CHEMISTRY COMMUNITY

vanessanguyen3I

Both are calculated the same way but Q is used when the reaction is not at equilibrium.

vanessanguyen3I

Ka, Kb, Kw, Kc, and Kp Ka: This is when you find the K value for a reaction with a compound that is acting as an acid. Kb: This is when you find the K value for a reaction with a : compound that is acting as a base. Kw: This is a constant!! At 25 degrees Celscius, it is 10ˆ-14 but it changes as temp...

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You need to choose the X value so that when you subtract it from the initial concentrations, you do not get a negative value. Be careful because the X value does not always have to be bigger than the initial concentration for this to happen!

vanessanguyen3I

I think its content from week 1-week 3 so depending where we finish next week.

vanessanguyen3I

Just wanted to note that when you find the equilibrium constant for this problem, you need to find the equilibrium constant for the reverse reaction, since adding more NO will cause the reaction to shift to the right. You should get K=.25 and then set up an ICE table from there. Hope this helps!!

vanessanguyen3I

I think Le Chatelier's Principle is simply referring to the fact that when you make a change in the system (like changing the pressure by changing the volume or changing the temperature), the system will respond by either producing more reactants or products to return to equilibrium.

vanessanguyen3I

Honestly something I'm trying to be better about this semester is asking for help. In 14a, if I ever struggled with a concept I never asked for help from anyone when there's so many resources. I just kept on trying to figure it out by myself. I finally talked to a TA once and it made figuring out th...

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First, click on the link in the textbook to find the value of K at 1000 K. You need to use the Kc value which is 1.2 x 10^-7. Then set up an ICE table. To find the initial concentration of Cl2, divide the amount of mols by the amount of liters to get .001. The initial concentration of Cl is 0. The c...

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Adding an inert gas will change the pressure but not the volume. As a result, adding an inert gas does not change the concentration of products or reactants.

vanessanguyen3I

To determine which is more stable, you can look at the value of K. The reaction with the larger K value has a higher concentration of product at equilibrium, while a reaction with a smaller K value has a higher concentration of reaction. The reactant with a higher concentration is the one that is mo...

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It depends on the difference in electronegativity between the atoms bonded! Also, it's useful to remember that most molecules have some form of ionic character (polarity), and most ionic compounds have some form of covalent character (depolarization), so it's not entirely black and white. Hope this ...

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To solve at the oxidation state, first look at the charge on the whole compound/molecule. Since transition metals can have different oxidation states, it's important to look at the other atoms its bonded to. For example, if the TM is bonded to one Cl and the whole compound has no charge, then its re...

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Going off of what Caelin said, acids also typically have H in their name at the beginning like HCl. If not that, they also usually have -COOH in their names to represent the carboxyl group in weak acids. For bases, they're typically metal oxides or hydroxides.

vanessanguyen3I

I think Dr. Lavelle said that for our class, dipole arrows go towards the more electronegative atom! However, I think it can be different depending on what textbooks you're looking at. That's just good to know

vanessanguyen3I

I struggled so much with textbook problems on my own and so next quarter I really plan to go to more TA office hours and workshops! I feel like those would really make the learning process so much easier for me.

vanessanguyen3I

The Roman numerals is dependent on the charge of the transition metal. For example, if your transition metal was iron and its charge was 2+, your roman numeral would be II

vanessanguyen3I

Constance Newell wrote:Are we still going to need a separate device for a camera

I think so, yes. Does this mean we're all taking the final at the same time?

vanessanguyen3I

I think we are supposed to know the prefixes from mono to hexa since those are the molecular shapes we studied in class. He also gave us other prefixes that we use if the ligand already has a prefix in its name. These are bis, tris, tetrakis, penakis. However, he said these are very uncommon.

vanessanguyen3I

I think so, yes. Molecules that only have nonpolar bonds will always be nonpolar. Molecules with polar bonds that have dipole moments that cancel out because they are in equal and opposite directions are also nonpolar. Molecules with polar bonds with dipole moments that do not cancel out are polar. ...

vanessanguyen3I

I have been having the SAME problem!!! I'm even on the VPN and it still doesn't work... fingers crossed it's fixed soon..

vanessanguyen3I

I definitely feel this way too. It's really easy for me to fall behind on lectures because they're asynchronous. I found that the best thing for is to PLAN a lot ahead with ample time to get everything done. That way, if I fall behind one day because I am not motivated, I know that I still have time...

vanessanguyen3I

Since a sigma bond is present in every covalent bond, but a pi bond is only present in double and triple bonds, I would assume that the bond between atoms with a pi bond is stronger since a double bond is stronger than a single bond. I'm not sure if a sigma bond is individually weaker than a pi bond...

