CHEMISTRY COMMUNITY

AlbertGu_2C

In one of the lectures, Lavelle gave us a list of things to keep in mind when going into a problem like the ICE table ones, and it included tips like finding what you know, what you don't know, etc. and I think that it's helpful to go through a mental inventory of these things before filling out you...

AlbertGu_2C

Total pressure is all of the PARTIAL pressures added together, which can usually be solved using Kp and the rest of the given partial pressures

AlbertGu_2C

the products over reactants should equal the Kc you need to find for the question

AlbertGu_2C

One thing that tripped me up when I tried to solve this question was how I had to add x to the initial concentrations in order to get the answer that you plug into the program. Hope this helps!

AlbertGu_2C

When solving with Kp, you can set up your ICE table to solve with the equation you use to solve for concentrations to solve for individual partial pressures. From there you can find all of them and add them together for total pressure.

AlbertGu_2C

For this question I chose the first equation and multiplied K by 3, and took the inverse of the 3rd K, but it still comes up as wrong. Is there a problem with how I calculated it or is there a problem with the question?

AlbertGu_2C

AlyssaMaynard1C wrote:Is the week 1 homework due tonight or next sunday, I'm slightly confused.

The homework is due on the 17th, hope this helps!

AlbertGu_2C

Kc is just K that uses the concentrations of the variables involved. Hope this helps!

AlbertGu_2C

As a follow-up question, if real and ideal gasses are so different than each other, why is it helpful to treat real gasses the same as ideal gasses? Would this assumption not cause problems? I have the same question, because wouldn't the violations of the ideal gas law that the real gas law present...

AlbertGu_2C

You would use Kc if the moles are given, and if the partial pressures are given, use Kp. If both are given then you would have to use the Ideal Gas Law along with given information to convert between the two. Hope this helps!

AlbertGu_2C

In a more general sense, Avogadro's number is just the amount of objects in one mole. So if you were to set up a dimensional analysis table to solve a question, you should be using Avogadro's number as (6.0221*10^23)objects/mole

AlbertGu_2C

I think that UA/TA step ups and workshops are what made my final experience bearable in this class. These people are seriously so knowledgeable and so helpful. I'm going to at least one every week starting in 14B!

AlbertGu_2C

I had enough time, but only because I rushed the questions and probably got a lot more wrong than is good for me

AlbertGu_2C

Probably trying to fix the "sleep schedule" that I currently am running

AlbertGu_2C

The [OH-] is actually the concentration of OH- ions in the solution, and the pOH is the indication of the "power" of the hydroxide ions, so the number for pOH will be the negative log of [OH]-, which means it will be bigger

AlbertGu_2C

In order to find the empirical formula given the molecular formula, the best way is to divide each number of atoms by the greatest common denominator. Ex. C3H6O3 becomes CH2O

AlbertGu_2C

The ligand's "other name" will take priority in ordering the compounds in alphabetical order. Hope this helps!

AlbertGu_2C

These elements are just differently named due to the original Roman name. Elements like Lead are instead Plumbum, and Tin Stannum. Hope this helps!

AlbertGu_2C

As said above, I think the best way is to partition the ligands, transition metal, then the anion afterwards, and then arrange/add the extra parts afterwards.

AlbertGu_2C

Only use -ate for negative charge, and neutral/positive compounds will just end in the name of the last part

AlbertGu_2C

I think this is still aqua and not hydroxo, since the OH2 would signify water just in a different order. Hope this helps!

AlbertGu_2C

2 lp and 2 bonds is bent, as well as 2 lp and 1 lp. Hope this helps!

AlbertGu_2C

I think the general rule is that there is less crowding in equatorial bonds rather than axial ones, so atoms will tend towards equatorial first, but I'm not sure and he will probably cover it in lecture at some point. Hope this helps!

AlbertGu_2C

to find the hybridization for the oxygen, you have to count the bonded ATOMS (not just bonds!) and the lone pairs as well, so it would come out to 2 atoms and 2 lone pairs, which is an sp3 orbital

AlbertGu_2C

This example specifies hybridization for the Phosphorus atoms, so you would count the three bonded atoms, the three O atoms, and then the lone pair, and that would lead to 4 groups, which means it's an sp3 orbital

AlbertGu_2C

Single bonds are only composed of Sigma bonds, however they also exist in double and triple bonds

AlbertGu_2C

Phosphate is in the 3p orbital, so it has access to the 3d orbital of electrons, so it can fill that shell to compensate.

AlbertGu_2C

Oxidation numbers are hypothetical values, so using formal charge for specific resonance structures is probably better.

AlbertGu_2C

A general rule to also remember could be that going down the table decreases more than going towards the left. Hope this helps!

AlbertGu_2C

Ionic bonds are the highest energy bonds that we know of in this class up until now, so it would have a higher melting point.

AlbertGu_2C

The parentheses on the NH2 denote that the 2 hydrogens are only bonded to the N, which is bonded to the first C, which is denoted by how COOH is after the NH2

AlbertGu_2C

If somebody could help me make a list of all of the exceptions for this midterm, that would help a lot. Right now I have the first four elements, elements past period 3 can exceed, and group 13 only need 6. Thanks!

