CHEMISTRY COMMUNITY

Megan Lu 3D

Hi! For these kinds of problems, we can do ∑H˚m(bonds broken) - ∑H˚m(bonds formed). Like Lea said, I usually just break up all the bonds and add them together, so that I’m essentially just doing [reactant bonds] - [product bonds]. Hope this helps!

Megan Lu 3D

Hi! I agree with what Brennan said. We want our galvanic cells to have some solid electrode on both sides, so in the absence of a solid electrode, we often add in Pt(s) or some other conducting solid. An exception to this is liquid mercury, which can serve as an electrode. Hope this helps!

Megan Lu 3D

Hi! Like others have stated above, I believe that we will be told which elementary steps are slow or fast. As far as comparing the speed of reactions with each other goes, we can examine the rate laws. Hope this helps!

Megan Lu 3D

Hi! I agree with what Emily wrote above. For this problem, I considered the rate laws for the two different mechanisms. (i) would likely be rate = k[C12H22O11], as the slow rate of step 1 limits the overall rate. (ii) would likely be rate = k[C12H22O11][H2O]. Typically, we would expect H2O to be exc...

Megan Lu 3D

Hi! Like others mentioned above, for part A, I used the Arrhenius equation k = Ae^(-Ea/RT). Because we are comparing the two different pathways (one without the catalyst and one with the catalyst), we can compare k1 and k2, where k1 is the reaction rate for the catalyst pathway and k2 is the reactio...

Megan Lu 3D

Hi! I’m still a little confused by cell notation myself, but I believe that this specific order tends to take place because we write the species in the order that electrons move between them / the order they are in contact. For example, Sapling gives this solution for one of the homework problems: P...

Megan Lu 3D

Hi! I agree with Jayasree. Because HCl is a strong acid, it dissociates into H+ and Cl-. So this would only occur for compounds that completely dissociate, not all aqueous solutions. Hope this helps!

Megan Lu 3D

Hi! I used Appendix 2B in the textbook for these problems; these are the same ones used in 6L. Hope this helps!

Megan Lu 3D

Hi! I’m also not sure if this is the fastest way, but I also started by identifying Mn and Fe as the species most likely involved in the redox reaction. I then looked up the cell potentials in Appendix 2B, balanced the half-reactions. +1.23 V corresponds to Mn and +0.77 V to Fe, so subtracting E˚(ca...

Megan Lu 3D

For 6L.5B in the textbook, the solution manual gives shorthand notation: Pt(s)|I-(aq)|I2(s)||Ce4+(aq),Ce3+(aq)|Pt(s). I thought that we were supposed to write aqueous species closest to the double lines (salt bridge). Is that correct? And if so, why is solid I2 closer to the double lines than aqueou...

Megan Lu 3D

Hi Ryan! Like others mentioned above, we can look at the polyatomic ion ClO3-, which has a charge of -1. If O is -2, Cl must be +5 to allow the ion to reach a charge of -1. Hope this helps!

Megan Lu 3D

Hi! Like Ryan explained above, we can use the Nernst equation E = E˚- (RT/nF)*ln(Q) for this problem. Finding E˚: We can find the E˚ of the cell by doing E˚(cathode) - E˚(anode). Based on the fact that Zn is being oxidized, we know it is the anode and Sn is the cathode. Finding Q: We have the molar ...

Megan Lu 3D

Hi! For this problem, I referred to Toolbox 6K.1 and just went through the steps to balance the equations. Here are the steps, in short. 1- Identify what is being oxidized and what is being reduced. 2- Write the two unbalanced oxidation and reduction half-reactions. 3- Balance all the non-hydrogen o...

Megan Lu 3D

Hi! For this problem, you are correct to use ∆G=-nFE˚. You can use the standard reduction potentials you found in E˚(cathode) - E˚(anode) to get E˚(cell). Then, consider how many electrons are being transferred to get n. After plugging in these values to ∆G˚=-nFE˚, you should get your answer in joul...

Megan Lu 3D

Hi! As mentioned above, potassium is not included in either of the half reactions because it is solely responsible for keeping the solution neutral; potassium would not react in the redox reaction. Hope this helps!

Megan Lu 3D

Hi! Like others have mentioned above, I believe that the anode corresponds to the side losing electrons (oxidation) and the cathode corresponds to the side gaining electrons (reduction). So we can consider which direction the electrons are moving to figure it out. Hope this helps!

