CHEMISTRY COMMUNITY

Crystal Yu 1D

Break down the main reaction into its smaller parts and you will find that the other reactions given will make more sense! Also remember that when a reaction is multiplied by a coefficient n, the Ka will change to Ka^n. When two reactions are added together, the Ka values are multiplied. I hope this...

Crystal Yu 1D

I believe in the textbook, the water is liquid not gaseous in question 5.61. Then, it would make sense because the water is simply a solvent and is not involved in the compression. I hope this helps! :)

Crystal Yu 1D

Probably a typo! If it is not on the Solution Manual Errors yet, maybe bring it up to Professor Lavelle during office hours so he can confirm and edit the Solution Manual Errors document! :)

Crystal Yu 1D

Another good way to tell if Kc is small is to think of the number comparatively. If Kc is 10^-4 and all the other values are just 0.3, then Kc is relatively much smaller meaning that you can approximate and make the calculation easier. Hope this helps! :)

Crystal Yu 1D

I think what really works for me is doing the textbook problems after reading. (as many problems as i need to feel confident in the material). This really forces you to apply concepts that you've just learned through reading or lecture videos. The application also really reinforces the learning! Add...

Crystal Yu 1D

I think it would be the same as 14A, but since you use the same account, the total number of posts will be greater.

Crystal Yu 1D

I think it would be helpful for you to look up Le Chatlier's Principle. To summarize, a increase in volume would cause the equilibrium to shift in a direction that favors the production of more gas molecules. More gas molecules = more volume occupied. You can then just calculate the effect on K, bec...

Crystal Yu 1D

When the partial pressure of CO2 is decreased, the reaction will shift towards the reactants (the side of the equation that CO2 is located), causing the partial pressure of O2 to decrease. Adding water, which is also a reactant, will cause the a shift to the products side, resulting in an increase o...

Crystal Yu 1D

50% of all points (the overall grade) in the class would be passing.

Crystal Yu 1D

I think in lecture, Professor Lavelle said we just needed to know:
UV corresponds to n = 1 (Lyman goes from n=1 to n=2)
visible corresponds to n = 2 (Balmer goes from n=2 to n=3)
infrared corresponds to n = 3
I hope this helps! :)

Crystal Yu 1D

Yes, a stronger conjugate base means a weaker acid. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid.

Crystal Yu 1D

I think the strong acids/bases and weak acids/bases should be memorized based off the information Professor Lavelle has given us in lecture. For reference, I think he lists the examples in Lecture #26 so I would review that and get familiar with them. :)

Crystal Yu 1D

Both technically mean the same thing, but I would note that H+ is a shorthand of H3O+. The actual interaction causes the formation of H3O+ when water reacts with the H+ from an acid, but both should be acceptable. I hope this helps! :)

Crystal Yu 1D

I've seen a couple posts like this and I also had a similar problem. I recommend trying both links, because one or the other has worked better for some people. For me, both wouldn't load so I restarted my computer and waited a bit. After an hour, I checked back and the video loaded. Hopefully this h...

Crystal Yu 1D

This can also be solved without knowing the Kb values. Morphine is a weak base because it contains a nitrogen with a lone pair. BrO- would be the stronger base, because it is the conjugate base of a weak acid, so we know it is relatively strong. I hope this helps! :)

Crystal Yu 1D

I think generally Group 1 and 2 cations just don't have a strong enough attraction to break the bonds in water molecules. For the examples he gave with Fe and Cu, because they are transition metals, they will also have more protons and thus exert a stronger attraction on the water molecules, making ...

Crystal Yu 1D

I also had this problem, but I think it definitely loads faster if you use the second link because it has a lower bitrate.
Sometimes, I just had to reload or come back to watch it at another time. I think the issue was more with my wifi connection though.
Hopefully this helps! :)

Crystal Yu 1D

I think the best way is to get familiar with which cations and anions contribute to a reaction with water. For example, in part a, Na+ is not an acid or a base, so it would not react with water. Therefore, only the C6H5O is involved in the reaction. Another way to think about it is that the reaction...

