CHEMISTRY COMMUNITY

Sjeffrey_1C

Decomposition is when a substance is used up. Composition is when it is formed.

Sjeffrey_1C

You look at the rate law to determine the order. So if the rate law is not given, they will just tell you the order so that you know which equations to use.

Sjeffrey_1C

Based on the email sent earlier with a message from someone who had taken it, it seems like you write out your answers. So I'm guessing you do your work on paper and submit pictures? So in that case you'd be fine. Not sure though if that's the case.

Sjeffrey_1C

To find the equilibrium constant, you use the step who's product is the intermediate. So yes, this would always (I think) be the first step. This K will allow you to solve for the concentration of the intermediate in other terms so that you eventually solve for the rate of formation of the final pro...

Sjeffrey_1C

Since the slow step reaction mechanism is rate limiting, I believe this is true.

Sjeffrey_1C

Additionally, an elementary reaction's rate law can be determined by using the stoichiometric coefficients, while a general reaction cannot be.

Sjeffrey_1C

It would probably be good to review, but since we have access to the textbook if needed, I wouldn't focus on it.

Sjeffrey_1C

I'm confused as to when we would use a steady state approximation or equilibrium approximation, versus when we would just use the rate of the slow step? For example, question 7C.7 asks that we find the rate law for the formation of the product, and the answer provided is just the rate law of the slo...

Sjeffrey_1C

The email says "Winter Quarter final exams will be offered remotely." so it seems like he has to find some alternative to having it in person.

Sjeffrey_1C

During the discussion before a test/midterm, it's seemed like my TA has not seen the tests/midterm or hasn't been asked for input. This might be because my discussion is early in the week though, so maybe the exams aren't ready by that point.

Sjeffrey_1C

These weren't covered so I'd assume no.

Sjeffrey_1C

I believe its [anode]/[cathode]

Sjeffrey_1C

Platinum is added to any side of a cell that is missing a solid conductor. This could be both sides, one side, or neither side depending.

Sjeffrey_1C

The equation takes into consideration that the values are reduction potential values, even though half of the reaction is actually in oxidation. Thus, you don't need to convert the values to oxidation potential.

Sjeffrey_1C

The question asks you to predict the cell potential given this diagram: Pt(s)|Cl2 (g, 250 torr)| HCl (aq, 1M) || HCl (aq, .85M) | H2 (g, 125 torr) Based on the left being the anode and right being the cathode, I got 2Cl- -> 2e- + Cl2 and 2H+ + 2e- -> H2. But this doesn't seem to make sense with each...

Sjeffrey_1C

The question asks to list metals in order of increasing strength as reducing agents. I understand you determine this by looking at the standard reduction potentials, but am confused since there are multiple values for each metal depending on the charge. For example, Fe3+ + 3e- -> Fe is -0.04V but Fe...

Sjeffrey_1C

I think I saw some example where they did something similar, so I did this
Oxidation: H2 -> 2H+ + 2e-
Reduction: Cl2 + 2e- -> 2Cl-

Sjeffrey_1C

I was also very confused about this. I think the book's answer may be a mistake.

Sjeffrey_1C

For the equilibrium constant part, this is evident in the equation delta G naught = -RTlnK. I'm not sure how pressure plays a part. Maybe because the equilibrium constant can be calculated with the partial pressures of the products and reactants.

Sjeffrey_1C

At 150 C, I used the equation ln (k2/k1) = delta H naught/R * ((1/T1)-(1/T2)) You plug in the values you already know... For example for part a K1 = 1.07 x 10^-16. T1 = 298. K and T2 = 423. K You solve for delta H naught using the values in the appendix. R is a constant. This leaves you with an equa...

Sjeffrey_1C

Generally it is written as
electrode material (anode) | aqueous solution || aqueous solution | electrode material (cathode)

It seems like the book isn't consistent about this though.

Sjeffrey_1C

The problem is that you are subtracting. Add delta G naught and RTlnQ and you will get the correct answer!

Sjeffrey_1C

I know Test #2 covers the objectives from the second page of the thermodynamics outline, but which homework problems from the thermodynamics outline apply to test #2 content? Thanks!

Sjeffrey_1C

The importance of it is a visual representation of the contents of a galvanic cell, so you can do calculations/write half-reactions/balance redox reactions, etc.