Megan Lu 3D

Hi! Like others have mentioned above, I believe that we are supposed to consider the sign of the standard electrode potential for the reaction. A positive value would be spontaneous; a negative value would not be spontaneous. You can also think of this in terms of ∆G˚, as a positive E˚ would yield a...

Megan Lu 3D

Hi! Like others have mentioned above, I think that we can just assume that the anode is on the left and the cathode on the right for cell diagrams in this class, although this is likely just an arbitrary choice to set up the model.

Megan Lu 3D

Hi! Like others have mentioned above, a salt bridge helps maintain the neutrality of each system. The salt bridge essentially transfers electrons to counteract the buildup of positive and negative charges, which would eventually lead to a charge imbalance that would “halt” the reaction. Hope this he...

Megan Lu 3D

Hi! Like others have explained above, a salt bridge serves to keep the systems neutral by transferring electrons to prevent positive and negative charges from building up and “halting” the electrode reaction as a result of an imbalance in charge. Hope this helps!

Megan Lu 3D

Hi! Like others have mentioned above, we can calculate the mass difference to determine the amount of substance that has vaporized. After dtermining the mols of vaporized substance, we can divide the energy supplied by the heater by this value, giving us our ∆Hvap. Hope this helps!

Megan Lu 3D

Hi! Like others have mentioned above, N2O has residual entropy because it is not symmetrical, and thus can be rearranged into other unique positions. Based on formal charge, we know that N2O tends to be arranged as N≡N-O, but we could also write O-N≡N. Because there are possible variations in N2O’s ...

Megan Lu 3D

Hi! Like others have mentioned above, we can start by looking at ∆G. Because this reaction is spontaneous, ∆G must be negative. Because the water molecules are moving from a more mobile form (gaseous) to being bound to the solid silica gel, ∆S must be negative. We can then apply this to the Gibbs eq...

Megan Lu 3D

Hi! Like Peter mentioned above, I started by determining the amount of energy was made available by the heater in the given 6.31 minutes. We know that the heater supplies 475.0 W, or 475.0 J/s. From there, we can do (6.31 min)(60 s/min)(475.0J/s)(1 kJ/1000 J)--putting this value in units of kJ is he...

Megan Lu 3D

Hi! Like Sophia explained above, determining whether compounds have residual entropies involves considering how many possible unique positions there are in their crystalline forms. If there is only one possible position, then we would not expect to find residual entropy. For example, CO2 is symmetri...

Megan Lu 3D

Hi! Like others have explained above, we need to find: 1. q(ice), 2. q(water), 3. deltaH. We can find q(ice) and q(water) by using m*C*(Tf-Ti). After plugging in values, we should have q(ice) and q(water) in terms of Tf. To find deltaH, we can take the number of grams of the ice cube, convert it to ...

Megan Lu 3D

Hi! Like others have explained above, we can set q(ice) + deltaHfusion = - qwater, and then solve for Tf. This honestly took me a couple tries to not mess up the numbers in my calculator haha but it should work out!

Megan Lu 3D

Hi! Yes, we would use the same constant C found in part A, since this describes the heat capacity of the calorimeter (calorimeter constant), and we are using the same calorimeter for part B. Hope this helps!

Megan Lu 3D

Hi! Based on your work, I think you lost your Tf term on the ice cube side of the question. When you did
(46.8/18.02)(6.01x10^3) + (46.8 x 4.18)(Tf - 0) = 1.58x10^4,
the Tf term seems to disappear. Instead, it should be 1.58x10^-4 * Tf. Hope this helps!

Megan Lu 3D

Hi! Like Allyssa and Aydin stated already, I also believe that we will be given the values for specific and molar heat capacities in the problem! It might be helpful to be familiar with the specific heat of liquid water, though, since it does come up quite frequently.

Megan Lu 3D

Hi! For the second part of this question, we can take the 169 g of carbon and convert to mols using the molar mass. From there, we can do the stoichiometry to determine the amount of energy released, given that there are 358.8 kJ per 4 mols of carbon (based on the equation). For the third part, we c...

Megan Lu 3D

Hi! Like Rita stated above, a combustion reaction simply describes a reactant reacting with oxygen gas (O2). Typically, this can be simplified to [reactant] + O2 —> CO2 + H2O. However, there are some exceptions to this general rule of thumb; for example, the combustion of hydrogen gas (H2) produces ...

