CHEMISTRY COMMUNITY

Tiao Tan 3C

Degeneracy does not have a unit assigned!

Tiao Tan 3C

For part A of 6M.5, we are asked to write a cell diagram for the following reaction: 2 NO3-(aq) + 8 H+(aq) + 6 Hg(l) -> 3 Hg2 (2+) (aq) + 2NO (g) + 4H2O (l) Why is that in the answer key, we put platinum at the end of the right side of the cell diagram but not on the left side? I now understand thi...

Tiao Tan 3C

Thank you!! That helped a lot.

Tiao Tan 3C

Hi. Could someone walk me through how to identify the half reactions in the anode and cathode? I am aware that the left hand sides should contain Al and O2 but I don't know how to proceed. The “aluminum–air fuel cell” is used as a reserve battery in remote locations. In this cell, aluminum reacts wi...

Tiao Tan 3C

Hi. I used Eº=-0.445 for this question instead of rounding it to -0.45 like what the solution does. The lnK I calculated was about -34.66. But the textbook used -35. Therefore my answer is 8.86*10^-16 while the answer's is 6*10^-6. This is pretty off. How do we decide whether or not to round our ans...

Tiao Tan 3C

I am also confused by why there isn't a solid. Also for question 6L.5, why does the solution add Pt(s) when there is already I2(s) in the half reaction?

Tiao Tan 3C

For this question the solution is Pt(s)|I-(aq)|I2(s)||Ce4+(aq),Ce3+(aq)|Pt(s)

Why is Pt(s) necessary on the left hand side? I thought that since we already have I2(s) we cna just use it as an electrode instead?
Is it because I2 is the product of 2I-(aq) ----> I2(s) + 2e- ?

Tiao Tan 3C

Oxidation reactions do not always have a change in oxidation number. The oxygen on both sides of the equation are not balanced. Try adding H2O to the right side of the equation to balance the oxygen first. And then balance the H by adding H+ and OH+ (because it is in a basic solution). Before balanc...

Tiao Tan 3C

Hi. I'm confused by sapling's solution.
How do we know that combining the concentration of H2O with the rate constants kslow, kforward, and kreverse will yield the overall rate constant k?

Tiao Tan 3C

qrev is the of heat produced in a reversible process.

Tiao Tan 3C

I usually pay attention to the units given in the question and the units the answer want us to use. If the question gives volume in liters and pressure in atm, then I would use the R constant that has units L and atm. If the questions further asks for joules, remember to convert your answer in L*atm...

Tiao Tan 3C

Thank you for the summary! It's very useful.

Tiao Tan 3C

∆Gº is the gibbs free energy under standard conditions 298K and 1atm. ∆G is not under standard conditions and can vary.

Tiao Tan 3C

I usually look at the units given in the problems to determine which R value to use. For example, if we want our answer to be in J, use 8.314.

Tiao Tan 3C

When delta H and delta S are both positive, then delta G would be nagtative at high temperatures because T*∆S would be greater than ∆H. The reaction will be spontaneous
delta G would be positive at low temperatures because T*∆S would be smaller than ∆H. The reaction is non-spontaneous.

Tiao Tan 3C

Q and K are both ratios of [products]/[reactants], but K is the ratio at equilibrium.
When Q is less than K the reaction favors to the products. When Q is greater than K the reaction favors to the reactants.

Tiao Tan 3C

∆Gº is ∆G at standard conditions, while ∆G can be under any conditions and they vary with the conditions.

Tiao Tan 3C

Hi! Multiplying P (in atm) with V (in L) using the w=-PV equation will result in L*atm. We then use the conversion 1 L*atm = 101.325J to convert the units to J or further to kJ. You can find the conversion on the equation sheet.

Tiao Tan 3C

I think this is because the reaction is happening at 25ºC, at which water is liquid.

Tiao Tan 3C

Why is the energy per one mole of TNT positive in the solution?
The energy released per mol of reaction (-13168kJ) is negative.

Tiao Tan 3C

Thanks for recommending all these youtube channels!! I always enjoyed the relationship between cooking and chemistry but never got to systematically learn about them :)

Tiao Tan 3C

When Q<K, ∆G would be negative and the reaction is spontaneous. The reaction tends to proceed in the forward direction forming more products. When Q>K, ∆G would be positive then the reaction is nonspontaneous (or spontaneous in the reverse direction). The reaction tends to proceed in the reverse dir...