AlbertGu_2C

AlyssaMaynard1C wrote:Will we be able to look at a chart on the midterm for the electronegativity and other descriptions?

I'm pretty sure electronegativity will be given but all the other trends are probably just comparisons

AlbertGu_2C

I would suggest learning and memorizing the trends given, and also the fact that electronegativity differences between 0.4 and 1.7 are polar bonds. They should be given on the test. Hope this helps!

AlbertGu_2C

Any element past the 3rd period of the periodic table is able to have an expanded octet of electrons due to having access to more values of l.

AlbertGu_2C

The negative sign in this unit usually means the subject has an attractive force. This would mean that the part of the atom that has the highest electronegativity would have a delta negative. Hope this helped!

AlbertGu_2C

Noble Gases are exceptions from the rules because they are out of the running so to speak for gaining/losing electrons

AlbertGu_2C

Electron affinity measures how inclined an atom is to take on another electron

AlbertGu_2C

More protons equals more positive charge to attract electrons

AlbertGu_2C

Any element past period 3 of the periodic table can form more bonds than accounted for by the octet rule as dictated by its d orbital having the capacity to take in more electrons

AlbertGu_2C

I think any charge that sums up to +2 should be counted the same in any configuration just based on the fact that it has charge as opposed to being neutral.

AlbertGu_2C

I think we only need to know ranges for prominent ones like visible and UV. Hope this helps!

AlbertGu_2C

Adding on to other answers, the Balmer series would be n=2 and the Paschen series would be n=3. Hope this helps!

AlbertGu_2C

It would depend on the direction relative to which the electron moved up and down the energy levels.

AlbertGu_2C

I haven't seen chem community in a while, sorry about that! To answer your question if you still need it, I used a different variable of the Rydberg equation that uses a different R constant. So if you wanted to do the question using the equation that he gives us you would have to find out the frequ...

AlbertGu_2C

Hydrogen bonds are only able to be established on those 3 elements due to how electronegative they are relative to Hydrogen

AlbertGu_2C

Usually, the question will ask for something related to the magnitude of energy that was released. However, when you use the Rydberg Constant, you will usually be solving for the energy released by one single electron, so relatively, the energy will be negative. Hope this helps!

AlbertGu_2C

Because when the denominator is infinity, the entire fraction equates to zero, since it is an infinitesimally small fraction. Hope this helps!

AlbertGu_2C

It would be D. The question states that there are 1,650,763.73 wavelengths in one meter, so in order to find the length of one wavelength, you have to divide 1 meter by 1,650,763.73, in order to get the wavelength, 6.058*10^-7. From there you can use E=h(c/wavelength) in order to find the energy req...

AlbertGu_2C

Whereas electrostatic potential energy is the energy between two charges, quantized energy of an electron is only exclusively dealing with one electron. Hope this helps!

AlbertGu_2C

Rather than the starting point, the Balmer and Lyman series denote the endings of an energy change. The Balmer series consists of visible light, ending in n=2, and the Lyman series consists of UV light, ending in n=1. Each lower energy level (Ex. n=2 to n=1 as opposed to n=3 to n=2), has a higher en...

AlbertGu_2C

Avogadro's number is just the amount of OBJECTS in a mole of x. So if I were to say that there is 1 mol of carrots in Billy's garden, then there'd be 6.0221*10^23 carrots in his garden. Hope this helps!

AlbertGu_2C

Thanks for the clear explanation, I understand now!

AlbertGu_2C

On Dr. Lavelle's lectures, he uses 3.29*10^15, but when I use it in the Rydberg equation it's 1.07*10^7, could anyone explain the difference to me? Thanks!

AlbertGu_2C

For this question, you first have to look at the given wavelength, which is 6.563 *10^-7 m (converted from nanometers). Matching this with the numbers given on the EM spectrum tells us that this light is on the wavelength of visible light, which fits in the Balmer series, which means that the light ...

AlbertGu_2C

I like to think of it this way: the electrons at lower energy levels have lower energy because they're essentially being held still due to being closer to the nucleus. Hope this helps!

AlbertGu_2C

As people have stated above, I use the sig figs presented in the problem and only round at the end. If there are multiple numbers with different amounts of sig figs in the problem, then I just round to 4 decimal points. Hopefully Dr. Lavelle will mention this in lecture soon.

AlbertGu_2C

I think in one of the modules Prof. Lavelle specifically states that you need to simplify out fraction stoichiometric ratios. Hope this helps!

AlbertGu_2C

The textbook states that Avogadro's CONSTANT is a pure number with no units, while Avogadro's NUMBER can have any unit pertaining to objects. Hope this helps!

AlbertGu_2C

Given: 1.5g Sn 26.45g Crucible Mass 28.35g Mass of Crucible + Product From here, you can calculate the mass of Oxygen, 28.35g (total) - 26.45g (crucible) = 1.9g product 1.9g product - 1.5g Sn = .4g O Use this .4g O in order to find out the moles of O .4g O/16.11g O/mol = .025 Use given 1.5g Sn in or...

AlbertGu_2C

As VSU_2G stated above, you would only be using a molar mass conversion of 2 butanone, or C4H8O, which is roughly 72 g/mol, hopefully this helps!

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