Megan Lu 3D

Hi! For this part of the problem, consider that the delta H can be considered “stoichiometrically”; that is, it is proportional to the molar coefficient laid out in the reaction. As such, we can say that there is +358.8 kJ / 1 mol S8(s). From there, the problem tells us that 1.31 mol S8 is reacting,...

Megan Lu 3D

Hi! For this problem, we can consider the equation deltaH rxn = [sum of enthalpy of formation (products)] - [sum of enthalpy of formation (reactants)]. If we were to use enthalpy of formation, deltaH rxn would equal [enthalpy of formation of CH4(g)] - [enthalpy of formation of C(g) + 4 * enthalpy of...

Megan Lu 3D

Hi! As others have specified above, the given values are standard enthalpies of combustion. Thus, we can write out the combustion reactions, where each molecule/compound essentially reacts with oxygen gas to form H2O and/or CO2, using the enthalpies of combustion as reaction enthalpies per mol of th...

Megan Lu 3D

Hi! Like Joyce said above, I would also use the dilution equation M1V1 = M2V2 to find the actual concentration, getting (0.025)(200.0) = [HCl](250.0), where [HCl] = 0.020 mol/L. Because there is a 1:1 ratio between HCl and H3O+, [HCl] = [H3O+] = 0.020 mol/L. Then, we can use pH = -log[H3O+]. Hope th...

Megan Lu 3D

HI! For part C, I simply considered how “likely” Cl2 and F2 were to break into their monatomic forms relative to each other. Because the molar concentration of Cl in part a was 1.1 x 10^-5 mol/L and the molar concentration of F in part b was 3.2 x 10^-4 mol/L, [Cl] < [F], despite the two both starti...

Megan Lu 3D

Hi! For these problems, I usually start off by writing out the equations, and then figure out where to go from there. It’s important to note that all the solutions in this question are strong acids/bases, so they will go to completion. The main objective is to determine [H3O+] or [OH-], and from the...

Megan Lu 3D

Hi! For this problem, it’s helpful to know that because Ba(OH)2 is an alkaline earth hydroxide, it is a strong base. As such, we would expect the reaction Ba(OH)2(aq) —> Ba2+(aq) + 2OH-(aq) to go to completion. To start, we can convert the 0.43g Ba(OH)2 to its molar concentration (use molar mass and...

Megan Lu 3D

Hi! I also had some trouble finding the Ka value in the textbook at first, but it is available in Table 6D.1 (page 477). I approached this problem by writing out the equation Al(H2O)3+(aq) + H2O(l) <—> H3O+(aq) + Al(H2O)2(OH-)2+(aq) , where Al3+ is a small, highly-charged metal cation that can act a...

Megan Lu 3D

Hi! Like others have stated above, we want to start with an ICE table.  (I) The initial pressures of PCl3 and Cl2 will be 0, as the container only consists of PCl5. The initial pressure of PCl5 is given as 0.0750 bar.  (C) Because the system will favor the reverse reaction to reach equilibrium, PCl5...

Megan Lu 3D

Hi! I agree that this concept is likely an important one to remember, as it serves as the foundation for understanding pH, pOH, acids, bases, etc., as well as the relationships between all these concepts. I would definitely try to remember it for both problem solving and conceptual questions. Hope t...

Megan Lu 3D

Hi! For exothermic reactions, heat is released for the forward reaction; thus, the addition of heat would favor the reverse reaction instead to produce more reactants, and the K value would decrease. For endothermic reactions, heat is absorbed for the forward reaction; thus, the addition of heat wou...

Megan Lu 3D

Hi! For the final part of this question, we can set up an ICE box.  (I) The initial concentration of N2O4 is unchanged, so we can leave it at 0.373 mol/L; the initial concentration of NO2 needs to account for the additional 1.00 mol added, so that would be 2.04 + 1.00 mol/L.  (C) Because the product...

Megan Lu 3D

Hi! Like others have pointed out above, Kw changes with temperature. Because the reaction is endothermic, the forward reaction will absorb heat; an increase in temperature will thus favor the formation of products more, and the equilibrium constant will increase.

Megan Lu 3D

Hi! Like others have explained above, I believe that the two terms can be used interchangeably; however, the mixture is usually used to describe the whole reaction/system, whereas the concentration is slightly more specific in describing a particular product or reactant.