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Draw arrows (to represent vectors) to see where there the electrons are pulled. If the arrows are equal and opposite, then they cancel. In this situation, this nonpolar molecule has polar bonds, but since all its dipole moments cancel out, it is nonpolar. However, you must be very careful to take in...

vanessanguyen3I

My biggest tip is: Do ALL the textbook problems and make sure understand all of them!!! There were SO many on the last midterm!! the textbook problems I find can be a little bit more difficult than the problems on the tests/examples he shows in class, so if you understand how to do them then you def...

vanessanguyen3I

I feel pretty okay about this class! I just usually get super stressed and anxious around midterms and finals time unfortunately )): other than that though, I feel like there's a lot of resources in this class so I always feel like I have help if I ask for it.

vanessanguyen3I

No, because no electrons are donated/accepted. H bonding is simply an intermolecular force that results between polar molecules! Hope this helps.

vanessanguyen3I

DO ALL THE TEXTBOOK PROBLEMS!! so many of them showed up on the last midterm! the textbook also goes into the content in greater depth which can really help, even if it is time consuming

vanessanguyen3I

GC is stronger because it has an extra hydrogen bond. This means that more heat is needed to break the bonds between GC versus the bonds between AT.

vanessanguyen3I

I usually give myself a lot of time to rest the following weekend. I do not force myself to study or work like I usually do to reward myself. I also try to forget about it and just relax with my friends and family! Watch a show or a holiday movie!

vanessanguyen3I

I think because ionic bonds are stronger, they will probably have a higher boiling point.

vanessanguyen3I

There's a girl on YouTube that posts videos where she creates a dance routine out of simple moves. She uses popular songs and it is really good cardio!

vanessanguyen3I

This problem's answer states that both energy and atomic radius are functions of n. Were we ever given this as an equation/function or is this just a principle we should know? I think it was mentioned in lecture, but to clarify: the number n an atom has is the number of shells it has. If an atom ha...

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You have to compare the electronegativity levels between the two elements that are bonded. As a guideline, Professor Lavelle stated that if there is a difference greater than 2 it is ionic, and if it has a difference of less than 1.5 it is covalent.

vanessanguyen3I

DO NOT TAKE CLASSES WHERE THE FINALS OVERLAP!! Your professors are NOT obligated to schedule a separate final time for you!! I would suggest just taking Chem BL next quarter or during the summer as I heard labs can be really time consuming.

vanessanguyen3I

While doing the textbook problems I encountered a problem: 1E 5 True or False: Electrons in an s-orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge because an electron in an s-orbital can penetrate to the nucleus of the atom. and the answer w...

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Hi! I was going over the homework for this section and I realized I don't really understand part b of this problem: Give the notation for the valence-shell configuration (including the outermost d-electrons) of (a) the alkali metals; (b) Group 15 elements; (c) Group 5 transition metals; (d) the “coi...

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Just wanted to add that for ionic Lewis structures, we also tend to write them like we do chemical equations. For example, to write the ionic Lewis structure for KCL, we write: K + CL ---> KCL while adding each element's valence electrons! When we write the KCL, we can see that K is drawn with no ou...

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The more reactive an element is, the more likely it is to give/receive electrons and the more likely it is to form bonds! The more reactive an element is, the more likely it is to undergo a chemical reaction to form bonds and release energy. S block elements are more likely to be reactive because th...

vanessanguyen3I

I used to have really bad test anxiety to the point where sometimes I simply couldn't complete problems because I kept on flipping through the pages of the exams and panicking! What's helped me though is taking a deep breath and taking a couple seconds to ground myself and go through one problem at ...

vanessanguyen3I

I was also really confused about Shrodinger equation, but basically it's just an equation that gives us math functions that represent electrons (orbitals). When you apply the Hamiltonian derivative to a certain wavefunction, it will give you that wavefunction and an energy value. We compare that ene...

vanessanguyen3I

I think it has to do also with the fact that delocalized electrons are shared amongst more atoms, and that their electron density is the same/even throughout the molecule.

vanessanguyen3I

Since the electron affinity is the difference between the incident light (energy of a photon) and the energy of the ejected electron (KE), I am assuming electron affinity is just another way of saying work function. I am not entirely sure though. Hope this helps

vanessanguyen3I

I think Dr. Lavelle said that based on experimental data, that's how they determined that Potassium fills its 4s shell first. The periodic table should simply help you remember!

vanessanguyen3I

Just adding to what Gieselle said: An ion that loses an electron is a cation. An ion that gains an electron is an anion.

