CHEMISTRY COMMUNITY

Arielle Sass 2A

E°, similar to ΔG°, doesn't change for a reaction because it only refers to when all the reactants at the anode and cathode are at standard conditions: 298 K, 1 Mol/Liter or 1 atm (maybe bar actually). If the reaction were to actually progress at all, it would no longer be in standard conditions sin...

Arielle Sass 2A

Hi! For this problem, you have to look at the list of reduction potentials to see which metals of the ones given have the lowest reduction potential, and thus the highest/most positive oxidation potential (because if the metal is reducing then it is being oxidized). For the metals that have multiple...

Arielle Sass 2A

Hi! Ba(OH)2 is a strong base, so we assume it will completely dissociate in water. Every mole of Ba(OH)2, when dissociated, contributes one mole of Ba2+ ions and two moles of OH- ions, so those ratios determine their concentrations. Don't forget though that [Ba(OH)2] will equal zero at equilibrium, ...

Arielle Sass 2A

Hi! You're right, E° only refers to the standard voltage for a cell, meaning that you can only use this to determine spontaneity for the cell at standard conditions. It's the same way that ΔG° only tells us the spontaneity at standard conditions, but we really need to find ΔG to see if a specific re...

Arielle Sass 2A

Hi! You're right; because the equation for spontaneity is ΔG°=-nFE°, if E° is negative, you know that ΔG° has to be positive and thus the reaction isn't spontaneous. So you could calculate ΔG° if you wanted to, but I believe the question only asks you to calculate it for the reactions that are spont...

Arielle Sass 2A

Hi! In terms of the relationship btw work and cell potential, I believe all you need to know is that the max work that can be done by a cell is equal to the ΔG of the cell, which is equal to -nFE, so if you can find the cell potential (E) you can find the maximum work. To answer this question, I fir...

Arielle Sass 2A

Hi! You probably just missed it, but it's on there below Pt: Pu 4+ +e- -> Pu 3+ E⁰=+0.97 In terms of how to solve, for a, first identify what's being reduced and what's being oxidized. N's oxidation state goes from +5 to +2 so it's being reduced, and Hg's oxidation state is going from 0 to +1 so it'...

Arielle Sass 2A

Hi, I might be wrong but I believe that all ions will be aqueous in all of these reactions, for any element. Also all non-charged metal species, like Cobalt we can assume are solid. The only metal that isn't solid at standard conditions is mercury, so I'd expect that it wouldn't be found in a galvan...

Arielle Sass 2A

Hi, you should include Pt(s) when the half reaction doesn't include any solids. A solid metal on each side of the cell is necessary for the electrons to pass from the anode to the cathode.

Arielle Sass 2A

Hi! What I personally do is work through every other assigned textbook problem, because like you said there are a lot of them. I always check my answers right after completing the problem, and if I get it wrong, then I try to figure out what I did wrong so I know how to answer it for next time. Then...

Arielle Sass 2A

Hi! Yes, I would balance the oxygen and any other elements first, and save hydrogen for last because the way that you balance it varies according to whether the reaction is in acidic or basic solution but you can always start out balancing the same way. I think there may be multiple ways to balance ...

Arielle Sass 2A

Hi, for this problem the answer that sapling seems to accept is without the parenthesis, but leaving the coefficient of 3 in front of the H2O, like 2Fe2O3*3H2O. I guess changing the coefficient of the first term "applies" to the hydrate part.

Arielle Sass 2A

Hi! For this question, I started out with the half reaction Cl 2 O 7 -> ClO 2 - . First balance the Chlorine atoms, adding the coefficient 2 to ClO 2 - . Then, there are 7 oxygen atoms on the left and 4 on the right, so I added 3 H 2 O to the right to balance that. After that to balance the hydrogen...

Arielle Sass 2A

Hi!
The one reduced is the oxidizing agent because it causes the other species to be oxidized by pulling electrons away from it. The one oxidized is the reducing agent because it causes the other species to be reduced by giving/pushing electrons to it.

Arielle Sass 2A

According to the textbook, an electrochemical cell is a device in which an electrical current is either produced by a spontaneous reaction or is used to cause a nonspontaneous reaction. A galvanic cell is one where a spontaneous rxn is used to generate a current. In the redox reactions we're looking...