Megan Lu 3D

Hi! Like others have said above, the addition of an inert gas to a closed container would likely result in an increase in pressure, but the volume would remain unchanged.

Megan Lu 3D

Hi! Like others have explained above, K does not change in response to changes in pressure because the reaction will simply shift to minimize any change in concentration (as specified in Le Chatelier’s principle). However, in the case of temperature, the reaction will change in the way it favors pro...

Megan Lu 3D

Hi! Like others have detailed above, stability can point to whether the reaction will favor forming a reactant or a product. When the equilibrium constant is large, there is a greater proportion of products, suggesting that the products are more stable and the reaction favors the forward direction. ...

Megan Lu 3D

Hi! Like others have mentioned above, when the equilibrium constant is 1, the concentration/pressure of the products and reactants will be equal. To my understanding, this just happens to be rare because there is usually a tendency for reactions to favor one direction over another.

Megan Lu 3D

Hi! Like others have said above, an oxyacid is an acid that contains oxygen, hydrogen, and another element. At least one acidic hydrogen is bonded to an oxygen atom. The more oxygen atoms (or the higher the oxidation number), the stronger the acid.

Megan Lu 3D

Hi! Like others have said above, coordination compounds can definitely have multiple types of ligands, which we list alphabetically in naming.

Megan Lu 3D

A weak acid will still donate protons. It does not dissociate completely, so there will be some H3PO4 left at equilibrium, but there will also be some of the salt present. if weak acids/bases didn't dissociate at all, they wouldn't be able to participate in an A/B reaction. Hi Zach, I also had this...

Megan Lu 3D

Hi! To my understanding, SO3 is not a Bronsted acid or base, as it cannot donate or accept a proton. You are right that it is a Lewis acid in water because the sulfur accepts an electron pair, but this electron pair should come from the oxygen atom, not H+ (depicted in the attached image). Perhaps y...

Megan Lu 3D

For Fundamentals #9b, I was wondering why the weak acid reactant H3PO4 in the neutralization reaction is written in its dissociated form [3H+(aq) + PO4 3-(aq)], as I thought that only strong acids were supposed to be written out as such. For example, the bromous acid (also a weak acid) in part c is ...

Megan Lu 3D

Hi! Like others have explained above, (en) is the ethylenediamine ligand, or NH2CH2CH2NH2 (you can find this in Table 9C.1), which is a bidentate ligand. To calculate the coordination number, we would consider the number of bonds that the metal (Co) can form with the ligands (en and CO). There are t...

Megan Lu 3D

Hi! When I did this problem, I drew out the Lewis structures for each ligand, and then determined the number of lone pairs on different atoms. For example, the Lewis structure for (a) HN(CH2CH2NH2)2 reveals three nitrogen atoms with one lone pair each. Because there are three separate lone pairs tha...

Megan Lu 3D

Hi! Like others have said above, the potassium should be located at the beginning, in front of the brackets, because it is a cation (more specifically, a positively-charged metal) with a 1+ charge. Potassium hexacyanidochromate (III) is a compound, so we name the cation first and then the anion.

Megan Lu 3D

Hi! To my understanding, a monodentate ligand describes a ligand that binds at one site of a central metal atom, donating one electron pair. A bidentate would be a ligand that binds at two sites, donating two electron pairs, and so on. To determine what kind of dentate a ligand is, we can consider t...

Megan Lu 3D

Hi! Like others mentioned, Toolbox 9C.1 in the textbook and Table 9C.1 are pretty helpful for this question. But this is how I approached the problem: We would start with the part inside the brackets, [Co(NH3)5Cl]. There are three separate components: Co, NH3, and Cl. We know that Co is the transiti...

Megan Lu 3D

Hi! Like others have explained above, quickly determining the hybridization primarily consists of counting the number of regions of electron density. Even in a complex structure like the one above, this holds true; we can see that each phosphorous has one lone pair and three bonds. This adds up to p...

Megan Lu 3D

Hi! For all atoms to exist on the same plane, we would be essentially be able to connect them all on a 2-D plane (ex: they all could lie on the same sheet of paper). To answer the Sapling problem, we would need to consider whether there are an even or odd number of atoms. When there is an even numbe...

