Hello! I believe that the reduction reaction is spontaneous if the reduction potential is positive. I went for reduction potential and using E(cathode) - E(anode) > 0. I'm unsure about oxidation potential, but assuming that you can calculate it by flipping the reduction potential sign, that means th...
Hello! For Week 8 Sapling, I had this question: What would the potential of a standard hydrogen electrode (S.H.E.) be under the given conditions? [H+]=0.64 M P(H2)=4.9 atm T=298 K After a lot of attempts, I was able to finally get the answer. I knew my Q was a little off, BUT when I tried to read th...
Hello! I don't see the answer choices, but I can explain the reasoning for this problem. The reduction reaction will occur if the overall reduction potential, calculated by E(cathode) - E(anode), is positive. Since oxidation occurs at the anode, the phrase "oxidize Cr to Cr3+ , but not Fe to Fe...
The problem specifies copper (I), and I believe you used the standard reduction potential for copper (II). To save you the search, here it is: Cu+(aq) + e– → Cu(s) +0.52 Hope this helps!
Hi there! I was stuck on this too haha. Your first part is completely correct, but for the second part, the order you place the elements/compounds in the line diagram matters. The shared ion (Cl-) should occupy the middle two spots of the diagram (both of them). The very left should have the anode (...
Like a few other people on this thread, I'm planning on taking Chem 14BL next quarter (Spring) and continuing the chem series next year. I was planning on taking one chem course at a time, but depending on how much better it is to take Chem 14C and Chem 14CL together, I might just do that.
I don't really have a definite answer other than there's no other way to do the problem. I think this is just one of those situations where you have to make the correct assumptions or act on the provided information.
I felt that. I've been trying to get into a good 3-4 day long schedule to review concepts and practice questions before the midterm, but keeping up with the ongoing material is like an added challenge.
Hello! I would use lewis structures to find formal charges and chemical formulas to find oxidation numbers—formal charge divides bonding pairs equally, and oxidation numbers assign both the electrons in a bonding pair to the more electronegative atom in the pair. However, both should add up to the o...
Hello! On the Unit 4 Outline, it says: Calculate changes in entropy due to changes in volume (ΔS = nRln(V2/V1)) = constant pressure Calculate changes in entropy due to changes in temperature (ΔS = nCln(T2/T1)) = constant volume Also, there are two different molar heat capacities depending on constan...
Hello! This is just in response to what decimal place we should use. I would use the entire 273.15 because the past Sapling homework, I kept using 273 and getting the problem wrong. I will say in that case the numbers were huge (something like 10^40), so the .15 really made a difference. However, no...
Hello! The way I thought about it was the higher the change in entropy, the more disordered the molecule becomes when transitioning from liquid state to gas state. Therefore, if molecule A had a higher change in entropy than molecule B, the liquid form of A was originally more ordered than the liqui...
Hello! Looking at the formulas, I think we would have to use ΔS = nC*ln(T2/T1). We are given T1, T2, and C. We are also given the information necessary to solve for n using the ideal gas law PV=nRT. PV = nRT (8.00 kPa)(12.00 L) = n (8.314 kPa*L/K*mol)(25+273.15)K Note that the R used was 8.314 kPa*L...
For thermodynamics and equilibrium constants, there is a formula relating them: ΔG = -RTlnK. You can calculate ΔG using the link for the table on Sapling..
Thank you for sharing tips! I do the same things (especially doing the problems in the days leading up to the exams)! I tend to prioritize sapling, then textbook, then lecture. I also wanted to point out that you should always check your answers several times if you have time left in the exams! I do...
Hi there! Since deltaU = q + w, we also need to calculate the work done by the system. I found w using w = - Pex*deltaV = -(1 atm)(deltaV). To calculate delta V (Vf - Vi), I use V = nRT/P. delta V = Vf - Vi = nRTf/P - nRTi/P (n, R, and P do not change) = nR/P (Tf - Ti) = 0.981 mol * 8.3145 J/(mol*K)...