Arielle Sass 2A

Hi, for this question q is equal to the enthalpy of fusion of ice + the heat needed to bring liquid water from 0 degrees C to 20 degrees C. But I believe that we are supposed to assume pressure is constant because nothing is otherwise stated, and to calculate the heat required for a phase change, yo...

Arielle Sass 2A

Hi, I'm also confused by this-- in a constant volume process, w=0, so I thought ΔH would just equal ΔU (since nRΔT refers to the amount of work done). The solution manual does this as well, saying qv=765 J. Could that be a mistake in the correction maybe?

Arielle Sass 2A

I believe you're right, it just depends on the units and what's given. Specific heat capacity has the units J/C*g and molar heat capacity Cp has the units J/K*mol

Arielle Sass 2A

Hi, the way that I think about this is that, since delta S =q/T, an addition of heat makes a much bigger difference for the thermal vibration of a very cold substance than for an already hot one. For example, holding a flame up to a cup of hot tea won't really change its entropy much, but holding a ...

Arielle Sass 2A

Can someone explain why the ΔS of the system is the same for both but ΔS surr and ΔS tot are different for this problem? I'm generally confused on how entropy changes differ based on whether a process is reversible/irreversible, and how you know which equation(s) to use in each situation. If someone...

Arielle Sass 2A

Hi, if it's stated that a substance is heated to a temperature above its boiling point, then you can assume it has changed phase and you should include the enthalpy of vaporization in the total enthalpy change, but if the question just says a substance is heated up to its boiling point and doesn't s...

Arielle Sass 2A

Hi! You’re right, the answer at the end of the textbook seems to be incorrect (it says 8 joules) but I believe the solutions manual booklet has the correct answer, which is 28J (for both w and delta U) Hi where do you find the solutions manual booklet? If you go to the Chem 14B page on the Sapling ...

Arielle Sass 2A

Hi, for this problem you first have to set up the correct reaction equation: C 6 H 6 (l) +7.5O 2 (g) ->3H 2 O (l) +6CO 2(g) Using this balanced reaction, you can find the change in the number of moles of gas, which is -1.5. Then substitute this as Δn in the ideal gas ...

Arielle Sass 2A

Hi, I believe thermal entropy is the entropy a system has from the movement of its particles, so it depends on the temperature of the system. Residual entropy, however, is the entropy a system has when you ignore thermal entropy, and it's only based on the amount of possible positions or states that...

Arielle Sass 2A

Hi, so for this question I first found the enthalpy needed for the liquid sample to go from 0 to 5 C using the equation q=mCΔT and the known specific heat capacity for water, 4.184 J/g -> (150)(4.184)(5-0)=3138 J. If we divide this by 0.5 hours, we get the rate at which energy is entering the sample...

Arielle Sass 2A

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25 C and 1.00 bar In the solution manual for this question, it says that the combustion reaction for C6H6 produces liquid water. Doesn't combustion produ...

Arielle Sass 2A

Hi! I was also having this issue, but its because of your units for R. You have to use a different "R" for the PV=nRT part, the one that is 8.206x10^-2 L*atm/K*mol. Then later when using the equation w=-nRTln(Vf/Vi), use R=8.314 J/K*mol. If you look closely at the units involved in each eq...

Arielle Sass 2A

Hi, I believe you could use this for different types of reactions if the question supplies you with enthalpy per mole for a reaction at constant pressure, but that is something that will depend question to question. I'd say just pay attention to the values that they give you and see what equations y...

Arielle Sass 2A

Hi, you're right! The system expands because of the production of N2 gas, so we have to find how many moles of gas there are to determine the volume change, and then use that to determine w.

Arielle Sass 2A

Hi! You’re right, the answer at the end of the textbook seems to be incorrect (it says 8 joules) but I believe the solutions manual booklet has the correct answer, which is 28J (for both w and delta U)

Arielle Sass 2A

Hi! So for the first step, the volume remains constant so there is 0 work being done. In the second step you could use the w=-(Pexternal)(deltaV). Your answer will come out with the units of L x atm, so you want to convert that into Joules. Then since the work from the first step was 0, your total ...

