CHEMISTRY COMMUNITY

ramiro_romero

I do not unerstand why the cell potential for 10 is 0.43 volts when the oxidation reaction of silver is reduced. I thought when we reversed an oxidation reaction we changed he sign of its cell potential. Therefore, I thought the answer was (1.23V)-(-0.80V)= 2.03V

ramiro_romero

Maybe double check the numbers you're using again? when I do 0.15*-2*96485/(8.314*298) I get the right answer How? I am doing the same calculation and I keep getting lnQ= -11.6. And for the concentration of Fe 2+ I get 9.2 x 10^-6. I've gone through the problem several times and I keep getting this...

ramiro_romero

Hi all, In book problem 7.17, the textbook gives us the illustration of a reaction (A->D) and asks whether using a catalyst to accelerate the third step only would change the overall reaction rate. According to the solutions manual, it will not. I understand that the first step is the rate-determin...

ramiro_romero

Yes, the exam starts at 11:15am and ends at 2:45pm. Our final is 3 hours and 30 minutes. Thanks to Lavelle, the exam will be through CCLE, so no need to install apps or print anything

ramiro_romero

well we know the ideal gas law inversely relates pressure with volume, pv=nRT, meaning as one increases the other decreases.

ramiro_romero

How does oxidation number of an element relate to how many electrons it will lose or gain? I am confused on how to calculate when in acidic or basic solutions it changes the net charges. When balancing half reactions of a redox reaction, an elements change in oxidation number will tell you whether ...

ramiro_romero

I think about individual particles travelling in a jar. The chances of 4 particles simultanesouly colliding with each other is far less likely than if 2 particles were to collide with each other.

ramiro_romero

A catalyst does not affect the value of the equilibrium constant or the position of equilibrium. This is because adding a catalyst to a chemical reaction speeds both the forward and reverse reaction by the same rate.

ramiro_romero

I'm pretty sure it's going to be taken by everyone at the same time. Lavelle sent an email about a simulated exam taken today.

ramiro_romero

I believe you are able to use any resources from lecture notes/discussions, and the textbook.

ramiro_romero

You'll need especially when determing equilibrium constants. Whether it be understanding logs and antilogs, or determining the value of x from an ice table.

ramiro_romero

This is correct, at equilibrium delta G equals zero and so does the cell potential.

ramiro_romero

The relationship between cell potential and gibbs free energy can be shown through the expression, deltaG = -nFEcell. You can also determine E cell by dividing deltaG/-nF.

ramiro_romero

Thank you! What I did was I reversed the half-reaction for Fe and changed the sign of its cell potential. So I did (1.61V) - (+0.44V) to get 1.17v for the cell potential.

ramiro_romero

A sat bridge is usually comprised by a strong electrolyte which is further made up of ions. This salt bridge allows for the neutralization and flow of electrons between a cathode and anode solution. Without a proper salt bridge, the cathode solution will become negatively charged while the anode sol...

ramiro_romero

For question 2b, I got E cell = 1.17V. delta G = -226 kJ/mol. and for K I got 91.1. I just want to make sure these answers correlate with what my peers got.

ramiro_romero

Thank you for everything and congratulations! Is there a specific time when we should expect the solutions?

ramiro_romero

When work is done on the system, it is positive. When work is done by the system onto surroundings, it is negative. If the system exerts force on its surroundings, the system itself loses energy, explaining the negative value.

ramiro_romero

Diluting the cathode solution decreases the value of the cell potential, while diluting the anode solution increases the value of the cell potential. This is because of the differences between the anode and cathode when finding Ecell=Ecathode-Eanode.

ramiro_romero

I specifically remember Lavelle mentioning in one of his lectures that if the value for K is less than (10^-3), then it is okay to approximate for x. Usually, just consider your initial moles of reactant and that the x value should leave no significant difference after subtracted from the moles of r...

ramiro_romero

Probably not, usually you need to use an ice table to find moles or concentration of products given the moles or concentration of reactants. You can also sometimes just multiply it according to the ratio of their coefficients.

ramiro_romero

Yes, you want your cell potential to be positive. Therefore, the largest value will be the cathode. Keep in mind the formula, Ecell = E cathode - E anode.