Hello! I believe we use deltaU = q + w. First, I used the formula q = nC deltaT, given that C is the constant‑pressure molar specific heat, to find q. q = 0.305 mol * 29.1 J/(mol*K) * 19.9K = 176.6 J. Then, I find w using w = - Pex*deltaV = -(1 atm)(deltaV). To calculate delta V (Vf - Vi), I use V =...
Hello! I normally compare the lewis structures of the products and the reactants, noting which bonds change. Sometimes, I miss bonds though when I try to see which bonds form and break. Otherwise, I count all the bonds for both sides and calculate the enthalpy of the reactants and products using bon...
4D.11 The oxidation of nitrogen in the hot exhaust of jet engines and automobiles occurs by the reaction N2(g)+O2(g)→2NO(g)ΔH°=+180.6kJ (c) When the oxidation of N2 to NO was completed in a bomb calorimeter, the heat absorbed was measured as 492 J. What mass of nitrogen gas was oxidized? First, we n...
Hello! Since the problems we've done so far (textbook + sapling + lecture examples) have provided the Ka/Kb values, I'm 99% sure we don't have to memorize them. The only things we'll probably have to remember is Kw (10^-14) and know the strong acids/bases because we assume that they dissociate compl...
I don't think so? As long as you make sure to use the correct units when calculating (including accounting for other constants and variables) + reporting the unit with your answer. I would get used to unit conversion just in case you're asked for a specific unit, though.
I'm think you're asking where they got the Kb value? My best guess is that the book tells you at the top of the exercises to refer to table 6C.1 and 6C.2 for the Ka and Kb values b/c we definitely do not have enough information to calculate it ourselves.
Hi there! First off, we know that the only contributor of heat is the liquid water in the cup (since we're ignoring the surroundings), and the heat will flow from the liquid water to the ice cube. When the ice cube fully melts, it will still be liquid water at 0 C, so we need to account for the rise...
Hi there! First off, we know that the only contributor of heat is the liquid water in the cup (since we're ignoring the surroundings), and the heat will flow from the liquid water to the ice cube. When the ice cube fully melts, it will still be liquid water at 0 C, so we need to account for the rise...
When the problem says the system is at constant pressure, do we assume that the pressure is the standard 1 atm? Hi there! All we know is that the system is at constant pressure, so the pressure does not change. I don't think we can assume that pressure is the standard, but some problems do say to a...
Hi there! I'll just put down my thoughts here and maybe someone can add on? I believe those two values are mathematically the same, and there are conversions to prove this. However, I don't fully understand this myself because there's a unit of time in one of those R constants (in the 8.314 J/mol*K...
Hi there! We do use an ICE table, but after we find the stand-in values for each concentration, we have to set up the Ka or Kb equation. I believe the book says you have to use Tables 6C.1 and 6C.2 to solve these problems, as these tables give you the Ka and Kb values. You'll need the values for all...
Hi there! I'll just put down my thoughts here and maybe someone can add on? I believe those two values are mathematically the same, and there are conversions to prove this. However, I don't fully understand this myself because there's a unit of time in one of those R constants (in the 8.314 J/mol*K,...
Hi there! Skimming what others have said, I definitely agree that practice is the best way to go. Before you start the review, make sure to really and I mean REALLY read each bullet point on the outlines. It kind of feels like the worst when you miss a problem because you forgot to review that one c...
Hi there! As a rule of thumb, I like to split the salt into the anion and cation and write the acid/base forms. Then, I compare the strengths of the acid/base. For example, K3PO4 splits into K and PO4, which would make the acid H3PO4 and base KOH. Since H3PO4 is a weak acid and KOH is a strong base,...
Hey there! I was part of Dr. Lavelle's Chem 14A class last quarter, and all of our exams were multiple choice. The answer choices/values were also different from each other, so there was pretty much no way that you could miss a question due to significant figures alone. I'm not sure if he will keep ...
Hey there! I'm not sure if this is always valid but my rule of thumb is to check the compound's anion and cation, writing the acid/base form of the anion and cation and comparing the strengths. If the base of the cation is stronger than the acid of the anion, the salt is acidic. If the base of the c...