Tiao Tan 3C

Because N2O has 3 different resonance structures, there would be multiple ways to arrange these molecules! Also, the structures with 1 double bond and 1 triple bond are asymmetrical.

Tiao Tan 3C

We use these different equations depending on the conditions! From the two different V values in ∆S = nRln(V2/V1), we can tell that volumn is changing. Temperature is constant. When temperature is changing and volume and pressure are constant, we would either use ∆S = nCvln(T2/T1) (Cv is the specifi...

Tiao Tan 3C

Given the partial pressures of the reactants and products, we can solve for Q (the reaction quotient), and then apply it into the equation ∆G = ∆Gº + RTlnQ. It's interesting how we connected things learnt ealier in this class together with what we're learning now haha

Tiao Tan 3C

I'm not sure which part you are struggling on but I had trouble composing the desired equation from a bunch of others. The way I do it now is try to figure out which components we would like to keep and which components we want to cancel out. Then move the ones that cancel each other out to opposite...

Tiao Tan 3C

Thank you! That cleared everything up.

Tiao Tan 3C

I agree with Madilyn! So far, I've only seen K representing the equilibrium constant in thermodynamics questions. The lowercase k representing the Boltzmann's constant will appear in the future.

Tiao Tan 3C

Hi. I'm aware that there are 3 steps to approach this question: "1. heating the water from 41.0 °C to its normal boiling point, 100.0 °C. 2. allow it to vaporize at its normal boiling point. 3. cool the water vapor back to 41.0 °C" How did we figure out the third step that we have to cool ...

Tiao Tan 3C

I agree with others that it is helpful to keep track of the units! For example, if you are using pressure together with R (probably in PV=nRT), you would want to use the R value that has a pressure unit inside, which is 8.206*10^-2 L*atm*K-1*mol-1

Tiao Tan 3C

How do you know whether you use the initial or final V and P? I have used the same formula and ideas as Olivia so I was wondering how you know which V and P values to use. I'm not sure but I think for pathway A, when calculating the n with PV=nRT, we're using initial values of P and V because the c...

Tiao Tan 3C

Did you use 8.314 J/mol/K throughout your calculation? I did that too but then realized we have to use 0.08206Latm/mol/K in the PV=nRT equation when calculating for n. Because for P we are using the pressure given in atm. And then we should use 8.314 J/mol/K in w=-nRTln(V2/V1) to calculate w. Becaus...

Tiao Tan 3C

I agree with the other response. In this sapling question we could also tell that we're expected to use q=c∆T by looking at the units. For the heat capacity (calorimeter constant) of the calorimeter, the unit is kJ/°C. This means that we do not consider its mass (which also makes sense since mass of...

Tiao Tan 3C

Because we are mixing cold water with hot water, the heat gained by cold water would equal to the heat released by hot water. We then apply the equation q=cm∆T mcold water * specific heat capacity of water* (final temperature - initial temp of cold water) = mhot water * specific heat capacity of wat...

Tiao Tan 3C

If you look at the textbook, bond energies are often given as positive values. When forming C-H bonds, the reaction is exogonic, which means that the energy should be negative. Hence, we need to add a negative sign to it.

Tiao Tan 3C

Adding on to what everyone else is saying, the bond strength of the products is weaker, this means that the system requires energy to form and maintain weaker bonds from stronger (more stable) bonds.

Tiao Tan 3C

As far as we've learnt, just remember that H2SO4 is the only exception that the first deprotonation is complete and we have to use the Ka2 value in calculations. For other polyprotic acids whose Ka1 value>Ka2, we ignore the Ka2 value.

Tiao Tan 3C

Yes you're right! Just remember changes in pressure do not affect liquids, solids, and aqueous solutions.

Tiao Tan 3C

There are 2 types of representations for the benzene molecule, one is 3 C=C bonds and 3 C-C bonds, and an alternative one uses a circle inside the hexagon (accounting for the resonance in structure). For the alternative structure, we see the ring as composed of C⋯––C bonds.