Arielle Sass 2A

Hi!
I think you’re right that most of the time with equilibrium equations the arrow is double sided, so I would say to write it out that way, but I wouldn’t overthink it if the textbook uses a one sided arrow. They both essentially mean the same thing.

Arielle Sass 2A

Hi! I solved this by looked at the approximate change of each substance until they reached equilibrium and assuming the one that changes the least, B, changes by one mole. So if B increases by five and C increases by ten, then B must be increasing by 1 mol and C must be increasing by 2 moles. A decr...

Arielle Sass 2A

Hi!
This is actually a textbook error. The K value the table gives is incorrect. It should actually be k=6.1x10^-3 so you are on the right track with finding the inverse. All of the textbook errors are on a doc on the Chem 14B website if you want to check it out!

Arielle Sass 2A

Hi!
I believe you are just expected to reference the table with K values for this question. You’re right, without knowing K there’s no way to answer this.

Arielle Sass 2A

Hi!
I was also confused by this but it’s because the graph they provide is in kPa, kilopascals, which has to be converted to atm and the conversion rate is about 100 kPa to 1 atm (the actual value is 101 something but they just use the approximation to get the answer)

Arielle Sass 2A

Hi! Method 2 is based on bond enthalpies, so to find the ΔH of a reaction you add the energy required to break the bonds of the reactants to the energy released from the new bonds of the products forming. For example, if you wanted to find the ΔH for the formation of CH4, using the reaction 2H2+C->C...

Arielle Sass 2A

Hi! Hess's law is that the overall enthalpy of a reaction (ΔHnet) is the sum of the component reaction enthalpies. This is possible because enthalpy is a state function so you can add enthalpy changes together. The example from the lecture asks us to find the ΔH for nitrogen dioxide formation and gi...

Arielle Sass 2A

Hi! I haven't done this question so I could be misunderstanding but if HI is your acid, then it will dissociate to H+ and I- so it will be a 1:1 ratio. Like HI+H2O<-->I-+H3O+

Arielle Sass 2A

Hi! I think it will depend on what information the question provides you with, but the textbook also contains tables for many bond enthalpies and standard enthalpies of formation so if the info you need is in those tables, I think any method will work.

Arielle Sass 2A

Hi! If enthalpy of fusion is the energy required to melt a solid, and enthalpy of vaporization is the energy required to vaporize a liquid, then adding these two values together will give you the enthalpy for a solid to become vapor (sort of like multiplying the K constants when you combine two part...

Arielle Sass 2A

Hi- there is a trend for these values. Stronger acids have lower pKa and thus a higher Ka. Stronger bases have a lower pKb and a higher Kb. You can compare relative strengths of acids and bases using these values!

Arielle Sass 2A

That looks right to me-- maybe you have a different value that's off?

Arielle Sass 2A

Hi!
I believe your "b" term, 24.24, should be negative. With that, I got X=0.2375 which works out with the initial concentrations :)

Arielle Sass 2A

Hi, Last quarter all calculators were allowed (well, regular ones including scientific and graphic calculators like TI-84) so I would assume the same policy will be in place. I also am curious though if we are allowed to use a Quadratic Formula program on our calculators to solve equilibrium express...

Arielle Sass 2A

A higher Kb means more products in a base's reaction with water, so a higher Kb value (and a lower pKb value) means a stronger base.

Arielle Sass 2A

Kc is the equilibrium constant based on concentration, and the values given to you are in concentration (mol/L) so you don't have to worry about converting anything to pressure. If they gave you Kp, then you would have to.

Arielle Sass 2A

The equilibrium constant is K=4, in the question :)

Arielle Sass 2A

Hi,
I believe Kp just wouldn't be relevant for a reaction that includes an aqueous compound. You'd only be able to use (and probably would only want to use) Kc. Plus I don't think Dr. Lavelle would test us on that anyway, thankfully ^^

Arielle Sass 2A

Solids and liquids do change in reactions, however their concentration does not, and equilibrium constants are calculated based off of concentration. Solids don't change concentration b/c they have a fixed volume, and liquids/solvents' concentrations don't significantly change enough to include in t...