Crystal Yu 1D

The two concepts are not exactly the same, but there are many instances in which the formal charge and the oxidation number may be the same. This link is pretty useful because it goes into detail for both. I hope it helps! :) https://chem.libretexts.org/Courses/Oregon_Institute_of_Technology/OIT%3A_...

Crystal Yu 1D

I would say it's classified as a strong acid! :)

Crystal Yu 1D

Different types of bonds all ultimately mean that there are electrons in that ONE area connecting the two atoms. Therefore, despite a bond being single, double, or triple, the electrons are connecting the atoms in that one area, meaning there is ONE region of electron density. Single, double, and tr...

Crystal Yu 1D

In the outline for this section, the goals include: "Be familiar with the rules for naming coordination compounds." "Describe the biological functions of naturally occurring coordination compounds (e.g., hemoglobin, myoglobin, vitamin B12)." "Discuss well-known examples of c...

Crystal Yu 1D

Generally, neutral ligands have the same name as the molecule, but there are a few exceptions.

H2O - aqua
NH3 - ammine
CO - carbonyl
NO - nitrosyl

You would just have to memorize these exceptions. There's a nice explanation for all the naming rules on page 723 of the textbook! :)

Crystal Yu 1D

I think it would be helpful in the long run to memorize them all, since we may need to know them in future chem classes. Also, one of the goals for this section that we are on is "Be familiar with the rules for naming coordination compounds" so I would assume that we have to be able to nam...

Crystal Yu 1D

I think the prefixes are applied in the order of the ligands as they come in the name. Because the ligands are listed alphabetically, the first ligand will use the di-, tri-, tetra-,... prefixes and the ligands following will use the bis-, tris-, tetrakis-, ... prefixes. I hope this helps! :)

Crystal Yu 1D

I think you are correct that it should be in alphabetical order, so maybe it is a solution manual error. BUT also note that in the textbook, it says that the rules given have been recently changed, so maybe that's why the solution manual and the textbook rules don't match? I think it shouldn't be to...

Crystal Yu 1D

Firstly, a pi bond occurs between two unhybridized p-orbitals that overlap side-to-side. Delocalized bonding electrons mean that the electrons are not always moving between 2 bonded atoms, but can move between various different pairs of bonded atoms. (Think resonance structures!) When you have unhyb...

Crystal Yu 1D

I think the reason is that sp2-hybridized atoms only use 2 of the p orbitals in hybridization. Essentially, think s + p + p → sp². For this reason, the third p orbital was unused for the hybridization. BUT that leaves it unused p orbital free to form a pi (π) bond with another atom. This pi bond exp...

Crystal Yu 1D

When atoms bond, their orbitals become hybridized to represent the "mixing" of different orbitals. The orbitals mix because the molecule seeks to have the most stable molecular shape. When two s orbitals overlap, they form a sigma (σ) bond. When two p orbitals overlap side to side, they fo...

Crystal Yu 1D

Hybridized orbitals represent a mixture of unhybridized orbitals, thereby showing what actually happens to the electrons when atoms bond. Unhybridized orbitals are what we have been working with this whole time (s, p, d, f). They represent the orbitals of an unbonded atom. I hope this helps! :)

Crystal Yu 1D

I don't think Professor Lavelle has specifically told us a date when they would be released. My TA mentioned that after the midterms, they want to check to make sure there were no problems with the questions or answers online so we don't have problems with wrong grades. This will usually take a few ...

Crystal Yu 1D

Hello, so VSEPR is meant to describe the shape with relation to one atom, usually the central atom. In the case that there are multiple central atoms, you would be given a specific atom to reference the shape to. This way you will still be able to describe the shape. For example, C2H4 is trigonal pl...

Crystal Yu 1D

For the midterm, you won't need to know nomenclature, but you will need to know how to draw Lewis Structures for specific molecules. The formulas of the molecules are always given, but nomenclature is useful for the future so I recommend learning it when you have time! :) Khan Academy has a few shor...