Sjeffrey_1C

In the book, for the purpose of some of the hw problems, we also learn how to indicate the presence of a salt bridge in a cell diagram.

Sjeffrey_1C

Standard reduction potential indicates the likelihood of a substance being reduced. So after reduction happens, I don't think the value would be applicable anymore because the substance has already been reduced. Not sure 100% that's the case though.

Sjeffrey_1C

High potential means a large difference in voltage, low potential means a small difference in voltage.

Sjeffrey_1C

High potential means a large difference in voltage, low potential means a small difference in voltage.

Sjeffrey_1C

F stands for Faraday's constant. Ecell can be calculated by subtracting the two values or will be given.

Sjeffrey_1C

For C, a polymerized substance generally has less disorder, thus less entropy.

Sjeffrey_1C

I definitely agree. I felt very prepared going in and then was very surprised by some of the questions. As said, a lot of it was high level problem solving. So I think it's important to just practice any sort of problem solving, and it will come easier during the tests. Next time, I'm going to go th...

Sjeffrey_1C

The change in temperature is associated with the change in K.

Sjeffrey_1C

Since you're using ΔH naught and ΔS naught, assume conditions are standard.

Sjeffrey_1C

I'm pretty sure they're interchangeable.

Sjeffrey_1C

Yes, since the equation is in equilibrium the concentrations will affect other concentrations to maintain equilibrium.

Sjeffrey_1C

If pressure increases, according to Boyle's law, volume decreases. Thus, there would be more reactant per the amount of space, thus an increase in concentration.

Sjeffrey_1C

Fatemah Yacoub 1F wrote:You must assume that it is constant pressure because otherwise you a reversible system which requires another set of equations.

In this case, you'd use work = -nRT ln (v2/v1)

Sjeffrey_1C

I'm a little confused what you're asking. C refers to specific heat capacity, while R is a constant involved in calculating specific heat capacity.

Sjeffrey_1C

I think the order you'd look would be complexity, volume, temperature, then state. Not sure though. I assume it will be obvious when comparing two molecules what characteristic they're testing for your knowledge of.

Sjeffrey_1C

You use different formulas depending if it is reversible or irreversible.

Reversible: work = -nRT ln (v1/v2)

Irreversible: work = -Pext ΔV

Sjeffrey_1C

Enthalpy in this case is referring to heat at a specific pressure, not a specific volume. Thus, the p subscript instead of v.

Sjeffrey_1C

You set up an ICE box so you end up with equilibrium values of N(CH3CH2)3 = .2 - x NH(CH3CH2)3 = x OH- = x Since the pKa is 11.04, Ka is 10^-11.04 Using Kw = (Ka)(Kb) = 10^-14, you solve for Kb and get .0010964 Then you set Kb = .0010964 - X^2/.2-x to solve for x. Once you have x = .014808, you know...

Sjeffrey_1C

A calorimeter is just used to measure heat. Do you have a specific situation in which you were thinking the mass would be included?

Sjeffrey_1C

For an open system, energy can be changed by adding/removing reactants or products.

For a closed system, energy can be changed by heating/cooling or compressing/expanding the system.

For an isolated system, energy cannot be changed.

Sjeffrey_1C

I believe this is because work is being done by the system to the surroundings, so work is negative.

Sjeffrey_1C

So in this case, visually the reaction's energy looks like this.

Sjeffrey_1C

Because in a stable state, there isn't a release/absorption of energy during the reaction, or it is net 0.

Sjeffrey_1C

PranaviKolla2B wrote:What is the most accurate way to calculate H?

I assume Hess's method is most accurate.

Sjeffrey_1C

I don't think that is the case. I may be wrong, but I remember in lecture it being said that you can use bond enthalpies for liquids or solids but you need to add the enthalpy of the phase change, or something along those lines.

Sjeffrey_1C

PranaviKolla2B wrote:What does each term relate to in the equation deltaG= deltaH-T*deltaS?

I believe delta G is gibbs free energy, Delta H is enthalpy, and delta S is entropy. T is temperature.

Sjeffrey_1C

You subtract or add x from the original concentration. X is either positive or negative depending on the direction of the reaction.