Arielle Sass 2A

To solve this I first found how much BrCl there is at equilibrium: (1.84x10^-4)*0.183= 3.367x10^-5 Subtracting that from the original amount of BrCl, we find that 1.503X10^-4 moles of the gas reacted and turned into Br2 or Cl2. Because the stoichiometric ratio is 2BrCl to 1Br2 (or Cl2), then half th...

Arielle Sass 2A

Hi, this is true because the more products there are, the more reactants there will be, regardless of which is more than the other. The equilibrium constant is the ratio of products to reactants so if one increases, the other has to increase as well, until the equilibrium ratio is reached again.

Arielle Sass 2A

Hi, I believe this is because as the amount of oxygens increases, the more evenly a negative charge on the molecule (after H+ is removed) can be distributed amongst the atoms, making the molecules with more Oxygens more stable as anions than the molecules with fewer oxygens. And when the molecule is...

Arielle Sass 2A

Hi, I was also confused by this before but I think the general idea is that when comparing two acids and deciding which is more acidic, like you said you first want to compare the bond strength. For example, when comparing the acidity of HCl and HBr, the H-Br bond is longer since Br is bigger than C...

Arielle Sass 2A

Hi, I think an "excited" electron means one that is already in the "excited" state, as in it is at a higher energy level n than it originally was, before it absorbed any light. So when the excited atom emits light, the electron moves back down to a lower energy level and becomes,...

Arielle Sass 2A

I'm having the same issue! I hope this is something that can get resolved soon because with finals coming up we'll be needing to watch the lectures even more...

Arielle Sass 2A

I understand from today's lecture that when a weak acid is in a solution with greater pH, it will deprotonate because it is more acidic than the solution it is in. However I thought that a characteristic of weak acids was that they only somewhat deprotonate/ionize in solution... Does the pH have to ...

Arielle Sass 2A

I noticed that the textbook says when writing a coordination compound formula from a name, the chemical symbols should be written in alphabetical order, but it also says that the answer to 9C.5(d) is Na[Fe(OH2)2(C2O4)2] where OH2 is written before C2O4. Also an example Dr. L gave in his lecture was ...

Arielle Sass 2A

Hi! In order to determine pH, we have to determine the concentration of H+ ions, which acids dissociate into (as well as some other anion). Because strong acids are characterized by releasing a lot of H+ ions, strong acids break apart very easily in water (since the pull of the dipole of the water m...

Arielle Sass 2A

Hi!
Dr. Lavelle said that in 14A we will only be learning about pH calculations for strong acids and bases, which only involves something like 1 mol acid= 1 mol H+ + 1 mol X-, so we won't have to do calculations involving the equilibrium constant K (since that's only needed for weak acids/bases).

Arielle Sass 2A

Hi, I think maybe you meant that there was an equilibrium pointing towards H3O+ in the reaction with HCl and H2O. This is because in the chemical reaction HCl(aq) + H2O <-> H3O+(aq) + Cl- , HCl is considered a strong acid, which means it dissociates/ionizes a lot when it's put in water (because it h...

Arielle Sass 2A

Hi, you're right that HBr is a stronger acid than HCl, because it has a weaker bond so it dissociates more easily, causing it to give off more H+ ions in the solution. In the scheme of all acids though, I think both HBr and HCl are considered strong.

Arielle Sass 2A

Hi, I don't think we're allowed to view our exams but if you go to any TA's office hours they can view your test and tell you which questions you got wrong!

Arielle Sass 2A

Hi,
If you are talking about (ClO2)- the ion, I believe this is a bent shape with a tetrahedral arrangement. The regions of electron density determine arrangement (4 regions=tetrahedral) however because there are two lone pairs, the “formula” is X2E2 so the actual shape would be bent.

Arielle Sass 2A

Hi, I believe we only move some electrons to the higher p-orbital when the hybridization doesn’t involve all three p-orbitals. For example, with sp3 hybridized orbitals, all three p-orbitals are at the same energy level as the s-orbital so there would be no electrons in the regular p-orbital. In the...

Arielle Sass 2A

Hi, I saw earlier in the quarter from Dr. Lavelle that the class is curved, so there isn’t a certain point range for each grade. It just depends on how well everyone does and the grades are distributed so that only a few students get A’s and D’s and more students get more average grades. Unfortunate...