Crystal Yu 1D

Also, just to clarify, if you were to form a double bond between O and F in the first molecule, the electrons would be taken from F. If you were to draw it out, there would be a double bond between O and F and 2 lone pairs (4 total electrons) around F. The oxygen atom would look the same and still h...

Crystal Yu 1D

Double bonds and triple bonds are made when the molecule is electron deficient. When a molecule is electron deficient, there aren't enough electrons to satisfy an octet for every atom, so the double and triple bonds are formed to allow atoms to share electrons and complete their octet. In the exampl...

Crystal Yu 1D

I think this would be an error in the answer key. CaCO3 should have higher boiling point because ion-ion attractions are stronger than Hydrogen bonding. :)

Crystal Yu 1D

Nandhini 1J wrote:Do we need to know Oxidation number for Midterm 2? I didn't see it mentioned on the outlines but it was on the Sapling homework.

No, I don't think they will be on Midterm 2, but we will probably learn the oxidation number rules in the future when we do redox reactions.

Crystal Yu 1D

Firstly, all molecules will experience London Dispersion Forces due to the temporary, induced dipoles caused by the changing electron density in a molecule. Large molecules will have greater London Dispersion Forces because they have more electrons and thus their induced dipoles will be stronger. Th...

Crystal Yu 1D

Generally speaking, a double bond is formed when the molecule is electron deficient (thus the need to share electrons and form a bond). In the cause of a polar double bond, one of the atoms involved in the bond is more electronegative and exerts a greater attraction on the electrons, pulling the ele...

Crystal Yu 1D

Hydrogen bonding occurs between a positive dipole hydrogen and a negative dipole F, O, or N! :)

Crystal Yu 1D

Lewis Acids are electron acceptors and Lewis Bases are electron donors.

For OF2, when you draw the Lewis structure, you can see that there are excess electrons in the form of lone pairs on Oxygen. These excess electrons can be donated, thus making it a Lewis Base. I hope this helps! :)

Crystal Yu 1D

You have to use a combination of rules to find the oxidation number accurately. These are the rules I found online, with example included. I hope it helps! :) 1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8,...

Crystal Yu 1D

The structure of the molecule can affect the net polarity of the molecule, so I think that might be what you are referring to. For example, two dipole moments of the same strength pointing in opposite directions will cancel out, resulting in no net polarity. Professor Lavelle said that he would go o...

Crystal Yu 1D

Yes, I think it is going to be on the midterm. I recall that in Monday's lecture (lecture 17?), Professor Lavelle said that we would be covering molecular structure in the upcoming lecture, so we will probably go over it tomorrow! :) Also, he sent an email yesterday or today about the material that ...

Crystal Yu 1D

I don't think nomenclature will be tested on the exam, but Professor Lavelle may include the names of molecules on the exam. I recall in one of his lectures that he would generally include the chemical formula along with the name though. I still recommend knowing general nomenclature just because it...

Crystal Yu 1D

I think the best way to decide is by looking at the formal charges for each atom. Remember that Formal Charge = # of valence electrons - # of unpaired electrons - (# of paired electrons / 2). Generally, having a formal charge of 0 is more stable and preferred. The sum of all the formal charges shoul...

Crystal Yu 1D

A hydrogen bond occurs between a positive dipole H and a negative dipole F, O, or N. First determine polarity for each of the bonds using electronegativity. This will help you determine which atoms are available for Hydrogen bonding. I have included a picture below to help you visualize it! :) The b...

Crystal Yu 1D

Sorry, I forgot to answer the part about oxidation numbers! To solve this part, you need to know the general oxidation rules. I recommend looking them up on Google or in the textbook, because they will be useful to have memorized in the future. For this problem, you only need two rules. 1. Oxygen bo...

Crystal Yu 1D

The average bond length would be the sum of the actual bond lengths over the number of bonds. Consider 4 Cl-O bonds, two of which are single bonds, two of which are double bonds. Using the data given for the typical length of a single Cl-O bond (172 pm) and the typical length of a double Cl-O bond (...