Megan Lu 3D

Hi! Like others have explained above, I started by determining the formula of the molecule. I multiplied the mass percentages with the molar mass to determine the number of moles of each element, yielding a formula of CH4O. From here, I drew the structure, keeping in mind that carbon is the most ele...

Megan Lu 3D

Hi! Like Olivia mentioned above, I would also examine the charge of the molecule or ion. In this question, we want the formal charges of the atom to have a sum of -3, as the ion has a charge of -3. Thus, the structure that would give an overall formal charge of -3 would be if arsenic had one double ...

Megan Lu 3D

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you! Hi! Like others have detailed above...

Megan Lu 3D

Hi! Like others have said above, I believe the molecule you are missing is XeF2, which has Xe in the middle and F on either side of Xe, and the lone pairs are distributed fairly evenly.

Megan Lu 3D

Hi, going off of this, I understand how to determine the polarity of a molecule but I am a little confused how to determine if an individual bond is polar or non-polar. Is it simply a large difference in electronegativity between the two atoms? Hi! I agree with the response above. To my understandi...

Megan Lu 3D

Hi! Personally, I don’t have any specific mnemonics to remember the names of the different shapes (other than looking at the prefixes), but I usually memorize them simply by drawing them all out on a blank sheet of paper without my notes. Hope this helps!

Megan Lu 3D

Hi! Like others have said above, I believe you are correct. Sigma bonds and pi bonds are found in covalent bonds, with single bonds having only a sigma bond. For double bonds and beyond, both sigma and pi bonds are present.

Megan Lu 3D

Hi! Like others have mentioned, I would start by drawing the Lewis structure, and then look at the bonding/lone electron pairs, which should correspond to a specific molecular shape, specified by the VSEPR model/formulas. Hope this helps!

Megan Lu 3D

So to clarify, in conclusion, a hydrogen bond only exist between H and an N, O, or F molecule? H bonds are polar correct? Because the H is attracted to a more electronegative atom? Hi! I believe that hydrogen bonds exist between a hydrogen atom (covalently bonded specifically to F/O/N within the mo...

Megan Lu 3D

Hi! In the case of hydrogen bonding, electrons are not considered to be shared, as hydrogen “bonding” refers to an intermolecular force. Rather, there is a force of attraction between the partial positive hydrogen and the partial negative charge of another highly electronegative atom in a separate m...

Megan Lu 3D

Hi Linette! I’m not too sure, but I believe that we can use oxidation states to evaluate which Lewis structures are consistent with general oxidation state rules (oxygen usually has an oxidation state of -2 in compounds, monatomic ions have oxidation states equal to their ionic charges, etc.). If th...

Megan Lu 3D

Hi! I believe the question is simply asking how closely the expected values match the observed ones. If the bond has “ample” character of the expected, then it would show bond lengths only somewhat similar to the specified observed; if the bond has “overwhelming” character of the expected, then it w...

Megan Lu 3D

Hi! To my understanding, it is the combination of electronegativity and size that allow N, O, and F specifically to form hydrogen bonds. F, O, and N are some of the elements with the greatest electronegativity, and they are also unique in that they have relatively small sizes compared to other also-...

Megan Lu 3D

Hi! Like others have mentioned above, I personally try to understand the concepts to apply when looking at the periodic table, but for memorization, I just stick to general directions on the period table. For example, as you move upwards and to the right on the periodic table, the atomic radius beco...

Megan Lu 3D

Hi! I believe that electrons in 3d would have a greater effective nuclear charge than electrons in the 4s orbital, because as electrons exist in energy levels farther away from the nucleus, the greater distance reduces the attraction. So in the case of scandium, the electron in 3d should experience ...

Megan Lu 3D

Hi! I believe you are right. When ms is not specified, there will be two possible electrons with the same ml quantum number, as ms refers to the electron spin (either up or down).

Megan Lu 3D

Hi! Like other people have said, I believe that this means that the electrons fill up the 4s orbital before the 3d orbital because 4s has lower energy than 3d. Hope this helps!

Megan Lu 3D

Hi! I believe that of all the information we are responsible for regarding orbitals, we especially need to be familiar with information like the number of orbitals in s/p/d/f, the location of the spdf blocks on the periodic table, etc.  As for 2s, the 2 refers to the 2nd energy level, and the s refe...