vanessanguyen3I

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0237 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts? You have use de broglie's equation and the KE equation (KE=1/2mv^2). First, rearrange De Brogl...

vanessanguyen3I

To solve for V, you just have to rearrange the equation! In the end, you should get v = h/m(lambda)

vanessanguyen3I

Could someone explain how S, D, P, and F orbitals connect to Schrodinger's equation? Wednesday's lecture really confused me. First, we were talking about how the sin function is is a mathematical model for electrons and then somehow from the Schrodinger's equation we have s, p, d, f orbitals in all ...

vanessanguyen3I

if you go to "your posts," your replies to other people's posts will appear. Unfortunately, you will still have to scroll.

vanessanguyen3I

It emits a photon every single time it falls. For example, let's say there is an electron at n=4. If that electron falls from n=4 to n=2 in one go, it will emit a photon. If it then falls from n=2 to n=1 it will emit a photon. However, if it falls from n=4 to n=1 in one fall, it will only emit one p...

vanessanguyen3I

I completely understand your thinking and I got the same answer too when I first tried it! Let me try to clear things up. First, I think you are applying the concept that even if you dilute a solution, the amount of moles stays the same. This is because you are simply adding more solution, but the s...

vanessanguyen3I

I like to create a schedule when midterms are beginning to creep up. I split each class up either by topics, or week by week. I do tons of practice problem and try to imitate the environment of the test. I also try to take breaks which I believe is super necessary when you are concentrating so inten...

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I am going to explain this in terms of the photoelectric effect. To increase intensity is synonymous to making the light brighter. In a classical wave, the larger the amplitude of the wave, the more intense the light is. In a classical model, increasing the intensity of the light would increase the ...

vanessanguyen3I

Use the balmer series when dealing with visible light. Use the Lyman series when dealing with UV radiation!

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Spectral lines are basically lines that mark frequencies of light that have been absorbed/emitted, depending on whether you are looking at an absorption or emission spectrum. If a frequency of light matches the gap between the energy levels of electrons, then the electron will absorb that light, and...

vanessanguyen3I

Hi! I think your TA will be going over how to use Respondus and the process for exams! As for the webcam, I think it is supposed aimed at your hands and your paper.

vanessanguyen3I

Hi! Just wanted to explain the significance of previous answers. Like others have said, increasing the intensity will result in increasing the number of photons. In the classical model, increasing the intensity by brightening the light would increase the energy. However, in the photoelectric experim...

vanessanguyen3I

He gave us a lot but said that the most common ones are kilo, centi, milli, micro, nano, and pico. He also said to know what an Angstrom is: 1 x 10^-10. Hope this helps!

vanessanguyen3I

Hi! Basically, by the end of this week we are supposed to finish the outlined Fundamentals problems in the textbook and one Sapling assignment. The textbook problems are highly encouraged, but you do not turn them in. You should watch the audio visual focus modules (1-4) on Dr. Lavelle's website, as...

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Lecture 3 Discussion 3l: https://groupme.com/join_group/63090804/z1gdVw5M

vanessanguyen3I

E23a) (a) Calculate the mass, in grams, of one water molecule. First start by finding the molar mass of water (18.02 g). Now, you know how many that water is 18.02 grams per mole. However, you want to find out how many grams per molecule. To do this, simply divide by Avogadro's number. I am going to...

vanessanguyen3I

Everything everyone else said was true! You can also complete the homework as many times as you want and it's graded on completion!

vanessanguyen3I

As the volume is doubled, the concentration is halved, therefore the final concentration after 90 doublings is: 0.10M*(1/2^90)=8.08*10^-29M. The number of moles in 10ml of the final solution is: 8.08*10^-29M*0.010L=8.08*10^-31 mol. Then you can calculate the number of molecules present in 10ml of t...

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I just wanted to remind you that the chemistry community rules ask that we explain the question in our post and not just reference specific book problems. This allows future students to utilize this site even if the books change. I don't mean to be nit-picky but I don't want you to lose points! Tha...

vanessanguyen3I

So what I did first was check what the relationship between the molarity before and after if the volume was doubled. 0.10 M (10 mL) = x (20 mL) x = 0.05 Notice that the volume is 0.10 M before doubling the volume. After doubling the volume, you get 0.05 M which is half of the initial molarity. Now ...

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Can anyone post their work for G.25 (in the textbook)? I just want to see how others solved this problem! Thanks! "Concentrated hydrochloric acid is 37.50% HCl by mass and has a density of 1.205 gcm3 . What volume (in milliliters) of concentrated hydrochloric acid must be used to prepare 10.0...

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