Crystal Yu 1D

For a regular dipole-dipole, the molecules are originally polar. You can tell if it is polar or not based on the relative electronegativity of the atoms in the molecule. For an induced dipole-induced dipole, the molecules themselves are not originally polar, but due to the shifting electron density,...

Crystal Yu 1D

This question depends on the oxidation number you solved for Cl. The oxidation number should be +7 for that question, and for this case, the oxidation number assumes that all 7 electrons are assigned or given to the more electronegative atom (O). In this case, none of the structures have a formal ch...

Crystal Yu 1D

I think it would be okay if you looked it up when solving the question since we didn't learn it in class, but I recommend you memorize them because it will be useful information in the future!

Crystal Yu 1D

I believe that some of the exceptions will simply have to be memorized, but a good rule of thumb when determining the structure is to check formal charges. When the formal charges are 0, the atom is in a stable state. For BF3, the formal charges are 0, so despite not having an octet for B, the struc...

Crystal Yu 1D

Based on what I understood from the lecture, resonance structures are all the possible structures when a single Lewis structure can't show the possible structure of the molecule. A resonance hybrid is the combination of all those possible resonance structures. Hope this helps :)

Crystal Yu 1D

Two exceptions to filling up the electron orbitals would be Chromium and Copper.
The electron configurations would be:
Cr: [Ar] 3d5 4s1
Cu: [Ar] 3d10 4s1

These exceptions are made because it is more favorable to have full or half full orbitals. Hope this helps! :)

Crystal Yu 1D

b. Outer electrons experience an effective nuclear charge because of electron-electron repulsion with inner electrons. This is also known as electron shielding . c. An electron in the 3s state will experience a larger Zeff than an electron in the 3d state. To clarify part b, electrons have like char...

Crystal Yu 1D

Br and F both have 7 valence electrons so the total should be 28. Maybe the book made an error or you counted them wrong?

Edit: If you are looking at the solutions manual, it does show 28 electrons. Hope this helps! :)

Crystal Yu 1D

It is false because electrons with l=2 are located in the d orbitals. Electrons with l=1 are located in the p orbitals. For electrons with the same principle quantum number (n), the p orbitals would be located closer to the nucleus and thus be better at shielding because it has more opportunities to...

Crystal Yu 1D

For Manganese 4+, the valence electrons left would be the 3 from the 3d orbital. D orbitals are counted as part of valence electrons because they are part of the outer most electrons this element. The difference between valence and non valence electrons is mainly which ones are on the outer part of ...

Crystal Yu 1D

I think specific exceptions are generally just going to have to be memorized. Professor Lavelle will probably go over rules more in the future lectures so the exceptions make more sense though! :)

Crystal Yu 1D

Yes, I think you conceptually are thinking the same thing. For resonance structures, the true bond lengths are all equal because the delocalized electrons are shared equally. They are something between a double and a single bond in the nitrate ion. I believe you have a 4/3 bond order because there a...

Crystal Yu 1D

The nitrate ion has a negative 1 charge which is why you have to add another electron to the total number of electrons calculated based on the valence electrons of each atom in the molecule.

Crystal Yu 1D

The previous post is correct. In the lecture, this was referring to the circular standing wave which showed that electrons have discrete energies with only certain specific wavelengths allowed in an atom. I would skim through this article because it was a good description about what Professor Lavell...

Crystal Yu 1D

Yes, 20.00 has 4 sig figs.

Crystal Yu 1D

The principle quantum number (n) is the refers to the electron shell. The angular momentum quantum number (l) refers to subshell of n. This means a specific s, p, d, f, etc. subshell of the n-shell that the electron is in. The magnetic quantum number ( m_{l} ) refers to the orbital of the subshell l...

Crystal Yu 1D

Generally, I think we will just use E=hv because it finds the energy of the photon using Plank's constant and the frequency of light. E=pc also finds the energy of a photon, but using the speed of light and momentum of the photon instead. It probably won't be used for calculation, but it's important...

