CHEMISTRY COMMUNITY

LReedy_3C

Electrons will always flow from the anode to the cathode, and movement through the salt bridge depends on of the charge of molecules getting added to the solution (cations will flow towards increasing negative charge/decreasing positive molecules, and anions will flow towards increasing positive cha...

LReedy_3C

14a came out about a week after the quarter ended so it will probably be by the end of spring break

LReedy_3C

This is a great explanation, it ultimately comes down to plugging both into that equation and turning it into a ratio

LReedy_3C

As long as you recognize what the ratio of k1/k2 would be, you don't necessarily need the exact values.

LReedy_3C

I was also very confused on this one, thank you for the explanation.

LReedy_3C

Does anyone have any advice on how to know which equations/variations to use? I'm having trouble remembering them just from the given equation sheet.

LReedy_3C

Women in Science society started last year! They'll probably get going again when we're back in person.

LReedy_3C

I think it might have something to do with it being an aqueous solution but I'm not quite sure. I ended up leaving it in because I didn't think I could solve for it.

LReedy_3C

For me it just comes out to moving carefully and writing absolutely everything down. After splitting into half reactions, I find it a lot easier but getting to that point takes a lot of thought for me.

LReedy_3C

Water can be added because the solutions are aqueous I believe. Hydrogen atoms or OH molecules can be added because if it is in an acidic/basic solution these molecules should be in abundance.

LReedy_3C

I'm reading Lonely Planets right now and its really good, all about the possibilities of life in space

LReedy_3C

Handwritten is great for me, because I can organize it how I want to and I have to be aware of what I'm writing. I also like being able to highlight past notes when I'm reviewing, and physically doing that helps me study more than on a computer.

LReedy_3C

Set activities that aren't school have been super helpful for me, and setting time as a break to do something like go for a walk or watch an entire movie at one time instead of just going on my phone randomly while I'm working has really helped me be more productive during time when I'm working and ...

LReedy_3C

Review sessions are super helpful! Definitely go if you can, there was an email with all of them sent out this weekend.

LReedy_3C

Why do 2H+ and 1 OH- combine to form one water molecule? wouldn't that leave hydrogen still

LReedy_3C

The list of basic oxidation rules helps me a lot while balancing.

LReedy_3C

I'm also having issues with this one, I'm going to try changing the order of reactants next.

LReedy_3C

I guess they could relate to biological cells in that they are a single electrochemical unit, and cannot be split into smaller parts and retain function. Also because multiple cells can be used to make up batteries.

LReedy_3C

I can't figure out where I'm going wrong with this question: For a particular redox reaction, NO is oxidized to NO−3 and Ag+ is reduced to Ag . Complete and balance the equation for this reaction in basic solution. The phases are optional. Currently I have: 4OH−+3Ag3++NO⟶NO−3+2H2O+3Ag Why is this in...

LReedy_3C

I know that Q=0, but I don't know if work can still be done in other means

LReedy_3C

I was also confused by this, I think it's just a flaw in the question

LReedy_3C

I think it's just relative to the way the reaction is presented to you

LReedy_3C

That turns the answer into 3000J because when you take two sig figs calculating the answer in KJ, it changes the number of joules when you multiply KJ by 1000.

LReedy_3C

Concentration of products (to the power of their coefficients) over concentrations of reactants (to the power of their coefficients), or I think you can do the same with partial pressures. It's the same as calculating the equilibrium constant the system's just not at equilibrium

LReedy_3C

I agree, I've read that the anode is always on the left and the cathode is always on the right

LReedy_3C

I think any system open to the atmosphere you can assume p = 1 atm because that's the standard

LReedy_3C

This question involved sorting phase changes by delta h and delta s, and I was wondering if there would ever be a situation where a phase change would have a delta h and a delta s with different signs?

LReedy_3C

I've been using the equations worksheet on Lavelle's website!

LReedy_3C

I have her right now, and she really cares a lot about student learning, but it's really easy to loose points over little things. Definitely a bit of work, but overall she wants you to do well.

LReedy_3C

Yep! perfect explanation

LReedy_3C

I think the end of this week would probably be a good time to have them done by

LReedy_3C

I'm pretty sure kJ/C and kJ/K are interchangeable because they have the same unit value just with different zero points.

LReedy_3C

I think it's because you can consider the organism as the system and the atmosphere as the surroundings, so its like an open system.

LReedy_3C

I'm still kind of confused on it, but residual entropy is independent of temperature and present at absolute zero, so thermal entropy and residual entropy combined make up total entropy. It is also defined as the difference in entropy between a non-equilibrium state and crystal state of a substance ...

LReedy_3C

and open system will have constant pressure because any increase can leave the system and any decrease can be replaced

LReedy_3C

The second part involves a molar conversion. 4 mols of carbon react with 1 mol S8 in the reaction, and 151g of carbon is reacting, so you can convert this to mols, divide by 4, and multiply by the delta H. For the third part, you can do the same process but in the other direction, dividing 397 by th...