ramiro_romero

I have a lot of trouble balancing the reactions in a basic solution all throughout. I remember my TA showing us a method of adding hydrogen protons the same way as if it were an acidic solution, but in the end, swapping out the protons for hydroxide ions and balancing the reaction from there. Does t...

ramiro_romero

n is how many moles of electrons are transferred. First you need to determine the half rxns, then balance the oxidation and reducing reactions so that the electrons cancel out (and don't appear in overall rxn). Finally, once both half rxns are balanced, the coeffecient of both electrons is your valu...

ramiro_romero

I'm not sure if he got it from a specific place in our course material. I don't recognize it from the textbook.

ramiro_romero

K is going to become inverse, so 1/K.

ramiro_romero

Thank you, I got the answer 1.54 but the textbook says 1.54 x 10^-2. I'm pretty sure it's a flaw in the solutions

ramiro_romero

You wrote the question for ii wrong. The question asks what will happen to the amount of NO2 if SO2 is removed by condensation (both these compounds are products of the rxn). Removing SO2 requires the rxn to move forward (favoring and producing further product) to return to equilibrium.

ramiro_romero

For problem 5.35 in the textbook, how do we calculate the equilibrium constant from the given plot? I got the balanced rxn [2A -> B + 2C] If somebody could help for the equilibrium constant in part B I would really appreciate it

ramiro_romero

Each order rxn will use a different formula for the determination of its rate. Of course, they will also use different formulas to determine its half life

ramiro_romero

Yes, but it is important that you have AT LEAST 50 for last quarter and 50 for this quarter since your TA will be checking the dates posted. So don't try to pass with only completing 40 this quarter because you did an extra of 60 last quarter.

ramiro_romero

Correct, if the anode value is larger than the cathode, the reactants/ reverse reaction will be favored. You should also know that this means a negative cell potential, and a positive delta G

ramiro_romero

For example 6L.2 and 6L.2A in the 7th edition textbook, we are asked to write a cell reaction from an electrochemical cell. After writing the reaction, we are asked to determine whether or not the reaction is spontaneous dependent upon a positive cell potential. I don't understand how to answer this...

ramiro_romero

You need the moles of electrons transferred in a cell in order to cancel out the moles in Faraday's constant (mol x C/mol = C). To find the moles of electrons transferred, you write out both the oxidation and reduction half reactions, balance the reactions so the electrons cancel on both sides, and ...

ramiro_romero

Cl2 is both the oxidizing and reducing agent in this reaction.

ramiro_romero

Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O...

ramiro_romero

Part D of this question asks us to "combine the half-reactions to produce a balanced redox equation". From the oxidation reaction, I got 2 electrons on the product side and from the reduction half-reaction, I got 6 electrons on the reactant side. The correct answer for part d multiplies th...

ramiro_romero

Hey can someone link or direct me to the step-up sessions. Couldn't find them on the website, thanks

ramiro_romero

I do not think so, if its a single constant value for volume you would be able to find moles. However, I do not see how you could find initial and final concentrations of a reaction only from that information.

ramiro_romero

Compressing a gas means increasing the concentration of each gas. The reaction would favor the side with less moles of gas because it reduces the number of particles in the gas. Since volume and pressure are inversely related, when compressing a gas you are decreasing the volume and this is why the ...

ramiro_romero

The Question states "Water expands when it freezes. How much work does 100. g of water do when it freezes at 0degreeC and pushes back the metal wall of a pipe that exerts an opposing pressure of 1070 atm? The densities of water and ice at 0degreeC are 1.00 g/cm^3 and 0.92 g/cm^3, respectively.&...

ramiro_romero

[H+]= 10^-pH [OH-}= 10^-pOH pH=-log[H+] pH= 14-(-log[OH-])

ramiro_romero

Lewis acids/bases accept/donate electron pair. Bronsted acids/bases accept/donate hydrogen protons

ramiro_romero

You find a conjugate base/acid by donating/accepting the acidic hydrogen proton in the reaction.

ramiro_romero

Yes, and also when describing the hybridization of pi bonds they will be the same. Ex. in a triple bond, the descriptions of both pi bonds would be the same and would exist in the p orbital

ramiro_romero

it will create a bent molecular shape

ramiro_romero

Yes, you count all regions of e- density when considering the orbitals of hybridization. Say an atom has 2 bonds and 1 lone pair, then the hybridization would be sp2.