Megan Lu 3D

Hi! Yes, it depends on whether the electron is emitting or absorbing energy; if it is emitting energy, the electron will drop energy levels, and if it is absorbing energy, the opposite will be true. To identify the initial energy state, consider whether we know if we are referring to the Lyman or Ba...

Megan Lu 3D

Hi! I think that the bottom two in the right column would not be compatible—in order for the wave patterns to be compatible, they should be continuous and consistent in amplitude and wavelength. I try to imagine copy-pasting the graphs side by side to see if it would create a function resembling a s...

Megan Lu 3D

Hi! I believe that for the wave patterns to be compatible, they must be continuous and consistent in amplitude/wavelength. Personally, I imagined copy-pasting the graph side by side; if this would create a continuous and consistent function resembling a sin function, it would be compatible.

Megan Lu 3D

Hi! I believe that the ionic radius simply refers to the distance between the nucleus and outermost electrons. For ionic compounds made up of two or more different kinds of ions, this definition should still hold true.

Megan Lu 3D

Hi! Personally, I just memorize the trend directionally; as you move to the right and upwards on the periodic table, the atomic radius decreases. The opposite is true for electron affinity and ionization energy.

Megan Lu 3D

Hi! I think the question is just asking for the energy work function, given that the frequency is 2.50 x 10^16 Hz. To solve for the energy, you would probably just use E = hv.

Megan Lu 3D

Hi! I don’t believe that amplitude is related to frequency or wavelength. To my understanding, increasing amplitude simply means increasing intensity.

Megan Lu 3D

Hi! I’m not entirely sure because I used a different approach than the ones described above, but I got the same answer as Xinyu^. These were my steps: 1. Convert the work function from eV to joules (1.7624x10^-19 J). 2. Use the wavelength of the incident light to solve for the frequency of the incom...

Megan Lu 3D

^ Hi! I believe the correct answer should be C because frequency in EM radiation refers to the number of cycles that occur within a given amount of time. If the frequency decreases, there are fewer cycles taking place in this interval of time and the waves broaden. Thus, the extent of change that ha...

Megan Lu 3D

Hi! I’m not sure about memorizing all the specific wavelengths/frequencies, but I think it would probably be helpful to know at least the order of the EM spectrum, and perhaps a general estimate of wavelengths in the context of the Lyman and Balmer series.

Megan Lu 3D

Hi! Yes, I believe that it was only the brightness (intensity) of the light that was increasing, not the actual frequency/wavelength.

Megan Lu 3D

Hi! I’m not sure that we need to memorize the names of the equations—I believe they are given. However, it’s probably a good move to be familiar enough with each one to be able to quickly identify and apply them if the question asks for a specific one.

Megan Lu 3D

Hi! I believe either is fine, unless the question specifically asks for one over the other. As others have said, though, I think many of the test questions will be multiple choice, so it might not be an issue at all.

Megan Lu 3D

Hi! I was planning on taking notes as I watched the lectures and then annotating my notes after all the lectures to see which parts I am least familiar with. (Unfortunately, my iPad deleted all my notes so now I have no lecture notes to review.) I think that doing practice problems is the most effec...

Megan Lu 3D

Hi! To my understanding, the higher energy of a shorter wavelength refers to the energy of the photon :)

Megan Lu 3D

Hi! In order to calculate between kilograms and grams by hand, you can multiply by the conversion factor of 1 kg/1000 g or 1000 g/1 kg. You can think about it in terms of canceling out units. So to go from kg to g, we would multiply by (1000 g/1 kg); to go from g to kg, we would multiply by (1 kg/10...

Megan Lu 3D

Hi! I don’t think we are required to know how to name and identify compounds for exams just yet (based on what Dr. Lavelle mentioned before), but it would probably be helpful to brush up on some basic nomenclature anyway :)

Megan Lu 3D

Hi! I think it would be extremely improbable for a chemical reaction to not have a limiting reactant. To do so would require having exact quantities of reactants that perfectly align with the stoichiometric ratios in a given equation, and reaching this point of precision is pretty unlikely. As such,...

Megan Lu 3D

Hi! Molecules are atoms that are bound together in a group. A mole is essentially a unit representing set quantity of anything (6.022x10^23 “things”).

Megan Lu 3D

Hi! Like others, I’ve been using as many decimal points as detailed in the period table posted on the class website. I assume that that is the periodic table we use for this class, so I stick to those values.

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