertainty is small for one then it must be large for the other. A problem would t...

Julie_Reyes1B

Hi! You are correct in your understanding that Lyman usually corresponds to UV light and Balmer usually corresponds to visible light. The main difference (which is what the book attempts to portray, I think, but in a confusing way) is that in the Lyman series, the electron comes to rest at n=1. In t...

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