Tiao Tan 3C

I agree that C and H2 are omitted because they are at their standard state. You can also tell by checking appendix 2A of our textbook. The ∆Hfº of these two substances are 0, which is why they are omitted.

Tiao Tan 3C

For part b, I don't seem to find any Ka value for Al(H20)6^3+ in the textbook. Am I looking for the wrong one? Where did you guys find the Ka value?

Tiao Tan 3C

For the conversion from pKa to Ka, just do 10^(-pKa).
In general, for -log(a)=b, a would equal to 10^(-b).

Tiao Tan 3C

I'm confused by why the solution is dividing the partial pressure given in the plot (in kPa) by 100. If this step is a unit conversion, shouldn't we divide by 1000? If not, what is this step doing?

Tiao Tan 3C

We can tell by looking at the signs of the enthalpy ∆H. If ∆H is positive, the rxn will be endothermic. If ∆H is negative, the rxn will be exothermic.

Tiao Tan 3C

So does ionization apply to both acids and bases but protonation and deprotonation only refers to either one?

Tiao Tan 3C

I would compare the pKa values with pH. In this case we are dealing with an acid: if pH<pKa, the environment(represented by pH) is more acidic. acid is protonated, it will remain as HA and therefore the reaction is neutral. if pH>pKa, the environment is more alkaline. The acid will give off a proton...

Tiao Tan 3C

For me I would compare the pKa values with pH. If we are dealing with an acid (like in question #9): if pH<pKa, the environment(represented by pH) is more acidic. acid is protonated, it will remain as HA and therefore the reaction is neutral. if pH>pKa, the environment is more alkaline. The acid wil...

Tiao Tan 3C

Yes you're correct. Q is measured at any instant of the reaction but K is only measured after the reaction reaches equilibrium.

Tiao Tan 3C

Accoding to the ideal gas law, PV=nRT. If we divide both sides with V, we could have the equation for pressure: P=(nRT)/V Because n (in moles) divided by volume V(in liters) would result in concentration (in moles per liter), we could substitute n/V in the equation with concentration. Therefore, P =...

Tiao Tan 3C

Hi. Dr. Lavelle brought this up to demonstrate that the change caused by adding/deducting pressure isn't strictly associated by moles of gases. It's associated with concentration of gases. Adding an inert gas to a container with fixed volume only changes pressure. It does not participate in the reac...

Tiao Tan 3C

This only applies to gases. Liquid pressure has very little association with volume.

Tiao Tan 3C

K would also be inversed. It will become 1/K because the equilibrium constant for the inversed reaction is adopting a reversed product-reactant relationship.

Tiao Tan 3C

T must always be in Kelvin. But if you are given in Celcius, you can convert it into Kelvin with the equation Kelvin = Celsius + 273.15.

Tiao Tan 3C

You're right M means is molarity and represents moles per Liter.

Tiao Tan 3C

Yes, if the denominator is the greater concentration, equilibrium constant K would be less than 1. The equilibrium lies more to the left, favoring reactants.

Tiao Tan 3C

Hi. The solution for 6.21 says that protons can only bind to the lone pairs on nitrogen. And Dr. Lavelle's reply in another post explains that: "Oxygen, as the more electronegative element, holds more tightly to its lone pair than the nitrogen. The nitrogen lone pair, therefore, is more availab...

Tiao Tan 3C

For question b), I drew SO2 with 2 double bonds, 2 lone pairs on each of the two O atoms, and a S with expanded octet. Everything else is like what the solution has shown.
Is my answer also correct? Does the structure of SO2 have to have 1 single and 1 double bond in this case?

Tiao Tan 3C

Can someone explain why carbonato (CO3^2-) can be either monodentate or bidentate?
The solution says it can bind a metal ion through either one or two of the oxygen ions.
How does this happen? And why is the third oxygen ion left out?

Tiao Tan 3C

Dr. Lavelle mentions soft water is alkaline and hard water is acidic in lecture.
But all my google results tell me that "hard water is more alkaline and has a higher mineral content than soft".
Which one should I believe in? Or does it depend on specific situations?

Tiao Tan 3C

Adding on, if pH > pKa, it means that the environmnet (solution) is more alkaline. The acid would give out a proton to form an anion to neutralize it (acid is deprotonated).
if pH < pKa, the environment is more acidic than the acid. The acid does not change anything (acid is protonated).