Arielle Sass 2A

Hi,
This questions just combining concepts we learned back at the beginning of the quarter with what we’re learning now! So yes, convert grams to moles and find the mole ratios to figure out the empirical formula of the molecule.

Arielle Sass 2A

I scored exactly one point higher... so maybe a bit of progress? Haha
Will Dr. Lavelle announce the score distribution?

Arielle Sass 2A

Hi! You can recognize these as ionic compounds because out of their two components, one is a cation and one is an anion. For example, with ammonium chloride, ammonium is a cation since its formula is NH4+ and chloride is an anion since its formula is Cl-. Unfortunately you kind of have to know the f...

Arielle Sass 2A

Hi, and yes! Since bonds or lone pairs both have to have two electrons, all molecules that aren't radicals have to have an even number of electrons.

Arielle Sass 2A

I would guess by "strength of the molecule" it is referring to intramolecular forces. I think if it was referring to IMF it would be more explicit, because the strength of the molecule on its own doesn't really mean anything in terms of how strongly the molecules are attracted to one anoth...

Arielle Sass 2A

To draw this molecule, you just have to look at the formula they give us. With a lot of molecules, like ClONO2, they can tell you the structure just in the ordering of the atoms. With ClONO2, you have a central O atom with a Cl and N attached to it, and the other two O atoms are attached to the N. I...

Arielle Sass 2A

We have not gone over it, so no! I don't think Dr. L is mean enough to do that haha

Arielle Sass 2A

Hi! The copper atom is one of those examples where the full d orbital is lower energy and “preferred” over the full s orbital, like in the case of chromium. So, regular copper has 1 electron in the 4s-orbital and 10 electrons in the 3d-orbital. When an electron is removed, it’s removed from the oute...

Arielle Sass 2A

Hi! I think you might mean CCl4 and CH4 (rather than HCl4). You are correct in that CCl4 has stronger Intermolecular forces (induced dipole-induced dipole interactions) because it has a lot more electrons. With more electrons, the molecule is bigger and “holds onto” its electrons less tightly than i...

Arielle Sass 2A

Hi! I believe we would get full points for writing it either way (unless specified) but the second way with the noble gas usually just takes less time so I think that way is better. Also, when looking at atoms bonding, we really only care about the valence electrons so writing out the e- configurati...

Arielle Sass 2A

Yes, I’m sure that would still count for participation points (considering even posting on the chemistry jokes board counts :)

Arielle Sass 2A

Hi! I don’t think there’s an exact way to know if a bond is a coordinate covalent bond just from looking at it, but I can imagine that if you have a good idea of some common Lewis acids and bases (H+, NH3, etc) you might be able to know that a molecule was a result of these other molecules bonding. ...

Arielle Sass 2A

Hi, "Degenerate" in our class so far has been used to refer to orbitals, and it means that they have the same energy. Not every orbital has the same energy but in some cases, like in single-electron atoms, different subshells of the same principal energy level are degenerate (eg, the 2s an...

Arielle Sass 2A

Hi, I believe when it says "ample" bond character it means it somewhat has that bond character, as opposed to the other options which say the bonds have an "overwhelming" bond character of some type. The bond lengths that they tell us are found within the molecule are mostly in t...

Arielle Sass 2A

Hi! The atoms within the molecules themselves are covalently bonded, however the molecule overall is an ion because it has that extra charge that would attract it to other (oppositely charged) ions.

Arielle Sass 2A

Hi! I believe the electron configuration is [Xe]4f 14 5d 10 6s 2 the f-subshell does contain 14 electrons (if you think about the fact that l=3, then there are seven possible values for m l , thus 14 electrons in all). I think that Lanthanum and Actinium are part of the d-block elements, so that is ...

Arielle Sass 2A

Yes, you're right. Every orbital (with three quantum numbers) has the "space" for two electrons, these electrons just must have opposite spins. I must admit I also don't completely understand exactly why there are two and only two electrons per orbital, but it seems to make sense when you ...

Arielle Sass 2A

Hi! I believe you'd find this by first finding the energy per photon (of the radiation shining on the metal), using a frequency or wavelength given in the problem. Then in order to find the amount of photons, you would use the total energy given and divide that by the energy per photon. Assuming tha...