Crystal Yu 1D

If you look at the principle quantum numbers and the diagrams that Professor Lavelle drew in today's lecture, the 3d orbital in the n=3 subshell. The 4s orbital is in the n=4 subshell. The n=4 subshell has a higher energy level than the n=3 subshell. I hope this helps! :)

Crystal Yu 1D

I think that Professor Lavelle may go over this more in detail in later lectures, but despite having the same number of electrons, ions will have a different number of protons. The proton-electron attraction will affect the size of atomic radius, electronegativity, etc. This may be a reason for the ...

Crystal Yu 1D

Valence electrons refers to the electrons that are in the outer shell and involved in forming bonds. For Scandium, if you write the electron configuration in noble gas form, it is [Ar] 3d1 4s2. In the case of 3d and 4s orbitals in Scandium, they are very similar in energy levels and both easily able...

Crystal Yu 1D

The basic way to solve these problems is using the given value in part a and conversion factors and ratios to find out how many ml of the solution is needed to obtain the wanted amount. Based on what you have so far though, you can use the dilution formula M1V1 = M2V2, where M is the molarity and V ...

Crystal Yu 1D

I think we just have to understand where the equation comes from conceptually. If it helps, I included my notes from when he was deriving the equation. It's lecture #8 if you want to watch it again! :)

Crystal Yu 1D

The x, y, and z are just meant to name the three different orbitals of the subshell P. I'm pretty sure Professor Lavelle said in lecture #10 that we would not have to know which one is x, y, and z unless it is stated in the problem, so I wouldn't stress too much about it! If you are curious though, ...

Crystal Yu 1D

Electromagnetic radiation refers to energy in the form of electromagnetic waves. The extent of change in the electrical field just means the energy difference. In this case, we can think of light (which is a form of electromagnetic radiation. Using the equation, E = hv, the energy of a photon is pro...

Crystal Yu 1D

The energy of one photon is given by the equation E = hv. Therefore, even if the wavelength and the frequency don't change, when you have 100 photons, the E of one photon has to be multiplied by 100 to get the energy of 100 photons. Thus, the answer is 100E. I hope this makes sense! :)

Crystal Yu 1D

I think you are supposed to divide Plank's constant by the product of the mass and wavelength, so you have your numerator and denominator flipped around.
De Broglie's equation is λ = h/mv.
When you solve for v, the equation becomes v = h/mλ.

Crystal Yu 1D

I think in the lecture, Professor Lavelle said that he would generally make the numbers obvious when they were immeasurable. For example, 10^-25 or 10^-38 would be very small values that are difficult to measure and would thus not have no measurable wavelike properties. 10^-10 or 10^-15 would have m...

Crystal Yu 1D

I don't think we need to know the specific values of each line within the series. The important part is the series and the energy level it's associated with.
UV light (Lyman) n=1
Visible light (Balmer) n=2
Infrared Light (Paschen) n=3

Crystal Yu 1D

Usually in the equations, you have to use whatever units the constants have in order to cancel them out. For example, Joules are kg·m^2·s^-2, so when there's Joules in the equation (Plank's constant), you need to use kg, meters, and seconds. However, sometimes the question will ask you to convert fr...

Crystal Yu 1D

Electromagnetic radiation refers to energy in the form of electromagnetic waves. The extent of change in the electrical field just means the energy difference. In this case, we can think of light (which is a form of electromagnetic radiation. Using the equation, E = hv, the energy of a photon is pro...

Crystal Yu 1D

I don't think we have covered this in a lot of detail, but simply put atomic spectroscopy is the transition of the electrons to different energy levels within an atom. Molecular spectroscopy is the transition of electrons in molecules. Electrons in molecules can still be excited. In molecules, the v...

Crystal Yu 1D

The question is asking about the uncertainty in your speed. When calculated, you end up with a very small number. Because the uncertainty in speed is small, then the likelihood of error calculated by the radar gun is also very low. If the radar gun has low error, then the trap will work most likely,...

Crystal Yu 1D

You can solve this question by using the Heisenberg Uncertainty equation. I think the most important part is solving for the correct delta x (uncertainty in position). From there, you can just plug in the numbers to get the delta v. You might be getting the answer wrong because your delta x is incor...