ramiro_romero

Ionic bonds will always create the highest boiling points, hydrogen bonds are just exceptionally strong dipole-dipole interactions. Follow the general rule ionic>hydrogen bonding>dipole dipole>London dispersion

ramiro_romero

Just remember to write them in numerical order, this is because of the energy levels.

ramiro_romero

Ml can be within the range of -l to l. You look at the placement of the last electron to determine the value of Ml. For example, if an electron is in the p orbital in the 2nd slot, its Ml would be 0 because it is in the middle of the range from -1 to 1.

ramiro_romero

No, it will be given on the reference sheet so you don't have to concern yourself with memorizing the number.

ramiro_romero

For strength in bonds will be most likely for the final. To the left and down are the trends for increasing atomic radii. The larger the distance between the centers of the atoms, the weaker the bonds. The weaker the bond, the stronger the acid and base (because a strong acid and base will completel...

ramiro_romero

Focus on the last electron when calculating quantum numbers. N and L can be calculated by simply looking at the primary quantum number and the angular orbital number. Ml and Ms will tell you the spin and specific orbital of that electron.

ramiro_romero

The graph looks the way it does because once the threshold frequency is met, electrons are instantly emitted. The same number of electrons are emitted assuming the intensity is constant. An increasing frequency (after the threshold frequency) has no effect on the number of photons emitted, they will...

ramiro_romero

please quote my question so i get a notification, thanks in advance! i know there is something wrong with my thought process, but i cannot figure it out. my thought process is that if a base is weaker, less oh- will be dissociated out. taking the -log of that smaller number will result in a smaller...

ramiro_romero

Intensity means the number a photons, a higher intensity means a higher number of photons. Unless the threshold frequency is met, electrons will not be emitted regardless of the lights intensity (number of photons). If the threshold frequency is met, increasing intensity will then increase the numbe...

ramiro_romero

Thank you all for the replies -- I get why it goes across now, but I still don't understand why it goes straight up? When the threshold frequency is met, the electrons are emitted instantly. If it were to be linear it would assume that electrons are emitted as frequency increases which is not the c...

ramiro_romero

Liner shapes will have 2 bonding atoms and 1 or 2 lone pairs on the central atoms. Linear shapes are also 2 bonding atoms but will have either zero or 3 lone pairs on the central atom. AX2E, AX2E2 = bent. AX2, AX2E3= linear. Also, bent shapes are more likely to be polar than linear shapes

ramiro_romero

E=hv will give you the energy of a photon. You should use it when finding frequency given energy or to find energy when given frequency.

ramiro_romero

Just make sure you eat an apple, glucose molecules will store in your body and fuel your brain ability!

ramiro_romero

You should attend office hours and review sessions. They really make all the difference!

ramiro_romero

Why is it that molecules with AX2E2 VSEPR formula are bent with the electrons beside each other? For example, ClO2 is bent, but I'm wondering why the Lone pairs on Cl aren't opposite of each other. Instead, they are closer together and push the Oxygen atoms closer together. I don't understand why lo...

ramiro_romero

Dissociation energy is always positive because energy is always absorbed for dissociation. To break a bond requires an additional absorption of energy

ramiro_romero

Yes, ionization and electronegativity of atoms follow the same periodic trends. Remember however that there are exceptions to ionization trends.

ramiro_romero

From periodic trends, it can be incorrectly assumed that Oxygen has a higher ionization energy than Nitrogen. However, Oxygen wants a full shell so is more willing to give up its electrons (requiring less energy to its electrons and therefore giving Oxygen a lower ionization energy than Nitrogen).

ramiro_romero

When relating energy of a photon to frequency, we use Einsteins equation. And if kinetic energy=0, we can also use it for the energy required to remove an electron. So, E = hv is used for relating the energy of a photon to its frequency, (remember that energy of a photon and its frequency are direct...

ramiro_romero

Clara Cho wrote:If I'm solving for the energy of a photon, would I use J/mol or J/electron?