Arielle Sass 2A

The nuclear charge of an atom is the total of its protons, so for example Helium's nuclear charge is +2. The effective nuclear charge only applies to outer electrons, and it is the nuclear charge or pull on those electrons. Because of electron shielding, outer electrons (in multi-electron atoms) don...

Arielle Sass 2A

Hi! So, whatever the value for n is, the value for ℓ can be anything from 0 up to n-1. mℓ can take on any value from -ℓ to +ℓ. And every set of three quantum numbers applies to two electrons (as in, every orbital mℓ holds 2 electrons). For the first problem, n=3, ℓ can either be 0, 1, or 2. If ℓ is ...

Arielle Sass 2A

This is actually because n 2 means the initial energy and n 1 means the final energy level. I was also confused at first because it might seem to make more sense that n 2 = the "second" energy level but now I just think about it in terms of, 2 is further from the nucleus and 1 is closer to...

Arielle Sass 2A

Hi,
it seems like you did everything correctly but I noticed you converted oz to grams- don't forget that the standard unit for plugging into equations is Kilograms. That should give you the correct answer!

Arielle Sass 2A

Use the value 3.29x10^15 Hz for Rydberg's formula when using the equation v=R[1/n12-1/n22]. I noticed that the equation you used was "1/λ = RZ^2(1/n1^2 - 1/n2^2) and so maybe for that equation, R is different (though I've never seen that equation so I don't exactly know where it comes from). Bu...

Arielle Sass 2A

Hi! So these quantum numbers signify an orbital that an electron can be in. Each energy level, also called a "shell," has a certain amount of "subshells" and then within those subshells are individual orbitals. n is the symbol for the energy level or shell and it can have any val...

Arielle Sass 2A

Hi, you're right! That's how I solved this equation. Using v=R[1/n 1 2 -1/n 2 2 ], I first input n 1 =5 with n 2 =6, and then I input n 1 =1 with n 2 =6. I initially kept getting this question wrong, but I realized it was because of a mistake with the Sig Figs that Sapling used... the only way it ac...

Arielle Sass 2A

So for this question, because the wavelength is 656 nm, we know that the light is in the visible region, and on the emission/absorption spectrum for hydrogen, the spectral lines of the visible region (aka Balmer series) are a result of energy released from an electron that jumped from some n to n=2....

Arielle Sass 2A

Electrons lose energy when they get closer to the nucleus because when an electron falls from some n to a lower n (energy level), it has to release the energy that it initially absorbed. Electrons only move to higher energy levels when they've absorbed electromagnetic radiation, so the only way for ...

Arielle Sass 2A

I don't know if this is more of a question or a warning, but when I was working out problem 1B.27 I noticed that in order to get the correct answer, you have to insert 5.0 m/s for Δv, however the question states that the speed of the object is 5.00 ± 5.0 m/s... and that should mean that Δv is 10 m/s...

Arielle Sass 2A

Hi, I kept getting this question wrong but it turns out that it was just the sig figs that were incorrect! With the question, not us :( The question only gives one value, n=6, so I put both of my final answers with one sig fig, but it counted my second nm answer as incorrect until I input it with 2 ...

Arielle Sass 2A

I was also confused about this question because I did get n=1 and n=3, but the textbook answer says "n₁=1 to n₂=3 making it sound like the electron moved from n=1 to n=3...

Arielle Sass 2A

The uncertainty equation for Kinetic Energy is the same as the one for kinetic energy, just with the uncertainty symbol in front of the variables that you are uncertain about: ΔE(k)=1/2m(Δv)^2 So if you have the uncertainty for velocity (which you can find with Δp=ΔV*m), then you can find the uncert...

Arielle Sass 2A

For this question I first used the equation c=λv to figure out what the frequency the radiation is with the wavelength given, and then used E=hv to calculate the energy per photon for that frequency. In order to find how many photons are generated in total, you just need to divide the 11 total Joule...

Arielle Sass 2A

I also got this, it seems we will need to get one but I think it's best to wait for your next discussion section to ask your TA about it before you actually spend the money

Arielle Sass 2A

It works because no matter how large or small of a sample of something you take, the percent composition of its components will always be the same. And what matters when finding empirical formula is the ratio of each component, not the actual amount of each element in the sample. So if you have the ...

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