CHEMISTRY COMMUNITY

Elizabeth Bamishaye 2I

Varsha Swamy 2J wrote:Catalysts go from reactant to product, while intermediates go from product to reactant.

Yes! Intermediates are not in the overall reaction too; they are produced then consumed. Catalysts are consumed then produced.

Elizabeth Bamishaye 2I

I know that for part E, what would make the statement true would be if the "Ecell will decrease if a substance is added to the cathode side that precipitates Cu2+."

What would help to understand this is the Le Chatelier's Principle.

Elizabeth Bamishaye 2I

Yeah just focus on what he gave us in lecture and what they contain.

Elizabeth Bamishaye 2I

When the molecules in a reaction collide with each other more, or if there is a higher frequency, then there is a greater chance that a reaction will occur. The molecules need to collide in order for the reaction to occur, and the having a high frequency increases collisions. Therefore, because mor...

Elizabeth Bamishaye 2I

Although Dr. Lavelle hasn't gone over it in class yet, the Arrhenius equation k=Ae^{-E_{a}/RT} helps determine what factors change the K value. The K increases if the activation energy decreases or if the temperature increases. I agree with this! But now that he has gone over this we now know this ...

Elizabeth Bamishaye 2I

From my understanding, fast steps are the only ones really that could be in equilibrium.

Elizabeth Bamishaye 2I

The main difference is the y-axis for both, which is [A] for zero order and ln[A] for the first order.

Elizabeth Bamishaye 2I

An important note is to make sure to not confuse the rate constant "k" with the equilibrium constant "K".

Elizabeth Bamishaye 2I

Intermediates are the species that are not in overall reaction. You could find them by viewing a species that is produced then it is consumed.

Elizabeth Bamishaye 2I

All of these listed above would help. Sometimes even searching a topic on youtube could lead you to other good and helpful videos.

Elizabeth Bamishaye 2I

Erik Khong 2E wrote:I am sure it doesn't really matter for this class, but I personally like to keep them in the order of importance. I tend to have them in the order of the species being reduced, the species being oxidized, then H2O and/or H+.

Yeah, I like to do this as well.

Elizabeth Bamishaye 2I

Yeah, this question needs to be more specific, but I would go by what the person above said.

Elizabeth Bamishaye 2I

I would go by the smallest one you see in the problem, to be honest.

Elizabeth Bamishaye 2I

The Nernst equation is used when finding the E under non-standard conditions. There are multiple versions of the equation, but they all mean the same thing and can be used interchangeably. I believe the equation you are talking about is E°=(RTlnK)/nF, which is pretty much the same thing again. This...

Elizabeth Bamishaye 2I

We would be told what to find. If it is not asked in the problem then I wouldn't worry much about it.

Elizabeth Bamishaye 2I

I agree with Vincent.

Elizabeth Bamishaye 2I

I think it is a possibility that it will be on the midterm.

Elizabeth Bamishaye 2I

To add onto the previous comments, the second law of thermodynamics states that changes in entropy can never be negative.

Elizabeth Bamishaye 2I

Yes, we will most likely be learning new material next week that would be on the midterm as well. He will make announcements about what would be on the exam, but what we learned already will definitely be on the midterm.

Elizabeth Bamishaye 2I

In lecture, Dr. Lavelle mentioned "Maximum work done by a process at constant Temperature & Pressure; also dG for a spontaneous process represents the energy that is free to do useful work."

Elizabeth Bamishaye 2I

Christina Cen 3E wrote:There isn't any exchange in heat or matter with the surroundings of an isolated system, but heat can still be generated inside the system itself. This change is isolated because it is contained within the system, and the heat is not released into the surroundings.

Yes, I agree with this.

Elizabeth Bamishaye 2I

Yes, I agree with all of the responses above because at some point I was thinking about this as well. It really depends on what you are given in the problem and what you have to look for.

Elizabeth Bamishaye 2I

Yes, I think when Dr. Lavelle says disorder it gets a bit confusing.

Basically, entropy increases when the temperature increases. Therefore, something with larger volume and lower pressure has more entropy; whereas something with lower volume and the larger press has less entropy.

Elizabeth Bamishaye 2I

To add to this, entropy is described to measure the disorder in a system. With that comes a basis of different available orders of molecules in a given system.

Elizabeth Bamishaye 2I

A state function is independent of pathways taken to get to a specific value, such as energy, temperature, enthalpy, and entropy. Enthalpy is the amount of heat released or absorbed at a constant pressure. Heat is not a state function because it is only to transfer energy in or out of a system; it d...

Elizabeth Bamishaye 2I

The transferring of heat occurs because of thermal energy because thermal energy is the driving force for heat transfer.

Elizabeth Bamishaye 2I

Yes, I agree. We will be given all the formulas that we need to know for the class but sometimes you have to combine multiple equations for one problem and you are expected to know how to use them accurately.

Elizabeth Bamishaye 2I

Yes, I think so as well because you are working out the problems and practicing but when it comes to the test, you must write everything in pen.

Elizabeth Bamishaye 2I

Yes, all you have to do is basically choose any problems to turn into your discussion section, but because this is week 1 and the first discussion sections start on Tuesday, it wouldn't give enough time and exposure to what we have learned so we have to turn in 14 problems to our discussion section ...

Elizabeth Bamishaye 2I

Also to add on the previous answers: qv is the heat absorbed or released at constant volume and uses bomb calorimetry; while qp is the heat absorbed or released at constant pressure and gives enthalpy values.

Elizabeth Bamishaye 2I

From lecture, I remember that it is the triple bond that would consist of 1 sigma bond and 2 pi bonds, which would be the time where there are 2 pi bonds.