Sjeffrey_1C

To be safe, I'd say just don't approximate on the test. But I'm not sure what you'd do in the case that you do.

Sjeffrey_1C

(a) in an experiment, 2.0 mmol Cl2 (g) was sealed in a reaction vessel of volume 2.0L and heated to 1000. K to study its dissociation into Cl atoms. Use info in table 5g.2 to calculate the equilibrium composition of the mixture. I keep getting the wrong answer. The table gives the k value to be 1.0 ...

Sjeffrey_1C

Liquids and solids are essentially incompressible, unlike gas, which is compressible.

Sjeffrey_1C

It basically explains how a reaction will maintain stability.

Sjeffrey_1C

In class, he explained that if the pressure of a reaction vessel is increased by adding inert gas, then moles of reactant, product and volume are constant, so there's no change in [R] and [P].

Sjeffrey_1C

He basically said that if volume decreases and there are more moles of gas on left, then the reaction shifts left, or if volume decreases and there are more moles of gas on the right, then the reaction shifts left. But this logic is incorrect.

Sjeffrey_1C

Additionally, if K and Q aren't equal, you know the equation is not in equilibrium. So If Q < K at some time during reaction, then [R] > [P] and the forward reaction is favored. If Q > K at some time during reaction, then [P] > [R] and the reverse reaction is favored.

Sjeffrey_1C

Lewis refers to eLectrons. Bronstead refers to protons. Bronstead acids or bases are always a Lewis acid or base (respectively). Aka a bronstead base is also a lewis base. But the vice-versa is not true. Ex. BF3 is a lewis acid (it accepts electrons) but is not a bronstead acid (doesn't have protons...

Sjeffrey_1C

A ligand forms a coordinate covalent bond with a transition metal cation, so I guess if CO did this it would be considered a ligand. A ligand is anything directly bound to the central atom.

Sjeffrey_1C

In the case of H2SO4, these are the reactions that show this:

H2SO4 + H2O -> HSO4- + H3O+

HSO4- + H2O -> SO42- + H3O+

overall equation: H2SO4 + 2H2O -> SO42- + 2H30+

Sjeffrey_1C

Also don't forget if a cation is present outside the brackets... like sodium in Na[Fe(Ox)2(OH2)2)]... name it first. Ex. Sodium diaquabisoxalato ferrate (III)

Sjeffrey_1C

An electron withdrawing group pulls electrons towards it from another source. It is a way of determining the relative strength of a conjugate base. E- with drawing groups delocalize a neg charge.

For example HClO or HBrO

Sjeffrey_1C

I'm assuming information after Test #2 will be the most emphasized (since it's the only content we haven't been tested on), but I am not 100% sure. Maybe ask a TA during discussion.

Sjeffrey_1C

The strong acid I see most often is HCl!

Sjeffrey_1C

If you draw the Lewis structures, look at which have lone pairs. This will give you an indication of which will donate electrons (lewis base) and which will accept electrons (lewis acid)!

Sjeffrey_1C

You definitely will have to know molecular geometry names and bond angles!

Sjeffrey_1C

A bronsted base accepts a proton and a bonsted donates a proton. Thus, NH3, in most reactions, is a bronsted base.

Sjeffrey_1C

Generally atoms with polarizing power have small atomic radii and are partial negative. Atoms that are polarizable have large atomic radii and are partial positive.

Sjeffrey_1C

Also keep in mind that the dipole moment is a permanent dipole moment, meaning its polar.

Sjeffrey_1C

Yes, and since it's polar that means there's a separation of charges. The S is partial positive and the Os are partial negative.

Sjeffrey_1C

Also keep in mind that although the trend is the same for both, ionization dips at oxygen (meaning oxygen is an exception to the trend) while electronegativity trend continues through oxygen.

Sjeffrey_1C

Each IMF has defining characteristics. This image might help.

Sjeffrey_1C

Look at the charges. If there are partial positive or partial negative components, it's likely that dipole moments will occur.

Sjeffrey_1C

Also resonance structures are used to determine the true most stable structure. So if multiple resonance structures contribute equally, the structure will be a hybrid of all the resonance structures. This is why sometimes if there's resonance, bond lengths will be between that of a single bond and a...