Hey there! I'm not sure if this is always valid but my rule of thumb is to check the compound's anion and cation, writing the acid/base form of the anion and cation and comparing the strengths. If the base of the cation is stronger than the acid of the anion, the salt is acidic. If the base of the c...
Hi there! I actually had a lot of issues with this problem as well, but I realized that I made a lot of small mistakes. Other than that, I used your exact logic. For reference, I'm going to list out the strong acids and bases: STRONG ACIDS: HCl, HNO3, H2SO4, HBr, HI, HClO4, HClO3 LiClO4 , KCl , NaF ...
Hi there! You could use the process of elimination to reach the answer (that's actually probably what I would do). However, you can also consider that the reaction is a decomposition reaction where X2 separates into two separate X's. Therefore, the main change is that a bond was broken, which would ...
Hi there! Reaction quotient (Q) and equilibrium constant (K) have the same overall formula: [products]/[reactants]. However, the equilibrium constant specifically tells us what the ratio should be when the reaction hits equilibrium, and it only changes if temperature changes. The Q tells us the rati...
Hello! Yes, I tend to think of enthalpy as a relative measure of the energies in the products and reactants. Also, the way I understand enthalpy is that it focuses on the molecules/atoms/compounds in the reaction instead of the system. For exothermic reactions, the reaction would release heat. Altho...
Hello! Since water in an aqueous reaction is technically a pure liquid (the solvent), we can ignore it in the K calculations. Alternatively, we could write it in the K calculations, but the concentration of water would be 1 because it is the solvent. Therefore, the K value would not change. Hope tha...
Hi there! Q and K both have the same formula ([products]/[reactants]), but K specifically describes when the reaction is in equilibrium and Q can be used to describe the reaction whenever. K is also a set value (as long as temperature does not change), but Q can change. Reactions want to approach eq...
I planned to crochet a cardigan or sweater, so I look forward to doing that! I might also draw a few stickers and upload them to my RedBubble because it's been a while. I will also probably binge New Girl or the Crown idk which one yet. Long story short, I'll be catching up with my hobbies and relax...
Hi there! When you said "last minute studying," I felt that. If you already know which topic you are struggling with, the first thing I would do is ACTIVELY review notes on the topic if you have them. Otherwise, I would go reread the section in the textbook. The last resort is the lectures...
Hi there! A buffer is a solution that can resist pH changes. They're super important for biological systems because these systems need to maintain homeostasis and avoid dramatic changes. Therefore, a lot of biological systems will have buffers. One example of these systems is the bicarbonate buffer ...
NH3 is a weak base since it can donate electrons or accept protons from the nitrogen lone pairs. However, NaCl is not an acid or a base; it's a salt. I think you'd have to put it in the other category.
Hi! I try to do practice problems from the textbook or online to practice the formulas. Occasionally, I skim the textbook, but the practice is overall a lot more important to me. I really only consult the textbook if I fail to understand something. If I want to practice concepts, I go to Chemistry C...
Hi there! In terms of what they mean, the roots match up with the other roots we're familiar with. di- = bis- tri- = tris- tetra- = tetrakis- However, we only use the bis- tris- tetrakis- when the ligand already has a root inside (like ethyleneDIamine) or the compound is polydentate, meaning it is a...
Hi there! You could calculate the oxidation number either way. The important thing to note is that the overall compound should be neutral (so the total oxidation number for the elements in the bracket AND outside the bracket should be zero aka all the elements in the entire compound), but the coordi...
Hi! For naming the compound, you arrange the ligand in alphabetical order based on the name, not the prefix. For example, tetraammine would come before dichloro because the A in ammine comes before the C in chloro. If they give you the name and you have to write out the formula, make sure to write i...
Hi! I guess my first thought would be what fields am I interested in + am I interested in the point of majoring/double majoring or minoring (though I think you could also just minor first and then switch to majoring if the requirements for major/minor aren't that off). Personally, I love biology and...
I would normally just memorize the list, but it might help to relate the weak/strong acid/base to chemistry concepts. For example, to find if acids is comparatively weaker or stronger, you can find the elements on the periodic table and compare them using trends. I think in his lecture today Dr. Lav...