J/mol of photon

ramiro_romero

For the sake of Heisendbergs equation, the change in velocity is important. A singular value for velocity is more relevant to debroglies

ramiro_romero

No, the threshold is not a constant value. The question should specify your threshold

ramiro_romero

If you are relating wavelength to either mass or velocity, then you know you should use DeBroglies equation. Please keeep in mind that momentum uses Kg for mass and m/s for velocity.

ramiro_romero

On the midterm we were asked to provide "all" quantum numbers. To my understanding, can someone just sum up the quantum numbers we were supposed to use (Not the actual qualitative answer).

ramiro_romero

Should we be writing our midterm with pen or pencil?

ramiro_romero

No testing on this information, it's pretty complex too so I wouldn't spend too much time trying to deeply understand it if you won't be tested on it. Basic concept of it should be enough for the sake of your general chemistry understanding

ramiro_romero

To find photons, you first multiply Watts by 2 seconds and get 64 J/2sec as the Total Energy. Then, you find the energy per photon by using the formula E=(hc)/lambda, (energy per photon should be 4.717 x 10^-19). Divide energy per photon by total energy, and you get the number of photons [1.35 x 10^...

ramiro_romero

You convert from nm to m by mnultiplying by 10^-9. However, I am much more concerned as to how you got your value for the energy of a photon. Your kinetic energy is specified at a value of zero. You should've gotten the work function equal to the energy of the photon (since there's no excess energy)...

ramiro_romero

Adding on, once the orbitals in a subshell are filled, the electrons will then start spinning in pairs. It would be done in a manner minimizing energy and it'll probably make more sense when we discuss valence electrons role in bonding.

ramiro_romero

Most definitely, hw problems were specified to quantum world questions last week and no update has been sent out regarding hw questions. So it's safe for us to assume that the same applies for this week.

ramiro_romero

The intensity of the light does not depend on the location where light hits. Increasing the intensity of light is simply increasing the number of photons.

ramiro_romero

Yes, at the very least understand its purpose/application.

ramiro_romero

I converted KeV directly to Joules so I also don't understand the ned to multiply KeV by 10^3. However, I am also very confused as to how these units cancel and my final answer was lambda=8.831pm. I multiplied plancks constant by c and got 1.9878x10^-25. I then divided the product by energy in joule...

ramiro_romero

Constructive diffraction results from waves in phase and destructive results from waves out of phase. Diffraction patterns result from interacting waves (the 2 types of interference previously mentioned), meaning that e- must have wavelike properties.

ramiro_romero

A complete combustion will always be in the following format. (Fuel) + O2 -> CO2 + H2O.

ramiro_romero

You should be fine not including the units for calculations (simply because it consumes too much of your limited time in tests). However, you should include units for every result of calculation so that you and your T.A. can clearly identify where your numbers are coming from

ramiro_romero

I undertstand that diffraction patterns result from constructive and destructive interference but what is the difference between the 2 and how do I identify them?

ramiro_romero

Is there a set time when we should complete each topic or are we allowed to study them whenever we feel fit?

ramiro_romero

I only have the 6th edition and the class website specifies a requirement of the 7th edition. However, comparing mine to the other textbooks the HW questions are exactly the same and there don't seem to be any major discrepancies. If you have the 6th edition you can assume you'll be fine in terms of...

ramiro_romero

I own the 6th edition and I didn't see chloride mentioned anywhere in the problem. However, this is how I answered. First, I subtracted the molar mass of (OH) from the molar mass of the metal hydroxide: (74.1) g/mol metal hydroxide - (32) g/mol (OH)= (42.1) g/mol metal Next, I added the molar mass o...

ramiro_romero

Usually a question will ask you for the yield in grams. In any case, both theoretical and actual yields must be in the same units then multiplied by 100 for the correct percent yield.

ramiro_romero

An empirical formula shows the relevant number of atoms a molecule has *ratio between atoms*, while a molecular formula shows the actual number of atoms *existing atoms*. The molecular formula will always be a multiple of the empirical.

ramiro_romero

When given wavelength and moles of atoms, how do I find energy emitted by that atom? specific question asks to find energy emitted by 1.00 mol of sodium atoms emitting light at a wavelength of 589 nm, but i'd prefer a response with the procedure instead of the answer to the specific question.

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