Sjeffrey_1C

I think when he says there isn't a curve he means a bell curve, meaning there isn't a specific grade distribution he wants. But it does seem like he will be curving the class, or not using a standard grading scale. This way if the overall grades aren't very high, he will shift the cutoffs so that pe...

Sjeffrey_1C

Just to clarify so there's no confusion, someone above said next week but it is actually in 2 weeks! Its the week of the 18th - 22nd!

Sjeffrey_1C

Additionally, these are basically the same trends as electron affinity and ionization energy. (Although ionization energy does dip for oxygen)

Sjeffrey_1C

The main goal is to minimize charges. So however you can get the charges to be lowest, while still adding up to the total charge and following octet rules etc. Placing larger charges on more electronegative elements is also important.

Sjeffrey_1C

Could someone give me a quick reminder of the units different things should be in for the constants and equations sheet? Like when to use nm vs m or kg vs g. Just for the most important equations! Thank you!

Sjeffrey_1C

When it says questions come from hw and online, does this mean word for word or just similar format/topic?

Sjeffrey_1C

Would someone mind posting the midterm review session times? I didn't write them down.

Sjeffrey_1C

This confused me at first too. If writing the electron configuration for Ga itself you would first fill 4S before 3D, which made me believe that when Ga loses 3 electrons to become 3+, electrons would be removed from 3D since it was filled last. This though, is not the case. Electrons are removed fr...

Sjeffrey_1C

You can tell that all of these have 0 unpaired electrons because the last orbitals are full in each case.

Sjeffrey_1C

Generally, if there is a singular of one element and multiple of another, the singular one will go in the middle. For example, CCl4, NF3, H2O, etc.

Sjeffrey_1C

https://www.khanacademy.org/science/che ... structures This video is also helpful for drawing lewis diagrams!

Sjeffrey_1C

These are important because the more you know about the position (delta x) of an electron, the less you know about its momentum (delta p), and vice-versa. Thus, it's impossible to know both an electron's position and momentum.

Sjeffrey_1C

As said, we can probably do some quantum problems. But since the past 3 weeks have been quantum, it might be good to move onto some new content in preparation.

Sjeffrey_1C

I think only two non-metals. A non-metal and a metal would form an ionic bond. Two metals would form a metallic bond.

Sjeffrey_1C

My TA mentioned that the 8th group is composed on octets. These are noble gases I believe.

Sjeffrey_1C

I'm not sure about the order they fill. But yes, the 4f subshell can hold 14 electrons as it has 7 orbitals, and each can hold 2 electrons. (And each electron has to have an opposite spin bc of Pauli)

Sjeffrey_1C

This is really important, thanks! I feel like most high school teachers didn't make those distinctions very clear, so it's easy to get confused.

Sjeffrey_1C

A wave function I believe gives you specific info regarding the wave's properties.

Sjeffrey_1C

On that note, does anyone know which topics will be covered which weeks? I expected to see weeks on the syllabus next to each new topic, but didn't notice anything like that.

Sjeffrey_1C

I believe we went over it earlier last week but I might have just read about it in the textbook. It refers to something that can absorb all light frequencies.

Sjeffrey_1C

Also during lecture, he provided some specific things you just need to know regarding orbitals. I'd recommend going through those and making sure you know those as they might help.

Sjeffrey_1C

I had a similar issue where I kept getting an answer that was slightly off. Make sure you check all your conversions. For the mass, make sure you convert to grams, and then kg... you should get .145745kg. For the velocity, you need to convert miles to meters and hours to seconds. Remember that 1 met...

Sjeffrey_1C

To add on, the concepts mentioned above are that in constructive interference, the amplitudes of the two waves summate (add together). In destructive interference, the amplitude's of the two waves cancel each other out.

Sjeffrey_1C

As others explained, electrons have wave-like properties, which are explained in the textbook, as well as particle properties. This concept as a whole is know as wave-particle duality, meaning electrons cannot be explained as merely a wave, or merely a particle, but as a combination of the two.

Sjeffrey_1C

As said above, you shouldn't convert to sig figs till the end so storing in your calculator as you go is very helpful. You can generally assign values to a letter so you can call it back later. For example, on the TI-30X, once you have a value you click "STO" and select a a letter to assig...

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