LReedy_3C

To start this one, I wrote out which bonds were breaking or forming for each reaction, and in the case of the one you added, you have to recognize that its a combustion reaction because its reacting with oxygen

LReedy_3C

To start with this one, remember that weak bonds take a little energy to break while strong bonds take a lot, and forming weak bonds gives off a little energy while forming strong bonds gives off more energy

LReedy_3C

He said he was doing thermochemistry and thermodynamics a little out of order (at least compared to the book), so I think they are going to overlap a little over the next few classes.

LReedy_3C

I don't think he's said anything yet!

LReedy_3C

Percent ionization is how much has disassociated, and I believe percent deprotonation reflects the amount of acid donating protons, and percent protonation the amount of base protonated (accepting an electron)

LReedy_3C

He just sent out an email listing 4D.15, 4D.17, 4D.19, 4D.21, 4D.23, 4E.5, 4E.7, 4E.9, 4.29, and 4.31 as some questions that we should know!

LReedy_3C

His latest email also listed 4D.23, 4E.5, 4E.7, 4E.9, 4.29, 4.31 as relevant questions

LReedy_3C

I used the graph given to get the equilibrium concentrations and solved for the coefficients and K

LReedy_3C

If they come up on an exam, conversions will probably be provided

LReedy_3C

I think we were just told to assume that a liquid is a pure substance while we can't make that assumption about a gas, and that's why aqueous solutions are considered but the liquid they are assumed to be in isn't

LReedy_3C

Solids and liquids are omitted from equilibrium equations, so you never have to consider their concentrations

LReedy_3C

I'm having a similar problem, unsure if we are supposed to be able to calculate it somehow or if I'm looking in the wrong place

LReedy_3C

I had trouble with this one too, and my error was in forgetting to use 2x instead of just x when calculating the final concentration

LReedy_3C

Once you make that change, you go about solving for x like we did in class using the quadratic formula

LReedy_3C

I was also confused, but from what I understand a solid is considered a pure substance, so the concentration can't meaningfully change. With aqueous solutions, they are counted but the h2o it is assumed to be in (what makes it aqueous) is not included.

LReedy_3C

yes, but only if k is 10^-4 or smaller

LReedy_3C

I believe so, in lecture he said that it is the only thing that changes k, and our textbook describes k as temperature dependent.

LReedy_3C

additionally, you cant have a value for the change in concentration that is larger than the initial concentration of reactants

LReedy_3C

I believe so! You would calculate the concentrations of each and then you could use an ice table to get to the final answer

LReedy_3C

I think the biggest thing is that pi bonds can't rotate, so molecules with pi bonds can't change shape, and can't rotate to bond with things unless the pi bond is broken.

LReedy_3C

Chelating is a specific shape polydentate complexes can form, in which two bonding sites are positioned in order to be able to form a ring. It's common in nature, and I think some chelating molecules (EDTA) can be used to purify water because they can bond cations tightly.

LReedy_3C

I think it has to do with the types of bonds holding them. Sigma bonds are able to rotate freely, but pi bonds cannot rotate.

LReedy_3C

Atomic radius is half the difference between neighboring atoms, while ionic radius is one half the distance of two atoms joined by an ionic bond.

LReedy_3C

Greater electronegative atoms delocalize the charge more effectively across the molecule, and as a result the bonds are easier to break bonds, and therefore are more acidic.

LReedy_3C

Inorganic acids also usually involve more electronegative atoms

LReedy_3C

Cl is better at delocalizing the charge of the molecule. The high electronegativity of Cl stabilizes the negative charge.

LReedy_3C

Following periodic trends is a good way to determine this. Atoms with smaller radii will form weaker acids because the bonds are stronger.

LReedy_3C

Additionally, the bond strength affects an acids strength, because strong acids will have weaker bonds that allow them to lose protons more easily, and fully dissociates in water.

LReedy_3C

I think most polydentates form chelate shapes, so if you see this shape there is more than one bonding site being occupied. I'm not sure how to determine the exact number.

LReedy_3C

In the example of water, the positive H atoms can bond to either pair of lone electrons, meaning there are four points that can participate in hydrogen bonding.

LReedy_3C

Stronger IMFs mean a higher melting point, so higher london forcers would increase boiling point. But London forces are sill weak compared to dipole-dipole or hydrogen bonding.

LReedy_3C

Cis- molecules refer to a layout of a molecule in which like atoms are on the same side, while trans- refers to when they are on opposite sides of the molecule.

LReedy_3C

The shape affects the type and strength of forces in the molecules. Because NO2 is polar, it has dipole-dipole interactions rather than just London forces, and these stronger forces make the boiling point higher

LReedy_3C

In part C of question 3F.5, why exactly are the London forces in CHI3 stronger than those in CHF3? Is it because of the size of the atoms or the polarizability?