Crystal Yu 1D

You would first solve for the Energy of photon. Then, you can use conversion factors to find the number of photons required for the energy given. I have attached a photo of my work below. The numbers are different, but I think it will help you visualize the steps! Hope this helps :)

Crystal Yu 1D

I think this is an error in the solution guide. If you check on the website, there is a link called "Solution Manual Errors".

Crystal Yu 1D

Photons do not have a rest mass, but they do have momentum. This is why the De Broglie equation still works for photons. I'm not sure exactly which part of the lecture you were referring to, but I hope this helps! :)

Crystal Yu 1D

I think in the lecture, Professor Lavelle was referring to how each atom/molecule only absorb/emit a specific, discrete quanta of energy. This will make more sense when he lectures on atomic spectra. Since each atom/molecule will only absorb/emit those specific energies, their atomic/molecular spect...

Crystal Yu 1D

The work function just means the energy needed to remove an electron. The kinetic energy is 0 if the electron isn't actually removed and emitted. The work function will still stay the same even if the electron is emitted, because the energy that was required to remove the electron did not change. Th...

Crystal Yu 1D

Each energy level n represents a different amount of energy. When the light and its energy level matches the difference in energy between two specific energy levels, the electron can be excited from the energy level that it is currently on to the other energy level. For example, if n=2 has an energy...

Crystal Yu 1D

There are two models used in the lecture. One is the classical model (wave type model) and the other is the quantum model (particle type model). Light doesn't act ONLY as a particle or ONLY as a wave, but it has the properties of both and the way we model light depends on the scale of the behavior. ...

Crystal Yu 1D

I would also add that by "quanta" of light, it means that the light represents a discrete and specific amount of energy.

Crystal Yu 1D

The threshold energy is equivalent to the work function or the energy needed to remove an electron. Sometimes the work function is given (like in the example during today's lecture). Otherwise, you can always solve for it, if you are given wavelength or frequency using the equation E = hv.

Crystal Yu 1D

To find $\Delta x$ , you would have to find +/- 1% accuracy of the hydrogen radius. So mathematically, this would be 2(0.01)(5 x 10^11 meters).

Crystal Yu 1D

For the Rydberg formula, $\nu = -R\left (\frac{1}{n_{2}}-\frac{1}{n_{1}} \right )$ such that $n_{2}>n_{1}$ .

Crystal Yu 1D

For 30, you have to solve for the energy of the photon first. This can be done using the equation Kinetic Energy = Energy of photon - Energy to remove electron. After finding the energy of the photon, you can solve for the frequency using the equation Energy of photon = hv, where h is Plank's consta...

Crystal Yu 1D

For 29, you use the work function and use a series of ratios to solve for the energy to remove on atom of Sodium. Because the work function is in the unit kJ/mole, you can convert from moles to atoms using Avogadro's number and from kJ to J by using conversion factors. I have attached my work below!...

Crystal Yu 1D

For 28, the kinetic energy of an ejected electron is given by the equation KE = 0.5 * m * v.
The mass of an electron is 9.11 x 10^-31 kg according to the textbook and the velocity is given, so when you plug in the values to the equation, it gives you the kinetic energy.

Crystal Yu 1D

Yes, intensity of light refers more to the amount of energy per unit of area, so how many photons of light there are. Wavelength changes the energy of the wave and the photons.

Crystal Yu 1D

I try not to round until the very end so that my answer can be more accurate! :) If your calculator doesn't keep the values, I would just write them down with at least 3 more sig figs than you need so when you have the final answer, it is closer to the correct value.

Crystal Yu 1D

An example of light acting as a wave is the experiment Professor Lavelle mentioned in one of his modules where light was shone at a wall with two slits in it. I think it's called the double-slit experiment, so if you google it, you can also find more specific explanations of the experiment! :)

Crystal Yu 1D

The frequency variable is actually nu ( \nu ) which looks a little different from v. I think what you have in the equations you mentioned are correct! The letters look the same in print, so I understand if it's confusing. I just recommend checking what each variable stands for when you learn a new f...

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