Tiao Tan 3C

Could someone please explain why would the strength of the bond to hydrogen decrease as the number of oxygen atoms increases?
Thank you!

Tiao Tan 3C

I am stuck with this problem. Could someone tell me what I did wrong?
The feedback says I have not identified all the basic salts.
Much appreciated

Tiao Tan 3C

Hi. I had the same question before and found this response by varlam17 on https://www.reddit.com/r/chemistry/comments/c5z6be/how_is_hydrogen_cyanide_hcn_defined_as_a_lewis/. "If a compound has a lone pair that does not guarantee that it's a base. Base and acid are relative, not absolute. A mole...

Tiao Tan 3C

Here's a tip for memorizing strong acids that I learnt from Michael's UA session.
All the capitalized letters will remind you of a strong acid:

SO I BRought NO CLean ClOthes
SO: H2SO4
I: HI
BR: HBr
NO: HNO3
CL: HCl
CLO: HClO4 & HClO3

I don't know if there's one for strong bases though.

Tiao Tan 3C

You can approach this problem like this.
Since the coordination compound is neutral, it has a net charge of 0.
Co+5(NH3)+5Cl+2Cl=0
Co+5(NH3)+5Cl= - 2Cl
Co+5(NH3)+5Cl= +2
Hope this helps!

Tiao Tan 3C

Adding on, there is also a coordination compound called diethylenetriamine (dien) that is on the naming coordination compound sheet: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf I memorize this together with en by looking at the number of N in the...

Tiao Tan 3C

I usually find coordination # directly from the molecular formula by counting the number of ligands in the bracket. Is this always correct? Are there any special cases where we need to draw the lewis structure to find out?

Tiao Tan 3C

Dr. Lavelle mentioned in class that if coordination number is 4, the shape would be either square planar or tetrahedral.
Why isn't the seesaw shape included?

Tiao Tan 3C

You are correct. It can be called angular or bent.

Tiao Tan 3C

AsF3 has a higher boiling point because it has both dipole-dipole and LDF, while AsF5 only has LDF.
AsF3 is a trigonal pyramidal shape while AsF5 is tribgonal bipyramidal (symmetrical).
Therefore, dipole moments of AsF3 do not cancel, making it a polar molecule.

Tiao Tan 3C

Adding on, if all the atoms except for the central atom are the same, and if the shape is a base geometry (meaning that there are no lone pairs), then the molecule would be always non-polar and dipole moments always cancel out!

Tiao Tan 3C

Both are correct and I prefer to use sp3d because it is easier to memorize!

Tiao Tan 3C

I am confused with how we can determine if a molecule satisfies the requirements for hydrogen bonding just by looking at the molecular formula. I know we need to draw lewis structures, but I don't know under what circumstances an atom does NOT bond to the central atom. For example, H2SeO4 in textboo...

Tiao Tan 3C

Why is this structure wrong for arsenic ion (first pic)? The text book solutions included only 1 double bond.

Tiao Tan 3C

Hi. For chloric acid, how do we know if H is bonded to O instead of being bonded to the central atom Cl?

Tiao Tan 3C

Benjamin_Hugh_1G wrote:The electron configuration is actually [Ar] 3d10 4s2 4p1. I believe the reason why Ga3+ is preferred over Ga+ is that removing 3 electrons will give Gallium a full shell, making it more stable.

Gotcha. thank you! I meant Ga+: [Ar]3d10 4s2 in my question. Sorry about the typo.

Tiao Tan 3C

Write the most likely charge for the ions formed by each of the following elements: d) Ga

Why is the answer Ga3+ instead of Ga+? Isn't it better to remove 1 electrons and form Ga: [Ar]3d10 4s2?

Tiao Tan 3C

Hi. Can someone please tell me why is this lewis structure wrong for nitrate ion?

Tiao Tan 3C

Dipole means that electrons are not shared equally! So they only exist in covalent bonds where atoms share electrons to form bonds.

Tiao Tan 3C

Basically electronegativity, electron affinity, and ionization energy all follow the same trend: increasing across a period and increasing up a group. I like to memorize the increasing trends with an arrow pointing to the upper right corner of the periodic table if that makes sense. Atomic radius is...