LReedy_3C

Also, if the shape/geometry of a molecule is even (like CH4 where they are spread equally around the central atom) it will not be polar because the charge/electronegativity is equal on all sides

LReedy_3C

It doesn't matter because all bonds are treated as equivalent, and the resonance would affect all the bonds equally in a real molecule anyway

LReedy_3C

There is no calculation involved, just recognizing that because it is a central atom with 5 regions of electron density, it is a trigonal pyramidal shape, and so all the OSCL bonds would be about 109.5 degrees

LReedy_3C

Being able to visualize how the molecules look 3 dimensionally is really helpful, and being able to apply that to the Lewis structure to understand where in the molecule each atom it is. There are diagrams in the book for each shape we've talked about

LReedy_3C

For part c), would the bonds be exactly 109.5 degrees because it's trigonal pyramidal, or does the fact that its 3 bonds and a lone pair affect the angle?

LReedy_3C

I have heard that one strategy when drawing a Lewis structure for a radical is to start filling in electrons on the most electronegative atom first, and then moving on from there, so the least electronegative atom gets the radical because the more electronegative atoms want to pull the electrons more.

LReedy_3C

The difference in electronegativity is the main thing that makes Mg-O more polar. This difference causes the electrons to be more strongly pulled to one nucleus rather than the other, giving the bond ionic character and making it polar.

LReedy_3C

induced dipole happens in London Dispersion forces, but it is very weak and short lived. It is much stronger in a dipole-induced-dipole scenario, where a polar molecule interacts with a non-polar molecule and creates an induced dipole.

LReedy_3C

To go about solving all of these, you would first draw a structure that obeys the octet rule, and then use double bonds to lower the formal charges to as close to zero as they can be. Then draw resonance structures for all of them.

LReedy_3C

You have to use the periodic trend of electronegativity. The trend is the same as ionization energy (growing left to right, bottom to top). Comparing which elements are farther apart will be the way to tell which has the larger electronegativity difference.

LReedy_3C

You can also count the number of electrons in the highest s and p orbitals to find the number of valence electrons

LReedy_3C

It's because the d block is valence in c), but in a) it is full, so the energy is lower and it is not a valence shell. In a), only the s and p orbitals contain the valence electrons.

LReedy_3C

I think it has to do with the d orbital being a lower energy level when it is full, so its no longer a valence shell.

LReedy_3C

The two aren't the same because once one valence shell is empty, the next shell becomes the valence shell. The only time an ion would have zero valence electrons is if it lost all its electrons, like in the case of H+

LReedy_3C

4s has a higher energy than 3d, so the s orbital loses its electrons first.

LReedy_3C

atomic radii is calculate by half the distance between two nuclei, and the other two are the same concept except for elements in ionic and covalent bonds.While the trend for all of these is the same, it isn't systematic in the d-block. Also, atomic radii of cations are always smaller than their pare...

LReedy_3C

The number is the principle quantum number, and the letter corresponds to the orbital angular momentum quantum number. If l=0 its s, l=1 its p, if l=2 its d, and if l=3 its f. From there you can find the answers for all of them.

LReedy_3C

It requires less energy for the electrons to be opposite, so because they are both up it must be excited.

LReedy_3C

Although they have the same number of electrons, the charge is different, and the interaction with the nucleus is different because some will have more protons and hold the electrons tighter.

LReedy_3C

A higher electron affinity means a lot of energy will be released if an electron is added. Because of this, high EA means it's more likely to add an electron.

LReedy_3C

4g and 2d cannot exist, because the value of l can be 0,1,...n-1, so for both of these the l value is too high to exist for that n.

LReedy_3C

The notation is first the n value, and then s, p, d, f, ect based on the l value (eg l=0 is s). The number of orbitals is found by options for the magnetic quantum number, which are -l,...0,....+l, so they change based on the l value

LReedy_3C

It's the electron density, which is what is drawn in diagrams

LReedy_3C

A more complicated formula is needed for many electron atoms because the electrons shield other electrons in the atom, so the energies of the different levels are different.

LReedy_3C

Also, the number of nodal planes is related to the value of l. As it increases, there are more conical nodes.

LReedy_3C

When plugging into the equation, I found it was much easier if I used the fact that the emission was in the UV region to assume that it was part of the Lyman series and n1=1. From there I plugged into the equation and solved for n2.

LReedy_3C

Because the wavelength is in the UV spectrum, I knew it was in the Lyman series and so at some point n = 1. Then, using c=λv, I calculated the frequency to be 2.924x10^15 Hz. From there it was just plugging all these numbers into the Rydberg equation and solving for n1.

LReedy_3C

All of the information for this problem can be found using E = hv and λv = c. First, find the frequency using E = hv, by plugging in planck's constant and 3.3x10^-19 J for E. This gives the frequency as 5.0x10^14Hz, which can then be plugged into λv=c to find the wavelength (600nm).

LReedy_3C

The double slit experiment demonstrates light behaving as a wave. When diffraction patterns appear when particles such electrons pass through a crystal it also demonstrates them acting as a wave.

LReedy_3C

Is question 1A.15 referring to an emission or absorption spectrum, and would it affect the final answer? I got n(final) to be 3, but if it is an emission spectrum would this be n(initial) = 3 and n(final) = 1?

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