I wouldn't say this is a joke, but I remember thinking it was kinda funny so I just wanted to share a chem experience! When I was first learning nomenclature in my chemistry class, the teacher joked that our local lake was full of dihydrogen monoxide. My whole class started saying things like "...
Yes! For significant figures, they follow these rules: adding/subtracting: round to the nearest place after the decimal point (least decimal places) (ex. 0.1 + 0.10 + 0.100 = 0.3, you should round to the tenths because of the three added numbers, 0.1 has the closest spot to the decimal point after t...
Hi there! Like the people above me have stated, I think it's something we have to just know and memorize. However, if you're asking about how we knew the bonds could or could not rotate, I can try to explain conceptually. If you consider that the sigma bond is kind of like two balls (molecules) atta...
Hi there! I'm not completely sure how to explain it, so I just wrote out my thought process. First, the hybrid orbitals result from combining the s and p (maybe d) orbitals in a way that the hybrid orbitals are considered equal to one another. Example: combining one s and two p makes three sp2 orbit...
If we consider electron domain geometry (where we include the lone pair electron clouds), the geometry would be trigonal bipyramidal. With this in mind, we remove the electron clouds because the question asks for the molecular geometry. The three electron clouds would be situated in the equator part...
First off definitions: Lewis acid: accepts electron pairs Lewis base: donates electron pairs The easiest way to tell if an atom is a lewis acid or base is to look at the charge (if it has one). If it is an anion (negative, like S2-), it can probably donate electrons (lewis base). If it is a cation (...
For a molecule to have H bonding, the H atom must be bonded to an N, O, or F atom. Also, this polar bond would only attract other atoms with lone pairs (or other hydrogens), but these atoms have to be bonded to N, O, F as well. When you draw out the lewis structures for butanol and diethyl ether, yo...
Hi! It's true that the 4d subshell has 5 orbitals (-2, -1, 0, +1, +2). However, this information only fits with n = 4 and l = 2. The question adds on that the magnetic quantum number must be -2. This refers to only one of the 5 orbitals (the one marked -2), so the answer is one orbital (only one orb...
Hi! Formal charge refers to the charge that an atom would have if the electrons in their shared bonds were split evenly. Therefore, electronegativity and polarity have no bearing on the formal charge. In comparison, oxidation number takes this electronegativity and polarity in mind when dividing th...
Hi! Formal charge refers to the charge that an atom would have if the electrons in their shared bonds were split evenly. Therefore, electronegativity and polarity have no bearing on the formal charge. In comparison, oxidation number takes this electronegativity and polarity in mind when dividing the...
Hi there! The best way I can explain this is to look at the Periodic Table. The row Manganese is on includes 4s, 3d, and 4p, so we only look at those subshells (not the 3s, 3d, 3p ones, even if they share the same principal number). However, the subshells for 4s and 4p are empty for Mn4+. Therefore...
Hi there! The best way I can explain this is to look at the Periodic Table. The row Manganese is on includes 4s, 3d, and 4p, so we only look at those subshells (not the 3s, 3d, 3p ones, even if they share the same principal number). However, the subshells for 4s and 4p are empty for Mn4+. Therefore,...
Hi! We have to use the [Ar] noble gas core because Zinc is losing electrons, pushing it backwards in terms of electron configuration. In this case, Zinc lost 4 electrons to become Zn4+. This normally "pushes" Zinc back to Iron's location on the Periodic Table in terms of number of electron...
Hello! The lone pairs refer to the unbonded electrons attached to the central atom. The bonded pairs refer to the electrons within the chemical bonds between the central atom and the other atoms. Since electrons are negative, they repel each other (like-like charge). The textbook states that the lon...
I believe that the amount of electrons plays a role. Fluorine has a lot of valence electrons, and the repulsions between them weaken the bond. At least those are my initial thoughts (?)
Hello! The second ionization energy will always be higher than the first b/c it's harder to remove an electron from a positively charged ion vs. a neutral ion. When you take your second electron, you consider that you have already removed one electron from the neutral atom, turning it into a cation....