Elizabeth Bamishaye 2I

Amphoteric compounds have both basic and acidic character, and the prime example of this is water.
Amphiprotic is a molecule that can act as a proton donor or proton acceptor.

Elizabeth Bamishaye 2I

A monodentate ligand has one donor atom used to bond to the central atom or ion as they bind to the central atom or ion at one point. A bidentate ligand has 2 donor atoms used to bond to a central atom or ion at 2 points. A polydentate ligand is characterized by having more than 2 sites used to bond...

Elizabeth Bamishaye 2I

- If reaction requires heat (endothermic) while forming a product, then increasing the temperature will favor products and equilibrium will be shifted to the right (K would increase). - If reaction gives off heat (exothermic) while forming a product, then increasing the temperature will favor reacta...

Elizabeth Bamishaye 2I

Yes, it is when naming coordination compounds. My TA also had in my discussion section that, "for some metals, the Latin names are used in the complex anions (e.g. Fe is called ferrate, not ironate)."

Elizabeth Bamishaye 2I

Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next. Yes, so spin-pair in the s-orbital and then parallel spins in the p-orbital before spin-pairing in the p orbital. (Right?) Yes, that's ...

Elizabeth Bamishaye 2I

You are to find one or the other, not both at the same time.

In order to be considered a equilibrium constant of pressure ( $K_P$ ), it should be a gaseous reaction; p can have units of pressure (e.g., atm or bar).

Elizabeth Bamishaye 2I

It would be just like using the rules of VSEPR orientations; in order to bond, you have to hybridize orbitals to have areas of electron density. If it's tetrahedral then there are 4 equivalent orbitals and its more stable (due to VSEPR), therefore it is sp3 and is in a tetrahedral orientation.

Elizabeth Bamishaye 2I

Radicals is a molecule, ion or atom that have an unpaired electron; these species are highly reactive towards themselves and/or other substances.

Elizabeth Bamishaye 2I

Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next.

Elizabeth Bamishaye 2I

A covalent bond is a bond in compounds that result from sharing one or more pairs of electrons. Therefore, in the powerpoint, it was basically saying that atoms are able to combine in order to create an octet of valence electrons by sharing electrons.

(Hopefully this makes sense.)

Elizabeth Bamishaye 2I

Yes, it is the complete opposite. Electropositivity is more for metals while electronegativity is more for non-metals. It increases in a group from top to bottom and decreases across a period from left to right.

Elizabeth Bamishaye 2I

With ionic bonds, the distance and charge affect the strength of the bond. For example, if there are more electrons being transferred then the bond would be stronger. With covalent bonds, the length of the bond and electronegativity affect the strength of the bond. Moreover, triple bonds > double bo...

Elizabeth Bamishaye 2I

A way to remember too is that when an electron is being removed because has to form a +1 ion, you have to take 1 electron off the outermost energy level. For Ba, the electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1^0 4p^6 5s^2 4d^1^0 5p^6 6s^2 ; the outermost energy level happens to be t...

Elizabeth Bamishaye 2I

Will the Heisenberg Indeterminacy Equation be given to us for Test 3 or should we know it for the test? I think that we should know it because it is at the start of Section 1.6. Also, for Test 3 we could get anything from section 1.6 to all of chapter 2. It's on the constant sheet too so you don't ...

Elizabeth Bamishaye 2I

Yes, it is true. Covalent bonds are formed when two nonmetals or a nonmetal and a metalloid share electrons.

Elizabeth Bamishaye 2I

Coulomb potential energy is the potential energy of an electric charge associated with another electric charge in a certain configuration. The potential energy depends on the amount of charge that each object contains, how far apart the charges are (distance), and Coulomb's constant. In the book, it...

Elizabeth Bamishaye 2I

For further clarification, the d orbital starts with 3d and goes to 3d^10. The d orbital specifically starts with the transition metals because those orbitals are what give them their properties. The order for electron configurations would go 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^10, 4p^6, and so ...

Elizabeth Bamishaye 2I

The $(n_{1})^{2}$ would be 2 because that is the $(n_{final})^2$ while $(n_{2})^{2}$ is the $(n_{initial})^2$ .

The equation you are to use is: $v=\Re (\frac{1}{(n_{final})^2}-\frac{1}{(n_{initial})^2}{})$ then you find the wavelength.

Elizabeth Bamishaye 2I

For part A, you determine the mass of electrons, which is 9.109\times 10^-31 kg . Then you change the given velocity from kilometers to meters and after that use the equation \lambda =h(m\nu )^{-1} or basically \frac{h}{m\nu} to get the answer for part A. For part b, you use E= h\nu to get t...

Elizabeth Bamishaye 2I

Mr. Lavelle mentioned that we do need to know them, but I feel like the most important are from kilo to pico. The bolded ones are the ones that I have tended to see more often by the problems we have done so far. Prefixes and their meaning: Giga 10^9 Mega 10^6 _______________________ Kilo 10^3 Deci ...

Elizabeth Bamishaye 2I

Basically, you are solving for the theoretical yield and once you get that you divide it from the 25.2g of hydrazine produced then multiply by 100 to receive the percentage yield of hydrazine.

Elizabeth Bamishaye 2I

I am also a little rusty on chemistry but Crash Course Chemistry on YouTube is definitely something helpful to watch.

Elizabeth Bamishaye 2I

I used four sig figs for the whole problem. I generally always use 1.008 as the molar mass of hydrogen so I did not have this problem. It would also help if you calculated it all and rounded the hydrogen to thousandths if you used 1.0078 as the molar mass of hydrogen so you would get 100% at once in...

Search found 54 matches