Tiao Tan 3C

Hi! To determine the valence electrons I usually count from the first element in the period. In this case, C is the 4th element in the period (Li, Be, B, C). Therefore, 4 valence electrons.

Tiao Tan 3C

I usually just rule out the ones that have other intermolecular forces. Look out for polar molecules and ions. If there is a polar molecule, then we can identify either ion-dipole, dipole-dipole, dipole-induced dipole and hydrogen bonding. If there are two ions, we can identify ion-ion. So we are lo...

Tiao Tan 3C

I agree with everyone else that bigger anions have larger polarizability. I just want to add that this is because bigger anions' electron clouds are more easily distorted because they are farther from the neucleus and less tightly held.

Tiao Tan 3C

Dr. Lavelle used BF3 to discuss an exception of the octet rule: in this case B only has 6 electrons.
and he used BF4– as an example of when B completes its octet with both 2 electrons provided by F–

Tiao Tan 3C

Delocalized electrons are electrons that is not fixed occur in a specific bond, and have equal probability of being located in any of the bonds involved in the resonance structure.

Tiao Tan 3C

It is okay for all group 13 elements to not have an octet because they all have 3 valence electrons that requires an unrealistic amount of electrons to form octet.

Tiao Tan 3C

L represents the number of lone pairs and you can count the number of dots around each atom from the lewis structure.

Tiao Tan 3C

Double bonds (4 e) have more electrons than single bonds (2 e), and have a stronger negative charge, atrracting the nuclei of the bonding atoms that are charged positively. The attraction is stronger, thus the half distance between centers of the bonding atoms is shorter.

Tiao Tan 3C

Incident light is the incoming light.
For example, in the photoelectric effect experiment, energy of the incident light = work function + kinetic energy of electrons ejected.

Tiao Tan 3C

I agree with everyone that as long as conversion is correct, you're fine.
But if the unit for the anwser is already implied in the question, do what the question asks.

Tiao Tan 3C

The Lyman series is the UV region. Excited electrons return to the n=1 shell.
The Balmer series is the visual light region. Excited electrons return to the n=2 shell. If you see the light has colors in the question, this means that they fall into the Balmer series (e.g. blue light, red light etc).

Tiao Tan 3C

Hi.
I was wondering why is the emission spectrum dark with colored lines, and why does the absorption specturm have black lines?
What caused this difference?

Tiao Tan 3C

"What is the minimum uncertainty in the speed of an electron confined within a lead atom of diameter 350. pm? Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom." The solution manual assumed that "the uncertainty in the position of the elect...

Tiao Tan 3C

The lyman series is the UV light region, with electrons transitions to ground state (shell n=1). They have the high energy because of the large gap between n=2 and n=1.
The Balmer seires is the visual light region, with electrons transitions to shell n=2.

Tiao Tan 3C

We are using the empirical equation for H-atom. En= - hR/n^2
Professor Lavelle was using an example with transitions from n=4 to n=2.
So it would be (-1/16)*hR and (-1/4)*hR.

Tiao Tan 3C

The energy of a photon is equal to the required energy to remove an electron only if there is no excess energy (i.e. kinetic energy of the electron = 0).
If there is excess kinetic energy, than the energy of a photon is greater than the required energy to remove an electron.
Hope that makes sense.

Tiao Tan 3C

So an electron has to be excited first for it to be ejected out of the shell? Or are they completely unrelated?

Tiao Tan 3C

Hi! From what I saw on Sapling, the question said "the electron in a hydrogen atom is excited to the n=6 shell", so I assume there was a typo in your question. If it is currently excited to the n=6 shell, it could return to either n=5, n=4, n=3, n=2, n=1. So there are 5 possible outcomes o...

Tiao Tan 3C

For Part C, I followed through those steps, but I still got the wrong answer. Can someone point out where I went wrong? Ephoton= 1.506 x 10^5 + 1.99x10^-19 (answer from part a) Ephoton= 150600 You haven't converted from Joule per mole to Joule per atom 1.506*10^5 is still in Joules per mole. Just d...

Tiao Tan 3C

It is the work function/threshold energy: the energy to remove electrons. It is usually given in the question so you don't have to worry! If not, then you would have to find it using the equation hv=Ek + Φ. I am sure the other conditions will be given for you to apply this equation.

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