Hello! Before we start writing electron configurations, we need to remember this sequence of orbitals: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, and so on and so forth. Then, you use the Periodic Table to locate the atom you want to write an electron configuration for. I think the easiest way f...
Hi! So hydrogen bonds have to include a N, O, F molecule b/c these atoms are extremely electronegative, meaning that in a bond with hydrogen, they will hog the electrons. This is why the bond becomes polar, and partial charges form on the atoms. These partial charges are the ones that cause H bonds ...
I think I figured it out (?) I think the central atom in the square planar shape has two lone pairs in addition to the four atoms, which could push the atoms to the four square-like corners. If anyone could confirm that'd be great
Hello! I was doing the 2.E. exercises in the book when I realized that I don't really know the difference between tetrahedral and square planar concerning lewis structures. When I drew the lewis structures of CH2F2, for example, I had no idea why it was tetrahedral vs. square planar. Can anyone tell...
Hello! I believe an ion-dipole interaction would still occur BUT it would be between the sodium ion and the individual chlorine atoms. Even though the molecule is symmetrical, this just means that the bond dipoles cancel each other out --> the net dipole is zero --> the molecule is nonpolar. All we ...
Hello! I think the rule that you used for determining the central atom in a lewis structure depends on the ionization energy (the central atom has the lowest IE). Technically, nitrogen does have a higher IE than chlorine, so why is it the central atom? Because of weird lewis structures and exception...
Hello! Not gonna lie, this question was pretty tricky so I used google to help me. I checked your lewis structure first, and it technically fits all the rules for lewis structures, so nice job :) The problem is carbamate is CH2NO2-. This formula automatically hints that the nitrogen should be bonded...
Hello! A lewis acid will ACCEPT electrons, and a lewis base will DONATE these electrons. For your question, it's important to note that once a lewis base donates the electrons, they technically could accept these electrons back (like a lewis acid). Therefore, the molecule that was previously a lewis...
Hello! First off, I'm not sure if I misinterpreted your question, but here are my thoughts! I don't believe partial dipole moments exist, so I wasn't sure if you meant a mix of partial charge and dipole moments OR temporary dipole moments. If you were asking about the partial charges + dipole moment...
Hello! In terms of the level of electron distortion, an element with high polarizability (alpha) means that its electron cloud will have a large tendency to distort. So if polarizability increases, the level of electron distortion increases. Now, we relate polarizability to the formula: potential en...
All of the above is correct! I just wanted to clarify that a coordinate covalent bond is just a bond that forms when one atom (of the two in the bond) contributes both of the electrons necessary to form that bond. Hope y'all have a nice day!
Hi everyone! I think Dr. Lavelle mentioned ionic radii and covalent radii during his lecture on periodic trends, but I'm not too sure about van der waal radii. All three should be a measurement of half the distance between two bonded atoms, but depending on the nature of the bond, we label the dista...
First off, I've watched a LOT of the recommendations above + I would say all of them are really good! Second, I just wanted to add a recommendation of my own, especially if you like crime/forensics. It's called Bones, and I think it's on Hulu (?)
Hi there! I'm not absolutely sure if this is the answer you're looking for, but here are my thoughts! The electrons that would be removed from an atom to form a cation are the valence electrons. The valence electrons are in the orbital furthest from the nucleus. To find the answer to your question, ...
Hey there! I believe the central atom has the lowest ionization energy, which means it's more willing to give up its electrons vs. the other atoms in the ion/molecule. Therefore, the central atom is more willing to form bonds, as bonds are created due to interactions between two atoms' electrons. Th...
Hi guys! I just wanted to hop into this discussion and ask for some clarification. I understand the differences between atomic spectroscopy and molecular spectroscopy, but how does it relate to the electronic transitions? Thanks!
Hi there! I thought you were asking if the +1/2 and -1/2 corresponded to a specific arrow (up/down). Generally, I believe the up arrow, which corresponds to spin up, represents a spin number of +1/2. Therefore, the down arrow, which corresponds to spin down, represents a spin number of -1/2. Hope th...
Hello! First off, the principal quantum number (n) indicates the shell, the angular quantum number (l) represents the subshell within the shell, and the magnetic quantum number (ml) represents the orbital within the subshell within the shell. With that in mind, let's look at the problems. A. n = 2 T...
Hello! I also just wanted to add on a little bit to everyone's answers. The R = 3.28984 x10^15 Hz is used for the Rydberg equation with frequency, or v = R(1/n1^2 - 1/n2^2). In comparison, the R = 1.097x10^7 m is used for the Rydberg equation with wavelength, or 1/wavelength = R(1/n1^2 - 1/n2^2). As...
Hello! Concerning the formula with 1/wavelength vs. the formula with frequency, I believe they're the same. In fact, they can be connected through the c = v * wavelength formula. However, I do believe that the Rydberg constant (R) changes depending on the formula. For the 1/wavelength one, R = 1.097...
Hi there! In the Balmer series, n1 = 2, and in the Lyman series, n1 = 1. For the most part, I find this helpful in solving problems, especially when they want us to find n1 and n2. If the problem states that the light wavelength, frequency, or radiation type, you can immediately find n1 using the Ba...
I am not sure if my answer is correct but I am leaning toward "A". Because frequency("v") is related to the speed("c") in the equation c=h*v a decrease in the frequency would result in a decrease in the speed of the radiation. But I am not 100% so if someone could doub...
A. is false because the speed of light (electromagnetic radiation) is constant, c=3x10^8 m/s. B. is false because frequency and wavelength are inversely proportional as c=wavelength x frequency. So if frequency decreases, then wavelength would have to increase. C. is true because since the electric...
Hi there! For part two, you are trying to see how many photons you can remove with the energy given. To find the number of emitted photons, you need to divide the amount of energy given (7.40×10^−7 J) by the work function (2.726 - 10^-19 J/photon), the amount of energy needed to remove one photon. U...
Hi there! NOTE: After I typed this up, I realized that I technically solved for n2. However, since the light is part of the visible light spectrum, it is part of the Balmer series and n1 should always equal 2. I'm unsure if this is just a typo, but otherwise, I got n2 = 4. We would start this proble...
Hello! I noticed you only asked about part c, so I only included an explanation for part c. I hope you don't mind! In part b, the question states that the frequency of the radiation is 2.50*10^16 Hz. You can use the formula c=v*wavelength to determine the wavelength. 3.00*10^8m/s = 2.50*10^16s-1 * w...
Hi there! I ran into the same issue when I did this exact homework problem! I got the correct wavelength for the n2 = 1, but I could not get the right wavelength for n2 = 4. When I looked back, I just realized I made a small error when I pressed the wrong buttons on my calculator, putting (1/1^2 - 1...
Hello! For me, photon and quantum are both extremely small units, but I always remember small differences between the two. When I think about photon, I think of the individual particle in light that carries a quantum of energy. When I use the term "quantum," I use it to describe a small, d...
Hi there! I'm not sure if I'm misinterpreting the question, so sorry in advance if I did. To determine the correct final answer in scientific notation, be careful with your dimensional analysis. All of the formulas use the SI units, so you should use the kilograms since the kilogram (not grams!) is ...
Hi there! First off, I think you're asking about Balmer and Lyman series? I'm sorry in advance if I've misinterpreted the question. So the Balmer series is a set of lines within the VISIBLE LIGHT region of the electromagnetic spectrum. In comparison, the Lyman series is from the UV region. If you're...
Hello! So I don't take notes on an iPad, but I have also wondered whether or not to convert to taking notes on the iPad. Here's what I think: PROS FOR iPad -It's a perfect mix between paper notes and typing. I get to physically write out the words and phrases (which personally helps me a lot), and I...
Hi there! So first off, you want to find the molar mass of the metal M. To do this, you use the given molar mass of M(OH)2 (74.10g/mol) and subtract the molar masses of the substances other than M (the hydroxides OH). Note that there are two hydroxides, and the molar mass of one hydroxide is 17.02 g...